My question comes from a section on wave guides from Richard Haberman's Applied Partial Differential Equations. We are given the PDE:
$$ \frac{\partial^2 u}{\partial t^2} = c^2\Big( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \Big) + e^{-i \omega t} \delta(x - x_0)\delta(y - y_0)\delta(z - z_0) $$
with $u = 0$ on the boundary of a rectangular wave guide, where $0 \leq y \leq L$ and $0 \leq z \leq H$ and $\omega$ is the forcing frequency. I understand why the boundary conditions would suggest a solution of the form
$$ u(x,y,z,t) = \sum_{n=1}^\infty \sum_{m=1}^\infty A_{nm}(x,t) \sin\frac{n \pi y}{L} \sin\frac{m \pi z}{H}. $$
My confusion arises from the claim that the amplitudes $A_{nm}$ then satisfy
$$ \frac{\partial^2 A_{nm}}{\partial t^2} = c^2\Big( \frac{\partial^2 A_{nm}}{\partial x^2} - \frac{n^2 \pi^2}{L^2}A_{nm} - \frac{m^2 \pi^2}{H^2}A_{nm} \Big) + \frac{4}{LH}e^{-i \omega t} \delta(x - x_0)\sin\frac{n \pi y_0}{L} \sin\frac{m \pi z_0}{H} $$
I see why the terms in parentheses on the right hand side would be required to hold for anywhere other than $(x_0,y_0,z_0)$, but I do not know how to work with the Dirac delta functions in this context. My only guess is that there a property of the Dirac delta function regarding infinite series that I am unaware, or perhaps some use of a Green's function. Any ideas would be much appreciated.
The function set $\{ \sin\frac{n\pi y}{L} \mid n=1,2,3,\ldots \}$ constitutes a basis of a Hilbert space on the interval $[0,L],$ and the function set $\{ \sin\frac{n\pi y}{L} \sin\frac{m\pi z}{H} \mid m,n=1,2,3,\ldots \}$ constitutes a basis of a Hilbert space on the rectangle $[0,L]\times[0,H].$ We can thus write a function $f:[0,L]\times[0,H]\to\mathbb C$ as a series $$ f(y,z) = \sum_{n=1}^\infty \sum_{m=1}^\infty \hat f_{nm} \sin\frac{n\pi y}{L} \sin\frac{m\pi z}{H}. $$
The values of the coefficients $\hat f_{nm}$ are given by $$ \hat f_{nm} = \frac{2}{L}\frac{2}{H} \iint_{[0,L]\times[0,H]} f(y,z) \sin\frac{n\pi y}{L}\sin\frac{m\pi z}{H} \, dy \, dz. $$
If we apply this formula on both sides of the given differential equation we get the terms that you understand (and which are easier to see using the series) and for the delta term we get $$ \frac{2}{L}\frac{2}{H} \iint_{[0,L]\times[0,H]} e^{-i \omega t} \delta(x - x_0)\delta(y - y_0)\delta(z - z_0) \sin\frac{n\pi y}{L}\sin\frac{m\pi z}{H} \, dy \, dz \\ = \frac{4}{LH} e^{-i \omega t} \delta(x - x_0) \sin\frac{n\pi y_0}{L}\sin\frac{m\pi z_0}{H} . $$