I am trying to find the vanishing viscosity limit of
$\begin{alignat}{3} \begin{cases} u^{\epsilon}_t + (|u^{\epsilon}|)_x &= \epsilon u^{\epsilon}_{xx} &\mathrm{in} &\mathbb{R} \times (0, \infty) \\ u^{\epsilon}(x, 0) &= u_0^{\epsilon}(x) &\mathrm{on} &\mathbb{R} \end{cases} \end{alignat}$
where
$u^{\epsilon}_0(x)= \begin{cases} \phantom{-} 1, \; x>0 \\ -1, \; x<0 \end{cases}$
For that I first look at a solution of the heat equation
$\begin{alignat}{3} \begin{cases} v_t^{\epsilon, +} - \epsilon v_{xx}^{\epsilon, +} &= 0 &\mathrm{in} &\mathbb{R} \times (0, \infty) \\ v^{\epsilon, +}(x, 0) &= v_0^+(x) &\mathrm{on} &\mathbb{R} \end{cases} \end{alignat}$
with initial data
$v_0^+(x)= \begin{cases} 1, \; x>0 \\ 0, \; x<0 \end{cases}$
From the solution of the heat equation we then construction a solution of the first problem with initial data $v_0^+$. After that we do the same for mirrored initial conditions
$v_0^-(x)= \begin{cases} \phantom{-} 0, \; x>0 \\ -1, \; x<0 \end{cases}$
The vanishing viscosity limit should then arise from a comparison principle and stability of the initial conditions $v_0^+, \, v_0^-$.
I know the solution to the first heat equation problem should be
$\frac{1}{2} + \frac{1}{2} \mathrm{erf}\left(\frac{x}{\sqrt{4t}}\right)$
where erf denotes the Gauss error function. But I don't know how to use this result to construct a solution to the first problem.