Conditions for Unique Solution for this PDE $$
U_{xy}+\frac{2}{x+y}\left(U_{x}-U_{y}\right)=0
$$
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.
 A: If $u \in C^1(\Omega)$, taking the change of variables $x = \xi + \eta$ and $y = \xi - \eta$ you have that
\begin{align}
\frac{\partial}{\partial x} &= \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta}\\
\frac{\partial}{\partial y} &= \frac{\partial}{\partial \xi} - \frac{\partial}{\partial \eta}
\end{align}
and
$$
\frac{\partial^2}{\partial x \partial y} = \frac{\partial^2}{\partial \xi^2} + \frac{\partial^2}{\partial \eta^2}
$$
Hence, the PDE transforms to
$$
\Delta u + \frac{2}{\xi} u_{\eta} = 0
$$
with
\begin{align}
u\left(x_0, \xi - \eta\right) &= k\left(x_0 - \xi - \eta\right)^3 \\
u\left(\xi + \eta, y_0\right) &= k\left(\xi + \eta - y_0\right)^3
\end{align}
where $\Delta$ is the Laplacian operator in the new variables $(\xi,\eta)$.
In this new coordinates, the equation is (linear) elliptic, and standard existence and uniqueness theorems apply. Now, given that the transformation $(x,y) \, \longrightarrow \, (\xi,\eta)$ is linear and invertible, the result holds in the $(x,y)$ coordinates.
