Smooth-solution L2 error bound for upwind DG transport

Multiplying the exact PDE by $w\in {\mathbb{P}}_p(K)$ and integrating by parts with [F3] when $d\ge 2$ gives its weak equation. When $d=1$, apply [F12] to the product $\beta uw$ on each interval to obtain the same endpoint weak equation. In both branches the exact inflow trace is $g$ and the outflow trace is the exact upwind trace.

For $d\ge 2$, each cell is a nondegenerate simplex whose interior is dense in the cell by [F6]. Every nonempty relatively open subset therefore intersects the interior in a nonempty Euclidean-open set, which contains a Euclidean ball. For $d=1$, each cell is an open interval by [F7], so every nonempty relatively open subset contains a Euclidean ball. In either branch [F5] and [A1] give that ball positive Lebesgue measure. Thus Lebesgue measure has full support on every cell.

Each cell is bounded and Borel: this follows from the interval description [F7] when $d=1$ and from the closed affine-simplex description [F6] when $d\ge 2$. Choose a finite-radius ball containing the cell. By [F5] the ball is Lebesgue measurable with finite measure, while [F8] makes the cell measurable; [F9] and [F10] then give ${\lambda }_d(K)\le {\lambda }_d(B)<\mathrm{\infty }$.

The full-support condition in step 1.2 and finite-measure condition in step 1.3 let us invoke the cellwise projection lemma [F11] on every cell. Since $u(t,\cdot )$ is continuous and each cell has finite measure, it belongs to $L^2(K)$; hence ${\mathrm{\Pi }}_{h}u$ is well defined.

Set $\theta =u_h-{\mathrm{\Pi }}_{h}u$, so $u_h-u=\theta +\eta$. Subtract the exact weak equation in step 1.1 from [F1] and test with $v_h=\theta$. To justify differentiating projection orthogonality [F2], choose a monomial basis $\{{\varphi }_i\}$ of ${\mathbb{P}}_p(K)$ and its fixed Gram matrix. By the construction in [F11], the projection coefficients solve that fixed finite linear system with moments ${\int }_{K}u(t){\varphi }_{i}dx$. Since $u_t$ is continuous on the compact space-time cell, [F12] and uniform continuity show that the difference quotients of $u$ converge uniformly in $x$ to $u_t$; the bounded test polynomials and finite measure from step 1.3 therefore allow passage through each moment integral. The coefficients, and hence ${\mathrm{\Pi }}_{K}u$, are differentiable in time, so differentiating [F2] gives $({\eta }_t,\theta {)}_K=0$. If $p=0$, then $\mathrm{\nabla }\theta =0$; if $p\ge 1$, then $\mathrm{\nabla }\theta \in {\mathbb{P}}_{p-1}(K{)}^d$. Since $\beta$ is constant, in either case $\beta \cdot \mathrm{\nabla }\theta \in {\mathbb{P}}_p(K)$ and $(\eta ,\beta \cdot \mathrm{\nabla }\theta {)}_K=0$. This leaves only the transport error and the upwind face-flux differences.

Integrate the remaining $-(\beta \theta ,\mathrm{\nabla }\theta {)}_K$ by parts using [F3] for $d\ge 2$ and [F12] on intervals for $d=1$, then pair cells using [F4] for $d\ge 2$ or the endpoint convention in [F7] for $d=1$. At an interior face the homogeneous upwind terms combine to $\frac{1}{2}{\int }_F\mathrm{\mid }\beta \cdot n_F\mathrm{\mid }[\theta {]}^2$; at inflow the volume term is $\frac{1}{2}{\int }_{{\mathrm{\Gamma }}_{-}}\mathrm{\mid }\beta \cdot n\mathrm{\mid }{\theta }^2$, and at outflow the homogeneous term is $\frac{1}{2}{\int }_{{\mathrm{\Gamma }}_{+}}(\beta \cdot n){\theta }^2$. By the face flux differences from step 3.1, inflow contributes no projection residual because both fluxes use $g$; interior and outflow residuals move to the right with signs $+{\int }_F(\beta \cdot n_F){\eta }^{\mathrm{u}\mathrm{p}}[\theta ]$ and $-{\int }_{{\mathrm{\Gamma }}_{+}}(\beta \cdot n)\eta \theta$, respectively. Hence $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }\theta {\mathrm{\parallel }}_2^2+\frac{1}{2}{\sum }_{F\in {\mathcal{F}}_h^{\mathrm{i}\mathrm{n}\mathrm{t}}}{\int }_F\mathrm{\mid }\beta \cdot n_F\mathrm{\mid }[\theta {]}^2+\frac{1}{2}{\int }_{{\mathrm{\Gamma }}_{-}}\mathrm{\mid }\beta \cdot n\mathrm{\mid }{\theta }^2+\frac{1}{2}{\int }_{{\mathrm{\Gamma }}_{+}}(\beta \cdot n){\theta }^2={\sum }_{F\in {\mathcal{F}}_h^{\mathrm{i}\mathrm{n}\mathrm{t}}}{\int }_F(\beta \cdot n_F){\eta }^{\mathrm{u}\mathrm{p}}[\theta ]-{\int }_{{\mathrm{\Gamma }}_{+}}(\beta \cdot n)\eta \theta \mathrm{.}$

For each interior face use $\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }{\eta }^{\mathrm{u}\mathrm{p}}\mathrm{\mid }\mathrm{\mid }[\theta ]\mathrm{\mid }\le \frac{\mathrm{\mid }b\mathrm{\mid }}{4}[\theta {]}^2+\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }{\eta }^{\mathrm{u}\mathrm{p}}{\mathrm{\mid }}^2$, and on outflow use $b\mathrm{\mid }\eta \mathrm{\mid }\mathrm{\mid }\theta \mathrm{\mid }\le \frac{b}{4}{\theta }^2+b\mathrm{\mid }\eta {\mathrm{\mid }}^2$. Absorbing the first terms into the half-dissipation terms of step 4.1 and dropping the remaining nonnegative boundary and jump terms gives $\frac{d}{dt}\mathrm{\parallel }\theta {\mathrm{\parallel }}_2^2\le 2R_h(t{)}^2$.

Integrating the differential inequality gives $\mathrm{\parallel }\theta (T){\mathrm{\parallel }}_2^2\le \mathrm{\parallel }\theta (0){\mathrm{\parallel }}_2^2+2{\int }_0^T{R}_h(t{)}^{2}dt$. Since $u_h-u=\theta +\eta$, the triangle inequality at $T$ gives the stated estimate.