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Upwind DG transport satisfies the semidiscrete L2 energy identity

Statement

Let $d \geq 1$ . If $d = 1$ , let $T_{h}$ be a finite oriented interval mesh of $Ω = (a, b)$ , $a < b$ , from Oriented interval meshes and broken polynomial spaces; if $d \geq 2$ , let $T_{h}$ be a finite matching affine simplicial mesh of $Ω$ . Let $β \in R^{d}$ be constant, and let $u_{h}$ solve Upwind discontinuous Galerkin discretization of constant transport with inflow data $g$ . Assume Countable Choice when $d \geq 2$ , as required by the affine-simplex Green supplier; no choice assumption is needed for the interval branch. Write $Γ_{+} = {x \in \partial Ω : β \cdot n > 0}$ and $Γ_{-} = {x \in \partial Ω : β \cdot n < 0}$ . For an interior facet $F = K^{-} \cap K^{+}$ choose a unit normal $n_{F}$ from $K^{-}$ to $K^{+}$ and write $[u_{h}] = u_{h}^{+} - u_{h}^{-}$ . In dimension one, integrals over facets and boundary points mean sums with respect to zero-dimensional surface measure. Then $\frac{1}{2} \frac{d}{d t} \sum_{K \in T_{h}} ∥ u_{h} ∥_{L^{2} (K)}^{2} + \frac{1}{2} \sum_{F interior} \int_{F} ∣ β \cdot n_{F} ∣ [u_{h}]^{2} d S + \frac{1}{2} \int_{Γ_{+}} (β \cdot n) u_{h}^{2} d S + \frac{1}{2} \int_{Γ_{-}} ∣ β \cdot n ∣ (u_{h} - g)^{2} d S = \frac{1}{2} \int_{Γ_{-}} ∣ β \cdot n ∣ g^{2} d S .$ The equality is understood in $R \cup {+ \infty}$ . It is a finite energy balance whenever $\int_{Γ_{-}} ∣ β \cdot n ∣ g^{2} d S < \infty$ at the time in question; otherwise both sides are $+ \infty$ and give no finite inflow bound. In particular, zero inflow data give a nonincreasing semidiscrete $L^{2}$ norm. This result is semidiscrete and makes no claim about a time integrator.

Facts & Assumptions

Given: The upwind DG weak form with constant $β$ , either the oriented interval mesh or the matching affine simplicial mesh according to $d$ , smooth cellwise polynomial states, and the stated inflow trace. Assume Countable Choice only for the affine-simplex branch.

[F1]

Upwind discontinuous Galerkin discretization of constant transport: on each cell, $(\partial_{t} u_{h}, v)_{K} - (β u_{h}, \nabla v)_{K} + \sum_{F \subset \partial K} ⟨ (β \cdot n_{K, F}) {\hat{u}}_{K, F}, v ⟩_{F} = 0$ .

[F2]

When $d \geq 2$ , Gauss–Green formula on an affine simplex gives $\int_{K} div G = \sum_{F \subset \partial K} \int_{F} G \cdot n_{K, F}$ for $G \in C^{1} (\overline{K}; R^{d})$ .

[F3]

If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz: for a continuously differentiable function on a closed interval, the integral of its derivative is the difference of its endpoint values.

[F4]

For $d \geq 2$ , Interior facet orientations cancel in a matching mesh says every interior facet of a matching affine mesh occurs twice with opposite induced orientations.

[F5]

Oriented interval meshes and broken polynomial spaces: adjacent interval cells have opposite outward normals at their common endpoint.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is used only for [F2] on the $d \geq 2$ branch; it is not needed in dimension one.

Proof

1.1givenF1F2F3F4F5A1algebra

In the DG equation on each cell take $v = u_{h}$ . The time term is $\frac{1}{2} \frac{d}{d t} ∥ u_{h} ∥_{L^{2} (K)}^{2}$ . When $d \geq 2$ , since $β$ is constant, $- \int_{K} β u_{h} \cdot \nabla u_{h} = - \frac{1}{2} \int_{K} div (β u_{h}^{2}) = - \frac{1}{2} \sum_{F \subset \partial K} \int_{F} (β \cdot n_{K, F}) u_{h}^{2}$ by [F2] and [A1]. When $d = 1$ and $K = (x_{j - 1}, x_{j})$ , [F3] gives the same endpoint identity $- \int_{K} β u_{h} \partial_{x} u_{h} = - \frac{1}{2} \sum_{e \in \partial K} (β n_{K, e}) u_{h} (e)^{2}$ . Sum over cells; [F4] pairs opposite normals on each interior facet for $d \geq 2$ , and [F5] gives opposite normals at every shared interval endpoint.

2.1F1step 1.1algebra

Fix an interior facet oriented from $K^{-}$ to $K^{+}$ and put $b = β \cdot n_{F}$ . The two face terms from step 1.1 and the two upwind numerical-flux terms combine to $- \frac{b}{2} ((u^{-})^{2} - (u^{+})^{2}) + b \hat{u} (u^{-} - u^{+})$ . If $b > 0$ , then $\hat{u} = u^{-}$ and this is $\frac{b}{2} (u^{+} - u^{-})^{2}$ ; if $b < 0$ , then $\hat{u} = u^{+}$ and this is $\frac{∣ b ∣}{2} (u^{+} - u^{-})^{2}$ . For $b = 0$ it vanishes. This scalar calculation also applies to the single endpoint quadrature in dimension one.

2.2F1step 1.1algebra

On an outflow boundary face, $b = β \cdot n > 0$ and $\hat{u} = u_{h}$ , so the volume and flux terms combine to $\frac{b}{2} u_{h}^{2}$ . On an inflow face, $b < 0$ and $\hat{u} = g$ , so they combine to $\frac{∣ b ∣}{2} u_{h}^{2} - ∣ b ∣ g u_{h} = \frac{∣ b ∣}{2} (u_{h} - g)^{2} - \frac{∣ b ∣}{2} g^{2}$ . Faces with $b = 0$ contribute zero.

3.1givenstep 2.1step 2.2algebra∎

If $\int_{Γ_{-}} ∣ β \cdot n ∣ g^{2} d S < \infty$ , all boundary terms in step 2.2 are integrable, since the polynomial traces are bounded on the finitely many bounded facets. Substituting steps 2.1 and 2.2 into the tested DG sum gives the finite displayed identity. If that inflow integral is infinite, the inequality $g^{2} \leq 2 (g - u_{h})^{2} + 2 u_{h}^{2}$ and finiteness of the weighted polynomial-trace integral force $\int_{Γ_{-}} ∣ β \cdot n ∣ (u_{h} - g)^{2} d S = + \infty$ ; the displayed identity then reads $+ \infty = + \infty$ , without subtracting infinite integrals. For $g = 0$ all dissipation terms are nonnegative and the finite identity proves the stability statement.

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Sources

Bernardo Cockburn, An Introduction to the Discontinuous Galerkin Method for Convection-Dominated Problems, CIME lecture notes (standard reference, not scraped)
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)