Solving Laplace's Equation with Mass

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

[tex] \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial \phi}{\partial r}+\mu^2 \phi=0[/tex]

I 'know' the solutions which are linear combinations of [tex]e^{\pm\mu r}/r[/tex] 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 [tex]\mu[/tex] 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
 
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AlephZero said:
Expand the differentials so you have a 2nd order equation.

Then substitute [itex]\psi = r \phi[/itex]

awesome! was that inspired by experience or a general rule of thumb?
 
You said the solution was something like [itex]e^{kr}/r[/itex]. I know how to solve a DE with solutions like [itex]e^{kr}[/itex] so factoring out the r seemed like a good thing to try :smile:

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.