A consistent nonmonotone flux selects a nonentropy expansion shock

The polynomial $F$ is continuous. On the diagonal, its correction term vanishes and $F(s,s)=s^2\mathrm{/}2=f_B(s)$, so it is consistent. Its first partial derivative is ${\mathrm{\partial }}_{a}F(a,b)=\frac{a}{2}+\frac{1}{16}((a^2-1)(b^2-1)+2a(a-b)(b^2-1))\mathrm{.}$ At $(-1,0)$ this equals $-1\mathrm{/}2-2\mathrm{/}16=-5\mathrm{/}8<0$, so it is not nondecreasing in its first coordinate.

On the uniform cells $I_j=[jh,(j+1)h)$ with $h>0$, the initial averages are $U_j^0=-1$ for $j<0$ and $U_j^0=1$ for $j\ge 0$. Every same-state face flux is $F(-1,-1)=F(1,1)=1\mathrm{/}2$. The only mixed pair is $(-1,1)$; both correction factors vanish and $F(-1,1)=1\mathrm{/}2$. Thus every face flux equals $1\mathrm{/}2$, and the recurrence leaves every cell average unchanged at every step.

The datum has $\mathrm{TV}(u_0)=2$ by [F5]. For any sequence $h_m,{\tau }_m\downarrow 0$ with this aligned grid convention, the piecewise-constant reconstruction is exactly the stationary step function almost everywhere. It therefore converges strongly in local space-time $L^1$ to $u(x,t)=-1$ for $x<0$ and $u(x,t)=1$ for $x>0$. Its initial trace is $u_0$.

Since $f_B(u)=1\mathrm{/}2$ almost everywhere, for any $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$ one has ${\int }_0^{T}u(x,t){\phi }_t(x,t)dt=-u_0(x)\phi (x,0)$ for almost every $x$, because $u(x,t)=u_0(x)$ is stationary and $\phi$ vanishes near $t=T$. Integrating in $x$ cancels the initial term in [F2]. The flux contribution is $\frac{1}{2}{\int }_{\mathbb{R}}{\phi }_x(x,t)dx=0$ for each $t$, since $\phi$ has compact spatial support. Thus $u$ satisfies the weak identity and is a locally bounded distributional weak solution.

Let $R:=T\mathrm{/}8$ and choose the cutoff $\chi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ of [F4] with $\chi =1$ on the unit ball and support in the radius-two ball. The function $\phi (x,t):=\chi (x\mathrm{/}R,(t-T\mathrm{/}2)\mathrm{/}R)$ is nonnegative, smooth, supported in $\mathbb{R}\times (0,T)$, and satisfies ${\int }_0^T\phi (0,t)dt>0$. For $k=0$, $\mathrm{\mid }u\mathrm{\mid }=1$ is stationary, so $\int \mathrm{\mid }u\mathrm{\mid }{\phi }_t=0$, while $q_0(-1)=-1\mathrm{/}2$ and $q_0(1)=1\mathrm{/}2$ give ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathbb{R}}{q}_0(u){\phi }_{x}dxdt=-{\int }_0^T\phi (0,t)dt<0.$ This contradicts the required nonnegative entropy pairing in [F3].