$$

U_{xy}+\frac{2}{x+y}\left(U_{x}-U_{y}\right)=0

$$

with the boundary conditions

$$

U(x_{0},y)=k(x_{0}-y)^{3}\\

U(x,y_{0})=k(x-y_{0})^{3}

$$

where $k$ is a constant given by $k=U_{0}(x_{0}-y_{0})^{3}$. $x_{0}$, $y_{0}$ and $U(x_{0},y_{0})=U_{0}$ are known. The solution for the PDE is given by

$$

U(x,y)=(x-y)^{5}\frac{\partial ^{4}}{\partial x^{2}\partial y^{2}}\left(\frac{f(x)-g(y)}{x-y}\right)

$$

After some simplifications I get

$$

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 to be determined. I am looking for conditions that ensure uniqueness for the solution of this PDE. Any help will be appreciated.

Thanks, Abiyo

p.s I tried the following approach but it didn't work.

$$

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)

$$ There are six unknowns $f(x_{0}),f'(x_{0}),f''(x_{0}),g(x_{0}),g'(x_{0})$ and $g''(y_{0})$. Assume $5$ values and the sixth one is determined. From there I proceed to find two ODEs and can find a solution to the PDE. The solution depends on my choice of these constants and hence I am looking for a constraint on this constants.