Conservative finite-volume weak limits — Examples

The cell averages are $1$ for $j<0$ and $-1$ for $j\ge 0$. The only face pairs are $(1,1)$, $(1,-1)$, and $(-1,-1)$; [F3] gives flux $1\mathrm{/}2$ in all three cases.

Every cell therefore has equal left and right fluxes at every step, so the update leaves every $U_j^{m,n}$ unchanged. Induction on $n$ gives exactly the stationary step profile displayed in the Example.

By [F2] and step 2.1, every reconstruction equals $u(x,t)=1$ for $x<0$ and $u(x,t)=-1$ for $x\ge 0$ almost everywhere, so ${\mathcal{U}}_m\to u$ strongly in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ and $\mathrm{\mid }u\mathrm{\mid }\le 1$. For any $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$, compact support makes $\phi$ vanish near $t=T$, and $u(x,t)=u_0(x)$ for $t>0$; hence ${\int }_0^T{\int }_{\mathbb{R}}u{\phi }_{t}dxdt=-{\int }_{\mathbb{R}}{u}_0(x)\phi (x,0)dx$. Also $f_B(u)=1\mathrm{/}2$ almost everywhere, so ${\int }_0^T{\int }_{\mathbb{R}}{f}_B(u){\phi }_{x}dxdt=0$. Thus the weak identity holds, and [F4] gives the claimed local weak solution without an additional choice axiom.

The traces are $u^{-}=1$ and $u^{+}=-1$. Thus [F5] gives $s=(f_B(-1)-f_B(1))\mathrm{/}(-1-1)=(1\mathrm{/}2-1\mathrm{/}2)\mathrm{/}(-2)=0$, agreeing with the stationary interface.

Use the compact rectangle $K=[0,1]\times [1\mathrm{/}4,3\mathrm{/}4]$ inside the space-time domain. Each $V_k$ takes only the values $-1,1$ on $[0,1)\times (0,1)$ and zero outside, so $\mathrm{\mid }{V}_k\mathrm{\mid }\le 1$ and $\mathrm{\parallel }{V}_k{\mathrm{\parallel }}_{L^1(K)}=1\mathrm{/}2$.

If $k<\mathrm{\ell }$, every $k$-cell in $[0,1)$ is the union of $2^{\mathrm{\ell }-k}$ finer $\mathrm{\ell }$-cells, an even number. The signs of $V_{\mathrm{\ell }}$ alternate on those finer cells, so their integral on each coarse cell is zero; since $V_k$ is constant on that coarse cell, ${\int }_K{V}_k{V}_{\mathrm{\ell }}=0$.

For signs $a,b\in \{-1,1\}$, $\mathrm{\mid }a-b\mathrm{\mid }=1-ab$. Thus for $k\mathrm{\ne }\mathrm{\ell }$, $\mathrm{\parallel }{V}_k-V_{\mathrm{\ell }}{\mathrm{\parallel }}_{L^1(K)}=1\mathrm{/}2-{\int }_K{V}_k{V}_{\mathrm{\ell }}=1\mathrm{/}2.$ Every subsequence is therefore non-Cauchy in $L^1(K)$, so no subsequence can converge strongly in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$.

This example supplies only a bounded sequence without strong compactness; it makes no finite-volume recurrence claim. Therefore the convergence premise in [F1] remains a separate hypothesis.

Cell averages of $u_0$ are $U_j^{m,0}=0$ for $j<0$ and $1$ for $j\ge 0$. The only face pairs along the resulting step are $(0,0)$, $(0,1)$ and $(1,1)$; [F3] gives their fluxes as $0,0,1\mathrm{/}2$. The ratio ${\tau }_m\mathrm{/}{h}_m=2$, while ${\max }_{0\le s\le 1}\mathrm{\mid }{f}_B^{\mathrm{\prime }}(s)\mathrm{\mid }=1$, so the usual noninteraction bound $({\tau }_m\mathrm{/}{h}_m)\max \mathrm{\mid }{f}_B^{\mathrm{\prime }}\mathrm{\mid }\le 1$ fails.

Assume inductively $U_j^{m,n}=0$ for $j<n$ and $1$ for $j\ge n$. If $j<n$, both face fluxes are $0$, so the state remains $0$. At $j=n$, the left flux is $0$ and the right flux is $1\mathrm{/}2$, hence $U_n^{m,n+1}=1-2(1\mathrm{/}2-0)=0$. For $j>n$ both fluxes are $1\mathrm{/}2$, so those states remain $1$. Therefore the step shifts to $U_j^{m,n+1}=0$ for $j<n+1$ and $1$ for $j\ge n+1$.

On $[2n{h}_m,2(n+1)h_m)$, the numerical jump is at $n{h}_m$, while the limit jump is at $t\mathrm{/}2\in [n{h}_m,(n+1)h_m)$. The profiles differ only on a strip of width at most $h_m$ and height $1$, so the $L^1$ error on every bounded space-time rectangle is at most $h_m{T}^{\mathrm{\prime }}\to 0$. All states are bounded by $1$. By [F4], the limit $u(x,t)=0$ for $x<t\mathrm{/}2$ and $1$ for $x>t\mathrm{/}2$ is a locally bounded distributional weak solution with those initial data.

Take $k=1\mathrm{/}2$. Then ${\eta }_k(0)={\eta }_k(1)=1\mathrm{/}2$, while $q_k(0)=1\mathrm{/}8$ and $q_k(1)=3\mathrm{/}8$. Choose nonnegative $\alpha \in C_c^{\mathrm{\infty }}((0,T))$ with $\int \alpha >0$ and nonnegative $\beta \in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\beta (0)>0$, and set $\psi (x,t)=\alpha (t)\beta (x-t\mathrm{/}2)$. This is an admissible nonnegative test. To evaluate its time term, put $B={\int }_{\mathbb{R}}\beta (y)dy$. Differentiating the smooth compactly supported test and changing variables gives ${\int }_{\mathbb{R}}{\psi }_t(x,t)dx=B{\alpha }^{\mathrm{\prime }}(t)-\frac{1}{2}\alpha (t){\int }_{\mathbb{R}}{\beta }^{\mathrm{\prime }}(y)dy=B{\alpha }^{\mathrm{\prime }}(t)$; the last integral is zero because $\beta$ is compactly supported. Therefore ${\int }_0^T{\int }_{\mathbb{R}}{\eta }_k(u){\psi }_{t}dxdt=\frac{1}{2}B(\alpha (T)-\alpha (0))=0$. Splitting the flux integral at $x=t\mathrm{/}2$ and applying the fundamental theorem of calculus gives ${\int }_{\mathbb{R}}{q}_k(u){\psi }_{x}dx=(1\mathrm{/}8-3\mathrm{/}8)\psi (t\mathrm{/}2,t)=-\frac{1}{4}\alpha (t)\beta (0)$. Thus the integral in [F5] equals $-\frac{1}{4}\beta (0)\int \alpha <0$, contradicting the required entropy inequality. The weak limit is not Kruzhkov entropic.