Conservative finite-volume weak limits

On each open region cut out by $a=b$, $a=0$, and $b=0$, the displayed formula in [F1] is either a polynomial, zero, or $(a^2+b^2+\mathrm{\mid }{a}^2-b^2\mathrm{\mid })\mathrm{/}4$, so it is continuous there.

At a boundary with $a=b>0$ both adjacent formulas tend to $a^2\mathrm{/}2$; at $a=b<0$ both tend to $a^2\mathrm{/}2$; at $a=0\le b$ the left and right limits are zero; at $a\le 0=b$ the limits are zero. On the remaining axis portions $a=0>b$ and $b=0<a$, a neighborhood lies in the single shock branch $b<a$, whose formula is continuous there. At $(a,b)=(0,0)$ every branch has absolute value at most $\max \{a^2,b^2\}\mathrm{/}2$, which tends to zero. These checks cover every branch boundary and prove continuity on all of ${\mathbb{R}}^2$.

On the diagonal, if $a\ge 0$ the first branch gives $G(a,a)=a^2\mathrm{/}2$; if $a<0$ the second branch gives $G(a,a)=a^2\mathrm{/}2$. Thus $G(a,a)=f_B(a)$ for all $a$, and [F2] makes $G$ a consistent numerical flux.

Multiply [F1] by ${\mathrm{\Phi }}_j^{m,n}$ and sum over $j\in \mathbb{Z}$ and $n=0,\dots ,N-1$. Only finitely many terms are nonzero, so the summed update is an exact finite sum equal to zero.

For each fixed $j$, telescoping gives ${\sum }_{n=0}^{N-1}(U_j^{m,n+1}-U_j^{m,n}){\mathrm{\Phi }}_j^{m,n}=-U_j^{m,0}{\mathrm{\Phi }}_j^{m,0}+{\sum }_{n=1}^{N-1}{U}_j^{m,n}({\mathrm{\Phi }}_j^{m,n-1}-{\mathrm{\Phi }}_j^{m,n}),$ since ${\mathrm{\Phi }}_j^{m,N-1}=0$ by the terminal support condition.

Reindex the spatial flux sum: ${\sum }_j(Q_{j+1\mathrm{/}2}^{m,n}-Q_{j-1\mathrm{/}2}^{m,n}){\mathrm{\Phi }}_j^{m,n}={\sum }_j{Q}_{j+1\mathrm{/}2}^{m,n}({\mathrm{\Phi }}_j^{m,n}-{\mathrm{\Phi }}_{j+1}^{m,n})\mathrm{.}$ Substitute this and step 2.1 into step 1.1, then multiply the zero identity by $-1$ to obtain exactly the displayed formula.

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 $I\times (0,T^{\mathrm{\prime }}]$.

Fix $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$. Its time support on $[0,T)$ is contained in $[0,T_{\phi }]$ for some $T_{\phi }<T$. For all sufficiently large $m$, choose $N_m\ge 3$ with $T_{\phi }<t_{N_m-2}^m<t_{N_m-1}^m<t_{N_m}^m<T$. By [F1] and [F2] these are precisely the scheme states and face fluxes of the reconstruction, so [F3] gives the exact summed update identity for this test.

The initial term in [F3] converges to the desired boundary term. Indeed, $h_m{U}_j^{m,0}={\int }_{I_j^m}{u}_0$, while ${\mathrm{\Phi }}_j^{m,0}$ is the average of $\phi$ on $I_j^m\times [0,{\tau }_m)$. Smoothness gives $\mathrm{\mid }{\mathrm{\Phi }}_j^{m,0}-\phi (x,0)\mathrm{\mid }\le C_{\phi }(h_m+{\tau }_m)$ for $x\in I_j^m$; summing the resulting errors over the finitely many cells meeting the support bounds the total by $C_{\phi }(h_m+{\tau }_m){\int }_K\mathrm{\mid }{u}_0\mathrm{\mid }\to 0$ on a fixed compact $K$.

For $n\ge 1$, the difference ${\mathrm{\Phi }}_j^{m,n}-{\mathrm{\Phi }}_j^{m,n-1}$ differs from $h_m^{-1}{\int }_{I_j^m}{\int }_{J_n^m}{\phi }_t$ by at most $C_{\phi }{\tau }_m^2$, by Taylor's formula in time for the smooth function $t\mapsto h_m^{-1}{\int }_{I_j^m}\phi (x,t)dx$. After multiplication by $h_m{U}_j^{m,n}$ and summation over $O(h_m^{-1})$ cells and $O({\tau }_m^{-1})$ time slabs in the fixed support, the total error is $O({\tau }_m)$. The missing first-slab integral is also $O({\tau }_m)$ because $\mathrm{\mid }{\mathcal{U}}_m\mathrm{\mid }$ is uniformly bounded on the compact support and ${\phi }_t$ is bounded. Hence the time sum in [F3] differs from ${\int }_0^T\int {\mathcal{U}}_m{\phi }_t$ by a quantity tending to zero.

For each $j,n$, translation of a cell interval and the fundamental theorem of calculus give ${\tau }_m({\mathrm{\Phi }}_{j+1}^{m,n}-{\mathrm{\Phi }}_j^{m,n})={\int }_{J_n^m}{\int }_{I_j^m}\frac{1}{h_m}{\int }_0^{h_m}{\phi }_x(x+s,t)dsdxdt$. Its difference from ${\int }_{J_n^m}{\int }_{I_j^m}{\phi }_x$ is at most $C_{\phi }{h}_m^2{\tau }_m$, by the bound on ${\phi }_{xx}$. The numerical face fluxes are uniformly bounded on cells meeting the test support because $F$ is continuous on the compact state square supplied by local boundedness. Summing over the finitely many cells and time slabs meeting the test support gives total error $O(h_m)$. Thus the flux sum in [F3] differs from ${\int }_0^T\int {\mathcal{Q}}_m{\phi }_x$ by a quantity tending to zero.

Strong local $L^1$ convergence gives $\int {\mathcal{U}}_m{\phi }_t\to \int u{\phi }_t$ on times bounded away from zero. On $I\times (0,ϵ)$, the local uniform bound for both ${\mathcal{U}}_m$ and $u$ makes the integral difference at most a constant times $\mathrm{\mid }I\mathrm{\mid }ϵ$; letting $ϵ\downarrow 0$ proves convergence on the full test support. By [F4], $\int {\mathcal{Q}}_m{\phi }_x\to \int f(u){\phi }_x$ on that support. Step 2.1 gives the initial term limit, and steps 2.2–2.3 identify the limits of the discrete sums. Taking the limit in [F3] proves the weak identity. By [F4], $u$ is locally essentially bounded, and [F5] identifies it as the stated local weak solution.