Confirming Green's function for homogeneous Helmholtz equation (3D)

[PhDeezNutz]

Homework Statement

The 3D Helmholtz equation is

$\left(\nabla^2 + k^2 \right) \Psi \left( r \right)= 0$

Supposedly the Green's function for this equation is

$G\left(r \right) = - \frac{1}{4 \pi} \frac{e^{ikr}}{r}$

Relevant Equations

A green's function is defined as the solution to the following

$\left( \nabla^2 + k^2 \right) G = \delta \left( r \right)$

The laplacian in spherical coordinates (for purely radial dependence) is

$\nabla^2 t = \frac{1}{r^2} \frac{\partial }{\partial r} \left(\frac{1}{r^2}\frac{\partial t}{\partial r} \right)$

Plugging in the supposed $G$ into the delta function equation

$\nabla^2 G = -\frac{1}{4 \pi} \frac{1}{r^2} \frac{\partial}{\partial r} \left(\frac{r^2 \left(ikr e^{ikr} - e^{ikr} \right)}{r^2} \right)$

$= -\frac{1}{4 \pi} \frac{1}{r^2} \left[ike^{ikr} - rk^2 e^{ikr} - ike^{ikr} \right]$

$= \frac{k^2 e^{ikr}}{4 \pi r}$

$k^2 G$ is simply

$k^2 G = - \frac{k^2 e^{ikr}}{4 \pi r}$

So we get

$\left( \nabla^2 + k^2 \right) G = 0$

I know the delta function is zero everywhere else besides r = 0 where it is infinity, but I'm getting 0 across the board instead of a delta function.

Thanks for any help in advanced.

 

[Orodruin]

but I'm getting 0 across the board instead of a delta function

No you are not. What you are doing is only valid for r>0 as your coordinate system is singular at r=0.

Try integrating the differential equation over a small sphere.

Edit: Also consider that the special case of Poisson’s equation (ie, Helmholtz with k=0) has a solution proportional to 1/r. This is known to have a delta inhomogeneity at r=0.