Solving Laplace's Equation with Mass

[homology]

Hi folks, I'm drawing a blank on the following:

<br /> \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial \phi}{\partial r}+\mu^2 \phi=0<br />

I 'know' the solutions which are linear combinations of e^{\pm\mu r}/r as you can check, but I've been trying to see if I could show the fact (as opposed to just checking that it works, which it does).

Context: The above, as I'm sure you realize, is 'almost' Laplace's equation. "If" a photon had mass (the term \mu would be proportional to it) the above equation would result for a region with no charge.

I'm guessing that the answer is simple so please be vague with your hints (I suppose that's almost asking for a smartass comment) as I'd like to figure it out myself.

Thanks

 

[AlephZero]

Expand the differentials so you have a 2nd order equation.

Then substitute \psi = r \phi

 

[homology]

Expand the differentials so you have a 2nd order equation.

Then substitute \psi = r \phi

awesome! was that inspired by experience or a general rule of thumb?

 

[AlephZero]

You said the solution was something like e^{kr}/r. I know how to solve a DE with solutions like e^{kr} so factoring out the r seemed like a good thing to try

In cylindrical and spherical coordinates, solutions often have a factor of 1/r or 1/r^2 compared with cartesian coordinates, because they represent something being spread out over a plane or solid angle, compared with a strip of constant width.

The DEs often have the form 1/r^n d/dr (r^n times something) as well.