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

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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.

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  • $\begingroup$ is the delta-function well defined at the origin? $\endgroup$ Nov 26, 2022 at 0:02
  • $\begingroup$ 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? $\endgroup$ Nov 26, 2022 at 0:15
  • $\begingroup$ @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 $\endgroup$
    – Irene
    Nov 26, 2022 at 0:22

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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".

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