Schrödinger model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.690983, and at B to obtain 0.690983. Subtract the starting value from the ending value: the signed change is 3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.690983)−(0.690983) = 3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Particle-in-a-box model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.690983, and at B to obtain 0.690983. Subtract the starting value from the ending value: the signed change is 3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.690983)−(0.690983) = 3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Dirac model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.16619, and at B to obtain 2.6. Subtract the starting value from the ending value: the signed change is 1.43381. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.6)−(1.16619) = 1.43381. The magnitude of the response change is 1.43381; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Born–Oppenheimer approximation · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.72, and at B to obtain 0.72. Subtract the starting value from the ending value: the signed change is 2.22045e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.72)−(0.72) = 2.22045e-16. The magnitude of the response change is 2.22045e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hartree–Fock model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.2 with condition B at x = 4.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.522535, and at B to obtain 0.00624188. Subtract the starting value from the ending value: the signed change is -0.516294. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.00624188)−(0.522535) = -0.516294. The magnitude of the response change is 0.516294; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Density functional theory (DFT) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.2 with condition B at x = 4.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.522535, and at B to obtain 0.00624188. Subtract the starting value from the ending value: the signed change is -0.516294. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.00624188)−(0.522535) = -0.516294. The magnitude of the response change is 0.516294; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Time-dependent DFT (TDDFT) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.345492, and at B to obtain 0.345492. Subtract the starting value from the ending value: the signed change is 1.66533e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.345492)−(0.345492) = 1.66533e-16. The magnitude of the response change is 1.66533e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Tight-binding model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.88496 with condition B at x = 1.88496. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.618034, and at B to obtain 0.618034. Subtract the starting value from the ending value: the signed change is 0. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.618034)−(0.618034) = 0. The magnitude of the response change is 0; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hubbard model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.4 with condition B at x = 9.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -1.13238, and at B to obtain -0.4. Subtract the starting value from the ending value: the signed change is 0.732381. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.4)−(-1.13238) = 0.732381. The magnitude of the response change is 0.732381; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Heisenberg spin model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.628319 with condition B at x = 2.51327. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain 0.809017. Subtract the starting value from the ending value: the signed change is 1.61803. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.809017)−(-0.809017) = 1.61803. The magnitude of the response change is 1.61803; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Ising model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.946806, and at B to obtain 0.946806. Subtract the starting value from the ending value: the signed change is 1.89361. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.946806)−(-0.946806) = 1.89361. The magnitude of the response change is 1.89361; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Quantum harmonic oscillator · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0220959, and at B to obtain 0.0220959. Subtract the starting value from the ending value: the signed change is -6.93889e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0220959)−(0.0220959) = -6.93889e-17. The magnitude of the response change is 6.93889e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Classical molecular dynamics (MD) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.309017, and at B to obtain 0.309017. Subtract the starting value from the ending value: the signed change is -2.22045e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.309017)−(0.309017) = -2.22045e-16. The magnitude of the response change is 2.22045e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Ab initio molecular dynamics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.309017, and at B to obtain 0.309017. Subtract the starting value from the ending value: the signed change is -2.22045e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.309017)−(0.309017) = -2.22045e-16. The magnitude of the response change is 2.22045e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lennard–Jones potential · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.36 with condition B at x = 2.59. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.532253, and at B to obtain -0.0132075. Subtract the starting value from the ending value: the signed change is 0.519045. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.0132075)−(-0.532253) = 0.519045. The magnitude of the response change is 0.519045; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
SPC/E water model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.36 with condition B at x = 2.59. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.532253, and at B to obtain -0.0132075. Subtract the starting value from the ending value: the signed change is 0.519045. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.0132075)−(-0.532253) = 0.519045. The magnitude of the response change is 0.519045; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
TIP4P water-model family · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.36 with condition B at x = 2.59. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.532253, and at B to obtain -0.0132075. Subtract the starting value from the ending value: the signed change is 0.519045. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.0132075)−(-0.532253) = 0.519045. The magnitude of the response change is 0.519045; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Martini coarse-grained model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.36 with condition B at x = 2.59. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.532253, and at B to obtain -0.0132075. Subtract the starting value from the ending value: the signed change is 0.519045. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.0132075)−(-0.532253) = 0.519045. The magnitude of the response change is 0.519045; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Morse potential · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.32 with condition B at x = 3.08. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.925006, and at B to obtain -0.0898063. Subtract the starting value from the ending value: the signed change is 0.835199. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.0898063)−(-0.925006) = 0.835199. The magnitude of the response change is 0.835199; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Embedded-atom method (EAM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.808 with condition B at x = 3.202. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.898888, and at B to obtain -1.78941. Subtract the starting value from the ending value: the signed change is -0.890525. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-1.78941)−(-0.898888) = -0.890525. The magnitude of the response change is 0.890525; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Modified embedded-atom method (MEAM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.808 with condition B at x = 3.202. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.898888, and at B to obtain -1.78941. Subtract the starting value from the ending value: the signed change is -0.890525. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-1.78941)−(-0.898888) = -0.890525. The magnitude of the response change is 0.890525; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Tersoff bond-order potential · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.696761, and at B to obtain -0.0798629. Subtract the starting value from the ending value: the signed change is 0.616899. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.0798629)−(-0.696761) = 0.616899. The magnitude of the response change is 0.616899; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Stillinger–Weber potential · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.628319 with condition B at x = 2.51327. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.30496, and at B to obtain 0.226275. Subtract the starting value from the ending value: the signed change is -1.07869. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.226275)−(1.30496) = -1.07869. The magnitude of the response change is 1.07869; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
ReaxFF reactive force field · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.18, and at B to obtain 0.18. Subtract the starting value from the ending value: the signed change is 5.55112e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.18)−(0.18) = 5.55112e-17. The magnitude of the response change is 5.55112e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
AMBER force-field family · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.18, and at B to obtain 0.18. Subtract the starting value from the ending value: the signed change is 5.55112e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.18)−(0.18) = 5.55112e-17. The magnitude of the response change is 5.55112e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
CHARMM force-field family · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.18, and at B to obtain 0.18. Subtract the starting value from the ending value: the signed change is 5.55112e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.18)−(0.18) = 5.55112e-17. The magnitude of the response change is 5.55112e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Machine-learned interatomic potential · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.18, and at B to obtain 0.18. Subtract the starting value from the ending value: the signed change is 5.55112e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.18)−(0.18) = 5.55112e-17. The magnitude of the response change is 5.55112e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
