Helmholtz in spherical co-ordinates - Boundary Conditions

[Gwinterz]

Hello,

I was just after an explanation of how people get to this conclusion:

Say you are looking at the Helmholtz equation in spherical co-ordinates.

You use separation of variables, you solve for the polar and azimuthal components.

Now you solve for the radial, you will find that the radial equation can be written in the form of the spherical bessel equation after a slight change of variables.

The solution to the radial part is then:

R(r) = a j_l (z) + b y_l (z)

where z(r).

I often see people do this:

Inside the sphere:

R(r) = a j_l (z)

This is fair enough, the bessel y diverges at z = 0.

However I don't understand why people say that outside the sphere:

R(r) = b y_l (z)

Why is the bessel j not involved here?

Thanks

 

[Greg Bernhardt]

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?

 

[jtbell]

This thread has now been moved to Differential Equations, where it might be more likely to get a response than in General Math.

 

[Gwinterz]

Hey,

Thanks for your replies,

Yea, this is a tough one, there really isn't any more information I can give. I have found that the literature is split, half of the time people do it one way, and the other half another way.

What I would think is the correct way is to say that the solution OUTSIDE the sphere is:
R(r) = a j_l (z) + b y_l (z)

Then apply boundary conditions to get a and b.

No body does this, either they set a = 0, which is what I mentioned in the first post, OR, they say:

R(r) = c h_l (z)

where h_l is either the hankel function of the first/second kind or j_l, or y_l. This approach is slightly better, but I still don't see how this is correct. While it's still a solution, it just doesn't seem right...

 

[blue_raver22]

for your question. The reason why the Bessel function j is not involved in the solution outside the sphere is due to the boundary conditions of the problem.

In spherical coordinates, the Helmholtz equation represents waves propagating in a spherical medium. When solving for the radial component, we are essentially looking at the behavior of the wave as it travels from the center of the sphere (r = 0) to the outer boundary (r = R). Therefore, the boundary conditions for the radial component are determined by the behavior of the wave at these two points.

Inside the sphere, at r = 0, the wave must be well-behaved and finite. This means that the solution for the radial component, R(r), must also be finite at r = 0. This is why we use the Bessel function j in the solution, as it is well-behaved and finite at z = 0.

Outside the sphere, at r = R, the wave encounters a different boundary condition. Here, the wave is no longer confined by the spherical medium and can propagate infinitely. This means that the solution for the radial component, R(r), must also be infinite at r = R. This is why we use the Bessel function y in the solution, as it diverges at z = 0 and is therefore able to satisfy this boundary condition.

I hope this helps to clarify why the Bessel function j is not involved in the solution outside the sphere. It is all determined by the specific boundary conditions of the problem. Let me know if you have any further questions.