LemmaStatement: AI-adaptedProof: 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.

Jacobi identity for the ALE mesh Jacobian

Statement

For the ALE moving-domain map $A_{t}$ of ALE moving-domain map and mesh velocity, let $F (ξ, t) : = D_{ξ} A_{t} (ξ)$ and $J (ξ, t) : = \det F (ξ, t)$ . Then

$\partial_{t} J (ξ, t) = J (ξ, t) {div}_{x} w_{m} (A_{t} (ξ), t) .$

This is the continuous geometric conservation identity. It makes no discrete or time-integration claim.

Facts & Assumptions

Given: The $C^{2}$ orientation-preserving ALE map $A_{t}$ , its Jacobian $J > 0$ , and mesh velocity $w_{m}$ as defined in ALE moving-domain map and mesh velocity.

[F1]

ALE moving-domain map and mesh velocity: $A$ is $C^{2}$ in space and time, $F = D_{ξ} A_{t}$ is invertible, $J = \det F$ , and $w_{m} (A_{t} (ξ), t) = \partial_{t} A_{t} (ξ)$ .

Proof

1.1givenF1algebra

Fix $(ξ, t)$ and write $H = \partial_{t} F (ξ, t)$ . Differentiability gives $F (ξ, t + h) = F + h H + o (h)$ . Since $F$ is invertible, factor this as $F [I + h F^{- 1} H + o (h)]$ . The permutation expansion of the determinant gives $\det (I + h K + o (h)) = 1 + h tr K + o (h)$ : the identity permutation contributes $1 + h \sum_{j} K_{j j} + o (h)$ , while every nonidentity permutation contains at least two off-diagonal factors of order $h$ . Hence $\partial_{t} J = J tr (F^{- 1} H)$ .

2.1F1step 1.1algebra∎

Differentiate $w_{m} (A_{t} (ξ), t) = \partial_{t} A_{t} (ξ)$ in $ξ$ . The chain rule and equality of the mixed derivatives of the $C^{2}$ map give $H = (D_{x} w_{m}) (A_{t} (ξ), t) F$ . Therefore $tr (F^{- 1} H) = tr (F^{- 1} (D_{x} w_{m}) F) = tr (D_{x} w_{m}) = {div}_{x} w_{m}$ , using cyclic invariance of trace. Substitution into step 1.1 proves the identity.

Depends on

ALE moving-domain map and mesh velocity

Used by

Dependency tree · two levels

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Sources

J. Donea, A. Huerta, J.-Ph. Ponthot, and A. Rodríguez-Ferran, Arbitrary Lagrangian–Eulerian Methods, Encyclopedia of Computational Mechanics, Vol. 1, Chapter 14 (2004) (standard reference, not scraped)