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Worked graphical examples

Three worked problems for every catalog entry: an analytical illustration, a two-condition comparison, and an interpolation prediction checked against the analytical response. Axes show physical units or stated dimensionless normalizations.

Schrödinger model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A particle is confined by infinite walls at x = 0 and L. Find the normalized ground-state probability density. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Probability density × L (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

L∣ψ1∣2=2sin⁡2(πx/L)L|\psi_1|^2=2\sin^2(\pi x/L)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=5.55112×10−16\frac{q(b)-q(a)}{b-a}=5.55112\times10^{-16}qlin(c)=0.690983,q(c)=1.84433q_{\rm lin}(c)=0.690983,\quad q(c)=1.84433e=qlin(c)−q(c)=−1.15334e=q_{\rm lin}(c)-q(c)=-1.15334

Solution. The endpoint responses are q(A) = 0.690983 and q(B) = 0.690983. Their secant slope is 5.55112e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.690983. Direct evaluation gives 1.84433; subtracting it from the prediction gives signed error -1.15334.

Schrödinger model: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Probability density × L (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Probability density × L (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.691 Worked point: (0.41, 1.844)
At x = 0.41, the orange circle is the analytical value 1.84433; the purple diamond is the interpolated value 0.690983. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.690983; analytical reference = 1.84433; absolute interpolation error = 1.15334. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Particle-in-a-box model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A particle is confined by infinite walls at x = 0 and L. Find the normalized ground-state probability density. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Probability density × L (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

L∣ψ1∣2=2sin⁡2(πx/L)L|\psi_1|^2=2\sin^2(\pi x/L)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=5.55112×10−16\frac{q(b)-q(a)}{b-a}=5.55112\times10^{-16}qlin(c)=0.690983,q(c)=1.84433q_{\rm lin}(c)=0.690983,\quad q(c)=1.84433e=qlin(c)−q(c)=−1.15334e=q_{\rm lin}(c)-q(c)=-1.15334

Solution. The endpoint responses are q(A) = 0.690983 and q(B) = 0.690983. Their secant slope is 5.55112e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.690983. Direct evaluation gives 1.84433; subtracting it from the prediction gives signed error -1.15334.

Particle-in-a-box model: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Probability density × L (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Probability density × L (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.691 Worked point: (0.41, 1.844)
At x = 0.41, the orange circle is the analytical value 1.84433; the purple diamond is the interpolated value 0.690983. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.690983; analytical reference = 1.84433; absolute interpolation error = 1.15334. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Dirac model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a free massive Dirac particle, evaluate the positive-energy branch versus momentum. For this calculation, x denotes the plotted horizontal coordinate (Momentum p / mc (dimensionless)), and q(x) denotes the plotted response (Energy E / mc² (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

E/(mc2)=1+(p/mc)2E/(mc^2)=\sqrt{1+(p/mc)^2}a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.796561\frac{q(b)-q(a)}{b-a}=0.796561qlin(c)=1.66802,q(c)=1.58521q_{\rm lin}(c)=1.66802,\quad q(c)=1.58521e=qlin(c)−q(c)=0.0828108e=q_{\rm lin}(c)-q(c)=0.0828108

Solution. The endpoint responses are q(A) = 1.16619 and q(B) = 2.6. Their secant slope is 0.796561. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.66802. Direct evaluation gives 1.58521; subtracting it from the prediction gives signed error 0.0828108.

Dirac model: Intermediate-condition prediction and exact check. Horizontal axis: Momentum p / mc (dimensionless). Vertical axis: Energy E / mc² (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Momentum p / mc (dimensionless) 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 Energy E / mc² (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.668 Worked point: (1.23, 1.585)
At x = 1.23, the orange circle is the analytical value 1.58521; the purple diamond is the interpolated value 1.66802. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.66802; analytical reference = 1.58521; absolute interpolation error = 0.0828108. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Born–Oppenheimer approximation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Approximate one Born–Oppenheimer potential-energy surface near its minimum by a spring of stiffness k. For this calculation, x denotes the plotted horizontal coordinate (Bond displacement / length scale (dimensionless)), and q(x) denotes the plotted response (Energy above minimum / kℓ² (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

(U−U0)/(kℓ2)=12q2,q=(R−Re)/ℓ(U-U_0)/(k\ell^2)=\tfrac12q^2,\quad q=(R-R_e)/\ella=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=9.25186×10−17\frac{q(b)-q(a)}{b-a}=9.25186\times10^{-17}qlin(c)=0.72,q(c)=0.0648q_{\rm lin}(c)=0.72,\quad q(c)=0.0648e=qlin(c)−q(c)=0.6552e=q_{\rm lin}(c)-q(c)=0.6552

Solution. The endpoint responses are q(A) = 0.72 and q(B) = 0.72. Their secant slope is 9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.72. Direct evaluation gives 0.0648; subtracting it from the prediction gives signed error 0.6552.

Born–Oppenheimer approximation: Intermediate-condition prediction and exact check. Horizontal axis: Bond displacement / length scale (dimensionless). Vertical axis: Energy above minimum / kℓ² (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Bond displacement / length scale (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Energy above minimum / kℓ² (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.72 Worked point: (-0.36, 0.0648)
At x = -0.36, the orange circle is the analytical value 0.0648; the purple diamond is the interpolated value 0.72. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.72; analytical reference = 0.0648; absolute interpolation error = 0.6552. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Local harmonic approximation on a single adiabatic surface; electronic crossings and nonadiabatic coupling are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hartree–Fock model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the normalized hydrogen 1s state for one electron in a Coulomb potential. Plot probability per radial interval. For this calculation, x denotes the plotted horizontal coordinate (Radius r / a₀ (dimensionless)), and q(x) denotes the plotted response (Radial probability density × a₀ (dimensionless)). Suppose only the analytical endpoint responses at x = 1.2 and x = 4.8 are tabulated. Use linear interpolation to predict the response at x = 2.46, then check it against the original equation.

a0P(r)=4(r/a0)2e−2r/a0a_0P(r)=4(r/a_0)^2e^{-2r/a_0}a=1.2,b=4.8,c=2.46a=1.2,\quad b=4.8,\quad c=2.46qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.143415\frac{q(b)-q(a)}{b-a}=-0.143415qlin(c)=0.341833,q(c)=0.176686q_{\rm lin}(c)=0.341833,\quad q(c)=0.176686e=qlin(c)−q(c)=0.165147e=q_{\rm lin}(c)-q(c)=0.165147

Solution. The endpoint responses are q(A) = 0.522535 and q(B) = 0.00624188. Their secant slope is -0.143415. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.341833. Direct evaluation gives 0.176686; subtracting it from the prediction gives signed error 0.165147.

Hartree–Fock model: Intermediate-condition prediction and exact check. Horizontal axis: Radius r / a₀ (dimensionless). Vertical axis: Radial probability density × a₀ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Radius r / a₀ (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 Radial probability density × a₀ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3418 Worked point: (2.46, 0.1767)
At x = 2.46, the orange circle is the analytical value 0.176686; the purple diamond is the interpolated value 0.341833. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.341833; analytical reference = 0.176686; absolute interpolation error = 0.165147. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Hartree–Fock is exact for this one-electron case. For DFT this is an exact-functional reference; approximate functionals need not reproduce it exactly. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Density functional theory (DFT) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the normalized hydrogen 1s state for one electron in a Coulomb potential. Plot probability per radial interval. For this calculation, x denotes the plotted horizontal coordinate (Radius r / a₀ (dimensionless)), and q(x) denotes the plotted response (Radial probability density × a₀ (dimensionless)). Suppose only the analytical endpoint responses at x = 1.2 and x = 4.8 are tabulated. Use linear interpolation to predict the response at x = 2.46, then check it against the original equation.

a0P(r)=4(r/a0)2e−2r/a0a_0P(r)=4(r/a_0)^2e^{-2r/a_0}a=1.2,b=4.8,c=2.46a=1.2,\quad b=4.8,\quad c=2.46qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.143415\frac{q(b)-q(a)}{b-a}=-0.143415qlin(c)=0.341833,q(c)=0.176686q_{\rm lin}(c)=0.341833,\quad q(c)=0.176686e=qlin(c)−q(c)=0.165147e=q_{\rm lin}(c)-q(c)=0.165147

Solution. The endpoint responses are q(A) = 0.522535 and q(B) = 0.00624188. Their secant slope is -0.143415. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.341833. Direct evaluation gives 0.176686; subtracting it from the prediction gives signed error 0.165147.

Density functional theory (DFT): Intermediate-condition prediction and exact check. Horizontal axis: Radius r / a₀ (dimensionless). Vertical axis: Radial probability density × a₀ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Radius r / a₀ (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 Radial probability density × a₀ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3418 Worked point: (2.46, 0.1767)
At x = 2.46, the orange circle is the analytical value 0.176686; the purple diamond is the interpolated value 0.341833. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.341833; analytical reference = 0.176686; absolute interpolation error = 0.165147. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Hartree–Fock is exact for this one-electron case. For DFT this is an exact-functional reference; approximate functionals need not reproduce it exactly. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Time-dependent DFT (TDDFT) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Consider a resonantly driven, noninteracting two-level reference starting in its lower state. For this calculation, x denotes the plotted horizontal coordinate (Rabi angle Ωt (radian)), and q(x) denotes the plotted response (Excited-state population (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

P2(t)=sin⁡2(Ωt/2)P_2(t)=\sin^2(\Omega t/2)a=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.41744×10−17\frac{q(b)-q(a)}{b-a}=4.41744\times10^{-17}qlin(c)=0.345492,q(c)=0.922164q_{\rm lin}(c)=0.345492,\quad q(c)=0.922164e=qlin(c)−q(c)=−0.576672e=q_{\rm lin}(c)-q(c)=-0.576672

Solution. The endpoint responses are q(A) = 0.345492 and q(B) = 0.345492. Their secant slope is 4.41744e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.345492. Direct evaluation gives 0.922164; subtracting it from the prediction gives signed error -0.576672.

Time-dependent DFT (TDDFT): Intermediate-condition prediction and exact check. Horizontal axis: Rabi angle Ωt (radian). Vertical axis: Excited-state population (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Rabi angle Ωt (radian) 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Excited-state population (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3455 Worked point: (2.576, 0.9222)
At x = 2.57611, the orange circle is the analytical value 0.922164; the purple diamond is the interpolated value 0.345492. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.345492; analytical reference = 0.922164; absolute interpolation error = 0.576672. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Two-level rotating-wave reference for time-dependent electronic calculations; not a general TDDFT solution. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Tight-binding model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. An infinite one-orbital chain has nearest-neighbor hopping tₕ and zero on-site energy. For this calculation, x denotes the plotted horizontal coordinate (Wave number × lattice spacing ka (radian)), and q(x) denotes the plotted response (Band energy / hopping tₕ (dimensionless)). Suppose only the analytical endpoint responses at x = -1.88496 and x = 1.88496 are tabulated. Use linear interpolation to predict the response at x = -0.565487, then check it against the original equation.

E(k)/th=−2cos⁡(ka)E(k)/t_h=-2\cos(ka)a=−1.88496,b=1.88496,c=−0.565487a=-1.88496,\quad b=1.88496,\quad c=-0.565487qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0\frac{q(b)-q(a)}{b-a}=0qlin(c)=0.618034,q(c)=−1.68866q_{\rm lin}(c)=0.618034,\quad q(c)=-1.68866e=qlin(c)−q(c)=2.30669e=q_{\rm lin}(c)-q(c)=2.30669

Solution. The endpoint responses are q(A) = 0.618034 and q(B) = 0.618034. Their secant slope is 0. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.618034. Direct evaluation gives -1.68866; subtracting it from the prediction gives signed error 2.30669.

Tight-binding model: Intermediate-condition prediction and exact check. Horizontal axis: Wave number × lattice spacing ka (radian). Vertical axis: Band energy / hopping tₕ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Wave number × lattice spacing ka (radian) −2.0 −1.5 −1.0 −0.5 0.0 0.5 Band energy / hopping tₕ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.618 Worked point: (-0.5655, -1.689)
At x = -0.565487, the orange circle is the analytical value -1.68866; the purple diamond is the interpolated value 0.618034. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.618034; analytical reference = -1.68866; absolute interpolation error = 2.30669. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hubbard model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Find the two-electron singlet ground energy of a two-site Hubbard dimer with hopping tₕ > 0 and repulsion U. For this calculation, x denotes the plotted horizontal coordinate (Repulsion U / hopping tₕ (dimensionless)), and q(x) denotes the plotted response (Ground energy E₀ / tₕ (dimensionless)). Suppose only the analytical endpoint responses at x = 2.4 and x = 9.6 are tabulated. Use linear interpolation to predict the response at x = 4.92, then check it against the original equation.

E0/th=12[u−u2+16],u=U/thE_0/t_h=\tfrac12[u-\sqrt{u^2+16}],\quad u=U/t_ha=2.4,b=9.6,c=4.92a=2.4,\quad b=9.6,\quad c=4.92qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.10172\frac{q(b)-q(a)}{b-a}=0.10172qlin(c)=−0.876047,q(c)=−0.710426q_{\rm lin}(c)=-0.876047,\quad q(c)=-0.710426e=qlin(c)−q(c)=−0.165622e=q_{\rm lin}(c)-q(c)=-0.165622

Solution. The endpoint responses are q(A) = -1.13238 and q(B) = -0.4. Their secant slope is 0.10172. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.876047. Direct evaluation gives -0.710426; subtracting it from the prediction gives signed error -0.165622.

Hubbard model: Intermediate-condition prediction and exact check. Horizontal axis: Repulsion U / hopping tₕ (dimensionless). Vertical axis: Ground energy E₀ / tₕ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 Repulsion U / hopping tₕ (dimensionless) −1.2 −1.0 −0.8 −0.6 −0.4 Ground energy E₀ / tₕ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.876 Worked point: (4.92, -0.7104)
At x = 4.92, the orange circle is the analytical value -0.710426; the purple diamond is the interpolated value -0.876047. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.876047; analytical reference = -0.710426; absolute interpolation error = 0.165622. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Heisenberg spin model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Two classical unit spins interact through −J s₁·s₂ with J > 0. For this calculation, x denotes the plotted horizontal coordinate (Relative spin angle θ (radian)), and q(x) denotes the plotted response (Energy / exchange J (dimensionless)). Suppose only the analytical endpoint responses at x = 0.628319 and x = 2.51327 are tabulated. Use linear interpolation to predict the response at x = 1.28805, then check it against the original equation.

E/J=−cos⁡θE/J=-\cos\thetaa=0.628319,b=2.51327,c=1.28805a=0.628319,\quad b=2.51327,\quad c=1.28805qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.858394\frac{q(b)-q(a)}{b-a}=0.858394qlin(c)=−0.242705,q(c)=−0.278991q_{\rm lin}(c)=-0.242705,\quad q(c)=-0.278991e=qlin(c)−q(c)=0.036286e=q_{\rm lin}(c)-q(c)=0.036286

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = 0.809017. Their secant slope is 0.858394. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.242705. Direct evaluation gives -0.278991; subtracting it from the prediction gives signed error 0.036286.

Heisenberg spin model: Intermediate-condition prediction and exact check. Horizontal axis: Relative spin angle θ (radian). Vertical axis: Energy / exchange J (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 Relative spin angle θ (radian) −1.00 −0.75 −0.50 −0.25 0.00 0.25 0.50 0.75 1.00 Energy / exchange J (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.2427 Worked point: (1.288, -0.279)
At x = 1.28805, the orange circle is the analytical value -0.278991; the purple diamond is the interpolated value -0.242705. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.242705; analytical reference = -0.278991; absolute interpolation error = 0.036286. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Classical two-spin special case; quantum spin spectra require a different treatment. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Ising model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A single spin s = ±1 has energy −hs at inverse temperature β. Find its thermal mean. For this calculation, x denotes the plotted horizontal coordinate (Field / thermal energy βh (dimensionless)), and q(x) denotes the plotted response (Mean spin (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

⟨s⟩=tanh⁡(βh)\langle s\rangle=\tanh(\beta h)a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.526003\frac{q(b)-q(a)}{b-a}=0.526003qlin(c)=−0.284042,q(c)=−0.492988q_{\rm lin}(c)=-0.284042,\quad q(c)=-0.492988e=qlin(c)−q(c)=0.208946e=q_{\rm lin}(c)-q(c)=0.208946

Solution. The endpoint responses are q(A) = -0.946806 and q(B) = 0.946806. Their secant slope is 0.526003. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.284042. Direct evaluation gives -0.492988; subtracting it from the prediction gives signed error 0.208946.

Ising model: Intermediate-condition prediction and exact check. Horizontal axis: Field / thermal energy βh (dimensionless). Vertical axis: Mean spin (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Field / thermal energy βh (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Mean spin (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.284 Worked point: (-0.54, -0.493)
At x = -0.54, the orange circle is the analytical value -0.492988; the purple diamond is the interpolated value -0.284042. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.284042; analytical reference = -0.492988; absolute interpolation error = 0.208946. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Quantum harmonic oscillator · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use oscillator length ℓ = √(ℏ/mω) and find the normalized ground-state density. For this calculation, x denotes the plotted horizontal coordinate (Position x / oscillator length ℓ (dimensionless)), and q(x) denotes the plotted response (Probability density × ℓ (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

ℓ∣ψ0∣2=π−1/2e−(x/ℓ)2\ell|\psi_0|^2=\pi^{-1/2}e^{-(x/\ell)^2}a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.92747×10−17\frac{q(b)-q(a)}{b-a}=-1.92747\times10^{-17}qlin(c)=0.0220959,q(c)=0.421488q_{\rm lin}(c)=0.0220959,\quad q(c)=0.421488e=qlin(c)−q(c)=−0.399392e=q_{\rm lin}(c)-q(c)=-0.399392

Solution. The endpoint responses are q(A) = 0.0220959 and q(B) = 0.0220959. Their secant slope is -1.92747e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0220959. Direct evaluation gives 0.421488; subtracting it from the prediction gives signed error -0.399392.

Quantum harmonic oscillator: Intermediate-condition prediction and exact check. Horizontal axis: Position x / oscillator length ℓ (dimensionless). Vertical axis: Probability density × ℓ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Position x / oscillator length ℓ (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Probability density × ℓ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0221 Worked point: (-0.54, 0.4215)
At x = -0.54, the orange circle is the analytical value 0.421488; the purple diamond is the interpolated value 0.0220959. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0220959; analytical reference = 0.421488; absolute interpolation error = 0.399392. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Classical molecular dynamics (MD) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take one isolated coordinate with potential kq²/2, initial displacement A, and zero initial velocity. For this calculation, x denotes the plotted horizontal coordinate (Time × natural frequency ωt (radian)), and q(x) denotes the plotted response (Bond displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

q(τ)=cos⁡τq(\tau)=\cos\taua=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−5.88992×10−17\frac{q(b)-q(a)}{b-a}=-5.88992\times10^{-17}qlin(c)=0.309017,q(c)=−0.844328q_{\rm lin}(c)=0.309017,\quad q(c)=-0.844328e=qlin(c)−q(c)=1.15334e=q_{\rm lin}(c)-q(c)=1.15334

Solution. The endpoint responses are q(A) = 0.309017 and q(B) = 0.309017. Their secant slope is -5.88992e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.309017. Direct evaluation gives -0.844328; subtracting it from the prediction gives signed error 1.15334.

Classical molecular dynamics (MD): Intermediate-condition prediction and exact check. Horizontal axis: Time × natural frequency ωt (radian). Vertical axis: Bond displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time × natural frequency ωt (radian) −1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 Bond displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.309 Worked point: (2.576, -0.8443)
At x = 2.57611, the orange circle is the analytical value -0.844328; the purple diamond is the interpolated value 0.309017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.309017; analytical reference = -0.844328; absolute interpolation error = 1.15334. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Harmonic force benchmark for MD or locally harmonic ab initio dynamics; real many-atom trajectories are not generally sinusoidal. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Ab initio molecular dynamics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take one isolated coordinate with potential kq²/2, initial displacement A, and zero initial velocity. For this calculation, x denotes the plotted horizontal coordinate (Time × natural frequency ωt (radian)), and q(x) denotes the plotted response (Bond displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

q(τ)=cos⁡τq(\tau)=\cos\taua=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−5.88992×10−17\frac{q(b)-q(a)}{b-a}=-5.88992\times10^{-17}qlin(c)=0.309017,q(c)=−0.844328q_{\rm lin}(c)=0.309017,\quad q(c)=-0.844328e=qlin(c)−q(c)=1.15334e=q_{\rm lin}(c)-q(c)=1.15334

Solution. The endpoint responses are q(A) = 0.309017 and q(B) = 0.309017. Their secant slope is -5.88992e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.309017. Direct evaluation gives -0.844328; subtracting it from the prediction gives signed error 1.15334.

Ab initio molecular dynamics: Intermediate-condition prediction and exact check. Horizontal axis: Time × natural frequency ωt (radian). Vertical axis: Bond displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time × natural frequency ωt (radian) −1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 Bond displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.309 Worked point: (2.576, -0.8443)
At x = 2.57611, the orange circle is the analytical value -0.844328; the purple diamond is the interpolated value 0.309017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.309017; analytical reference = -0.844328; absolute interpolation error = 1.15334. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Harmonic force benchmark for MD or locally harmonic ab initio dynamics; real many-atom trajectories are not generally sinusoidal. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lennard–Jones potential · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate an unshifted 12–6 pair potential at reduced separation r/σ. For this calculation, x denotes the plotted horizontal coordinate (Separation r / σ (dimensionless)), and q(x) denotes the plotted response (Pair energy U / ε (dimensionless)). Suppose only the analytical endpoint responses at x = 1.36 and x = 2.59 are tabulated. Use linear interpolation to predict the response at x = 1.7905, then check it against the original equation.

U/ε=4[(σ/r)12−(σ/r)6]U/\varepsilon=4[(\sigma/r)^{12}-(\sigma/r)^6]a=1.36,b=2.59,c=1.7905a=1.36,\quad b=2.59,\quad c=1.7905qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.421988\frac{q(b)-q(a)}{b-a}=0.421988qlin(c)=−0.350587,q(c)=−0.117714q_{\rm lin}(c)=-0.350587,\quad q(c)=-0.117714e=qlin(c)−q(c)=−0.232873e=q_{\rm lin}(c)-q(c)=-0.232873

Solution. The endpoint responses are q(A) = -0.532253 and q(B) = -0.0132075. Their secant slope is 0.421988. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.350587. Direct evaluation gives -0.117714; subtracting it from the prediction gives signed error -0.232873.

Lennard–Jones potential: Intermediate-condition prediction and exact check. Horizontal axis: Separation r / σ (dimensionless). Vertical axis: Pair energy U / ε (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.4 1.6 1.8 2.0 2.2 2.4 2.6 Separation r / σ (dimensionless) −0.5 −0.4 −0.3 −0.2 −0.1 0.0 Pair energy U / ε (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.3506 Worked point: (1.79, -0.1177)
At x = 1.7905, the orange circle is the analytical value -0.117714; the purple diamond is the interpolated value -0.350587. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.350587; analytical reference = -0.117714; absolute interpolation error = 0.232873. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For water and Martini entries, this is only a Lennard–Jones interaction contribution; electrostatics, constraints, and other sites are not included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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SPC/E water model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate an unshifted 12–6 pair potential at reduced separation r/σ. For this calculation, x denotes the plotted horizontal coordinate (Separation r / σ (dimensionless)), and q(x) denotes the plotted response (Pair energy U / ε (dimensionless)). Suppose only the analytical endpoint responses at x = 1.36 and x = 2.59 are tabulated. Use linear interpolation to predict the response at x = 1.7905, then check it against the original equation.

U/ε=4[(σ/r)12−(σ/r)6]U/\varepsilon=4[(\sigma/r)^{12}-(\sigma/r)^6]a=1.36,b=2.59,c=1.7905a=1.36,\quad b=2.59,\quad c=1.7905qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.421988\frac{q(b)-q(a)}{b-a}=0.421988qlin(c)=−0.350587,q(c)=−0.117714q_{\rm lin}(c)=-0.350587,\quad q(c)=-0.117714e=qlin(c)−q(c)=−0.232873e=q_{\rm lin}(c)-q(c)=-0.232873

Solution. The endpoint responses are q(A) = -0.532253 and q(B) = -0.0132075. Their secant slope is 0.421988. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.350587. Direct evaluation gives -0.117714; subtracting it from the prediction gives signed error -0.232873.

SPC/E water model: Intermediate-condition prediction and exact check. Horizontal axis: Separation r / σ (dimensionless). Vertical axis: Pair energy U / ε (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.4 1.6 1.8 2.0 2.2 2.4 2.6 Separation r / σ (dimensionless) −0.5 −0.4 −0.3 −0.2 −0.1 0.0 Pair energy U / ε (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.3506 Worked point: (1.79, -0.1177)
At x = 1.7905, the orange circle is the analytical value -0.117714; the purple diamond is the interpolated value -0.350587. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.350587; analytical reference = -0.117714; absolute interpolation error = 0.232873. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For water and Martini entries, this is only a Lennard–Jones interaction contribution; electrostatics, constraints, and other sites are not included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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TIP4P water-model family · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate an unshifted 12–6 pair potential at reduced separation r/σ. For this calculation, x denotes the plotted horizontal coordinate (Separation r / σ (dimensionless)), and q(x) denotes the plotted response (Pair energy U / ε (dimensionless)). Suppose only the analytical endpoint responses at x = 1.36 and x = 2.59 are tabulated. Use linear interpolation to predict the response at x = 1.7905, then check it against the original equation.

U/ε=4[(σ/r)12−(σ/r)6]U/\varepsilon=4[(\sigma/r)^{12}-(\sigma/r)^6]a=1.36,b=2.59,c=1.7905a=1.36,\quad b=2.59,\quad c=1.7905qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.421988\frac{q(b)-q(a)}{b-a}=0.421988qlin(c)=−0.350587,q(c)=−0.117714q_{\rm lin}(c)=-0.350587,\quad q(c)=-0.117714e=qlin(c)−q(c)=−0.232873e=q_{\rm lin}(c)-q(c)=-0.232873

Solution. The endpoint responses are q(A) = -0.532253 and q(B) = -0.0132075. Their secant slope is 0.421988. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.350587. Direct evaluation gives -0.117714; subtracting it from the prediction gives signed error -0.232873.

TIP4P water-model family: Intermediate-condition prediction and exact check. Horizontal axis: Separation r / σ (dimensionless). Vertical axis: Pair energy U / ε (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.4 1.6 1.8 2.0 2.2 2.4 2.6 Separation r / σ (dimensionless) −0.5 −0.4 −0.3 −0.2 −0.1 0.0 Pair energy U / ε (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.3506 Worked point: (1.79, -0.1177)
At x = 1.7905, the orange circle is the analytical value -0.117714; the purple diamond is the interpolated value -0.350587. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.350587; analytical reference = -0.117714; absolute interpolation error = 0.232873. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For water and Martini entries, this is only a Lennard–Jones interaction contribution; electrostatics, constraints, and other sites are not included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Martini coarse-grained model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate an unshifted 12–6 pair potential at reduced separation r/σ. For this calculation, x denotes the plotted horizontal coordinate (Separation r / σ (dimensionless)), and q(x) denotes the plotted response (Pair energy U / ε (dimensionless)). Suppose only the analytical endpoint responses at x = 1.36 and x = 2.59 are tabulated. Use linear interpolation to predict the response at x = 1.7905, then check it against the original equation.

U/ε=4[(σ/r)12−(σ/r)6]U/\varepsilon=4[(\sigma/r)^{12}-(\sigma/r)^6]a=1.36,b=2.59,c=1.7905a=1.36,\quad b=2.59,\quad c=1.7905qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.421988\frac{q(b)-q(a)}{b-a}=0.421988qlin(c)=−0.350587,q(c)=−0.117714q_{\rm lin}(c)=-0.350587,\quad q(c)=-0.117714e=qlin(c)−q(c)=−0.232873e=q_{\rm lin}(c)-q(c)=-0.232873

Solution. The endpoint responses are q(A) = -0.532253 and q(B) = -0.0132075. Their secant slope is 0.421988. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.350587. Direct evaluation gives -0.117714; subtracting it from the prediction gives signed error -0.232873.

Martini coarse-grained model: Intermediate-condition prediction and exact check. Horizontal axis: Separation r / σ (dimensionless). Vertical axis: Pair energy U / ε (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.4 1.6 1.8 2.0 2.2 2.4 2.6 Separation r / σ (dimensionless) −0.5 −0.4 −0.3 −0.2 −0.1 0.0 Pair energy U / ε (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.3506 Worked point: (1.79, -0.1177)
At x = 1.7905, the orange circle is the analytical value -0.117714; the purple diamond is the interpolated value -0.350587. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.350587; analytical reference = -0.117714; absolute interpolation error = 0.232873. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For water and Martini entries, this is only a Lennard–Jones interaction contribution; electrostatics, constraints, and other sites are not included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Morse potential · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate a Morse bond with its dissociation limit set to zero. For this calculation, x denotes the plotted horizontal coordinate (Bond extension a(r−rₑ) (dimensionless)), and q(x) denotes the plotted response (Energy U / Dₑ (dimensionless)). Suppose only the analytical endpoint responses at x = 0.32 and x = 3.08 are tabulated. Use linear interpolation to predict the response at x = 1.286, then check it against the original equation.

U/De=[1−e−q]2−1,q=a(r−re)U/D_e=[1-e^{-q}]^2-1,\quad q=a(r-r_e)a=0.32,b=3.08,c=1.286a=0.32,\quad b=3.08,\quad c=1.286qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.302608\frac{q(b)-q(a)}{b-a}=0.302608qlin(c)=−0.632686,q(c)=−0.476366q_{\rm lin}(c)=-0.632686,\quad q(c)=-0.476366e=qlin(c)−q(c)=−0.15632e=q_{\rm lin}(c)-q(c)=-0.15632

Solution. The endpoint responses are q(A) = -0.925006 and q(B) = -0.0898063. Their secant slope is 0.302608. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.632686. Direct evaluation gives -0.476366; subtracting it from the prediction gives signed error -0.15632.

Morse potential: Intermediate-condition prediction and exact check. Horizontal axis: Bond extension a(r−rₑ) (dimensionless). Vertical axis: Energy U / Dₑ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.5 1.0 1.5 2.0 2.5 3.0 Bond extension a(r−rₑ) (dimensionless) −1.0 −0.8 −0.6 −0.4 −0.2 0.0 Energy U / Dₑ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.6327 Worked point: (1.286, -0.4764)
At x = 1.286, the orange circle is the analytical value -0.476366; the purple diamond is the interpolated value -0.632686. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.632686; analytical reference = -0.476366; absolute interpolation error = 0.15632. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Embedded-atom method (EAM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose the illustrative embedding function F = −E*√(ρ/ρ*). Plot its density dependence. For this calculation, x denotes the plotted horizontal coordinate (Local density ρ / ρ* (dimensionless)), and q(x) denotes the plotted response (Embedding energy F / E* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.808 and x = 3.202 are tabulated. Use linear interpolation to predict the response at x = 1.6459, then check it against the original equation.

F(ρ)/E∗=−ρ/ρ∗F(\rho)/E_*=-\sqrt{\rho/\rho_*}a=0.808,b=3.202,c=1.6459a=0.808,\quad b=3.202,\quad c=1.6459qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.371982\frac{q(b)-q(a)}{b-a}=-0.371982qlin(c)=−1.21057,q(c)=−1.28293q_{\rm lin}(c)=-1.21057,\quad q(c)=-1.28293e=qlin(c)−q(c)=0.0723544e=q_{\rm lin}(c)-q(c)=0.0723544

Solution. The endpoint responses are q(A) = -0.898888 and q(B) = -1.78941. Their secant slope is -0.371982. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -1.21057. Direct evaluation gives -1.28293; subtracting it from the prediction gives signed error 0.0723544.

Embedded-atom method (EAM): Intermediate-condition prediction and exact check. Horizontal axis: Local density ρ / ρ* (dimensionless). Vertical axis: Embedding energy F / E* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Local density ρ / ρ* (dimensionless) −1.8 −1.6 −1.4 −1.2 −1.0 −0.8 Embedding energy F / E* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -1.211 Worked point: (1.646, -1.283)
At x = 1.6459, the orange circle is the analytical value -1.28293; the purple diamond is the interpolated value -1.21057. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -1.21057; analytical reference = -1.28293; absolute interpolation error = 0.0723544. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative EAM-type embedding function; not a fitted material parameterization. MEAM angular screening and density corrections are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Modified embedded-atom method (MEAM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose the illustrative embedding function F = −E*√(ρ/ρ*). Plot its density dependence. For this calculation, x denotes the plotted horizontal coordinate (Local density ρ / ρ* (dimensionless)), and q(x) denotes the plotted response (Embedding energy F / E* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.808 and x = 3.202 are tabulated. Use linear interpolation to predict the response at x = 1.6459, then check it against the original equation.

F(ρ)/E∗=−ρ/ρ∗F(\rho)/E_*=-\sqrt{\rho/\rho_*}a=0.808,b=3.202,c=1.6459a=0.808,\quad b=3.202,\quad c=1.6459qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.371982\frac{q(b)-q(a)}{b-a}=-0.371982qlin(c)=−1.21057,q(c)=−1.28293q_{\rm lin}(c)=-1.21057,\quad q(c)=-1.28293e=qlin(c)−q(c)=0.0723544e=q_{\rm lin}(c)-q(c)=0.0723544

Solution. The endpoint responses are q(A) = -0.898888 and q(B) = -1.78941. Their secant slope is -0.371982. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -1.21057. Direct evaluation gives -1.28293; subtracting it from the prediction gives signed error 0.0723544.

Modified embedded-atom method (MEAM): Intermediate-condition prediction and exact check. Horizontal axis: Local density ρ / ρ* (dimensionless). Vertical axis: Embedding energy F / E* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Local density ρ / ρ* (dimensionless) −1.8 −1.6 −1.4 −1.2 −1.0 −0.8 Embedding energy F / E* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -1.211 Worked point: (1.646, -1.283)
At x = 1.6459, the orange circle is the analytical value -1.28293; the purple diamond is the interpolated value -1.21057. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -1.21057; analytical reference = -1.28293; absolute interpolation error = 0.0723544. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative EAM-type embedding function; not a fitted material parameterization. MEAM angular screening and density corrections are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Tersoff bond-order potential · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. In a Tersoff-form pair term, hold cutoff and bond order at one and choose two exponential terms with coefficients 1 and 2. For this calculation, x denotes the plotted horizontal coordinate (Reduced separation q (dimensionless)), and q(x) denotes the plotted response (Pair energy / E* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

U/E∗=e−2q−2e−qU/E_*=e^{-2q}-2e^{-q}a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.257041\frac{q(b)-q(a)}{b-a}=0.257041qlin(c)=−0.480847,q(c)=−0.350332q_{\rm lin}(c)=-0.480847,\quad q(c)=-0.350332e=qlin(c)−q(c)=−0.130515e=q_{\rm lin}(c)-q(c)=-0.130515

Solution. The endpoint responses are q(A) = -0.696761 and q(B) = -0.0798629. Their secant slope is 0.257041. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.480847. Direct evaluation gives -0.350332; subtracting it from the prediction gives signed error -0.130515.

Tersoff bond-order potential: Intermediate-condition prediction and exact check. Horizontal axis: Reduced separation q (dimensionless). Vertical axis: Pair energy / E* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Reduced separation q (dimensionless) −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 Pair energy / E* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.4808 Worked point: (1.64, -0.3503)
At x = 1.64, the orange circle is the analytical value -0.350332; the purple diamond is the interpolated value -0.480847. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.480847; analytical reference = -0.350332; absolute interpolation error = 0.130515. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Toy fixed-environment pair contribution; this excludes environment-dependent bond order and cutoff transitions. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Stillinger–Weber potential · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold the radial factor of a Stillinger–Weber three-body term fixed and vary the included angle. For this calculation, x denotes the plotted horizontal coordinate (Bond angle θ (radian)), and q(x) denotes the plotted response (Angular energy / K (dimensionless)). Suppose only the analytical endpoint responses at x = 0.628319 and x = 2.51327 are tabulated. Use linear interpolation to predict the response at x = 1.28805, then check it against the original equation.

U3/K=(cos⁡θ+1/3)2U_3/K=(\cos\theta+1/3)^2a=0.628319,b=2.51327,c=1.28805a=0.628319,\quad b=2.51327,\quad c=1.28805qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.572262\frac{q(b)-q(a)}{b-a}=-0.572262qlin(c)=0.927423,q(c)=0.374941q_{\rm lin}(c)=0.927423,\quad q(c)=0.374941e=qlin(c)−q(c)=0.552482e=q_{\rm lin}(c)-q(c)=0.552482

Solution. The endpoint responses are q(A) = 1.30496 and q(B) = 0.226275. Their secant slope is -0.572262. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.927423. Direct evaluation gives 0.374941; subtracting it from the prediction gives signed error 0.552482.

Stillinger–Weber potential: Intermediate-condition prediction and exact check. Horizontal axis: Bond angle θ (radian). Vertical axis: Angular energy / K (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 Bond angle θ (radian) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 Angular energy / K (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.9274 Worked point: (1.288, 0.3749)
At x = 1.28805, the orange circle is the analytical value 0.374941; the purple diamond is the interpolated value 0.927423. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.927423; analytical reference = 0.374941; absolute interpolation error = 0.552482. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Angular contribution only, with fixed radial prefactor K > 0. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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ReaxFF reactive force field · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Near a stable isolated bond minimum, use the local quadratic energy with curvature k > 0. For this calculation, x denotes the plotted horizontal coordinate (Bond extension / ℓ (dimensionless)), and q(x) denotes the plotted response (Energy increment / kℓ² (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

ΔU/(kℓ2)=q2/2\Delta U/(k\ell^2)=q^2/2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.18,q(c)=0.0162q_{\rm lin}(c)=0.18,\quad q(c)=0.0162e=qlin(c)−q(c)=0.1638e=q_{\rm lin}(c)-q(c)=0.1638

Solution. The endpoint responses are q(A) = 0.18 and q(B) = 0.18. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.18. Direct evaluation gives 0.0162; subtracting it from the prediction gives signed error 0.1638.

ReaxFF reactive force field: Intermediate-condition prediction and exact check. Horizontal axis: Bond extension / ℓ (dimensionless). Vertical axis: Energy increment / kℓ² (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Bond extension / ℓ (dimensionless) 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 Energy increment / kℓ² (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.18 Worked point: (-0.18, 0.0162)
At x = -0.18, the orange circle is the analytical value 0.0162; the purple diamond is the interpolated value 0.18. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.18; analytical reference = 0.0162; absolute interpolation error = 0.1638. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Local Taylor benchmark, not the full force field or a trained potential prediction; reactive changes and other coordinates are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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AMBER force-field family · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Near a stable isolated bond minimum, use the local quadratic energy with curvature k > 0. For this calculation, x denotes the plotted horizontal coordinate (Bond extension / ℓ (dimensionless)), and q(x) denotes the plotted response (Energy increment / kℓ² (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

ΔU/(kℓ2)=q2/2\Delta U/(k\ell^2)=q^2/2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.18,q(c)=0.0162q_{\rm lin}(c)=0.18,\quad q(c)=0.0162e=qlin(c)−q(c)=0.1638e=q_{\rm lin}(c)-q(c)=0.1638

Solution. The endpoint responses are q(A) = 0.18 and q(B) = 0.18. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.18. Direct evaluation gives 0.0162; subtracting it from the prediction gives signed error 0.1638.

AMBER force-field family: Intermediate-condition prediction and exact check. Horizontal axis: Bond extension / ℓ (dimensionless). Vertical axis: Energy increment / kℓ² (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Bond extension / ℓ (dimensionless) 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 Energy increment / kℓ² (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.18 Worked point: (-0.18, 0.0162)
At x = -0.18, the orange circle is the analytical value 0.0162; the purple diamond is the interpolated value 0.18. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.18; analytical reference = 0.0162; absolute interpolation error = 0.1638. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Local Taylor benchmark, not the full force field or a trained potential prediction; reactive changes and other coordinates are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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CHARMM force-field family · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Near a stable isolated bond minimum, use the local quadratic energy with curvature k > 0. For this calculation, x denotes the plotted horizontal coordinate (Bond extension / ℓ (dimensionless)), and q(x) denotes the plotted response (Energy increment / kℓ² (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

ΔU/(kℓ2)=q2/2\Delta U/(k\ell^2)=q^2/2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.18,q(c)=0.0162q_{\rm lin}(c)=0.18,\quad q(c)=0.0162e=qlin(c)−q(c)=0.1638e=q_{\rm lin}(c)-q(c)=0.1638

Solution. The endpoint responses are q(A) = 0.18 and q(B) = 0.18. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.18. Direct evaluation gives 0.0162; subtracting it from the prediction gives signed error 0.1638.

CHARMM force-field family: Intermediate-condition prediction and exact check. Horizontal axis: Bond extension / ℓ (dimensionless). Vertical axis: Energy increment / kℓ² (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Bond extension / ℓ (dimensionless) 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 Energy increment / kℓ² (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.18 Worked point: (-0.18, 0.0162)
At x = -0.18, the orange circle is the analytical value 0.0162; the purple diamond is the interpolated value 0.18. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.18; analytical reference = 0.0162; absolute interpolation error = 0.1638. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Local Taylor benchmark, not the full force field or a trained potential prediction; reactive changes and other coordinates are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Machine-learned interatomic potential · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Near a stable isolated bond minimum, use the local quadratic energy with curvature k > 0. For this calculation, x denotes the plotted horizontal coordinate (Bond extension / ℓ (dimensionless)), and q(x) denotes the plotted response (Energy increment / kℓ² (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

ΔU/(kℓ2)=q2/2\Delta U/(k\ell^2)=q^2/2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.18,q(c)=0.0162q_{\rm lin}(c)=0.18,\quad q(c)=0.0162e=qlin(c)−q(c)=0.1638e=q_{\rm lin}(c)-q(c)=0.1638

Solution. The endpoint responses are q(A) = 0.18 and q(B) = 0.18. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.18. Direct evaluation gives 0.0162; subtracting it from the prediction gives signed error 0.1638.

Machine-learned interatomic potential: Intermediate-condition prediction and exact check. Horizontal axis: Bond extension / ℓ (dimensionless). Vertical axis: Energy increment / kℓ² (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Bond extension / ℓ (dimensionless) 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 Energy increment / kℓ² (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.18 Worked point: (-0.18, 0.0162)
At x = -0.18, the orange circle is the analytical value 0.0162; the purple diamond is the interpolated value 0.18. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.18; analytical reference = 0.0162; absolute interpolation error = 0.1638. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Local Taylor benchmark, not the full force field or a trained potential prediction; reactive changes and other coordinates are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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OPLS force-field family · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Retain only the first OPLS torsion coefficient V₁. For this calculation, x denotes the plotted horizontal coordinate (Dihedral angle φ (radian)), and q(x) denotes the plotted response (Torsion energy / V₁ (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

U/V1=12(1+cos⁡ϕ)U/V_1=\tfrac12(1+\cos\phi)a=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.94496×10−17\frac{q(b)-q(a)}{b-a}=-2.94496\times10^{-17}qlin(c)=0.654508,q(c)=0.077836q_{\rm lin}(c)=0.654508,\quad q(c)=0.077836e=qlin(c)−q(c)=0.576672e=q_{\rm lin}(c)-q(c)=0.576672

Solution. The endpoint responses are q(A) = 0.654508 and q(B) = 0.654508. Their secant slope is -2.94496e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.654508. Direct evaluation gives 0.077836; subtracting it from the prediction gives signed error 0.576672.

OPLS force-field family: Intermediate-condition prediction and exact check. Horizontal axis: Dihedral angle φ (radian). Vertical axis: Torsion energy / V₁ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Dihedral angle φ (radian) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Torsion energy / V₁ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.6545 Worked point: (2.576, 0.07784)
At x = 2.57611, the orange circle is the analytical value 0.077836; the purple diamond is the interpolated value 0.654508. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.654508; analytical reference = 0.077836; absolute interpolation error = 0.576672. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single torsional energy contribution, not the full molecular force field. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Drude polarizable model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A charged Drude oscillator has harmonic stiffness k and charge q. Find its static induced dipole. For this calculation, x denotes the plotted horizontal coordinate (Electric field E / E* (dimensionless)), and q(x) denotes the plotted response (Dipole p / αE* (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

p/(αE∗)=E/E∗p/(\alpha E_*)=E/E_*a=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=−0.36,q(c)=−0.36q_{\rm lin}(c)=-0.36,\quad q(c)=-0.36e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = -1.2 and q(B) = 1.2. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.36. Direct evaluation gives -0.36; subtracting it from the prediction gives signed error 0.