OPLS force-field family · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.654508, and at B to obtain 0.654508. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.654508)−(0.654508) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Drude polarizable model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -1.2, and at B to obtain 1.2. Subtract the starting value from the ending value: the signed change is 2.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.2)−(-1.2) = 2.4. The magnitude of the response change is 2.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Coarse-grained molecular model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.62, and at B to obtain 1.62. Subtract the starting value from the ending value: the signed change is 1.55431e-15. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.62)−(1.62) = 1.55431e-15. The magnitude of the response change is 1.55431e-15; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Dissipative particle dynamics (DPD) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Brownian dynamics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 2, and at B to obtain 8. Subtract the starting value from the ending value: the signed change is 6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (8)−(2) = 6. The magnitude of the response change is 6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Langevin dynamics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Kinetic Monte Carlo · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Cahn–Hilliard model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0418209, and at B to obtain 0.0418209. Subtract the starting value from the ending value: the signed change is -3.46945e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0418209)−(0.0418209) = -3.46945e-17. The magnitude of the response change is 3.46945e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Allen–Cahn model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0418209, and at B to obtain 0.0418209. Subtract the starting value from the ending value: the signed change is -3.46945e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0418209)−(0.0418209) = -3.46945e-17. The magnitude of the response change is 3.46945e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Phase-field crystal model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.00113681, and at B to obtain 0.00113681. Subtract the starting value from the ending value: the signed change is -8.67362e-19. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.00113681)−(0.00113681) = -8.67362e-19. The magnitude of the response change is 8.67362e-19; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Potts grain-growth model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.576117, and at B to obtain 0.964663. Subtract the starting value from the ending value: the signed change is 0.388546. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.964663)−(0.576117) = 0.388546. The magnitude of the response change is 0.388546; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Discrete dislocation dynamics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1, and at B to obtain 4. Subtract the starting value from the ending value: the signed change is 3. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4)−(1) = 3. The magnitude of the response change is 3; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Population balance model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.2 with condition B at x = 4.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0391639, and at B to obtain 0.0391639. Subtract the starting value from the ending value: the signed change is -1.249e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0391639)−(0.0391639) = -1.249e-16. The magnitude of the response change is 1.249e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Ideal gas equation of state · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.4 with condition B at x = 4.1. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.714286, and at B to obtain 0.243902. Subtract the starting value from the ending value: the signed change is -0.470383. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.243902)−(0.714286) = -0.470383. The magnitude of the response change is 0.470383; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Van der Waals equation of state · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.28 with condition B at x = 3.32. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.54923, and at B to obtain 0.799256. Subtract the starting value from the ending value: the signed change is -0.749971. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.799256)−(1.54923) = -0.749971. The magnitude of the response change is 0.749971; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Peng–Robinson equation of state · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.4 with condition B at x = 5.1. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.505081, and at B to obtain 0.1871. Subtract the starting value from the ending value: the signed change is -0.31798. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.1871)−(0.505081) = -0.31798. The magnitude of the response change is 0.31798; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Soave–Redlich–Kwong equation of state · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.4 with condition B at x = 5.1. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.469188, and at B to obtain 0.179614. Subtract the starting value from the ending value: the signed change is -0.289573. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.179614)−(0.469188) = -0.289573. The magnitude of the response change is 0.289573; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Virial equation of state · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.042, and at B to obtain 1.192. Subtract the starting value from the ending value: the signed change is 0.15. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.192)−(1.042) = 0.15. The magnitude of the response change is 0.15; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Gibbs-energy minimization · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2006 with condition B at x = 0.7994. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.501233, and at B to obtain -0.501233. Subtract the starting value from the ending value: the signed change is 0. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.501233)−(-0.501233) = 0. The magnitude of the response change is 0; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
CALPHAD model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2006 with condition B at x = 0.7994. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.501233, and at B to obtain -0.501233. Subtract the starting value from the ending value: the signed change is 0. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.501233)−(-0.501233) = 0. The magnitude of the response change is 0; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
NRTL activity model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.2, and at B to obtain 0.8. Subtract the starting value from the ending value: the signed change is 0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.8)−(0.2) = 0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
UNIQUAC activity model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.2, and at B to obtain 0.8. Subtract the starting value from the ending value: the signed change is 0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.8)−(0.2) = 0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Debye–Hückel model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.002 with condition B at x = 0.008. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.0227632, and at B to obtain -0.0455263. Subtract the starting value from the ending value: the signed change is -0.0227632. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.0455263)−(-0.0227632) = -0.0227632. The magnitude of the response change is 0.0227632; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Mass-action reaction kinetics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Batch reactor model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Arrhenius rate model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.28 with condition B at x = 0.82. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0281157, and at B to obtain 0.295374. Subtract the starting value from the ending value: the signed change is 0.267259. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.295374)−(0.0281157) = 0.267259. The magnitude of the response change is 0.267259; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Transition-state theory · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.28 with condition B at x = 0.82. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0281157, and at B to obtain 0.295374. Subtract the starting value from the ending value: the signed change is 0.267259. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.295374)−(0.0281157) = 0.267259. The magnitude of the response change is 0.267259; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Michaelis–Menten kinetics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.6 with condition B at x = 6.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.615385, and at B to obtain 0.864865. Subtract the starting value from the ending value: the signed change is 0.24948. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.864865)−(0.615385) = 0.24948. The magnitude of the response change is 0.24948; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Langmuir adsorption isotherm · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.6 with condition B at x = 6.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.615385, and at B to obtain 0.864865. Subtract the starting value from the ending value: the signed change is 0.24948. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.864865)−(0.615385) = 0.24948. The magnitude of the response change is 0.24948; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Langmuir–Hinshelwood kinetics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.6 with condition B at x = 6.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.236686, and at B to obtain 0.116874. Subtract the starting value from the ending value: the signed change is -0.119813. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.116874)−(0.236686) = -0.119813. The magnitude of the response change is 0.119813; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Fickian diffusion · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Maxwell–Stefan diffusion · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Advection–diffusion–reaction model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.4 with condition B at x = 3.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0113049, and at B to obtain 0.0113049. Subtract the starting value from the ending value: the signed change is -1.21431e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0113049)−(0.0113049) = -1.21431e-17. The magnitude of the response change is 1.21431e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Continuous stirred-tank reactor (CSTR) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.5, and at B to obtain 0.2. Subtract the starting value from the ending value: the signed change is -0.3. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.2)−(0.5) = -0.3. The magnitude of the response change is 0.3; its sign gives the direction relative to condition A.
