What Are the Conditions for Uniqueness in Solving This PDE?

[abiyo]

Consider the 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.

 

[jvicens]

Dear Abiyo,

Thank you for your post. Your approach using the six unknowns and setting five values to determine the sixth one is a good start. However, you are correct in looking for a constraint on these constants to ensure uniqueness of the solution.

One way to ensure uniqueness is to use the boundary conditions given in the problem. We can substitute the boundary conditions into the solution for the PDE and equate it to the given boundary conditions. This will give us two equations in terms of $f(x)$ and $g(y)$, which we can solve to determine the values of these functions.

For the first boundary condition, we have:
$$
U(x_{0},y)=k(x_{0}-y)^{3}=2\left(f''(x_{0})-g''(y)\right)(x_{0}-y)^{2}-12\left(f'(x_{0})+g'(y)\right)(x_{0}-y)+24\left(f(x_{0})-g(y)\right)
$$
Setting $y=y_{0}$ and using the fact that $k=U_{0}(x_{0}-y_{0})^{3}$, we get:
$$
U(x_{0},y_{0})=U_{0}(x_{0}-y_{0})^{3}=2\left(f''(x_{0})-g''(y_{0})\right)(x_{0}-y_{0})^{2}-12\left(f'(x_{0})+g'(y_{0})\right)(x_{0}-y_{0})+24\left(f(x_{0})-g(y_{0})\right)
$$
Simplifying this equation, we get:
$$
f(x_{0})-g(y_{0})=\frac{U_{0}}{24}
$$
Similarly, for the second boundary condition, we have:
$$
U(x,y_{0})=k(x-y_{0})^{3}=2\left(f''(x)-g''(y_{0})\right)(x-y_{0})^{2}-12\left(f'(x)+g'(y_{0})\right)(x-y_{0})+24\left(f(x)-g(y_{0})\right)
$$
Setting $x=x_{0}$ and using the fact that $k=