The function $1/r$ satisfies the Laplace equation?

Question

Hello. I have a question. The image is from the book Distribution theory by Gerri Van Dijk.

Question. In what sense does $1/r$ satisfy the equation outside of $0$?

Answer 1

In spherical coordinates the Laplace operator has the form $$ \Delta f = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial f}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial f}{\partial \theta} \right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 f}{\partial \varphi^2} . $$

With $f(r,\theta,\varphi)=1/r$ the second and third terms clearly vanish, but so does also the first term: $$ \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial (1/r)}{\partial r} \right) = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \left( -\frac{1}{r^2} \right) \right) = \frac{1}{r^2} \frac{\partial}{\partial r} \left(-1 \right) = 0. $$

Thus, $\Delta (1/r) = 0$ outside of origin, and also $\delta=0$ outside of origin.