Drude polarizable model: Intermediate-condition prediction and exact check. Horizontal axis: Electric field E / E* (dimensionless). Vertical axis: Dipole p / αE* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Electric field E / E* (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Dipole p / αE* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.36 Worked point: (-0.36, -0.36)
At x = -0.36, the orange circle is the analytical value -0.36; the purple diamond is the interpolated value -0.36. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.36; analytical reference = -0.36; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Coarse-grained molecular model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Let a coarse variable have Gaussian probability proportional to exp(−q²/2). For this calculation, x denotes the plotted horizontal coordinate (Coarse coordinate / standard deviation (dimensionless)), and q(x) denotes the plotted response (Free energy / kBT (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

F(q)/(kBT)=q2/2F(q)/(k_BT)=q^2/2a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.31753×10−16\frac{q(b)-q(a)}{b-a}=4.31753\times10^{-16}qlin(c)=1.62,q(c)=0.1458q_{\rm lin}(c)=1.62,\quad q(c)=0.1458e=qlin(c)−q(c)=1.4742e=q_{\rm lin}(c)-q(c)=1.4742

Solution. The endpoint responses are q(A) = 1.62 and q(B) = 1.62. Their secant slope is 4.31753e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.62. Direct evaluation gives 0.1458; subtracting it from the prediction gives signed error 1.4742.

Coarse-grained molecular model: Intermediate-condition prediction and exact check. Horizontal axis: Coarse coordinate / standard deviation (dimensionless). Vertical axis: Free energy / kBT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Coarse coordinate / standard deviation (dimensionless) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 Free energy / kBT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.62 Worked point: (-0.54, 0.1458)
At x = -0.54, the orange circle is the analytical value 0.1458; the purple diamond is the interpolated value 1.62. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.62; analytical reference = 0.1458; absolute interpolation error = 1.4742. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exactly solvable Gaussian coarse-graining example; it does not assert that arbitrary coarse models are harmonic. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Dissipative particle dynamics (DPD) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold pair distance and weight fixed; the mean relative velocity obeys dy/dτ = −y. Random force has zero mean. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Mean relative velocity / initial mean (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Dissipative particle dynamics (DPD): Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Mean relative velocity / initial mean (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Mean relative velocity / initial mean (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Mean of a linear frozen-geometry pair reduction. DPD sample trajectories fluctuate and require a stochastic integrator. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Brownian dynamics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For free Brownian motion in one dimension take D = 1 m²/s and initial position zero. For this calculation, x denotes the plotted horizontal coordinate (Time t (s)), and q(x) denotes the plotted response (Mean-square displacement (m²)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

⟨[x(t)−x(0)]2⟩=2Dt\langle[x(t)-x(0)]^2\rangle=2Dta=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2\frac{q(b)-q(a)}{b-a}=2qlin(c)=4.1,q(c)=4.1q_{\rm lin}(c)=4.1,\quad q(c)=4.1e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 2 and q(B) = 8. Their secant slope is 2. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 4.1. Direct evaluation gives 4.1; subtracting it from the prediction gives signed error 0.

Brownian dynamics: Intermediate-condition prediction and exact check. Horizontal axis: Time t (s). Vertical axis: Mean-square displacement (m²). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time t (s) 2 3 4 5 6 7 8 Mean-square displacement (m²) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 4.1 Worked point: (2.05, 4.1)
At x = 2.05, the orange circle is the analytical value 4.1; the purple diamond is the interpolated value 4.1. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 4.1; analytical reference = 4.1; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ensemble expectation, not a single random trajectory; illustrative diffusivity. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Langevin dynamics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A free Langevin particle has linear drag γ, mass m, mean initial speed v₀, and zero-mean thermal noise. Use τ = γt/m. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Mean velocity / initial mean velocity (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Langevin dynamics: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Mean velocity / initial mean velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Mean velocity / initial mean velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ensemble mean velocity; the plotted smooth decay is not an individual noisy trajectory. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Kinetic Monte Carlo · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A kinetic Monte Carlo process has one constant total escape rate λ. Find the probability that its first event has occurred. For this calculation, x denotes the plotted horizontal coordinate (Elapsed hazard λt (dimensionless)), and q(x) denotes the plotted response (Event probability (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

P(T≤t)=1−e−λtP(T\le t)=1-e^{-\lambda t}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Kinetic Monte Carlo: Intermediate-condition prediction and exact check. Horizontal axis: Elapsed hazard λt (dimensionless). Vertical axis: Event probability (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Elapsed hazard λt (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Event probability (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Waiting-time distribution for a fixed state and rate, not the entire evolving event network. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Cahn–Hilliard model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use dimensionless Cahn–Hilliard dynamics with M = a = κ = 1, quadratic free energy ac²/2, periodic boundaries, and initial perturbation cos x. Plot t = 1. For this calculation, x denotes the plotted horizontal coordinate (Position x (dimensionless)), and q(x) denotes the plotted response (Composition perturbation (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

c−cˉ=e−2cos⁡xc-\bar c=e^{-2}\cos xa=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.20299×10−18\frac{q(b)-q(a)}{b-a}=-9.20299\times10^{-18}qlin(c)=0.0418209,q(c)=−0.114267q_{\rm lin}(c)=0.0418209,\quad q(c)=-0.114267e=qlin(c)−q(c)=0.156088e=q_{\rm lin}(c)-q(c)=0.156088

Solution. The endpoint responses are q(A) = 0.0418209 and q(B) = 0.0418209. Their secant slope is -9.20299e-18. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0418209. Direct evaluation gives -0.114267; subtracting it from the prediction gives signed error 0.156088.

Cahn–Hilliard model: Intermediate-condition prediction and exact check. Horizontal axis: Position x (dimensionless). Vertical axis: Composition perturbation (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Position x (dimensionless) −0.150 −0.125 −0.100 −0.075 −0.050 −0.025 0.000 0.025 0.050 Composition perturbation (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.04182 Worked point: (2.576, -0.1143)
At x = 2.57611, the orange circle is the analytical value -0.114267; the purple diamond is the interpolated value 0.0418209. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0418209; analytical reference = -0.114267; absolute interpolation error = 0.156088. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact quadratic-free-energy special case, not nonlinear phase separation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Allen–Cahn model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take mobility, positive quadratic free-energy curvature, and gradient coefficient all equal to one, with initial cos x. For this calculation, x denotes the plotted horizontal coordinate (Position x (dimensionless)), and q(x) denotes the plotted response (Order parameter η (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

η(x,1)=e−2cos⁡x\eta(x,1)=e^{-2}\cos xa=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.20299×10−18\frac{q(b)-q(a)}{b-a}=-9.20299\times10^{-18}qlin(c)=0.0418209,q(c)=−0.114267q_{\rm lin}(c)=0.0418209,\quad q(c)=-0.114267e=qlin(c)−q(c)=0.156088e=q_{\rm lin}(c)-q(c)=0.156088

Solution. The endpoint responses are q(A) = 0.0418209 and q(B) = 0.0418209. Their secant slope is -9.20299e-18. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0418209. Direct evaluation gives -0.114267; subtracting it from the prediction gives signed error 0.156088.

Allen–Cahn model: Intermediate-condition prediction and exact check. Horizontal axis: Position x (dimensionless). Vertical axis: Order parameter η (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Position x (dimensionless) −0.150 −0.125 −0.100 −0.075 −0.050 −0.025 0.000 0.025 0.050 Order parameter η (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.04182 Worked point: (2.576, -0.1143)
At x = 2.57611, the orange circle is the analytical value -0.114267; the purple diamond is the interpolated value 0.0418209. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0418209; analytical reference = -0.114267; absolute interpolation error = 0.156088. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linear quadratic-free-energy special case; domain walls of a double-well model are not represented. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Phase-field crystal model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Linearize ∂tψ = ∇²[(r+(1+∇²)²)ψ+ψ³] about ψ = 0 with r = 1; initial amplitude A₀ = 0.01 and wave number one. For this calculation, x denotes the plotted horizontal coordinate (Position x (dimensionless)), and q(x) denotes the plotted response (Density perturbation δψ (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

δψ(x,1)=A0e−1cos⁡x\delta\psi(x,1)=A_0e^{-1}\cos xa=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.30075×10−19\frac{q(b)-q(a)}{b-a}=-2.30075\times10^{-19}qlin(c)=0.00113681,q(c)=−0.00310611q_{\rm lin}(c)=0.00113681,\quad q(c)=-0.00310611e=qlin(c)−q(c)=0.00424292e=q_{\rm lin}(c)-q(c)=0.00424292

Solution. The endpoint responses are q(A) = 0.00113681 and q(B) = 0.00113681. Their secant slope is -2.30075e-19. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.00113681. Direct evaluation gives -0.00310611; subtracting it from the prediction gives signed error 0.00424292.

Phase-field crystal model: Intermediate-condition prediction and exact check. Horizontal axis: Position x (dimensionless). Vertical axis: Density perturbation δψ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Position x (dimensionless) −0.004 −0.003 −0.002 −0.001 0.000 0.001 Density perturbation δψ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.001137 Worked point: (2.576, -0.003106)
At x = 2.57611, the orange circle is the analytical value -0.00310611; the purple diamond is the interpolated value 0.00113681. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.00113681; analytical reference = -0.00310611; absolute interpolation error = 0.00424292. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linearized small-perturbation solution; the cubic term is omitted. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Potts grain-growth model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a three-state two-site Potts pair with energy −J when the states agree, compute the equilibrium agreement probability. For this calculation, x denotes the plotted horizontal coordinate (Coupling / thermal energy J/kBT (dimensionless)), and q(x) denotes the plotted response (Alignment probability (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

Psame=eueu+q−1,q=3P_{\rm same}=\frac{e^u}{e^u+q-1},\quad q=3a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.129515\frac{q(b)-q(a)}{b-a}=0.129515qlin(c)=0.712108,q(c)=0.795248q_{\rm lin}(c)=0.712108,\quad q(c)=0.795248e=qlin(c)−q(c)=−0.0831396e=q_{\rm lin}(c)-q(c)=-0.0831396

Solution. The endpoint responses are q(A) = 0.576117 and q(B) = 0.964663. Their secant slope is 0.129515. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.712108. Direct evaluation gives 0.795248; subtracting it from the prediction gives signed error -0.0831396.

Potts grain-growth model: Intermediate-condition prediction and exact check. Horizontal axis: Coupling / thermal energy J/kBT (dimensionless). Vertical axis: Alignment probability (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Coupling / thermal energy J/kBT (dimensionless) 0.6 0.7 0.8 0.9 1.0 Alignment probability (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7121 Worked point: (2.05, 0.7952)
At x = 2.05, the orange circle is the analytical value 0.795248; the purple diamond is the interpolated value 0.712108. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.712108; analytical reference = 0.795248; absolute interpolation error = 0.0831396. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Finite equilibrium toy problem, not a simulated grain-growth history. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Discrete dislocation dynamics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take one straight segment, constant force per length f = 1 N/m and mobility M = 1 m²/(N·s), starting at x = 0. For this calculation, x denotes the plotted horizontal coordinate (Time t (s)), and q(x) denotes the plotted response (Dislocation displacement (m)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

x(t)=Mftx(t)=Mfta=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=2.05,q(c)=2.05q_{\rm lin}(c)=2.05,\quad q(c)=2.05e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 1 and q(B) = 4. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.05. Direct evaluation gives 2.05; subtracting it from the prediction gives signed error 0.

Discrete dislocation dynamics: Intermediate-condition prediction and exact check. Horizontal axis: Time t (s). Vertical axis: Dislocation displacement (m). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time t (s) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Dislocation displacement (m) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.05 Worked point: (2.05, 2.05)
At x = 2.05, the orange circle is the analytical value 2.05; the purple diamond is the interpolated value 2.05. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.05; analytical reference = 2.05; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative coefficients; interactions, pinning, and changing segment geometry are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Population balance model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For ∂tn+∂sn = 0 use n(s,0) = exp[−(s−2)²], constant growth G = 1, and compatible boundary inflow. Plot t = 1. For this calculation, x denotes the plotted horizontal coordinate (Particle size s (dimensionless)), and q(x) denotes the plotted response (Number-density profile (dimensionless)). Suppose only the analytical endpoint responses at x = 1.2 and x = 4.8 are tabulated. Use linear interpolation to predict the response at x = 2.46, then check it against the original equation.

n(s,1)=e−(s−3)2n(s,1)=e^{-(s-3)^2}a=1.2,b=4.8,c=2.46a=1.2,\quad b=4.8,\quad c=2.46qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.46945×10−17\frac{q(b)-q(a)}{b-a}=-3.46945\times10^{-17}qlin(c)=0.0391639,q(c)=0.747067q_{\rm lin}(c)=0.0391639,\quad q(c)=0.747067e=qlin(c)−q(c)=−0.707903e=q_{\rm lin}(c)-q(c)=-0.707903

Solution. The endpoint responses are q(A) = 0.0391639 and q(B) = 0.0391639. Their secant slope is -3.46945e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0391639. Direct evaluation gives 0.747067; subtracting it from the prediction gives signed error -0.707903.

Population balance model: Intermediate-condition prediction and exact check. Horizontal axis: Particle size s (dimensionless). Vertical axis: Number-density profile (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Particle size s (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Number-density profile (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.03916 Worked point: (2.46, 0.7471)
At x = 2.46, the orange circle is the analytical value 0.747067; the purple diamond is the interpolated value 0.0391639. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0391639; analytical reference = 0.747067; absolute interpolation error = 0.707903. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Ideal gas equation of state · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold temperature and amount of ideal gas fixed while varying its volume. For this calculation, x denotes the plotted horizontal coordinate (Volume V / V* (dimensionless)), and q(x) denotes the plotted response (Pressure pV* / nRT (dimensionless)). Suppose only the analytical endpoint responses at x = 1.4 and x = 4.1 are tabulated. Use linear interpolation to predict the response at x = 2.345, then check it against the original equation.

pV∗/(nRT)=1/(V/V∗)pV_*/(nRT)=1/(V/V_*)a=1.4,b=4.1,c=2.345a=1.4,\quad b=4.1,\quad c=2.345qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.174216\frac{q(b)-q(a)}{b-a}=-0.174216qlin(c)=0.549652,q(c)=0.426439q_{\rm lin}(c)=0.549652,\quad q(c)=0.426439e=qlin(c)−q(c)=0.123212e=q_{\rm lin}(c)-q(c)=0.123212

Solution. The endpoint responses are q(A) = 0.714286 and q(B) = 0.243902. Their secant slope is -0.174216. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.549652. Direct evaluation gives 0.426439; subtracting it from the prediction gives signed error 0.123212.

Ideal gas equation of state: Intermediate-condition prediction and exact check. Horizontal axis: Volume V / V* (dimensionless). Vertical axis: Pressure pV* / nRT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 Volume V / V* (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 Pressure pV* / nRT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5497 Worked point: (2.345, 0.4264)
At x = 2.345, the orange circle is the analytical value 0.426439; the purple diamond is the interpolated value 0.549652. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.549652; analytical reference = 0.426439; absolute interpolation error = 0.123212. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Van der Waals equation of state · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the reduced van der Waals equation at T/Tc = 1.2. For this calculation, x denotes the plotted horizontal coordinate (Molar volume Vₘ / Vc (dimensionless)), and q(x) denotes the plotted response (Pressure p / pc (dimensionless)). Suppose only the analytical endpoint responses at x = 1.28 and x = 3.32 are tabulated. Use linear interpolation to predict the response at x = 1.994, then check it against the original equation.

p/pc=8(1.2)3v−1−3v2p/p_c=\frac{8(1.2)}{3v-1}-\frac3{v^2}a=1.28,b=3.32,c=1.994a=1.28,\quad b=3.32,\quad c=1.994qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.367633\frac{q(b)-q(a)}{b-a}=-0.367633qlin(c)=1.28674,q(c)=1.17242q_{\rm lin}(c)=1.28674,\quad q(c)=1.17242e=qlin(c)−q(c)=0.11432e=q_{\rm lin}(c)-q(c)=0.11432

Solution. The endpoint responses are q(A) = 1.54923 and q(B) = 0.799256. Their secant slope is -0.367633. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.28674. Direct evaluation gives 1.17242; subtracting it from the prediction gives signed error 0.11432.

Van der Waals equation of state: Intermediate-condition prediction and exact check. Horizontal axis: Molar volume Vₘ / Vc (dimensionless). Vertical axis: Pressure p / pc (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 Molar volume Vₘ / Vc (dimensionless) 0.8 1.0 1.2 1.4 1.6 Pressure p / pc (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.287 Worked point: (1.994, 1.172)
At x = 1.994, the orange circle is the analytical value 1.17242; the purple diamond is the interpolated value 1.28674. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.28674; analytical reference = 1.17242; absolute interpolation error = 0.11432. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Peng–Robinson equation of state · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. At fixed temperature choose aα/(RTb) = 2 and evaluate the Peng–Robinson pressure. For this calculation, x denotes the plotted horizontal coordinate (Molar volume v = Vₘ / b (dimensionless)), and q(x) denotes the plotted response (Pressure pb / RT (dimensionless)). Suppose only the analytical endpoint responses at x = 2.4 and x = 5.1 are tabulated. Use linear interpolation to predict the response at x = 3.345, then check it against the original equation.

pb/(RT)=1v−1−2v2+2v−1pb/(RT)=\frac1{v-1}-\frac2{v^2+2v-1}a=2.4,b=5.1,c=3.345a=2.4,\quad b=5.1,\quad c=3.345qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.11777\frac{q(b)-q(a)}{b-a}=-0.11777qlin(c)=0.393788,q(c)=0.307949q_{\rm lin}(c)=0.393788,\quad q(c)=0.307949e=qlin(c)−q(c)=0.0858386e=q_{\rm lin}(c)-q(c)=0.0858386

Solution. The endpoint responses are q(A) = 0.505081 and q(B) = 0.1871. Their secant slope is -0.11777. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.393788. Direct evaluation gives 0.307949; subtracting it from the prediction gives signed error 0.0858386.

Peng–Robinson equation of state: Intermediate-condition prediction and exact check. Horizontal axis: Molar volume v = Vₘ / b (dimensionless). Vertical axis: Pressure pb / RT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2.5 3.0 3.5 4.0 4.5 5.0 Molar volume v = Vₘ / b (dimensionless) 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 Pressure pb / RT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3938 Worked point: (3.345, 0.3079)
At x = 3.345, the orange circle is the analytical value 0.307949; the purple diamond is the interpolated value 0.393788. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.393788; analytical reference = 0.307949; absolute interpolation error = 0.0858386. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative EOS parameters; not a fitted fluid or a phase-equilibrium calculation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Soave–Redlich–Kwong equation of state · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. At fixed temperature choose aα/(RTb) = 2 for the SRK equation. For this calculation, x denotes the plotted horizontal coordinate (Molar volume v = Vₘ / b (dimensionless)), and q(x) denotes the plotted response (Pressure pb / RT (dimensionless)). Suppose only the analytical endpoint responses at x = 2.4 and x = 5.1 are tabulated. Use linear interpolation to predict the response at x = 3.345, then check it against the original equation.

pb/(RT)=1v−1−2v(v+1)pb/(RT)=\frac1{v-1}-\frac2{v(v+1)}a=2.4,b=5.1,c=3.345a=2.4,\quad b=5.1,\quad c=3.345qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.107249\frac{q(b)-q(a)}{b-a}=-0.107249qlin(c)=0.367837,q(c)=0.288831q_{\rm lin}(c)=0.367837,\quad q(c)=0.288831e=qlin(c)−q(c)=0.0790059e=q_{\rm lin}(c)-q(c)=0.0790059

Solution. The endpoint responses are q(A) = 0.469188 and q(B) = 0.179614. Their secant slope is -0.107249. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.367837. Direct evaluation gives 0.288831; subtracting it from the prediction gives signed error 0.0790059.

Soave–Redlich–Kwong equation of state: Intermediate-condition prediction and exact check. Horizontal axis: Molar volume v = Vₘ / b (dimensionless). Vertical axis: Pressure pb / RT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2.5 3.0 3.5 4.0 4.5 5.0 Molar volume v = Vₘ / b (dimensionless) 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 Pressure pb / RT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3678 Worked point: (3.345, 0.2888)
At x = 3.345, the orange circle is the analytical value 0.288831; the purple diamond is the interpolated value 0.367837. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.367837; analytical reference = 0.288831; absolute interpolation error = 0.0790059. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative parameters; the temperature dependence of α is fixed for this isotherm. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Virial equation of state · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use scaled second and third virial coefficients 0.2 and 0.05 over a dilute density interval. For this calculation, x denotes the plotted horizontal coordinate (Reduced density ρ* (dimensionless)), and q(x) denotes the plotted response (Compressibility factor Z (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

Z=1+0.2ρ∗+0.05ρ∗2Z=1+0.2\rho_*+0.05\rho_*^2a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.25\frac{q(b)-q(a)}{b-a}=0.25qlin(c)=1.0945,q(c)=1.09041q_{\rm lin}(c)=1.0945,\quad q(c)=1.09041e=qlin(c)−q(c)=0.004095e=q_{\rm lin}(c)-q(c)=0.004095

Solution. The endpoint responses are q(A) = 1.042 and q(B) = 1.192. Their secant slope is 0.25. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.0945. Direct evaluation gives 1.09041; subtracting it from the prediction gives signed error 0.004095.

Virial equation of state: Intermediate-condition prediction and exact check. Horizontal axis: Reduced density ρ* (dimensionless). Vertical axis: Compressibility factor Z (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Reduced density ρ* (dimensionless) 1.025 1.050 1.075 1.100 1.125 1.150 1.175 1.200 Compressibility factor Z (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.095 Worked point: (0.41, 1.09)
At x = 0.41, the orange circle is the analytical value 1.09041; the purple diamond is the interpolated value 1.0945. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.0945; analytical reference = 1.09041; absolute interpolation error = 0.004095. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Truncated low-density illustrative expansion, not an extrapolation to dense fluids. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Gibbs-energy minimization · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take an ideal binary solution with equal pure-component reference energies. Find the composition dependence of its mixing free energy. For this calculation, x denotes the plotted horizontal coordinate (Mole fraction x (dimensionless)), and q(x) denotes the plotted response (Mixing free energy / RT (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2006 and x = 0.7994 are tabulated. Use linear interpolation to predict the response at x = 0.41018, then check it against the original equation.

Δg/(RT)=xln⁡x+(1−x)ln⁡(1−x)\Delta g/(RT)=x\ln x+(1-x)\ln(1-x)a=0.2006,b=0.7994,c=0.41018a=0.2006,\quad b=0.7994,\quad c=0.41018qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0\frac{q(b)-q(a)}{b-a}=0qlin(c)=−0.501233,q(c)=−0.676924q_{\rm lin}(c)=-0.501233,\quad q(c)=-0.676924e=qlin(c)−q(c)=0.175691e=q_{\rm lin}(c)-q(c)=0.175691

Solution. The endpoint responses are q(A) = -0.501233 and q(B) = -0.501233. Their secant slope is 0. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.501233. Direct evaluation gives -0.676924; subtracting it from the prediction gives signed error 0.175691.

Gibbs-energy minimization: Intermediate-condition prediction and exact check. Horizontal axis: Mole fraction x (dimensionless). Vertical axis: Mixing free energy / RT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Mole fraction x (dimensionless) −0.70 −0.65 −0.60 −0.55 −0.50 Mixing free energy / RT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.5012 Worked point: (0.4102, -0.6769)
At x = 0.41018, the orange circle is the analytical value -0.676924; the purple diamond is the interpolated value -0.501233. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.501233; analytical reference = -0.676924; absolute interpolation error = 0.175691. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ideal-solution Gibbs term; real CALPHAD databases include additional phase and interaction terms. Conserved bulk composition constrains accessible equilibria. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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CALPHAD model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take an ideal binary solution with equal pure-component reference energies. Find the composition dependence of its mixing free energy. For this calculation, x denotes the plotted horizontal coordinate (Mole fraction x (dimensionless)), and q(x) denotes the plotted response (Mixing free energy / RT (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2006 and x = 0.7994 are tabulated. Use linear interpolation to predict the response at x = 0.41018, then check it against the original equation.

Δg/(RT)=xln⁡x+(1−x)ln⁡(1−x)\Delta g/(RT)=x\ln x+(1-x)\ln(1-x)a=0.2006,b=0.7994,c=0.41018a=0.2006,\quad b=0.7994,\quad c=0.41018qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0\frac{q(b)-q(a)}{b-a}=0qlin(c)=−0.501233,q(c)=−0.676924q_{\rm lin}(c)=-0.501233,\quad q(c)=-0.676924e=qlin(c)−q(c)=0.175691e=q_{\rm lin}(c)-q(c)=0.175691

Solution. The endpoint responses are q(A) = -0.501233 and q(B) = -0.501233. Their secant slope is 0. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.501233. Direct evaluation gives -0.676924; subtracting it from the prediction gives signed error 0.175691.

CALPHAD model: Intermediate-condition prediction and exact check. Horizontal axis: Mole fraction x (dimensionless). Vertical axis: Mixing free energy / RT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Mole fraction x (dimensionless) −0.70 −0.65 −0.60 −0.55 −0.50 Mixing free energy / RT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.5012 Worked point: (0.4102, -0.6769)
At x = 0.41018, the orange circle is the analytical value -0.676924; the purple diamond is the interpolated value -0.501233. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.501233; analytical reference = -0.676924; absolute interpolation error = 0.175691. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ideal-solution Gibbs term; real CALPHAD databases include additional phase and interaction terms. Conserved bulk composition constrains accessible equilibria. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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NRTL activity model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Set NRTL interaction parameters to zero; for UNIQUAC also take identical molecular sizes and shapes with zero interaction energies. For this calculation, x denotes the plotted horizontal coordinate (Mole fraction x₁ (dimensionless)), and q(x) denotes the plotted response (Component activity a₁ (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

a1=x1,γ1=1a_1=x_1,\quad\gamma_1=1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.41,q(c)=0.41q_{\rm lin}(c)=0.41,\quad q(c)=0.41e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.2 and q(B) = 0.8. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.41. Direct evaluation gives 0.41; subtracting it from the prediction gives signed error 0.

NRTL activity model: Intermediate-condition prediction and exact check. Horizontal axis: Mole fraction x₁ (dimensionless). Vertical axis: Component activity a₁ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Mole fraction x₁ (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Component activity a₁ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.41 Worked point: (0.41, 0.41)
At x = 0.41, the orange circle is the analytical value 0.41; the purple diamond is the interpolated value 0.41. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.41; analytical reference = 0.41; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ideal-mixture limiting case only; unequal molecular sizes in UNIQUAC can retain a combinatorial contribution. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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UNIQUAC activity model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Set NRTL interaction parameters to zero; for UNIQUAC also take identical molecular sizes and shapes with zero interaction energies. For this calculation, x denotes the plotted horizontal coordinate (Mole fraction x₁ (dimensionless)), and q(x) denotes the plotted response (Component activity a₁ (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

a1=x1,γ1=1a_1=x_1,\quad\gamma_1=1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.41,q(c)=0.41q_{\rm lin}(c)=0.41,\quad q(c)=0.41e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.2 and q(B) = 0.8. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.41. Direct evaluation gives 0.41; subtracting it from the prediction gives signed error 0.

UNIQUAC activity model: Intermediate-condition prediction and exact check. Horizontal axis: Mole fraction x₁ (dimensionless). Vertical axis: Component activity a₁ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Mole fraction x₁ (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Component activity a₁ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.41 Worked point: (0.41, 0.41)
At x = 0.41, the orange circle is the analytical value 0.41; the purple diamond is the interpolated value 0.41. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.41; analytical reference = 0.41; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ideal-mixture limiting case only; unequal molecular sizes in UNIQUAC can retain a combinatorial contribution. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Debye–Hückel model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a monovalent ion in water near 25 °C use the Debye–Hückel limiting-law coefficient A = 0.509 (mol/L)⁻¹ᐟ². For this calculation, x denotes the plotted horizontal coordinate (Ionic strength I (mol/L)), and q(x) denotes the plotted response (log₁₀(activity coefficient) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.002 and x = 0.008 are tabulated. Use linear interpolation to predict the response at x = 0.0041, then check it against the original equation.

log⁡10γ=−0.509I\log_{10}\gamma=-0.509\sqrt Ia=0.002,b=0.008,c=0.0041a=0.002,\quad b=0.008,\quad c=0.0041qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.79386\frac{q(b)-q(a)}{b-a}=-3.79386qlin(c)=−0.0307303,q(c)=−0.0325919q_{\rm lin}(c)=-0.0307303,\quad q(c)=-0.0325919e=qlin(c)−q(c)=0.00186162e=q_{\rm lin}(c)-q(c)=0.00186162

Solution. The endpoint responses are q(A) = -0.0227632 and q(B) = -0.0455263. Their secant slope is -3.79386. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.0307303. Direct evaluation gives -0.0325919; subtracting it from the prediction gives signed error 0.00186162.

Debye–Hückel model: Intermediate-condition prediction and exact check. Horizontal axis: Ionic strength I (mol/L). Vertical axis: log₁₀(activity coefficient) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Ionic strength I (mol/L) −0.045 −0.040 −0.035 −0.030 −0.025 log₁₀(activity coefficient) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.03073 Worked point: (0.0041, -0.03259)
At x = 0.0041, the orange circle is the analytical value -0.0325919; the purple diamond is the interpolated value -0.0307303. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.0307303; analytical reference = -0.0325919; absolute interpolation error = 0.00186162. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Limiting-law illustration; specific ion interactions and concentrated solutions are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Mass-action reaction kinetics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a single irreversible first-order reaction A → products in a constant-volume batch, use τ = kt and y = cA/cA0. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Reactant concentration / initial concentration (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Mass-action reaction kinetics: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Reactant concentration / initial concentration (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Reactant concentration / initial concentration (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact one-mode reduction with constant coefficients; additional coupled physics is excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Batch reactor model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a single irreversible first-order reaction A → products in a constant-volume batch, use τ = kt and y = cA/cA0. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Reactant concentration / initial concentration (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Batch reactor model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Reactant concentration / initial concentration (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Reactant concentration / initial concentration (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact one-mode reduction with constant coefficients; additional coupled physics is excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Arrhenius rate model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold activation energy Ea > 0 and prefactor A constant. For this calculation, x denotes the plotted horizontal coordinate (Scaled temperature RT / Ea (dimensionless)), and q(x) denotes the plotted response (Rate constant / prefactor k/A (dimensionless)). Suppose only the analytical endpoint responses at x = 0.28 and x = 0.82 are tabulated. Use linear interpolation to predict the response at x = 0.469, then check it against the original equation.

k/A=e−1/θ,θ=RT/Eak/A=e^{-1/\theta},\quad\theta=RT/E_aa=0.28,b=0.82,c=0.469a=0.28,\quad b=0.82,\quad c=0.469qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.494923\frac{q(b)-q(a)}{b-a}=0.494923qlin(c)=0.121656,q(c)=0.118577q_{\rm lin}(c)=0.121656,\quad q(c)=0.118577e=qlin(c)−q(c)=0.00307956e=q_{\rm lin}(c)-q(c)=0.00307956

Solution. The endpoint responses are q(A) = 0.0281157 and q(B) = 0.295374. Their secant slope is 0.494923. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.121656. Direct evaluation gives 0.118577; subtracting it from the prediction gives signed error 0.00307956.

Arrhenius rate model: Intermediate-condition prediction and exact check. Horizontal axis: Scaled temperature RT / Ea (dimensionless). Vertical axis: Rate constant / prefactor k/A (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.3 0.4 0.5 0.6 0.7 0.8 Scaled temperature RT / Ea (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 Rate constant / prefactor k/A (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1217 Worked point: (0.469, 0.1186)
At x = 0.469, the orange circle is the analytical value 0.118577; the purple diamond is the interpolated value 0.121656. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.121656; analytical reference = 0.118577; absolute interpolation error = 0.00307956. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Transition-state theory · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take transmission coefficient one and treat the molar activation free energy as constant over the displayed interval. For this calculation, x denotes the plotted horizontal coordinate (Scaled temperature RT / ΔG‡ (dimensionless)), and q(x) denotes the plotted response (Scaled rate kh / kBT (dimensionless)). Suppose only the analytical endpoint responses at x = 0.28 and x = 0.82 are tabulated. Use linear interpolation to predict the response at x = 0.469, then check it against the original equation.

kh/(kBT)=e−1/θ,θ=RT/ΔG‡kh/(k_BT)=e^{-1/\theta},\quad\theta=RT/\Delta G^\ddaggera=0.28,b=0.82,c=0.469a=0.28,\quad b=0.82,\quad c=0.469qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.494923\frac{q(b)-q(a)}{b-a}=0.494923qlin(c)=0.121656,q(c)=0.118577q_{\rm lin}(c)=0.121656,\quad q(c)=0.118577e=qlin(c)−q(c)=0.00307956e=q_{\rm lin}(c)-q(c)=0.00307956

Solution. The endpoint responses are q(A) = 0.0281157 and q(B) = 0.295374. Their secant slope is 0.494923. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.121656. Direct evaluation gives 0.118577; subtracting it from the prediction gives signed error 0.00307956.

Transition-state theory: Intermediate-condition prediction and exact check. Horizontal axis: Scaled temperature RT / ΔG‡ (dimensionless). Vertical axis: Scaled rate kh / kBT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.3 0.4 0.5 0.6 0.7 0.8 Scaled temperature RT / ΔG‡ (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 Scaled rate kh / kBT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1217 Worked point: (0.469, 0.1186)
At x = 0.469, the orange circle is the analytical value 0.118577; the purple diamond is the interpolated value 0.121656. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.121656; analytical reference = 0.118577; absolute interpolation error = 0.00307956. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative fixed-barrier curve; real activation free energy can vary with temperature. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Michaelis–Menten kinetics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For Michaelis–Menten set x = substrate/Km and y = v/Vmax. For Langmuir adsorption set x = KP and y = occupied-site fraction. For this calculation, x denotes the plotted horizontal coordinate (Scaled concentration or pressure (dimensionless)), and q(x) denotes the plotted response (Fraction of saturation (dimensionless)). Suppose only the analytical endpoint responses at x = 1.6 and x = 6.4 are tabulated. Use linear interpolation to predict the response at x = 3.28, then check it against the original equation.

y=x1+xy=\frac{x}{1+x}a=1.6,b=6.4,c=3.28a=1.6,\quad b=6.4,\quad c=3.28qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.0519751\frac{q(b)-q(a)}{b-a}=0.0519751qlin(c)=0.702703,q(c)=0.766355q_{\rm lin}(c)=0.702703,\quad q(c)=0.766355e=qlin(c)−q(c)=−0.0636524e=q_{\rm lin}(c)-q(c)=-0.0636524

Solution. The endpoint responses are q(A) = 0.615385 and q(B) = 0.864865. Their secant slope is 0.0519751. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.702703. Direct evaluation gives 0.766355; subtracting it from the prediction gives signed error -0.0636524.

Michaelis–Menten kinetics: Intermediate-condition prediction and exact check. Horizontal axis: Scaled concentration or pressure (dimensionless). Vertical axis: Fraction of saturation (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2 3 4 5 6 Scaled concentration or pressure (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 Fraction of saturation (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7027 Worked point: (3.28, 0.7664)
At x = 3.28, the orange circle is the analytical value 0.766355; the purple diamond is the interpolated value 0.702703. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.702703; analytical reference = 0.766355; absolute interpolation error = 0.0636524. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single-substrate steady enzyme law or single-species equilibrium adsorption, as appropriate to the entry. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Langmuir adsorption isotherm · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For Michaelis–Menten set x = substrate/Km and y = v/Vmax. For Langmuir adsorption set x = KP and y = occupied-site fraction. For this calculation, x denotes the plotted horizontal coordinate (Scaled concentration or pressure (dimensionless)), and q(x) denotes the plotted response (Fraction of saturation (dimensionless)). Suppose only the analytical endpoint responses at x = 1.6 and x = 6.4 are tabulated. Use linear interpolation to predict the response at x = 3.28, then check it against the original equation.

y=x1+xy=\frac{x}{1+x}a=1.6,b=6.4,c=3.28a=1.6,\quad b=6.4,\quad c=3.28qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.0519751\frac{q(b)-q(a)}{b-a}=0.0519751qlin(c)=0.702703,q(c)=0.766355q_{\rm lin}(c)=0.702703,\quad q(c)=0.766355e=qlin(c)−q(c)=−0.0636524e=q_{\rm lin}(c)-q(c)=-0.0636524

Solution. The endpoint responses are q(A) = 0.615385 and q(B) = 0.864865. Their secant slope is 0.0519751. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.702703. Direct evaluation gives 0.766355; subtracting it from the prediction gives signed error -0.0636524.

Langmuir adsorption isotherm: Intermediate-condition prediction and exact check. Horizontal axis: Scaled concentration or pressure (dimensionless). Vertical axis: Fraction of saturation (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2 3 4 5 6 Scaled concentration or pressure (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 Fraction of saturation (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7027 Worked point: (3.28, 0.7664)
At x = 3.28, the orange circle is the analytical value 0.766355; the purple diamond is the interpolated value 0.702703. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.702703; analytical reference = 0.766355; absolute interpolation error = 0.0636524. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single-substrate steady enzyme law or single-species equilibrium adsorption, as appropriate to the entry. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Langmuir–Hinshelwood kinetics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the illustrative Langmuir–Hinshelwood rate r/r* = x/(1+x)², with other factors held constant. For this calculation, x denotes the plotted horizontal coordinate (Scaled reactant pressure x (dimensionless)), and q(x) denotes the plotted response (Scaled surface rate r / r* (dimensionless)). Suppose only the analytical endpoint responses at x = 1.6 and x = 6.4 are tabulated. Use linear interpolation to predict the response at x = 3.28, then check it against the original equation.

r/r∗=x(1+x)2r/r_*=\frac{x}{(1+x)^2}a=1.6,b=6.4,c=3.28a=1.6,\quad b=6.4,\quad c=3.28qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.024961\frac{q(b)-q(a)}{b-a}=-0.024961qlin(c)=0.194752,q(c)=0.179055q_{\rm lin}(c)=0.194752,\quad q(c)=0.179055e=qlin(c)−q(c)=0.015697e=q_{\rm lin}(c)-q(c)=0.015697

Solution. The endpoint responses are q(A) = 0.236686 and q(B) = 0.116874. Their secant slope is -0.024961. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.194752. Direct evaluation gives 0.179055; subtracting it from the prediction gives signed error 0.015697.

Langmuir–Hinshelwood kinetics: Intermediate-condition prediction and exact check. Horizontal axis: Scaled reactant pressure x (dimensionless). Vertical axis: Scaled surface rate r / r* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2 3 4 5 6 Scaled reactant pressure x (dimensionless) 0.12 0.14 0.16 0.18 0.20 0.22 0.24 Scaled surface rate r / r* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1948 Worked point: (3.28, 0.1791)
At x = 3.28, the orange circle is the analytical value 0.179055; the purple diamond is the interpolated value 0.194752. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.194752; analytical reference = 0.179055; absolute interpolation error = 0.015697. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One specified adsorption-limited rate law; the family contains many different mechanisms. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Fickian diffusion · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve ∂τu = ∂ξξu with u(0,τ)=u(1,τ)=0 and initial sin(πξ), then plot τ = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Concentration perturbation / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Fickian diffusion: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Concentration perturbation / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Concentration perturbation / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Fickian constant-diffusivity slab. Maxwell–Stefan reduces to this form for an ideal binary mixture with constant total concentration and diffusivity. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Maxwell–Stefan diffusion · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve ∂τu = ∂ξξu with u(0,τ)=u(1,τ)=0 and initial sin(πξ), then plot τ = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Concentration perturbation / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Maxwell–Stefan diffusion: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Concentration perturbation / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Concentration perturbation / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Fickian constant-diffusivity slab. Maxwell–Stefan reduces to this form for an ideal binary mixture with constant total concentration and diffusivity. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Advection–diffusion–reaction model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. On the infinite line solve ut+ux = 0.1uxx−0.2u with u(x,0)=exp(−x²). Plot t = 1. For this calculation, x denotes the plotted horizontal coordinate (Position x (dimensionless)), and q(x) denotes the plotted response (Concentration u (dimensionless)). Suppose only the analytical endpoint responses at x = -1.4 and x = 3.4 are tabulated. Use linear interpolation to predict the response at x = 0.28, then check it against the original equation.

u(x,1)=e−0.21.4exp⁡[−(x−1)2/1.4]u(x,1)=\frac{e^{-0.2}}{\sqrt{1.4}}\exp[-(x-1)^2/1.4]a=−1.4,b=3.4,c=0.28a=-1.4,\quad b=3.4,\quad c=0.28qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.52981×10−18\frac{q(b)-q(a)}{b-a}=-2.52981\times10^{-18}qlin(c)=0.0113049,q(c)=0.47782q_{\rm lin}(c)=0.0113049,\quad q(c)=0.47782e=qlin(c)−q(c)=−0.466515e=q_{\rm lin}(c)-q(c)=-0.466515

Solution. The endpoint responses are q(A) = 0.0113049 and q(B) = 0.0113049. Their secant slope is -2.52981e-18. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0113049. Direct evaluation gives 0.47782; subtracting it from the prediction gives signed error -0.466515.

Advection–diffusion–reaction model: Intermediate-condition prediction and exact check. Horizontal axis: Position x (dimensionless). Vertical axis: Concentration u (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1 0 1 2 3 Position x (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Concentration u (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0113 Worked point: (0.28, 0.4778)
At x = 0.28, the orange circle is the analytical value 0.47782; the purple diamond is the interpolated value 0.0113049. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0113049; analytical reference = 0.47782; absolute interpolation error = 0.466515. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Continuous stirred-tank reactor (CSTR) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. At steady state a well-mixed reactor consumes A by a first-order reaction at rate kcA. For this calculation, x denotes the plotted horizontal coordinate (Damköhler number kV/Q (dimensionless)), and q(x) denotes the plotted response (Outlet / inlet concentration (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

cout/cin=1/(1+Da)c_{\rm out}/c_{\rm in}=1/(1+\mathrm{Da})a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.1\frac{q(b)-q(a)}{b-a}=-0.1qlin(c)=0.395,q(c)=0.327869q_{\rm lin}(c)=0.395,\quad q(c)=0.327869e=qlin(c)−q(c)=0.0671311e=q_{\rm lin}(c)-q(c)=0.0671311

Solution. The endpoint responses are q(A) = 0.5 and q(B) = 0.2. Their secant slope is -0.1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.395. Direct evaluation gives 0.327869; subtracting it from the prediction gives signed error 0.0671311.

