Characteristic curves PDE $uu_x+u_y=u$ with $u(x,0)=-x$

Question

I have the following PDE:
$$uu_x+u_y=u\qquad , y>0$$
$$u(x,0)=-x$$
I need to find the ground curves, solve the PDE and find where the solution is valid.

First parametrize the initial data:
$$x=s, y=0, u=-s \qquad\text{at}\qquad \tau=0.$$
Then use the characteristic equations:
$$\frac{dx}{d\tau}=u$$
$$\frac{dy}{d\tau}=1$$
$$\frac{du}{d\tau}=u$$
Using these and the initial data,
$y=\tau$, $u=-se^{\tau}, x=-se^{\tau}+2s$.

Then $x=s(2-e^y)$. This is the one parameter family of characteristic ground curves.

Then the solution is $$u(x,y)=\frac{-xe^y}{2-e^y}.$$
I think the solution is valid when $2-e^y \neq 0$ and $y>0$, but the solution says that it is valid only for $0<y<\log2$ because beyond that the characteristic curves intersect.

My questions are: how do I find where the characteristic curves intersect (in this example and also in general)? How do I find out (in general), when solving a PDE with this method, where the solution is valid (apart from the obvious thing that the denominator should not vanish)?

Answer 1

The one parameter family of characteristic curves is $$x(y;s) = s(2 - e^y)$$ This means that the curve $x(y;0) = 0$ will cross the curve $x(y;s')$ for any $s' >0$ at $y= \log(2)$. In general the solution to the PDE will exist up to the lifetime of the characteristic ODE. This is when the characteristic cross or the flow fails to be invertible. In your case the flow is $$\mathcal{F} : (y,s) \mapsto (y,x(y;s)) = (y, s(2 - e^y)), \ s>0$$ so that $\mathcal{F}^{-1}$ exists up until a stopping time $y = \log(2)$. i.e. $$\mathcal{F}^{-1} :(y,x) \mapsto (y, \log(x)) $$ So this is valid until $x(y;s)=0$ or $y=\log(2)$