# finding explicit solution to PDE with given initial conditions

A solution to a PDE of my interest is

$u(x,y)=2\left(F''(x)-G''(y)\right)(x-y)^{2}-12\left(F'(x)+G'(y)\right)(x-y)+24\left(F(x)-G(y)\right)$

where $F(x)$ and $G(y)$ are arbitrary functions to be determined from initial conditions. In this case, I know

$u(x,y_{0})=\frac{u_{0}(x-y_{0})^{2}}{(x_{0}-y_{0})^{2}}$,

$u(x_{0},y)=\frac{u_{0}(x_{0}-y)^{2}}{(x_{0}-y_{0})^{2}}$.

I tried all sorts of things to find $F(x)$ and $G(y)$. I tried all sorts of things to no avail; any suggestions what steps/approaches I should take?

I made some progress on this so I thought I would share it. First let $u(x_{0},y_{0})=u_{0}$ then one can write the following

$u(x_{0},y_{0})=2\left(F''(x_{0}-G''(y_{0})\right)(x_{0}-y_{0})^{2}-12\left(F'(x_{0})+G'(y_{0})\right)(x_{0}-y_{0})+24\left(F(x_{0}-G(y_{0}\right)=u_{0}$

Now choose five values arbitrarily and the 6th value will be determined. I made the following choices: $F''(x_{0})=1\quad F'(x_{0})=1\quad F(x_{0})=1 \quad G''(y_{0})=?\quad G'(y_{0})=-1 \quad G(y_{0})=1$

Then find $G''(y_{0})$ as follows:

$u_{0}=2(1-G''(y_{0}))(x_{0}-y_{0})^{2}$ and this gives $G''(y_{0})=1-k$ where

$k=\frac{1}{2}\frac{u_{0}}{(x_{0}-y_{0})^{2}}$

$u(x,y_{0})=\frac{u_{0}(x-y_{0})^{2}}{(x_{0}-y_{0})^{2}}$
$(x-y_{0})^{2}F''(x)-6(x-y_{0})F'(x)+12F(x)=H(x)$
$H(x)=(x-y_{0})^{2}+6(x-y_{0})+12$