A conservative consistent finite-volume weak limit need not be entropic

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.