Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pass
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.

Entropy dissipation of the orthogonal finite-volume viscous scheme

Statement

Every differentiable solution of One-dimensional orthogonal finite-volume viscous scheme with Ui(t)∈G satisfies

ddt∑i=1Nhη(Ui)=−∑i=1N−1(vi+1−vi)TKi+1/2(vi+1−vi)h≤0,ddt∑i=1NhUi=0.

The endpoint viscous fluxes vanish by definition, and the face quadratic forms use the state block Ki+1/2=K11((Ui+Ui+1)/2).

Facts & Assumptions

Given: L>0, an integer N≥1, h=L/N, differentiable cell states Ui(t)∈G, and the finite-volume viscous ODE in One-dimensional orthogonal finite-volume viscous scheme.

[F1]

One-dimensional orthogonal finite-volume viscous scheme: U˙i=(Gi+1/2−Gi−1/2)/h, Gi+1/2=Ki+1/2(vi+1−vi)/h, and G1/2=GN+1/2=0.

[F2]

Positive-semidefinite entropy-variable factorization of viscous flux: each one-direction state block K11(U) is symmetric positive semidefinite.

Proof

1.1F1algebra

For the total conservative state, substitute [F1] and telescope the finite sum: ddt∑i=1NhUi=∑i=1N(Gi+1/2−Gi−1/2)=GN+1/2−G1/2=0.

1.2F1F3givenalgebra

Since vi=∇Uη(Ui) by [F3], the chain rule gives ddt∑ihη(Ui)=∑ihviTU˙i. Substituting [F1] cancels the factors h and gives ∑iviT(Gi+1/2−Gi−1/2).

2.1F1step 1.2algebra

At each interior face i+1/2, the two adjacent cells contribute viTGi+1/2−vi+1TGi+1/2=−(vi+1−vi)TGi+1/2. The boundary terms vanish by [F1]. Substituting the face flux gives the exact negative sum in the Statement.

3.1F1F2step 2.1algebra∎

Each midpoint state is admissible by the convexity built into [F1], and its matrix block is positive semidefinite by [F2]. Thus every face term (vi+1−vi)TKi+1/2(vi+1−vi)/h is nonnegative because h>0. This proves the entropy inequality, including the case N=1 when the face sum is empty.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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