How can the PDE $y^2u_{xx}+2xyu_{xy}+u_{yy}=0$ be reduced to canonical form in its hyperbolic region, namely $|x|>1,y\neq0$?
I know the required substitution $(\xi(x,y),\eta(x,y))$ should be given by the two solutions of $y^2(f_x)^2+2xyf_xf_y+(f_y)^2=0$, but after substituting $x=\cosh\theta$ I got the solution $f(x,y)=C\exp\Big(\frac{y^2}{2}\pm\cosh^{-1}x-\frac{1}{2}e^{\pm2\cosh^{-1}x}\Big)$. Substituting this sort of expression into the PDE looks way too ghastly for a problem that's supposed to take less than half an hour!
Maybe there's a different way of solving $y^2(f_x)^2+2xyf_xf_y+(f_y)^2=0$? It looks as though you should be able to spot a solution, but I can't quite manage! Or can we change this to get a different type of substitution which will still yield the canonical form?
Many thanks for any help with this!
Best Answer
$\xi$ and $\eta$ are null coordinates. In the $\xi,\eta$ coordinates the $1+1$ linear hyperbolic PDE is always equivalent to $u_{\xi\eta}(\xi,\eta) = 0$. (If you do the explicit change of coordinates, you will end up with a conformal factor $\Omega(\xi,\eta)$ multiplying $u_{\xi\eta}(\xi,\eta)$. But as long as your change of coordinates $(x,y) \to (\xi,\eta)$ is regular, the conformal factor $\Omega \neq 0$ and you can divide it away.)
That is sort of the point of the canonical form. Once you found the correct change of variables, you can just solve the equation $u_{\xi\eta} = 0$ and do the reverse coordinate transform at the end.