Viscous solutions contract spatial translates in L-one

The translate difference solves a linear equation. Fix $h$ and put $w(t,x)=u^{\epsilon }(t,x+h)-u^{\epsilon }(t,x)$, so that $w\in C([0,T];L^1)\cap C^{1,2}$ by [F1]. Define $a(t,x)={\int }_0^1{f}^{\mathrm{\prime }}(s{u}^{\epsilon }(t,x+h)+(1-s)u^{\epsilon }(t,x))ds$; then $a$ is $C^1$ and bounded by $L_M={\sup }_{\mathrm{\mid }s\mathrm{\mid }\le \mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}}\mathrm{\mid }{f}^{\mathrm{\prime }}(s)\mathrm{\mid }$, and $f(u^{\epsilon }(t,x+h))-f(u^{\epsilon }(t,x))=a(t,x)w(t,x)$. Subtracting the two pointwise viscous equations and using the chain rule gives $w_t+\mathrm{div}(aw)=\epsilon \mathrm{\Delta }w$.

The modulus balance. For $\delta >0$ let ${\eta }_{\delta }(r)=\sqrt{r^2+{\delta }^2}$, so that ${\eta }_{\delta }\in C^{\mathrm{\infty }}$, ${\eta }_{\delta }\ge \mathrm{\mid }r\mathrm{\mid }$, $\mathrm{\mid }{\eta }_{\delta }^{\mathrm{\prime }}\mathrm{\mid }\le 1$, and ${\eta }_{\delta }(r)-r{\eta }_{\delta }^{\mathrm{\prime }}(r)={\delta }^2\mathrm{/}{\eta }_{\delta }(r)$, ${\eta }_{\delta }^{\mathrm{\prime }\mathrm{\prime }}(r)={\delta }^2\mathrm{/}{\eta }_{\delta }(r{)}^3$. On compact subsets of ${\mathbb{R}}^n\times (0,T)$, using step 1.1 and [F2], ${\mathrm{\partial }}_t{\eta }_{\delta }(w)+\mathrm{div}(a{\eta }_{\delta }(w))-\epsilon \mathrm{\Delta }{\eta }_{\delta }(w)={\eta }_{\delta }^{\mathrm{\prime }}(w)(w_t+a\cdot \mathrm{\nabla }w-\epsilon \mathrm{\Delta }w)+(\mathrm{div}a){\eta }_{\delta }(w)-\epsilon {\eta }_{\delta }^{\mathrm{\prime }\mathrm{\prime }}(w)\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2=\frac{{\delta }^2}{{\eta }_{\delta }(w)}\mathrm{div}a-\epsilon \frac{{\delta }^2\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2}{{\eta }_{\delta }(w{)}^3}$.

The limit $\delta \downarrow 0$. The second term of step 2.1 is nonpositive, and the first is bounded in absolute value by $\delta \mathrm{\mid }\mathrm{div}a\mathrm{\mid }$, which tends to $0$ in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ as $\delta \downarrow 0$ because $\mathrm{div}a$ is bounded on compact sets; since ${\eta }_{\delta }(w)\to \mathrm{\mid }w\mathrm{\mid }$ pointwise and $\mathrm{\mid }{\eta }_{\delta }(w)\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }+\delta$, dominated convergence gives the distributional inequality ${\mathrm{\partial }}_t\mathrm{\mid }w\mathrm{\mid }+\mathrm{div}(a\mathrm{\mid }w\mathrm{\mid })\le \epsilon \mathrm{\Delta }\mathrm{\mid }w\mathrm{\mid }$ on ${\mathrm{\Pi }}_T$.

Cutoff estimate. Test step 3.1 with ${\chi }_R(x)$ times a nonnegative smooth time test. In distributions in time this gives $\frac{d}{dt}\int \mathrm{\mid }w(t)\mathrm{\mid }{\chi }_R\le C(L_M{R}^{-1}+\epsilon R^{-2})\mathrm{\parallel }w(t){\mathrm{\parallel }}_1$. Approximating the indicator of $(s,t)$ by smooth time cutoffs and using the $L^1$ continuity of $w$ gives, for all $0<s<t<T$, $\int \mathrm{\mid }w(t)\mathrm{\mid }{\chi }_R\le \int \mathrm{\mid }w(s)\mathrm{\mid }{\chi }_R+C(L_M{R}^{-1}+\epsilon R^{-2}){\int }_s^t\mathrm{\parallel }w(r){\mathrm{\parallel }}_{1}dr$. The time integral is finite by [F1]. Dominated convergence as $R\to \mathrm{\infty }$ yields $\mathrm{\parallel }w(t){\mathrm{\parallel }}_1\le \mathrm{\parallel }w(s){\mathrm{\parallel }}_1$; continuity includes $t=T$.

Conclusion. Letting $s\downarrow 0$ in step 4.1 and using the $C([0,T];L^1)$ continuity of $w$ and $w(0,\cdot )=u_0(\cdot +h)-u_0(\cdot )$ gives $\mathrm{\parallel }{u}^{\epsilon }(\cdot +h,t)-u^{\epsilon }(\cdot ,t){\mathrm{\parallel }}_1=\mathrm{\parallel }w(t){\mathrm{\parallel }}_1\le \mathrm{\parallel }w(0){\mathrm{\parallel }}_1=\mathrm{\parallel }{u}_0(\cdot +h)-u_0(\cdot ){\mathrm{\parallel }}_1$ for every $t\in [0,T]$. Every constant used depends on $f$ only through $L_M$, the bound for $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ on the solution range, so the estimate is uniform over such flux families.