Continuous stirred-tank reactor (CSTR): Intermediate-condition prediction and exact check. Horizontal axis: Damköhler number kV/Q (dimensionless). Vertical axis: Outlet / inlet concentration (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Damköhler number kV/Q (dimensionless) 0.20 0.25 0.30 0.35 0.40 0.45 0.50 Outlet / inlet concentration (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.395 Worked point: (2.05, 0.3279)
At x = 2.05, the orange circle is the analytical value 0.327869; the purple diamond is the interpolated value 0.395. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.395; analytical reference = 0.327869; absolute interpolation error = 0.0671311. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Constant-volume, isothermal, constant-flow reactor. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Plug-flow reactor (PFR) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For an isothermal PFR with constant velocity u and first-order consumption k, use τ = kz/u and y = c/cin. For this calculation, x denotes the plotted horizontal coordinate (Axial residence coordinate kz / u (dimensionless)), and q(x) denotes the plotted response (Reactant concentration / inlet concentration (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Plug-flow reactor (PFR): Intermediate-condition prediction and exact check. Horizontal axis: Axial residence coordinate kz / u (dimensionless). Vertical axis: Reactant concentration / inlet concentration (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Axial residence coordinate kz / u (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Reactant concentration / inlet concentration (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact axial concentration profile in ideal plug flow; the horizontal coordinate is residence time kz/u, not laboratory time. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Navier–Stokes model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take steady, fully developed incompressible flow with constant viscosity between fixed parallel plates. For Hagen–Poiseuille use the equivalent diameter cut through a round pipe. For this calculation, x denotes the plotted horizontal coordinate (Transverse position / half-width (dimensionless)), and q(x) denotes the plotted response (Axial velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Navier–Stokes model: Intermediate-condition prediction and exact check. Horizontal axis: Transverse position / half-width (dimensionless). Vertical axis: Axial velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Transverse position / half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Axial velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact laminar benchmark. Plate and pipe pressure-to-maximum-speed factors differ; the plotted normalized profile is identical. DNS here resolves this simple laminar case. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Stokes creeping-flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take steady, fully developed incompressible flow with constant viscosity between fixed parallel plates. For Hagen–Poiseuille use the equivalent diameter cut through a round pipe. For this calculation, x denotes the plotted horizontal coordinate (Transverse position / half-width (dimensionless)), and q(x) denotes the plotted response (Axial velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Stokes creeping-flow model: Intermediate-condition prediction and exact check. Horizontal axis: Transverse position / half-width (dimensionless). Vertical axis: Axial velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Transverse position / half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Axial velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact laminar benchmark. Plate and pipe pressure-to-maximum-speed factors differ; the plotted normalized profile is identical. DNS here resolves this simple laminar case. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lubrication approximation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take steady, fully developed incompressible flow with constant viscosity between fixed parallel plates. For Hagen–Poiseuille use the equivalent diameter cut through a round pipe. For this calculation, x denotes the plotted horizontal coordinate (Transverse position / half-width (dimensionless)), and q(x) denotes the plotted response (Axial velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Lubrication approximation: Intermediate-condition prediction and exact check. Horizontal axis: Transverse position / half-width (dimensionless). Vertical axis: Axial velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Transverse position / half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Axial velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact laminar benchmark. Plate and pipe pressure-to-maximum-speed factors differ; the plotted normalized profile is identical. DNS here resolves this simple laminar case. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hagen–Poiseuille model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take steady, fully developed incompressible flow with constant viscosity between fixed parallel plates. For Hagen–Poiseuille use the equivalent diameter cut through a round pipe. For this calculation, x denotes the plotted horizontal coordinate (Transverse position / half-width (dimensionless)), and q(x) denotes the plotted response (Axial velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Hagen–Poiseuille model: Intermediate-condition prediction and exact check. Horizontal axis: Transverse position / half-width (dimensionless). Vertical axis: Axial velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Transverse position / half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Axial velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact laminar benchmark. Plate and pipe pressure-to-maximum-speed factors differ; the plotted normalized profile is identical. DNS here resolves this simple laminar case. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Direct numerical simulation (DNS) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take steady, fully developed incompressible flow with constant viscosity between fixed parallel plates. For Hagen–Poiseuille use the equivalent diameter cut through a round pipe. For this calculation, x denotes the plotted horizontal coordinate (Transverse position / half-width (dimensionless)), and q(x) denotes the plotted response (Axial velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Direct numerical simulation (DNS): Intermediate-condition prediction and exact check. Horizontal axis: Transverse position / half-width (dimensionless). Vertical axis: Axial velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Transverse position / half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Axial velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact laminar benchmark. Plate and pipe pressure-to-maximum-speed factors differ; the plotted normalized profile is identical. DNS here resolves this simple laminar case. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Euler flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Linearize inviscid Euler flow about a uniform rest state and use a sinusoidal pressure perturbation. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Pressure perturbation / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Euler flow model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Pressure perturbation / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Pressure perturbation / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linear acoustic limit of Euler flow, not a finite-amplitude compressible flow solution. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Potential-flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Find surface pressure for incompressible, inviscid, irrotational uniform flow around a circular cylinder without circulation. For this calculation, x denotes the plotted horizontal coordinate (Cylinder surface angle θ (radian)), and q(x) denotes the plotted response (Pressure coefficient Cp (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

Cp=1−4sin⁡2θC_p=1-4\sin^2\thetaa=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.35597×10−16\frac{q(b)-q(a)}{b-a}=-2.35597\times10^{-16}qlin(c)=−2.61803,q(c)=−0.148441q_{\rm lin}(c)=-2.61803,\quad q(c)=-0.148441e=qlin(c)−q(c)=−2.46959e=q_{\rm lin}(c)-q(c)=-2.46959

Solution. The endpoint responses are q(A) = -2.61803 and q(B) = -2.61803. Their secant slope is -2.35597e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -2.61803. Direct evaluation gives -0.148441; subtracting it from the prediction gives signed error -2.46959.

Potential-flow model: Intermediate-condition prediction and exact check. Horizontal axis: Cylinder surface angle θ (radian). Vertical axis: Pressure coefficient Cp (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Cylinder surface angle θ (radian) −3 −2 −1 0 1 Pressure coefficient Cp (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -2.618 Worked point: (2.576, -0.1484)
At x = 2.57611, the orange circle is the analytical value -0.148441; the purple diamond is the interpolated value -2.61803. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -2.61803; analytical reference = -0.148441; absolute interpolation error = 2.46959. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. No viscosity or separation; this ideal model does not predict real cylinder drag. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Boundary-layer model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A flat wall suddenly moves at speed U beneath an initially stationary semi-infinite viscous fluid. For this calculation, x denotes the plotted horizontal coordinate (Similarity coordinate y / 2√(νt) (dimensionless)), and q(x) denotes the plotted response (Velocity u / wall speed U (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

u/U=erfc⁡(η),η=y/(2νt)u/U=\operatorname{erfc}(\eta),\quad\eta=y/(2\sqrt{\nu t})a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.219697\frac{q(b)-q(a)}{b-a}=-0.219697qlin(c)=0.257735,q(c)=0.0819499q_{\rm lin}(c)=0.257735,\quad q(c)=0.0819499e=qlin(c)−q(c)=0.175785e=q_{\rm lin}(c)-q(c)=0.175785

Solution. The endpoint responses are q(A) = 0.396144 and q(B) = 0.000688514. Their secant slope is -0.219697. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.257735. Direct evaluation gives 0.0819499; subtracting it from the prediction gives signed error 0.175785.

Boundary-layer model: Intermediate-condition prediction and exact check. Horizontal axis: Similarity coordinate y / 2√(νt) (dimensionless). Vertical axis: Velocity u / wall speed U (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Similarity coordinate y / 2√(νt) (dimensionless) 0.0 0.1 0.2 0.3 0.4 Velocity u / wall speed U (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2577 Worked point: (1.23, 0.08195)
At x = 1.23, the orange circle is the analytical value 0.0819499; the purple diamond is the interpolated value 0.257735. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.257735; analytical reference = 0.0819499; absolute interpolation error = 0.175785. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Stokes’ first problem, an unsteady boundary-layer benchmark; not the Blasius spatially developing solution. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Darcy–Weisbach model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold the Darcy friction factor f, pipe geometry, and density fixed. For this calculation, x denotes the plotted horizontal coordinate (Mean speed U / U* (dimensionless)), and q(x) denotes the plotted response (Scaled pressure drop (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

Δp/(fLρU∗2/2D)=(U/U∗)2\Delta p/(fL\rho U_*^2/2D)=(U/U_*)^2a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=3\frac{q(b)-q(a)}{b-a}=3qlin(c)=2.25,q(c)=1.5129q_{\rm lin}(c)=2.25,\quad q(c)=1.5129e=qlin(c)−q(c)=0.7371e=q_{\rm lin}(c)-q(c)=0.7371

Solution. The endpoint responses are q(A) = 0.36 and q(B) = 5.76. Their secant slope is 3. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.25. Direct evaluation gives 1.5129; subtracting it from the prediction gives signed error 0.7371.

Darcy–Weisbach model: Intermediate-condition prediction and exact check. Horizontal axis: Mean speed U / U* (dimensionless). Vertical axis: Scaled pressure drop (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Mean speed U / U* (dimensionless) 0 1 2 3 4 5 6 Scaled pressure drop (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.25 Worked point: (1.23, 1.513)
At x = 1.23, the orange circle is the analytical value 1.5129; the purple diamond is the interpolated value 2.25. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.25; analytical reference = 1.5129; absolute interpolation error = 0.7371. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Fixed-friction-factor illustration; f usually varies with Reynolds number and roughness. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Non-Newtonian power-law fluid · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose positive shear rates and power-law exponent n = 1/2, with reference stress K√(reference rate). For this calculation, x denotes the plotted horizontal coordinate (Shear rate / reference rate (dimensionless)), and q(x) denotes the plotted response (Shear stress / reference stress (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

τ/τ∗=(γ˙/γ˙∗)1/2\tau/\tau_*=(\dot\gamma/\dot\gamma_*)^{1/2}a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.372678\frac{q(b)-q(a)}{b-a}=0.372678qlin(c)=1.20748,q(c)=1.28062q_{\rm lin}(c)=1.20748,\quad q(c)=1.28062e=qlin(c)−q(c)=−0.0731481e=q_{\rm lin}(c)-q(c)=-0.0731481

Solution. The endpoint responses are q(A) = 0.894427 and q(B) = 1.78885. Their secant slope is 0.372678. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.20748. Direct evaluation gives 1.28062; subtracting it from the prediction gives signed error -0.0731481.

Non-Newtonian power-law fluid: Intermediate-condition prediction and exact check. Horizontal axis: Shear rate / reference rate (dimensionless). Vertical axis: Shear stress / reference stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Shear rate / reference rate (dimensionless) 0.8 1.0 1.2 1.4 1.6 1.8 Shear stress / reference stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.207 Worked point: (1.64, 1.281)
At x = 1.64, the orange circle is the analytical value 1.28062; the purple diamond is the interpolated value 1.20748. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.20748; analytical reference = 1.28062; absolute interpolation error = 0.0731481. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Steady shear constitutive evaluation; no low- or high-shear viscosity plateau is included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Bingham plastic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Increase a nonnegative applied shear stress on an ideal Bingham material. For this calculation, x denotes the plotted horizontal coordinate (Applied stress / yield stress (dimensionless)), and q(x) denotes the plotted response (Scaled shear rate μpγ̇ / τy (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

μpγ˙/τy=max⁡(s−1,0),s=τ/τy\mu_p\dot\gamma/\tau_y=\max(s-1,0),\quad s=\tau/\tau_ya=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.777778\frac{q(b)-q(a)}{b-a}=0.777778qlin(c)=0.49,q(c)=0.23q_{\rm lin}(c)=0.49,\quad q(c)=0.23e=qlin(c)−q(c)=0.26e=q_{\rm lin}(c)-q(c)=0.26

Solution. The endpoint responses are q(A) = 0 and q(B) = 1.4. Their secant slope is 0.777778. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.49. Direct evaluation gives 0.23; subtracting it from the prediction gives signed error 0.26.

Bingham plastic model: Intermediate-condition prediction and exact check. Horizontal axis: Applied stress / yield stress (dimensionless). Vertical axis: Scaled shear rate μpγ̇ / τy (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Applied stress / yield stress (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 Scaled shear rate μpγ̇ / τy (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.49 Worked point: (1.23, 0.23)
At x = 1.23, the orange circle is the analytical value 0.23; the purple diamond is the interpolated value 0.49. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.49; analytical reference = 0.23; absolute interpolation error = 0.26. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Herschel–Bulkley model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use exponent n = 1/2 and define g so that Kγ̇ⁿ/τy = √g. Evaluate the yielded branch. For this calculation, x denotes the plotted horizontal coordinate (Scaled positive shear rate g (dimensionless)), and q(x) denotes the plotted response (Shear stress / yield stress (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

τ/τy=1+g1/2\tau/\tau_y=1+g^{1/2}a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.372678\frac{q(b)-q(a)}{b-a}=0.372678qlin(c)=2.20748,q(c)=2.28062q_{\rm lin}(c)=2.20748,\quad q(c)=2.28062e=qlin(c)−q(c)=−0.0731481e=q_{\rm lin}(c)-q(c)=-0.0731481

Solution. The endpoint responses are q(A) = 1.89443 and q(B) = 2.78885. Their secant slope is 0.372678. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.20748. Direct evaluation gives 2.28062; subtracting it from the prediction gives signed error -0.0731481.

Herschel–Bulkley model: Intermediate-condition prediction and exact check. Horizontal axis: Scaled positive shear rate g (dimensionless). Vertical axis: Shear stress / yield stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Scaled positive shear rate g (dimensionless) 1.8 2.0 2.2 2.4 2.6 2.8 Shear stress / yield stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.207 Worked point: (1.64, 2.281)
At x = 1.64, the orange circle is the analytical value 2.28062; the purple diamond is the interpolated value 2.20748. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.20748; analytical reference = 2.28062; absolute interpolation error = 0.0731481. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Positive yielded branch only; at zero rate the unyielded model allows a range of stresses. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Oldroyd-B model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. After a small deformation, hold the fluid motionless. A homogeneous Oldroyd-B polymer shear stress obeys λdτp/dt+τp=0. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Polymer shear stress / initial stress (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Oldroyd-B model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Polymer shear stress / initial stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Polymer shear stress / initial stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Zero-velocity, homogeneous stress-relaxation subproblem; convected terms vanish and the solvent stress is zero. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Reynolds-averaged Navier–Stokes (RANS) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Verify the molecular-viscosity momentum equation using fully developed plane Poiseuille flow with turbulent or subgrid stresses disabled. For this calculation, x denotes the plotted horizontal coordinate (Position / channel half-width (dimensionless)), and q(x) denotes the plotted response (Velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Reynolds-averaged Navier–Stokes (RANS): Intermediate-condition prediction and exact check. Horizontal axis: Position / channel half-width (dimensionless). Vertical axis: Velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Position / channel half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Laminar-limit verification only; it neither models turbulence nor validates a RANS, LES, or DES closure. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Large-eddy simulation (LES) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Verify the molecular-viscosity momentum equation using fully developed plane Poiseuille flow with turbulent or subgrid stresses disabled. For this calculation, x denotes the plotted horizontal coordinate (Position / channel half-width (dimensionless)), and q(x) denotes the plotted response (Velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Large-eddy simulation (LES): Intermediate-condition prediction and exact check. Horizontal axis: Position / channel half-width (dimensionless). Vertical axis: Velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Position / channel half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Laminar-limit verification only; it neither models turbulence nor validates a RANS, LES, or DES closure. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Detached-eddy simulation (DES) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Verify the molecular-viscosity momentum equation using fully developed plane Poiseuille flow with turbulent or subgrid stresses disabled. For this calculation, x denotes the plotted horizontal coordinate (Position / channel half-width (dimensionless)), and q(x) denotes the plotted response (Velocity / center velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/Umax⁡=1−ξ2u/U_{\max}=1-\xi^2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.64,q(c)=0.9676q_{\rm lin}(c)=0.64,\quad q(c)=0.9676e=qlin(c)−q(c)=−0.3276e=q_{\rm lin}(c)-q(c)=-0.3276

Solution. The endpoint responses are q(A) = 0.64 and q(B) = 0.64. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.64. Direct evaluation gives 0.9676; subtracting it from the prediction gives signed error -0.3276.

Detached-eddy simulation (DES): Intermediate-condition prediction and exact check. Horizontal axis: Position / channel half-width (dimensionless). Vertical axis: Velocity / center velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Position / channel half-width (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Velocity / center velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.64 Worked point: (-0.18, 0.9676)
At x = -0.18, the orange circle is the analytical value 0.9676; the purple diamond is the interpolated value 0.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.64; analytical reference = 0.9676; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Laminar-limit verification only; it neither models turbulence nor validates a RANS, LES, or DES closure. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Spalart–Allmaras model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For nonnegative working variable χ evaluate the standard SA eddy-viscosity mapping with cv1 = 7.1. For this calculation, x denotes the plotted horizontal coordinate (Working variable χ = ν̃ / ν (dimensionless)), and q(x) denotes the plotted response (Eddy viscosity νt / ν (dimensionless)). Suppose only the analytical endpoint responses at x = 4 and x = 16 are tabulated. Use linear interpolation to predict the response at x = 8.2, then check it against the original equation.

νt/ν=χχ3χ3+7.13\nu_t/\nu=\chi\frac{\chi^3}{\chi^3+7.1^3}a=4,b=16,c=8.2a=4,\quad b=16,\quad c=8.2qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.17562\frac{q(b)-q(a)}{b-a}=1.17562qlin(c)=5.54439,q(c)=4.97231q_{\rm lin}(c)=5.54439,\quad q(c)=4.97231e=qlin(c)−q(c)=0.572076e=q_{\rm lin}(c)-q(c)=0.572076

Solution. The endpoint responses are q(A) = 0.606763 and q(B) = 14.7143. Their secant slope is 1.17562. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 5.54439. Direct evaluation gives 4.97231; subtracting it from the prediction gives signed error 0.572076.

Spalart–Allmaras model: Intermediate-condition prediction and exact check. Horizontal axis: Working variable χ = ν̃ / ν (dimensionless). Vertical axis: Eddy viscosity νt / ν (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 4 6 8 10 12 14 16 Working variable χ = ν̃ / ν (dimensionless) 0 2 4 6 8 10 12 14 16 Eddy viscosity νt / ν (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 5.544 Worked point: (8.2, 4.972)
At x = 8.2, the orange circle is the analytical value 4.97231; the purple diamond is the interpolated value 5.54439. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 5.54439; analytical reference = 4.97231; absolute interpolation error = 0.572076. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Algebraic closure contribution only, not a solution of the SA transport equation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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k–epsilon model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold dissipation ε = ε* fixed and use Cμ = 0.09. For this calculation, x denotes the plotted horizontal coordinate (Turbulent kinetic energy k / k* (dimensionless)), and q(x) denotes the plotted response (Scaled eddy viscosity (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

νtϵ∗/k∗2=0.09(k/k∗)2\nu_t\epsilon_*/k_*^2=0.09(k/k_*)^2a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.36\frac{q(b)-q(a)}{b-a}=0.36qlin(c)=0.36,q(c)=0.242064q_{\rm lin}(c)=0.36,\quad q(c)=0.242064e=qlin(c)−q(c)=0.117936e=q_{\rm lin}(c)-q(c)=0.117936

Solution. The endpoint responses are q(A) = 0.0576 and q(B) = 0.9216. Their secant slope is 0.36. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.36. Direct evaluation gives 0.242064; subtracting it from the prediction gives signed error 0.117936.

k–epsilon model: Intermediate-condition prediction and exact check. Horizontal axis: Turbulent kinetic energy k / k* (dimensionless). Vertical axis: Scaled eddy viscosity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Turbulent kinetic energy k / k* (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Scaled eddy viscosity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.36 Worked point: (1.64, 0.2421)
At x = 1.64, the orange circle is the analytical value 0.242064; the purple diamond is the interpolated value 0.36. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.36; analytical reference = 0.242064; absolute interpolation error = 0.117936. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Closure evaluation, not a prediction of k or ε from their coupled transport equations. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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k–omega model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold specific dissipation ω = ω* > 0 and use the basic νt = k/ω relation. For this calculation, x denotes the plotted horizontal coordinate (Turbulent kinetic energy k / k* (dimensionless)), and q(x) denotes the plotted response (Scaled eddy viscosity (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

νtω∗/k∗=k/k∗\nu_t\omega_*/k_* = k/k_*a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=1.64,q(c)=1.64q_{\rm lin}(c)=1.64,\quad q(c)=1.64e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.8 and q(B) = 3.2. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.64. Direct evaluation gives 1.64; subtracting it from the prediction gives signed error 0.

k–omega model: Intermediate-condition prediction and exact check. Horizontal axis: Turbulent kinetic energy k / k* (dimensionless). Vertical axis: Scaled eddy viscosity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Turbulent kinetic energy k / k* (dimensionless) 1.0 1.5 2.0 2.5 3.0 Scaled eddy viscosity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.64 Worked point: (1.64, 1.64)
At x = 1.64, the orange circle is the analytical value 1.64; the purple diamond is the interpolated value 1.64. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.64; analytical reference = 1.64; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Basic algebraic closure with fixed ω; model variants may include limiters. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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SST k–omega model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold positive k and ω fixed. Evaluate νt = a1k/max(a1ω,SF2) with a1 = 0.31. For this calculation, x denotes the plotted horizontal coordinate (Scaled strain SF₂ / ω (dimensionless)), and q(x) denotes the plotted response (Limited viscosity νtω / k (dimensionless)). Suppose only the analytical endpoint responses at x = 0.4 and x = 1.6 are tabulated. Use linear interpolation to predict the response at x = 0.82, then check it against the original equation.

νtω/k=0.31max⁡(0.31,s),s=SF2/ω\nu_t\omega/k=\frac{0.31}{\max(0.31,s)},\quad s=SF_2/\omegaa=0.4,b=1.6,c=0.82a=0.4,\quad b=1.6,\quad c=0.82qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.484375\frac{q(b)-q(a)}{b-a}=-0.484375qlin(c)=0.571562,q(c)=0.378049q_{\rm lin}(c)=0.571562,\quad q(c)=0.378049e=qlin(c)−q(c)=0.193514e=q_{\rm lin}(c)-q(c)=0.193514

Solution. The endpoint responses are q(A) = 0.775 and q(B) = 0.19375. Their secant slope is -0.484375. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.571562. Direct evaluation gives 0.378049; subtracting it from the prediction gives signed error 0.193514.

SST k–omega model: Intermediate-condition prediction and exact check. Horizontal axis: Scaled strain SF₂ / ω (dimensionless). Vertical axis: Limited viscosity νtω / k (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Scaled strain SF₂ / ω (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Limited viscosity νtω / k (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5716 Worked point: (0.82, 0.378)
At x = 0.82, the orange circle is the analytical value 0.378049; the purple diamond is the interpolated value 0.571562. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.571562; analytical reference = 0.378049; absolute interpolation error = 0.193514. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Algebraic SST limiter illustration; blending functions and transport equations are not solved. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Reynolds-stress transport model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a homogeneous Reynolds-stress anisotropy component use the reduced closure db/dt = −b/T with constant T. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Anisotropy component / initial component (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Reynolds-stress transport model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Anisotropy component / initial component (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Anisotropy component / initial component (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Isolated linear return-to-isotropy term; production, transport, and changing dissipation are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Smagorinsky subgrid model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use Cs = 0.1 and constant filter width Δ. For this calculation, x denotes the plotted horizontal coordinate (Resolved strain |S| / S* (dimensionless)), and q(x) denotes the plotted response (Scaled subgrid viscosity (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

νt/(Δ2S∗)=0.01(∣S∣/S∗)\nu_t/(\Delta^2 S_*)=0.01(|S|/S_*)a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.01\frac{q(b)-q(a)}{b-a}=0.01qlin(c)=0.0205,q(c)=0.0205q_{\rm lin}(c)=0.0205,\quad q(c)=0.0205e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.01 and q(B) = 0.04. Their secant slope is 0.01. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0205. Direct evaluation gives 0.0205; subtracting it from the prediction gives signed error 0.

Smagorinsky subgrid model: Intermediate-condition prediction and exact check. Horizontal axis: Resolved strain |S| / S* (dimensionless). Vertical axis: Scaled subgrid viscosity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Resolved strain |S| / S* (dimensionless) 0.010 0.015 0.020 0.025 0.030 0.035 0.040 Scaled subgrid viscosity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0205 Worked point: (2.05, 0.0205)
At x = 2.05, the orange circle is the analytical value 0.0205; the purple diamond is the interpolated value 0.0205. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0205; analytical reference = 0.0205; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Constant-coefficient closure; no dynamic procedure or wall damping is included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Volume-of-fluid (VOF) representation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Advect the initial smoothed interface α(x,0) = [1−tanh(5x)]/2 at unit velocity with no compression term. For this calculation, x denotes the plotted horizontal coordinate (Position x (dimensionless)), and q(x) denotes the plotted response (Phase volume fraction α (dimensionless)). Suppose only the analytical endpoint responses at x = -0.2 and x = 2.2 are tabulated. Use linear interpolation to predict the response at x = 0.64, then check it against the original equation.

α(x,1)=12[1−tanh⁡(5(x−1))]\alpha(x,1)=\tfrac12[1-\tanh(5(x-1))]a=−0.2,b=2.2,c=0.64a=-0.2,\quad b=2.2,\quad c=0.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.416662\frac{q(b)-q(a)}{b-a}=-0.416662qlin(c)=0.649998,q(c)=0.973403q_{\rm lin}(c)=0.649998,\quad q(c)=0.973403e=qlin(c)−q(c)=−0.323405e=q_{\rm lin}(c)-q(c)=-0.323405

Solution. The endpoint responses are q(A) = 0.999994 and q(B) = 6.14417e-06. Their secant slope is -0.416662. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.649998. Direct evaluation gives 0.973403; subtracting it from the prediction gives signed error -0.323405.

Volume-of-fluid (VOF) representation: Intermediate-condition prediction and exact check. Horizontal axis: Position x (dimensionless). Vertical axis: Phase volume fraction α (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.0 0.5 1.0 1.5 2.0 Position x (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Phase volume fraction α (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.65 Worked point: (0.64, 0.9734)
At x = 0.64, the orange circle is the analytical value 0.973403; the purple diamond is the interpolated value 0.649998. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.649998; analytical reference = 0.973403; absolute interpolation error = 0.323405. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact scalar-advection benchmark with a deliberately smooth interface; interface reconstruction and multiphase momentum are not solved. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Euler–Euler two-fluid model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For two homogeneous phases coupled only by linear interphase drag, scale time by the combined drag relaxation time and slip by its initial value. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Remaining fraction (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Euler–Euler two-fluid model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Remaining fraction (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Remaining fraction (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Subtract the two phase momentum balances to obtain a decaying relative velocity; spatial transport, pressure gradients, and phase change are absent. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lagrangian particle tracking · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A particle starts at rest in a uniform fluid of constant speed U and experiences linear drag only. For this calculation, x denotes the plotted horizontal coordinate (Time / particle relaxation time (dimensionless)), and q(x) denotes the plotted response (Particle speed / fluid speed (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

v/U=1−e−t/τpv/U=1-e^{-t/\tau_p}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Lagrangian particle tracking: Intermediate-condition prediction and exact check. Horizontal axis: Time / particle relaxation time (dimensionless). Vertical axis: Particle speed / fluid speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / particle relaxation time (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Particle speed / fluid speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Dilute isolated-particle Stokes-drag reduction; no gravity or feedback on the fluid. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Fourier heat conduction · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve u″ = 0 on 0 < ξ < 1 with u(0) = 1 and u(1) = 0, constant transport coefficient, and no source. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Temperature excess / imposed difference (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ)=1−ξu(\xi)=1-\xia=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1\frac{q(b)-q(a)}{b-a}=-1qlin(c)=0.59,q(c)=0.59q_{\rm lin}(c)=0.59,\quad q(c)=0.59e=qlin(c)−q(c)=1.11022×10−16e=q_{\rm lin}(c)-q(c)=1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.8 and q(B) = 0.2. Their secant slope is -1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.59. Direct evaluation gives 0.59; subtracting it from the prediction gives signed error 1.11022e-16.

Fourier heat conduction: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Temperature excess / imposed difference (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Temperature excess / imposed difference (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.59 Worked point: (0.41, 0.59)
At x = 0.41, the orange circle is the analytical value 0.59; the purple diamond is the interpolated value 0.59. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.59; analytical reference = 0.59; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For numerical-method entries this is the exact target to verify against, not a computed discretization or convergence claim. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Groundwater flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve u″ = 0 on 0 < ξ < 1 with u(0) = 1 and u(1) = 0, constant transport coefficient, and no source. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Hydraulic head excess / imposed difference (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ)=1−ξu(\xi)=1-\xia=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1\frac{q(b)-q(a)}{b-a}=-1qlin(c)=0.59,q(c)=0.59q_{\rm lin}(c)=0.59,\quad q(c)=0.59e=qlin(c)−q(c)=1.11022×10−16e=q_{\rm lin}(c)-q(c)=1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.8 and q(B) = 0.2. Their secant slope is -1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.59. Direct evaluation gives 0.59; subtracting it from the prediction gives signed error 1.11022e-16.

Groundwater flow model: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Hydraulic head excess / imposed difference (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Hydraulic head excess / imposed difference (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.59 Worked point: (0.41, 0.59)
At x = 0.41, the orange circle is the analytical value 0.59; the purple diamond is the interpolated value 0.59. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.59; analytical reference = 0.59; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For numerical-method entries this is the exact target to verify against, not a computed discretization or convergence claim. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Finite element method (FEM / FEA) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve u″ = 0 on 0 < ξ < 1 with u(0) = 1 and u(1) = 0, constant transport coefficient, and no source. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Normalized temperature or head (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ)=1−ξu(\xi)=1-\xia=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1\frac{q(b)-q(a)}{b-a}=-1qlin(c)=0.59,q(c)=0.59q_{\rm lin}(c)=0.59,\quad q(c)=0.59e=qlin(c)−q(c)=1.11022×10−16e=q_{\rm lin}(c)-q(c)=1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.8 and q(B) = 0.2. Their secant slope is -1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.59. Direct evaluation gives 0.59; subtracting it from the prediction gives signed error 1.11022e-16.

Finite element method (FEM / FEA): Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Normalized temperature or head (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Normalized temperature or head (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.59 Worked point: (0.41, 0.59)
At x = 0.41, the orange circle is the analytical value 0.59; the purple diamond is the interpolated value 0.59. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.59; analytical reference = 0.59; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For numerical-method entries this is the exact target to verify against, not a computed discretization or convergence claim. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Finite volume method (FVM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve u″ = 0 on 0 < ξ < 1 with u(0) = 1 and u(1) = 0, constant transport coefficient, and no source. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Normalized temperature or head (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ)=1−ξu(\xi)=1-\xia=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1\frac{q(b)-q(a)}{b-a}=-1qlin(c)=0.59,q(c)=0.59q_{\rm lin}(c)=0.59,\quad q(c)=0.59e=qlin(c)−q(c)=1.11022×10−16e=q_{\rm lin}(c)-q(c)=1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.8 and q(B) = 0.2. Their secant slope is -1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.59. Direct evaluation gives 0.59; subtracting it from the prediction gives signed error 1.11022e-16.

Finite volume method (FVM): Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Normalized temperature or head (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Normalized temperature or head (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.59 Worked point: (0.41, 0.59)
At x = 0.41, the orange circle is the analytical value 0.59; the purple diamond is the interpolated value 0.59. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.59; analytical reference = 0.59; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For numerical-method entries this is the exact target to verify against, not a computed discretization or convergence claim. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Finite difference method (FDM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve u″ = 0 on 0 < ξ < 1 with u(0) = 1 and u(1) = 0, constant transport coefficient, and no source. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Normalized temperature or head (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ)=1−ξu(\xi)=1-\xia=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1\frac{q(b)-q(a)}{b-a}=-1qlin(c)=0.59,q(c)=0.59q_{\rm lin}(c)=0.59,\quad q(c)=0.59e=qlin(c)−q(c)=1.11022×10−16e=q_{\rm lin}(c)-q(c)=1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.8 and q(B) = 0.2. Their secant slope is -1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.59. Direct evaluation gives 0.59; subtracting it from the prediction gives signed error 1.11022e-16.

Finite difference method (FDM): Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Normalized temperature or head (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Normalized temperature or head (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.59 Worked point: (0.41, 0.59)
At x = 0.41, the orange circle is the analytical value 0.59; the purple diamond is the interpolated value 0.59. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.59; analytical reference = 0.59; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For numerical-method entries this is the exact target to verify against, not a computed discretization or convergence claim. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Boundary element method (BEM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve u″ = 0 on 0 < ξ < 1 with u(0) = 1 and u(1) = 0, constant transport coefficient, and no source. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Normalized temperature or head (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ)=1−ξu(\xi)=1-\xia=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1\frac{q(b)-q(a)}{b-a}=-1qlin(c)=0.59,q(c)=0.59q_{\rm lin}(c)=0.59,\quad q(c)=0.59e=qlin(c)−q(c)=1.11022×10−16e=q_{\rm lin}(c)-q(c)=1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.8 and q(B) = 0.2. Their secant slope is -1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.59. Direct evaluation gives 0.59; subtracting it from the prediction gives signed error 1.11022e-16.

Boundary element method (BEM): Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Normalized temperature or head (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Normalized temperature or head (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.59 Worked point: (0.41, 0.59)
At x = 0.41, the orange circle is the analytical value 0.59; the purple diamond is the interpolated value 0.59. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.59; analytical reference = 0.59; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. For numerical-method entries this is the exact target to verify against, not a computed discretization or convergence claim. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Transient heat equation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use uτ = uξξ on the unit interval, zero end values, and u(ξ,0) = sin(πξ). Plot τ = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Temperature perturbation / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Transient heat equation: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Temperature perturbation / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Temperature perturbation / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact PDE benchmark. For reduced bases, PINNs, and neural operators, this is a reference target, not a claimed trained or computed prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Spectral method · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use uτ = uξξ on the unit interval, zero end values, and u(ξ,0) = sin(πξ). Plot τ = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Field / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Spectral method: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Field / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Field / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact PDE benchmark. For reduced bases, PINNs, and neural operators, this is a reference target, not a claimed trained or computed prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Reduced basis model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use uτ = uξξ on the unit interval, zero end values, and u(ξ,0) = sin(πξ). Plot τ = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Field / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Reduced basis model: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Field / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Field / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact PDE benchmark. For reduced bases, PINNs, and neural operators, this is a reference target, not a claimed trained or computed prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Physics-informed neural network (PINN) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use uτ = uξξ on the unit interval, zero end values, and u(ξ,0) = sin(πξ). Plot τ = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Field / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Physics-informed neural network (PINN): Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Field / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Field / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact PDE benchmark. For reduced bases, PINNs, and neural operators, this is a reference target, not a claimed trained or computed prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Neural operator · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use uτ = uξξ on the unit interval, zero end values, and u(ξ,0) = sin(πξ). Plot τ = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Field / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Neural operator: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Field / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Field / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact PDE benchmark. For reduced bases, PINNs, and neural operators, this is a reference target, not a claimed trained or computed prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lumped-capacitance thermal model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A thermal capacitance C connects through resistance R to fixed ambient temperature. Set τ = t/(RC) and y = (T−T∞)/(T0−T∞). For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Temperature excess / initial excess (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Lumped-capacitance thermal model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Temperature excess / initial excess (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Temperature excess / initial excess (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-node constant-property cooling example; multizone and multi-node networks have additional modes. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Thermal resistance-capacitance network · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A thermal capacitance C connects through resistance R to fixed ambient temperature. Set τ = t/(RC) and y = (T−T∞)/(T0−T∞). For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Temperature excess / initial excess (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Thermal resistance-capacitance network: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Temperature excess / initial excess (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Temperature excess / initial excess (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-node constant-property cooling example; multizone and multi-node networks have additional modes. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Newton cooling model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A thermal capacitance C connects through resistance R to fixed ambient temperature. Set τ = t/(RC) and y = (T−T∞)/(T0−T∞). For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Temperature excess / initial excess (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Newton cooling model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Temperature excess / initial excess (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Temperature excess / initial excess (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-node constant-property cooling example; multizone and multi-node networks have additional modes. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Building thermal-zone model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A thermal capacitance C connects through resistance R to fixed ambient temperature. Set τ = t/(RC) and y = (T−T∞)/(T0−T∞). For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Temperature excess / initial excess (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Building thermal-zone model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Temperature excess / initial excess (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Temperature excess / initial excess (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-node constant-property cooling example; multizone and multi-node networks have additional modes. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Radiative transfer equation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A steady beam traverses a homogeneous purely absorbing medium. Set τ = Σx and y = intensity / incident intensity. For this calculation, x denotes the plotted horizontal coordinate (Optical thickness Σx (dimensionless)), and q(x) denotes the plotted response (Beam intensity / incident intensity (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Radiative transfer equation: Intermediate-condition prediction and exact check. Horizontal axis: Optical thickness Σx (dimensionless). Vertical axis: Beam intensity / incident intensity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Optical thickness Σx (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Beam intensity / incident intensity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact absorption-only transport benchmark, without scattering or emission. For Monte Carlo transport this is the expected value, not a sampled realization. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Neutron transport model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A steady beam traverses a homogeneous purely absorbing medium. Set τ = Σx and y = intensity / incident intensity. For this calculation, x denotes the plotted horizontal coordinate (Optical thickness Σx (dimensionless)), and q(x) denotes the plotted response (Beam intensity / incident intensity (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Neutron transport model: Intermediate-condition prediction and exact check. Horizontal axis: Optical thickness Σx (dimensionless). Vertical axis: Beam intensity / incident intensity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Optical thickness Σx (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Beam intensity / incident intensity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact absorption-only transport benchmark, without scattering or emission. For Monte Carlo transport this is the expected value, not a sampled realization. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Monte Carlo transport · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A steady beam traverses a homogeneous purely absorbing medium. Set τ = Σx and y = intensity / incident intensity. For this calculation, x denotes the plotted horizontal coordinate (Optical thickness Σx (dimensionless)), and q(x) denotes the plotted response (Beam intensity / incident intensity (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Monte Carlo transport: Intermediate-condition prediction and exact check. Horizontal axis: Optical thickness Σx (dimensionless). Vertical axis: Beam intensity / incident intensity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Optical thickness Σx (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Beam intensity / incident intensity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact absorption-only transport benchmark, without scattering or emission. For Monte Carlo transport this is the expected value, not a sampled realization. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Stefan–Boltzmann surface model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A gray surface sees a large isothermal surrounding at T*, with constant emissivity. For this calculation, x denotes the plotted horizontal coordinate (Surface / surrounding temperature T / T* (dimensionless)), and q(x) denotes the plotted response (Scaled net radiative flux (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 1.7 are tabulated. Use linear interpolation to predict the response at x = 1.115, then check it against the original equation.

q/(ϵσT∗4)=θ4−1q/(\epsilon\sigma T_*^4)=\theta^4-1a=0.8,b=1.7,c=1.115a=0.8,\quad b=1.7,\quad c=1.115qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=8.825\frac{q(b)-q(a)}{b-a}=8.825qlin(c)=2.18948,q(c)=0.545608q_{\rm lin}(c)=2.18948,\quad q(c)=0.545608e=qlin(c)−q(c)=1.64387e=q_{\rm lin}(c)-q(c)=1.64387

Solution. The endpoint responses are q(A) = -0.5904 and q(B) = 7.3521. Their secant slope is 8.825. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.18948. Direct evaluation gives 0.545608; subtracting it from the prediction gives signed error 1.64387.

Stefan–Boltzmann surface model: Intermediate-condition prediction and exact check. Horizontal axis: Surface / surrounding temperature T / T* (dimensionless). Vertical axis: Scaled net radiative flux (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.8 1.0 1.2 1.4 1.6 Surface / surrounding temperature T / T* (dimensionless) 0 2 4 6 8 Scaled net radiative flux (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.189 Worked point: (1.115, 0.5456)
At x = 1.115, the orange circle is the analytical value 0.545608; the purple diamond is the interpolated value 2.18948. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.18948; analytical reference = 0.545608; absolute interpolation error = 1.64387. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Surface-to-surface radiosity model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Two infinite parallel diffuse-gray plates have emissivities ε and 0.8 and fixed unequal temperatures. For this calculation, x denotes the plotted horizontal coordinate (Surface 1 emissivity ε (dimensionless)), and q(x) denotes the plotted response (Radiative exchange factor (dimensionless)). Suppose only the analytical endpoint responses at x = 0.24 and x = 0.81 are tabulated. Use linear interpolation to predict the response at x = 0.4395, then check it against the original equation.

qσ(T14−T24)=11/ϵ+1/0.8−1\frac{q}{\sigma(T_1^4-T_2^4)}=\frac1{1/\epsilon+1/0.8-1}a=0.24,b=0.81,c=0.4395a=0.24,\quad b=0.81,\quad c=0.4395qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.784529\frac{q(b)-q(a)}{b-a}=0.784529qlin(c)=0.382929,q(c)=0.395991q_{\rm lin}(c)=0.382929,\quad q(c)=0.395991e=qlin(c)−q(c)=−0.0130619e=q_{\rm lin}(c)-q(c)=-0.0130619

Solution. The endpoint responses are q(A) = 0.226415 and q(B) = 0.673597. Their secant slope is 0.784529. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.382929. Direct evaluation gives 0.395991; subtracting it from the prediction gives signed error -0.0130619.

Surface-to-surface radiosity model: Intermediate-condition prediction and exact check. Horizontal axis: Surface 1 emissivity ε (dimensionless). Vertical axis: Radiative exchange factor (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.3 0.4 0.5 0.6 0.7 0.8 Surface 1 emissivity ε (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 Radiative exchange factor (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3829 Worked point: (0.4395, 0.396)
At x = 0.4395, the orange circle is the analytical value 0.395991; the purple diamond is the interpolated value 0.382929. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.382929; analytical reference = 0.395991; absolute interpolation error = 0.0130619. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Equal facing areas, view factor one, and a nonparticipating gap. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Stefan phase-change problem · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take a one-phase Stefan problem whose Stefan number selects similarity constant λ = 0.5. For this calculation, x denotes the plotted horizontal coordinate (Fourier time αt / L² (dimensionless)), and q(x) denotes the plotted response (Front position s / L (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

s/L=2λαt/L2,λ=0.5s/L=2\lambda\sqrt{\alpha t/L^2},\quad\lambda=0.5a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.372678\frac{q(b)-q(a)}{b-a}=0.372678qlin(c)=1.20748,q(c)=1.28062q_{\rm lin}(c)=1.20748,\quad q(c)=1.28062e=qlin(c)−q(c)=−0.0731481e=q_{\rm lin}(c)-q(c)=-0.0731481

Solution. The endpoint responses are q(A) = 0.894427 and q(B) = 1.78885. Their secant slope is 0.372678. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.20748. Direct evaluation gives 1.28062; subtracting it from the prediction gives signed error -0.0731481.

Stefan phase-change problem: Intermediate-condition prediction and exact check. Horizontal axis: Fourier time αt / L² (dimensionless). Vertical axis: Front position s / L (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Fourier time αt / L² (dimensionless) 0.8 1.0 1.2 1.4 1.6 1.8 Front position s / L (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.207 Worked point: (1.64, 1.281)
At x = 1.64, the orange circle is the analytical value 1.28062; the purple diamond is the interpolated value 1.20748. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.20748; analytical reference = 1.28062; absolute interpolation error = 0.0731481. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Semi-infinite, one-phase conduction limit with a fixed boundary temperature; λ must be consistent with the material and thermal data. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Enthalpy–porosity model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose the linear liquid-fraction law between solidus Ts and liquidus Tl. For this calculation, x denotes the plotted horizontal coordinate (Temperature within melting interval θ (dimensionless)), and q(x) denotes the plotted response (Liquid fraction (dimensionless)). Suppose only the analytical endpoint responses at x = -0.1 and x = 1.1 are tabulated. Use linear interpolation to predict the response at x = 0.32, then check it against the original equation.

fl=min⁡[1,max⁡(0,θ)],θ=(T−Ts)/(Tl−Ts)f_l=\min[1,\max(0,\theta)],\quad\theta=(T-T_s)/(T_l-T_s)a=−0.1,b=1.1,c=0.32a=-0.1,\quad b=1.1,\quad c=0.32qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.833333\frac{q(b)-q(a)}{b-a}=0.833333qlin(c)=0.35,q(c)=0.32q_{\rm lin}(c)=0.35,\quad q(c)=0.32e=qlin(c)−q(c)=0.03e=q_{\rm lin}(c)-q(c)=0.03

Solution. The endpoint responses are q(A) = 0 and q(B) = 1. Their secant slope is 0.833333. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.35. Direct evaluation gives 0.32; subtracting it from the prediction gives signed error 0.03.

Enthalpy–porosity model: Intermediate-condition prediction and exact check. Horizontal axis: Temperature within melting interval θ (dimensionless). Vertical axis: Liquid fraction (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.0 0.2 0.4 0.6 0.8 1.0 Temperature within melting interval θ (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Liquid fraction (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.35 Worked point: (0.32, 0.32)
At x = 0.32, the orange circle is the analytical value 0.32; the purple diamond is the interpolated value 0.35. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.35; analytical reference = 0.32; absolute interpolation error = 0.03. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Constitutive phase-fraction example only; momentum damping and the transient enthalpy equation are not solved. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Linear elasticity (Hooke model) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply uniform uniaxial strain to a homogeneous small-strain elastic bar with traction-free lateral surfaces. For this calculation, x denotes the plotted horizontal coordinate (Axial strain ε (dimensionless)), and q(x) denotes the plotted response (Axial stress / directional modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.002 and x = 0.008 are tabulated. Use linear interpolation to predict the response at x = 0.0041, then check it against the original equation.

σ/E=ε\sigma/E=\varepsilona=0.002,b=0.008,c=0.0041a=0.002,\quad b=0.008,\quad c=0.0041qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.0041,q(c)=0.0041q_{\rm lin}(c)=0.0041,\quad q(c)=0.0041e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.002 and q(B) = 0.008. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0041. Direct evaluation gives 0.0041; subtracting it from the prediction gives signed error 0.

