I'm studying PDE's and found this question:
Solve this equation and found a large area that the solution is well defined:
\begin{equation}xu_x+yu_y=-u , \ u(0,y)= h(y)\end{equation}
I used the Characteristics Method described here.
We parametrize a curve $\Gamma(x(s,r),y(s,r))$ that passes through the point $(0,y)$ as:
\begin{array}$x(0,r)=0&\\ y(0,r)=r&\\ \Gamma(0,y)=\Gamma(0,r)=h(r)\end{array}
Now we solve this system of ODE's: \begin{array}$\frac{\partial x}{\partial s}(s,r)=x \implies x(s,r)=c_1(r)e^s&\\ \frac{\partial y}{\partial s}(s,r)=y \implies y(s,r)=c_2(r)e^s&\\ \frac{\partial z}{\partial s}(s,r)= -z\implies z(s,r)=c_3(r)e^{-s}\end{array}
And find $c_1,c_2 \text{ and } c_3$ by the conditions of $\Gamma$.
The problem comes here because we found that \begin{array}$c_1=0 \implies x(s,r)=0&\\ c_2=r \implies y(s,r)=re^s&\\ c_3=h(r) \implies z(s,r)=h(r)e^{-s}\end{array}
Isolating $e^s$ in $y(s,r)$ we obtain that $e^s=\frac{y(s,r)}{r}$ and then $z(s,r)$ becomes: \begin{equation}z(s,r)=\frac{rh(r)}{y(s,r)} \end{equation}
Because $x(s,r)=0$ we can't find $s$ in the variables $x$ and $y$ and that is my problem.
Thinking geometricaly while $s$ goes near to $0$, $y(s,r)$ decreases while $z(s,r)$ increases very fast at a arbitrary point $(x_0,y_0)$. Using the conditions from the exercise the solution lies in the line $(0,y_0)$ for $y_0 \in \mathbb{R}$, and that is weird to me.
Can someone help with this?