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


1 Answer 1


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}


$$ \frac{\partial^2}{\partial x \partial y} = \frac{\partial^2}{\partial \xi^2} + \frac{\partial^2}{\partial \eta^2} $$

Hence, the PDE transforms to

$$ \Delta u + \frac{2}{\xi} u_{\eta} = 0 $$


\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.

  • $\begingroup$ thank you Pragabhava, this looks neat. I will apply this and see where it takes me. $\endgroup$
    – felasfa
    Jun 21, 2013 at 23:08

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