TITLE: Solve $ u_{xx}=x^2 u_{yy}$
QUESTION [0 upvotes]: I am trying to solve the following PDE: 
$$ u_{xx}=x^2 u_{yy} $$
I thought it is much easier to consider its canonical form, so I used the change of variable and got my new equation: 
$$  u_{vw}=\frac{1}{v-w}(u_v-u_w) $$
But even this seems a bit difficult for me. Can any one help me solving my equation.
Thank you in advance.

REPLY [3 votes]: $$u_{xx}=x^2u_{yy}$$
This is the continuation of the Charles Hudgins's answer.
$$
\frac{X''(x)}{x^2X(x)} = \frac{Y''(y)}{Y(y)}=K=\lambda^2
$$
With $\lambda$ real or complex. Note that setting $K=\lambda^2$ doesn't reduce the generality insofar $\lambda$ can be complex. This setting is in interest of simpler typing (avoiding square roots).
$Y''(y)-\lambda^2 Y(y)=0\quad$ gives $$ Y=c_1e^{\pm\lambda y}$$
which includes the sinusoidal cases when $\lambda$ is complex.
$X''(x)-\lambda^2x^2X(x)=0\quad$ gives 
$$X=c_2x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)$$
This involves the Modified Bessel function of the first kind $I_\nu(z)$ .
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
A family of particular solutions of the PDE is :
$$u=C\,x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)e^{\pm\lambda y}$$
Any linear combination of particular solutions is solution of the PDE. Thus on discret form :
$$u(x,y)=\sum_{\forall\lambda} C_\lambda\,x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)e^{\pm\lambda y}$$
where $C_\lambda$ are arbitrary constants, to be determined according to some boundary condition.
Or on integral form :
$$u(x,y)=\int C(\lambda)\,x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)e^{\pm\lambda y}d\lambda$$
where the function $C(\lambda)$ is arbitrary, to be determined according to some boundary condition.
Of course this is very complicated in general. But simplification might occur in some cases of favourable boundary conditions allowing to reduce to a limited number of standard functions. That is why it is very important that the boundary conditions be known.
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