Divergence not defined

I’m currently working on the practice problems in Introduction to Electrodynamics by Griffiths. I got confused by the solution to this problem.

What does “ill-defined divergence” even mean? I understand how and when to use delta function, but I don’t understand how divergence is not defined.

• is the delta-function well defined at the origin? Commented Nov 26, 2022 at 0:02
• I suggest that you evaluate the divergence of the ${\bf v}$ field. Is there any delta function? And what is the value of the divergence at the origin? Commented Nov 26, 2022 at 0:15
• @GiorgioP I think I got it. When $n<-2$, there’s a zero in the denominator if we want to calculate the divergence at the origin Commented Nov 26, 2022 at 0:22

I think we can use

$$\nabla \cdot (\psi \vec{a}) = \vec{a} \cdot \nabla \psi + \psi \nabla \cdot \vec{a}$$

to see what's happened for $$n \lt -2$$.

\begin{align*} \nabla \cdot (r^{-3} \hat{r}) &= \left( \frac{1}{r^2} \hat{r} \right) \cdot \nabla \frac{1}{r} + \frac{1}{r} \nabla \cdot \frac{1}{r^2} \hat{r} \\ &= - \frac{1}{r^4} + \frac{1}{r} \nabla \cdot \frac{1}{r^2} \hat{r} && \left( \nabla \frac{1}{r} = - \frac{1}{r^2} \right) \\ &= - \frac{1}{r^4} + \frac{4 \pi}{r} \delta^3(\vec{r}) \end{align*}

When $$r$$ toward $$0$$ (below misuses the delta function, delta function is meaning less outside of integral),

\begin{align*} \nabla \cdot (r^{-3} \hat{r}) &= - \frac{1}{r^4} + \frac{4 \pi}{r} \delta^3(\vec{r}) \\ &= -\infty + \infty \cdot \infty \end{align*}

We can not assign a meaningful value to $$\nabla \cdot (r^{-3} \hat{r})$$, so it is called "ill-defined".