Show that the vanishing viscosity solution for $u_t+au_x=\epsilon u_{xx}$ is equal to $u_0(x-at)$

Question

I am self-studying Numerical Methods for Conservation Laws by Leveque.

Background

Leveque introduces the advection equation with constant speed $a$:

$$u_t+au_x=0$$

Given smooth initial data $u(x,0)=u_0(x)$, the solution to the differential equation is $u_0(x-at)$. For nonsmooth data $u_0(x)$, the $u_0(x-at)$ satisfies the corresponding integral equations.

Another way of finding a general solution for nonsmooth initial data is to use the vanishing viscosity approach. Let $\epsilon>0$ and consider

$$ u_t + au_x = \epsilon u_{xx} $$

Let $u^\epsilon(x,t)$ be the solution to the advection diffusion equation with diffusivity constant $\epsilon$. Then it can be shown that for

$$ v^\epsilon(x,t)=u^\epsilon(x+at,t) $$

$v^\epsilon$ is a solution to the diffusion equation

$$ v_t^\epsilon(x,t)=\epsilon v_{xx}(x,t) $$

Using the known solution to the diffusion equation, we can find a solution to the original advection diffusion equation using $u^\epsilon(x,t)=v^\epsilon(x-at,t)$.

Question

The exercise Leveque poses is to show that the vanishing viscosity solution $\lim_{\epsilon \to 0} u^\epsilon(x,t)$ is equal to $u_0(x-at)$.

I tried doing this by first writing the solution to the diffusion equation

$$ v^\epsilon(x,t)=\frac{1}{\sqrt{4\pi\epsilon t}}\int_{-\infty}^{\infty}e^{\frac{{-(x-y)^2}}{4\epsilon t}}u_0(y)dy $$

so

$$ u^\epsilon(x,t)=v^\epsilon(x-at,t)=\frac{1}{\sqrt{4\pi\epsilon t}}\int_{-\infty}^{\infty}e^{\frac{{-(x-at-y)^2}}{4\epsilon t}}u_0(y)dy $$

I want to show that

$$ \lim_{\epsilon \to 0} \frac{1}{\sqrt{4\pi\epsilon t}}\int_{-\infty}^{\infty}e^{\frac{{-(x-at-y)^2}}{4\epsilon t}}u_0(y)dy = u_0(x-at) $$

But I'm having trouble. Reading through Partial Differential Equations: An Introduction by Strauss, I found that I can use integration by parts to show (I think) that this limit is

$$ \lim_{\epsilon\to 0}u^\epsilon(x,t)=u_0(x-at)+u_0(x)|_{-\infty}^\infty $$

But this doesn't work unless I make additional assumptions about the form of $u_0(x)$.

What is the right approach to showing $\lim_{\epsilon\to 0}u^\epsilon(x,t)=u_0(x-at)$?

Answer 1

Turns out I made a mistake evaluating the integration by parts! To solve this problem (from Leveque), I recommend following the procedure outlined in Strauss section 2.4.

Strauss develops the solution to the diffusion equation by considering

$$ Q(x,t)=\frac{1}{2}+\frac{1}{\sqrt{\pi}}\int_0^{x\sqrt{4kt}}e^{-p^2}dp $$

which has the properties $\lim_{t\to 0} Q =1$ if $x>0$ and $\lim_{t\to 0}Q=0$ if $x<0$. It turns out that the solution $v^\epsilon(x,t)$ to

$$ v^\epsilon_t=\epsilon v^\epsilon_{xx}~~~~~~\text{s.t.}~~~~~v^\epsilon(x,0)=\phi(x)=u_0(x) $$

can be written

$$ v^\epsilon(x,t)=\int_{-\infty}^{\infty}\frac{\partial Q}{\partial x}(x-y,t)\phi(y)dy $$

If you apply integration by parts and utilize the limiting properties of $Q$, then, as I mentioned in my question, you'll find

\begin{align*} \lim_{\epsilon\to 0}u^\epsilon(x,t)&=\lim_{\epsilon\to 0} v^\epsilon(x-at,t)\\ &=u_0(x-at)+u_0(-\infty)-u_0(-\infty)\\ &=u_0(x-at) \end{align*}

It's a little involved -- I missed a negative sign in my initial attempt. To check if everything works out, I plotted the solution for an initial condition that does not go to zero at infinity, the step function. $$ u_0(x)= \begin{cases} 1 & x< 0 \\ 0 & x\geq 0 \end{cases} $$

The movie below shows the solution with $\epsilon=0.1$ and $a=1$.