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The degree-zero upwind forward-Euler step preserves mass and convex entropy

Statement

Let $T_{h}$ be a finite periodic matching affine simplicial mesh in $d \geq 2$ , let $β \in R^{d}$ be constant, and apply Forward-Euler step for degree-zero upwind DG transport with $λ_{K} \leq 1$ for every cell. Assume Countable Choice for the imported affine-simplex Green formula. Then the cell-volume-weighted mean is conserved: $\sum_{K} V_{K} U_{K}^{+} = \sum_{K} V_{K} U_{K} .$ For every convex function $η : R \to R$ , the weighted discrete entropy does not increase: $\sum_{K} V_{K} η (U_{K}^{+}) \leq \sum_{K} V_{K} η (U_{K}) .$

Facts & Assumptions

Given: A finite periodic matching affine simplicial mesh, constant transport velocity, the degree-zero upwind forward-Euler update, the cellwise face CFL $λ_{K} \leq 1$ , and a convex scalar function $η$ .

[F1]

Forward-Euler step for degree-zero upwind DG transport: $U_{K}^{+} = U_{K} - \frac{Δ t}{V_{K}} \sum_{F} A_{F} (b_{K, F}^{+} U_{K} + b_{K, F}^{-} U_{K_{F}})$ with $λ_{K} = \frac{Δ t}{V_{K}} \sum_{F} A_{F} b_{K, F}^{+} \leq 1$ .

[F2]

Gauss–Green formula on an affine simplex: $\sum_{F \subset \partial K} ∣ F ∣ n_{K, F} = 0$ , obtained by applying the affine-simplex Green formula to each constant vector field.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is used only through [F2].

Proof

1.1F1F2A1algebra

By [F2] with the constant vector field $β$ , $\sum_{F} A_{F} b_{K, F} = 0$ , hence $\sum_{F} A_{F} (- b_{K, F})^{+} = \sum_{F} A_{F} b_{K, F}^{+}$ . Rewriting [F1] gives $U_{K}^{+} = (1 - λ_{K}) U_{K} + \sum_{F \subset \partial K} \frac{Δ t A_{F} (- b_{K, F})^{+}}{V_{K}} U_{K_{F}}$ . All coefficients are nonnegative by $λ_{K} \leq 1$ , and their sum is $(1 - λ_{K}) + λ_{K} = 1$ .

2.1givenstep 1.1algebra

Convexity of $η$ and the convex-combination formula in step 1.1 give $V_{K} η (U_{K}^{+}) \leq V_{K} (1 - λ_{K}) η (U_{K}) + Δ t \sum_{F} A_{F} (- b_{K, F})^{+} η (U_{K_{F}})$ . Sum over cells. Each incoming coefficient for an upwind cell $J$ is paired with the equal outgoing coefficient $Δ t A_{F} b_{J, F}^{+}$ on that same periodic face, so these terms exactly fill the subtracted $V_{J} λ_{J} η (U_{J})$ contributions. Therefore $\sum_{K} V_{K} η (U_{K}^{+}) \leq \sum_{K} V_{K} η (U_{K})$ .

3.1F1givenalgebra∎

Multiply the update [F1] by $V_{K}$ and sum over cells. Every periodic interior face contributes one shared flux with each of its two opposite outward normals, so its two signed contributions cancel. This proves $\sum_{K} V_{K} U_{K}^{+} = \sum_{K} V_{K} U_{K}$ .

Depends on

Used by

SSPRK(3,3) preserves the degree-zero upwind entropy bound Theorem

Dependency tree · two levels

16 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

Bernardo Cockburn and Chi-Wang Shu, Runge–Kutta Discontinuous Galerkin Methods for Convection-Dominated Problems, Journal of Scientific Computing 16 (2001) (standard reference, not scraped)