TITLE: How to solve a second order partial differential equation involving a delta Dirac function?
QUESTION [7 upvotes]: In a mathematical physical problem, I came across the following partial differential equation involving a delta Dirac function:
$$
a \, \frac{\partial^2 w}{\partial x^2}
+ b \, \frac{\partial^2 w}{\partial y^2}
+ \delta^2(x,y) = 0 \, , 
$$
subject to the boundary conditions $w(x = \pm 1, y) = w(x, y = \pm 1) = 0$.
Here $a, b \in \mathbb{R}_+$ and $\delta^2(x,y) = \delta(x)\delta(y)$ is the two-dimensional delta Dirac function.
While solutions for ODEs with delta Dirac functions can readily be obtained using the standard approach, I am not aware of any resolution recipe for PDEs with delta Dirac functions.
Any help or hint is highly desirable and appreciated.
Thank you

REPLY [2 votes]: The method of images with Dirichlet boundary conditions for a square region $[-1,1]^2$ implies that the 2D Dirac delta distribution $$\delta(x)\delta(y)$$ is part of an alternating Dirac comb/Shah function $$A(x)A(y),$$ where
$$\begin{align}A(x)&~=~\sum_{n\in\mathbb{Z}}(-1)^n\delta(x\!-\!2n)~=~III_4(x)-III_4(x+2)
~=~\frac{1}{4}\sum_{n\in\mathbb{Z}}e^{i\pi n x/2}
-\frac{1}{4}\sum_{n\in\mathbb{Z}}e^{i\pi n(x+2)/2}\cr
&~=~\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1-(-1)^n}{2}e^{i\pi n x/2}
~\stackrel{n=2k-1}=~\frac{1}{2}\sum_{k\in\mathbb{Z}}e^{i\pi(k-1/2) x}
~=~\sum_{k\in\mathbb{N}}\cos(\pi(k\!-\!1/2) x)\cr 
&~=~\sum_{k\in\mathbb{N}-\frac{1}{2}}\cos(\pi k x).\end{align}$$
Therefore the solution to OP's BVP becomes
$$ w(x,y)~=~\frac{1}{\pi^2}\sum_{n,m\in\mathbb{N}-\frac{1}{2}}\frac{\cos(\pi n x)\cos(\pi m y)}{a n^2+b m^2}.$$
We leave it to the reader to analyze convergence properties of the double sum.