Show that the vanishing viscosity solution for $u_t+au_x=\epsilon u_{xx}$ is equal to $u_0(x-at)$ 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 $\lim_{\epsilon\to 0}u^\epsilon(x,t)=u_0(x-at)$?
 A: 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 $a=1$.