Linear elasticity (Hooke model): Intermediate-condition prediction and exact check. Horizontal axis: Axial strain ε (dimensionless). Vertical axis: Axial stress / directional modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial strain ε (dimensionless) 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial stress / directional modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0041 Worked point: (0.0041, 0.0041)
At x = 0.0041, the orange circle is the analytical value 0.0041; the purple diamond is the interpolated value 0.0041. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0041; analytical reference = 0.0041; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Homogeneous linear reference for truss, RVE, and FE² entries; this is not a heterogeneous microscale simulation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Orthotropic elasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply uniform uniaxial strain to a homogeneous small-strain elastic bar with traction-free lateral surfaces. For this calculation, x denotes the plotted horizontal coordinate (Axial strain ε (dimensionless)), and q(x) denotes the plotted response (Axial stress / directional modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.002 and x = 0.008 are tabulated. Use linear interpolation to predict the response at x = 0.0041, then check it against the original equation.

σ/E=ε\sigma/E=\varepsilona=0.002,b=0.008,c=0.0041a=0.002,\quad b=0.008,\quad c=0.0041qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.0041,q(c)=0.0041q_{\rm lin}(c)=0.0041,\quad q(c)=0.0041e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.002 and q(B) = 0.008. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0041. Direct evaluation gives 0.0041; subtracting it from the prediction gives signed error 0.

Orthotropic elasticity: Intermediate-condition prediction and exact check. Horizontal axis: Axial strain ε (dimensionless). Vertical axis: Axial stress / directional modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial strain ε (dimensionless) 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial stress / directional modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0041 Worked point: (0.0041, 0.0041)
At x = 0.0041, the orange circle is the analytical value 0.0041; the purple diamond is the interpolated value 0.0041. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0041; analytical reference = 0.0041; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Homogeneous linear reference for truss, RVE, and FE² entries; this is not a heterogeneous microscale simulation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Truss model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply uniform uniaxial strain to a homogeneous small-strain elastic bar with traction-free lateral surfaces. For this calculation, x denotes the plotted horizontal coordinate (Axial strain ε (dimensionless)), and q(x) denotes the plotted response (Axial stress / directional modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.002 and x = 0.008 are tabulated. Use linear interpolation to predict the response at x = 0.0041, then check it against the original equation.

σ/E=ε\sigma/E=\varepsilona=0.002,b=0.008,c=0.0041a=0.002,\quad b=0.008,\quad c=0.0041qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.0041,q(c)=0.0041q_{\rm lin}(c)=0.0041,\quad q(c)=0.0041e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.002 and q(B) = 0.008. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0041. Direct evaluation gives 0.0041; subtracting it from the prediction gives signed error 0.

Truss model: Intermediate-condition prediction and exact check. Horizontal axis: Axial strain ε (dimensionless). Vertical axis: Axial stress / directional modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial strain ε (dimensionless) 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial stress / directional modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0041 Worked point: (0.0041, 0.0041)
At x = 0.0041, the orange circle is the analytical value 0.0041; the purple diamond is the interpolated value 0.0041. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0041; analytical reference = 0.0041; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Homogeneous linear reference for truss, RVE, and FE² entries; this is not a heterogeneous microscale simulation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Representative volume element (RVE) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply uniform uniaxial strain to a homogeneous small-strain elastic bar with traction-free lateral surfaces. For this calculation, x denotes the plotted horizontal coordinate (Axial strain ε (dimensionless)), and q(x) denotes the plotted response (Axial stress / directional modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.002 and x = 0.008 are tabulated. Use linear interpolation to predict the response at x = 0.0041, then check it against the original equation.

σ/E=ε\sigma/E=\varepsilona=0.002,b=0.008,c=0.0041a=0.002,\quad b=0.008,\quad c=0.0041qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.0041,q(c)=0.0041q_{\rm lin}(c)=0.0041,\quad q(c)=0.0041e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.002 and q(B) = 0.008. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0041. Direct evaluation gives 0.0041; subtracting it from the prediction gives signed error 0.

Representative volume element (RVE): Intermediate-condition prediction and exact check. Horizontal axis: Axial strain ε (dimensionless). Vertical axis: Axial stress / directional modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial strain ε (dimensionless) 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial stress / directional modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0041 Worked point: (0.0041, 0.0041)
At x = 0.0041, the orange circle is the analytical value 0.0041; the purple diamond is the interpolated value 0.0041. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0041; analytical reference = 0.0041; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Homogeneous linear reference for truss, RVE, and FE² entries; this is not a heterogeneous microscale simulation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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FE² computational homogenization · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply uniform uniaxial strain to a homogeneous small-strain elastic bar with traction-free lateral surfaces. For this calculation, x denotes the plotted horizontal coordinate (Axial strain ε (dimensionless)), and q(x) denotes the plotted response (Axial stress / directional modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.002 and x = 0.008 are tabulated. Use linear interpolation to predict the response at x = 0.0041, then check it against the original equation.

σ/E=ε\sigma/E=\varepsilona=0.002,b=0.008,c=0.0041a=0.002,\quad b=0.008,\quad c=0.0041qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.0041,q(c)=0.0041q_{\rm lin}(c)=0.0041,\quad q(c)=0.0041e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.002 and q(B) = 0.008. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0041. Direct evaluation gives 0.0041; subtracting it from the prediction gives signed error 0.

FE² computational homogenization: Intermediate-condition prediction and exact check. Horizontal axis: Axial strain ε (dimensionless). Vertical axis: Axial stress / directional modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial strain ε (dimensionless) 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Axial stress / directional modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0041 Worked point: (0.0041, 0.0041)
At x = 0.0041, the orange circle is the analytical value 0.0041; the purple diamond is the interpolated value 0.0041. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0041; analytical reference = 0.0041; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Homogeneous linear reference for truss, RVE, and FE² entries; this is not a heterogeneous microscale simulation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Neo-Hookean hyperelasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Stretch an incompressible neo-Hookean solid uniaxially with traction-free transverse faces. For this calculation, x denotes the plotted horizontal coordinate (Axial stretch λ (dimensionless)), and q(x) denotes the plotted response (Cauchy stress / shear modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.88 and x = 1.72 are tabulated. Use linear interpolation to predict the response at x = 1.174, then check it against the original equation.

σ/μ=λ2−λ−1\sigma/\mu=\lambda^2-\lambda^{-1}a=0.88,b=1.72,c=1.174a=0.88,\quad b=1.72,\quad c=1.174qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=3.26068\frac{q(b)-q(a)}{b-a}=3.26068qlin(c)=0.596675,q(c)=0.526487q_{\rm lin}(c)=0.596675,\quad q(c)=0.526487e=qlin(c)−q(c)=0.070188e=q_{\rm lin}(c)-q(c)=0.070188

Solution. The endpoint responses are q(A) = -0.361964 and q(B) = 2.377. Their secant slope is 3.26068. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.596675. Direct evaluation gives 0.526487; subtracting it from the prediction gives signed error 0.070188.

Neo-Hookean hyperelasticity: Intermediate-condition prediction and exact check. Horizontal axis: Axial stretch λ (dimensionless). Vertical axis: Cauchy stress / shear modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Axial stretch λ (dimensionless) −0.5 0.0 0.5 1.0 1.5 2.0 2.5 Cauchy stress / shear modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5967 Worked point: (1.174, 0.5265)
At x = 1.174, the orange circle is the analytical value 0.526487; the purple diamond is the interpolated value 0.596675. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.596675; analytical reference = 0.526487; absolute interpolation error = 0.070188. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Mooney–Rivlin hyperelasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use an incompressible two-parameter Mooney–Rivlin material with C10 = C01 and μ = 2(C10+C01). For this calculation, x denotes the plotted horizontal coordinate (Axial stretch λ (dimensionless)), and q(x) denotes the plotted response (Cauchy stress / initial shear modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.88 and x = 1.72 are tabulated. Use linear interpolation to predict the response at x = 1.174, then check it against the original equation.

σ/μ=12(1+λ−1)(λ2−λ−1)\sigma/\mu=\tfrac12(1+\lambda^{-1})(\lambda^2-\lambda^{-1})a=0.88,b=1.72,c=1.174a=0.88,\quad b=1.72,\quad c=1.174qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2.69778\frac{q(b)-q(a)}{b-a}=2.69778qlin(c)=0.406504,q(c)=0.487472q_{\rm lin}(c)=0.406504,\quad q(c)=0.487472e=qlin(c)−q(c)=−0.0809673e=q_{\rm lin}(c)-q(c)=-0.0809673

Solution. The endpoint responses are q(A) = -0.386643 and q(B) = 1.87949. Their secant slope is 2.69778. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.406504. Direct evaluation gives 0.487472; subtracting it from the prediction gives signed error -0.0809673.

Mooney–Rivlin hyperelasticity: Intermediate-condition prediction and exact check. Horizontal axis: Axial stretch λ (dimensionless). Vertical axis: Cauchy stress / initial shear modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Axial stretch λ (dimensionless) −0.5 0.0 0.5 1.0 1.5 2.0 Cauchy stress / initial shear modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.4065 Worked point: (1.174, 0.4875)
At x = 1.174, the orange circle is the analytical value 0.487472; the purple diamond is the interpolated value 0.406504. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.406504; analytical reference = 0.487472; absolute interpolation error = 0.0809673. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Ogden hyperelasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose W = (2μ/α²)(λ1^α+λ2^α+λ3^α−3), α = 4, and incompressible uniaxial tension. For this calculation, x denotes the plotted horizontal coordinate (Axial stretch λ (dimensionless)), and q(x) denotes the plotted response (Cauchy stress / initial shear modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.88 and x = 1.72 are tabulated. Use linear interpolation to predict the response at x = 1.174, then check it against the original equation.

σ/μ=12(λ4−λ−2)\sigma/\mu=\tfrac12(\lambda^4-\lambda^{-2})a=0.88,b=1.72,c=1.174a=0.88,\quad b=1.72,\quad c=1.174qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=5.42008\frac{q(b)-q(a)}{b-a}=5.42008qlin(c)=1.24769,q(c)=0.58705q_{\rm lin}(c)=1.24769,\quad q(c)=0.58705e=qlin(c)−q(c)=0.66064e=q_{\rm lin}(c)-q(c)=0.66064

Solution. The endpoint responses are q(A) = -0.345813 and q(B) = 4.20706. Their secant slope is 5.42008. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.24769. Direct evaluation gives 0.58705; subtracting it from the prediction gives signed error 0.66064.

Ogden hyperelasticity: Intermediate-condition prediction and exact check. Horizontal axis: Axial stretch λ (dimensionless). Vertical axis: Cauchy stress / initial shear modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Axial stretch λ (dimensionless) 0 1 2 3 4 Cauchy stress / initial shear modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.248 Worked point: (1.174, 0.5871)
At x = 1.174, the orange circle is the analytical value 0.58705; the purple diamond is the interpolated value 1.24769. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.24769; analytical reference = 0.58705; absolute interpolation error = 0.66064. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. The energy convention is stated explicitly because Ogden coefficient conventions vary. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Euler–Bernoulli beam model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A prismatic Euler–Bernoulli cantilever of length L carries a transverse tip force P. For this calculation, x denotes the plotted horizontal coordinate (Axial position ξ = x / L (dimensionless)), and q(x) denotes the plotted response (Deflection w / (PL³/EI) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

w/(PL3/EI)=ξ2(3−ξ)/6w/(PL^3/EI)=\xi^2(3-\xi)/6a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.36\frac{q(b)-q(a)}{b-a}=0.36qlin(c)=0.0942667,q(c)=0.0725632q_{\rm lin}(c)=0.0942667,\quad q(c)=0.0725632e=qlin(c)−q(c)=0.0217035e=q_{\rm lin}(c)-q(c)=0.0217035

Solution. The endpoint responses are q(A) = 0.0186667 and q(B) = 0.234667. Their secant slope is 0.36. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0942667. Direct evaluation gives 0.0725632; subtracting it from the prediction gives signed error 0.0217035.

Euler–Bernoulli beam model: Intermediate-condition prediction and exact check. Horizontal axis: Axial position ξ = x / L (dimensionless). Vertical axis: Deflection w / (PL³/EI) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Axial position ξ = x / L (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 Deflection w / (PL³/EI) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.09427 Worked point: (0.41, 0.07256)
At x = 0.41, the orange circle is the analytical value 0.0725632; the purple diamond is the interpolated value 0.0942667. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0942667; analytical reference = 0.0725632; absolute interpolation error = 0.0217035. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Timoshenko beam model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take an end-loaded Timoshenko cantilever with EI/(κGA L²) = 0.1. For this calculation, x denotes the plotted horizontal coordinate (Axial position x / L (dimensionless)), and q(x) denotes the plotted response (Scaled transverse deflection (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

w/(PL3/EI)=ξ2(3−ξ)/6+0.1ξw/(PL^3/EI)=\xi^2(3-\xi)/6+0.1\xia=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.46\frac{q(b)-q(a)}{b-a}=0.46qlin(c)=0.135267,q(c)=0.113563q_{\rm lin}(c)=0.135267,\quad q(c)=0.113563e=qlin(c)−q(c)=0.0217035e=q_{\rm lin}(c)-q(c)=0.0217035

Solution. The endpoint responses are q(A) = 0.0386667 and q(B) = 0.314667. Their secant slope is 0.46. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.135267. Direct evaluation gives 0.113563; subtracting it from the prediction gives signed error 0.0217035.

Timoshenko beam model: Intermediate-condition prediction and exact check. Horizontal axis: Axial position x / L (dimensionless). Vertical axis: Scaled transverse deflection (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Axial position x / L (dimensionless) 0.05 0.10 0.15 0.20 0.25 0.30 Scaled transverse deflection (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1353 Worked point: (0.41, 0.1136)
At x = 0.41, the orange circle is the analytical value 0.113563; the purple diamond is the interpolated value 0.135267. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.135267; analytical reference = 0.113563; absolute interpolation error = 0.0217035. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linear prismatic beam; κ is the shear correction factor. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Kirchhoff–Love plate model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply a single sinusoidal load mode to a simply supported rectangular plate. Plot its normalized centerline deflection. For this calculation, x denotes the plotted horizontal coordinate (Plate position x / a (dimensionless)), and q(x) denotes the plotted response (Centerline deflection / maximum (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

w(x,b/2)/wmax⁡=sin⁡(πx/a)w(x,b/2)/w_{\max}=\sin(\pi x/a)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.85037×10−16\frac{q(b)-q(a)}{b-a}=1.85037\times10^{-16}qlin(c)=0.587785,q(c)=0.960294q_{\rm lin}(c)=0.587785,\quad q(c)=0.960294e=qlin(c)−q(c)=−0.372508e=q_{\rm lin}(c)-q(c)=-0.372508

Solution. The endpoint responses are q(A) = 0.587785 and q(B) = 0.587785. Their secant slope is 1.85037e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.587785. Direct evaluation gives 0.960294; subtracting it from the prediction gives signed error -0.372508.

Kirchhoff–Love plate model: Intermediate-condition prediction and exact check. Horizontal axis: Plate position x / a (dimensionless). Vertical axis: Centerline deflection / maximum (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Plate position x / a (dimensionless) 0.6 0.7 0.8 0.9 1.0 Centerline deflection / maximum (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5878 Worked point: (0.41, 0.9603)
At x = 0.41, the orange circle is the analytical value 0.960294; the purple diamond is the interpolated value 0.587785. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.587785; analytical reference = 0.960294; absolute interpolation error = 0.372508. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Kirchhoff–Love and compatible Mindlin single-mode solutions share this normalized shape but have different bending/shear amplitude formulas. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Mindlin–Reissner plate model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply a single sinusoidal load mode to a simply supported rectangular plate. Plot its normalized centerline deflection. For this calculation, x denotes the plotted horizontal coordinate (Plate position x / a (dimensionless)), and q(x) denotes the plotted response (Centerline deflection / maximum (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

w(x,b/2)/wmax⁡=sin⁡(πx/a)w(x,b/2)/w_{\max}=\sin(\pi x/a)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.85037×10−16\frac{q(b)-q(a)}{b-a}=1.85037\times10^{-16}qlin(c)=0.587785,q(c)=0.960294q_{\rm lin}(c)=0.587785,\quad q(c)=0.960294e=qlin(c)−q(c)=−0.372508e=q_{\rm lin}(c)-q(c)=-0.372508

Solution. The endpoint responses are q(A) = 0.587785 and q(B) = 0.587785. Their secant slope is 1.85037e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.587785. Direct evaluation gives 0.960294; subtracting it from the prediction gives signed error -0.372508.

Mindlin–Reissner plate model: Intermediate-condition prediction and exact check. Horizontal axis: Plate position x / a (dimensionless). Vertical axis: Centerline deflection / maximum (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Plate position x / a (dimensionless) 0.6 0.7 0.8 0.9 1.0 Centerline deflection / maximum (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5878 Worked point: (0.41, 0.9603)
At x = 0.41, the orange circle is the analytical value 0.960294; the purple diamond is the interpolated value 0.587785. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.587785; analytical reference = 0.960294; absolute interpolation error = 0.372508. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Kirchhoff–Love and compatible Mindlin single-mode solutions share this normalized shape but have different bending/shear amplitude formulas. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Shell model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A thin spherical shell of radius R and thickness t carries uniform internal pressure p. For this calculation, x denotes the plotted horizontal coordinate (Pressure loading pR / Et (dimensionless)), and q(x) denotes the plotted response (Membrane stress σ / E (dimensionless)). Suppose only the analytical endpoint responses at x = 0.004 and x = 0.016 are tabulated. Use linear interpolation to predict the response at x = 0.0082, then check it against the original equation.

σ/(E)=12 [pR/(Et)]\sigma/(E)=\tfrac12\,[pR/(Et)]a=0.004,b=0.016,c=0.0082a=0.004,\quad b=0.016,\quad c=0.0082qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.5\frac{q(b)-q(a)}{b-a}=0.5qlin(c)=0.0041,q(c)=0.0041q_{\rm lin}(c)=0.0041,\quad q(c)=0.0041e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.002 and q(B) = 0.008. Their secant slope is 0.5. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0041. Direct evaluation gives 0.0041; subtracting it from the prediction gives signed error 0.

Shell model: Intermediate-condition prediction and exact check. Horizontal axis: Pressure loading pR / Et (dimensionless). Vertical axis: Membrane stress σ / E (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.004 0.006 0.008 0.010 0.012 0.014 0.016 Pressure loading pR / Et (dimensionless) 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Membrane stress σ / E (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0041 Worked point: (0.0082, 0.0041)
At x = 0.0082, the orange circle is the analytical value 0.0041; the purple diamond is the interpolated value 0.0041. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0041; analytical reference = 0.0041; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Thin-shell membrane approximation, away from supports and local bending disturbances. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Cable and membrane models · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. An ideal flexible cable supports its own uniform weight per arc length; choose a = horizontal tension / weight per length. For this calculation, x denotes the plotted horizontal coordinate (Horizontal distance x / a (dimensionless)), and q(x) denotes the plotted response (Height above lowest point y / a (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

y/a=cosh⁡(x/a)−1y/a=\cosh(x/a)-1a=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.85037×10−16\frac{q(b)-q(a)}{b-a}=1.85037\times10^{-16}qlin(c)=0.810656,q(c)=0.0655029q_{\rm lin}(c)=0.810656,\quad q(c)=0.0655029e=qlin(c)−q(c)=0.745153e=q_{\rm lin}(c)-q(c)=0.745153

Solution. The endpoint responses are q(A) = 0.810656 and q(B) = 0.810656. Their secant slope is 1.85037e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.810656. Direct evaluation gives 0.0655029; subtracting it from the prediction gives signed error 0.745153.

Cable and membrane models: Intermediate-condition prediction and exact check. Horizontal axis: Horizontal distance x / a (dimensionless). Vertical axis: Height above lowest point y / a (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Horizontal distance x / a (dimensionless) 0.0 0.2 0.4 0.6 0.8 Height above lowest point y / a (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.8107 Worked point: (-0.36, 0.0655)
At x = -0.36, the orange circle is the analytical value 0.0655029; the purple diamond is the interpolated value 0.810656. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.810656; analytical reference = 0.0655029; absolute interpolation error = 0.745153. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Self-weight catenary, not the parabolic approximation for uniform load per horizontal span. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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von Mises J2 plasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Load monotonically in uniaxial tension from an unstressed state, with no hardening. For this calculation, x denotes the plotted horizontal coordinate (Total strain Eε / σy (dimensionless)), and q(x) denotes the plotted response (Axial stress / yield stress (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

σ/σy=min⁡(Eε/σy,1)\sigma/\sigma_y=\min(E\varepsilon/\sigma_y,1)a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.222222\frac{q(b)-q(a)}{b-a}=0.222222qlin(c)=0.74,q(c)=1q_{\rm lin}(c)=0.74,\quad q(c)=1e=qlin(c)−q(c)=−0.26e=q_{\rm lin}(c)-q(c)=-0.26

Solution. The endpoint responses are q(A) = 0.6 and q(B) = 1. Their secant slope is 0.222222. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.74. Direct evaluation gives 1; subtracting it from the prediction gives signed error -0.26.

von Mises J2 plasticity: Intermediate-condition prediction and exact check. Horizontal axis: Total strain Eε / σy (dimensionless). Vertical axis: Axial stress / yield stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Total strain Eε / σy (dimensionless) 0.6 0.7 0.8 0.9 1.0 Axial stress / yield stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.74 Worked point: (1.23, 1)
At x = 1.23, the orange circle is the analytical value 1; the purple diamond is the interpolated value 0.74. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.74; analytical reference = 1; absolute interpolation error = 0.26. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Uniaxial case where J2 and Tresca coincide; multiaxial yield surfaces differ. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Tresca yield model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Load monotonically in uniaxial tension from an unstressed state, with no hardening. For this calculation, x denotes the plotted horizontal coordinate (Total strain Eε / σy (dimensionless)), and q(x) denotes the plotted response (Axial stress / yield stress (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

σ/σy=min⁡(Eε/σy,1)\sigma/\sigma_y=\min(E\varepsilon/\sigma_y,1)a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.222222\frac{q(b)-q(a)}{b-a}=0.222222qlin(c)=0.74,q(c)=1q_{\rm lin}(c)=0.74,\quad q(c)=1e=qlin(c)−q(c)=−0.26e=q_{\rm lin}(c)-q(c)=-0.26

Solution. The endpoint responses are q(A) = 0.6 and q(B) = 1. Their secant slope is 0.222222. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.74. Direct evaluation gives 1; subtracting it from the prediction gives signed error -0.26.

Tresca yield model: Intermediate-condition prediction and exact check. Horizontal axis: Total strain Eε / σy (dimensionless). Vertical axis: Axial stress / yield stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Total strain Eε / σy (dimensionless) 0.6 0.7 0.8 0.9 1.0 Axial stress / yield stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.74 Worked point: (1.23, 1)
At x = 1.23, the orange circle is the analytical value 1; the purple diamond is the interpolated value 0.74. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.74; analytical reference = 1; absolute interpolation error = 0.26. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Uniaxial case where J2 and Tresca coincide; multiaxial yield surfaces differ. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Drucker–Prager plasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Define the illustrative yield line q−0.5p−c=0 with compression-positive pressure p. For this calculation, x denotes the plotted horizontal coordinate (Compressive mean stress p / c (dimensionless)), and q(x) denotes the plotted response (Deviatoric strength q / c (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

q/c=1+0.5(p/c)q/c=1+0.5(p/c)a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.5\frac{q(b)-q(a)}{b-a}=0.5qlin(c)=1.82,q(c)=1.82q_{\rm lin}(c)=1.82,\quad q(c)=1.82e=qlin(c)−q(c)=−2.22045×10−16e=q_{\rm lin}(c)-q(c)=-2.22045\times10^{-16}

Solution. The endpoint responses are q(A) = 1.4 and q(B) = 2.6. Their secant slope is 0.5. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.82. Direct evaluation gives 1.82; subtracting it from the prediction gives signed error -2.22045e-16.

Drucker–Prager plasticity: Intermediate-condition prediction and exact check. Horizontal axis: Compressive mean stress p / c (dimensionless). Vertical axis: Deviatoric strength q / c (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Compressive mean stress p / c (dimensionless) 1.4 1.6 1.8 2.0 2.2 2.4 2.6 Deviatoric strength q / c (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.82 Worked point: (1.64, 1.82)
At x = 1.64, the orange circle is the analytical value 1.82; the purple diamond is the interpolated value 1.82. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.82; analytical reference = 1.82; absolute interpolation error = 2.22045e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. A specified pressure/deviatoric convention and slope; different parameter mappings to friction angle exist. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Mohr–Coulomb model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use cohesion c > 0 and friction angle 30 degrees. For this calculation, x denotes the plotted horizontal coordinate (Compressive normal stress σn / c (dimensionless)), and q(x) denotes the plotted response (Shear strength τf / c (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

τf/c=1+(σn/c)tan⁡(30∘)\tau_f/c=1+(\sigma_n/c)\tan(30^\circ)a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.57735\frac{q(b)-q(a)}{b-a}=0.57735qlin(c)=1.94685,q(c)=1.94685q_{\rm lin}(c)=1.94685,\quad q(c)=1.94685e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 1.46188 and q(B) = 2.84752. Their secant slope is 0.57735. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.94685. Direct evaluation gives 1.94685; subtracting it from the prediction gives signed error 0.

Mohr–Coulomb model: Intermediate-condition prediction and exact check. Horizontal axis: Compressive normal stress σn / c (dimensionless). Vertical axis: Shear strength τf / c (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Compressive normal stress σn / c (dimensionless) 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 Shear strength τf / c (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.947 Worked point: (1.64, 1.947)
At x = 1.64, the orange circle is the analytical value 1.94685; the purple diamond is the interpolated value 1.94685. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.94685; analytical reference = 1.94685; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Johnson–Cook model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Set B/A = 0.5, n = 0.5, strain rate equal to its reference value, and homologous temperature zero. For this calculation, x denotes the plotted horizontal coordinate (Equivalent plastic strain (dimensionless)), and q(x) denotes the plotted response (Flow stress / A (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

σ/A=1+0.5εp\sigma/A=1+0.5\sqrt{\varepsilon_p}a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.372678\frac{q(b)-q(a)}{b-a}=0.372678qlin(c)=1.30187,q(c)=1.32016q_{\rm lin}(c)=1.30187,\quad q(c)=1.32016e=qlin(c)−q(c)=−0.018287e=q_{\rm lin}(c)-q(c)=-0.018287

Solution. The endpoint responses are q(A) = 1.22361 and q(B) = 1.44721. Their secant slope is 0.372678. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.30187. Direct evaluation gives 1.32016; subtracting it from the prediction gives signed error -0.018287.

Johnson–Cook model: Intermediate-condition prediction and exact check. Horizontal axis: Equivalent plastic strain (dimensionless). Vertical axis: Flow stress / A (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Equivalent plastic strain (dimensionless) 1.20 1.25 1.30 1.35 1.40 1.45 Flow stress / A (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.302 Worked point: (0.41, 1.32)
At x = 0.41, the orange circle is the analytical value 1.32016; the purple diamond is the interpolated value 1.30187. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.30187; analytical reference = 1.32016; absolute interpolation error = 0.018287. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative constants, not a calibrated metal response. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Crystal plasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For positive resolved shear choose rate sensitivity m = 0.2 and fixed slip resistance g. For this calculation, x denotes the plotted horizontal coordinate (Resolved shear / slip resistance τ/g (dimensionless)), and q(x) denotes the plotted response (Slip rate / reference rate (dimensionless)). Suppose only the analytical endpoint responses at x = 0.3 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = 0.615, then check it against the original equation.

γ˙/γ˙0=(τ/g)5\dot\gamma/\dot\gamma_0=(\tau/g)^5a=0.3,b=1.2,c=0.615a=0.3,\quad b=1.2,\quad c=0.615qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2.7621\frac{q(b)-q(a)}{b-a}=2.7621qlin(c)=0.872492,q(c)=0.0879783q_{\rm lin}(c)=0.872492,\quad q(c)=0.0879783e=qlin(c)−q(c)=0.784513e=q_{\rm lin}(c)-q(c)=0.784513

Solution. The endpoint responses are q(A) = 0.00243 and q(B) = 2.48832. Their secant slope is 2.7621. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.872492. Direct evaluation gives 0.0879783; subtracting it from the prediction gives signed error 0.784513.

Crystal plasticity: Intermediate-condition prediction and exact check. Horizontal axis: Resolved shear / slip resistance τ/g (dimensionless). Vertical axis: Slip rate / reference rate (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.4 0.6 0.8 1.0 1.2 Resolved shear / slip resistance τ/g (dimensionless) 0.0 0.5 1.0 1.5 2.0 2.5 Slip rate / reference rate (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.8725 Worked point: (0.615, 0.08798)
At x = 0.615, the orange circle is the analytical value 0.0879783; the purple diamond is the interpolated value 0.872492. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.872492; analytical reference = 0.0879783; absolute interpolation error = 0.784513. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single-system constitutive evaluation; lattice rotation and hardening are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Maxwell viscoelastic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply a step strain ε₀ to a Maxwell spring–dashpot series element and hold it fixed. Scale stress by Eε₀ and time by η/E. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Stress / initial stress (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Maxwell viscoelastic model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Stress / initial stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Stress / initial stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact one-mode reduction with constant coefficients; additional coupled physics is excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Kelvin–Voigt model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply constant stress σ₀ at t = 0 to an initially undeformed parallel spring and dashpot. For this calculation, x denotes the plotted horizontal coordinate (Time Et / η (dimensionless)), and q(x) denotes the plotted response (Normalized creep strain Eε / σ₀ (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

Eε/σ0=1−e−Et/ηE\varepsilon/\sigma_0=1-e^{-Et/\eta}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Kelvin–Voigt model: Intermediate-condition prediction and exact check. Horizontal axis: Time Et / η (dimensionless). Vertical axis: Normalized creep strain Eε / σ₀ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time Et / η (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Normalized creep strain Eε / σ₀ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Standard linear solid · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply a fixed strain step to a standard linear solid with relaxed modulus E∞ = 0.4E0. For this calculation, x denotes the plotted horizontal coordinate (Time t / τ (dimensionless)), and q(x) denotes the plotted response (Stress / initial stress (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

σ/(E0ε0)=0.4+0.6e−t/τ\sigma/(E_0\varepsilon_0)=0.4+0.6e^{-t/\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.0699128\frac{q(b)-q(a)}{b-a}=-0.0699128qlin(c)=0.547319,q(c)=0.477241q_{\rm lin}(c)=0.547319,\quad q(c)=0.477241e=qlin(c)−q(c)=0.0700783e=q_{\rm lin}(c)-q(c)=0.0700783

Solution. The endpoint responses are q(A) = 0.620728 and q(B) = 0.410989. Their secant slope is -0.0699128. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.547319. Direct evaluation gives 0.477241; subtracting it from the prediction gives signed error 0.0700783.

Standard linear solid: Intermediate-condition prediction and exact check. Horizontal axis: Time t / τ (dimensionless). Vertical axis: Stress / initial stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time t / τ (dimensionless) 0.40 0.45 0.50 0.55 0.60 Stress / initial stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5473 Worked point: (2.05, 0.4772)
At x = 2.05, the orange circle is the analytical value 0.477241; the purple diamond is the interpolated value 0.547319. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.547319; analytical reference = 0.477241; absolute interpolation error = 0.0700783. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Norton creep law · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. At fixed temperature use Norton exponent n = 3 and reference rate Aσ*³. For this calculation, x denotes the plotted horizontal coordinate (Stress σ / σ* (dimensionless)), and q(x) denotes the plotted response (Creep rate / reference rate (dimensionless)). Suppose only the analytical endpoint responses at x = 0.4 and x = 1.6 are tabulated. Use linear interpolation to predict the response at x = 0.82, then check it against the original equation.

ε˙/ε˙∗=(σ/σ∗)3\dot\varepsilon/\dot\varepsilon_*=(\sigma/\sigma_*)^3a=0.4,b=1.6,c=0.82a=0.4,\quad b=1.6,\quad c=0.82qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=3.36\frac{q(b)-q(a)}{b-a}=3.36qlin(c)=1.4752,q(c)=0.551368q_{\rm lin}(c)=1.4752,\quad q(c)=0.551368e=qlin(c)−q(c)=0.923832e=q_{\rm lin}(c)-q(c)=0.923832

Solution. The endpoint responses are q(A) = 0.064 and q(B) = 4.096. Their secant slope is 3.36. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.4752. Direct evaluation gives 0.551368; subtracting it from the prediction gives signed error 0.923832.

Norton creep law: Intermediate-condition prediction and exact check. Horizontal axis: Stress σ / σ* (dimensionless). Vertical axis: Creep rate / reference rate (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Stress σ / σ* (dimensionless) 0 1 2 3 4 Creep rate / reference rate (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.475 Worked point: (0.82, 0.5514)
At x = 0.82, the orange circle is the analytical value 0.551368; the purple diamond is the interpolated value 1.4752. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.4752; analytical reference = 0.551368; absolute interpolation error = 0.923832. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Steady creep constitutive law at fixed material state and temperature. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Linear elastic fracture mechanics (LEFM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the leading mode-I elastic crack-tip field on θ = 0. For this calculation, x denotes the plotted horizontal coordinate (Distance ahead of tip r / ℓ (dimensionless)), and q(x) denotes the plotted response (Scaled opening stress (dimensionless)). Suppose only the analytical endpoint responses at x = 0.44 and x = 1.61 are tabulated. Use linear interpolation to predict the response at x = 0.8495, then check it against the original equation.

σyy/(KI/2πℓ)=(r/ℓ)−1/2\sigma_{yy}/(K_I/\sqrt{2\pi\ell})=(r/\ell)^{-1/2}a=0.44,b=1.61,c=0.8495a=0.44,\quad b=1.61,\quad c=0.8495qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.614911\frac{q(b)-q(a)}{b-a}=-0.614911qlin(c)=1.25575,q(c)=1.08497q_{\rm lin}(c)=1.25575,\quad q(c)=1.08497e=qlin(c)−q(c)=0.170779e=q_{\rm lin}(c)-q(c)=0.170779

Solution. The endpoint responses are q(A) = 1.50756 and q(B) = 0.78811. Their secant slope is -0.614911. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.25575. Direct evaluation gives 1.08497; subtracting it from the prediction gives signed error 0.170779.

Linear elastic fracture mechanics (LEFM): Intermediate-condition prediction and exact check. Horizontal axis: Distance ahead of tip r / ℓ (dimensionless). Vertical axis: Scaled opening stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.6 0.8 1.0 1.2 1.4 1.6 Distance ahead of tip r / ℓ (dimensionless) 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 Scaled opening stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.256 Worked point: (0.8495, 1.085)
At x = 0.8495, the orange circle is the analytical value 1.08497; the purple diamond is the interpolated value 1.25575. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.25575; analytical reference = 1.08497; absolute interpolation error = 0.170779. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Near-tip linear-elastic asymptotic field, outside the process zone; the singular tip itself is excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Cohesive-zone model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose peak traction at half the complete-separation opening and linear loading/softening branches. For this calculation, x denotes the plotted horizontal coordinate (Opening d = δ / δc (dimensionless)), and q(x) denotes the plotted response (Traction / peak traction (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

t/tmax⁡={2dd≤0.52(1−d)d>0.5t/t_{\max}=\begin{cases}2d&d\le0.5\\2(1-d)&d>0.5\end{cases}a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.85037×10−16\frac{q(b)-q(a)}{b-a}=-1.85037\times10^{-16}qlin(c)=0.4,q(c)=0.82q_{\rm lin}(c)=0.4,\quad q(c)=0.82e=qlin(c)−q(c)=−0.42e=q_{\rm lin}(c)-q(c)=-0.42

Solution. The endpoint responses are q(A) = 0.4 and q(B) = 0.4. Their secant slope is -1.85037e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.4. Direct evaluation gives 0.82; subtracting it from the prediction gives signed error -0.42.

Cohesive-zone model: Intermediate-condition prediction and exact check. Horizontal axis: Opening d = δ / δc (dimensionless). Vertical axis: Traction / peak traction (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Opening d = δ / δc (dimensionless) 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Traction / peak traction (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.4 Worked point: (0.41, 0.82)
At x = 0.41, the orange circle is the analytical value 0.82; the purple diamond is the interpolated value 0.4. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.4; analytical reference = 0.82; absolute interpolation error = 0.42. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Monotonic prescribed cohesive law; unloading and mixed-mode effects are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Phase-field fracture model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Minimize the isolated AT2 crack-surface functional with d(0)=1 and d→0 far from the crack, without mechanical driving away from x=0. For this calculation, x denotes the plotted horizontal coordinate (Distance from crack x / ℓ (dimensionless)), and q(x) denotes the plotted response (Damage d (dimensionless)). Suppose only the analytical endpoint responses at x = -3 and x = 3 are tabulated. Use linear interpolation to predict the response at x = -0.9, then check it against the original equation.

d(x)=e−∣x∣/ℓd(x)=e^{-|x|/\ell}a=−3,b=3,c=−0.9a=-3,\quad b=3,\quad c=-0.9qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0\frac{q(b)-q(a)}{b-a}=0qlin(c)=0.0497871,q(c)=0.40657q_{\rm lin}(c)=0.0497871,\quad q(c)=0.40657e=qlin(c)−q(c)=−0.356783e=q_{\rm lin}(c)-q(c)=-0.356783

Solution. The endpoint responses are q(A) = 0.0497871 and q(B) = 0.0497871. Their secant slope is 0. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0497871. Direct evaluation gives 0.40657; subtracting it from the prediction gives signed error -0.356783.

Phase-field fracture model: Intermediate-condition prediction and exact check. Horizontal axis: Distance from crack x / ℓ (dimensionless). Vertical axis: Damage d (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −3 −2 −1 0 1 2 3 Distance from crack x / ℓ (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Damage d (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.04979 Worked point: (-0.9, 0.4066)
At x = -0.9, the orange circle is the analytical value 0.40657; the purple diamond is the interpolated value 0.0497871. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0497871; analytical reference = 0.40657; absolute interpolation error = 0.356783. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Stationary isolated crack-profile benchmark, not a coupled fracture-growth solution. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Paris fatigue crack-growth law · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use da/dN=C(ΔK)² and ΔK=Δσ√(πa) with constant stress range and geometry factor one. For this calculation, x denotes the plotted horizontal coordinate (Cycle count N / N* (dimensionless)), and q(x) denotes the plotted response (Crack length a / a₀ (dimensionless)). Suppose only the analytical endpoint responses at x = 0.3 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = 0.615, then check it against the original equation.

a/a0=eN/N∗,N∗=(CΔσ2π)−1a/a_0=e^{N/N_*},\quad N_*=(C\Delta\sigma^2\pi)^{-1}a=0.3,b=1.2,c=0.615a=0.3,\quad b=1.2,\quad c=0.615qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2.18918\frac{q(b)-q(a)}{b-a}=2.18918qlin(c)=2.03945,q(c)=1.84966q_{\rm lin}(c)=2.03945,\quad q(c)=1.84966e=qlin(c)−q(c)=0.189793e=q_{\rm lin}(c)-q(c)=0.189793

Solution. The endpoint responses are q(A) = 1.34986 and q(B) = 3.32012. Their secant slope is 2.18918. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.03945. Direct evaluation gives 1.84966; subtracting it from the prediction gives signed error 0.189793.

Paris fatigue crack-growth law: Intermediate-condition prediction and exact check. Horizontal axis: Cycle count N / N* (dimensionless). Vertical axis: Crack length a / a₀ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.4 0.6 0.8 1.0 1.2 Cycle count N / N* (dimensionless) 1.5 2.0 2.5 3.0 3.5 Crack length a / a₀ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.039 Worked point: (0.615, 1.85)
At x = 0.615, the orange circle is the analytical value 1.84966; the purple diamond is the interpolated value 2.03945. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.03945; analytical reference = 1.84966; absolute interpolation error = 0.189793. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Only within the Paris regime; threshold, instability, and changing geometry are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Miner cumulative damage rule · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply constant-amplitude cycles with a fixed fatigue life Nf. For this calculation, x denotes the plotted horizontal coordinate (Applied cycles / failure cycles n/Nf (dimensionless)), and q(x) denotes the plotted response (Accumulated damage D (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

D=n/NfD=n/N_fa=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.41,q(c)=0.41q_{\rm lin}(c)=0.41,\quad q(c)=0.41e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.2 and q(B) = 0.8. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.41. Direct evaluation gives 0.41; subtracting it from the prediction gives signed error 0.

Miner cumulative damage rule: Intermediate-condition prediction and exact check. Horizontal axis: Applied cycles / failure cycles n/Nf (dimensionless). Vertical axis: Accumulated damage D (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Applied cycles / failure cycles n/Nf (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Accumulated damage D (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.41 Worked point: (0.41, 0.41)
At x = 0.41, the orange circle is the analytical value 0.41; the purple diamond is the interpolated value 0.41. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.41; analytical reference = 0.41; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linear accumulation hypothesis, not a physical guarantee of failure at exactly D=1. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Archard wear model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold wear coefficient k, normal force W, and hardness H constant. For this calculation, x denotes the plotted horizontal coordinate (Sliding distance s / s* (dimensionless)), and q(x) denotes the plotted response (Scaled wear volume VH / kWs* (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

VH/(kWs∗)=s/s∗VH/(kWs_*)=s/s_*a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=2.05,q(c)=2.05q_{\rm lin}(c)=2.05,\quad q(c)=2.05e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 1 and q(B) = 4. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.05. Direct evaluation gives 2.05; subtracting it from the prediction gives signed error 0.

Archard wear model: Intermediate-condition prediction and exact check. Horizontal axis: Sliding distance s / s* (dimensionless). Vertical axis: Scaled wear volume VH / kWs* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Sliding distance s / s* (dimensionless) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Scaled wear volume VH / kWs* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.05 Worked point: (2.05, 2.05)
At x = 2.05, the orange circle is the analytical value 2.05; the purple diamond is the interpolated value 2.05. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.05; analytical reference = 2.05; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Steady Archard wear regime with no changes in contact, debris, or material properties. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Newton–Euler rigid-body model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A rigid body starts at rest with constant net force-to-mass ratio 1 m/s² along one axis and zero net torque. For this calculation, x denotes the plotted horizontal coordinate (Elapsed time t (s)), and q(x) denotes the plotted response (Displacement x (m)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

x(t)=12at2,a=1  m/s2x(t)=\tfrac12at^2,\quad a=1\;\mathrm{m/s^2}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2.5\frac{q(b)-q(a)}{b-a}=2.5qlin(c)=3.125,q(c)=2.10125q_{\rm lin}(c)=3.125,\quad q(c)=2.10125e=qlin(c)−q(c)=1.02375e=q_{\rm lin}(c)-q(c)=1.02375

Solution. The endpoint responses are q(A) = 0.5 and q(B) = 8. Their secant slope is 2.5. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 3.125. Direct evaluation gives 2.10125; subtracting it from the prediction gives signed error 1.02375.

Newton–Euler rigid-body model: Intermediate-condition prediction and exact check. Horizontal axis: Elapsed time t (s). Vertical axis: Displacement x (m). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Elapsed time t (s) 0 2 4 6 8 Displacement x (m) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 3.125 Worked point: (2.05, 2.101)
At x = 2.05, the orange circle is the analytical value 2.10125; the purple diamond is the interpolated value 3.125. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 3.125; analytical reference = 2.10125; absolute interpolation error = 1.02375. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single translational degree of freedom; the remaining forces, torques, and rotational motion are set to zero. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Six-degree-of-freedom flight model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A rigid body starts at rest with constant net force-to-mass ratio 1 m/s² along one axis and zero net torque. For this calculation, x denotes the plotted horizontal coordinate (Elapsed time t (s)), and q(x) denotes the plotted response (Displacement x (m)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

x(t)=12at2,a=1  m/s2x(t)=\tfrac12at^2,\quad a=1\;\mathrm{m/s^2}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2.5\frac{q(b)-q(a)}{b-a}=2.5qlin(c)=3.125,q(c)=2.10125q_{\rm lin}(c)=3.125,\quad q(c)=2.10125e=qlin(c)−q(c)=1.02375e=q_{\rm lin}(c)-q(c)=1.02375

Solution. The endpoint responses are q(A) = 0.5 and q(B) = 8. Their secant slope is 2.5. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 3.125. Direct evaluation gives 2.10125; subtracting it from the prediction gives signed error 1.02375.

