# 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?

• 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. Nov 14, 2022 at 17:34
• 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. Nov 14, 2022 at 19:01

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.

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