Local $L^1$ contraction for two entropy solutions

Cutoff inequalities on a time slab. Fix $t_3\in (0,T)$ with $L{t}_3<R$ — for $L>0$ such $t_3$ exist by taking $t_3<\min \{T,R\mathrm{/}L\}$, and for $L=0$ every $t_3\in (0,T)$ works — and fix $\sigma \in (0,R-L{t}_3)$. Let ${\mathrm{\Phi }}_{\sigma }$ be as in [F3] and set $F_{\sigma }(t)={\int }_{{\mathbb{R}}^n}w(t,x){\mathrm{\Phi }}_{\sigma }(t,x)dx$ and $G_{\sigma }(t)={\int }_{{\mathbb{R}}^n}(w{\mathrm{\partial }}_t{\mathrm{\Phi }}_{\sigma }+q\cdot {\mathrm{\nabla }}_x{\mathrm{\Phi }}_{\sigma })(t,x)dx$ for $t\in (0,t_3)$. Because ${\beta }_{\sigma }^{\mathrm{\prime }}\le 0$ and [F3] holds, $G_{\sigma }\le {\int }_{{\mathbb{R}}^n}w({\mathrm{\partial }}_t{\mathrm{\Phi }}_{\sigma }+L\mathrm{\mid }{\mathrm{\nabla }}_x{\mathrm{\Phi }}_{\sigma }\mathrm{\mid })dx=0$ by [F1]. For nonnegative $\eta \in C_c^{\mathrm{\infty }}((0,t_3))$ the function $\phi =\eta {\mathrm{\Phi }}_{\sigma }$ is an admissible nonnegative test function in [F1], since ${\mathrm{\Phi }}_{\sigma }$ is smooth on the slab and compactly supported in $x$; hence $0\le {\int }_0^{t_3}(F_{\sigma }{\eta }^{\mathrm{\prime }}+G_{\sigma }\eta )\le {\int }_0^{t_3}{F}_{\sigma }{\eta }^{\mathrm{\prime }}$, that is, ${\int }_0^{t_3}{F}_{\sigma }{\eta }^{\mathrm{\prime }}\ge 0$.

Monotonicity in time. Fix a nonnegative smooth bump $\zeta$ supported in $(0,1)$ with ${\int }_0^1\zeta =1$ and put $H(r)={\int }_{-\mathrm{\infty }}^r\zeta$. For $0<s<t<t_3$ and small $\epsilon >0$, the function ${\eta }_{\epsilon }(\tau )=H\left(\frac{\tau -s}{\epsilon }\right)-H\left(\frac{\tau -t}{\epsilon }\right)$ is admissible in step 1.1 and ${\eta }_{\epsilon }^{\mathrm{\prime }}\to {\delta }_s-{\delta }_t$ as $\epsilon \downarrow 0$. At Lebesgue points $s<t$ of $F_{\sigma }$, step 1.1 gives $F_{\sigma }(s)-F_{\sigma }(t)={\lim }_{\epsilon \downarrow 0}{\int }_0^{t_3}{F}_{\sigma }{\eta }_{\epsilon }^{\mathrm{\prime }}\ge 0$, so $F_{\sigma }(s)\ge F_{\sigma }(t)$ for all Lebesgue points $0<s<t<t_3$ of $F_{\sigma }$, a full-measure set of pairs by [F4].

The limit as $s\downarrow 0$. We claim ${\mathrm{ess\hspace{0.17em}lim}}_{s\downarrow 0}{F}_{\sigma }(s)={\int }_{{\mathbb{R}}^n}\mathrm{\mid }{u}_0(x)-v_0(x)\mathrm{\mid }{\beta }_{\sigma }(\mathrm{\mid }x-x_0\mathrm{\mid })dx$. Indeed, the difference is bounded by ${\int }_{B(x_0,R)}\mathrm{\mid }u(s)-u_0\mathrm{\mid }{\beta }_{\sigma }+{\int }_{B(x_0,R)}\mathrm{\mid }v(s)-v_0\mathrm{\mid }{\beta }_{\sigma }+{\int }_{{\mathbb{R}}^n}\mathrm{\mid }{u}_0-v_0\mathrm{\mid }\mid {\beta }_{\sigma }(\mathrm{\mid }x-x_0\mathrm{\mid }+Ls)-{\beta }_{\sigma }(\mathrm{\mid }x-x_0\mathrm{\mid })\mid$; the first two terms tend to $0$ by [F2], since ${\beta }_{\sigma }$ is supported in $B(x_0,R)$, and the third tends to $0$ because ${\beta }_{\sigma }$ has bounded derivative and $\mathrm{\mid }{u}_0-v_0\mathrm{\mid }\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. Combining with step 2.1 and letting $s\downarrow 0$ through Lebesgue points of $F_{\sigma }$, for almost every $t\in (0,t_3)$, $F_{\sigma }(t)\le {\int }_{{\mathbb{R}}^n}\mathrm{\mid }{u}_0-v_0\mathrm{\mid }{\beta }_{\sigma }(\mathrm{\mid }x-x_0\mathrm{\mid })dx$.

Removing the cutoff. Let ${\sigma }_j\downarrow 0$ with ${\sigma }_j<R-L{t}_3$ and intersect the full-measure sets of step 3.1 over all $j$ using [F4]: for almost every $t\in (0,t_3)$ the inequality of step 3.1 with $\sigma ={\sigma }_j$ holds for every $j$. For such $t$, ${\beta }_{{\sigma }_j}(\mathrm{\mid }x-x_0\mathrm{\mid }+Lt)\to {\mathbf{1}}_{\{\mathrm{\mid }x-x_0\mathrm{\mid }<R-Lt\}}(x)$ pointwise away from the sphere $\mathrm{\mid }x-x_0\mathrm{\mid }=R-Lt$, and the corresponding integrands are dominated by $w(t,\cdot ){\mathbf{1}}_{B(x_0,R)}$ respectively $\mathrm{\mid }{u}_0-v_0\mathrm{\mid }{\mathbf{1}}_{B(x_0,R)}$, which are integrable; hence dominated convergence gives ${\int }_{B(x_0,R-Lt)}\mathrm{\mid }u-v\mathrm{\mid }(t,x)dx\le {\int }_{B(x_0,R)}\mathrm{\mid }{u}_0-v_0\mathrm{\mid }(x)dx$. Every $t\in (0,T)$ with $Lt<R$ lies in $(0,t_3)$ for some admissible $t_3$ — put $H=\min \{T,R\mathrm{/}L\}$ for $L>0$, and $H=T$ for $L=0$, and use the explicit sequence $t_3^{(j)}=H(1-1\mathrm{/}(j+1))\uparrow H$ — so the estimate holds for almost every such $t$, with exceptional set depending on $x_0,R$. If $u_0=v_0$ almost everywhere on $B(x_0,R)$ the right-hand side vanishes, so $u(t,\cdot )=v(t,\cdot )$ almost everywhere on $B(x_0,R-Lt)$ for almost every such $t$.