Green's Function for Helmholtz Eqn in Cube

In summary, the conversation discusses finding the Green's Function for the Helmholtz equation in a three-dimensional cube by solving the equation and enforcing orthogonality on the obtained functions. It is suggested to use Fourier transform to obtain the Green's function without boundary conditions and add a particular solution that satisfies the boundary conditions instead. This would result in a Green's function similar to the one-dimensional case, with the addition of any function that satisfies the boundary conditions.
  • #1
Herr Malus
28
0

Homework Statement


Find the Green's Function for the Helmholtz Eqn in the cube 0≤x,y,z≤L by solving the equation:
[itex]\nabla[/itex] 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 [itex]\nabla2[/itex]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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  • #2
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.
 
  • #3
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.
 

1. What is the Green's function for the Helmholtz equation in a cube?

The Green's function for the Helmholtz equation in a cube is a mathematical function that represents the response of a system to a point source of energy. It is used to solve the Helmholtz equation, a partial differential equation that describes the propagation of waves in a medium.

2. How is the Green's function for the Helmholtz equation in a cube derived?

The Green's function for the Helmholtz equation in a cube is derived using the method of images. This involves creating a mirror image of the point source inside the cube and solving for the electric potential at all points inside the cube. The resulting function is the Green's function.

3. What are the properties of the Green's function for the Helmholtz equation in a cube?

The Green's function for the Helmholtz equation in a cube has several important properties, including being symmetric, having a singularity at the location of the point source, and satisfying the Helmholtz equation itself. It also vanishes at the boundary of the cube, as well as at infinity.

4. How is the Green's function for the Helmholtz equation in a cube used in practical applications?

The Green's function for the Helmholtz equation in a cube is used in a variety of practical applications, including solving problems in electromagnetism, acoustics, and fluid dynamics. It is especially useful for finding the response of a system to a localized source of energy or for modeling wave propagation through a medium.

5. Are there any limitations to using the Green's function for the Helmholtz equation in a cube?

While the Green's function for the Helmholtz equation in a cube is a powerful tool for solving certain problems, it does have some limitations. For example, it is only applicable to systems that can be modeled as a cube and that satisfy the assumptions of the Helmholtz equation. Additionally, it may not provide an exact solution in more complex systems and may require numerical methods for accurate results.

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