# I why div [grad(1/r)]=-4πδ(r vec)

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1. Jul 3, 2017

### cristianbahena

I don't understand why grad(1/r)= (-1/r^2)ê_r. I know its derivate, but in cero the function is not definite and you can't find its derivative in cero.

And, why div [grad(1/r)]=-4πδ(r vec) ?

2. Jul 3, 2017

### Math_QED

This is simply applying definitions. Also,

$\frac{d|x|}{dx} = \frac{|x|}{x} = \begin{cases} 1 \quad \mathrm{if} \quad x >0\\-1 \quad \mathrm{if} \quad x <0 \end{cases}$

3. Jul 3, 2017

### hilbert2

You're probably supposed to show that

$\lim_{\epsilon \rightarrow 0+}\int_{\mathbb{R}^3}\nabla \cdot \left(\nabla \left(\frac{1}{|\mathbf{r}|+\epsilon}\right)\right) f(\mathbf{r})dV \propto f(\mathbf{0})$

for a relatively arbitrary function $f( \mathbf{r})$. Note that $\mathbf{r}$ is a position vector, $|\mathbf{r}|$ its norm and $dV$ a volume element.