I want to solve the following:
$$ (I) \ u_t+e^ue_x=0 \ \ \ \ ; x>0,t>0 \\ (II) \ u(0,x)=2 \ \ \ \ ; x>0 \\ (III) \ u(t,0)=a>0 \ \ \ \ ; t>0$$ I just can't get a solution which satiesfies all of the requirements.
Obviously, a constant function $2$ solves (I) and (II) and a constant function $a$ solves $(I)$ and $(III)$.
Using method of characteristics, I found an implicit solution $u(x,t)= C(x-e^u t)$ but this also doesn't work for $(II)$ and $(III)$.
I am also wondering: is it even possible for a continous function to satisfie both boundary conditions?
$u$ is constant along characteristic curves.
The characteristic curves of (II) with $u=2$ are $x-e^2t=x_0$ for $x_0>0$. They cross the line $x=0$ at $t=-e^{-2}x_0<0$, so never in the quadrant under consideration.
The curves of (III) with $u=a$ are $x-e^at=-e^at_0$ for $t_0>0$. They cross the line $t=0$ at $x=-e^at_0<0$, thus also not in the quadrant.
If $a<2$ the lines from (II) and (III) cross, so that there is a shock front on the line $x=\frac12(e^2+e^a)t$. For $a\ge2$ one gets a dispersion wave filling the remaining sector, if $2\le x/t\le a$ then this point has value $u(x,t)=\ln(x/t)$.