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?

  • 1
    $\begingroup$ For which $x$ and $t$ are those conditions supposed to hold? You can't expect them to hold for all $x$ and $t$, but if (for example) you are solving the PDE in the region $\{ x>0, \, t>0 \}$, then you could give conditions for $u$ on the positive $x$ and $t$ axes. $\endgroup$ Nov 14, 2022 at 17:34
  • $\begingroup$ I am trying to solve the problem for $t>0$ and $x\in [0,L]$ for some finite length - say $L=5$. I solved this numerically using an upwind scheme and I am getting growth in time of the solution, but I don't quite see why. $\endgroup$
    – Mjoseph
    Nov 14, 2022 at 19:01

1 Answer 1


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.

  • $\begingroup$ Shouldn't $e^0=1$? $\endgroup$
    – Mjoseph
    Nov 14, 2022 at 20:02
  • $\begingroup$ $f(t-1)=u_B=$constant implies $f(t)=u_B$ any value of $t$. $\endgroup$
    – JJacquelin
    Nov 15, 2022 at 11:15
  • $\begingroup$ Thank you for your help here and in my other related question! $\endgroup$
    – Mjoseph
    Nov 15, 2022 at 12:17

You must log in to answer this question.