The Kruzhkov doubling inequality for two entropy solutions

The doubled test function. For $h>0$ smaller than half the distance of the temporal support of $\phi$ from $\{0,T\}$, set $g_h(t,x,\tau ,y)=\phi (\frac{t+\tau }{2},\frac{x+y}{2}){\theta }_h(t-\tau ){\rho }_h(x-y)$; it is nonnegative and smooth with compact support in each pair of variables, and ${\mathrm{\partial }}_t{g}_h+{\mathrm{\partial }}_{\tau }{g}_h=({\mathrm{\partial }}_t\phi )(\frac{t+\tau }{2},\frac{x+y}{2}){\theta }_h(t-\tau ){\rho }_h(x-y)$, ${\mathrm{\nabla }}_x{g}_h+{\mathrm{\nabla }}_y{g}_h=(\mathrm{\nabla }\phi )(\frac{t+\tau }{2},\frac{x+y}{2}){\theta }_h(t-\tau ){\rho }_h(x-y)$.

The inequality for $u$ with state $k=v(\tau ,y)$. For almost every $(\tau ,y)$ the constant $k=v(\tau ,y)$ is admissible in [F1], and testing the entropy inequality for $u$ by the nonnegative function $g_h(\cdot ,\cdot ,\tau ,y)$ gives ${\int }_{{\mathrm{\Pi }}_T}(\mathrm{\mid }u(t,x)-v(\tau ,y)\mathrm{\mid }{\mathrm{\partial }}_t{g}_h+\mathrm{sgn}(u-v(\tau ,y))(f(u)-f(v(\tau ,y)))\cdot {\mathrm{\nabla }}_x{g}_h)dxdt\ge 0$; the integrand is bounded by a constant times the compactly supported smooth $g_h$ and its derivatives, so the left side is a bounded measurable function of $(\tau ,y)$.

Adding the symmetric inequality. Integrating the inequality of step 1.2 over $(\tau ,y)\in {\mathrm{\Pi }}_T$, and likewise testing the entropy inequality for $v$ with $k=u(t,x)$ and integrating over $(t,x)$, then adding, Fubini's theorem gives $0\le \int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_{{\mathrm{\Pi }}_T\times {\mathrm{\Pi }}_T}[\mathrm{\mid }u-v\mathrm{\mid }({\mathrm{\partial }}_t{g}_h+{\mathrm{\partial }}_{\tau }{g}_h)+\mathrm{sgn}(u-v)(f(u)-f(v))\cdot ({\mathrm{\nabla }}_x{g}_h+{\mathrm{\nabla }}_y{g}_h)]$; the two integrals have the same bounded integrand because of the symmetry of $g_h$.

Passing to the diagonal. Put $z=(t,x)$ and $z^{\mathrm{\prime }}=(\tau ,y)=z-h^{\mathrm{\prime }}$. On a fixed compact set containing the doubled supports, translation continuity gives $\mathrm{\parallel }v(\cdot -h^{\mathrm{\prime }})-v{\mathrm{\parallel }}_1\to 0$ uniformly for $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }\le 2h$. The map $Q(a,b)=\mathrm{sgn}(a-b)(f(a)-f(b))$ is Lipschitz in each variable on the common bounded range: when a variable crosses the other one, split the interval at that point and use $Q(b,b)=0$ and the flux Lipschitz bound. Hence replacing $v(z^{\mathrm{\prime }})$ by $v(z)$ changes the doubled integral by at most $C{\sup }_{\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }\le 2h}\mathrm{\parallel }v(\cdot -h^{\mathrm{\prime }})-v{\mathrm{\parallel }}_1$. Replacing $D\phi ((z+z^{\mathrm{\prime }})\mathrm{/}2)$ by $D\phi (z)$ has error $O(h)$ by smoothness, boundedness and unit kernel mass. Step 2.1 therefore converges to ${\int }_{{\mathrm{\Pi }}_T}(\mathrm{\mid }u-v\mathrm{\mid }{\phi }_t+Q(u,v)\cdot \mathrm{\nabla }\phi )\ge 0$.