The convex-flux Riemann formula fails for a nonconvex flux

Speed, weak solvability and the initial trace. With $u_L=1$, $u_R=-1$, $f(u)=u^3$: $[u]=u_R-u_L=-2$ and $[f]=f(u_R)-f(u_L)=-1-1=-2$, so the Rankine--Hugoniot speed is $s=[f]\mathrm{/}[u]=1$. By [F1] the weak residual of the jump profile is $([f]-s[u]){\delta }_{x=st}$, which vanishes because $s[u]=1\cdot (-2)=-2=[f]$; hence $u$ is a distributional weak solution. For $t>0$ the set where $u(t,x)\mathrm{\ne }{u}_0(x)$ is contained in the interval $(0,t)$ (the region swept by the moving discontinuity compared with the initial step at $0$), of length $t$ and amplitude at most $2$, so ${\int }_K\mathrm{\mid }u(t,x)-u_0(x)\mathrm{\mid }dx\le 2t\to 0$ for every compact $K$: the strong local $L^1$ trace is $u_0$.

Failure of the Kruzhkov inequality at $k=-\frac{1}{2}$. For $k=-\frac{1}{2}$, ${\eta }_k(1)=\mathrm{\mid }1+\frac{1}{2}\mathrm{\mid }=\frac{3}{2}$, ${\eta }_k(-1)=\mathrm{\mid }-1+\frac{1}{2}\mathrm{\mid }=\frac{1}{2}$, ${\eta }_k(1)-{\eta }_k(-1)=1$; and $q_k(s)=\mathrm{sgn}(s+\frac{1}{2})(s^3+\frac{1}{8})$ gives $q_k(1)=1\cdot \frac{9}{8}=\frac{9}{8}$, $q_k(-1)=(-1)\cdot (-\frac{7}{8})=\frac{7}{8}$, so $[q_k]=q_k(-1)-q_k(1)=-\frac{1}{4}$ and $[{\eta }_k]={\eta }_k(-1)-{\eta }_k(1)=-1$. By [F2] the entropy production measure is $([q_k]-s[{\eta }_k]){\delta }_{\mathrm{\Gamma }}=(-\frac{1}{4}+1){\delta }_{\mathrm{\Gamma }}=\frac{3}{4}{\delta }_{\mathrm{\Gamma }}>0$; testing against a nonnegative test function concentrated near the interface produces a strictly positive entropy production, so the Kruzhkov entropy inequality fails and $u$ is not a Kruzhkov entropy solution.

The composite weak solution. Put $c=3\mathrm{/}4$ and define $v(t,x)=1$ for $x<ct$, $v(t,x)=-\sqrt{x\mathrm{/}(3t)}$ for $ct<x<3t$, and $v(t,x)=-1$ for $x\ge 3t$. At the shock the traces are $1$ and $-1\mathrm{/}2$, with $[v]=-3\mathrm{/}2$ and $[f]=-9\mathrm{/}8=c[v]$. On the fan, $\psi (\xi )=-\sqrt{\xi \mathrm{/}3}$ satisfies $f\mathrm{\prime }(\psi (\xi ))=\xi$, so $v_t+f\mathrm{\prime }(v)v_x=t^{-1}\psi \mathrm{\prime }(\xi )(-\xi +f\mathrm{\prime }(\psi (\xi )))=0$. At $x=3t$ the traces match. Regionwise integration using [F1, F3] therefore gives zero weak residual. The discrepancy with the initial datum is confined to $(0,3t)$ and has amplitude at most $2$, so its local $L^1$ norm is at most $6t$; this supplies the strong trace and the Cauchy boundary term.

Chord condition and nonconvexity. With $u^{-}=1$, $u^{+}=-1$ and $s=1$, the chord residual is $F(z)=f(z)-f(1)-s(z-1)=z^3-1-(z-1)=z^3-z$, and the criterion of The convex entropy condition for a single shock is the chord condition requires $F(z)(u^{+}-u^{-})\ge 0$, that is, $F(z)\le 0$ on $[-1,1]$. But $F(z)=z(z-1)(z+1)>0$ for $z\in (-1,0)$, so the chord condition fails, independently confirming the entropy failure of step 1.2. Moreover $f^{\mathrm{\prime }\mathrm{\prime }}(u)=6u$ changes sign on $[-1,1]$, so $f^{\mathrm{\prime }}$ is not increasing and the hypotheses of the strictly convex Riemann solver The Riemann solver for a strictly convex flux are not satisfied.

Entropy admissibility of the composite. At its descending shock the chord residual is $z^3-1-c(z-1)=(z-1)(z+1\mathrm{/}2{)}^2\le 0$ for $-1\mathrm{/}2\le z\le 1$. Thus the chord criterion of [F2] gives $[q]-c[\eta ]\le 0$ for every convex $C^2$ pair. On the smooth fan the entropy residual is $\eta \mathrm{\prime }(v)(v_t+f\mathrm{\prime }(v)v_x)=0$, and it vanishes on both constant regions; matching traces at $x=3t$ give no interface measure. Hence every smooth convex entropy inequality holds. For each $k$, take ${\eta }_{\delta }(z)=\sqrt{(z-k{)}^2+{\delta }^2}-\delta$ and $q_{\delta }(z)={\int }_k^z{\eta }_{\delta }\mathrm{\prime }(r)f\mathrm{\prime }(r)dr$. On the bounded range, ${\eta }_{\delta }\to \mathrm{\mid }z-k\mathrm{\mid }$ uniformly. The fluxes converge uniformly to $\mathrm{sgn}(z-k)(f(z)-f(k))$: outside an arbitrarily small interval about $k$, the derivatives converge uniformly to the sign, and inside it the integral error is bounded by twice its length times a bound for $\mathrm{\mid }f\mathrm{\prime }\mathrm{\mid }$. Passing against compact tests gives every Kruzhkov inequality. With step 1.3 and [F4], $v$ is the unique entropy solution.

The concave hull and conclusion. On $[-1,-1\mathrm{/}2]$ the hull follows $z^3$; on $[-1\mathrm{/}2,1]$ it is $\mathrm{\ell }(z)=1+c(z-1)$. The residual in step 2.2 shows $\mathrm{\ell }\ge z^3$ on the latter interval. The arc is concave, and its derivative decreases to $3\mathrm{/}4$ at $-1\mathrm{/}2$, matching the slope of $\mathrm{\ell }$, so the joined function is a concave majorant. Any concave majorant lies above the cubic on the arc and above the line joining the values at $-1\mathrm{/}2$ and $1$ on the chord interval; it therefore lies above this function. This proves it is the least concave majorant. Its arc and chord yield exactly the fan and shock verified in steps 1.3--2.2. The single shock of step 1.1 is weak but non-entropic, whereas this composite is entropic, proving the claimed failure of the convex-flux formula and its replacement here.