Green's Function for Helmholtz Eqn in Cube

Herr Malus
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Homework Statement


Find the Green's Function for the Helmholtz Eqn in the cube 0≤x,y,z≤L by solving the equation:
\nabla 2 u+k 2 u=δ(x-x')
with u=0 on the surface of the cube
This is problem 9-4 in Mathews and Walker Mathematical Methods of Physics

Homework Equations


Sines, they have the properties we're looking for.

The Attempt at a Solution


So, if the solution is in one dimension we obviously have
G=asinkx x<x' and
G=bsink(x-L) for x>x'
And these are obtained by solving \nabla<sup>2</sup>u+k2u=0 and looking for a and b based on the matching condition at x'. I'm wondering if I can do the same thing for this three dimensional case, using separation of variables and solving for k^2 as the eigenvalue. Then enforcing orthogonality on the functions I get. My problem with this is just that it seems very awkward to have eight sets of equation for the Green's function, based on whether x<x' while y>y' and z<z' for example.

Any help appreciated.
 
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I guess you could do that. You can always use Fourier transform to obtain the Green's function without BC and add a particular solution of the homogeneous Helmholtz equation that satisfies the BC instead, which I imagine would be a less awkward approach.
 
So using the Fourier transform would just give you the green's function for helmholtz in 3d space which is something like an exponential divided by 4∏r followed by simply any function satisfying the BC? So in a 1-d case:
Helmholtz Green's function + sinkx for x<x' or
Helmholtz Green's function +sink(x-L) for x>x'

Thanks for the help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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