Expansion shock fails entropy selection

The two constant states solve the equation on their open sides. Their jump speed satisfies $s(b-a)=F(b)-F(a)$, so [F1] under [F5] makes the interior distributional residual zero. For a test $\phi \in C_c^{\mathrm{\infty }}([0,T)\times \mathbb{R})$ in the class fixed by [F4], choose smooth temporal cutoffs ${\chi }_{\delta }$ that vanish on $[0,\delta ]$, equal $1$ on $[2\delta ,T)$, satisfy ${\chi }_{\delta }^{\mathrm{\prime }}\ge 0$, and have ${\int }_0^T{\chi }_{\delta }^{\mathrm{\prime }}=1$. Apply the interior distributional identity to ${\chi }_{\delta }\phi$. The term with ${\chi }_{\delta }^{\mathrm{\prime }}$ tends to $\int u_0(x)\phi (0,x)dx$: on the compact spatial support, $u(t,\cdot )$ differs from $u_0$ only between $0$ and $st$, so $\mathrm{\parallel }u(t,\cdot )-u_0{\mathrm{\parallel }}_{L^1}\le \mathrm{\mid }a-b\mathrm{\mid }\mathrm{\mid }s\mathrm{\mid }t$, while $\phi (t,\cdot )\to \phi (0,\cdot )$ uniformly. The omitted integrals over $0<t<2\delta$ tend to zero because $u,F(u)$ and the test derivatives are bounded on that compact set. The limit is the initial weak identity.

For $\eta (z)=z^2\mathrm{/}2$ and $q(z)=z^3\mathrm{/}3$, compatibility holds because $q^{\mathrm{\prime }}(z)=z^2={\eta }^{\mathrm{\prime }}(z)F^{\mathrm{\prime }}(z)$, and $\eta$ is convex. The jump residual is $[q]-s[\eta ]=\frac{b^3-a^3}{3}-\frac{a+b}{2}\frac{b^2-a^2}{2}=\frac{(b-a{)}^3}{12}>0.$ By [F2] and [F4], a positive residual violates the entropy inequality. Thus this weak expansion shock is not entropy-admissible.

Since $F^{\mathrm{\prime }\mathrm{\prime }}=1>0$ on $[a,b]$, [F3] under [F5] gives the centered rarefaction profile as an entropy-admissible solution for the same initial data. It differs from the step profile on a set of positive spatial measure for every $t>0$ because the fan interval has width $t(b-a)>0$.