Maxwell's Eqn: Vanishing Laplacian of 1/r Explained

[Master J]

In my derivation of one of Maxwell's Equations, I needed the fact that the Laplacian of 1/r vanishes everywhere except at r=0, where r is the norm of a radial vector.

I cannot see how this is? I like to be solid in the math I use for a derivation, so this would really help if someone could clear this up for me. Thanks!

 

[HallsofIvy]

Putting the Laplacian in spherical coordinates, we have
\nabla^2 u= \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+ \frac{1}{r^2 sin^2(\phi)}\frac{\partial^2 u}{\partial\theta^2}+ \frac{1}{r^2 sin(\phi)}\frac{\partial }{\partial\phi}\left(sin(\phi)\frac{\partial u}{\partial \phi}\right).
(See http://mathworld.wolfram.com/SphericalCoordinates.html)

With u= 1/r, the derivatives with respect to \theta and \phi will be 0, of course.