Method of characteristics: $xu_t -tu_x = u$ I am just learning the method of characteristics. Suppose I want to solve
$$xu_t - tu_x = u \quad u(x,0) = h(x)$$
I write
$$\dot{t}(s) = x\\
\dot{x}(s) = -t \\
\dot{z}(s) = z$$
If $t(0) = 0$ and $x(0) = x_0$, we have (I think)
$$t(s) = x_0\sin(s)\\
x(s) = x_0\cos(s)\\
z(s) = h(x_0)e^s$$
Where do I go from here? First I'd like to get rid of the $s$, then I'd like to get rid of the $x_0$, but I'm not seeing how to do that.

Clearly, the domain of a solution of $xu_t-tu_x=u$, can not contain any circle centered at the origin. But this is the only restriction!
Take $\Omega=\mathbb R^2\smallsetminus C$, where $C$ is a curve starting from the origin and going to infinity, assume also that $C$ does not intersect to positive $x$-axis. Let $h=h(x), x\in(0,\infty)$ be a $C^1$ function. Then using the ideas of the previously exposed solution, you can obtain a unique solution $u=u(x,t)$ of the PDE defined in $\Omega$, such that $u(x,0)=h(x)$.
