I am required to solve the following PDE using the Method of Characteristics:
$$yu_x-xu_y=0$$
I was able to do that (hopefully successfully) and I got that $u(x,y)$ is given by:
$$u(x,y)=F(x^2+y^2)$$
Where $F\in C^1(\mathbb{R})$. Now, I need to find particular solutions for the following (seperate) initial conditions:
$$(a)\ u(x,0)=x^2 \implies \Gamma_1(s)=(s,0,s^2) \\ (b)\ u(x,1)=x^2+1 \implies \Gamma_2(s)=(s,1,s^2+1)$$
For $(a)$, I checked the Transversality Condition:
$$J(\Gamma_1)= \left(\begin{matrix} 0 & -s & 0 \\ 1 & 0 & 2s \end{matrix}\right)\implies J_{M}(\Gamma_1)=0-(-s)=s$$
This means that for every $s\neq0$ there exists a solution in the neighborhood of $\Gamma_1(s)$. Plugging $s=0$ in the Jacobian, one can see that the lines of the matrix are linearly independent. This means that for $s=0$ there is no solution. Since $s=x$, I concluded that $x=0$ is the problem. However, I cannot see what is the problem with $u(x,y)=x^2+y^2$. It seems to be continuously differentiable twice and to satisfy both the equation and the initial condition, for every $(x,y)\in\mathbb{R}^2$.
My main question therefore is - what do I do if $J_M=0$ for a finite amount of values of $s$. Is there a rule of thumb? Since for condition $(b)$ there seems to be an infinite amount of solutions (for some reason. Guess it has to do something with the fact that for that particular problematic $s$, the lines of the matrix would be linearly dependent). I know that if $J_M\neq0$ for every $s$, then there is a single solution, and if $J_M\equiv 0$ (for every $s$) then there is either an infinite amount of solutions, or no solution at all.
Thanks