Scope. Constant-volume, isothermal, constant-flow reactor. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Plug-flow reactor (PFR) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Stokes creeping-flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.64, and at B to obtain 0.64. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.64)−(0.64) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lubrication approximation · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.64, and at B to obtain 0.64. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.64)−(0.64) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hagen–Poiseuille model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.64, and at B to obtain 0.64. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.64)−(0.64) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Direct numerical simulation (DNS) · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.64, and at B to obtain 0.64. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.64)−(0.64) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Euler flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Potential-flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -2.61803, and at B to obtain -2.61803. Subtract the starting value from the ending value: the signed change is -8.88178e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-2.61803)−(-2.61803) = -8.88178e-16. The magnitude of the response change is 8.88178e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Boundary-layer model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.396144, and at B to obtain 0.000688514. Subtract the starting value from the ending value: the signed change is -0.395455. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.000688514)−(0.396144) = -0.395455. The magnitude of the response change is 0.395455; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Darcy–Weisbach model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.36, and at B to obtain 5.76. Subtract the starting value from the ending value: the signed change is 5.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (5.76)−(0.36) = 5.4. The magnitude of the response change is 5.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Non-Newtonian power-law fluid · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.894427, and at B to obtain 1.78885. Subtract the starting value from the ending value: the signed change is 0.894427. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.78885)−(0.894427) = 0.894427. The magnitude of the response change is 0.894427; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Bingham plastic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0, and at B to obtain 1.4. Subtract the starting value from the ending value: the signed change is 1.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.4)−(0) = 1.4. The magnitude of the response change is 1.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Herschel–Bulkley model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.89443, and at B to obtain 2.78885. Subtract the starting value from the ending value: the signed change is 0.894427. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.78885)−(1.89443) = 0.894427. The magnitude of the response change is 0.894427; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Oldroyd-B model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Large-eddy simulation (LES) · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.64, and at B to obtain 0.64. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.64)−(0.64) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Detached-eddy simulation (DES) · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.64, and at B to obtain 0.64. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.64)−(0.64) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Spalart–Allmaras model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 4 with condition B at x = 16. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.606763, and at B to obtain 14.7143. Subtract the starting value from the ending value: the signed change is 14.1075. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (14.7143)−(0.606763) = 14.1075. The magnitude of the response change is 14.1075; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
k–epsilon model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0576, and at B to obtain 0.9216. Subtract the starting value from the ending value: the signed change is 0.864. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.9216)−(0.0576) = 0.864. The magnitude of the response change is 0.864; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
k–omega model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.8, and at B to obtain 3.2. Subtract the starting value from the ending value: the signed change is 2.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (3.2)−(0.8) = 2.4. The magnitude of the response change is 2.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
SST k–omega model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.4 with condition B at x = 1.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.775, and at B to obtain 0.19375. Subtract the starting value from the ending value: the signed change is -0.58125. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.19375)−(0.775) = -0.58125. The magnitude of the response change is 0.58125; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Reynolds-stress transport model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Smagorinsky subgrid model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.01, and at B to obtain 0.04. Subtract the starting value from the ending value: the signed change is 0.03. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.04)−(0.01) = 0.03. The magnitude of the response change is 0.03; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Volume-of-fluid (VOF) representation · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.2 with condition B at x = 2.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.999994, and at B to obtain 6.14417e-06. Subtract the starting value from the ending value: the signed change is -0.999988. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (6.14417e-06)−(0.999994) = -0.999988. The magnitude of the response change is 0.999988; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Euler–Euler two-fluid model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lagrangian particle tracking · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Fourier heat conduction · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.8, and at B to obtain 0.2. Subtract the starting value from the ending value: the signed change is -0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.2)−(0.8) = -0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Groundwater flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.8, and at B to obtain 0.2. Subtract the starting value from the ending value: the signed change is -0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.2)−(0.8) = -0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Finite element method (FEM / FEA) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.8, and at B to obtain 0.2. Subtract the starting value from the ending value: the signed change is -0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.2)−(0.8) = -0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Finite volume method (FVM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.8, and at B to obtain 0.2. Subtract the starting value from the ending value: the signed change is -0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.2)−(0.8) = -0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Finite difference method (FDM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.8, and at B to obtain 0.2. Subtract the starting value from the ending value: the signed change is -0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.2)−(0.8) = -0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Boundary element method (BEM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.8, and at B to obtain 0.2. Subtract the starting value from the ending value: the signed change is -0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.2)−(0.8) = -0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Transient heat equation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Spectral method · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Reduced basis model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Physics-informed neural network (PINN) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Neural operator · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lumped-capacitance thermal model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Thermal resistance-capacitance network · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Newton cooling model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Building thermal-zone model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Radiative transfer equation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Neutron transport model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Monte Carlo transport · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Stefan–Boltzmann surface model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 1.7. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.5904, and at B to obtain 7.3521. Subtract the starting value from the ending value: the signed change is 7.9425. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (7.3521)−(-0.5904) = 7.9425. The magnitude of the response change is 7.9425; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Surface-to-surface radiosity model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.24 with condition B at x = 0.81. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.226415, and at B to obtain 0.673597. Subtract the starting value from the ending value: the signed change is 0.447182. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.673597)−(0.226415) = 0.447182. The magnitude of the response change is 0.447182; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Stefan phase-change problem · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.894427, and at B to obtain 1.78885. Subtract the starting value from the ending value: the signed change is 0.894427. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.78885)−(0.894427) = 0.894427. The magnitude of the response change is 0.894427; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Enthalpy–porosity model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.1 with condition B at x = 1.1. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0, and at B to obtain 1. Subtract the starting value from the ending value: the signed change is 1. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1)−(0) = 1. The magnitude of the response change is 1; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Linear elasticity (Hooke model) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.002 with condition B at x = 0.008. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.002, and at B to obtain 0.008. Subtract the starting value from the ending value: the signed change is 0.006. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.008)−(0.002) = 0.006. The magnitude of the response change is 0.006; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Orthotropic elasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.002 with condition B at x = 0.008. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.002, and at B to obtain 0.008. Subtract the starting value from the ending value: the signed change is 0.006. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.008)−(0.002) = 0.006. The magnitude of the response change is 0.006; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Truss model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.002 with condition B at x = 0.008. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.002, and at B to obtain 0.008. Subtract the starting value from the ending value: the signed change is 0.006. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.008)−(0.002) = 0.006. The magnitude of the response change is 0.006; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Representative volume element (RVE) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.002 with condition B at x = 0.008. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.002, and at B to obtain 0.008. Subtract the starting value from the ending value: the signed change is 0.006. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.008)−(0.002) = 0.006. The magnitude of the response change is 0.006; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