Six-degree-of-freedom flight model: Intermediate-condition prediction and exact check. Horizontal axis: Elapsed time t (s). Vertical axis: Displacement x (m). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Elapsed time t (s) 0 2 4 6 8 Displacement x (m) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 3.125 Worked point: (2.05, 2.101)
At x = 2.05, the orange circle is the analytical value 2.10125; the purple diamond is the interpolated value 3.125. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 3.125; analytical reference = 2.10125; absolute interpolation error = 1.02375. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single translational degree of freedom; the remaining forces, torques, and rotational motion are set to zero. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lagrangian mechanics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose a single unconstrained linear mode with zero damping, initial displacement A, and zero velocity. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 2.51327 and x = 10.0531 are tabulated. Use linear interpolation to predict the response at x = 5.15221, then check it against the original equation.

q/A=cos⁡(ωt)q/A=\cos(\omega t)a=2.51327,b=10.0531,c=5.15221a=2.51327,\quad b=10.0531,\quad c=5.15221qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−4.41744×10−17\frac{q(b)-q(a)}{b-a}=-4.41744\times10^{-17}qlin(c)=−0.809017,q(c)=0.425779q_{\rm lin}(c)=-0.809017,\quad q(c)=0.425779e=qlin(c)−q(c)=−1.2348e=q_{\rm lin}(c)-q(c)=-1.2348

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = -0.809017. Their secant slope is -4.41744e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.809017. Direct evaluation gives 0.425779; subtracting it from the prediction gives signed error -1.2348.

Lagrangian mechanics: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 10 Phase ωt (radian) −1.0 −0.5 0.0 0.5 1.0 Displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.809 Worked point: (5.152, 0.4258)
At x = 5.15221, the orange circle is the analytical value 0.425779; the purple diamond is the interpolated value -0.809017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.809017; analytical reference = 0.425779; absolute interpolation error = 1.2348. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact single harmonic mode; multibody constraints and other modal couplings are absent. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hamiltonian mechanics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose a single unconstrained linear mode with zero damping, initial displacement A, and zero velocity. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 2.51327 and x = 10.0531 are tabulated. Use linear interpolation to predict the response at x = 5.15221, then check it against the original equation.

q/A=cos⁡(ωt)q/A=\cos(\omega t)a=2.51327,b=10.0531,c=5.15221a=2.51327,\quad b=10.0531,\quad c=5.15221qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−4.41744×10−17\frac{q(b)-q(a)}{b-a}=-4.41744\times10^{-17}qlin(c)=−0.809017,q(c)=0.425779q_{\rm lin}(c)=-0.809017,\quad q(c)=0.425779e=qlin(c)−q(c)=−1.2348e=q_{\rm lin}(c)-q(c)=-1.2348

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = -0.809017. Their secant slope is -4.41744e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.809017. Direct evaluation gives 0.425779; subtracting it from the prediction gives signed error -1.2348.

Hamiltonian mechanics: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 10 Phase ωt (radian) −1.0 −0.5 0.0 0.5 1.0 Displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.809 Worked point: (5.152, 0.4258)
At x = 5.15221, the orange circle is the analytical value 0.425779; the purple diamond is the interpolated value -0.809017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.809017; analytical reference = 0.425779; absolute interpolation error = 1.2348. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact single harmonic mode; multibody constraints and other modal couplings are absent. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Mass–spring–damper model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose a single unconstrained linear mode with zero damping, initial displacement A, and zero velocity. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 2.51327 and x = 10.0531 are tabulated. Use linear interpolation to predict the response at x = 5.15221, then check it against the original equation.

q/A=cos⁡(ωt)q/A=\cos(\omega t)a=2.51327,b=10.0531,c=5.15221a=2.51327,\quad b=10.0531,\quad c=5.15221qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−4.41744×10−17\frac{q(b)-q(a)}{b-a}=-4.41744\times10^{-17}qlin(c)=−0.809017,q(c)=0.425779q_{\rm lin}(c)=-0.809017,\quad q(c)=0.425779e=qlin(c)−q(c)=−1.2348e=q_{\rm lin}(c)-q(c)=-1.2348

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = -0.809017. Their secant slope is -4.41744e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.809017. Direct evaluation gives 0.425779; subtracting it from the prediction gives signed error -1.2348.

Mass–spring–damper model: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 10 Phase ωt (radian) −1.0 −0.5 0.0 0.5 1.0 Displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.809 Worked point: (5.152, 0.4258)
At x = 5.15221, the orange circle is the analytical value 0.425779; the purple diamond is the interpolated value -0.809017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.809017; analytical reference = 0.425779; absolute interpolation error = 1.2348. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact single harmonic mode; multibody constraints and other modal couplings are absent. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Modal superposition model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose a single unconstrained linear mode with zero damping, initial displacement A, and zero velocity. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 2.51327 and x = 10.0531 are tabulated. Use linear interpolation to predict the response at x = 5.15221, then check it against the original equation.

q/A=cos⁡(ωt)q/A=\cos(\omega t)a=2.51327,b=10.0531,c=5.15221a=2.51327,\quad b=10.0531,\quad c=5.15221qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−4.41744×10−17\frac{q(b)-q(a)}{b-a}=-4.41744\times10^{-17}qlin(c)=−0.809017,q(c)=0.425779q_{\rm lin}(c)=-0.809017,\quad q(c)=0.425779e=qlin(c)−q(c)=−1.2348e=q_{\rm lin}(c)-q(c)=-1.2348

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = -0.809017. Their secant slope is -4.41744e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.809017. Direct evaluation gives 0.425779; subtracting it from the prediction gives signed error -1.2348.

Modal superposition model: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 10 Phase ωt (radian) −1.0 −0.5 0.0 0.5 1.0 Displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.809 Worked point: (5.152, 0.4258)
At x = 5.15221, the orange circle is the analytical value 0.425779; the purple diamond is the interpolated value -0.809017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.809017; analytical reference = 0.425779; absolute interpolation error = 1.2348. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact single harmonic mode; multibody constraints and other modal couplings are absent. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Multibody dynamics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose a single unconstrained linear mode with zero damping, initial displacement A, and zero velocity. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 2.51327 and x = 10.0531 are tabulated. Use linear interpolation to predict the response at x = 5.15221, then check it against the original equation.

q/A=cos⁡(ωt)q/A=\cos(\omega t)a=2.51327,b=10.0531,c=5.15221a=2.51327,\quad b=10.0531,\quad c=5.15221qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−4.41744×10−17\frac{q(b)-q(a)}{b-a}=-4.41744\times10^{-17}qlin(c)=−0.809017,q(c)=0.425779q_{\rm lin}(c)=-0.809017,\quad q(c)=0.425779e=qlin(c)−q(c)=−1.2348e=q_{\rm lin}(c)-q(c)=-1.2348

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = -0.809017. Their secant slope is -4.41744e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.809017. Direct evaluation gives 0.425779; subtracting it from the prediction gives signed error -1.2348.

Multibody dynamics: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 10 Phase ωt (radian) −1.0 −0.5 0.0 0.5 1.0 Displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.809 Worked point: (5.152, 0.4258)
At x = 5.15221, the orange circle is the analytical value 0.425779; the purple diamond is the interpolated value -0.809017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.809017; analytical reference = 0.425779; absolute interpolation error = 1.2348. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact single harmonic mode; multibody constraints and other modal couplings are absent. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Duffing oscillator · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For positive linear and cubic stiffness choose ℓ=√(k/β). Find the force needed to hold a static displacement. For this calculation, x denotes the plotted horizontal coordinate (Static displacement x / ℓ (dimensionless)), and q(x) denotes the plotted response (Static force / kℓ (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

F/(kℓ)=q+q3,q=x/ℓF/(k\ell)=q+q^3,\quad q=x/\ella=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2.44\frac{q(b)-q(a)}{b-a}=2.44qlin(c)=−0.8784,q(c)=−0.406656q_{\rm lin}(c)=-0.8784,\quad q(c)=-0.406656e=qlin(c)−q(c)=−0.471744e=q_{\rm lin}(c)-q(c)=-0.471744

Solution. The endpoint responses are q(A) = -2.928 and q(B) = 2.928. Their secant slope is 2.44. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.8784. Direct evaluation gives -0.406656; subtracting it from the prediction gives signed error -0.471744.

Duffing oscillator: Intermediate-condition prediction and exact check. Horizontal axis: Static displacement x / ℓ (dimensionless). Vertical axis: Static force / kℓ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Static displacement x / ℓ (dimensionless) −3 −2 −1 0 1 2 3 Static force / kℓ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.8784 Worked point: (-0.36, -0.4067)
At x = -0.36, the orange circle is the analytical value -0.406656; the purple diamond is the interpolated value -0.8784. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.8784; analytical reference = -0.406656; absolute interpolation error = 0.471744. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Static hardening equilibrium curve, not a forced nonlinear transient or resonance calculation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Linear acoustic wave model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional sinusoidal wave in a uniform, lossless linear medium. Plot the normalized field at time zero. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Acoustic pressure / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Linear acoustic wave model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Acoustic pressure / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Acoustic pressure / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. An acoustic, electromagnetic, elastic, or linear Alfvén-wave reference as appropriate. For MHD this is the small transverse perturbation of a uniform magnetized equilibrium; for FDTD it is an exact target, not a discretized result. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Transmission-line acoustic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional sinusoidal wave in a uniform, lossless linear medium. Plot the normalized field at time zero. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Acoustic pressure / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Transmission-line acoustic model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Acoustic pressure / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Acoustic pressure / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. An acoustic, electromagnetic, elastic, or linear Alfvén-wave reference as appropriate. For MHD this is the small transverse perturbation of a uniform magnetized equilibrium; for FDTD it is an exact target, not a discretized result. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Maxwell electromagnetic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional sinusoidal wave in a uniform, lossless linear medium. Plot the normalized field at time zero. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Electric-field component / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Maxwell electromagnetic model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Electric-field component / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Electric-field component / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. An acoustic, electromagnetic, elastic, or linear Alfvén-wave reference as appropriate. For MHD this is the small transverse perturbation of a uniform magnetized equilibrium; for FDTD it is an exact target, not a discretized result. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Transmission-line electrical model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional sinusoidal wave in a uniform, lossless linear medium. Plot the normalized field at time zero. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Line voltage / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Transmission-line electrical model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Line voltage / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Line voltage / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. An acoustic, electromagnetic, elastic, or linear Alfvén-wave reference as appropriate. For MHD this is the small transverse perturbation of a uniform magnetized equilibrium; for FDTD it is an exact target, not a discretized result. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Elastic seismic-wave model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional sinusoidal wave in a uniform, lossless linear medium. Plot the normalized field at time zero. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Elastic displacement / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Elastic seismic-wave model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Elastic displacement / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Elastic displacement / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. An acoustic, electromagnetic, elastic, or linear Alfvén-wave reference as appropriate. For MHD this is the small transverse perturbation of a uniform magnetized equilibrium; for FDTD it is an exact target, not a discretized result. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Magnetohydrodynamics (MHD) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional sinusoidal wave in a uniform, lossless linear medium. Plot the normalized field at time zero. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Transverse velocity perturbation / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Magnetohydrodynamics (MHD): Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Transverse velocity perturbation / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Transverse velocity perturbation / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. An acoustic, electromagnetic, elastic, or linear Alfvén-wave reference as appropriate. For MHD this is the small transverse perturbation of a uniform magnetized equilibrium; for FDTD it is an exact target, not a discretized result. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Finite-difference time-domain (FDTD) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional sinusoidal wave in a uniform, lossless linear medium. Plot the normalized field at time zero. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Electric-field component / amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Finite-difference time-domain (FDTD): Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Electric-field component / amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Electric-field component / amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. An acoustic, electromagnetic, elastic, or linear Alfvén-wave reference as appropriate. For MHD this is the small transverse perturbation of a uniform magnetized equilibrium; for FDTD it is an exact target, not a discretized result. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Helmholtz acoustic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve p″+k²p=0 with pressure-release endpoints and choose the first nonzero eigenmode. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Pressure amplitude / P (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

p(x)/P=sin⁡(πx/L),k=π/Lp(x)/P=\sin(\pi x/L),\quad k=\pi/La=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.85037×10−16\frac{q(b)-q(a)}{b-a}=1.85037\times10^{-16}qlin(c)=0.587785,q(c)=0.960294q_{\rm lin}(c)=0.587785,\quad q(c)=0.960294e=qlin(c)−q(c)=−0.372508e=q_{\rm lin}(c)-q(c)=-0.372508

Solution. The endpoint responses are q(A) = 0.587785 and q(B) = 0.587785. Their secant slope is 1.85037e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.587785. Direct evaluation gives 0.960294; subtracting it from the prediction gives signed error -0.372508.

Helmholtz acoustic model: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Pressure amplitude / P (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.6 0.7 0.8 0.9 1.0 Pressure amplitude / P (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5878 Worked point: (0.41, 0.9603)
At x = 0.41, the orange circle is the analytical value 0.960294; the purple diamond is the interpolated value 0.587785. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.587785; analytical reference = 0.960294; absolute interpolation error = 0.372508. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Electrostatic Poisson model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve φ″ = −ρ/ε for constant charge density between φ(0)=φ(L)=0. For this calculation, x denotes the plotted horizontal coordinate (Position ξ = x / L (dimensionless)), and q(x) denotes the plotted response (Scaled electrostatic potential (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

ϕ/(ρL2/ϵ)=12ξ(1−ξ)\phi/(\rho L^2/\epsilon)=\tfrac12\xi(1-\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−4.62593×10−17\frac{q(b)-q(a)}{b-a}=-4.62593\times10^{-17}qlin(c)=0.08,q(c)=0.12095q_{\rm lin}(c)=0.08,\quad q(c)=0.12095e=qlin(c)−q(c)=−0.04095e=q_{\rm lin}(c)-q(c)=-0.04095

Solution. The endpoint responses are q(A) = 0.08 and q(B) = 0.08. Their secant slope is -4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.08. Direct evaluation gives 0.12095; subtracting it from the prediction gives signed error -0.04095.

Electrostatic Poisson model: Intermediate-condition prediction and exact check. Horizontal axis: Position ξ = x / L (dimensionless). Vertical axis: Scaled electrostatic potential (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position ξ = x / L (dimensionless) 0.08 0.09 0.10 0.11 0.12 0.13 Scaled electrostatic potential (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.08 Worked point: (0.41, 0.121)
At x = 0.41, the orange circle is the analytical value 0.12095; the purple diamond is the interpolated value 0.08. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.08; analytical reference = 0.12095; absolute interpolation error = 0.04095. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Magnetostatic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Consider the exterior of a long straight wire of radius a carrying steady current I in vacuum. For this calculation, x denotes the plotted horizontal coordinate (Radius from wire r / a (dimensionless)), and q(x) denotes the plotted response (Magnetic field / surface value (dimensionless)). Suppose only the analytical endpoint responses at x = 1.8 and x = 4.2 are tabulated. Use linear interpolation to predict the response at x = 2.64, then check it against the original equation.

B/(μ0I/2πa)=a/rB/(\mu_0 I/2\pi a)=a/ra=1.8,b=4.2,c=2.64a=1.8,\quad b=4.2,\quad c=2.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.132275\frac{q(b)-q(a)}{b-a}=-0.132275qlin(c)=0.444444,q(c)=0.378788q_{\rm lin}(c)=0.444444,\quad q(c)=0.378788e=qlin(c)−q(c)=0.0656566e=q_{\rm lin}(c)-q(c)=0.0656566

Solution. The endpoint responses are q(A) = 0.555556 and q(B) = 0.238095. Their secant slope is -0.132275. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.444444. Direct evaluation gives 0.378788; subtracting it from the prediction gives signed error 0.0656566.

Magnetostatic model: Intermediate-condition prediction and exact check. Horizontal axis: Radius from wire r / a (dimensionless). Vertical axis: Magnetic field / surface value (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2.0 2.5 3.0 3.5 4.0 Radius from wire r / a (dimensionless) 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 Magnetic field / surface value (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.4444 Worked point: (2.64, 0.3788)
At x = 2.64, the orange circle is the analytical value 0.378788; the purple diamond is the interpolated value 0.444444. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.444444; analytical reference = 0.378788; absolute interpolation error = 0.0656566. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exterior field of an ideal long wire; end effects are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Eddy-current model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A sinusoidal magnetic field penetrates a homogeneous conducting half-space with skin depth δ. For this calculation, x denotes the plotted horizontal coordinate (Depth x / skin depth δ (dimensionless)), and q(x) denotes the plotted response (Magnetic-field amplitude fraction (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

∣B(x)∣/∣B(0)∣=e−x/δ|B(x)|/|B(0)|=e^{-x/\delta}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Eddy-current model: Intermediate-condition prediction and exact check. Horizontal axis: Depth x / skin depth δ (dimensionless). Vertical axis: Magnetic-field amplitude fraction (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Depth x / skin depth δ (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Magnetic-field amplitude fraction (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linear conductor with constant conductivity and permeability; displacement current neglected. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Magnetic-circuit model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a single magnetic circuit of fixed reluctance ℛ with no leakage. For this calculation, x denotes the plotted horizontal coordinate (Magnetomotive force / ℛΦ* (dimensionless)), and q(x) denotes the plotted response (Magnetic flux Φ / Φ* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

Φ/Φ∗=(NI)/(RΦ∗)\Phi/\Phi_*=(NI)/(\mathcal R\Phi_*)a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=1.23,q(c)=1.23q_{\rm lin}(c)=1.23,\quad q(c)=1.23e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.6 and q(B) = 2.4. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.23. Direct evaluation gives 1.23; subtracting it from the prediction gives signed error 0.

Magnetic-circuit model: Intermediate-condition prediction and exact check. Horizontal axis: Magnetomotive force / ℛΦ* (dimensionless). Vertical axis: Magnetic flux Φ / Φ* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Magnetomotive force / ℛΦ* (dimensionless) 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 Magnetic flux Φ / Φ* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.23 Worked point: (1.23, 1.23)
At x = 1.23, the orange circle is the analytical value 1.23; the purple diamond is the interpolated value 1.23. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.23; analytical reference = 1.23; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linear unsaturated material and fixed geometry. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Jiles–Atherton hysteresis model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate the Langevin-form anhysteretic component of a Jiles–Atherton model, using its zero-field limit M=0. For this calculation, x denotes the plotted horizontal coordinate (Effective field / anhysteretic scale (dimensionless)), and q(x) denotes the plotted response (Anhysteretic magnetization / saturation (dimensionless)). Suppose only the analytical endpoint responses at x = -3 and x = 3 are tabulated. Use linear interpolation to predict the response at x = -0.9, then check it against the original equation.

Man/Ms=coth⁡h−1/hM_{an}/M_s=\coth h-1/ha=−3,b=3,c=−0.9a=-3,\quad b=3,\quad c=-0.9qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.223879\frac{q(b)-q(a)}{b-a}=0.223879qlin(c)=−0.201491,q(c)=−0.284956q_{\rm lin}(c)=-0.201491,\quad q(c)=-0.284956e=qlin(c)−q(c)=0.0834652e=q_{\rm lin}(c)-q(c)=0.0834652

Solution. The endpoint responses are q(A) = -0.671636 and q(B) = 0.671636. Their secant slope is 0.223879. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.201491. Direct evaluation gives -0.284956; subtracting it from the prediction gives signed error 0.0834652.

Jiles–Atherton hysteresis model: Intermediate-condition prediction and exact check. Horizontal axis: Effective field / anhysteretic scale (dimensionless). Vertical axis: Anhysteretic magnetization / saturation (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −3 −2 −1 0 1 2 3 Effective field / anhysteretic scale (dimensionless) −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 Anhysteretic magnetization / saturation (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.2015 Worked point: (-0.9, -0.285)
At x = -0.9, the orange circle is the analytical value -0.284956; the purple diamond is the interpolated value -0.201491. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.201491; analytical reference = -0.284956; absolute interpolation error = 0.0834652. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Anhysteretic reference only, not the history-dependent hysteresis loop. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Geometrical optics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A ray crosses a plane interface from index 1 into index 1.5. For this calculation, x denotes the plotted horizontal coordinate (Incident angle θ₁ (degree)), and q(x) denotes the plotted response (Refracted angle θ₂ (degree)). Suppose only the analytical endpoint responses at x = 16 and x = 64 are tabulated. Use linear interpolation to predict the response at x = 32.8, then check it against the original equation.

θ2=arcsin⁡[sin⁡(θ1)/1.5]\theta_2=\arcsin[\sin(\theta_1)/1.5]a=16,b=64,c=32.8a=16,\quad b=64,\quad c=32.8qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.546325\frac{q(b)-q(a)}{b-a}=0.546325qlin(c)=19.767,q(c)=21.1702q_{\rm lin}(c)=19.767,\quad q(c)=21.1702e=qlin(c)−q(c)=−1.40315e=q_{\rm lin}(c)-q(c)=-1.40315

Solution. The endpoint responses are q(A) = 10.5887 and q(B) = 36.8123. Their secant slope is 0.546325. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 19.767. Direct evaluation gives 21.1702; subtracting it from the prediction gives signed error -1.40315.

Geometrical optics: Intermediate-condition prediction and exact check. Horizontal axis: Incident angle θ₁ (degree). Vertical axis: Refracted angle θ₂ (degree). image/svg+xml IICSM analytical illustration / Matplotlib 20 30 40 50 60 Incident angle θ₁ (degree) 10 15 20 25 30 35 Refracted angle θ₂ (degree) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 19.77 Worked point: (32.8, 21.17)
At x = 32.8, the orange circle is the analytical value 21.1702; the purple diamond is the interpolated value 19.767. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 19.767; analytical reference = 21.1702; absolute interpolation error = 1.40315. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Scalar diffraction model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Illuminate a slit of width a uniformly with monochromatic coherent light and observe the Fraunhofer pattern. For this calculation, x denotes the plotted horizontal coordinate (Diffraction coordinate u = πa sinθ / λ (dimensionless)), and q(x) denotes the plotted response (Intensity I / I₀ (dimensionless)). Suppose only the analytical endpoint responses at x = -4.8 and x = 4.8 are tabulated. Use linear interpolation to predict the response at x = -1.44, then check it against the original equation.

I/I0=[sin⁡uu]2I/I_0=\left[\frac{\sin u}{u}\right]^2a=−4.8,b=4.8,c=−1.44a=-4.8,\quad b=4.8,\quad c=-1.44qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.61401×10−18\frac{q(b)-q(a)}{b-a}=-3.61401\times10^{-18}qlin(c)=0.0430705,q(c)=0.47405q_{\rm lin}(c)=0.0430705,\quad q(c)=0.47405e=qlin(c)−q(c)=−0.430979e=q_{\rm lin}(c)-q(c)=-0.430979

Solution. The endpoint responses are q(A) = 0.0430705 and q(B) = 0.0430705. Their secant slope is -3.61401e-18. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0430705. Direct evaluation gives 0.47405; subtracting it from the prediction gives signed error -0.430979.

Scalar diffraction model: Intermediate-condition prediction and exact check. Horizontal axis: Diffraction coordinate u = πa sinθ / λ (dimensionless). Vertical axis: Intensity I / I₀ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −4 −2 0 2 4 Diffraction coordinate u = πa sinθ / λ (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Intensity I / I₀ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.04307 Worked point: (-1.44, 0.474)
At x = -1.44, the orange circle is the analytical value 0.47405; the purple diamond is the interpolated value 0.0430705. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0430705; analytical reference = 0.47405; absolute interpolation error = 0.430979. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Gaussian beam model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. At a fixed axial plane, take a fundamental paraxial Gaussian beam with 1/e² intensity radius w. For this calculation, x denotes the plotted horizontal coordinate (Transverse position / beam radius w (dimensionless)), and q(x) denotes the plotted response (Relative intensity (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

I(r)/I(0)=e−2(r/w)2I(r)/I(0)=e^{-2(r/w)^2}a=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.02384×10−17\frac{q(b)-q(a)}{b-a}=-2.02384\times10^{-17}qlin(c)=0.0561348,q(c)=0.771669q_{\rm lin}(c)=0.0561348,\quad q(c)=0.771669e=qlin(c)−q(c)=−0.715534e=q_{\rm lin}(c)-q(c)=-0.715534

Solution. The endpoint responses are q(A) = 0.0561348 and q(B) = 0.0561348. Their secant slope is -2.02384e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0561348. Direct evaluation gives 0.771669; subtracting it from the prediction gives signed error -0.715534.

Gaussian beam model: Intermediate-condition prediction and exact check. Horizontal axis: Transverse position / beam radius w (dimensionless). Vertical axis: Relative intensity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Transverse position / beam radius w (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Relative intensity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.05613 Worked point: (-0.36, 0.7717)
At x = -0.36, the orange circle is the analytical value 0.771669; the purple diamond is the interpolated value 0.0561348. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0561348; analytical reference = 0.771669; absolute interpolation error = 0.715534. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One transverse cut at a fixed plane; w changes with axial distance. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Drude–Lorentz optical model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take the free-electron Drude limit with zero collision rate, no Lorentz resonances, and background permittivity one. For this calculation, x denotes the plotted horizontal coordinate (Frequency ω / plasma frequency ωp (dimensionless)), and q(x) denotes the plotted response (Relative permittivity (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 2.5 are tabulated. Use linear interpolation to predict the response at x = 1.525, then check it against the original equation.

ϵr=1−(ωp/ω)2\epsilon_r=1-(\omega_p/\omega)^2a=1,b=2.5,c=1.525a=1,\quad b=2.5,\quad c=1.525qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.56\frac{q(b)-q(a)}{b-a}=0.56qlin(c)=0.294,q(c)=0.570008q_{\rm lin}(c)=0.294,\quad q(c)=0.570008e=qlin(c)−q(c)=−0.276008e=q_{\rm lin}(c)-q(c)=-0.276008

Solution. The endpoint responses are q(A) = 0 and q(B) = 0.84. Their secant slope is 0.56. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.294. Direct evaluation gives 0.570008; subtracting it from the prediction gives signed error -0.276008.

Drude–Lorentz optical model: Intermediate-condition prediction and exact check. Horizontal axis: Frequency ω / plasma frequency ωp (dimensionless). Vertical axis: Relative permittivity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 Frequency ω / plasma frequency ωp (dimensionless) 0.0 0.2 0.4 0.6 0.8 Relative permittivity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.294 Worked point: (1.525, 0.57)
At x = 1.525, the orange circle is the analytical value 0.570008; the purple diamond is the interpolated value 0.294. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.294; analytical reference = 0.570008; absolute interpolation error = 0.276008. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Lossless frequency-domain special case; the zero-frequency singular point is excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lumped RLC circuit model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Set inductance to zero and apply a voltage step Vs to a series resistor and initially uncharged capacitor. For this calculation, x denotes the plotted horizontal coordinate (Time t / RC (dimensionless)), and q(x) denotes the plotted response (Capacitor voltage / supply (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

VC/Vs=1−e−t/(RC)V_C/V_s=1-e^{-t/(RC)}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Lumped RLC circuit model: Intermediate-condition prediction and exact check. Horizontal axis: Time t / RC (dimensionless). Vertical axis: Capacitor voltage / supply (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time t / RC (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Capacitor voltage / supply (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. RC limiting circuit, not a general second-order RLC transient. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Shockley diode model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate the Shockley diode law without series resistance or reverse breakdown. For this calculation, x denotes the plotted horizontal coordinate (Voltage V / nVT (dimensionless)), and q(x) denotes the plotted response (Current I / saturation current (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

I/Is=eV/(nVT)−1I/I_s=e^{V/(nV_T)}-1a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.63454\frac{q(b)-q(a)}{b-a}=1.63454qlin(c)=1.22482,q(c)=−0.417252q_{\rm lin}(c)=1.22482,\quad q(c)=-0.417252e=qlin(c)−q(c)=1.64207e=q_{\rm lin}(c)-q(c)=1.64207

Solution. The endpoint responses are q(A) = -0.834701 and q(B) = 5.04965. Their secant slope is 1.63454. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.22482. Direct evaluation gives -0.417252; subtracting it from the prediction gives signed error 1.64207.

Shockley diode model: Intermediate-condition prediction and exact check. Horizontal axis: Voltage V / nVT (dimensionless). Vertical axis: Current I / saturation current (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Voltage V / nVT (dimensionless) −1 0 1 2 3 4 5 Current I / saturation current (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.225 Worked point: (-0.54, -0.4173)
At x = -0.54, the orange circle is the analytical value -0.417252; the purple diamond is the interpolated value 1.22482. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.22482; analytical reference = -0.417252; absolute interpolation error = 1.64207. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Ebers–Moll transistor model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the forward-active Ebers–Moll branch and neglect the reverse junction contribution. For this calculation, x denotes the plotted horizontal coordinate (Base–emitter voltage / thermal voltage (dimensionless)), and q(x) denotes the plotted response (Scaled collector current (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

IC/(αFIES)=eVBE/VT−1I_C/(\alpha_F I_{ES})=e^{V_{BE}/V_T}-1a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=9.29458\frac{q(b)-q(a)}{b-a}=9.29458qlin(c)=9.03299,q(c)=4.15517q_{\rm lin}(c)=9.03299,\quad q(c)=4.15517e=qlin(c)−q(c)=4.87782e=q_{\rm lin}(c)-q(c)=4.87782

Solution. The endpoint responses are q(A) = 1.22554 and q(B) = 23.5325. Their secant slope is 9.29458. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 9.03299. Direct evaluation gives 4.15517; subtracting it from the prediction gives signed error 4.87782.

Ebers–Moll transistor model: Intermediate-condition prediction and exact check. Horizontal axis: Base–emitter voltage / thermal voltage (dimensionless). Vertical axis: Scaled collector current (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Base–emitter voltage / thermal voltage (dimensionless) 0 5 10 15 20 25 Scaled collector current (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 9.033 Worked point: (1.64, 4.155)
At x = 1.64, the orange circle is the analytical value 4.15517; the purple diamond is the interpolated value 9.03299. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 9.03299; analytical reference = 4.15517; absolute interpolation error = 4.87782. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Forward-active approximation; no saturation, Early effect, or breakdown. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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MOSFET square-law model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a long-channel MOSFET in strong-inversion saturation with constant mobility and no channel-length modulation. For this calculation, x denotes the plotted horizontal coordinate (Gate overdrive / reference voltage (dimensionless)), and q(x) denotes the plotted response (Scaled drain current (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

ID/(βV∗2/2)=[(VGS−Vth)/V∗]2I_D/(\beta V_*^2/2)=[(V_{GS}-V_{th})/V_*]^2a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=3\frac{q(b)-q(a)}{b-a}=3qlin(c)=2.25,q(c)=1.5129q_{\rm lin}(c)=2.25,\quad q(c)=1.5129e=qlin(c)−q(c)=0.7371e=q_{\rm lin}(c)-q(c)=0.7371

Solution. The endpoint responses are q(A) = 0.36 and q(B) = 5.76. Their secant slope is 3. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.25. Direct evaluation gives 1.5129; subtracting it from the prediction gives signed error 0.7371.

MOSFET square-law model: Intermediate-condition prediction and exact check. Horizontal axis: Gate overdrive / reference voltage (dimensionless). Vertical axis: Scaled drain current (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Gate overdrive / reference voltage (dimensionless) 0 1 2 3 4 5 6 Scaled drain current (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.25 Worked point: (1.23, 1.513)
At x = 1.23, the orange circle is the analytical value 1.5129; the purple diamond is the interpolated value 2.25. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.25; analytical reference = 1.5129; absolute interpolation error = 0.7371. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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BSIM compact-model family · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use an ideal weak-inversion exponential trend at fixed drain bias as a compact-model check. For this calculation, x denotes the plotted horizontal coordinate (Scaled gate bias (VGS−V*) / nVT (dimensionless)), and q(x) denotes the plotted response (Drain current / reference current (dimensionless)). Suppose only the analytical endpoint responses at x = -3 and x = 0 are tabulated. Use linear interpolation to predict the response at x = -1.95, then check it against the original equation.

ID/I∗=e(VGS−V∗)/(nVT)I_D/I_*=e^{(V_{GS}-V_*)/(nV_T)}a=−3,b=0,c=−1.95a=-3,\quad b=0,\quad c=-1.95qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.316738\frac{q(b)-q(a)}{b-a}=0.316738qlin(c)=0.382362,q(c)=0.142274q_{\rm lin}(c)=0.382362,\quad q(c)=0.142274e=qlin(c)−q(c)=0.240088e=q_{\rm lin}(c)-q(c)=0.240088

Solution. The endpoint responses are q(A) = 0.0497871 and q(B) = 1. Their secant slope is 0.316738. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.382362. Direct evaluation gives 0.142274; subtracting it from the prediction gives signed error 0.240088.

BSIM compact-model family: Intermediate-condition prediction and exact check. Horizontal axis: Scaled gate bias (VGS−V*) / nVT (dimensionless). Vertical axis: Drain current / reference current (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.0 Scaled gate bias (VGS−V*) / nVT (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Drain current / reference current (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3824 Worked point: (-1.95, 0.1423)
At x = -1.95, the orange circle is the analytical value 0.142274; the purple diamond is the interpolated value 0.382362. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.382362; analytical reference = 0.142274; absolute interpolation error = 0.240088. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Asymptotic benchmark only; not the complete BSIM equations or a result from a foundry model card. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Drift–diffusion semiconductor model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take uniform electron density n, fixed mobility μ, and a low-field steady state. The density gradient is zero. For this calculation, x denotes the plotted horizontal coordinate (Electric field E / E* (dimensionless)), and q(x) denotes the plotted response (Current density / qnμE* (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

J/(qnμE∗)=E/E∗J/(qn\mu E_*)=E/E_*a=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=−0.36,q(c)=−0.36q_{\rm lin}(c)=-0.36,\quad q(c)=-0.36e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = -1.2 and q(B) = 1.2. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.36. Direct evaluation gives -0.36; subtracting it from the prediction gives signed error 0.

Drift–diffusion semiconductor model: Intermediate-condition prediction and exact check. Horizontal axis: Electric field E / E* (dimensionless). Vertical axis: Current density / qnμE* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Electric field E / E* (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Current density / qnμE* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.36 Worked point: (-0.36, -0.36)
At x = -0.36, the orange circle is the analytical value -0.36; the purple diamond is the interpolated value -0.36. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.36; analytical reference = -0.36; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Low-field isothermal drift limit; carrier heating and higher hydrodynamic moments are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hydrodynamic carrier model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take uniform electron density n, fixed mobility μ, and a low-field steady state. The density gradient is zero. For this calculation, x denotes the plotted horizontal coordinate (Electric field E / E* (dimensionless)), and q(x) denotes the plotted response (Current density / qnμE* (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

J/(qnμE∗)=E/E∗J/(qn\mu E_*)=E/E_*a=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=−0.36,q(c)=−0.36q_{\rm lin}(c)=-0.36,\quad q(c)=-0.36e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = -1.2 and q(B) = 1.2. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.36. Direct evaluation gives -0.36; subtracting it from the prediction gives signed error 0.

Hydrodynamic carrier model: Intermediate-condition prediction and exact check. Horizontal axis: Electric field E / E* (dimensionless). Vertical axis: Current density / qnμE* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Electric field E / E* (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Current density / qnμE* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.36 Worked point: (-0.36, -0.36)
At x = -0.36, the orange circle is the analytical value -0.36; the purple diamond is the interpolated value -0.36. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.36; analytical reference = -0.36; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Low-field isothermal drift limit; carrier heating and higher hydrodynamic moments are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Nernst equilibrium potential · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For Ox+ne− ⇌ Red use ideal specified activities and fixed temperature. For this calculation, x denotes the plotted horizontal coordinate (Oxidized / reduced activity ratio (dimensionless)), and q(x) denotes the plotted response (Scaled equilibrium potential (dimensionless)). Suppose only the analytical endpoint responses at x = 2.08 and x = 8.02 are tabulated. Use linear interpolation to predict the response at x = 4.159, then check it against the original equation.

nF(E−E∘)/(RT)=ln⁡(aox/ared)nF(E-E^\circ)/(RT)=\ln(a_{ox}/a_{red})a=2.08,b=8.02,c=4.159a=2.08,\quad b=8.02,\quad c=4.159qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.2272\frac{q(b)-q(a)}{b-a}=0.2272qlin(c)=1.20472,q(c)=1.42527q_{\rm lin}(c)=1.20472,\quad q(c)=1.42527e=qlin(c)−q(c)=−0.220557e=q_{\rm lin}(c)-q(c)=-0.220557

Solution. The endpoint responses are q(A) = 0.732368 and q(B) = 2.08194. Their secant slope is 0.2272. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.20472. Direct evaluation gives 1.42527; subtracting it from the prediction gives signed error -0.220557.

Nernst equilibrium potential: Intermediate-condition prediction and exact check. Horizontal axis: Oxidized / reduced activity ratio (dimensionless). Vertical axis: Scaled equilibrium potential (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2 3 4 5 6 7 8 Oxidized / reduced activity ratio (dimensionless) 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 Scaled equilibrium potential (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.205 Worked point: (4.159, 1.425)
At x = 4.159, the orange circle is the analytical value 1.42527; the purple diamond is the interpolated value 1.20472. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.20472; analytical reference = 1.42527; absolute interpolation error = 0.220557. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Butler–Volmer kinetics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Set anodic and cathodic transfer coefficients to one half, with one-electron charge convention. For this calculation, x denotes the plotted horizontal coordinate (Overpotential Fη / RT (dimensionless)), and q(x) denotes the plotted response (Current density / exchange current (dimensionless)). Suppose only the analytical endpoint responses at x = -2.4 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = -0.72, then check it against the original equation.

j/j0=2sinh⁡(η∗/2),η∗=Fη/(RT)j/j_0=2\sinh(\eta_*/2),\quad\eta_*=F\eta/(RT)a=−2.4,b=2.4,c=−0.72a=-2.4,\quad b=2.4,\quad c=-0.72qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.25788\frac{q(b)-q(a)}{b-a}=1.25788qlin(c)=−0.905677,q(c)=−0.735653q_{\rm lin}(c)=-0.905677,\quad q(c)=-0.735653e=qlin(c)−q(c)=−0.170024e=q_{\rm lin}(c)-q(c)=-0.170024

Solution. The endpoint responses are q(A) = -3.01892 and q(B) = 3.01892. Their secant slope is 1.25788. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.905677. Direct evaluation gives -0.735653; subtracting it from the prediction gives signed error -0.170024.

Butler–Volmer kinetics: Intermediate-condition prediction and exact check. Horizontal axis: Overpotential Fη / RT (dimensionless). Vertical axis: Current density / exchange current (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −2 −1 0 1 2 Overpotential Fη / RT (dimensionless) −3 −2 −1 0 1 2 3 Current density / exchange current (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.9057 Worked point: (-0.72, -0.7357)
At x = -0.72, the orange circle is the analytical value -0.735653; the purple diamond is the interpolated value -0.905677. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.905677; analytical reference = -0.735653; absolute interpolation error = 0.170024. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Tafel approximation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the anodic high-overpotential regime where the cathodic exponential is negligible. For this calculation, x denotes the plotted horizontal coordinate (Anodic current / exchange current (dimensionless)), and q(x) denotes the plotted response (Scaled overpotential αFη / RT (dimensionless)). Suppose only the analytical endpoint responses at x = 28 and x = 82 are tabulated. Use linear interpolation to predict the response at x = 46.9, then check it against the original equation.

αFη/(RT)=ln⁡(j/j0)\alpha F\eta/(RT)=\ln(j/j_0)a=28,b=82,c=46.9a=28,\quad b=82,\quad c=46.9qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.0198984\frac{q(b)-q(a)}{b-a}=0.0198984qlin(c)=3.70828,q(c)=3.84802q_{\rm lin}(c)=3.70828,\quad q(c)=3.84802e=qlin(c)−q(c)=−0.139733e=q_{\rm lin}(c)-q(c)=-0.139733

Solution. The endpoint responses are q(A) = 3.3322 and q(B) = 4.40672. Their secant slope is 0.0198984. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 3.70828. Direct evaluation gives 3.84802; subtracting it from the prediction gives signed error -0.139733.

Tafel approximation: Intermediate-condition prediction and exact check. Horizontal axis: Anodic current / exchange current (dimensionless). Vertical axis: Scaled overpotential αFη / RT (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 30 40 50 60 70 80 Anodic current / exchange current (dimensionless) 3.4 3.6 3.8 4.0 4.2 4.4 Scaled overpotential αFη / RT (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 3.708 Worked point: (46.9, 3.848)
At x = 46.9, the orange circle is the analytical value 3.84802; the purple diamond is the interpolated value 3.70828. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 3.70828; analytical reference = 3.84802; absolute interpolation error = 0.139733. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Asymptotic approximation, plotted well above j/j0=1; not valid near equilibrium. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Poisson–Nernst–Planck model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. At zero ionic flux, linearize a symmetric dilute electrolyte near equilibrium next to a planar wall. For this calculation, x denotes the plotted horizontal coordinate (Distance / Debye length (dimensionless)), and q(x) denotes the plotted response (Potential / wall potential (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

ϕ/ϕ0=e−x/λD\phi/\phi_0=e^{-x/\lambda_D}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Poisson–Nernst–Planck model: Intermediate-condition prediction and exact check. Horizontal axis: Distance / Debye length (dimensionless). Vertical axis: Potential / wall potential (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Distance / Debye length (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Potential / wall potential (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Debye–Hückel equilibrium limit of PNP, requiring |zFφ|≪RT; no driven ionic transport. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Doyle–Fuller–Newman (DFN/P2D) model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Start with a spherical active particle of radius R and mean concentration c*. Impose constant outward molar flux jout. For this calculation, x denotes the plotted horizontal coordinate (Extraction coordinate jout t / Rc* (dimensionless)), and q(x) denotes the plotted response (Particle-average concentration / c* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.05 and x = 0.2 are tabulated. Use linear interpolation to predict the response at x = 0.1025, then check it against the original equation.

cˉ/c∗=1−3τ,τ=joutt/(Rc∗)\bar c/c_* =1-3\tau,\quad\tau=j_{out}t/(Rc_*)a=0.05,b=0.2,c=0.1025a=0.05,\quad b=0.2,\quad c=0.1025qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3\frac{q(b)-q(a)}{b-a}=-3qlin(c)=0.6925,q(c)=0.6925q_{\rm lin}(c)=0.6925,\quad q(c)=0.6925e=qlin(c)−q(c)=−1.11022×10−16e=q_{\rm lin}(c)-q(c)=-1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.85 and q(B) = 0.4. Their secant slope is -3. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.6925. Direct evaluation gives 0.6925; subtracting it from the prediction gives signed error -1.11022e-16.

Doyle–Fuller–Newman (DFN/P2D) model: Intermediate-condition prediction and exact check. Horizontal axis: Extraction coordinate jout t / Rc* (dimensionless). Vertical axis: Particle-average concentration / c* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 Extraction coordinate jout t / Rc* (dimensionless) 0.4 0.5 0.6 0.7 0.8 0.9 Particle-average concentration / c* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.6925 Worked point: (0.1025, 0.6925)
At x = 0.1025, the orange circle is the analytical value 0.6925; the purple diamond is the interpolated value 0.6925. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.6925; analytical reference = 0.6925; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact particle mass balance shared by DFN, SPM, and SPMe. It does not give the radial profile, terminal voltage, electrolyte dynamics, or a usable-capacity prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Single-particle battery model (SPM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Start with a spherical active particle of radius R and mean concentration c*. Impose constant outward molar flux jout. For this calculation, x denotes the plotted horizontal coordinate (Extraction coordinate jout t / Rc* (dimensionless)), and q(x) denotes the plotted response (Particle-average concentration / c* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.05 and x = 0.2 are tabulated. Use linear interpolation to predict the response at x = 0.1025, then check it against the original equation.

cˉ/c∗=1−3τ,τ=joutt/(Rc∗)\bar c/c_* =1-3\tau,\quad\tau=j_{out}t/(Rc_*)a=0.05,b=0.2,c=0.1025a=0.05,\quad b=0.2,\quad c=0.1025qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3\frac{q(b)-q(a)}{b-a}=-3qlin(c)=0.6925,q(c)=0.6925q_{\rm lin}(c)=0.6925,\quad q(c)=0.6925e=qlin(c)−q(c)=−1.11022×10−16e=q_{\rm lin}(c)-q(c)=-1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.85 and q(B) = 0.4. Their secant slope is -3. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.6925. Direct evaluation gives 0.6925; subtracting it from the prediction gives signed error -1.11022e-16.

