TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Semidiscrete cell entropy inequality for Tadmor-dissipative fluxes

Statement

Let $(η, q)$ be a convex entropy pair on an open set $A$ , with entropy variable $v$ and potential $ψ$ as in Convex entropy pairs, entropy variables, and entropy potentials for systems. On a periodic mesh of $N \geq 2$ cells of width $Δ x > 0$ , let differentiable states $U_{i} (t) \in A$ satisfy ${\dot{U}}_{i} = - \frac{F_{i + 1 / 2}^{e s} - F_{i - 1 / 2}^{e s}}{Δ x},$ where every face flux is the entropy-stable dissipative flux of Entropy-stable flux from positive-semidefinite entropy-variable dissipation. Define ${\hat{q}}_{i + 1 / 2} : = {(\frac{v_{i} + v_{i + 1}}{2})}^{T} F_{i + 1 / 2}^{e s} - \frac{ψ_{i} + ψ_{i + 1}}{2} .$ Then each cell satisfies $\frac{d}{d t} η (U_{i}) + \frac{{\hat{q}}_{i + 1 / 2} - {\hat{q}}_{i - 1 / 2}}{Δ x} \leq 0.$ Summing over the periodic mesh gives nonincrease of total discrete entropy. This is a semidiscrete result; it asserts no entropy inequality for a chosen fully discrete time integrator.

Facts & Assumptions

Given: A periodic semidiscrete finite-volume system, states in $A$ , a convex entropy pair, and the dissipative flux definition. Prove the local cell entropy inequality.

[F1]

Convex entropy pairs, entropy variables, and entropy potentials for systems: $v = \nabla η$ and $ψ = v^{T} F - q$ .

[F2]

Tadmor entropy-conservative two-point flux: $[v]^{T} F^{e c} = [ψ]$ at every face.

[F3]

Entropy-stable flux from positive-semidefinite entropy-variable dissipation: $F^{e s} = F^{e c} - \frac{1}{2} D [v]$ with $D = D^{T}$ and $z^{T} D z \geq 0$ .

Proof

1.1givenF2F3algebra

At a face, [F2]–[F3] give $[v]^{T} F^{e s} = [ψ] - \frac{1}{2} [v]^{T} D [v] \leq [ψ]$ , since $[v]^{T} D [v] \geq 0$ . This uses only positive semidefiniteness, including the case where the jump or dissipation vanishes.

2.1givenF1F2F3step 1.1algebra

Write $[v]_{i + 1 / 2} = v_{i + 1} - v_{i}$ and $[ψ]_{i + 1 / 2} = ψ_{i + 1} - ψ_{i}$ , and use the arithmetic means in the definition of $\hat{q}$ . Then $- v_{i}^{T} F_{i + 1 / 2}^{e s} + {\hat{q}}_{i + 1 / 2} = \frac{1}{2} [v]_{i + 1 / 2}^{T} F_{i + 1 / 2}^{e s} - \frac{1}{2} (ψ_{i} + ψ_{i + 1})$ and $v_{i}^{T} F_{i - 1 / 2}^{e s} - {\hat{q}}_{i - 1 / 2} = \frac{1}{2} [v]_{i - 1 / 2}^{T} F_{i - 1 / 2}^{e s} + \frac{1}{2} (ψ_{i - 1} + ψ_{i})$ . At equal states the same formula gives $\hat{q} = q$ by [F1] and the consistency in [F2], so this is a consistent numerical entropy flux. The right and left interface terms now combine with the entropy-jump defects from step 1.1.

3.1givenF1step 1.1step 2.1algebra∎

By the chain rule and the semidiscrete update, $Δ x \dot{η} (U_{i}) = - v_{i}^{T} F_{i + 1 / 2}^{e s} + v_{i}^{T} F_{i - 1 / 2}^{e s}$ . Add ${\hat{q}}_{i + 1 / 2} - {\hat{q}}_{i - 1 / 2}$ and use step 2.1. The potential terms combine as $- \frac{1}{2} (ψ_{i} + ψ_{i + 1}) + \frac{1}{2} (ψ_{i - 1} + ψ_{i}) = - \frac{1}{2} ([ψ]_{i + 1 / 2} + [ψ]_{i - 1 / 2})$ . Hence $Δ x$ times the cell residual equals $\frac{1}{2} ([v]_{i + 1 / 2}^{T} F_{i + 1 / 2}^{e s} - [ψ]_{i + 1 / 2}) + \frac{1}{2} ([v]_{i - 1 / 2}^{T} F_{i - 1 / 2}^{e s} - [ψ]_{i - 1 / 2})$ , which is nonpositive by step 1.1. This proves the local cell inequality; summing it on the periodic mesh cancels all numerical entropy fluxes. No time discretization was used.

Depends on

Used by

An entropy-stable Burgers flux still smears an under-resolved shock Example

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Tadmor, Entropy Stable Schemes for Hyperbolic Conservation Laws, Handbook of Numerical Methods for Hyperbolic Problems, Vol. 17 (standard reference, not scraped)
R. Bürger, P. E. Méndez, and C. Parés, On Entropy Stable Schemes for Degenerate Parabolic Multispecies Kinematic Flow Models (standard reference, not scraped)