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PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)
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Entropy jump residual

Statement

Assume the Axiom of Countable Choice (ACω).

Let I⊆R and Ω⊆Rd be open, let O⊆Rm be open and convex, and let Fj∈C1(O;Rm), 1≤j≤d. Let n∈Rd have ∣n∣=1, let s∈R, and set Σ={(t,x)∈I×Ω:n⋅x=st}. Suppose U is piecewise C1 across Σ, satisfies Ut+∑jDFj(U)∂xjU=0 classically on each open side, and the state, flux, and entropy compositions are locally integrable. Let η∈C2(O;R) be convex and let qj∈C1(O;R) satisfy Dqj(V)=Dη(V)DFj(V) for every V∈O. With [G]=GR−GL, the distributional entropy residual is ∂tη(U)+∑j=1d∂xjqj(U)=(∑j=1dnj[qj(U)]−s[η(U)])δΣ. Here δΣ uses the plane measure defined in the vector jump-residual theorem: integrate in t and in z∈Rd−1 with x=Ez+stn∈Ω, where E:Rd−1→n⊥ is a linear isometry onto n⊥, using Lebesgue measure λd−1 and singleton mass one when d=1. The entropy inequality ∂tη(U)+∑j∂xjqj(U)≤0 holds near the interface if and only if ∑jnj[qj(U)]−s[η(U)]≤0 at every interface point. Entropy admissibility is an additional condition; the conservation jump equality alone does not imply it.

Facts & Assumptions

Given: Countable Choice; the state field, entropy pair, and planar interface in the Statement.

[F1]

Distributional systems and entropy flux pairs defines a convex entropy pair by Dqj=Dη DFj and fixes the entropy inequality sign by requiring the nonnegative-test integral to be nonnegative.

[F2]

Planar shocks and piecewise smooth traces fixes the one-sided traces and jump orientation [G]=GR−GL.

[F3]

Vector Rankine–Hugoniot condition proves the interface distribution calculation from the distributional derivative definition and one-dimensional integration by parts in a normal coordinate.

[F4]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a) states D(q∘U)=Dq(U)∘DU when U and q are totally differentiable.

[F5]

The Axiom of Countable Choice (ACω): Countable Choice is the hypothesis required by the Rankine–Hugoniot supplier in [F3].

[A1]

On each open side, U satisfies the conservation system classically and has the piecewise C1 extensions from [F2].

Proof

1.1givenA1F1F4algebra

On either open side, [F4] and entropy compatibility give ∂tη(U)+∑j∂xjqj(U)=Dη(U)(∂tU+∑jDFj(U)∂xjU)=0.

2.1givenF2F3F5step 1.1

The fields η(U) and qj(U) have C1 extensions to each side of Σ. Apply the normal-coordinate distribution calculation from [F3] under the stated Countable Choice assumption [F5] to this scalar density and its scalar fluxes; the bulk residuals vanish by step 1.1, leaving the interface coefficient ∑jnj[qj(U)]−s[η(U)]. The interface measure is the time integral of the isometric pushforward of Lebesgue measure specified in [F3].

3.1givenF2step 2.1∎

If this coefficient is nonpositive everywhere, pairing the residual with any nonnegative test gives a nonpositive value, which is the distributional inequality. If the coefficient is positive at a point, continuity makes it positive on a smaller interface patch; a nonnegative test supported there has positive pairing, contradicting the inequality. Thus the distributional inequality holds exactly under the stated pointwise condition.

Depends on

Dependency tree · 2 levels

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