FE² computational homogenization · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.002 with condition B at x = 0.008. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.002, and at B to obtain 0.008. Subtract the starting value from the ending value: the signed change is 0.006. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.008)−(0.002) = 0.006. The magnitude of the response change is 0.006; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Neo-Hookean hyperelasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.88 with condition B at x = 1.72. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.361964, and at B to obtain 2.377. Subtract the starting value from the ending value: the signed change is 2.73897. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.377)−(-0.361964) = 2.73897. The magnitude of the response change is 2.73897; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Mooney–Rivlin hyperelasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.88 with condition B at x = 1.72. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.386643, and at B to obtain 1.87949. Subtract the starting value from the ending value: the signed change is 2.26614. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.87949)−(-0.386643) = 2.26614. The magnitude of the response change is 2.26614; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Ogden hyperelasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.88 with condition B at x = 1.72. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.345813, and at B to obtain 4.20706. Subtract the starting value from the ending value: the signed change is 4.55287. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4.20706)−(-0.345813) = 4.55287. The magnitude of the response change is 4.55287; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Euler–Bernoulli beam model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0186667, and at B to obtain 0.234667. Subtract the starting value from the ending value: the signed change is 0.216. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.234667)−(0.0186667) = 0.216. The magnitude of the response change is 0.216; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Timoshenko beam model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0386667, and at B to obtain 0.314667. Subtract the starting value from the ending value: the signed change is 0.276. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.314667)−(0.0386667) = 0.276. The magnitude of the response change is 0.276; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Kirchhoff–Love plate model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.587785, and at B to obtain 0.587785. Subtract the starting value from the ending value: the signed change is 1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.587785)−(0.587785) = 1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Mindlin–Reissner plate model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.587785, and at B to obtain 0.587785. Subtract the starting value from the ending value: the signed change is 1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.587785)−(0.587785) = 1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Shell model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.004 with condition B at x = 0.016. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.002, and at B to obtain 0.008. Subtract the starting value from the ending value: the signed change is 0.006. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.008)−(0.002) = 0.006. The magnitude of the response change is 0.006; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Cable and membrane models · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.810656, and at B to obtain 0.810656. Subtract the starting value from the ending value: the signed change is 4.44089e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.810656)−(0.810656) = 4.44089e-16. The magnitude of the response change is 4.44089e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
von Mises J2 plasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.6, and at B to obtain 1. Subtract the starting value from the ending value: the signed change is 0.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1)−(0.6) = 0.4. The magnitude of the response change is 0.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Tresca yield model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.6, and at B to obtain 1. Subtract the starting value from the ending value: the signed change is 0.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1)−(0.6) = 0.4. The magnitude of the response change is 0.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Drucker–Prager plasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.4, and at B to obtain 2.6. Subtract the starting value from the ending value: the signed change is 1.2. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.6)−(1.4) = 1.2. The magnitude of the response change is 1.2; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Mohr–Coulomb model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.46188, and at B to obtain 2.84752. Subtract the starting value from the ending value: the signed change is 1.38564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.84752)−(1.46188) = 1.38564. The magnitude of the response change is 1.38564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Johnson–Cook model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.22361, and at B to obtain 1.44721. Subtract the starting value from the ending value: the signed change is 0.223607. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.44721)−(1.22361) = 0.223607. The magnitude of the response change is 0.223607; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Crystal plasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.3 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.00243, and at B to obtain 2.48832. Subtract the starting value from the ending value: the signed change is 2.48589. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.48832)−(0.00243) = 2.48589. The magnitude of the response change is 2.48589; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Maxwell viscoelastic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Kelvin–Voigt model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Standard linear solid · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.620728, and at B to obtain 0.410989. Subtract the starting value from the ending value: the signed change is -0.209738. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.410989)−(0.620728) = -0.209738. The magnitude of the response change is 0.209738; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Norton creep law · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.4 with condition B at x = 1.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.064, and at B to obtain 4.096. Subtract the starting value from the ending value: the signed change is 4.032. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4.096)−(0.064) = 4.032. The magnitude of the response change is 4.032; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Linear elastic fracture mechanics (LEFM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.44 with condition B at x = 1.61. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.50756, and at B to obtain 0.78811. Subtract the starting value from the ending value: the signed change is -0.719446. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.78811)−(1.50756) = -0.719446. The magnitude of the response change is 0.719446; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Cohesive-zone model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.4, and at B to obtain 0.4. Subtract the starting value from the ending value: the signed change is -1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.4)−(0.4) = -1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Phase-field fracture model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -3 with condition B at x = 3. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0497871, and at B to obtain 0.0497871. Subtract the starting value from the ending value: the signed change is 0. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0497871)−(0.0497871) = 0. The magnitude of the response change is 0; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Paris fatigue crack-growth law · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.3 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.34986, and at B to obtain 3.32012. Subtract the starting value from the ending value: the signed change is 1.97026. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (3.32012)−(1.34986) = 1.97026. The magnitude of the response change is 1.97026; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Miner cumulative damage rule · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.2, and at B to obtain 0.8. Subtract the starting value from the ending value: the signed change is 0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.8)−(0.2) = 0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Archard wear model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1, and at B to obtain 4. Subtract the starting value from the ending value: the signed change is 3. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4)−(1) = 3. The magnitude of the response change is 3; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Newton–Euler rigid-body model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.5, and at B to obtain 8. Subtract the starting value from the ending value: the signed change is 7.5. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (8)−(0.5) = 7.5. The magnitude of the response change is 7.5; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Six-degree-of-freedom flight model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.5, and at B to obtain 8. Subtract the starting value from the ending value: the signed change is 7.5. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (8)−(0.5) = 7.5. The magnitude of the response change is 7.5; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lagrangian mechanics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.51327 with condition B at x = 10.0531. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain -0.809017. Subtract the starting value from the ending value: the signed change is -3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.809017)−(-0.809017) = -3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hamiltonian mechanics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.51327 with condition B at x = 10.0531. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain -0.809017. Subtract the starting value from the ending value: the signed change is -3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.809017)−(-0.809017) = -3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Mass–spring–damper model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.51327 with condition B at x = 10.0531. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain -0.809017. Subtract the starting value from the ending value: the signed change is -3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.809017)−(-0.809017) = -3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Modal superposition model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.51327 with condition B at x = 10.0531. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain -0.809017. Subtract the starting value from the ending value: the signed change is -3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.809017)−(-0.809017) = -3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Multibody dynamics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.51327 with condition B at x = 10.0531. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain -0.809017. Subtract the starting value from the ending value: the signed change is -3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.809017)−(-0.809017) = -3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Duffing oscillator · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -2.928, and at B to obtain 2.928. Subtract the starting value from the ending value: the signed change is 5.856. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.928)−(-2.928) = 5.856. The magnitude of the response change is 5.856; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Linear acoustic wave model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Transmission-line acoustic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Maxwell electromagnetic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Transmission-line electrical model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Elastic seismic-wave model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Magnetohydrodynamics (MHD) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Finite-difference time-domain (FDTD) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Helmholtz acoustic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.587785, and at B to obtain 0.587785. Subtract the starting value from the ending value: the signed change is 1.11022e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.587785)−(0.587785) = 1.11022e-16. The magnitude of the response change is 1.11022e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Electrostatic Poisson model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.08, and at B to obtain 0.08. Subtract the starting value from the ending value: the signed change is -2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.08)−(0.08) = -2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Magnetostatic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.8 with condition B at x = 4.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.555556, and at B to obtain 0.238095. Subtract the starting value from the ending value: the signed change is -0.31746. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.238095)−(0.555556) = -0.31746. The magnitude of the response change is 0.31746; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Eddy-current model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Magnetic-circuit model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.6, and at B to obtain 2.4. Subtract the starting value from the ending value: the signed change is 1.8. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.4)−(0.6) = 1.8. The magnitude of the response change is 1.8; its sign gives the direction relative to condition A.
Scope. Linear unsaturated material and fixed geometry. The physical assumptions, coefficients, and normalization from Example 1 remain fixed; only the stated horizontal coordinate changes. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Jiles–Atherton hysteresis model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -3 with condition B at x = 3. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.671636, and at B to obtain 0.671636. Subtract the starting value from the ending value: the signed change is 1.34327. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.671636)−(-0.671636) = 1.34327. The magnitude of the response change is 1.34327; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Geometrical optics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 16 with condition B at x = 64. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 10.5887, and at B to obtain 36.8123. Subtract the starting value from the ending value: the signed change is 26.2236. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (36.8123)−(10.5887) = 26.2236. The magnitude of the response change is 26.2236; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Scalar diffraction model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -4.8 with condition B at x = 4.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0430705, and at B to obtain 0.0430705. Subtract the starting value from the ending value: the signed change is -3.46945e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0430705)−(0.0430705) = -3.46945e-17. The magnitude of the response change is 3.46945e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Gaussian beam model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0561348, and at B to obtain 0.0561348. Subtract the starting value from the ending value: the signed change is -4.85723e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0561348)−(0.0561348) = -4.85723e-17. The magnitude of the response change is 4.85723e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Drude–Lorentz optical model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 2.5. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0, and at B to obtain 0.84. Subtract the starting value from the ending value: the signed change is 0.84. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.84)−(0) = 0.84. The magnitude of the response change is 0.84; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lumped RLC circuit model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Shockley diode model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.834701, and at B to obtain 5.04965. Subtract the starting value from the ending value: the signed change is 5.88435. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (5.04965)−(-0.834701) = 5.88435. The magnitude of the response change is 5.88435; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Ebers–Moll transistor model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.22554, and at B to obtain 23.5325. Subtract the starting value from the ending value: the signed change is 22.307. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (23.5325)−(1.22554) = 22.307. The magnitude of the response change is 22.307; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
MOSFET square-law model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.36, and at B to obtain 5.76. Subtract the starting value from the ending value: the signed change is 5.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (5.76)−(0.36) = 5.4. The magnitude of the response change is 5.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
BSIM compact-model family · Example 2
Two-condition response comparison
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)). Compare condition A at x = -3 with condition B at x = 0. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0497871, and at B to obtain 1. Subtract the starting value from the ending value: the signed change is 0.950213. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1)−(0.0497871) = 0.950213. The magnitude of the response change is 0.950213; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Drift–diffusion semiconductor model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -1.2, and at B to obtain 1.2. Subtract the starting value from the ending value: the signed change is 2.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.2)−(-1.2) = 2.4. The magnitude of the response change is 2.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hydrodynamic carrier model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -1.2, and at B to obtain 1.2. Subtract the starting value from the ending value: the signed change is 2.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.2)−(-1.2) = 2.4. The magnitude of the response change is 2.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Nernst equilibrium potential · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.08 with condition B at x = 8.02. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.732368, and at B to obtain 2.08194. Subtract the starting value from the ending value: the signed change is 1.34957. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.08194)−(0.732368) = 1.34957. The magnitude of the response change is 1.34957; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Butler–Volmer kinetics · Example 2
Two-condition response comparison