Single-particle battery model (SPM): Intermediate-condition prediction and exact check. Horizontal axis: Extraction coordinate jout t / Rc* (dimensionless). Vertical axis: Particle-average concentration / c* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 Extraction coordinate jout t / Rc* (dimensionless) 0.4 0.5 0.6 0.7 0.8 0.9 Particle-average concentration / c* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.6925 Worked point: (0.1025, 0.6925)
At x = 0.1025, the orange circle is the analytical value 0.6925; the purple diamond is the interpolated value 0.6925. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.6925; analytical reference = 0.6925; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact particle mass balance shared by DFN, SPM, and SPMe. It does not give the radial profile, terminal voltage, electrolyte dynamics, or a usable-capacity prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Single-particle model with electrolyte (SPMe) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Start with a spherical active particle of radius R and mean concentration c*. Impose constant outward molar flux jout. For this calculation, x denotes the plotted horizontal coordinate (Extraction coordinate jout t / Rc* (dimensionless)), and q(x) denotes the plotted response (Particle-average concentration / c* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.05 and x = 0.2 are tabulated. Use linear interpolation to predict the response at x = 0.1025, then check it against the original equation.

cˉ/c∗=1−3τ,τ=joutt/(Rc∗)\bar c/c_* =1-3\tau,\quad\tau=j_{out}t/(Rc_*)a=0.05,b=0.2,c=0.1025a=0.05,\quad b=0.2,\quad c=0.1025qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3\frac{q(b)-q(a)}{b-a}=-3qlin(c)=0.6925,q(c)=0.6925q_{\rm lin}(c)=0.6925,\quad q(c)=0.6925e=qlin(c)−q(c)=−1.11022×10−16e=q_{\rm lin}(c)-q(c)=-1.11022\times10^{-16}

Solution. The endpoint responses are q(A) = 0.85 and q(B) = 0.4. Their secant slope is -3. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.6925. Direct evaluation gives 0.6925; subtracting it from the prediction gives signed error -1.11022e-16.

Single-particle model with electrolyte (SPMe): Intermediate-condition prediction and exact check. Horizontal axis: Extraction coordinate jout t / Rc* (dimensionless). Vertical axis: Particle-average concentration / c* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 Extraction coordinate jout t / Rc* (dimensionless) 0.4 0.5 0.6 0.7 0.8 0.9 Particle-average concentration / c* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.6925 Worked point: (0.1025, 0.6925)
At x = 0.1025, the orange circle is the analytical value 0.6925; the purple diamond is the interpolated value 0.6925. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.6925; analytical reference = 0.6925; absolute interpolation error = 1.11022e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact particle mass balance shared by DFN, SPM, and SPMe. It does not give the radial profile, terminal voltage, electrolyte dynamics, or a usable-capacity prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Equivalent-circuit battery model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply a constant current I to an initially relaxed single-RC battery polarization branch. For this calculation, x denotes the plotted horizontal coordinate (Time / polarization RC constant (dimensionless)), and q(x) denotes the plotted response (Polarization voltage / IRp (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

Vp/(IRp)=1−e−t/(RpCp)V_p/(IR_p)=1-e^{-t/(R_pC_p)}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Equivalent-circuit battery model: Intermediate-condition prediction and exact check. Horizontal axis: Time / polarization RC constant (dimensionless). Vertical axis: Polarization voltage / IRp (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / polarization RC constant (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Polarization voltage / IRp (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One branch with fixed parameters; state of charge and open-circuit voltage are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Darcy porous-flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Let G=−dp/dx be positive, and hold permeability k and viscosity μ constant. For this calculation, x denotes the plotted horizontal coordinate (Driving pressure gradient G / G* (dimensionless)), and q(x) denotes the plotted response (Scaled Darcy velocity (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

u/(kG∗/μ)=G/G∗u/(kG_*/\mu)=G/G_*a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=1.23,q(c)=1.23q_{\rm lin}(c)=1.23,\quad q(c)=1.23e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.6 and q(B) = 2.4. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.23. Direct evaluation gives 1.23; subtracting it from the prediction gives signed error 0.

Darcy porous-flow model: Intermediate-condition prediction and exact check. Horizontal axis: Driving pressure gradient G / G* (dimensionless). Vertical axis: Scaled Darcy velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Driving pressure gradient G / G* (dimensionless) 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 Scaled Darcy velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.23 Worked point: (1.23, 1.23)
At x = 1.23, the orange circle is the analytical value 1.23; the purple diamond is the interpolated value 1.23. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.23; analytical reference = 1.23; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single-phase creeping flow in a homogeneous porous medium. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Brinkman porous-flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve μe u″−μu/k+G=0 between no-slip walls ±H. Choose screening length ℓ=√(μe k/μ) and H/ℓ=2. For this calculation, x denotes the plotted horizontal coordinate (Transverse position x / H (dimensionless)), and q(x) denotes the plotted response (Velocity / Darcy bulk velocity (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

u/(kG/μ)=1−cosh⁡(x/ℓ)/cosh⁡(H/ℓ),H/ℓ=2u/(kG/\mu)=1-\cosh(x/\ell)/\cosh(H/\ell),\quad H/\ell=2a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.85037×10−16\frac{q(b)-q(a)}{b-a}=-1.85037\times10^{-16}qlin(c)=0.518724,q(c)=0.716787q_{\rm lin}(c)=0.518724,\quad q(c)=0.716787e=qlin(c)−q(c)=−0.198063e=q_{\rm lin}(c)-q(c)=-0.198063

Solution. The endpoint responses are q(A) = 0.518724 and q(B) = 0.518724. Their secant slope is -1.85037e-16. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.518724. Direct evaluation gives 0.716787; subtracting it from the prediction gives signed error -0.198063.

Brinkman porous-flow model: Intermediate-condition prediction and exact check. Horizontal axis: Transverse position x / H (dimensionless). Vertical axis: Velocity / Darcy bulk velocity (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Transverse position x / H (dimensionless) 0.50 0.55 0.60 0.65 0.70 0.75 Velocity / Darcy bulk velocity (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5187 Worked point: (-0.18, 0.7168)
At x = -0.18, the orange circle is the analytical value 0.716787; the purple diamond is the interpolated value 0.518724. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.518724; analytical reference = 0.716787; absolute interpolation error = 0.198063. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Forchheimer model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose velocity and gradient scales so that the linear and quadratic drag coefficients are both one. For this calculation, x denotes the plotted horizontal coordinate (Scaled positive velocity v (dimensionless)), and q(x) denotes the plotted response (Scaled pressure gradient (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

G/G∗=v+v2G/G_*=v+v^2a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4\frac{q(b)-q(a)}{b-a}=4qlin(c)=3.48,q(c)=2.7429q_{\rm lin}(c)=3.48,\quad q(c)=2.7429e=qlin(c)−q(c)=0.7371e=q_{\rm lin}(c)-q(c)=0.7371

Solution. The endpoint responses are q(A) = 0.96 and q(B) = 8.16. Their secant slope is 4. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 3.48. Direct evaluation gives 2.7429; subtracting it from the prediction gives signed error 0.7371.

Forchheimer model: Intermediate-condition prediction and exact check. Horizontal axis: Scaled positive velocity v (dimensionless). Vertical axis: Scaled pressure gradient (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Scaled positive velocity v (dimensionless) 1 2 3 4 5 6 7 8 9 Scaled pressure gradient (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 3.48 Worked point: (1.23, 2.743)
At x = 1.23, the orange circle is the analytical value 2.7429; the purple diamond is the interpolated value 3.48. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 3.48; analytical reference = 2.7429; absolute interpolation error = 0.7371. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Richards equation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Linearize moisture capacity and hydraulic conductivity about a uniform reference state, neglect gravity, and solve the resulting diffusion equation on a slab. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Pressure-head perturbation / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Richards equation: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Pressure-head perturbation / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Pressure-head perturbation / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Constant-coefficient linearization of Richards’ equation. The nonlinear retention and conductivity changes are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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van Genuchten retention model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose n=2 and m=1−1/n=1/2 for a drying retention curve. For this calculation, x denotes the plotted horizontal coordinate (Scaled suction α|h| (dimensionless)), and q(x) denotes the plotted response (Effective saturation Se (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

Se=[1+(α∣h∣)2]−1/2S_e=[1+(\alpha|h|)^2]^{-1/2}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.154857\frac{q(b)-q(a)}{b-a}=-0.154857qlin(c)=0.544507,q(c)=0.438424q_{\rm lin}(c)=0.544507,\quad q(c)=0.438424e=qlin(c)−q(c)=0.106083e=q_{\rm lin}(c)-q(c)=0.106083

Solution. The endpoint responses are q(A) = 0.707107 and q(B) = 0.242536. Their secant slope is -0.154857. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.544507. Direct evaluation gives 0.438424; subtracting it from the prediction gives signed error 0.106083.

van Genuchten retention model: Intermediate-condition prediction and exact check. Horizontal axis: Scaled suction α|h| (dimensionless). Vertical axis: Effective saturation Se (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Scaled suction α|h| (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 Effective saturation Se (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5445 Worked point: (2.05, 0.4384)
At x = 2.05, the orange circle is the analytical value 0.438424; the purple diamond is the interpolated value 0.544507. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.544507; analytical reference = 0.438424; absolute interpolation error = 0.106083. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Retention relation only; hysteresis and conductivity are not evaluated. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Biot poroelasticity · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use one-dimensional linear consolidation with drained ends and an initial excess pore-pressure mode sin(πx/L). Plot cvt/L²=0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Excess pore pressure / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Biot poroelasticity: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Excess pore pressure / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Excess pore pressure / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact single-mode Terzaghi solution and a compatible one-dimensional poroelastic reduction; not an arbitrary initial loading history. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Terzaghi consolidation model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use one-dimensional linear consolidation with drained ends and an initial excess pore-pressure mode sin(πx/L). Plot cvt/L²=0.1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Excess pore pressure / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Terzaghi consolidation model: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Excess pore pressure / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Excess pore pressure / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact single-mode Terzaghi solution and a compatible one-dimensional poroelastic reduction; not an arbitrary initial loading history. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Modified Cam-Clay model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold preconsolidation pressure pc and critical-state slope M fixed. Plot the compression-positive yield locus. For this calculation, x denotes the plotted horizontal coordinate (Mean effective pressure p / pc (dimensionless)), and q(x) denotes the plotted response (Deviatoric stress q / Mpc (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

q/(Mpc)=(p/pc)(1−p/pc)q/(Mp_c)=\sqrt{(p/p_c)(1-p/p_c)}a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−9.25186×10−17\frac{q(b)-q(a)}{b-a}=-9.25186\times10^{-17}qlin(c)=0.4,q(c)=0.491833q_{\rm lin}(c)=0.4,\quad q(c)=0.491833e=qlin(c)−q(c)=−0.0918333e=q_{\rm lin}(c)-q(c)=-0.0918333

Solution. The endpoint responses are q(A) = 0.4 and q(B) = 0.4. Their secant slope is -9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.4. Direct evaluation gives 0.491833; subtracting it from the prediction gives signed error -0.0918333.

Modified Cam-Clay model: Intermediate-condition prediction and exact check. Horizontal axis: Mean effective pressure p / pc (dimensionless). Vertical axis: Deviatoric stress q / Mpc (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Mean effective pressure p / pc (dimensionless) 0.40 0.42 0.44 0.46 0.48 0.50 Deviatoric stress q / Mpc (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.4 Worked point: (0.41, 0.4918)
At x = 0.41, the orange circle is the analytical value 0.491833; the purple diamond is the interpolated value 0.4. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.4; analytical reference = 0.491833; absolute interpolation error = 0.0918333. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Yield-surface geometry only; hardening and stress-path evolution are not solved. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Saint-Venant shallow-water model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Linearize shallow-water dynamics about rest at constant depth H; the wave speed is √(gH). For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Surface elevation / wave amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Saint-Venant shallow-water model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Surface elevation / wave amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Surface elevation / wave amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Small free-surface perturbation in a constant-depth channel, without friction or dispersion. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Kinematic-wave routing · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the linear routing equation ht+hx=0 with initial Gaussian pulse exp(−x²). For this calculation, x denotes the plotted horizontal coordinate (Channel position x (dimensionless)), and q(x) denotes the plotted response (Flow-depth perturbation h (dimensionless)). Suppose only the analytical endpoint responses at x = -1.4 and x = 3.4 are tabulated. Use linear interpolation to predict the response at x = 0.28, then check it against the original equation.

h(x,1)=e−(x−1)2h(x,1)=e^{-(x-1)^2}a=−1.4,b=3.4,c=0.28a=-1.4,\quad b=3.4,\quad c=0.28qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.17455×10−18\frac{q(b)-q(a)}{b-a}=-1.17455\times10^{-18}qlin(c)=0.00315111,q(c)=0.595473q_{\rm lin}(c)=0.00315111,\quad q(c)=0.595473e=qlin(c)−q(c)=−0.592321e=q_{\rm lin}(c)-q(c)=-0.592321

Solution. The endpoint responses are q(A) = 0.00315111 and q(B) = 0.00315111. Their secant slope is -1.17455e-18. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.00315111. Direct evaluation gives 0.595473; subtracting it from the prediction gives signed error -0.592321.

Kinematic-wave routing: Intermediate-condition prediction and exact check. Horizontal axis: Channel position x (dimensionless). Vertical axis: Flow-depth perturbation h (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1 0 1 2 3 Channel position x (dimensionless) 0.0 0.2 0.4 0.6 0.8 1.0 Flow-depth perturbation h (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.003151 Worked point: (0.28, 0.5955)
At x = 0.28, the orange circle is the analytical value 0.595473; the purple diamond is the interpolated value 0.00315111. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.00315111; analytical reference = 0.595473; absolute interpolation error = 0.592321. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Constant-celerity reduction; nonlinear depth-dependent routing can distort or steepen the pulse. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Rainfall–runoff model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. After rainfall stops, let storage S obey S′=−S/K and outflow Q=S/K. Normalize either by its initial value. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Reservoir outflow / initial outflow (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Rainfall–runoff model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Reservoir outflow / initial outflow (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Reservoir outflow / initial outflow (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-reservoir rainfall–runoff component; no new rain, infiltration, or additional routing stores. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Advection–dispersion groundwater model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. On an infinite line take velocity one, dispersion coefficient 0.1, and initial concentration exp(−x²). For this calculation, x denotes the plotted horizontal coordinate (Distance x (dimensionless)), and q(x) denotes the plotted response (Tracer concentration (dimensionless)). Suppose only the analytical endpoint responses at x = -1.4 and x = 3.4 are tabulated. Use linear interpolation to predict the response at x = 0.28, then check it against the original equation.

c(x,1)=11.4e−(x−1)2/1.4c(x,1)=\frac1{\sqrt{1.4}}e^{-(x-1)^2/1.4}a=−1.4,b=3.4,c=0.28a=-1.4,\quad b=3.4,\quad c=0.28qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.89121×10−18\frac{q(b)-q(a)}{b-a}=-2.89121\times10^{-18}qlin(c)=0.0138078,q(c)=0.58361q_{\rm lin}(c)=0.0138078,\quad q(c)=0.58361e=qlin(c)−q(c)=−0.569802e=q_{\rm lin}(c)-q(c)=-0.569802

Solution. The endpoint responses are q(A) = 0.0138078 and q(B) = 0.0138078. Their secant slope is -2.89121e-18. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0138078. Direct evaluation gives 0.58361; subtracting it from the prediction gives signed error -0.569802.

Advection–dispersion groundwater model: Intermediate-condition prediction and exact check. Horizontal axis: Distance x (dimensionless). Vertical axis: Tracer concentration (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1 0 1 2 3 Distance x (dimensionless) 0.0 0.2 0.4 0.6 0.8 Tracer concentration (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.01381 Worked point: (0.28, 0.5836)
At x = 0.28, the orange circle is the analytical value 0.58361; the purple diamond is the interpolated value 0.0138078. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0138078; analytical reference = 0.58361; absolute interpolation error = 0.569802. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Homogeneous advection–dispersion with no reactions or sorption. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Numerical weather prediction · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use an ideal gas at constant temperature and constant gravity, with density ρ0 at height zero. For this calculation, x denotes the plotted horizontal coordinate (Height / pressure scale height H (dimensionless)), and q(x) denotes the plotted response (Density / base density (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

ρ(z)/ρ0=e−z/H\rho(z)/\rho_0=e^{-z/H}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Numerical weather prediction: Intermediate-condition prediction and exact check. Horizontal axis: Height / pressure scale height H (dimensionless). Vertical axis: Density / base density (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Height / pressure scale height H (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Density / base density (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Hydrostatic column benchmark only. For stellar structure this approximates a thin isothermal layer, not an entire star; radiation, convection, and dynamics are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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General circulation model (GCM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use an ideal gas at constant temperature and constant gravity, with density ρ0 at height zero. For this calculation, x denotes the plotted horizontal coordinate (Height / pressure scale height H (dimensionless)), and q(x) denotes the plotted response (Density / base density (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

ρ(z)/ρ0=e−z/H\rho(z)/\rho_0=e^{-z/H}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

General circulation model (GCM): Intermediate-condition prediction and exact check. Horizontal axis: Height / pressure scale height H (dimensionless). Vertical axis: Density / base density (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Height / pressure scale height H (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Density / base density (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Hydrostatic column benchmark only. For stellar structure this approximates a thin isothermal layer, not an entire star; radiation, convection, and dynamics are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Stellar structure model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use an ideal gas at constant temperature and constant gravity, with density ρ0 at height zero. For this calculation, x denotes the plotted horizontal coordinate (Height / pressure scale height H (dimensionless)), and q(x) denotes the plotted response (Density / base density (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

ρ(z)/ρ0=e−z/H\rho(z)/\rho_0=e^{-z/H}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Stellar structure model: Intermediate-condition prediction and exact check. Horizontal axis: Height / pressure scale height H (dimensionless). Vertical axis: Density / base density (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Height / pressure scale height H (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Density / base density (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Hydrostatic column benchmark only. For stellar structure this approximates a thin isothermal layer, not an entire star; radiation, convection, and dynamics are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Earth system model (ESM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a constant radiative-forcing step F, use CΔT′=F−λΔT with positive linear feedback parameter λ and initially zero anomaly. For this calculation, x denotes the plotted horizontal coordinate (Time λt / heat capacity C (dimensionless)), and q(x) denotes the plotted response (Temperature change / equilibrium change (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

ΔT/(F/λ)=1−e−λt/C\Delta T/(F/\lambda)=1-e^{-\lambda t/C}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Earth system model (ESM): Intermediate-condition prediction and exact check. Horizontal axis: Time λt / heat capacity C (dimensionless). Vertical axis: Temperature change / equilibrium change (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time λt / heat capacity C (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Temperature change / equilibrium change (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Reduced global-mean energy balance. For ESM this is an illustrative diagnostic reduction, not a full Earth-system forecast. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Energy-balance climate model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a constant radiative-forcing step F, use CΔT′=F−λΔT with positive linear feedback parameter λ and initially zero anomaly. For this calculation, x denotes the plotted horizontal coordinate (Time λt / heat capacity C (dimensionless)), and q(x) denotes the plotted response (Temperature change / equilibrium change (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

ΔT/(F/λ)=1−e−λt/C\Delta T/(F/\lambda)=1-e^{-\lambda t/C}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Energy-balance climate model: Intermediate-condition prediction and exact check. Horizontal axis: Time λt / heat capacity C (dimensionless). Vertical axis: Temperature change / equilibrium change (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time λt / heat capacity C (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Temperature change / equilibrium change (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Reduced global-mean energy balance. For ESM this is an illustrative diagnostic reduction, not a full Earth-system forecast. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Ocean circulation model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a constant-depth, nonrotating, inviscid shallow-water reduction of ocean circulation. For this calculation, x denotes the plotted horizontal coordinate (Position / wavelength (dimensionless)), and q(x) denotes the plotted response (Surface elevation / wave amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,0)=sin⁡(2πξ)u(\xi,0)=\sin(2\pi\xi)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−3.17019\frac{q(b)-q(a)}{b-a}=-3.17019qlin(c)=0.285317,q(c)=0.535827q_{\rm lin}(c)=0.285317,\quad q(c)=0.535827e=qlin(c)−q(c)=−0.25051e=q_{\rm lin}(c)-q(c)=-0.25051

Solution. The endpoint responses are q(A) = 0.951057 and q(B) = -0.951057. Their secant slope is -3.17019. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.285317. Direct evaluation gives 0.535827; subtracting it from the prediction gives signed error -0.25051.

Ocean circulation model: Intermediate-condition prediction and exact check. Horizontal axis: Position / wavelength (dimensionless). Vertical axis: Surface elevation / wave amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position / wavelength (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Surface elevation / wave amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2853 Worked point: (0.41, 0.5358)
At x = 0.41, the orange circle is the analytical value 0.535827; the purple diamond is the interpolated value 0.285317. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.285317; analytical reference = 0.535827; absolute interpolation error = 0.25051. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single linear barotropic mode; rotation, stratification, mixing, and realistic boundaries are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Sea-ice thermodynamic-dynamic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Assume zero initial thickness, fixed surface-to-freezing temperature difference ΔT, and conductive flux kΔT/h through the ice. For this calculation, x denotes the plotted horizontal coordinate (Elapsed time / growth time t* (dimensionless)), and q(x) denotes the plotted response (Ice thickness h / ℓ (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

h/ℓ=t/t∗h/\ell=\sqrt{t/t_*}a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.372678\frac{q(b)-q(a)}{b-a}=0.372678qlin(c)=1.20748,q(c)=1.28062q_{\rm lin}(c)=1.20748,\quad q(c)=1.28062e=qlin(c)−q(c)=−0.0731481e=q_{\rm lin}(c)-q(c)=-0.0731481

Solution. The endpoint responses are q(A) = 0.894427 and q(B) = 1.78885. Their secant slope is 0.372678. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.20748. Direct evaluation gives 1.28062; subtracting it from the prediction gives signed error -0.0731481.

Sea-ice thermodynamic-dynamic model: Intermediate-condition prediction and exact check. Horizontal axis: Elapsed time / growth time t* (dimensionless). Vertical axis: Ice thickness h / ℓ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Elapsed time / growth time t* (dimensionless) 0.8 1.0 1.2 1.4 1.6 1.8 Ice thickness h / ℓ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.207 Worked point: (1.64, 1.281)
At x = 1.64, the orange circle is the analytical value 1.28062; the purple diamond is the interpolated value 1.20748. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.20748; analytical reference = 1.28062; absolute interpolation error = 0.0731481. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Stefan growth limit with no ocean heat flux, snow insulation, or ice dynamics. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lifting-line model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use lifting-line theory for an ideal elliptically loaded wing of aspect ratio eight and two-dimensional slope 2π per radian. For this calculation, x denotes the plotted horizontal coordinate (Angle of attack α (degree)), and q(x) denotes the plotted response (Lift coefficient CL (dimensionless)). Suppose only the analytical endpoint responses at x = -3.6 and x = 3.6 are tabulated. Use linear interpolation to predict the response at x = -1.08, then check it against the original equation.

CL=2πα1+2/(e AR),e=1, AR=8C_L=\frac{2\pi\alpha}{1+2/(e\,AR)},\quad e=1,\ AR=8a=−3.6,b=3.6,c=−1.08a=-3.6,\quad b=3.6,\quad c=-1.08qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.0877298\frac{q(b)-q(a)}{b-a}=0.0877298qlin(c)=−0.0947482,q(c)=−0.0947482q_{\rm lin}(c)=-0.0947482,\quad q(c)=-0.0947482e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = -0.315827 and q(B) = 0.315827. Their secant slope is 0.0877298. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.0947482. Direct evaluation gives -0.0947482; subtracting it from the prediction gives signed error 0.

Lifting-line model: Intermediate-condition prediction and exact check. Horizontal axis: Angle of attack α (degree). Vertical axis: Lift coefficient CL (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −3 −2 −1 0 1 2 3 Angle of attack α (degree) −0.3 −0.2 −0.1 0.0 0.1 0.2 0.3 Lift coefficient CL (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.09475 Worked point: (-1.08, -0.09475)
At x = -1.08, the orange circle is the analytical value -0.0947482; the purple diamond is the interpolated value -0.0947482. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.0947482; analytical reference = -0.0947482; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Small-angle attached-flow approximation; no stall prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Blade-element momentum model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the ideal nonrotating actuator-disk limit underlying axial momentum theory. For this calculation, x denotes the plotted horizontal coordinate (Axial induction factor a (dimensionless)), and q(x) denotes the plotted response (Power coefficient CP (dimensionless)). Suppose only the analytical endpoint responses at x = 0.1 and x = 0.4 are tabulated. Use linear interpolation to predict the response at x = 0.205, then check it against the original equation.

CP=4a(1−a)2C_P=4a(1-a)^2a=0.1,b=0.4,c=0.205a=0.1,\quad b=0.4,\quad c=0.205qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.84\frac{q(b)-q(a)}{b-a}=0.84qlin(c)=0.4122,q(c)=0.51826q_{\rm lin}(c)=0.4122,\quad q(c)=0.51826e=qlin(c)−q(c)=−0.10606e=q_{\rm lin}(c)-q(c)=-0.10606

Solution. The endpoint responses are q(A) = 0.324 and q(B) = 0.576. Their secant slope is 0.84. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.4122. Direct evaluation gives 0.51826; subtracting it from the prediction gives signed error -0.10606.

Blade-element momentum model: Intermediate-condition prediction and exact check. Horizontal axis: Axial induction factor a (dimensionless). Vertical axis: Power coefficient CP (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Axial induction factor a (dimensionless) 0.30 0.35 0.40 0.45 0.50 0.55 0.60 Power coefficient CP (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.4122 Worked point: (0.205, 0.5183)
At x = 0.205, the orange circle is the analytical value 0.51826; the purple diamond is the interpolated value 0.4122. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.4122; analytical reference = 0.51826; absolute interpolation error = 0.10606. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Momentum-theory benchmark for BEM; blade geometry, swirl, drag, tip losses, and high-induction corrections are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Bicycle vehicle model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Assume low-speed rolling without tire slip for a vehicle of wheelbase L. For this calculation, x denotes the plotted horizontal coordinate (Steering angle δ (degree)), and q(x) denotes the plotted response (Path curvature × wheelbase κL (dimensionless)). Suppose only the analytical endpoint responses at x = -18 and x = 18 are tabulated. Use linear interpolation to predict the response at x = -5.4, then check it against the original equation.

κL=tan⁡δ\kappa L=\tan\deltaa=−18,b=18,c=−5.4a=-18,\quad b=18,\quad c=-5.4qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.0180511\frac{q(b)-q(a)}{b-a}=0.0180511qlin(c)=−0.0974759,q(c)=−0.0945278q_{\rm lin}(c)=-0.0974759,\quad q(c)=-0.0945278e=qlin(c)−q(c)=−0.00294808e=q_{\rm lin}(c)-q(c)=-0.00294808

Solution. The endpoint responses are q(A) = -0.32492 and q(B) = 0.32492. Their secant slope is 0.0180511. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.0974759. Direct evaluation gives -0.0945278; subtracting it from the prediction gives signed error -0.00294808.

Bicycle vehicle model: Intermediate-condition prediction and exact check. Horizontal axis: Steering angle δ (degree). Vertical axis: Path curvature × wheelbase κL (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −15 −10 −5 0 5 10 15 Steering angle δ (degree) −0.4 −0.3 −0.2 −0.1 0.0 0.1 0.2 0.3 0.4 Path curvature × wheelbase κL (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.09748 Worked point: (-5.4, -0.09453)
At x = -5.4, the orange circle is the analytical value -0.0945278; the purple diamond is the interpolated value -0.0974759. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.0974759; analytical reference = -0.0945278; absolute interpolation error = 0.00294808. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Kinematic limit, not a high-speed dynamic tire-force model. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Quarter-car suspension model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Hold the unsprung mass fixed, set damping to zero, and release the sprung mass from displacement A. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Sprung displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 2.51327 and x = 10.0531 are tabulated. Use linear interpolation to predict the response at x = 5.15221, then check it against the original equation.

z/A=cos⁡(ωt)z/A=\cos(\omega t)a=2.51327,b=10.0531,c=5.15221a=2.51327,\quad b=10.0531,\quad c=5.15221qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−4.41744×10−17\frac{q(b)-q(a)}{b-a}=-4.41744\times10^{-17}qlin(c)=−0.809017,q(c)=0.425779q_{\rm lin}(c)=-0.809017,\quad q(c)=0.425779e=qlin(c)−q(c)=−1.2348e=q_{\rm lin}(c)-q(c)=-1.2348

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = -0.809017. Their secant slope is -4.41744e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.809017. Direct evaluation gives 0.425779; subtracting it from the prediction gives signed error -1.2348.

Quarter-car suspension model: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Sprung displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 10 Phase ωt (radian) −1.0 −0.5 0.0 0.5 1.0 Sprung displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.809 Worked point: (5.152, 0.4258)
At x = 5.15221, the orange circle is the analytical value 0.425779; the purple diamond is the interpolated value -0.809017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.809017; analytical reference = 0.425779; absolute interpolation error = 1.2348. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single-mode constrained reduction, not the full two-degree-of-freedom road response. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Pacejka tire model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose B=10, C=1.3, E=0, zero offsets, and fixed load in the basic Pacejka Magic Formula. For this calculation, x denotes the plotted horizontal coordinate (Slip ratio s (dimensionless)), and q(x) denotes the plotted response (Force / peak-scale D (dimensionless)). Suppose only the analytical endpoint responses at x = -0.18 and x = 0.18 are tabulated. Use linear interpolation to predict the response at x = -0.054, then check it against the original equation.

F/D=sin⁡[1.3arctan⁡(10s)]F/D=\sin[1.3\arctan(10s)]a=−0.18,b=0.18,c=−0.054a=-0.18,\quad b=0.18,\quad c=-0.054qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=5.45768\frac{q(b)-q(a)}{b-a}=5.45768qlin(c)=−0.294715,q(c)=−0.600138q_{\rm lin}(c)=-0.294715,\quad q(c)=-0.600138e=qlin(c)−q(c)=0.305423e=q_{\rm lin}(c)-q(c)=0.305423

Solution. The endpoint responses are q(A) = -0.982382 and q(B) = 0.982382. Their secant slope is 5.45768. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.294715. Direct evaluation gives -0.600138; subtracting it from the prediction gives signed error 0.305423.

Pacejka tire model: Intermediate-condition prediction and exact check. Horizontal axis: Slip ratio s (dimensionless). Vertical axis: Force / peak-scale D (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.15 −0.10 −0.05 0.00 0.05 0.10 0.15 Slip ratio s (dimensionless) −1.0 −0.5 0.0 0.5 1.0 Force / peak-scale D (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.2947 Worked point: (-0.054, -0.6001)
At x = -0.054, the orange circle is the analytical value -0.600138; the purple diamond is the interpolated value -0.294715. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.294715; analytical reference = -0.600138; absolute interpolation error = 0.305423. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Illustrative coefficients, not a calibrated tire or a combined-slip model. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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AC power-flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use two fixed voltage magnitudes connected by a purely reactive line; for a generator use the analogous fixed internal-voltage coupling. For this calculation, x denotes the plotted horizontal coordinate (Electrical angle difference δ (radian)), and q(x) denotes the plotted response (Transferred power / coupling scale (dimensionless)). Suppose only the analytical endpoint responses at x = -0.942478 and x = 0.942478 are tabulated. Use linear interpolation to predict the response at x = -0.282743, then check it against the original equation.

P/(V1V2/X)=sin⁡δP/(V_1V_2/X)=\sin\deltaa=−0.942478,b=0.942478,c=−0.282743a=-0.942478,\quad b=0.942478,\quad c=-0.282743qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.858394\frac{q(b)-q(a)}{b-a}=0.858394qlin(c)=−0.242705,q(c)=−0.278991q_{\rm lin}(c)=-0.242705,\quad q(c)=-0.278991e=qlin(c)−q(c)=0.036286e=q_{\rm lin}(c)-q(c)=0.036286

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = 0.809017. Their secant slope is 0.858394. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.242705. Direct evaluation gives -0.278991; subtracting it from the prediction gives signed error 0.036286.

AC power-flow model: Intermediate-condition prediction and exact check. Horizontal axis: Electrical angle difference δ (radian). Vertical axis: Transferred power / coupling scale (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.75 −0.50 −0.25 0.00 0.25 0.50 0.75 Electrical angle difference δ (radian) −1.00 −0.75 −0.50 −0.25 0.00 0.25 0.50 0.75 1.00 Transferred power / coupling scale (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.2427 Worked point: (-0.2827, -0.279)
At x = -0.282743, the orange circle is the analytical value -0.278991; the purple diamond is the interpolated value -0.242705. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.242705; analytical reference = -0.278991; absolute interpolation error = 0.036286. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Steady electrical-power term; the swing-equation rotor transient and voltage dynamics are not solved. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Swing-equation generator model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use two fixed voltage magnitudes connected by a purely reactive line; for a generator use the analogous fixed internal-voltage coupling. For this calculation, x denotes the plotted horizontal coordinate (Electrical angle difference δ (radian)), and q(x) denotes the plotted response (Transferred power / coupling scale (dimensionless)). Suppose only the analytical endpoint responses at x = -0.942478 and x = 0.942478 are tabulated. Use linear interpolation to predict the response at x = -0.282743, then check it against the original equation.

P/(V1V2/X)=sin⁡δP/(V_1V_2/X)=\sin\deltaa=−0.942478,b=0.942478,c=−0.282743a=-0.942478,\quad b=0.942478,\quad c=-0.282743qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.858394\frac{q(b)-q(a)}{b-a}=0.858394qlin(c)=−0.242705,q(c)=−0.278991q_{\rm lin}(c)=-0.242705,\quad q(c)=-0.278991e=qlin(c)−q(c)=0.036286e=q_{\rm lin}(c)-q(c)=0.036286

Solution. The endpoint responses are q(A) = -0.809017 and q(B) = 0.809017. Their secant slope is 0.858394. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.242705. Direct evaluation gives -0.278991; subtracting it from the prediction gives signed error 0.036286.

Swing-equation generator model: Intermediate-condition prediction and exact check. Horizontal axis: Electrical angle difference δ (radian). Vertical axis: Transferred power / coupling scale (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.75 −0.50 −0.25 0.00 0.25 0.50 0.75 Electrical angle difference δ (radian) −1.00 −0.75 −0.50 −0.25 0.00 0.25 0.50 0.75 1.00 Transferred power / coupling scale (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.2427 Worked point: (-0.2827, -0.279)
At x = -0.282743, the orange circle is the analytical value -0.278991; the purple diamond is the interpolated value -0.242705. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.242705; analytical reference = -0.278991; absolute interpolation error = 0.036286. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Steady electrical-power term; the swing-equation rotor transient and voltage dynamics are not solved. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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DC power-flow approximation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use nearly equal fixed bus voltage magnitudes, negligible resistance, and small angle difference. For this calculation, x denotes the plotted horizontal coordinate (Small angle difference δ (radian)), and q(x) denotes the plotted response (Scaled active power (dimensionless)). Suppose only the analytical endpoint responses at x = -0.12 and x = 0.12 are tabulated. Use linear interpolation to predict the response at x = -0.036, then check it against the original equation.

P/(V1V2/X)≈δP/(V_1V_2/X)\approx\deltaa=−0.12,b=0.12,c=−0.036a=-0.12,\quad b=0.12,\quad c=-0.036qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=−0.036,q(c)=−0.036q_{\rm lin}(c)=-0.036,\quad q(c)=-0.036e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = -0.12 and q(B) = 0.12. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.036. Direct evaluation gives -0.036; subtracting it from the prediction gives signed error 0.

DC power-flow approximation: Intermediate-condition prediction and exact check. Horizontal axis: Small angle difference δ (radian). Vertical axis: Scaled active power (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.10 −0.05 0.00 0.05 0.10 Small angle difference δ (radian) −0.10 −0.05 0.00 0.05 0.10 Scaled active power (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.036 Worked point: (-0.036, -0.036)
At x = -0.036, the orange circle is the analytical value -0.036; the purple diamond is the interpolated value -0.036. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.036; analytical reference = -0.036; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. DC power-flow approximation; it does not calculate reactive power or voltage magnitudes. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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State-space model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the scalar state equation y′+y=1 with y(0)=0, or transfer function 1/(s+1). For this calculation, x denotes the plotted horizontal coordinate (Time / system time constant (dimensionless)), and q(x) denotes the plotted response (Output / final value (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=1−e−τy(\tau)=1-e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

State-space model: Intermediate-condition prediction and exact check. Horizontal axis: Time / system time constant (dimensionless). Vertical axis: Output / final value (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / system time constant (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Output / final value (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact linear plant reference. For bond graphs/electrical analogs use a single storage-and-resistance element; for HIL this is a reference trajectory, not measured hardware data. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Transfer-function model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the scalar state equation y′+y=1 with y(0)=0, or transfer function 1/(s+1). For this calculation, x denotes the plotted horizontal coordinate (Time / system time constant (dimensionless)), and q(x) denotes the plotted response (Output / final value (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=1−e−τy(\tau)=1-e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Transfer-function model: Intermediate-condition prediction and exact check. Horizontal axis: Time / system time constant (dimensionless). Vertical axis: Output / final value (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / system time constant (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Output / final value (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact linear plant reference. For bond graphs/electrical analogs use a single storage-and-resistance element; for HIL this is a reference trajectory, not measured hardware data. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Bond-graph model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the scalar state equation y′+y=1 with y(0)=0, or transfer function 1/(s+1). For this calculation, x denotes the plotted horizontal coordinate (Time / system time constant (dimensionless)), and q(x) denotes the plotted response (Output / final value (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=1−e−τy(\tau)=1-e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Bond-graph model: Intermediate-condition prediction and exact check. Horizontal axis: Time / system time constant (dimensionless). Vertical axis: Output / final value (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / system time constant (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Output / final value (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact linear plant reference. For bond graphs/electrical analogs use a single storage-and-resistance element; for HIL this is a reference trajectory, not measured hardware data. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Electrical analog model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the scalar state equation y′+y=1 with y(0)=0, or transfer function 1/(s+1). For this calculation, x denotes the plotted horizontal coordinate (Time / system time constant (dimensionless)), and q(x) denotes the plotted response (Output / final value (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=1−e−τy(\tau)=1-e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Electrical analog model: Intermediate-condition prediction and exact check. Horizontal axis: Time / system time constant (dimensionless). Vertical axis: Output / final value (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / system time constant (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Output / final value (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact linear plant reference. For bond graphs/electrical analogs use a single storage-and-resistance element; for HIL this is a reference trajectory, not measured hardware data. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hardware-in-the-loop model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the scalar state equation y′+y=1 with y(0)=0, or transfer function 1/(s+1). For this calculation, x denotes the plotted horizontal coordinate (Time / system time constant (dimensionless)), and q(x) denotes the plotted response (Output / final value (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=1−e−τy(\tau)=1-e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Hardware-in-the-loop model: Intermediate-condition prediction and exact check. Horizontal axis: Time / system time constant (dimensionless). Vertical axis: Output / final value (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / system time constant (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Output / final value (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact linear plant reference. For bond graphs/electrical analogs use a single storage-and-resistance element; for HIL this is a reference trajectory, not measured hardware data. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hybrid dynamical model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Drop a ball from 1 m with g=9.81 m/s². At first ground contact reverse velocity and multiply its magnitude by restitution e=0.8. Plot before the second impact. For this calculation, x denotes the plotted horizontal coordinate (Time t (s)), and q(x) denotes the plotted response (Ball height h (m)). Suppose only the analytical endpoint responses at x = 0.22 and x = 0.88 are tabulated. Use linear interpolation to predict the response at x = 0.451, then check it against the original equation.

h(t)={1−12gt2t≤tiegti(t−ti)−12g(t−ti)2t>tih(t)=\begin{cases}1-\tfrac12gt^2&t\le t_i\\ egt_i(t-t_i)-\tfrac12g(t-t_i)^2&t>t_i\end{cases}a=0.22,b=0.88,c=0.451a=0.22,\quad b=0.88,\quad c=0.451qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.219373\frac{q(b)-q(a)}{b-a}=-0.219373qlin(c)=0.711923,q(c)=0.00231809q_{\rm lin}(c)=0.711923,\quad q(c)=0.00231809e=qlin(c)−q(c)=0.709605e=q_{\rm lin}(c)-q(c)=0.709605

Solution. The endpoint responses are q(A) = 0.762598 and q(B) = 0.617812. Their secant slope is -0.219373. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.711923. Direct evaluation gives 0.00231809; subtracting it from the prediction gives signed error 0.709605.

Hybrid dynamical model: Intermediate-condition prediction and exact check. Horizontal axis: Time t (s). Vertical axis: Ball height h (m). image/svg+xml IICSM analytical illustration / Matplotlib 0.3 0.4 0.5 0.6 0.7 0.8 Time t (s) 0.0 0.2 0.4 0.6 0.8 Ball height h (m) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7119 Worked point: (0.451, 0.002318)
At x = 0.451, the orange circle is the analytical value 0.00231809; the purple diamond is the interpolated value 0.711923. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.711923; analytical reference = 0.00231809; absolute interpolation error = 0.709605. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ideal instantaneous first bounce; air resistance and contact deformation are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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System-dynamics stock-flow model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. An initially empty stock receives constant inflow q and drains at rate kS. For this calculation, x denotes the plotted horizontal coordinate (Time kt (dimensionless)), and q(x) denotes the plotted response (Stock / equilibrium stock (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

S/(q/k)=1−e−ktS/(q/k)=1-e^{-kt}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

System-dynamics stock-flow model: Intermediate-condition prediction and exact check. Horizontal axis: Time kt (dimensionless). Vertical axis: Stock / equilibrium stock (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time kt (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Stock / equilibrium stock (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single stock, constant coefficients, and no delays or saturation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Discrete-event simulation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Identical events occur at Δt,2Δt,… with zero events completed at t=0. For this calculation, x denotes the plotted horizontal coordinate (Time / event interval Δt (dimensionless)), and q(x) denotes the plotted response (Cumulative completed events (count)). Suppose only the analytical endpoint responses at x = 1.2 and x = 4.8 are tabulated. Use linear interpolation to predict the response at x = 2.46, then check it against the original equation.

N(t)=⌊t/Δt⌋N(t)=\lfloor t/\Delta t\rfloora=1.2,b=4.8,c=2.46a=1.2,\quad b=4.8,\quad c=2.46qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.833333\frac{q(b)-q(a)}{b-a}=0.833333qlin(c)=2.05,q(c)=2q_{\rm lin}(c)=2.05,\quad q(c)=2e=qlin(c)−q(c)=0.05e=q_{\rm lin}(c)-q(c)=0.05

Solution. The endpoint responses are q(A) = 1 and q(B) = 4. Their secant slope is 0.833333. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.05. Direct evaluation gives 2; subtracting it from the prediction gives signed error 0.05.

Discrete-event simulation: Intermediate-condition prediction and exact check. Horizontal axis: Time / event interval Δt (dimensionless). Vertical axis: Cumulative completed events (count). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Time / event interval Δt (dimensionless) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Cumulative completed events (count) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.05 Worked point: (2.46, 2)
At x = 2.46, the orange circle is the analytical value 2; the purple diamond is the interpolated value 2.05. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.05; analytical reference = 2; absolute interpolation error = 0.05. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Simple scheduled-event benchmark; a discrete-event model need not have periodic arrivals. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere. For this discrete count, a fractional interpolated value is not an attainable count; the exact event schedule determines the answer.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Agent-based physical-system model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Agents start at mean position zero, have constant mean velocity 1 m/s, and do not interact. For this calculation, x denotes the plotted horizontal coordinate (Time t (s)), and q(x) denotes the plotted response (Ensemble mean position (m)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

⟨x(t)⟩=x0+vt\langle x(t)\rangle=x_0+vta=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=2.05,q(c)=2.05q_{\rm lin}(c)=2.05,\quad q(c)=2.05e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 1 and q(B) = 4. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.05. Direct evaluation gives 2.05; subtracting it from the prediction gives signed error 0.

Agent-based physical-system model: Intermediate-condition prediction and exact check. Horizontal axis: Time t (s). Vertical axis: Ensemble mean position (m). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time t (s) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Ensemble mean position (m) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.05 Worked point: (2.05, 2.05)
At x = 2.05, the orange circle is the analytical value 2.05; the purple diamond is the interpolated value 2.05. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.05; analytical reference = 2.05; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Noninteracting kinematic benchmark; not an emergent many-agent simulation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Markov state model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Two states exchange population at equal rate k. Initially all probability is in state one. For this calculation, x denotes the plotted horizontal coordinate (Time kt (dimensionless)), and q(x) denotes the plotted response (Probability of state 1 (dimensionless)). Suppose only the analytical endpoint responses at x = 0.8 and x = 3.2 are tabulated. Use linear interpolation to predict the response at x = 1.64, then check it against the original equation.

