Suppose I have the scalar conservation law $u_{t}+(e^u)_{x}=0$. I want to determine the exact solution with the following initial data:
$$ \mathring{u}(x) = \left\lbrace\begin{aligned} &2 & &\text{if } 0<x < 1, \\ &0 & &\text{otherwise } \end{aligned}\right. $$
I think a rarefaction ran forms at $x=0$ and a shock forms at $x=1$. The Rankine-Hugoniot shock speed for this equation is given by: $$ s=\frac {e^{u_{r}}-e^{u_{l}}}{u_{r}-u_{l}}=\frac {e^{2}-1}{2}. $$ The edges of the rarefaction travel at speed $2$. Therefore, it will not intersect the shock at some point. Is this correct?
Note that $u_t+e^uu_x=0$, so the edges of the rarefaction are the lines $x=t$ and $x=e^2 t$. The shock wave follows the path $x=1+st$, where the Rankine-Hugoniot shock speed $s$ in OP is correct. The shock wave interacts with the rarefaction at time $t^*$ if $$ e^2 t^* = 1 + s t^*, \qquad s = \tfrac{e^2 - 1}2 \approx 3.2. $$ Therefore, we find $t^* = \frac{2}{e^2+1} \approx 0.24$. Here is a plot of these lines: