Vanishing viscosity selects the Hopf--Lax solution for bounded data

The Cole--Hopf family solves the viscous equation. Write $W^{\epsilon }(x,t):={\int }_{\mathbb{R}}\mathrm{\Gamma }(x-y,\epsilon t)e^{-\sin y\mathrm{/}(2\epsilon )}dy$. The substitution $z=2\sqrt{\epsilon t}r$ and $\int e^{-r^2}dr=\sqrt{\pi }$ show $\int \mathrm{\Gamma }(z,\epsilon t)dz=1$, and the same substitution with the Gaussian tail shows that $\mathrm{\Gamma }(\cdot ,\epsilon t)$ is an approximate identity as $t\downarrow 0$. Differentiating the kernel gives ${\mathrm{\partial }}_t\mathrm{\Gamma }(\cdot ,\epsilon t)=\epsilon {\mathrm{\partial }}_{xx}\mathrm{\Gamma }(\cdot ,\epsilon t)$, so by [F2] $W_t^{\epsilon }=\epsilon W_{xx}^{\epsilon }$, $W^{\epsilon }>0$, and $u^{\epsilon }=-2\epsilon \log W^{\epsilon }$ satisfies $u_t^{\epsilon }=\epsilon u_{xx}^{\epsilon }-\mathrm{\mid }{u}_x^{\epsilon }{\mathrm{\mid }}^2\mathrm{/}2$, that is $u_t^{\epsilon }+\mathrm{\mid }{u}_x^{\epsilon }{\mathrm{\mid }}^2\mathrm{/}2=\epsilon u_{xx}^{\epsilon }$, on $t>0$. The approximate identity applied to the bounded uniformly continuous function $e^{-\sin (\cdot )\mathrm{/}(2\epsilon )}$ gives $W^{\epsilon }\to e^{-\sin x\mathrm{/}(2\epsilon )}$ uniformly as $t\downarrow 0$, hence $u^{\epsilon }(x,t)\to \sin x$ uniformly; and $-1\le u^{\epsilon }\le 1$ because $e^{-1\mathrm{/}(2\epsilon )}\le e^{-\sin y\mathrm{/}(2\epsilon )}\le e^{1\mathrm{/}(2\epsilon )}$ and the kernel has unit mass.

Uniform comparison with Hopf--Lax. Fix $x$, $t>0$, put $g(y)=\sin y+(x-y{)}^2\mathrm{/}(2t)$ and $m=\min g=Q_t{u}_0(x)$, and take one minimiser $y_{\ast }$. Then $-1\le m\le 1$, $g^{\mathrm{\prime }}(y_{\ast })=0$, and $g^{\mathrm{\prime }\mathrm{\prime }}(y)\le A:=1+1\mathrm{/}t$. Applying the mean value theorem [F2] to $g^{\mathrm{\prime }}$ shows that the derivative of $g(y_{\ast }+r)-m-A{r}^2\mathrm{/}2$ is nonpositive for $r>0$ and nonnegative for $r<0$; a second application gives $g(y_{\ast }+r)\le m+A{r}^2\mathrm{/}2$. The full Gaussian integral therefore gives $W^{\epsilon }(x,t)\ge e^{-m\mathrm{/}(2\epsilon )}\mathrm{/}\sqrt{1+t}$, hence $u^{\epsilon }-m\le \epsilon \log (1+t)$. Conversely, $g-m\ge 0$ everywhere, and $g-m\ge (x-y{)}^2\mathrm{/}(2t)-2$. If $\mathrm{\mid }y-x\mathrm{\mid }\ge \sqrt{8t}$, the latter is at least $(x-y{)}^2\mathrm{/}(4t)$. Splitting the integral at this radius and using [F2] bounds its normalized ratio by $e^{m\mathrm{/}(2\epsilon )}{W}^{\epsilon }(x,t)\le (4\pi \epsilon t{)}^{-1\mathrm{/}2}(2\sqrt{8t}+{\int }_{\mathbb{R}}{e}^{-(x-y{)}^2\mathrm{/}(8\epsilon t)}dy)=\sqrt{\frac{8}{\pi \epsilon }}+\sqrt{2}\mathrm{.}$ Thus $u^{\epsilon }-m\ge -2\epsilon \log (\sqrt{8\mathrm{/}(\pi \epsilon )}+\sqrt{2})$. These two bounds hold for every $x$ and $t>0$; at $t=0$ the functions agree. Taking their maximum for $0\le t\le T$ gives the stated absolute error bound. Its right-hand side tends to zero: writing $s=1\mathrm{/}\epsilon \ge 1$, the integral formula for the logarithm in [F2] gives $\log s={\int }_1^{s}dr\mathrm{/}r\le {\int }_1^{s}dr\mathrm{/}\sqrt{r}=2(\sqrt{s}-1)$, hence $(\log s)\mathrm{/}s\to 0$.

The limit is the viscosity solution. By [F3] the function $Q_t{u}_0$ is a viscosity solution of $u_t+\mathrm{\mid }{u}_x{\mathrm{\mid }}^2\mathrm{/}2=0$; independently, since $u_0$ is 1-Lipschitz with $\mathrm{\mid }{u}_0\mathrm{\mid }\le 1$, the bounds $u_0(x)-t\mathrm{/}2\le Q_t{u}_0(x)\le u_0(x)$ hold (competitor $y=x$ and the Lipschitz bound for $\sin$), so $Q_t{u}_0$ has the initial trace $u_0$ and is bounded and uniformly continuous. The uniform estimate of step 1.2, whose right-hand side tends to zero, then gives locally uniform convergence of the viscous family to this viscosity solution without invoking a general vanishing-viscosity theorem and without a momentum-Lipschitz hypothesis.