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Upwind SBP-SAT transport satisfies the semidiscrete L2 energy identity

Statement

Let $d \geq 1$ . If $d = 1$ , let the cells be a finite oriented interval mesh from Oriented interval meshes and broken polynomial spaces; if $d \geq 2$ , let them form a finite matching affine simplicial mesh. On every cell assume diagonal-norm SBP data of Diagonal-norm summation-by-parts element data, with paired interior face nodes, equal paired weights, and opposite outward normals. Let $β \in R^{d}$ be constant, and let the differentiable nodal state $U_{K} (t)$ solve the upwind SBP-SAT method of Upwind SBP-SAT discretization of constant transport with prescribed inflow values $g_{F, q}$ where $b_{K, F} : = β \cdot n_{K, F} < 0$ . Write $ω_{F, q}$ for the diagonal entries of the face norm. On an interior face oriented from $K^{-}$ to $K^{+}$ , set $b_{F} = β \cdot n_{F}$ and $[U]_{F, q} = U_{F, q}^{+} - U_{F, q}^{-}$ . Let $Γ_{+}$ and $Γ_{-}$ denote the boundary faces with positive and negative outward speed, respectively. For $d = 1$ , each face has one endpoint node and its face sum is a one-point sum. The discrete energy $E_{h} = \frac{1}{2} \sum_{K} U_{K}^{T} H_{K} U_{K}$ satisfies $\frac{d E_{h}}{d t} + \frac{1}{2} \sum_{F interior} \sum_{q} ω_{F, q} ∣ b_{F} ∣ [U]_{F, q}^{2} + \frac{1}{2} \sum_{F \subset Γ_{+}} \sum_{q} ω_{F, q} b_{K, F} U_{F, q}^{2} + \frac{1}{2} \sum_{F \subset Γ_{-}} \sum_{q} ω_{F, q} ∣ b_{K, F} ∣ (U_{F, q} - g_{F, q})^{2} = \frac{1}{2} \sum_{F \subset Γ_{-}} \sum_{q} ω_{F, q} ∣ b_{K, F} ∣ g_{F, q}^{2} .$ Consequently, zero inflow data give a nonincreasing SBP norm.

Facts & Assumptions

Given: Positive diagonal SBP norm matrices, derivative and face restriction matrices satisfying the exact SBP identity, the upwind SAT sign convention, constant $β$ , and matching interior face nodes, weights, and opposite normals.

[F1]

Diagonal-norm summation-by-parts element data: $Q_{K, ℓ} + Q_{K, ℓ}^{T} = \sum_{F \subset \partial K} n_{K, F, ℓ} R_{K, F}^{T} M_{K, F} R_{K, F}$ , with positive diagonal volume and face norms and matching face nodes and weights.

[F2]

Upwind SBP-SAT discretization of constant transport: $H_{K} {\dot{U}}_{K} + \sum_{ℓ = 1}^{d} β_{ℓ} Q_{K, ℓ} U_{K} + \sum_{F} R_{K, F}^{T} M_{K, F} b_{K, F} ({\hat{U}}_{K, F} - U_{K, F}) = 0$ with the upwind trace at every face.

Proof

1.1F1F2algebra

Multiply the cell equation in [F2] on the left by $U_{K}^{T}$ . The derivative term is $\frac{1}{2} \frac{d}{d t} (U_{K}^{T} H_{K} U_{K})$ because $H_{K}$ is symmetric. For each direction, the scalar identity $U_{K}^{T} Q_{K, ℓ} U_{K} = \frac{1}{2} U_{K}^{T} (Q_{K, ℓ} + Q_{K, ℓ}^{T}) U_{K}$ and [F1] give the face contribution $\frac{1}{2} \sum_{F} b_{K, F} U_{K, F}^{T} M_{K, F} U_{K, F}$ after summing with coefficients $β_{ℓ}$ . Including the SAT term, the cell energy equation is $\frac{1}{2} \frac{d}{d t} (U_{K}^{T} H_{K} U_{K}) + \sum_{F \subset \partial K} (b_{K, F} {\hat{U}}_{K, F}^{T} M_{K, F} U_{K, F} - \frac{b_{K, F}}{2} U_{K, F}^{T} M_{K, F} U_{K, F}) = 0.$

2.1F1F2step 1.1algebra

Fix an interior face oriented from $K^{-}$ to $K^{+}$ and a paired node $q$ . Its paired face weights are equal by [F1]. If $b_{F} > 0$ , both upwind traces are $U_{F, q}^{-}$ , so the two contributions in step 1.1 sum to $\frac{ω_{F, q} b_{F}}{2} (U_{F, q}^{-} - U_{F, q}^{+})^{2}$ . If $b_{F} < 0$ , the upwind traces are $U_{F, q}^{+}$ and the sum is $\frac{ω_{F, q} ∣ b_{F} ∣}{2} (U_{F, q}^{+} - U_{F, q}^{-})^{2}$ . If $b_{F} = 0$ , both contributions vanish.

2.2F2step 1.1algebra

At a boundary face with $b_{K, F} > 0$ , ${\hat{U}}_{K, F} = U_{K, F}$ , so each node contributes $ω_{F, q} b_{K, F} U_{F, q}^{2} / 2$ . When $b_{K, F} < 0$ , ${\hat{U}}_{K, F} = g_{F, q}$ , so the contribution is $ω_{F, q} ∣ b_{K, F} ∣ (U_{F, q} - g_{F, q})^{2} / 2 - ω_{F, q} ∣ b_{K, F} ∣ g_{F, q}^{2} / 2$ . Zero-speed faces contribute zero.

3.1givenstep 2.1step 2.2algebra∎

Sum the contributions from steps 2.1 and 2.2 over all cells. Each paired interior node appears once in the jump sum, and boundary nodes appear once in the corresponding inflow or outflow sum, yielding the displayed identity. Positivity of the face weights makes every dissipation term nonnegative; hence zero inflow data imply $d E_{h} / d t \leq 0$ .

Depends on

Used by

The SBP inflow SAT produces the boundary energy term Example

Dependency tree · two levels

5 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

Travis C. Fisher and Mark H. Carpenter, High-Order Entropy Stable Finite Difference Schemes for Nonlinear Conservation Laws: Finite Domains, NASA/TM-2013-217971 (standard reference, not scraped)
Bernardo Cockburn, An Introduction to the Discontinuous Galerkin Method for Convection-Dominated Problems, CIME lecture notes (standard reference, not scraped)