Strictly convex scalar Riemann construction

If $a=b$, the profile and initial data are constant. For every compactly supported test, integration of ${\phi }_t$ over $[0,T)$ gives $-\phi (0,x)$ and integration of ${\phi }_x$ over $\mathbb{R}$ gives zero. The weak identity follows; the same calculation with the constants $\eta (a)$ and $q(a)$ gives equality in every entropy identity.

Suppose $a>b$. Apply [F5] to $F$ on $[b,a]$ to obtain $s=F^{\mathrm{\prime }}(\xi )$ for some $\xi \in (b,a)$. Apply [F5] to $F^{\mathrm{\prime }}$ on any subinterval of $[b,a]$; since its derivative $F^{\mathrm{\prime }\mathrm{\prime }}$ is positive, $F^{\mathrm{\prime }}$ is strictly increasing there. Hence $F^{\mathrm{\prime }}(b)<s<F^{\mathrm{\prime }}(a)$, which is the scalar Lax pattern in [F4].

Suppose $a<b$. Applying [F5] to $F^{\mathrm{\prime }}$ on any subinterval of $[a,b]$ shows it is strictly increasing, so it has an inverse $g$ on $[F^{\mathrm{\prime }}(a),F^{\mathrm{\prime }}(b)]$. This inverse is continuous: for each interior $z$, strict inequalities $F^{\mathrm{\prime }}(z-\delta )<F^{\mathrm{\prime }}(z)<F^{\mathrm{\prime }}(z+\delta )$ trap $g(y)$ near $z$ whenever $y$ is sufficiently near $F^{\mathrm{\prime }}(z)$, and the endpoint argument is one-sided. For $y=F^{\mathrm{\prime }}(z)$, the difference quotient $(g(y+h)-z)\mathrm{/}h$ equals $(x-z)\mathrm{/}(F^{\mathrm{\prime }}(x)-F^{\mathrm{\prime }}(z))$ with $x=g(y+h)$; as $h\to 0$, continuity gives $x\to z$, and differentiability of $F^{\mathrm{\prime }}$ gives limit $1\mathrm{/}{F}^{\mathrm{\prime }\mathrm{\prime }}(z)$. Thus $g^{\mathrm{\prime }}(y)=1\mathrm{/}{F}^{\mathrm{\prime }\mathrm{\prime }}(g(y))$ on the interior. Set $\alpha =F^{\mathrm{\prime }}(a)$ and $\beta =F^{\mathrm{\prime }}(b)$. Since $F^{\mathrm{\prime }\mathrm{\prime }}$ is continuous and positive on $[a,b]$, the interior derivative has one-sided limits $1\mathrm{/}{F}^{\mathrm{\prime }\mathrm{\prime }}(a)$ at $\alpha$ and $1\mathrm{/}{F}^{\mathrm{\prime }\mathrm{\prime }}(b)$ at $\beta$; the endpoint difference quotients have those same limits because, for $y=F^{\mathrm{\prime }}(z)$ with $z\downarrow a$, $(g(y)-a)\mathrm{/}(y-\alpha )=(z-a)\mathrm{/}(F^{\mathrm{\prime }}(z)-F^{\mathrm{\prime }}(a))\to 1\mathrm{/}{F}^{\mathrm{\prime }\mathrm{\prime }}(a)$, and analogously at $b$. Hence $g$ is $C^1$ on $[\alpha ,\beta ]$ with one-sided endpoint derivatives. Explicitly extend it by $\tilde{g}(y)=a+(y-\alpha )\mathrm{/}{F}^{\mathrm{\prime }\mathrm{\prime }}(a)$ for $y<\alpha$, $\tilde{g}(y)=g(y)$ for $\alpha \le y\le \beta$, and $\tilde{g}(y)=b+(y-\beta )\mathrm{/}{F}^{\mathrm{\prime }\mathrm{\prime }}(b)$ for $y>\beta$; this is $C^1$ on $\mathbb{R}$. At each fan edge with $t>0$, composing $\tilde{g}$ with $x\mathrm{/}t$ gives a $C^1$ extension of the fan-side profile to a neighborhood of that edge point, while the adjacent constant branch has its constant $C^1$ extension. Applying [F7] to $u(t,x)=g(x\mathrm{/}t)$ gives $u_t=-x\mathrm{/}(t^2{F}^{\mathrm{\prime }\mathrm{\prime }}(u))$ and $u_x=1\mathrm{/}(t{F}^{\mathrm{\prime }\mathrm{\prime }}(u))$, whence $u_t+F^{\mathrm{\prime }}(u)u_x=0$. The constant branches solve the equation too, and the profile is continuous at both fan edges.

