# Solution that doesn't diverge at origin

1. Jun 4, 2014

### jimjam1

Hi - wondering if you can help me find a solution of:

$\nabla^{2}u-\frac{u}{\lambda^{2}}=a\delta(r)$

for spherical symmetry in 3D with the condition that $\lim_{r\rightarrow \infty}u=0$. It can be rewritten in spherical coordinates as

$\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial u}{\partial r}\right)-\frac{u}{\lambda^{2}}=a\delta(r)$.

Any help would be much appreciated! :)

2. Jun 4, 2014

### pasmith

You are solving $$\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial u}{\partial r}\right)-\frac{u}{\lambda^{2}} = 0$$ in $r > 0$. Setting $$u(r) = \frac{f(r)}r$$ yields $$f'' - \lambda^{-2} f = 0$$ and the only way to not have $u$ diverge at the origin is to take $f(0) = 0$, which yields $f(r) = A\sinh(\lambda^{-1} r)$ and thus $$u(r) = \frac{A\sinh(\lambda^{-1} r)}{r}.$$ L'hopital confirms that $$\lim_{r \to 0} u(r) = \lim_{r \to 0} \frac{A\lambda^{-1}\cosh(\lambda^{-1} r)}{1} = A\lambda^{-1}.$$ Unfortunately $|u| \to \infty$ as $r \to \infty$. To get a solution which decays at infinity you must take $f(r) = e^{-r/\lambda}$, and the resulting $u$ diverges at the origin.

(Usually this setup is an abstraction of "there is a small sphere at the origin". Within the sphere you use a solution which is bounded at the origin, and outside the sphere you use a solution which decays as $|r| \to \infty$.)

Last edited: Jun 4, 2014
3. Jul 13, 2014

### rigetFrog

Bessel or Hankel functions.

4. Jul 17, 2014

### jasonRF

I wouldn't expect a physical solution that does not diverge at the origin. That is the exact equation for the electric potential of a point charge in a hot plasma, where $\lambda$ would be the Debye length. Potentials of point charges always diverge at the location of the charge...