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)). Compare condition A at x = -2.4 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -3.01892, and at B to obtain 3.01892. Subtract the starting value from the ending value: the signed change is 6.03785. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (3.01892)−(-3.01892) = 6.03785. The magnitude of the response change is 6.03785; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Tafel approximation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 28 with condition B at x = 82. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 3.3322, and at B to obtain 4.40672. Subtract the starting value from the ending value: the signed change is 1.07451. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4.40672)−(3.3322) = 1.07451. The magnitude of the response change is 1.07451; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Poisson–Nernst–Planck model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Doyle–Fuller–Newman (DFN/P2D) model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.05 with condition B at x = 0.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.85, and at B to obtain 0.4. Subtract the starting value from the ending value: the signed change is -0.45. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.4)−(0.85) = -0.45. The magnitude of the response change is 0.45; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Single-particle battery model (SPM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.05 with condition B at x = 0.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.85, and at B to obtain 0.4. Subtract the starting value from the ending value: the signed change is -0.45. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.4)−(0.85) = -0.45. The magnitude of the response change is 0.45; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Single-particle model with electrolyte (SPMe) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.05 with condition B at x = 0.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.85, and at B to obtain 0.4. Subtract the starting value from the ending value: the signed change is -0.45. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.4)−(0.85) = -0.45. The magnitude of the response change is 0.45; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Equivalent-circuit battery model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Darcy porous-flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.6, and at B to obtain 2.4. Subtract the starting value from the ending value: the signed change is 1.8. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.4)−(0.6) = 1.8. The magnitude of the response change is 1.8; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Brinkman porous-flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.518724, and at B to obtain 0.518724. Subtract the starting value from the ending value: the signed change is -2.22045e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.518724)−(0.518724) = -2.22045e-16. The magnitude of the response change is 2.22045e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Forchheimer model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.96, and at B to obtain 8.16. Subtract the starting value from the ending value: the signed change is 7.2. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (8.16)−(0.96) = 7.2. The magnitude of the response change is 7.2; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Richards equation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
van Genuchten retention model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.707107, and at B to obtain 0.242536. Subtract the starting value from the ending value: the signed change is -0.464571. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.242536)−(0.707107) = -0.464571. The magnitude of the response change is 0.464571; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Biot poroelasticity · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Terzaghi consolidation model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Modified Cam-Clay model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.4, and at B to obtain 0.4. Subtract the starting value from the ending value: the signed change is -5.55112e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.4)−(0.4) = -5.55112e-17. The magnitude of the response change is 5.55112e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Saint-Venant shallow-water model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Kinematic-wave routing · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.4 with condition B at x = 3.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.00315111, and at B to obtain 0.00315111. Subtract the starting value from the ending value: the signed change is -5.63785e-18. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.00315111)−(0.00315111) = -5.63785e-18. The magnitude of the response change is 5.63785e-18; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Rainfall–runoff model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Advection–dispersion groundwater model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.4 with condition B at x = 3.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0138078, and at B to obtain 0.0138078. Subtract the starting value from the ending value: the signed change is -1.38778e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0138078)−(0.0138078) = -1.38778e-17. The magnitude of the response change is 1.38778e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Numerical weather prediction · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
General circulation model (GCM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Stellar structure model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Earth system model (ESM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Energy-balance climate model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Ocean circulation model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.951057, and at B to obtain -0.951057. Subtract the starting value from the ending value: the signed change is -1.90211. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.951057)−(0.951057) = -1.90211. The magnitude of the response change is 1.90211; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Sea-ice thermodynamic-dynamic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.894427, and at B to obtain 1.78885. Subtract the starting value from the ending value: the signed change is 0.894427. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.78885)−(0.894427) = 0.894427. The magnitude of the response change is 0.894427; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lifting-line model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -3.6 with condition B at x = 3.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.315827, and at B to obtain 0.315827. Subtract the starting value from the ending value: the signed change is 0.631655. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.315827)−(-0.315827) = 0.631655. The magnitude of the response change is 0.631655; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Blade-element momentum model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.1 with condition B at x = 0.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.324, and at B to obtain 0.576. Subtract the starting value from the ending value: the signed change is 0.252. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.576)−(0.324) = 0.252. The magnitude of the response change is 0.252; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Bicycle vehicle model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -18 with condition B at x = 18. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.32492, and at B to obtain 0.32492. Subtract the starting value from the ending value: the signed change is 0.649839. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.32492)−(-0.32492) = 0.649839. The magnitude of the response change is 0.649839; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Quarter-car suspension model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.51327 with condition B at x = 10.0531. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain -0.809017. Subtract the starting value from the ending value: the signed change is -3.33067e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.809017)−(-0.809017) = -3.33067e-16. The magnitude of the response change is 3.33067e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Pacejka tire model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.18 with condition B at x = 0.18. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.982382, and at B to obtain 0.982382. Subtract the starting value from the ending value: the signed change is 1.96476. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.982382)−(-0.982382) = 1.96476. The magnitude of the response change is 1.96476; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
AC power-flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.942478 with condition B at x = 0.942478. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain 0.809017. Subtract the starting value from the ending value: the signed change is 1.61803. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.809017)−(-0.809017) = 1.61803. The magnitude of the response change is 1.61803; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Swing-equation generator model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.942478 with condition B at x = 0.942478. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.809017, and at B to obtain 0.809017. Subtract the starting value from the ending value: the signed change is 1.61803. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.809017)−(-0.809017) = 1.61803. The magnitude of the response change is 1.61803; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
DC power-flow approximation · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.12 with condition B at x = 0.12. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.12, and at B to obtain 0.12. Subtract the starting value from the ending value: the signed change is 0.24. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.12)−(-0.12) = 0.24. The magnitude of the response change is 0.24; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
State-space model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Transfer-function model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Bond-graph model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Electrical analog model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hardware-in-the-loop model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hybrid dynamical model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.22 with condition B at x = 0.88. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.762598, and at B to obtain 0.617812. Subtract the starting value from the ending value: the signed change is -0.144786. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.617812)−(0.762598) = -0.144786. The magnitude of the response change is 0.144786; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
