In three dimensions, the Laplacian of $1/r$ is $0$ outside the origin

Question

Why does the following hold?

$$\Delta_{3}\frac{1}{r}\Bigg\vert_{\mathbb{R}^3 \setminus \left\lbrace 0 \right\rbrace}=0$$

Answer 1

Note: $\Delta 1/|x| = -4\pi\delta(0)$.

$$ \begin{align} I(x) = \int\limits_D \Delta_x \frac{1}{|x - x'|} dx' &= \int\limits_D \mbox{div}_x\,\mbox{grad}_x \frac{1}{|x-x'|} dx' = - \int\limits_D \mbox{div}_x\frac{x-x'}{|x-x'|^3} dx' \\ &= - \int\limits_{\partial D}\frac{1}{|x-x'|^2} e_r \cdot dA' \end{align} $$ where I used $$ \left(\mbox{grad}_x \frac{1}{|x|} \right)_i = \frac{\partial}{\partial x_i} \frac{1}{|x|} = \partial_i \left(\sum_j x_j^2\right)^{-1/2} = -\frac{1}{2} \left(\sum_j x_j^2\right)^{-3/2} 2 x_i= -\frac{x_i}{|x|^3} $$

For $x = 0$ for $D$ being a ball of radius $R$ without inner ball of radius $\epsilon$ one gets $$ I(0) = -4\pi \left(\frac{R^2}{R^2}- \frac{\epsilon^2}{\epsilon^2}\right) = 0 $$

The limit $R\to\infty$ doesn't change the result.

The surface area of the ball is proportional to $R^2$ while it gets compensated by the constant $1/R^2$ term.

This is also consistent with the Gauss theorem of electrostatics, as there is no point source included in $D$.

Answer 2

Hint: use the expression for the Laplacian in spherical coordinates.