Solution to 1st Order PDE using Method of Characteristics I want to solve the following equation:
$$ u_t + e^{-x} u_x = e^{-x} u $$
with  initial condition $u(x,0)=u_0(x)$ and boundary condition $u(0,t)=u_B$.
Using the method of characteristics gives:
$$ \frac{dt}{1}=\frac{dx}{e^{-x}}=\frac{du}{e^{-x}u} $$
Of these relations, I first solved:
$$ \frac{dt}{dx}=e^{x} \implies t=e^{x} + A $$
Then I solved
$$ \frac{du}{dx}=u \implies u=B\ e^x $$
Then using the fact that $B=f(A)=f(t-e^x)$, I was able to find that the boundary and initial conditions imply that:
$$ u(0,t)=f(t-1) =u_B \quad\&\quad u(x,0) = f(-e^x)=e^{-x} u_0(x) $$
But this is where I am a bit confused. How can I solve both of these for $f$? It feels like I can't pose both an IC and BC but I don't quite see why that would be? have I just made a mistake somewhere?
 A: OK. up to the characteristic equations :
$$t-e^x=A\quad\text{and}\quad e^{-x}u=B$$
General solution with $B=f(A)$ :
$$\boxed{u(x,t)=e^xf(t-e^x)}$$
Conditions $\quad u(x,0)=e^x\quad$ and $\quad u(0,t)=u_B=$constant (I suppose).
Note that first condition $\quad u(0,0)=1\quad$ and second condition $\quad u(0,0)=u_B$.
If $u_B\neq 1$ there is a discontinuity at $(x=0\:,\:t=0)$. This suggests that the  function $u(x,t)$ is a piecewise function. This will be confirmed latter.
First condition :
$$u(x,0)=u_0(x)=e^xf(0-e^x)=e^x f(-e^x)$$
Let $X=-e^x\quad \implies \quad x=\ln|-X| \quad \implies \quad f(X)=e^{-x} u_0\big(\ln|-X|\big)$
$$f(X)=e^{-(\ln|-X|)}u_0\big(\ln|-X|\big)=\frac{1}{-X}u_0\big(\ln|-X|\big)$$
Now the function $f$ is known. We put it into the above general solution where the argument is $t-e^x$.
$$f(t-e^x)=\frac{1}{-(t-e^x)}u_0\big(\ln|-(t-e^x)|\big)$$
$$\boxed{u(x,t)=\frac{e^x}{e^x-t}u_0\big(\ln|e^x-t|\big)}\tag 1$$
Second condition :
$$u(0,t)=u_B=e^0 f(t-e^0)$$
$$f(t-1)=u_B\quad\implies\quad f=u_B$$
Now the function $f$ is known. We put it into the above general solution :
$$\boxed{u(x,t)=e^x u_B}\tag 2$$
The solution satisfying both conditions is a picewise fuction made of the funcions $(1)$ and $(2)$ each one on a distinc domain separated from one to the other by a border  which implicit equation is
$$\frac{e^x}{e^x-t}u_0\big(\ln|e^x-t|\big)=e^x u_B$$
One need to know what is the function $u_0(x)$ to say if the implicit equation can be transformed to an explicit form.