The shock is constant on each open side, so the classical equation holds there. Its traces are $u_L=a$ and $u_R=b$, and the Rankine–Hugoniot coefficient is $[F]-s[u]=F(b)-F(a)-s(b-a)=0$. By [F2] under [F8], the distributional equation holds on $(0,T)\times \mathbb{R}$.

The traces at each fan edge agree, so both the state and flux jumps vanish there. By [F2], the distributional conservation residual is zero; hence the weak equation holds on $(0,T)\times \mathbb{R}$.

On any compact spatial set, $u(t,\cdot )$ differs from $u_0$ only between $0$ and $st$, an interval of length $\mathrm{\mid }s\mathrm{\mid }t$, and $\mathrm{\mid }u-u_0\mathrm{\mid }\le \mathrm{\mid }a-b\mathrm{\mid }$. Hence $\mathrm{\parallel }u(t,\cdot )-u_0{\mathrm{\parallel }}_{L^1(K)}\le \mathrm{\mid }a-b\mathrm{\mid }\mathrm{\mid }s\mathrm{\mid }t$. For an arbitrary initial-boundary test $\phi$, take smooth cutoffs ${\chi }_{ϵ}$ that vanish on $[0,ϵ]$, equal $1$ for $t\ge 2ϵ$, and have ${\chi }_{ϵ}^{\mathrm{\prime }}\ge 0$ with integral $1$. Apply the interior weak identity from step 2.1 to ${\chi }_{ϵ}\phi$. The term $\int u{\chi }_{ϵ}^{\mathrm{\prime }}\phi$ tends to $\int u_0\phi (0,\cdot )$ by the displayed local $L^1$ estimate and uniform continuity of $\phi$; the terms over $0<t<2ϵ$ tend to zero because $u,F(u)$ and the test derivatives are bounded on the compact support. Taking the limit gives the initial weak identity.

Let $\eta \in C^2(\mathbb{R})$ be convex with compatible flux $q$, and let $C(z)=F(b)+s(z-b)$ be the chord joining $(b,F(b))$ to $(a,F(a))$. For $b<z<a$, [F5] gives points ${\xi }_1\in (b,z)$ and ${\xi }_2\in (z,a)$ whose derivatives equal the secant slopes on $[b,z]$ and $[z,a]$. Applying [F5] to $F^{\mathrm{\prime }}$ between ${\xi }_1$ and ${\xi }_2$ shows $F^{\mathrm{\prime }}({\xi }_1)<F^{\mathrm{\prime }}({\xi }_2)$; rearranging the secant-slope inequality gives $F(z)<C(z)$. At the endpoints $F=C$. Convexity of $\eta$ gives ${\eta }^{\mathrm{\prime }\mathrm{\prime }}\ge 0$, since its nonnegative symmetric second differences divided by $h^2$ converge to ${\eta }^{\mathrm{\prime }\mathrm{\prime }}$. The jump coefficient is $[q]-s[\eta ]=-{\int }_b^a{\eta }^{\mathrm{\prime }}(z)(F^{\mathrm{\prime }}(z)-s)dz={\int }_b^a{\eta }^{\mathrm{\prime }\mathrm{\prime }}(z)(F(z)-C(z))dz\le 0.$ The integral identity follows by applying [F6] to the $C^1$ function ${\eta }^{\mathrm{\prime }}(F-C)$; its endpoint values are zero. By [F3] under [F8] the entropy residual is nonpositive. Apply the cutoff argument of step 3.1 to this interior entropy inequality. The entropy density is Lipschitz on $[b,a]$, so the local $L^1$ trace estimate in step 3.1 gives the initial entropy term; boundedness makes the omitted time strip vanish.

Write $\alpha =F^{\mathrm{\prime }}(a)$ and $\beta =F^{\mathrm{\prime }}(b)$, so $\alpha <\beta$ by [A1]. The fan and the initial step can differ only on $[\min (0,\alpha t),\max (0,\beta t)]$, whose length is $t(\max (0,\beta )-\min (0,\alpha ))\le t(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })$. On this interval both values lie in $[a,b]$, so $\mathrm{\parallel }u(t,\cdot )-u_0{\mathrm{\parallel }}_{L^1(K)}\le (b-a)t(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })$ for every compact $K$. Since $F$ is Lipschitz on $[a,b]$, its composition has the same local strong-trace property. Apply the same cutoff argument as in step 3.1 to the interior weak identity of step 2.2; the local $L^1$ bound makes the cutoff boundary term converge to the prescribed initial value, and boundedness makes the omitted time strip vanish. This proves the weak identity with the prescribed initial data.

For each $k$, the compositions ${\eta }_k(u)$ and $q_k(u)$ are constant on each open side of the shock; splitting an interior test at $x=st$ and integrating by parts on the two sides therefore gives their distributional residual coefficient $[q_k]-s[{\eta }_k]$. If $k<b$ or $k>a$, the sign of $z-k$ is constant for $z\in [b,a]$, so $({\eta }_k(z),q_k(z))$ is, up to additive constants, the same signed multiple of $(z,F(z))$; its coefficient vanishes by the Rankine–Hugoniot equality from step 2.1. If $b\le k\le a$, direct substitution gives $[q_k]-s[{\eta }_k]=2(F(k)-C(k))\le 0$, with equality at $k=b,a$, since the strict convexity calculation in step 4.1 puts $F$ below its chord $C$ on $[b,a]$. Thus the Kružkov residual is nonpositive without applying the smooth entropy-jump result to the nonsmooth pair. Apply the cutoff argument of step 3.1 to this interior inequality; ${\eta }_k$ and $q_k$ are Lipschitz on $[b,a]$, so the local trace and bounded-strip limits give the required initial term.

For each compatible convex $C^2$ pair, [F7] and $q^{\mathrm{\prime }}={\eta }^{\mathrm{\prime }}{F}^{\mathrm{\prime }}$ give ${\mathrm{\partial }}_t\eta (u)+{\mathrm{\partial }}_{x}q(u)={\eta }^{\mathrm{\prime }}(u)(u_t+F^{\mathrm{\prime }}(u)u_x)=0$ in the fan. Outside the fan both compositions are constant. Their traces agree at the edges, so [F3] gives zero interface residual. Apply the cutoff argument of step 4.2 to this interior identity; the Lipschitz continuity of $\eta$ and $q$ on $[a,b]$ gives the required initial entropy term.

Fix $k\in \mathbb{R}$. If $k\mathrm{\notin }[a,b]$, ${\eta }_k$ is affine on the range of $u$ and the chain-rule computation of step 5.2 applies on each smooth piece. If $a<k<b$, split the fan additionally at ${\mathrm{\Gamma }}_k=\{(t,x):t>0,x=t{F}^{\mathrm{\prime }}(k)\}$; on each resulting open piece the sign of $u-k$ is fixed, both Kružkov compositions are $C^1$, and [F7] gives zero classical entropy residual. Direct integration by parts on these pieces shows that each interface of speed $c$ contributes $\int ([q_k]-c[{\eta }_k])\psi (t,ct)dt$ for an interior test $\psi$. At ${\mathrm{\Gamma }}_k$ both traces of ${\eta }_k(u)$ and $q_k(u)$ are $0$, since $u=k$ there; at each fan edge the traces of both compositions agree because the state traces agree. Thus every interface contribution is zero, and there is no singular entropy residual; this direct calculation does not apply [F2] to a nonsmooth transformed pair. If $k=a$ or $k=b$, the possible kink lies at a fan edge, where the same matching-trace calculation applies. The one-sided $C^1$ extensions established in step 1.3 justify the piecewise integrations at both fan edges. Apply the cutoff argument of step 4.2 to the resulting interior identity; both compositions are Lipschitz on $[a,b]$, so the local trace estimate supplies the initial term. This proves every stated inequality, with equality for the fan.