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

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The Roe flux is consistent and conservative

Statement

For any supplied Roe matrix, the Roe flux is consistent: ${\hat{f}}_{R o e} (u, u) = f (u)$ . For a finite periodic array define $U_{j}^{n e w} = U_{j} - ν ({\hat{f}}_{j + 1 / 2} - {\hat{f}}_{j - 1 / 2}),$ where ${\hat{f}}_{j + 1 / 2} = {\hat{f}}_{R o e} (U_{j}, U_{j + 1})$ and indices are modulo $n$ , with $n \geq 1$ . For any scalar $ν$ , the update preserves the sum of cell states. The Roe jump property is the defining matrix identity in Roe linearization and Roe numerical flux and is not an entropy assertion.

Facts & Assumptions

Given: A Roe matrix for each state pair and the flux definition.

[F1]

Roe linearization and Roe numerical flux: ${\hat{f}}_{R o e} = \frac{1}{2} (f_{L} + f_{R}) - \frac{1}{2} ∣ \tilde{A} ∣ (u_{R} - u_{L})$ and $\tilde{A} (u, u) = D f (u)$ .

Proof

1.1givenF1algebra

If $u_{L} = u_{R} = u$ , the jump is zero, so [F1] gives ${\hat{f}}_{R o e} (u, u) = \frac{1}{2} (f (u) + f (u)) = f (u)$ .

2.1givenF1algebra∎

Substituting the displayed periodic update into the sum of all cells gives $- ν \sum_{j} ({\hat{f}}_{j + 1 / 2} - {\hat{f}}_{j - 1 / 2})$ . Each interface flux occurs once with each sign, so the sum is zero. The Roe jump identity is not used in this telescoping argument and no entropy inequality follows from it.

Depends on

Roe linearization and Roe numerical flux

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Randall J. LeVeque, Numerical Methods for Conservation Laws (standard reference, not scraped)