P1(t)=12(1+e−2kt)P_1(t)=\tfrac12(1+e^{-2kt})a=0.8,b=3.2,c=1.64a=0.8,\quad b=3.2,\quad c=1.64qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.0417156\frac{q(b)-q(a)}{b-a}=-0.0417156qlin(c)=0.565907,q(c)=0.518814q_{\rm lin}(c)=0.565907,\quad q(c)=0.518814e=qlin(c)−q(c)=0.047093e=q_{\rm lin}(c)-q(c)=0.047093

Solution. The endpoint responses are q(A) = 0.600948 and q(B) = 0.500831. Their secant slope is -0.0417156. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.565907. Direct evaluation gives 0.518814; subtracting it from the prediction gives signed error 0.047093.

Markov state model: Intermediate-condition prediction and exact check. Horizontal axis: Time kt (dimensionless). Vertical axis: Probability of state 1 (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 Time kt (dimensionless) 0.50 0.52 0.54 0.56 0.58 0.60 Probability of state 1 (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5659 Worked point: (1.64, 0.5188)
At x = 1.64, the orange circle is the analytical value 0.518814; the purple diamond is the interpolated value 0.565907. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.565907; analytical reference = 0.518814; absolute interpolation error = 0.047093. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Kalman state estimator · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For one scalar measurement with observation coefficient one, hold the positive prior variance fixed. For this calculation, x denotes the plotted horizontal coordinate (Measurement / prior variance R/P⁻ (dimensionless)), and q(x) denotes the plotted response (Scalar Kalman gain (dimensionless)). Suppose only the analytical endpoint responses at x = 2 and x = 8 are tabulated. Use linear interpolation to predict the response at x = 4.1, then check it against the original equation.

K=P−P−+R=11+R/P−K=\frac{P^-}{P^-+R}=\frac1{1+R/P^-}a=2,b=8,c=4.1a=2,\quad b=8,\quad c=4.1qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.037037\frac{q(b)-q(a)}{b-a}=-0.037037qlin(c)=0.255556,q(c)=0.196078q_{\rm lin}(c)=0.255556,\quad q(c)=0.196078e=qlin(c)−q(c)=0.0594771e=q_{\rm lin}(c)-q(c)=0.0594771

Solution. The endpoint responses are q(A) = 0.333333 and q(B) = 0.111111. Their secant slope is -0.037037. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.255556. Direct evaluation gives 0.196078; subtracting it from the prediction gives signed error 0.0594771.

Kalman state estimator: Intermediate-condition prediction and exact check. Horizontal axis: Measurement / prior variance R/P⁻ (dimensionless). Vertical axis: Scalar Kalman gain (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2 3 4 5 6 7 8 Measurement / prior variance R/P⁻ (dimensionless) 0.10 0.15 0.20 0.25 0.30 0.35 Scalar Kalman gain (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2556 Worked point: (4.1, 0.1961)
At x = 4.1, the orange circle is the analytical value 0.196078; the purple diamond is the interpolated value 0.255556. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.255556; analytical reference = 0.196078; absolute interpolation error = 0.0594771. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Single measurement update; not a full dynamic filter trajectory. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Model predictive control · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Let xnext=x+u and minimize (x+u)²+ρu² with ρ=1 and no constraints. For this calculation, x denotes the plotted horizontal coordinate (Initial state x (dimensionless)), and q(x) denotes the plotted response (Optimal input u* (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

u∗=−x1+ρ,ρ=1u_*=-\frac{x}{1+\rho},\quad\rho=1a=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.5\frac{q(b)-q(a)}{b-a}=-0.5qlin(c)=0.18,q(c)=0.18q_{\rm lin}(c)=0.18,\quad q(c)=0.18e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.6 and q(B) = -0.6. Their secant slope is -0.5. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.18. Direct evaluation gives 0.18; subtracting it from the prediction gives signed error 0.

Model predictive control: Intermediate-condition prediction and exact check. Horizontal axis: Initial state x (dimensionless). Vertical axis: Optimal input u* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Initial state x (dimensionless) −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Optimal input u* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.18 Worked point: (-0.36, 0.18)
At x = -0.36, the orange circle is the analytical value 0.18; the purple diamond is the interpolated value 0.18. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.18; analytical reference = 0.18; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical horizon-one MPC example; longer horizons and constraints change the feedback law. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hodgkin–Huxley membrane model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Set sodium and potassium conductances to zero, hold leak reversal potential EL fixed, and normalize V−EL by its initial value. Use τ=gLt/Cm. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Membrane voltage excess / initial excess (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Hodgkin–Huxley membrane model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Membrane voltage excess / initial excess (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Membrane voltage excess / initial excess (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Passive leak-only reduction of Hodgkin–Huxley; action potentials and voltage-dependent gates are deliberately excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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FitzHugh–Nagumo model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For v′=v−v³/3−w+I set I=0 and find the zero-fast-derivative curve. For this calculation, x denotes the plotted horizontal coordinate (Fast variable v (dimensionless)), and q(x) denotes the plotted response (Recovery variable w (dimensionless)). Suppose only the analytical endpoint responses at x = -1.5 and x = 1.5 are tabulated. Use linear interpolation to predict the response at x = -0.45, then check it against the original equation.

w=v−v3/3,I=0w=v-v^3/3,\quad I=0a=−1.5,b=1.5,c=−0.45a=-1.5,\quad b=1.5,\quad c=-0.45qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.25\frac{q(b)-q(a)}{b-a}=0.25qlin(c)=−0.1125,q(c)=−0.419625q_{\rm lin}(c)=-0.1125,\quad q(c)=-0.419625e=qlin(c)−q(c)=0.307125e=q_{\rm lin}(c)-q(c)=0.307125

Solution. The endpoint responses are q(A) = -0.375 and q(B) = 0.375. Their secant slope is 0.25. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.1125. Direct evaluation gives -0.419625; subtracting it from the prediction gives signed error 0.307125.

FitzHugh–Nagumo model: Intermediate-condition prediction and exact check. Horizontal axis: Fast variable v (dimensionless). Vertical axis: Recovery variable w (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Fast variable v (dimensionless) −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 Recovery variable w (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.1125 Worked point: (-0.45, -0.4196)
At x = -0.45, the orange circle is the analytical value -0.419625; the purple diamond is the interpolated value -0.1125. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.1125; analytical reference = -0.419625; absolute interpolation error = 0.307125. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. A phase-plane nullcline, not a trajectory or the complete system equilibrium; equilibria also lie on the recovery nullcline. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hill muscle model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use (F+a)(v+b)=(F0+a)b, with a/F0=0.25 and vmax=bF0/a. For this calculation, x denotes the plotted horizontal coordinate (Shortening speed / unloaded speed (dimensionless)), and q(x) denotes the plotted response (Muscle force / isometric force (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

F/F0=0.25(1−v/vmax⁡)0.25+v/vmax⁡F/F_0=\frac{0.25(1-v/v_{\max})}{0.25+v/v_{\max}}a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.661376\frac{q(b)-q(a)}{b-a}=-0.661376qlin(c)=0.305556,q(c)=0.223485q_{\rm lin}(c)=0.305556,\quad q(c)=0.223485e=qlin(c)−q(c)=0.0820707e=q_{\rm lin}(c)-q(c)=0.0820707

Solution. The endpoint responses are q(A) = 0.444444 and q(B) = 0.047619. Their secant slope is -0.661376. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.305556. Direct evaluation gives 0.223485; subtracting it from the prediction gives signed error 0.0820707.

Hill muscle model: Intermediate-condition prediction and exact check. Horizontal axis: Shortening speed / unloaded speed (dimensionless). Vertical axis: Muscle force / isometric force (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Shortening speed / unloaded speed (dimensionless) 0.0 0.1 0.2 0.3 0.4 Muscle force / isometric force (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.3056 Worked point: (0.41, 0.2235)
At x = 0.41, the orange circle is the analytical value 0.223485; the purple diamond is the interpolated value 0.305556. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.305556; analytical reference = 0.223485; absolute interpolation error = 0.0820707. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Steady concentric shortening only; activation and length effects are held fixed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Windkessel circulation model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. With zero inflow, a two-element Windkessel discharges through resistance R from compliance C. Use τ=t/(RC) and normalize pressure above venous pressure. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Pressure above venous level / initial excess (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Windkessel circulation model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Pressure above venous level / initial excess (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Pressure above venous level / initial excess (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Constant-compliance diastolic interval, not a full pulsatile cardiac cycle. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Pennes bioheat model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take spatially uniform tissue with constant heat source Q, blood heat-exchange coefficient W>0, and initial tissue temperature equal to arterial temperature Ta. For this calculation, x denotes the plotted horizontal coordinate (Perfusion relaxation time Wt / ρc (dimensionless)), and q(x) denotes the plotted response (Temperature rise / Q/W (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

(T−Ta)/(Q/W)=1−e−Wt/(ρc)(T-T_a)/(Q/W)=1-e^{-Wt/(\rho c)}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.116521\frac{q(b)-q(a)}{b-a}=0.116521qlin(c)=0.754468,q(c)=0.871265q_{\rm lin}(c)=0.754468,\quad q(c)=0.871265e=qlin(c)−q(c)=−0.116797e=q_{\rm lin}(c)-q(c)=-0.116797

Solution. The endpoint responses are q(A) = 0.632121 and q(B) = 0.981684. Their secant slope is 0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.754468. Direct evaluation gives 0.871265; subtracting it from the prediction gives signed error -0.116797.

Pennes bioheat model: Intermediate-condition prediction and exact check. Horizontal axis: Perfusion relaxation time Wt / ρc (dimensionless). Vertical axis: Temperature rise / Q/W (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Perfusion relaxation time Wt / ρc (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Temperature rise / Q/W (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7545 Worked point: (2.05, 0.8713)
At x = 2.05, the orange circle is the analytical value 0.871265; the purple diamond is the interpolated value 0.754468. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.754468; analytical reference = 0.871265; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Uniform-temperature reduction; no spatial conduction, temperature-dependent perfusion, or safety prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Reaction–diffusion morphogenesis model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Solve ut=uxx−u with zero ends and initial sin(πx), then plot t=1. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Activator perturbation (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(x,1)=e−(π2+1)sin⁡(πx)u(x,1)=e^{-(\pi^2+1)}\sin(\pi x)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=5.64689×10−21\frac{q(b)-q(a)}{b-a}=5.64689\times10^{-21}qlin(c)=1.11843×10−5,q(c)=1.82724×10−5q_{\rm lin}(c)=1.11843\times10^{-5},\quad q(c)=1.82724\times10^{-5}e=qlin(c)−q(c)=−7.08805×10−6e=q_{\rm lin}(c)-q(c)=-7.08805\times10^{-6}

Solution. The endpoint responses are q(A) = 1.11843e-05 and q(B) = 1.11843e-05. Their secant slope is 5.64689e-21. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.11843e-05. Direct evaluation gives 1.82724e-05; subtracting it from the prediction gives signed error -7.08805e-06.

Reaction–diffusion morphogenesis model: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Activator perturbation (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 1.2 1.4 1.6 1.8 Activator perturbation (dimensionless) 1e−5 Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.118e-05 Worked point: (0.41, 1.827e-05)
At x = 0.41, the orange circle is the analytical value 1.82724e-05; the purple diamond is the interpolated value 1.11843e-05. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.11843e-05; analytical reference = 1.82724e-05; absolute interpolation error = 7.08805e-06. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-species linear stable subproblem, not a two-species Turing pattern or nonlinear morphogenesis prediction. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Monod growth model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate growth rate at prescribed substrate concentration with fixed Monod parameters. For this calculation, x denotes the plotted horizontal coordinate (Substrate concentration S / Ks (dimensionless)), and q(x) denotes the plotted response (Growth rate / maximum rate (dimensionless)). Suppose only the analytical endpoint responses at x = 1.6 and x = 6.4 are tabulated. Use linear interpolation to predict the response at x = 3.28, then check it against the original equation.

μ/μmax⁡=S/(Ks+S)\mu/\mu_{\max}=S/(K_s+S)a=1.6,b=6.4,c=3.28a=1.6,\quad b=6.4,\quad c=3.28qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.0519751\frac{q(b)-q(a)}{b-a}=0.0519751qlin(c)=0.702703,q(c)=0.766355q_{\rm lin}(c)=0.702703,\quad q(c)=0.766355e=qlin(c)−q(c)=−0.0636524e=q_{\rm lin}(c)-q(c)=-0.0636524

Solution. The endpoint responses are q(A) = 0.615385 and q(B) = 0.864865. Their secant slope is 0.0519751. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.702703. Direct evaluation gives 0.766355; subtracting it from the prediction gives signed error -0.0636524.

Monod growth model: Intermediate-condition prediction and exact check. Horizontal axis: Substrate concentration S / Ks (dimensionless). Vertical axis: Growth rate / maximum rate (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2 3 4 5 6 Substrate concentration S / Ks (dimensionless) 0.60 0.65 0.70 0.75 0.80 0.85 Growth rate / maximum rate (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7027 Worked point: (3.28, 0.7664)
At x = 3.28, the orange circle is the analytical value 0.766355; the purple diamond is the interpolated value 0.702703. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.702703; analytical reference = 0.766355; absolute interpolation error = 0.0636524. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Growth-rate relation only; substrate depletion and biomass evolution are not integrated. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Physiologically based compartment model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. After an initial dose, use one well-mixed compartment with first-order elimination, no further input, and τ=kt. For this calculation, x denotes the plotted horizontal coordinate (Time / relaxation time (dimensionless)), and q(x) denotes the plotted response (Compartment concentration / initial concentration (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

y(τ)=e−τy(\tau)=e^{-\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Physiologically based compartment model: Intermediate-condition prediction and exact check. Horizontal axis: Time / relaxation time (dimensionless). Vertical axis: Compartment concentration / initial concentration (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time / relaxation time (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Compartment concentration / initial concentration (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-compartment limiting case; interorgan exchange, binding, and nonlinear metabolism are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Boltzmann kinetic equation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take a spatially uniform equilibrium with zero drift and the normalized Gaussian velocity marginal. For collisionless plasma use zero fields and a neutralizing background. For this calculation, x denotes the plotted horizontal coordinate (Velocity v / thermal speed (dimensionless)), and q(x) denotes the plotted response (Marginal probability density × thermal speed (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

vthf(v)=π−1/2e−(v/vth)2v_{th}f(v)=\pi^{-1/2}e^{-(v/v_{th})^2}a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.92747×10−17\frac{q(b)-q(a)}{b-a}=-1.92747\times10^{-17}qlin(c)=0.0220959,q(c)=0.421488q_{\rm lin}(c)=0.0220959,\quad q(c)=0.421488e=qlin(c)−q(c)=−0.399392e=q_{\rm lin}(c)-q(c)=-0.399392

Solution. The endpoint responses are q(A) = 0.0220959 and q(B) = 0.0220959. Their secant slope is -1.92747e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0220959. Direct evaluation gives 0.421488; subtracting it from the prediction gives signed error -0.399392.

Boltzmann kinetic equation: Intermediate-condition prediction and exact check. Horizontal axis: Velocity v / thermal speed (dimensionless). Vertical axis: Marginal probability density × thermal speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Velocity v / thermal speed (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Marginal probability density × thermal speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0221 Worked point: (-0.54, 0.4215)
At x = -0.54, the orange circle is the analytical value 0.421488; the purple diamond is the interpolated value 0.0220959. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0220959; analytical reference = 0.421488; absolute interpolation error = 0.399392. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Equilibrium distribution or exact kinetic benchmark. DSMC and PIC would estimate it using particles; this plot is not a finite-particle sample. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Vlasov–Poisson model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take a spatially uniform equilibrium with zero drift and the normalized Gaussian velocity marginal. For collisionless plasma use zero fields and a neutralizing background. For this calculation, x denotes the plotted horizontal coordinate (Velocity v / thermal speed (dimensionless)), and q(x) denotes the plotted response (Marginal probability density × thermal speed (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

vthf(v)=π−1/2e−(v/vth)2v_{th}f(v)=\pi^{-1/2}e^{-(v/v_{th})^2}a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.92747×10−17\frac{q(b)-q(a)}{b-a}=-1.92747\times10^{-17}qlin(c)=0.0220959,q(c)=0.421488q_{\rm lin}(c)=0.0220959,\quad q(c)=0.421488e=qlin(c)−q(c)=−0.399392e=q_{\rm lin}(c)-q(c)=-0.399392

Solution. The endpoint responses are q(A) = 0.0220959 and q(B) = 0.0220959. Their secant slope is -1.92747e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0220959. Direct evaluation gives 0.421488; subtracting it from the prediction gives signed error -0.399392.

Vlasov–Poisson model: Intermediate-condition prediction and exact check. Horizontal axis: Velocity v / thermal speed (dimensionless). Vertical axis: Marginal probability density × thermal speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Velocity v / thermal speed (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Marginal probability density × thermal speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0221 Worked point: (-0.54, 0.4215)
At x = -0.54, the orange circle is the analytical value 0.421488; the purple diamond is the interpolated value 0.0220959. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0220959; analytical reference = 0.421488; absolute interpolation error = 0.399392. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Equilibrium distribution or exact kinetic benchmark. DSMC and PIC would estimate it using particles; this plot is not a finite-particle sample. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Vlasov–Maxwell model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take a spatially uniform equilibrium with zero drift and the normalized Gaussian velocity marginal. For collisionless plasma use zero fields and a neutralizing background. For this calculation, x denotes the plotted horizontal coordinate (Velocity v / thermal speed (dimensionless)), and q(x) denotes the plotted response (Marginal probability density × thermal speed (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

vthf(v)=π−1/2e−(v/vth)2v_{th}f(v)=\pi^{-1/2}e^{-(v/v_{th})^2}a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.92747×10−17\frac{q(b)-q(a)}{b-a}=-1.92747\times10^{-17}qlin(c)=0.0220959,q(c)=0.421488q_{\rm lin}(c)=0.0220959,\quad q(c)=0.421488e=qlin(c)−q(c)=−0.399392e=q_{\rm lin}(c)-q(c)=-0.399392

Solution. The endpoint responses are q(A) = 0.0220959 and q(B) = 0.0220959. Their secant slope is -1.92747e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0220959. Direct evaluation gives 0.421488; subtracting it from the prediction gives signed error -0.399392.

Vlasov–Maxwell model: Intermediate-condition prediction and exact check. Horizontal axis: Velocity v / thermal speed (dimensionless). Vertical axis: Marginal probability density × thermal speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Velocity v / thermal speed (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Marginal probability density × thermal speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0221 Worked point: (-0.54, 0.4215)
At x = -0.54, the orange circle is the analytical value 0.421488; the purple diamond is the interpolated value 0.0220959. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0220959; analytical reference = 0.421488; absolute interpolation error = 0.399392. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Equilibrium distribution or exact kinetic benchmark. DSMC and PIC would estimate it using particles; this plot is not a finite-particle sample. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Direct simulation Monte Carlo (DSMC) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take a spatially uniform equilibrium with zero drift and the normalized Gaussian velocity marginal. For collisionless plasma use zero fields and a neutralizing background. For this calculation, x denotes the plotted horizontal coordinate (Velocity v / thermal speed (dimensionless)), and q(x) denotes the plotted response (Marginal probability density × thermal speed (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

vthf(v)=π−1/2e−(v/vth)2v_{th}f(v)=\pi^{-1/2}e^{-(v/v_{th})^2}a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.92747×10−17\frac{q(b)-q(a)}{b-a}=-1.92747\times10^{-17}qlin(c)=0.0220959,q(c)=0.421488q_{\rm lin}(c)=0.0220959,\quad q(c)=0.421488e=qlin(c)−q(c)=−0.399392e=q_{\rm lin}(c)-q(c)=-0.399392

Solution. The endpoint responses are q(A) = 0.0220959 and q(B) = 0.0220959. Their secant slope is -1.92747e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0220959. Direct evaluation gives 0.421488; subtracting it from the prediction gives signed error -0.399392.

Direct simulation Monte Carlo (DSMC): Intermediate-condition prediction and exact check. Horizontal axis: Velocity v / thermal speed (dimensionless). Vertical axis: Marginal probability density × thermal speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Velocity v / thermal speed (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Marginal probability density × thermal speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0221 Worked point: (-0.54, 0.4215)
At x = -0.54, the orange circle is the analytical value 0.421488; the purple diamond is the interpolated value 0.0220959. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0220959; analytical reference = 0.421488; absolute interpolation error = 0.399392. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Equilibrium distribution or exact kinetic benchmark. DSMC and PIC would estimate it using particles; this plot is not a finite-particle sample. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Particle-in-cell (PIC) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take a spatially uniform equilibrium with zero drift and the normalized Gaussian velocity marginal. For collisionless plasma use zero fields and a neutralizing background. For this calculation, x denotes the plotted horizontal coordinate (Velocity v / thermal speed (dimensionless)), and q(x) denotes the plotted response (Marginal probability density × thermal speed (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

vthf(v)=π−1/2e−(v/vth)2v_{th}f(v)=\pi^{-1/2}e^{-(v/v_{th})^2}a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.92747×10−17\frac{q(b)-q(a)}{b-a}=-1.92747\times10^{-17}qlin(c)=0.0220959,q(c)=0.421488q_{\rm lin}(c)=0.0220959,\quad q(c)=0.421488e=qlin(c)−q(c)=−0.399392e=q_{\rm lin}(c)-q(c)=-0.399392

Solution. The endpoint responses are q(A) = 0.0220959 and q(B) = 0.0220959. Their secant slope is -1.92747e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0220959. Direct evaluation gives 0.421488; subtracting it from the prediction gives signed error -0.399392.

Particle-in-cell (PIC): Intermediate-condition prediction and exact check. Horizontal axis: Velocity v / thermal speed (dimensionless). Vertical axis: Marginal probability density × thermal speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Velocity v / thermal speed (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Marginal probability density × thermal speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0221 Worked point: (-0.54, 0.4215)
At x = -0.54, the orange circle is the analytical value 0.421488; the purple diamond is the interpolated value 0.0220959. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0220959; analytical reference = 0.421488; absolute interpolation error = 0.399392. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Equilibrium distribution or exact kinetic benchmark. DSMC and PIC would estimate it using particles; this plot is not a finite-particle sample. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Neutron diffusion approximation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. On a slab solve nτ=nξξ with zero extrapolated-end values and initial sin(πξ), ignoring reactions in this illustrative diffusion subproblem. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Neutron-density perturbation / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(πξ)e−π2τ,τ=0.1u(\xi,\tau)=\sin(\pi\xi)e^{-\pi^2\tau},\quad\tau=0.1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.62593×10−17\frac{q(b)-q(a)}{b-a}=4.62593\times10^{-17}qlin(c)=0.219072,q(c)=0.357909q_{\rm lin}(c)=0.219072,\quad q(c)=0.357909e=qlin(c)−q(c)=−0.138837e=q_{\rm lin}(c)-q(c)=-0.138837

Solution. The endpoint responses are q(A) = 0.219072 and q(B) = 0.219072. Their secant slope is 4.62593e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.219072. Direct evaluation gives 0.357909; subtracting it from the prediction gives signed error -0.138837.

Neutron diffusion approximation: Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Neutron-density perturbation / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Neutron-density perturbation / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2191 Worked point: (0.41, 0.3579)
At x = 0.41, the orange circle is the analytical value 0.357909; the purple diamond is the interpolated value 0.219072. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.219072; analytical reference = 0.357909; absolute interpolation error = 0.138837. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Diffusion-only benchmark; absorption and fission terms would modify the mode growth/decay rate. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Point reactor kinetics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Set delayed-neutron fraction and external source to zero, take constant negative reactivity ρ, and prompt generation time Λ. For this calculation, x denotes the plotted horizontal coordinate (Subcritical prompt time |ρ|t / Λ (dimensionless)), and q(x) denotes the plotted response (Neutron population / initial population (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

n/n0=e−τ,τ=∣ρ∣t/Λn/n_0=e^{-\tau},\quad\tau=|\rho|t/\Lambdaa=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Point reactor kinetics: Intermediate-condition prediction and exact check. Horizontal axis: Subcritical prompt time |ρ|t / Λ (dimensionless). Vertical axis: Neutron population / initial population (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Subcritical prompt time |ρ|t / Λ (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Neutron population / initial population (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Prompt-only idealization, not a realistic startup, shutdown, or reactor-safety calculation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Bateman decay-chain model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Initially N1=N10 and N2=0. Let the parent decay to the daughter with λ2=2λ1 and unit branching fraction. For this calculation, x denotes the plotted horizontal coordinate (Time λ₁t (dimensionless)), and q(x) denotes the plotted response (Daughter population / initial parent (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

N2/N10=e−λ1t−e−2λ1tN_2/N_{10}=e^{-\lambda_1t}-e^{-2\lambda_1t}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.0715213\frac{q(b)-q(a)}{b-a}=-0.0715213qlin(c)=0.157447,q(c)=0.112162q_{\rm lin}(c)=0.157447,\quad q(c)=0.112162e=qlin(c)−q(c)=0.0452845e=q_{\rm lin}(c)-q(c)=0.0452845

Solution. The endpoint responses are q(A) = 0.232544 and q(B) = 0.0179802. Their secant slope is -0.0715213. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.157447. Direct evaluation gives 0.112162; subtracting it from the prediction gives signed error 0.0452845.

Bateman decay-chain model: Intermediate-condition prediction and exact check. Horizontal axis: Time λ₁t (dimensionless). Vertical axis: Daughter population / initial parent (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time λ₁t (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 Daughter population / initial parent (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1574 Worked point: (2.05, 0.1122)
At x = 2.05, the orange circle is the analytical value 0.112162; the purple diamond is the interpolated value 0.157447. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.157447; analytical reference = 0.112162; absolute interpolation error = 0.0452845. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical solution or constitutive evaluation for the stated special case. It is not a general solution of the full model family. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Newtonian gravitational N-body model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Reduce an isolated gravitational system to two point masses with total mass M in a circular relative orbit of radius a. For this calculation, x denotes the plotted horizontal coordinate (Orbital phase nt (radian)), and q(x) denotes the plotted response (Relative orbital x coordinate / radius (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

x/a=cos⁡(nt),n=GM/a3x/a=\cos(nt),\quad n=\sqrt{GM/a^3}a=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−5.88992×10−17\frac{q(b)-q(a)}{b-a}=-5.88992\times10^{-17}qlin(c)=0.309017,q(c)=−0.844328q_{\rm lin}(c)=0.309017,\quad q(c)=-0.844328e=qlin(c)−q(c)=1.15334e=q_{\rm lin}(c)-q(c)=1.15334

Solution. The endpoint responses are q(A) = 0.309017 and q(B) = 0.309017. Their secant slope is -5.88992e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.309017. Direct evaluation gives -0.844328; subtracting it from the prediction gives signed error 1.15334.

Newtonian gravitational N-body model: Intermediate-condition prediction and exact check. Horizontal axis: Orbital phase nt (radian). Vertical axis: Relative orbital x coordinate / radius (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Orbital phase nt (radian) −1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 Relative orbital x coordinate / radius (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.309 Worked point: (2.576, -0.8443)
At x = 2.57611, the orange circle is the analytical value -0.844328; the purple diamond is the interpolated value 0.309017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.309017; analytical reference = -0.844328; absolute interpolation error = 1.15334. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact two-body circular orbit, not a general many-body solution; a one-coordinate time trace is shown. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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General relativity model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a stationary observer outside a nonrotating spherical mass, compare proper time with Schwarzschild coordinate time at infinity. For this calculation, x denotes the plotted horizontal coordinate (Schwarzschild radius ratio r / rs (dimensionless)), and q(x) denotes the plotted response (Static clock rate dτ/dt (dimensionless)). Suppose only the analytical endpoint responses at x = 2.44 and x = 6.61 are tabulated. Use linear interpolation to predict the response at x = 3.8995, then check it against the original equation.

dτ/dt=1−rs/rd\tau/dt=\sqrt{1-r_s/r}a=2.44,b=6.61,c=3.8995a=2.44,\quad b=6.61,\quad c=3.8995qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.0366992\frac{q(b)-q(a)}{b-a}=0.0366992qlin(c)=0.821784,q(c)=0.862297q_{\rm lin}(c)=0.821784,\quad q(c)=0.862297e=qlin(c)−q(c)=−0.0405137e=q_{\rm lin}(c)-q(c)=-0.0405137

Solution. The endpoint responses are q(A) = 0.768221 and q(B) = 0.921257. Their secant slope is 0.0366992. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.821784. Direct evaluation gives 0.862297; subtracting it from the prediction gives signed error -0.0405137.

General relativity model: Intermediate-condition prediction and exact check. Horizontal axis: Schwarzschild radius ratio r / rs (dimensionless). Vertical axis: Static clock rate dτ/dt (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 Schwarzschild radius ratio r / rs (dimensionless) 0.750 0.775 0.800 0.825 0.850 0.875 0.900 0.925 Static clock rate dτ/dt (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.8218 Worked point: (3.9, 0.8623)
At x = 3.8995, the orange circle is the analytical value 0.862297; the purple diamond is the interpolated value 0.821784. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.821784; analytical reference = 0.862297; absolute interpolation error = 0.0405137. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exterior vacuum Schwarzschild solution, r>rs. A static observer cannot remain at the horizon. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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FLRW cosmological model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take a spatially flat FLRW universe with pressureless matter only and zero cosmological constant. For this calculation, x denotes the plotted horizontal coordinate (Cosmic time t / t* (dimensionless)), and q(x) denotes the plotted response (Scale factor a / a* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.616 and x = 2.404 are tabulated. Use linear interpolation to predict the response at x = 1.2418, then check it against the original equation.

a(t)/a(t∗)=(t/t∗)2/3a(t)/a(t_*)=(t/t_*)^{2/3}a=0.616,b=2.404,c=1.2418a=0.616,\quad b=2.404,\quad c=1.2418qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.59876\frac{q(b)-q(a)}{b-a}=0.59876qlin(c)=1.09867,q(c)=1.15532q_{\rm lin}(c)=1.09867,\quad q(c)=1.15532e=qlin(c)−q(c)=−0.0566429e=q_{\rm lin}(c)-q(c)=-0.0566429

Solution. The endpoint responses are q(A) = 0.72397 and q(B) = 1.79455. Their secant slope is 0.59876. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.09867. Direct evaluation gives 1.15532; subtracting it from the prediction gives signed error -0.0566429.

FLRW cosmological model: Intermediate-condition prediction and exact check. Horizontal axis: Cosmic time t / t* (dimensionless). Vertical axis: Scale factor a / a* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Cosmic time t / t* (dimensionless) 0.6 0.8 1.0 1.2 1.4 1.6 1.8 Scale factor a / a* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.099 Worked point: (1.242, 1.155)
At x = 1.2418, the orange circle is the analytical value 1.15532; the purple diamond is the interpolated value 1.09867. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.09867; analytical reference = 1.15532; absolute interpolation error = 0.0566429. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Matter-only special case, not a fit to the present universe. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Smoothed particle hydrodynamics (SPH) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Evaluate the standard one-dimensional cubic-spline smoothing kernel of support radius 2h. For this calculation, x denotes the plotted horizontal coordinate (Kernel coordinate x / h (dimensionless)), and q(x) denotes the plotted response (Kernel weight hW (dimensionless)). Suppose only the analytical endpoint responses at x = -1.2 and x = 1.2 are tabulated. Use linear interpolation to predict the response at x = -0.36, then check it against the original equation.

hW(q)=23{1−1.5q2+0.75q3q<1(2−q)3/41≤q≤2hW(q)=\frac23\begin{cases}1-1.5q^2+0.75q^3&q<1\\(2-q)^3/4&1\le q\le2\end{cases}a=−1.2,b=1.2,c=−0.36a=-1.2,\quad b=1.2,\quad c=-0.36qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.89121×10−17\frac{q(b)-q(a)}{b-a}=-2.89121\times10^{-17}qlin(c)=0.0853333,q(c)=0.560395q_{\rm lin}(c)=0.0853333,\quad q(c)=0.560395e=qlin(c)−q(c)=−0.475061e=q_{\rm lin}(c)-q(c)=-0.475061

Solution. The endpoint responses are q(A) = 0.0853333 and q(B) = 0.0853333. Their secant slope is -2.89121e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0853333. Direct evaluation gives 0.560395; subtracting it from the prediction gives signed error -0.475061.

Smoothed particle hydrodynamics (SPH): Intermediate-condition prediction and exact check. Horizontal axis: Kernel coordinate x / h (dimensionless). Vertical axis: Kernel weight hW (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 Kernel coordinate x / h (dimensionless) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Kernel weight hW (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.08533 Worked point: (-0.36, 0.5604)
At x = -0.36, the orange circle is the analytical value 0.560395; the purple diamond is the interpolated value 0.0853333. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0853333; analytical reference = 0.560395; absolute interpolation error = 0.475061. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Kernel evaluation, not a complete SPH flow or solid simulation. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Discrete element method (DEM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Choose a linear frictionless normal-contact spring with stiffness k, no damping, and positive overlap. For this calculation, x denotes the plotted horizontal coordinate (Positive overlap δ / δ* (dimensionless)), and q(x) denotes the plotted response (Normal force / kδ* (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

F/(kδ∗)=δ/δ∗F/(k\delta_*)=\delta/\delta_*a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=0.41,q(c)=0.41q_{\rm lin}(c)=0.41,\quad q(c)=0.41e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 0.2 and q(B) = 0.8. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.41. Direct evaluation gives 0.41; subtracting it from the prediction gives signed error 0.

Discrete element method (DEM): Intermediate-condition prediction and exact check. Horizontal axis: Positive overlap δ / δ* (dimensionless). Vertical axis: Normal force / kδ* (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Positive overlap δ / δ* (dimensionless) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Normal force / kδ* (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.41 Worked point: (0.41, 0.41)
At x = 0.41, the orange circle is the analytical value 0.41; the purple diamond is the interpolated value 0.41. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.41; analytical reference = 0.41; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One elastic contact contribution; many-particle dynamics and tangential friction are excluded. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Lattice Boltzmann method (LBM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a small-amplitude periodic transverse shear wave in the low-Mach hydrodynamic limit; plot νt/L²=0.02. For this calculation, x denotes the plotted horizontal coordinate (Periodic position x / L (dimensionless)), and q(x) denotes the plotted response (Transverse speed / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(ξ,τ)=sin⁡(2πξ)e−4π2τ,τ=0.02u(\xi,\tau)=\sin(2\pi\xi)e^{-4\pi^2\tau},\quad\tau=0.02a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.43939\frac{q(b)-q(a)}{b-a}=-1.43939qlin(c)=0.129546,q(c)=0.243287q_{\rm lin}(c)=0.129546,\quad q(c)=0.243287e=qlin(c)−q(c)=−0.113742e=q_{\rm lin}(c)-q(c)=-0.113742

Solution. The endpoint responses are q(A) = 0.431818 and q(B) = -0.431818. Their secant slope is -1.43939. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.129546. Direct evaluation gives 0.243287; subtracting it from the prediction gives signed error -0.113742.

Lattice Boltzmann method (LBM): Intermediate-condition prediction and exact check. Horizontal axis: Periodic position x / L (dimensionless). Vertical axis: Transverse speed / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Periodic position x / L (dimensionless) −0.4 −0.2 0.0 0.2 0.4 Transverse speed / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1295 Worked point: (0.41, 0.2433)
At x = 0.41, the orange circle is the analytical value 0.243287; the purple diamond is the interpolated value 0.129546. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.129546; analytical reference = 0.243287; absolute interpolation error = 0.113742. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact continuum benchmark for LBM, not a finite-lattice prediction; compressibility and lattice errors must be checked separately. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Material point method (MPM) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Apply a uniform small axial strain of 0.01 to a homogeneous elastic bar. For this calculation, x denotes the plotted horizontal coordinate (Reference position x / L (dimensionless)), and q(x) denotes the plotted response (Displacement u / L (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(x)/L=0.01(x/L)u(x)/L=0.01(x/L)a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.01\frac{q(b)-q(a)}{b-a}=0.01qlin(c)=0.0041,q(c)=0.0041q_{\rm lin}(c)=0.0041,\quad q(c)=0.0041e=qlin(c)−q(c)=−8.67362×10−19e=q_{\rm lin}(c)-q(c)=-8.67362\times10^{-19}

Solution. The endpoint responses are q(A) = 0.002 and q(B) = 0.008. Their secant slope is 0.01. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0041. Direct evaluation gives 0.0041; subtracting it from the prediction gives signed error -8.67362e-19.

Material point method (MPM): Intermediate-condition prediction and exact check. Horizontal axis: Reference position x / L (dimensionless). Vertical axis: Displacement u / L (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Reference position x / L (dimensionless) 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Displacement u / L (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.0041 Worked point: (0.41, 0.0041)
At x = 0.41, the orange circle is the analytical value 0.0041; the purple diamond is the interpolated value 0.0041. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0041; analytical reference = 0.0041; absolute interpolation error = 8.67362e-19. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact continuum target for MPM; grid transfer and particle quadrature errors are not represented. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Homogenization · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Two perfectly bonded parallel axial bars share the same strain, with modulus ratio E2/E1=4. For this calculation, x denotes the plotted horizontal coordinate (Stiff-phase volume fraction f (dimensionless)), and q(x) denotes the plotted response (Effective modulus / soft modulus (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

Eeff/E1=(1−f)+4fE_{eff}/E_1=(1-f)+4fa=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=3\frac{q(b)-q(a)}{b-a}=3qlin(c)=2.23,q(c)=2.23q_{\rm lin}(c)=2.23,\quad q(c)=2.23e=qlin(c)−q(c)=4.44089×10−16e=q_{\rm lin}(c)-q(c)=4.44089\times10^{-16}

Solution. The endpoint responses are q(A) = 1.6 and q(B) = 3.4. Their secant slope is 3. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.23. Direct evaluation gives 2.23; subtracting it from the prediction gives signed error 4.44089e-16.

Homogenization: Intermediate-condition prediction and exact check. Horizontal axis: Stiff-phase volume fraction f (dimensionless). Vertical axis: Effective modulus / soft modulus (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Stiff-phase volume fraction f (dimensionless) 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 Effective modulus / soft modulus (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.23 Worked point: (0.41, 2.23)
At x = 0.41, the orange circle is the analytical value 2.23; the purple diamond is the interpolated value 2.23. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.23; analytical reference = 2.23; absolute interpolation error = 4.44089e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact iso-strain parallel-bar construction; generally an upper-bound estimate for other microstructures. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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QM/MM coupling · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. As a consistency check, choose a common harmonic coordinate whose total coupled-region energy is kq²/2 and whose effective mass is m. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Coordinate / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

q/A=cos⁡(ωt)q/A=\cos(\omega t)a=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−5.88992×10−17\frac{q(b)-q(a)}{b-a}=-5.88992\times10^{-17}qlin(c)=0.309017,q(c)=−0.844328q_{\rm lin}(c)=0.309017,\quad q(c)=-0.844328e=qlin(c)−q(c)=1.15334e=q_{\rm lin}(c)-q(c)=1.15334

Solution. The endpoint responses are q(A) = 0.309017 and q(B) = 0.309017. Their secant slope is -5.88992e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.309017. Direct evaluation gives -0.844328; subtracting it from the prediction gives signed error 1.15334.

QM/MM coupling: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Coordinate / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Phase ωt (radian) −1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 Coordinate / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.309 Worked point: (2.576, -0.8443)
At x = 2.57611, the orange circle is the analytical value -0.844328; the purple diamond is the interpolated value 0.309017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.309017; analytical reference = -0.844328; absolute interpolation error = 1.15334. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Prescribed harmonic reference only; no electronic calculation, interface force transfer, or adaptive region simulation is performed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Atomistic–continuum coupling · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. As a consistency check, choose a common harmonic coordinate whose total coupled-region energy is kq²/2 and whose effective mass is m. For this calculation, x denotes the plotted horizontal coordinate (Phase ωt (radian)), and q(x) denotes the plotted response (Coordinate / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 1.25664 and x = 5.02655 are tabulated. Use linear interpolation to predict the response at x = 2.57611, then check it against the original equation.

q/A=cos⁡(ωt)q/A=\cos(\omega t)a=1.25664,b=5.02655,c=2.57611a=1.25664,\quad b=5.02655,\quad c=2.57611qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−5.88992×10−17\frac{q(b)-q(a)}{b-a}=-5.88992\times10^{-17}qlin(c)=0.309017,q(c)=−0.844328q_{\rm lin}(c)=0.309017,\quad q(c)=-0.844328e=qlin(c)−q(c)=1.15334e=q_{\rm lin}(c)-q(c)=1.15334

Solution. The endpoint responses are q(A) = 0.309017 and q(B) = 0.309017. Their secant slope is -5.88992e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.309017. Direct evaluation gives -0.844328; subtracting it from the prediction gives signed error 1.15334.

Atomistic–continuum coupling: Intermediate-condition prediction and exact check. Horizontal axis: Phase ωt (radian). Vertical axis: Coordinate / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Phase ωt (radian) −1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 Coordinate / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.309 Worked point: (2.576, -0.8443)
At x = 2.57611, the orange circle is the analytical value -0.844328; the purple diamond is the interpolated value 0.309017. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.309017; analytical reference = -0.844328; absolute interpolation error = 1.15334. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Prescribed harmonic reference only; no electronic calculation, interface force transfer, or adaptive region simulation is performed. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Fluid–structure interaction (FSI) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Approximate fluid loading as a constant added mass ma=m on an undamped spring-supported body. For this calculation, x denotes the plotted horizontal coordinate (Dry structural phase τ = √(k/m)t (radian)), and q(x) denotes the plotted response (Displacement / initial amplitude (dimensionless)). Suppose only the analytical endpoint responses at x = 2.51327 and x = 10.0531 are tabulated. Use linear interpolation to predict the response at x = 5.15221, then check it against the original equation.

q/A=cos⁡(τ1+ma/m),ma/m=1q/A=\cos\left(\frac{\tau}{\sqrt{1+m_a/m}}\right),\quad m_a/m=1a=2.51327,b=10.0531,c=5.15221a=2.51327,\quad b=10.0531,\quad c=5.15221qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.11713\frac{q(b)-q(a)}{b-a}=0.11713qlin(c)=0.104203,q(c)=−0.876828q_{\rm lin}(c)=0.104203,\quad q(c)=-0.876828e=qlin(c)−q(c)=0.981031e=q_{\rm lin}(c)-q(c)=0.981031

Solution. The endpoint responses are q(A) = -0.204895 and q(B) = 0.678243. Their secant slope is 0.11713. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.104203. Direct evaluation gives -0.876828; subtracting it from the prediction gives signed error 0.981031.

Fluid–structure interaction (FSI): Intermediate-condition prediction and exact check. Horizontal axis: Dry structural phase τ = √(k/m)t (radian). Vertical axis: Displacement / initial amplitude (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 3 4 5 6 7 8 9 10 Dry structural phase τ = √(k/m)t (radian) −1.0 −0.5 0.0 0.5 1.0 Displacement / initial amplitude (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1042 Worked point: (5.152, -0.8768)
At x = 5.15221, the orange circle is the analytical value -0.876828; the purple diamond is the interpolated value 0.104203. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.104203; analytical reference = -0.876828; absolute interpolation error = 0.981031. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Linear added-mass reduction of FSI; no viscous drag, free-surface, or flow-field solution. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Thermomechanical coupling · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. A one-dimensional elastic bar is prevented from expanding while its temperature rises uniformly. For this calculation, x denotes the plotted horizontal coordinate (Temperature rise / reference rise (dimensionless)), and q(x) denotes the plotted response (Scaled axial stress (dimensionless)). Suppose only the analytical endpoint responses at x = 0.4 and x = 1.6 are tabulated. Use linear interpolation to predict the response at x = 0.82, then check it against the original equation.

σ/(EαΔT∗)=−ΔT/ΔT∗\sigma/(E\alpha\Delta T_*)=-\Delta T/\Delta T_*a=0.4,b=1.6,c=0.82a=0.4,\quad b=1.6,\quad c=0.82qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1\frac{q(b)-q(a)}{b-a}=-1qlin(c)=−0.82,q(c)=−0.82q_{\rm lin}(c)=-0.82,\quad q(c)=-0.82e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = -0.4 and q(B) = -1.6. Their secant slope is -1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.82. Direct evaluation gives -0.82; subtracting it from the prediction gives signed error 0.