System-dynamics stock-flow model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Discrete-event simulation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.2 with condition B at x = 4.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1, and at B to obtain 4. Subtract the starting value from the ending value: the signed change is 3. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4)−(1) = 3. The magnitude of the response change is 3; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Agent-based physical-system model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1, and at B to obtain 4. Subtract the starting value from the ending value: the signed change is 3. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4)−(1) = 3. The magnitude of the response change is 3; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Markov state model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.8 with condition B at x = 3.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.600948, and at B to obtain 0.500831. Subtract the starting value from the ending value: the signed change is -0.100117. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.500831)−(0.600948) = -0.100117. The magnitude of the response change is 0.100117; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Kalman state estimator · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2 with condition B at x = 8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.333333, and at B to obtain 0.111111. Subtract the starting value from the ending value: the signed change is -0.222222. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.111111)−(0.333333) = -0.222222. The magnitude of the response change is 0.222222; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Model predictive control · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.6, and at B to obtain -0.6. Subtract the starting value from the ending value: the signed change is -1.2. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.6)−(0.6) = -1.2. The magnitude of the response change is 1.2; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hodgkin–Huxley membrane model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
FitzHugh–Nagumo model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.5 with condition B at x = 1.5. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.375, and at B to obtain 0.375. Subtract the starting value from the ending value: the signed change is 0.75. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.375)−(-0.375) = 0.75. The magnitude of the response change is 0.75; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hill muscle model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.444444, and at B to obtain 0.047619. Subtract the starting value from the ending value: the signed change is -0.396825. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.047619)−(0.444444) = -0.396825. The magnitude of the response change is 0.396825; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Windkessel circulation model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Pennes bioheat model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.632121, and at B to obtain 0.981684. Subtract the starting value from the ending value: the signed change is 0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.981684)−(0.632121) = 0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Reaction–diffusion morphogenesis model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.11843e-05, and at B to obtain 1.11843e-05. Subtract the starting value from the ending value: the signed change is 3.38813e-21. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.11843e-05)−(1.11843e-05) = 3.38813e-21. The magnitude of the response change is 3.38813e-21; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Monod growth model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.6 with condition B at x = 6.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.615385, and at B to obtain 0.864865. Subtract the starting value from the ending value: the signed change is 0.24948. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.864865)−(0.615385) = 0.24948. The magnitude of the response change is 0.24948; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Physiologically based compartment model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Boltzmann kinetic equation · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0220959, and at B to obtain 0.0220959. Subtract the starting value from the ending value: the signed change is -6.93889e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0220959)−(0.0220959) = -6.93889e-17. The magnitude of the response change is 6.93889e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Vlasov–Poisson model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0220959, and at B to obtain 0.0220959. Subtract the starting value from the ending value: the signed change is -6.93889e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0220959)−(0.0220959) = -6.93889e-17. The magnitude of the response change is 6.93889e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Vlasov–Maxwell model · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0220959, and at B to obtain 0.0220959. Subtract the starting value from the ending value: the signed change is -6.93889e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0220959)−(0.0220959) = -6.93889e-17. The magnitude of the response change is 6.93889e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Direct simulation Monte Carlo (DSMC) · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0220959, and at B to obtain 0.0220959. Subtract the starting value from the ending value: the signed change is -6.93889e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0220959)−(0.0220959) = -6.93889e-17. The magnitude of the response change is 6.93889e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Particle-in-cell (PIC) · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0220959, and at B to obtain 0.0220959. Subtract the starting value from the ending value: the signed change is -6.93889e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0220959)−(0.0220959) = -6.93889e-17. The magnitude of the response change is 6.93889e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Neutron diffusion approximation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.219072, and at B to obtain 0.219072. Subtract the starting value from the ending value: the signed change is 2.77556e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.219072)−(0.219072) = 2.77556e-17. The magnitude of the response change is 2.77556e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Point reactor kinetics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Bateman decay-chain model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.232544, and at B to obtain 0.0179802. Subtract the starting value from the ending value: the signed change is -0.214564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0179802)−(0.232544) = -0.214564. The magnitude of the response change is 0.214564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Newtonian gravitational N-body model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.309017, and at B to obtain 0.309017. Subtract the starting value from the ending value: the signed change is -2.22045e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.309017)−(0.309017) = -2.22045e-16. The magnitude of the response change is 2.22045e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
General relativity model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.44 with condition B at x = 6.61. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.768221, and at B to obtain 0.921257. Subtract the starting value from the ending value: the signed change is 0.153036. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.921257)−(0.768221) = 0.153036. The magnitude of the response change is 0.153036; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
FLRW cosmological model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.616 with condition B at x = 2.404. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.72397, and at B to obtain 1.79455. Subtract the starting value from the ending value: the signed change is 1.07058. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.79455)−(0.72397) = 1.07058. The magnitude of the response change is 1.07058; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Smoothed particle hydrodynamics (SPH) · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.2 with condition B at x = 1.2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0853333, and at B to obtain 0.0853333. Subtract the starting value from the ending value: the signed change is -6.93889e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0853333)−(0.0853333) = -6.93889e-17. The magnitude of the response change is 6.93889e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Discrete element method (DEM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.2, and at B to obtain 0.8. Subtract the starting value from the ending value: the signed change is 0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.8)−(0.2) = 0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Lattice Boltzmann method (LBM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.431818, and at B to obtain -0.431818. Subtract the starting value from the ending value: the signed change is -0.863637. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-0.431818)−(0.431818) = -0.863637. The magnitude of the response change is 0.863637; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Material point method (MPM) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.002, and at B to obtain 0.008. Subtract the starting value from the ending value: the signed change is 0.006. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.008)−(0.002) = 0.006. The magnitude of the response change is 0.006; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Homogenization · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.6, and at B to obtain 3.4. Subtract the starting value from the ending value: the signed change is 1.8. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (3.4)−(1.6) = 1.8. The magnitude of the response change is 1.8; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
QM/MM coupling · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.309017, and at B to obtain 0.309017. Subtract the starting value from the ending value: the signed change is -2.22045e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.309017)−(0.309017) = -2.22045e-16. The magnitude of the response change is 2.22045e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Atomistic–continuum coupling · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.25664 with condition B at x = 5.02655. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.309017, and at B to obtain 0.309017. Subtract the starting value from the ending value: the signed change is -2.22045e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.309017)−(0.309017) = -2.22045e-16. The magnitude of the response change is 2.22045e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Fluid–structure interaction (FSI) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 2.51327 with condition B at x = 10.0531. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.204895, and at B to obtain 0.678243. Subtract the starting value from the ending value: the signed change is 0.883138. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.678243)−(-0.204895) = 0.883138. The magnitude of the response change is 0.883138; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Thermomechanical coupling · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.4 with condition B at x = 1.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -0.4, and at B to obtain -1.6. Subtract the starting value from the ending value: the signed change is -1.2. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (-1.6)−(-0.4) = -1.2. The magnitude of the response change is 1.2; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Proper orthogonal decomposition (POD) · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.216234, and at B to obtain 0.216234. Subtract the starting value from the ending value: the signed change is 5.55112e-17. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.216234)−(0.216234) = 5.55112e-17. The magnitude of the response change is 5.55112e-17; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Gaussian-process surrogate · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1.8 with condition B at x = 1.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.197899, and at B to obtain 0.197899. Subtract the starting value from the ending value: the signed change is -3.05311e-16. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.197899)−(0.197899) = -3.05311e-16. The magnitude of the response change is 3.05311e-16; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Digital twin framework · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.367879, and at B to obtain 0.0183156. Subtract the starting value from the ending value: the signed change is -0.349564. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0183156)−(0.367879) = -0.349564. The magnitude of the response change is 0.349564; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Bayesian model calibration · Example 2
