# Helmholtz in spherical co-ordinates - Boundary Conditions

1. Aug 14, 2014

### 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

2. Aug 27, 2014

### Greg Bernhardt

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

3. Aug 28, 2014

### Staff: Mentor

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

4. Sep 8, 2014

### Gwinterz

Hey,

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...