The Question:
In the region $y>0$, reduce the PDE
$$y \frac{\partial ^2u}{\partial x^2} = \frac{\partial ^2u}{\partial y^2}$$
to canonical form, and sketch the characteristic curves.
My Attempt:
$0^2-y(-1)=y>0$ so this PDE is hyperbolic.
$yt^2-1=0 \implies t(x,y)=\pm \frac{1}{\sqrt y}$, and $\frac{dy}{dx} = \pm \frac{1}{\sqrt y} \implies 3x \pm 2y^{3/2}=constant$
So the canonical variables are $\phi(x,y) = 3x - 2y^{3/2}$ and $\psi (x,y) = 3x + 2y^{3/2}$.
Changing the PDE into canonical variables gives us
$$\frac{\partial ^2 u}{\partial \phi \partial \psi} = \frac{1}{6(\psi - \phi)} \biggl (\frac{\partial u}{\partial \psi} - \frac{\partial u }{\partial \phi } \biggr )$$
and I have no idea how to solve this...
Did I make some sort of mistake somewhere, or is it possible to sketch the characteristic curves without explicitly computing the solutions?
Any hints would be much appreciated.