Two-condition response comparison
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)). Compare condition A at x = -1 with condition B at x = 2. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0594651, and at B to obtain 0.0594651. Subtract the starting value from the ending value: the signed change is 0. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0594651)−(0.0594651) = 0. The magnitude of the response change is 0; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Polynomial chaos expansion · Example 2
Two-condition response comparison
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)). Compare condition A at x = -0.6 with condition B at x = 0.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.7, and at B to obtain 2.3. Subtract the starting value from the ending value: the signed change is 0.6. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.3)−(1.7) = 0.6. The magnitude of the response change is 0.6; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Geometrically scaled physical model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.2 with condition B at x = 0.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.008, and at B to obtain 0.512. Subtract the starting value from the ending value: the signed change is 0.504. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.512)−(0.008) = 0.504. The magnitude of the response change is 0.504; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Wind-tunnel model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.4 with condition B at x = 1.6. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.16, and at B to obtain 2.56. Subtract the starting value from the ending value: the signed change is 2.4. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.56)−(0.16) = 2.4. The magnitude of the response change is 2.4; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hydraulic flume model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.208 with condition B at x = 0.802. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.45607, and at B to obtain 0.895545. Subtract the starting value from the ending value: the signed change is 0.439474. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.895545)−(0.45607) = 0.439474. The magnitude of the response change is 0.439474; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Dimensional-analysis similarity model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.208 with condition B at x = 0.802. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.45607, and at B to obtain 0.895545. Subtract the starting value from the ending value: the signed change is 0.439474. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.895545)−(0.45607) = 0.439474. The magnitude of the response change is 0.439474; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Shake-table structural model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.16 with condition B at x = 0.64. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.02627, and at B to obtain 1.69377. Subtract the starting value from the ending value: the signed change is 0.667494. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.69377)−(1.02627) = 0.667494. The magnitude of the response change is 0.667494; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Photoelastic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1 with condition B at x = 4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1, and at B to obtain 4. Subtract the starting value from the ending value: the signed change is 3. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (4)−(1) = 3. The magnitude of the response change is 3; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Ornstein-Zernike equation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.58904, and at B to obtain 0.996859. Subtract the starting value from the ending value: the signed change is 0.407818. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.996859)−(0.58904) = 0.407818. The magnitude of the response change is 0.407818; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Percus-Yevick closure · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.09 with condition B at x = 0.36. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.26192, and at B to obtain 2.88086. Subtract the starting value from the ending value: the signed change is 1.61893. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (2.88086)−(1.26192) = 1.61893. The magnitude of the response change is 1.61893; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Hypernetted-chain (HNC) closure · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.6 with condition B at x = 2.4. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.497741, and at B to obtain 0.996854. Subtract the starting value from the ending value: the signed change is 0.499113. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.996854)−(0.497741) = 0.499113. The magnitude of the response change is 0.499113; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Carnahan-Starling hard-sphere equation of state · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.09 with condition B at x = 0.36. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 1.45623, and at B to obtain 5.50439. Subtract the starting value from the ending value: the signed change is 4.04817. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (5.50439)−(1.45623) = 4.04817. The magnitude of the response change is 4.04817; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Stokes-Einstein diffusion relation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 48 with condition B at x = 162. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 4.54734e-12, and at B to obtain 1.34736e-12. Subtract the starting value from the ending value: the signed change is -3.19998e-12. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.34736e-12)−(4.54734e-12) = -3.19998e-12. The magnitude of the response change is 3.19998e-12; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Green-Kubo viscosity relation · Example 2
Two-condition response comparison
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)). Compare condition A at x = 1.2 with condition B at x = 4.8. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.698806, and at B to obtain 0.99177. Subtract the starting value from the ending value: the signed change is 0.292964. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.99177)−(0.698806) = 0.292964. The magnitude of the response change is 0.292964; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Einstein crystal heat-capacity model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.48 with condition B at x = 1.62. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.705082, and at B to obtain 0.968843. Subtract the starting value from the ending value: the signed change is 0.263761. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.968843)−(0.705082) = 0.263761. The magnitude of the response change is 0.263761; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Debye phonon model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.014 with condition B at x = 0.041. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.000641497, and at B to obtain 0.0161125. Subtract the starting value from the ending value: the signed change is 0.015471. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.0161125)−(0.000641497) = 0.015471. The magnitude of the response change is 0.015471; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Sommerfeld free-electron model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.0108 with condition B at x = 0.0402. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.0532959, and at B to obtain 0.198379. Subtract the starting value from the ending value: the signed change is 0.145083. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.198379)−(0.0532959) = 0.145083. The magnitude of the response change is 0.145083; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Tight-binding electronic model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.628319 with condition B at x = 2.51327. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain -1.61803, and at B to obtain 1.61803. Subtract the starting value from the ending value: the signed change is 3.23607. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (1.61803)−(-1.61803) = 3.23607. The magnitude of the response change is 3.23607; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Nearly-free-electron model · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.06 with condition B at x = 0.24. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.847395, and at B to obtain 0.567294. Subtract the starting value from the ending value: the signed change is -0.280101. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.567294)−(0.847395) = -0.280101. The magnitude of the response change is 0.280101; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.
Harmonic lattice dynamics · Example 2
Two-condition response comparison
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)). Compare condition A at x = 0.628319 with condition B at x = 2.51327. Find the signed response change q(B)−q(A).
Solution. Evaluate the original analytical expression at A to obtain 0.309017, and at B to obtain 0.951057. Subtract the starting value from the ending value: the signed change is 0.64204. The graph subtracts q(A) from every response, so its starting value is zero.
Worked evaluation. At condition B, q(B)−q(A) = (0.951057)−(0.309017) = 0.64204. The magnitude of the response change is 0.64204; its sign gives the direction relative to condition A.
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. A signed difference is reported; it is not automatically a time rate, a relative percentage, or a change of physical parameters.
Use the model entry’s references and assumptions for the full formulation. This curve is an analytical illustration, not measured product data.