Thermomechanical coupling: Intermediate-condition prediction and exact check. Horizontal axis: Temperature rise / reference rise (dimensionless). Vertical axis: Scaled axial stress (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Temperature rise / reference rise (dimensionless) −1.6 −1.4 −1.2 −1.0 −0.8 −0.6 −0.4 Scaled axial stress (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.82 Worked point: (0.82, -0.82)
At x = 0.82, the orange circle is the analytical value -0.82; the purple diamond is the interpolated value -0.82. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.82; analytical reference = -0.82; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Small-strain constant-property axial model, with tension positive; uniform heating produces compression. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Proper orthogonal decomposition (POD) · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. All snapshots are scalar multiples of sin(πx). Reconstruct the snapshot at dimensionless time one using one POD mode. For this calculation, x denotes the plotted horizontal coordinate (Position x / L (dimensionless)), and q(x) denotes the plotted response (Reconstructed field (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

u(x,t)=e−tsin⁡(πx),t=1u(x,t)=e^{-t}\sin(\pi x),\quad t=1a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=9.25186×10−17\frac{q(b)-q(a)}{b-a}=9.25186\times10^{-17}qlin(c)=0.216234,q(c)=0.353272q_{\rm lin}(c)=0.216234,\quad q(c)=0.353272e=qlin(c)−q(c)=−0.137038e=q_{\rm lin}(c)-q(c)=-0.137038

Solution. The endpoint responses are q(A) = 0.216234 and q(B) = 0.216234. Their secant slope is 9.25186e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.216234. Direct evaluation gives 0.353272; subtracting it from the prediction gives signed error -0.137038.

Proper orthogonal decomposition (POD): Intermediate-condition prediction and exact check. Horizontal axis: Position x / L (dimensionless). Vertical axis: Reconstructed field (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Position x / L (dimensionless) 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 Reconstructed field (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2162 Worked point: (0.41, 0.3533)
At x = 0.41, the orange circle is the analytical value 0.353272; the purple diamond is the interpolated value 0.216234. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.216234; analytical reference = 0.353272; absolute interpolation error = 0.137038. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact rank-one constructed data set; real POD truncation can incur substantial error. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Gaussian-process surrogate · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a zero-mean, unit-variance squared-exponential GP, unit length scale, and one noiseless observation y(0)=1. For this calculation, x denotes the plotted horizontal coordinate (Input / kernel length scale (dimensionless)), and q(x) denotes the plotted response (Posterior mean (dimensionless)). Suppose only the analytical endpoint responses at x = -1.8 and x = 1.8 are tabulated. Use linear interpolation to predict the response at x = -0.54, then check it against the original equation.

μ(x)=e−x2/2\mu(x)=e^{-x^2/2}a=−1.8,b=1.8,c=−0.54a=-1.8,\quad b=1.8,\quad c=-0.54qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−8.48087×10−17\frac{q(b)-q(a)}{b-a}=-8.48087\times10^{-17}qlin(c)=0.197899,q(c)=0.864331q_{\rm lin}(c)=0.197899,\quad q(c)=0.864331e=qlin(c)−q(c)=−0.666432e=q_{\rm lin}(c)-q(c)=-0.666432

Solution. The endpoint responses are q(A) = 0.197899 and q(B) = 0.197899. Their secant slope is -8.48087e-17. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.197899. Direct evaluation gives 0.864331; subtracting it from the prediction gives signed error -0.666432.

Gaussian-process surrogate: Intermediate-condition prediction and exact check. Horizontal axis: Input / kernel length scale (dimensionless). Vertical axis: Posterior mean (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 Input / kernel length scale (dimensionless) 0.2 0.4 0.6 0.8 1.0 Posterior mean (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1979 Worked point: (-0.54, 0.8643)
At x = -0.54, the orange circle is the analytical value 0.864331; the purple diamond is the interpolated value 0.197899. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.197899; analytical reference = 0.864331; absolute interpolation error = 0.666432. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical posterior mean under the stated kernel; it is not a physical law, and posterior uncertainty is not shown. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Digital twin framework · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a lumped thermal model with a known constant cooling time as an ideal reference for a thermal digital twin. For this calculation, x denotes the plotted horizontal coordinate (Elapsed time / thermal constant (dimensionless)), and q(x) denotes the plotted response (Temperature excess / initial excess (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

(T−T∞)/(T0−T∞)=e−t/τ(T-T_\infty)/(T_0-T_\infty)=e^{-t/\tau}a=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−0.116521\frac{q(b)-q(a)}{b-a}=-0.116521qlin(c)=0.245532,q(c)=0.128735q_{\rm lin}(c)=0.245532,\quad q(c)=0.128735e=qlin(c)−q(c)=0.116797e=q_{\rm lin}(c)-q(c)=0.116797

Solution. The endpoint responses are q(A) = 0.367879 and q(B) = 0.0183156. Their secant slope is -0.116521. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.245532. Direct evaluation gives 0.128735; subtracting it from the prediction gives signed error 0.116797.

Digital twin framework: Intermediate-condition prediction and exact check. Horizontal axis: Elapsed time / thermal constant (dimensionless). Vertical axis: Temperature excess / initial excess (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Elapsed time / thermal constant (dimensionless) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 Temperature excess / initial excess (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.2455 Worked point: (2.05, 0.1287)
At x = 2.05, the orange circle is the analytical value 0.128735; the purple diamond is the interpolated value 0.245532. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.245532; analytical reference = 0.128735; absolute interpolation error = 0.116797. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Reference physics only. No sensors, online updates, or actual equipment measurements are included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Bayesian model calibration · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use prior θ~Normal(0,1) and one measurement y=1 with independent Normal(0,1) measurement noise. For this calculation, x denotes the plotted horizontal coordinate (Unknown parameter θ (dimensionless)), and q(x) denotes the plotted response (Posterior density (per unit θ)). Suppose only the analytical endpoint responses at x = -1 and x = 2 are tabulated. Use linear interpolation to predict the response at x = 0.05, then check it against the original equation.

p(θ∣y=1)=π−1/2e−(θ−0.5)2p(\theta\mid y=1)=\pi^{-1/2}e^{-(\theta-0.5)^2}a=−1,b=2,c=0.05a=-1,\quad b=2,\quad c=0.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0\frac{q(b)-q(a)}{b-a}=0qlin(c)=0.0594651,q(c)=0.460766q_{\rm lin}(c)=0.0594651,\quad q(c)=0.460766e=qlin(c)−q(c)=−0.401301e=q_{\rm lin}(c)-q(c)=-0.401301

Solution. The endpoint responses are q(A) = 0.0594651 and q(B) = 0.0594651. Their secant slope is 0. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.0594651. Direct evaluation gives 0.460766; subtracting it from the prediction gives signed error -0.401301.

Bayesian model calibration: Intermediate-condition prediction and exact check. Horizontal axis: Unknown parameter θ (dimensionless). Vertical axis: Posterior density (per unit θ). image/svg+xml IICSM analytical illustration / Matplotlib −1.0 −0.5 0.0 0.5 1.0 1.5 2.0 Unknown parameter θ (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 Posterior density (per unit θ) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.05947 Worked point: (0.05, 0.4608)
At x = 0.05, the orange circle is the analytical value 0.460766; the purple diamond is the interpolated value 0.0594651. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.0594651; analytical reference = 0.460766; absolute interpolation error = 0.401301. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact conjugate scalar calibration example; not a calibrated engineering system. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Polynomial chaos expansion · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a linear response to a uniform random input. Expand in the first two Legendre polynomials. For this calculation, x denotes the plotted horizontal coordinate (Uniform random input ξ (dimensionless)), and q(x) denotes the plotted response (Response Y (dimensionless)). Suppose only the analytical endpoint responses at x = -0.6 and x = 0.6 are tabulated. Use linear interpolation to predict the response at x = -0.18, then check it against the original equation.

Y(ξ)=2+0.5ξ,ξ∼U[−1,1]Y(\xi)=2+0.5\xi,\quad\xi\sim U[-1,1]a=−0.6,b=0.6,c=−0.18a=-0.6,\quad b=0.6,\quad c=-0.18qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.5\frac{q(b)-q(a)}{b-a}=0.5qlin(c)=1.91,q(c)=1.91q_{\rm lin}(c)=1.91,\quad q(c)=1.91e=qlin(c)−q(c)=−2.22045×10−16e=q_{\rm lin}(c)-q(c)=-2.22045\times10^{-16}

Solution. The endpoint responses are q(A) = 1.7 and q(B) = 2.3. Their secant slope is 0.5. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.91. Direct evaluation gives 1.91; subtracting it from the prediction gives signed error -2.22045e-16.

Polynomial chaos expansion: Intermediate-condition prediction and exact check. Horizontal axis: Uniform random input ξ (dimensionless). Vertical axis: Response Y (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 Uniform random input ξ (dimensionless) 1.7 1.8 1.9 2.0 2.1 2.2 2.3 Response Y (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.91 Worked point: (-0.18, 1.91)
At x = -0.18, the orange circle is the analytical value 1.91; the purple diamond is the interpolated value 1.91. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.91; analytical reference = 1.91; absolute interpolation error = 2.22045e-16. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact degree-one expansion for the chosen response, not a surrogate fitted to arbitrary simulation data. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Geometrically scaled physical model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Scale all dimensions of a shape by the same positive length ratio. For this calculation, x denotes the plotted horizontal coordinate (Model / prototype length (dimensionless)), and q(x) denotes the plotted response (Model / prototype volume (dimensionless)). Suppose only the analytical endpoint responses at x = 0.2 and x = 0.8 are tabulated. Use linear interpolation to predict the response at x = 0.41, then check it against the original equation.

Vm/Vp=(Lm/Lp)3V_m/V_p=(L_m/L_p)^3a=0.2,b=0.8,c=0.41a=0.2,\quad b=0.8,\quad c=0.41qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.84\frac{q(b)-q(a)}{b-a}=0.84qlin(c)=0.1844,q(c)=0.068921q_{\rm lin}(c)=0.1844,\quad q(c)=0.068921e=qlin(c)−q(c)=0.115479e=q_{\rm lin}(c)-q(c)=0.115479

Solution. The endpoint responses are q(A) = 0.008 and q(B) = 0.512. Their secant slope is 0.84. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.1844. Direct evaluation gives 0.068921; subtracting it from the prediction gives signed error 0.115479.

Geometrically scaled physical model: Intermediate-condition prediction and exact check. Horizontal axis: Model / prototype length (dimensionless). Vertical axis: Model / prototype volume (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Model / prototype length (dimensionless) 0.0 0.1 0.2 0.3 0.4 0.5 Model / prototype volume (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1844 Worked point: (0.41, 0.06892)
At x = 0.41, the orange circle is the analytical value 0.068921; the purple diamond is the interpolated value 0.1844. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.1844; analytical reference = 0.068921; absolute interpolation error = 0.115479. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Geometric similarity alone does not ensure force, material, or dynamic similarity. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Wind-tunnel model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use fixed air density and reference dynamic pressure q*=ρU*²/2. For this calculation, x denotes the plotted horizontal coordinate (Tunnel speed / reference speed (dimensionless)), and q(x) denotes the plotted response (Dynamic pressure / reference pressure (dimensionless)). Suppose only the analytical endpoint responses at x = 0.4 and x = 1.6 are tabulated. Use linear interpolation to predict the response at x = 0.82, then check it against the original equation.

q/q∗=(U/U∗)2q/q_*=(U/U_*)^2a=0.4,b=1.6,c=0.82a=0.4,\quad b=1.6,\quad c=0.82qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=2\frac{q(b)-q(a)}{b-a}=2qlin(c)=1,q(c)=0.6724q_{\rm lin}(c)=1,\quad q(c)=0.6724e=qlin(c)−q(c)=0.3276e=q_{\rm lin}(c)-q(c)=0.3276

Solution. The endpoint responses are q(A) = 0.16 and q(B) = 2.56. Their secant slope is 2. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1. Direct evaluation gives 0.6724; subtracting it from the prediction gives signed error 0.3276.

Wind-tunnel model: Intermediate-condition prediction and exact check. Horizontal axis: Tunnel speed / reference speed (dimensionless). Vertical axis: Dynamic pressure / reference pressure (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Tunnel speed / reference speed (dimensionless) 0.0 0.5 1.0 1.5 2.0 2.5 Dynamic pressure / reference pressure (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1 Worked point: (0.82, 0.6724)
At x = 0.82, the orange circle is the analytical value 0.6724; the purple diamond is the interpolated value 1. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1; analytical reference = 0.6724; absolute interpolation error = 0.3276. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Test-planning relation, not measured wind-tunnel data; Reynolds and Mach similarity require separate checks. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hydraulic flume model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the same gravitational acceleration and match Froude number U/√(gL) between a model and prototype. For this calculation, x denotes the plotted horizontal coordinate (Model / prototype length (dimensionless)), and q(x) denotes the plotted response (Model / prototype speed (dimensionless)). Suppose only the analytical endpoint responses at x = 0.208 and x = 0.802 are tabulated. Use linear interpolation to predict the response at x = 0.4159, then check it against the original equation.

Um/Up=Lm/LpU_m/U_p=\sqrt{L_m/L_p}a=0.208,b=0.802,c=0.4159a=0.208,\quad b=0.802,\quad c=0.4159qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.739856\frac{q(b)-q(a)}{b-a}=0.739856qlin(c)=0.609886,q(c)=0.644903q_{\rm lin}(c)=0.609886,\quad q(c)=0.644903e=qlin(c)−q(c)=−0.0350169e=q_{\rm lin}(c)-q(c)=-0.0350169

Solution. The endpoint responses are q(A) = 0.45607 and q(B) = 0.895545. Their secant slope is 0.739856. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.609886. Direct evaluation gives 0.644903; subtracting it from the prediction gives signed error -0.0350169.

Hydraulic flume model: Intermediate-condition prediction and exact check. Horizontal axis: Model / prototype length (dimensionless). Vertical axis: Model / prototype speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Model / prototype length (dimensionless) 0.5 0.6 0.7 0.8 0.9 Model / prototype speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.6099 Worked point: (0.4159, 0.6449)
At x = 0.4159, the orange circle is the analytical value 0.644903; the purple diamond is the interpolated value 0.609886. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.609886; analytical reference = 0.644903; absolute interpolation error = 0.0350169. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Gravity-dominated similarity appropriate to free-surface flumes; Reynolds, Weber, and other dimensionless groups may not also match. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Dimensional-analysis similarity model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the same gravitational acceleration and match Froude number U/√(gL) between a model and prototype. For this calculation, x denotes the plotted horizontal coordinate (Model / prototype length (dimensionless)), and q(x) denotes the plotted response (Model / prototype speed (dimensionless)). Suppose only the analytical endpoint responses at x = 0.208 and x = 0.802 are tabulated. Use linear interpolation to predict the response at x = 0.4159, then check it against the original equation.

Um/Up=Lm/LpU_m/U_p=\sqrt{L_m/L_p}a=0.208,b=0.802,c=0.4159a=0.208,\quad b=0.802,\quad c=0.4159qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.739856\frac{q(b)-q(a)}{b-a}=0.739856qlin(c)=0.609886,q(c)=0.644903q_{\rm lin}(c)=0.609886,\quad q(c)=0.644903e=qlin(c)−q(c)=−0.0350169e=q_{\rm lin}(c)-q(c)=-0.0350169

Solution. The endpoint responses are q(A) = 0.45607 and q(B) = 0.895545. Their secant slope is 0.739856. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.609886. Direct evaluation gives 0.644903; subtracting it from the prediction gives signed error -0.0350169.

Dimensional-analysis similarity model: Intermediate-condition prediction and exact check. Horizontal axis: Model / prototype length (dimensionless). Vertical axis: Model / prototype speed (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Model / prototype length (dimensionless) 0.5 0.6 0.7 0.8 0.9 Model / prototype speed (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.6099 Worked point: (0.4159, 0.6449)
At x = 0.4159, the orange circle is the analytical value 0.644903; the purple diamond is the interpolated value 0.609886. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.609886; analytical reference = 0.644903; absolute interpolation error = 0.0350169. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Gravity-dominated similarity appropriate to free-surface flumes; Reynolds, Weber, and other dimensionless groups may not also match. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Shake-table structural model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For an undamped single-degree-of-freedom oscillator with sinusoidal base motion, calculate the steady absolute displacement below resonance. For this calculation, x denotes the plotted horizontal coordinate (Excitation / natural frequency (dimensionless)), and q(x) denotes the plotted response (Absolute displacement amplitude ratio (dimensionless)). Suppose only the analytical endpoint responses at x = 0.16 and x = 0.64 are tabulated. Use linear interpolation to predict the response at x = 0.328, then check it against the original equation.

∣X/Y∣=1/∣1−r2∣,r=Ω/ωn|X/Y|=1/|1-r^2|,\quad r=\Omega/\omega_na=0.16,b=0.64,c=0.328a=0.16,\quad b=0.64,\quad c=0.328qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.39061\frac{q(b)-q(a)}{b-a}=1.39061qlin(c)=1.2599,q(c)=1.12055q_{\rm lin}(c)=1.2599,\quad q(c)=1.12055e=qlin(c)−q(c)=0.139342e=q_{\rm lin}(c)-q(c)=0.139342

Solution. The endpoint responses are q(A) = 1.02627 and q(B) = 1.69377. Their secant slope is 1.39061. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.2599. Direct evaluation gives 1.12055; subtracting it from the prediction gives signed error 0.139342.

Shake-table structural model: Intermediate-condition prediction and exact check. Horizontal axis: Excitation / natural frequency (dimensionless). Vertical axis: Absolute displacement amplitude ratio (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.2 0.3 0.4 0.5 0.6 Excitation / natural frequency (dimensionless) 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Absolute displacement amplitude ratio (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.26 Worked point: (0.328, 1.121)
At x = 0.328, the orange circle is the analytical value 1.12055; the purple diamond is the interpolated value 1.2599. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.2599; analytical reference = 1.12055; absolute interpolation error = 0.139342. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Ideal steady reference, not shake-table measurements; the undamped resonance singularity is outside the plotted range. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Photoelastic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a transparent specimen of thickness t, stress-optic coefficient C, and monochromatic wavelength λ. For this calculation, x denotes the plotted horizontal coordinate (Stress–optic retardation CtΔσ / λ (dimensionless)), and q(x) denotes the plotted response (Fringe order N (dimensionless)). Suppose only the analytical endpoint responses at x = 1 and x = 4 are tabulated. Use linear interpolation to predict the response at x = 2.05, then check it against the original equation.

N=Ct(σ1−σ2)/λN=Ct(\sigma_1-\sigma_2)/\lambdaa=1,b=4,c=2.05a=1,\quad b=4,\quad c=2.05qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1\frac{q(b)-q(a)}{b-a}=1qlin(c)=2.05,q(c)=2.05q_{\rm lin}(c)=2.05,\quad q(c)=2.05e=qlin(c)−q(c)=0e=q_{\rm lin}(c)-q(c)=0

Solution. The endpoint responses are q(A) = 1 and q(B) = 4. Their secant slope is 1. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.05. Direct evaluation gives 2.05; subtracting it from the prediction gives signed error 0.

Photoelastic model: Intermediate-condition prediction and exact check. Horizontal axis: Stress–optic retardation CtΔσ / λ (dimensionless). Vertical axis: Fringe order N (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Stress–optic retardation CtΔσ / λ (dimensionless) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Fringe order N (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.05 Worked point: (2.05, 2.05)
At x = 2.05, the orange circle is the analytical value 2.05; the purple diamond is the interpolated value 2.05. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.05; analytical reference = 2.05; absolute interpolation error = 0. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Uniform stress through thickness and linear stress-optic law; this is not a fringe photograph. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Ornstein-Zernike equation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Assume rho times the Fourier-transformed direct correlation is −exp[−(kℓ)²]. Find S(k) from the OZ relation. For this calculation, x denotes the plotted horizontal coordinate (Wavevector magnitude kℓ (dimensionless)), and q(x) denotes the plotted response (Structure factor S(k) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

ρc^(k)=−e−(kℓ)2,S(k)=11+e−(kℓ)2\rho\widehat c(k)=-e^{-(k\ell)^2},\quad S(k)=\frac{1}{1+e^{-(k\ell)^2}}a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.226566\frac{q(b)-q(a)}{b-a}=0.226566qlin(c)=0.731777,q(c)=0.819491q_{\rm lin}(c)=0.731777,\quad q(c)=0.819491e=qlin(c)−q(c)=−0.0877137e=q_{\rm lin}(c)-q(c)=-0.0877137

Solution. The endpoint responses are q(A) = 0.58904 and q(B) = 0.996859. Their secant slope is 0.226566. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.731777. Direct evaluation gives 0.819491; subtracting it from the prediction gives signed error -0.0877137.

Ornstein-Zernike equation: Intermediate-condition prediction and exact check. Horizontal axis: Wavevector magnitude kℓ (dimensionless). Vertical axis: Structure factor S(k) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Wavevector magnitude kℓ (dimensionless) 0.6 0.7 0.8 0.9 1.0 Structure factor S(k) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7318 Worked point: (1.23, 0.8195)
At x = 1.23, the orange circle is the analytical value 0.819491; the purple diamond is the interpolated value 0.731777. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.731777; analytical reference = 0.819491; absolute interpolation error = 0.0877137. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. A prescribed-correlation algebraic benchmark, not a self-consistent closure solution or measured scattering spectrum. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Percus-Yevick closure · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use the analytical three-dimensional, monodisperse hard-sphere PY solution to evaluate its contact pair distribution as packing fraction varies. For this calculation, x denotes the plotted horizontal coordinate (Hard-sphere packing fraction φ (dimensionless)), and q(x) denotes the plotted response (Contact pair distribution g(σ+) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.09 and x = 0.36 are tabulated. Use linear interpolation to predict the response at x = 0.1845, then check it against the original equation.

g(σ+)=1+ϕ/2(1−ϕ)2g(\sigma^+)=\frac{1+\phi/2}{(1-\phi)^2}a=0.09,b=0.36,c=0.1845a=0.09,\quad b=0.36,\quad c=0.1845qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=5.99605\frac{q(b)-q(a)}{b-a}=5.99605qlin(c)=1.82855,q(c)=1.64238q_{\rm lin}(c)=1.82855,\quad q(c)=1.64238e=qlin(c)−q(c)=0.18617e=q_{\rm lin}(c)-q(c)=0.18617

Solution. The endpoint responses are q(A) = 1.26192 and q(B) = 2.88086. Their secant slope is 5.99605. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 1.82855. Direct evaluation gives 1.64238; subtracting it from the prediction gives signed error 0.18617.

Percus-Yevick closure: Intermediate-condition prediction and exact check. Horizontal axis: Hard-sphere packing fraction φ (dimensionless). Vertical axis: Contact pair distribution g(σ+) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.10 0.15 0.20 0.25 0.30 0.35 Hard-sphere packing fraction φ (dimensionless) 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 Contact pair distribution g(σ+) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 1.829 Worked point: (0.1845, 1.642)
At x = 0.1845, the orange circle is the analytical value 1.64238; the purple diamond is the interpolated value 1.82855. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 1.82855; analytical reference = 1.64238; absolute interpolation error = 0.18617. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Contact-value evaluation of the PY approximation; the analytical OZ/PY solution is taken as the starting result. Thermodynamic routes are not identical. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Hypernetted-chain (HNC) closure · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take the zero-density limit of an equilibrium soft Gaussian-core fluid with beta epsilon=1. Find its pair distribution. For this calculation, x denotes the plotted horizontal coordinate (Separation r / σ (dimensionless)), and q(x) denotes the plotted response (Pair distribution g(r) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.6 and x = 2.4 are tabulated. Use linear interpolation to predict the response at x = 1.23, then check it against the original equation.

βu(r)=e−(r/σ)2,g(r)=exp⁡[−e−(r/σ)2]\beta u(r)=e^{-(r/\sigma)^2},\quad g(r)=\exp[-e^{-(r/\sigma)^2}]a=0.6,b=2.4,c=1.23a=0.6,\quad b=2.4,\quad c=1.23qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.277285\frac{q(b)-q(a)}{b-a}=0.277285qlin(c)=0.67243,q(c)=0.802302q_{\rm lin}(c)=0.67243,\quad q(c)=0.802302e=qlin(c)−q(c)=−0.129872e=q_{\rm lin}(c)-q(c)=-0.129872

Solution. The endpoint responses are q(A) = 0.497741 and q(B) = 0.996854. Their secant slope is 0.277285. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.67243. Direct evaluation gives 0.802302; subtracting it from the prediction gives signed error -0.129872.

Hypernetted-chain (HNC) closure: Intermediate-condition prediction and exact check. Horizontal axis: Separation r / σ (dimensionless). Vertical axis: Pair distribution g(r) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 Separation r / σ (dimensionless) 0.5 0.6 0.7 0.8 0.9 1.0 Pair distribution g(r) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.6724 Worked point: (1.23, 0.8023)
At x = 1.23, the orange circle is the analytical value 0.802302; the purple diamond is the interpolated value 0.67243. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.67243; analytical reference = 0.802302; absolute interpolation error = 0.129872. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact dilute two-particle limit for this specified potential; at finite liquid density, solve the coupled HNC/OZ equations instead. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Carnahan-Starling hard-sphere equation of state · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For a monodisperse hard-sphere fluid, compute pressure relative to ideal-gas pressure from packing fraction using Carnahan-Starling. For this calculation, x denotes the plotted horizontal coordinate (Hard-sphere packing fraction φ (dimensionless)), and q(x) denotes the plotted response (Compressibility factor Z = p / (ρkBT) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.09 and x = 0.36 are tabulated. Use linear interpolation to predict the response at x = 0.1845, then check it against the original equation.

Z(ϕ)=1+ϕ+ϕ2−ϕ3(1−ϕ)3Z(\phi)=\frac{1+\phi+\phi^2-\phi^3}{(1-\phi)^3}a=0.09,b=0.36,c=0.1845a=0.09,\quad b=0.36,\quad c=0.1845qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=14.9932\frac{q(b)-q(a)}{b-a}=14.9932qlin(c)=2.87309,q(c)=2.23524q_{\rm lin}(c)=2.87309,\quad q(c)=2.23524e=qlin(c)−q(c)=0.637848e=q_{\rm lin}(c)-q(c)=0.637848

Solution. The endpoint responses are q(A) = 1.45623 and q(B) = 5.50439. Their secant slope is 14.9932. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 2.87309. Direct evaluation gives 2.23524; subtracting it from the prediction gives signed error 0.637848.

Carnahan-Starling hard-sphere equation of state: Intermediate-condition prediction and exact check. Horizontal axis: Hard-sphere packing fraction φ (dimensionless). Vertical axis: Compressibility factor Z = p / (ρkBT) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.10 0.15 0.20 0.25 0.30 0.35 Hard-sphere packing fraction φ (dimensionless) 1 2 3 4 5 Compressibility factor Z = p / (ρkBT) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 2.873 Worked point: (0.1845, 2.235)
At x = 0.1845, the orange circle is the analytical value 2.23524; the purple diamond is the interpolated value 2.87309. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 2.87309; analytical reference = 2.23524; absolute interpolation error = 0.637848. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Constitutive evaluation of the approximate fluid EOS; no attractive forces, mixture effects, or solid phase are included. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Stokes-Einstein diffusion relation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take T=298 K and solvent viscosity eta=0.001 Pa s. Estimate D for dilute no-slip spherical probes with radii between 10 and 200 nm. For this calculation, x denotes the plotted horizontal coordinate (Hydrodynamic radius R (nm)), and q(x) denotes the plotted response (Translational diffusivity D (m²/s)). Suppose only the analytical endpoint responses at x = 48 and x = 162 are tabulated. Use linear interpolation to predict the response at x = 87.9, then check it against the original equation.

D(R)=(1.380649×10−23)(298)6π(10−3)(R×10−9)  m2 s−1D(R)=\frac{(1.380649\times10^{-23})(298)}{6\pi(10^{-3})(R\times10^{-9})}\;\mathrm{m^2\,s^{-1}}a=48,b=162,c=87.9a=48,\quad b=162,\quad c=87.9qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−2.807×10−14\frac{q(b)-q(a)}{b-a}=-2.807\times10^{-14}qlin(c)=3.42735×10−12,q(c)=2.48319×10−12q_{\rm lin}(c)=3.42735\times10^{-12},\quad q(c)=2.48319\times10^{-12}e=qlin(c)−q(c)=9.44157×10−13e=q_{\rm lin}(c)-q(c)=9.44157\times10^{-13}

Solution. The endpoint responses are q(A) = 4.54734e-12 and q(B) = 1.34736e-12. Their secant slope is -2.807e-14. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 3.42735e-12. Direct evaluation gives 2.48319e-12; subtracting it from the prediction gives signed error 9.44157e-13.

Stokes-Einstein diffusion relation: Intermediate-condition prediction and exact check. Horizontal axis: Hydrodynamic radius R (nm). Vertical axis: Translational diffusivity D (m²/s). image/svg+xml IICSM analytical illustration / Matplotlib 60 80 100 120 140 160 Hydrodynamic radius R (nm) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Translational diffusivity D (m²/s) 1e−12 Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 3.427e-12 Worked point: (87.9, 2.483e-12)
At x = 87.9, the orange circle is the analytical value 2.48319e-12; the purple diamond is the interpolated value 3.42735e-12. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 3.42735e-12; analytical reference = 2.48319e-12; absolute interpolation error = 9.44157e-13. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Chosen constant solvent viscosity, not a measured water-property curve. Continuum, no-slip, dilute-sphere assumptions apply. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Green-Kubo viscosity relation · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Assume the equilibrium intensive shear-pressure autocorrelation C(t)=C0 exp(−t/τ), with C0>0. Calculate the running Green-Kubo viscosity integral. For this calculation, x denotes the plotted horizontal coordinate (Integration time t / τ (dimensionless)), and q(x) denotes the plotted response (Running viscosity η(t) / η∞ (dimensionless)). Suppose only the analytical endpoint responses at x = 1.2 and x = 4.8 are tabulated. Use linear interpolation to predict the response at x = 2.46, then check it against the original equation.

C(t)=C0e−t/τ,η(t)η∞=1−e−t/τ,η∞=VC0τkBTC(t)=C_0e^{-t/\tau},\quad\frac{\eta(t)}{\eta_\infty}=1-e^{-t/\tau},\quad\eta_\infty=\frac{VC_0\tau}{k_{\mathrm B}T}a=1.2,b=4.8,c=2.46a=1.2,\quad b=4.8,\quad c=2.46qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.081379\frac{q(b)-q(a)}{b-a}=0.081379qlin(c)=0.801343,q(c)=0.914565q_{\rm lin}(c)=0.801343,\quad q(c)=0.914565e=qlin(c)−q(c)=−0.113222e=q_{\rm lin}(c)-q(c)=-0.113222

Solution. The endpoint responses are q(A) = 0.698806 and q(B) = 0.99177. Their secant slope is 0.081379. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.801343. Direct evaluation gives 0.914565; subtracting it from the prediction gives signed error -0.113222.

Green-Kubo viscosity relation: Intermediate-condition prediction and exact check. Horizontal axis: Integration time t / τ (dimensionless). Vertical axis: Running viscosity η(t) / η∞ (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Integration time t / τ (dimensionless) 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Running viscosity η(t) / η∞ (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.8013 Worked point: (2.46, 0.9146)
At x = 2.46, the orange circle is the analytical value 0.914565; the purple diamond is the interpolated value 0.801343. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.801343; analytical reference = 0.914565; absolute interpolation error = 0.113222. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Analytical exponential-correlation benchmark; real liquid stress correlations may oscillate or have long tails. This is not a molecular-dynamics measurement. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Einstein crystal heat-capacity model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. For 3N identical oscillators, calculate the normalized heat capacity versus temperature. For this calculation, x denotes the plotted horizontal coordinate (Temperature T / ΘE (dimensionless)), and q(x) denotes the plotted response (Heat capacity CV / (3NkB) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.48 and x = 1.62 are tabulated. Use linear interpolation to predict the response at x = 0.879, then check it against the original equation.

θ=T/ΘE,CV3NkB=θ−2e−1/θ(1−e−1/θ)2\theta=T/\Theta_{\mathrm E},\quad\frac{C_V}{3Nk_{\mathrm B}}=\frac{\theta^{-2}e^{-1/\theta}}{(1-e^{-1/\theta})^2}a=0.48,b=1.62,c=0.879a=0.48,\quad b=1.62,\quad c=0.879qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.231369\frac{q(b)-q(a)}{b-a}=0.231369qlin(c)=0.797398,q(c)=0.898782q_{\rm lin}(c)=0.797398,\quad q(c)=0.898782e=qlin(c)−q(c)=−0.101384e=q_{\rm lin}(c)-q(c)=-0.101384

Solution. The endpoint responses are q(A) = 0.705082 and q(B) = 0.968843. Their secant slope is 0.231369. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.797398. Direct evaluation gives 0.898782; subtracting it from the prediction gives signed error -0.101384.

Einstein crystal heat-capacity model: Intermediate-condition prediction and exact check. Horizontal axis: Temperature T / ΘE (dimensionless). Vertical axis: Heat capacity CV / (3NkB) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.6 0.8 1.0 1.2 1.4 1.6 Temperature T / ΘE (dimensionless) 0.70 0.75 0.80 0.85 0.90 0.95 1.00 Heat capacity CV / (3NkB) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7974 Worked point: (0.879, 0.8988)
At x = 0.879, the orange circle is the analytical value 0.898782; the purple diamond is the interpolated value 0.797398. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.797398; analytical reference = 0.898782; absolute interpolation error = 0.101384. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact evaluation within the single-frequency harmonic Einstein model; the acoustic low-temperature cubic law is absent. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Debye phonon model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. In the regime T much smaller than ThetaD, estimate lattice heat capacity using the leading Debye asymptote. For this calculation, x denotes the plotted horizontal coordinate (Temperature T / ΘD (dimensionless)), and q(x) denotes the plotted response (Lattice heat capacity CV / (NkB) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.014 and x = 0.041 are tabulated. Use linear interpolation to predict the response at x = 0.02345, then check it against the original equation.

CVNkB≃12π45(TΘD)3\frac{C_V}{Nk_{\mathrm B}}\simeq\frac{12\pi^4}{5}\left(\frac{T}{\Theta_{\mathrm D}}\right)^3a=0.014,b=0.041,c=0.02345a=0.014,\quad b=0.041,\quad c=0.02345qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.572999\frac{q(b)-q(a)}{b-a}=0.572999qlin(c)=0.00605634,q(c)=0.00301467q_{\rm lin}(c)=0.00605634,\quad q(c)=0.00301467e=qlin(c)−q(c)=0.00304167e=q_{\rm lin}(c)-q(c)=0.00304167

Solution. The endpoint responses are q(A) = 0.000641497 and q(B) = 0.0161125. Their secant slope is 0.572999. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.00605634. Direct evaluation gives 0.00301467; subtracting it from the prediction gives signed error 0.00304167.

Debye phonon model: Intermediate-condition prediction and exact check. Horizontal axis: Temperature T / ΘD (dimensionless). Vertical axis: Lattice heat capacity CV / (NkB) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.015 0.020 0.025 0.030 0.035 0.040 Temperature T / ΘD (dimensionless) 0.0000 0.0025 0.0050 0.0075 0.0100 0.0125 0.0150 0.0175 Lattice heat capacity CV / (NkB) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.006056 Worked point: (0.02345, 0.003015)
At x = 0.02345, the orange circle is the analytical value 0.00301467; the purple diamond is the interpolated value 0.00605634. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.00605634; analytical reference = 0.00301467; absolute interpolation error = 0.00304167. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Low-temperature analytical asymptote only; the plotted range stops at T/ThetaD=0.05. Use the finite-cutoff integral outside this regime. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Sommerfeld free-electron model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Find the leading electronic heat capacity of a three-dimensional free-electron gas at fixed electron number and low temperature. For this calculation, x denotes the plotted horizontal coordinate (Temperature T / TF (dimensionless)), and q(x) denotes the plotted response (Electronic heat capacity Ce / (NkB) (dimensionless)). Suppose only the analytical endpoint responses at x = 0.0108 and x = 0.0402 are tabulated. Use linear interpolation to predict the response at x = 0.02109, then check it against the original equation.

CeNkB≃π22TTF\frac{C_e}{Nk_{\mathrm B}}\simeq\frac{\pi^2}{2}\frac{T}{T_{\mathrm F}}a=0.0108,b=0.0402,c=0.02109a=0.0108,\quad b=0.0402,\quad c=0.02109qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=4.9348\frac{q(b)-q(a)}{b-a}=4.9348qlin(c)=0.104075,q(c)=0.104075q_{\rm lin}(c)=0.104075,\quad q(c)=0.104075e=qlin(c)−q(c)=1.38778×10−17e=q_{\rm lin}(c)-q(c)=1.38778\times10^{-17}

Solution. The endpoint responses are q(A) = 0.0532959 and q(B) = 0.198379. Their secant slope is 4.9348. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.104075. Direct evaluation gives 0.104075; subtracting it from the prediction gives signed error 1.38778e-17.

Sommerfeld free-electron model: Intermediate-condition prediction and exact check. Horizontal axis: Temperature T / TF (dimensionless). Vertical axis: Electronic heat capacity Ce / (NkB) (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.015 0.020 0.025 0.030 0.035 0.040 Temperature T / TF (dimensionless) 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 Electronic heat capacity Ce / (NkB) (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.1041 Worked point: (0.02109, 0.1041)
At x = 0.02109, the orange circle is the analytical value 0.104075; the purple diamond is the interpolated value 0.104075. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.104075; analytical reference = 0.104075; absolute interpolation error = 1.38778e-17. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Leading low-temperature contribution of ideal electrons only; excludes lattice heat capacity, band corrections, interactions, and superconductivity. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Tight-binding electronic model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Use a one-dimensional chain with one orbital per site and positive nearest-neighbor hopping t. Find its band over half the Brillouin zone. For this calculation, x denotes the plotted horizontal coordinate (Crystal wavevector ka (radian)), and q(x) denotes the plotted response (Band energy (E − ε0) / t (dimensionless)). Suppose only the analytical endpoint responses at x = 0.628319 and x = 2.51327 are tabulated. Use linear interpolation to predict the response at x = 1.28805, then check it against the original equation.

E(k)−ϵ0t=−2cos⁡(ka)\frac{E(k)-\epsilon_0}{t}=-2\cos(ka)a=0.628319,b=2.51327,c=1.28805a=0.628319,\quad b=2.51327,\quad c=1.28805qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=1.71679\frac{q(b)-q(a)}{b-a}=1.71679qlin(c)=−0.48541,q(c)=−0.557982q_{\rm lin}(c)=-0.48541,\quad q(c)=-0.557982e=qlin(c)−q(c)=0.072572e=q_{\rm lin}(c)-q(c)=0.072572

Solution. The endpoint responses are q(A) = -1.61803 and q(B) = 1.61803. Their secant slope is 1.71679. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = -0.48541. Direct evaluation gives -0.557982; subtracting it from the prediction gives signed error 0.072572.

Tight-binding electronic model: Intermediate-condition prediction and exact check. Horizontal axis: Crystal wavevector ka (radian). Vertical axis: Band energy (E − ε0) / t (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 Crystal wavevector ka (radian) −2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 2.0 Band energy (E − ε0) / t (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: -0.4854 Worked point: (1.288, -0.558)
At x = 1.28805, the orange circle is the analytical value -0.557982; the purple diamond is the interpolated value -0.48541. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = -0.48541; analytical reference = -0.557982; absolute interpolation error = 0.072572. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-orbital, orthonormal, noninteracting chain. The other half-zone follows by inversion symmetry; real semiconductor bands generally need multiple orbitals. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Nearly-free-electron model · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Let ER=hbar²(G/2)²/(2m) and a real lattice Fourier coupling VG=0.1 ER. Calculate the lower branch near k=G/2. For this calculation, x denotes the plotted horizontal coordinate (Offset q = 2k/G − 1 (dimensionless)), and q(x) denotes the plotted response (Lower band energy E− / ER (dimensionless)). Suppose only the analytical endpoint responses at x = 0.06 and x = 0.24 are tabulated. Use linear interpolation to predict the response at x = 0.123, then check it against the original equation.

HER=((1+q)20.10.1(q−1)2),E−ER=1+q2−4q2+0.01\frac{H}{E_R}=\begin{pmatrix}(1+q)^2&0.1\\0.1&(q-1)^2\end{pmatrix},\quad\frac{E_-}{E_R}=1+q^2-\sqrt{4q^2+0.01}a=0.06,b=0.24,c=0.123a=0.06,\quad b=0.24,\quad c=0.123qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=−1.55612\frac{q(b)-q(a)}{b-a}=-1.55612qlin(c)=0.74936,q(c)=0.749581q_{\rm lin}(c)=0.74936,\quad q(c)=0.749581e=qlin(c)−q(c)=−0.000220866e=q_{\rm lin}(c)-q(c)=-0.000220866

Solution. The endpoint responses are q(A) = 0.847395 and q(B) = 0.567294. Their secant slope is -1.55612. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.74936. Direct evaluation gives 0.749581; subtracting it from the prediction gives signed error -0.000220866.

Nearly-free-electron model: Intermediate-condition prediction and exact check. Horizontal axis: Offset q = 2k/G − 1 (dimensionless). Vertical axis: Lower band energy E− / ER (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.075 0.100 0.125 0.150 0.175 0.200 0.225 Offset q = 2k/G − 1 (dimensionless) 0.55 0.60 0.65 0.70 0.75 0.80 0.85 Lower band energy E− / ER (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.7494 Worked point: (0.123, 0.7496)
At x = 0.123, the orange circle is the analytical value 0.749581; the purple diamond is the interpolated value 0.74936. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.74936; analytical reference = 0.749581; absolute interpolation error = 0.000220866. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. Exact two-state diagonalization, approximate nearly-free-electron physics. Only the lower branch on one side of the Bragg plane is plotted; remote plane waves are omitted. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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Harmonic lattice dynamics · Example 3

Intermediate-condition prediction and exact check

Problem & parameters. Take identical masses m separated by a, joined by nearest-neighbor springs K. Find the normal-mode dispersion over half the Brillouin zone. For this calculation, x denotes the plotted horizontal coordinate (Phonon wavevector qa (radian)), and q(x) denotes the plotted response (Frequency ω / [2√(K/m)] (dimensionless)). Suppose only the analytical endpoint responses at x = 0.628319 and x = 2.51327 are tabulated. Use linear interpolation to predict the response at x = 1.28805, then check it against the original equation.

ω(q)2K/m=sin⁡qa2(0≤qa≤π)\frac{\omega(q)}{2\sqrt{K/m}}=\sin\frac{qa}{2}\quad(0\le qa\le\pi)a=0.628319,b=2.51327,c=1.28805a=0.628319,\quad b=2.51327,\quad c=1.28805qlin(x)=q(a)+q(b)−q(a)b−a(x−a)q_{\rm lin}(x)=q(a)+\frac{q(b)-q(a)}{b-a}(x-a)q(b)−q(a)b−a=0.340613\frac{q(b)-q(a)}{b-a}=0.340613qlin(c)=0.533731,q(c)=0.60042q_{\rm lin}(c)=0.533731,\quad q(c)=0.60042e=qlin(c)−q(c)=−0.0666894e=q_{\rm lin}(c)-q(c)=-0.0666894

Solution. The endpoint responses are q(A) = 0.309017 and q(B) = 0.951057. Their secant slope is 0.340613. The target lies 35% of the way from A to B, so interpolation gives 0.65 q(A) + 0.35 q(B) = 0.533731. Direct evaluation gives 0.60042; subtracting it from the prediction gives signed error -0.0666894.

Harmonic lattice dynamics: Intermediate-condition prediction and exact check. Horizontal axis: Phonon wavevector qa (radian). Vertical axis: Frequency ω / [2√(K/m)] (dimensionless). image/svg+xml IICSM analytical illustration / Matplotlib 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 Phonon wavevector qa (radian) 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Frequency ω / [2√(K/m)] (dimensionless) Intermediate-condition prediction and exact check Exact stated model response Two-point linear interpolation Prediction: 0.5337 Worked point: (1.288, 0.6004)
At x = 1.28805, the orange circle is the analytical value 0.60042; the purple diamond is the interpolated value 0.533731. The dashed line connects the two exact endpoint responses.

Worked evaluation. Prediction = 0.533731; analytical reference = 0.60042; absolute interpolation error = 0.0666894. All response values use the vertical-axis units or normalization. The error is for this chosen interval and condition only.

Scope. One-dimensional harmonic monatomic chain; no optical branch, anharmonic scattering, or measured material parameters. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. The dashed line is a two-point approximation, not a second physical solution. It can miss curvature, extrema, or jumps; zero error here does not validate interpolation elsewhere.

Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.

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