# Laplacian solution(page 2) to Jackson 1.5

1. Aug 14, 2008

### daudaudaudau

Hi.

In http://www-personal.umich.edu/~pran/jackson/P505/p1s.pdf" solution(page 2) to Jackson 1.5 it is stated that

$$\nabla^2 \left(\frac{1}{r}\right)=-4\pi\delta^3(\mathbf r)$$.

But why is this true?

$$\nabla^2\left(\frac{1}{r}\right)=\frac{1}{r^2}\frac{d}{d r}\left(r^2\frac{d}{dr}\frac{1}{r}\right)=\frac{1}{r^2}\frac{d}{dr}(-1)$$

Last edited by a moderator: Apr 23, 2017
2. Aug 14, 2008

### Anthony

Re: Laplacian

This equality is to be understood in the distributional sense. It should be read as:

$$\int \frac{\Delta \phi}{|x|}\, \mathrm{d}x = -4\pi \phi (0), \qquad \forall \phi \in C^{\infty}_c (\mathbf{R}^3)$$

:)