# finding $F$ and $G$ for this PDE

I know that a solution to a pde I am interested is in the form: $u(x,y)=(x-y)^{5}\frac{\partial^{4}}{\partial x^{2} \partial y^{2}}\left(\frac{F(x)-G(y)}{x-y}\right)$ where $F$ and $G$ are arbitrary functions to be determined.

For my case I know $u(x,y_{0})$ and $u(x_{0},y)$ explicitly. What is the best way to compute $F$ and $G$ in this case?

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Let $U^{xx}(x):=\int_{x_0}^{x}dx'\int_{x_0}^{x'}dx''\frac{u(x'',y_0)}{(x''-y_0)^5}$ and $U^{yy}(y):=\int_{y_0}^{y}dy'\int_{y_0}^{y'}dy''\frac{u(x_0,y'')}{(x_0 -y'')^5}$. Because $u(x,y_0)$ and $u(x_0,y)$ are known explicitly, also $U^{xx}(x)$ and $U^{yy}(y)$ are known explicitly, at least in principle. From

$U^{xx}(x)=\int_{x_0}^{x}dx'\left[\frac{\partial^{3}}{\partial x \partial y^{2}} \left(\frac{F(x)-G(y)}{x-y}\right) |_{y=y_0} \right]_{x_0}^{x'}$

$U^{yy}(y)=\int_{y_0}^{y}dy'\left[\frac{\partial^{3}}{\partial y\partial x^{2}} \left(\frac{G(y)-F(x)}{y-x}\right) |_{x=x_0} \right]_{y_0}^{y'}$

it follows that

$U^{xx}(x)=2\frac{F(x)-G(y_0)}{(x-y_0)^3}-\frac{2G'(y_0)}{(x-y_0)^2}-\frac{G''(y_0)}{x-y_0}-a(x-x_0)-b$

$U^{yy}(y)=2\frac{G(y)-F(x_0)}{(y-x_0)^3}-\frac{2F'(x_0)}{(y-x_0)^2}-\frac{F''(x_0)}{y-x_0}-c(y-y_0)-d$

with

$a=\frac{\partial}{\partial x} \left(\frac{F(x)}{(x-y_0)^3}\right)|_{x=x_0}+\frac{\partial^{2}}{\partial y^{2}} \left(\frac{G(y)}{(x-y)^2}\right)|_{y=y_0}$

$b=\frac{\partial^{2}}{\partial y^{2}} \left(\frac{F(x_0)-G(y)}{x_0-y}\right) |_{y=y_0}$

$c=\frac{\partial}{\partial y} \left(\frac{G(y)}{(y-x_0)^3}\right)|_{y=y_0}+\frac{\partial^{2}}{\partial x^{2}} \left(\frac{F(x)}{(y-x)^2}\right)|_{x=x_0}$

$d=\frac{\partial^{2}}{\partial x^{2}} \left(\frac{G(y_0)-F(x)}{y_0-x}\right) |_{x=x_0}$

Note that $a$, $b$, $c$ and $d$ are actual linear combinations of $F(x_0)$, $F'(x_0)$, $F''(x_0)$, $G(y_0)$, $G'(y_0)$ and $G''(y_0)$. By differentiating the equations for $U^{xx}(x)$ and $U^{yy}(y)$ two times, and plugging $x_0$ and $y_0$ into all obtained equations, we seem to obtain 6 different linear equations for $F(x_0)$, $F'(x_0)$, $F''(x_0)$, $G(y_0)$, $G'(y_0)$ and $G''(y_0)$. But two of these equations are just the original equations, so we just have 5 different equations. However, this is no problem, because only $F(x)-G(y)$ is important, so $F$ and $G$ are only determined up to a constant.

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