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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
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Different Fourier coefficients give different model fluxes

Physical scope

Comparison of two stipulated isotropic Fourier models on the same temperature profile; no material measurement or universal constitutive law is asserted.

Example

On O=(0,L)3 in metre coordinates and I=(0,τ) in second coordinates, with L>0 m and τ>0 s, prescribe T(x,t)=T0+ΔT x1/L, where T0>0 K and ΔT>0 K. In the chosen Cartesian coordinates, define e1:=(1,0,0)∈R3. Compare two isotropic model choices κ1,κ2>0, κ1≠κ2, each using jQ=−κi∇T. The coefficients have units W m−1 K−1.

Facts & Assumptions

Given: The Fourier-model definition and the stipulated fields and coefficients.

[F1]

Scope of transport constitutive laws: for a temperature field with continuous first spatial partial derivatives, ∇T=(∂x1T,∂x2T,∂x3T) in the physical one-based coordinate display (function indices i−1), and a Fourier model may postulate jQ=−K∇T, where K has entries in W m−1 K−1; its isotropic form is K=κI3 with κ≥0.

[A1]

T0>0 K, ΔT>0 K, L>0 m; e1=(1,0,0) in the chosen Cartesian coordinates; κ1,κ2>0 in W m−1 K−1 with κ1≠κ2; both models use the same profile T=T0+ΔT x1/L; and model i adopts jQ,i=−κi∇T.

Verification

Given: [F1] and [A1].

1.1A1constructalgebra

The common temperature gradient is ∇T=(ΔT/L)e1. The two assigned model fluxes are therefore jQ,1=−κ1(ΔT/L)e1 and jQ,2=−κ2(ΔT/L)e1.

2.1A1step 1.1algebra

Their difference is jQ,1−jQ,2=−(κ1−κ2)(ΔT/L)e1≠0. The same field profile therefore has different predicted fluxes under these two constitutive choices.

3.1F1A1step 2.1∎

The calculation uses the Fourier relation as a stated model input and does not determine either κi from equilibrium entropy data. It is a model comparison, not a measured result.

Depends on

Dependency tree · 2 levels