Strong local convergence of consistent finite-volume face fluxes

Fix a closed bounded interval $I$ and $T^{\mathrm{\prime }}<T$, with common state bound $M$ on $I\times [0,T^{\mathrm{\prime }}]$. For $0<ϵ<T^{\mathrm{\prime }}$, let $K=I\times [ϵ,T^{\mathrm{\prime }}]$. If $E=\{(x,t)\in K:\mathrm{\mid }u(x,t)\mathrm{\mid }>M\}$ had positive measure, then $E={\bigcup }_{r\ge 1}{E}_r$, where $E_r=\{(x,t)\in K:\mathrm{\mid }u(x,t)\mathrm{\mid }\ge M+1\mathrm{/}r\}$. By countable subadditivity in [F3], some $E_r$ has positive measure. Since $\mathrm{\mid }{\mathcal{U}}_m\mathrm{\mid }\le M$ almost everywhere on $K$, it follows that $\mathrm{\parallel }{\mathcal{U}}_m-u{\mathrm{\parallel }}_{L^1(K)}\ge r^{-1}\mathrm{\mid }{E}_r\mathrm{\mid }>0$ for every $m$, contradicting strong $L^1$ convergence on $K$. Thus $\mathrm{\mid }u\mathrm{\mid }\le M$ almost everywhere on $K$. Taking $ϵ=1\mathrm{/}r$ for sufficiently large integers $r$ covers $I\times (0,T^{\mathrm{\prime }})$; any bounded interval is contained in a closed bounded one, so $u$ is locally essentially bounded up to $t=0$.

Fix $0<ϵ<T^{\mathrm{\prime }}$ and a compact rectangle $K\subset \mathbb{R}\times [ϵ,T^{\mathrm{\prime }}]$. Enlarge its spatial interval by $1$; step 1.1 bounds $u$ there. The cutoff $v=u$ on this enlarged bounded rectangle and $v=0$ elsewhere is bounded and measurable, so [F3] makes $v\in L^1({\mathbb{R}}^2)$. For all large $m$, $h_m<1$. By [F2], $\mathrm{\parallel }u(\cdot +h_m,\cdot )-u{\mathrm{\parallel }}_{L^1(K)}\to 0$: the cutoff agrees with $u$ on $K$ and on $K+h_m$. Also $K+h_m$ is contained in one fixed compact enlargement $K^{\mathrm{\prime }}$ for all large $m$, so strong local convergence gives $\mathrm{\parallel }{\mathcal{U}}_m(\cdot +h_m,\cdot )-u(\cdot +h_m,\cdot ){\mathrm{\parallel }}_{L^1(K)}=\mathrm{\parallel }{\mathcal{U}}_m-u{\mathrm{\parallel }}_{L^1(K+h_m)}\le \mathrm{\parallel }{\mathcal{U}}_m-u{\mathrm{\parallel }}_{L^1(K^{\mathrm{\prime }})}\to 0$. The equality follows by applying [F7] to the restriction of ${\mathcal{U}}_m-u$ to the compact set $K+h_m$. The triangle inequality gives ${\mathcal{U}}_m(\cdot +h_m,\cdot )\to u$ in $L^1(K)$.

Let $K$ be as in step 2.1, and choose a common bound $M$ on the enlarged rectangle. The square $S=[-M,M{]}^2$ is a nonempty closed box, hence compact by [F4]. Continuity of $F$ and [F5] bound $F$ on $S$; [F6] makes it uniformly continuous there. Put $a_m={\mathcal{U}}_m(x,t)$ and $b_m={\mathcal{U}}_m(x+h_m,t)$. Steps 1.1–2.1 show both converge to $u$ in measure on $K$, and all three values lie in $[-M,M]$ almost everywhere. For $\delta >0$, the set where either input error is at least $\delta$ has measure at most ${\delta }^{-1}$ times the sum of the two $L^1$ errors. Since $K$ has finite measure by [F3], uniform continuity makes $\mathrm{\mid }F(a_m,b_m)-F(u,u)\mathrm{\mid }$ arbitrarily small in integral off this set, while boundedness of $F$ controls its integral on the set. Thus $F(a_m,b_m)\to F(u,u)=f(u)$ in $L^1(K)$. By [F1], ${\mathcal{Q}}_m\to f(u)$ in $L^1(K)$.

On $I\times (0,ϵ)$, for all sufficiently large $m$ we have $h_m<1$, so the local state bound on the interval enlarged by $1$ bounds both flux inputs uniformly in $m$; step 1.1 bounds $u$. Continuity of $F$ on the compact square $[-M,M{]}^2$ and continuity of $f$ on the compact interval $[-M,M]$ bound ${\mathcal{Q}}_m$ and $f(u)$ there by one constant. The integral of $\mathrm{\mid }{\mathcal{Q}}_m-f(u)\mathrm{\mid }$ over this strip is at most a constant times ${\lambda }_2(I\times (0,ϵ))=\mathrm{\mid }I\mathrm{\mid }ϵ$ by [F8]. Step 3.1 gives convergence on $I\times [ϵ,T^{\mathrm{\prime }}]$. Letting first $m\to \mathrm{\infty }$ and then $ϵ\downarrow 0$ proves convergence on $I\times (0,T^{\mathrm{\prime }}]$.