If $\vec{u} = \vec{r}/r$, find $\mathrm{div} ~(\nabla \vec{u})$?

Question

If $\vec{u} = \vec{r}/r$, find $\mathrm{div} ~(\nabla \vec{u})$.

Please help me to solve this problem.

Answer 1

Note that we have

$$\nabla \vec u=\hat x_i\hat x_j\partial_i (u_j) \tag 1$$

Taking the divergence of $(1)$ yields

$$\begin{align} \nabla \cdot \left(\nabla \vec u\right)&=\partial_i\left(\hat x_j\partial_i(u_j)\right)\\\\ &=\nabla^2\vec u\\\\ &=\nabla \nabla \cdot \vec u-\nabla \times \nabla \times \vec u \end{align}$$

Then, we have for $r\ne 0$

$$\nabla \times \left(\frac{\vec r}{r}\right)=0$$

and

$$\nabla \cdot \left(\frac{\vec r}{r}\right)=\frac{2}{r}$$

so that

$$\nabla^2 \left(\frac{\vec r}{r}\right)=-2\frac{\vec r}{r^3}$$

Answer 2

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$\ds{\,\mathrm{div}\pars{\nabla\pars{\vec{r} \over r}} = \nabla^{2}\pars{\vec{r} \over r}}$. You can use the identity:

$\ds{\nabla^{2}\pars{ab} = \pars{\nabla^{2}a}b + 2\pars{\nabla a}\cdot\pars{\nabla b} + a\nabla^{2}b}$ which is a ' sort of Newton Binomial '.

\begin{align} \color{#f00}{\nabla^{2}\pars{x \over r}} & = \nabla^{2}\pars{x\,{1 \over r}}= \pars{\nabla^{2}x}\,{1 \over r} + 2\pars{\nabla x}\cdot\nabla\pars{1 \over r} + x\,\nabla^{2}\pars{1 \over r} \\[4mm] & = 0 \times {1 \over r} + 2\,\hat{x}\cdot\pars{-\,{1 \over r^{2}}\,{\vec{r} \over r}} + x\bracks{-4\pi\delta\pars{\vec{r}}} = \color{#f00}{-\,{2x \over r^{3}}} \end{align}

Note that $\ds{\nabla\,\mathrm{f}\pars{r} = \,\mathrm{f}\,'\pars{r}\,{\vec{r} \over r}}$.

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Similarly, for the $\ds{y}$ and $\ds{z}$ components: $$ \color{#f00}{\nabla^{2}\pars{\vec{r} \over r}} = \color{#f00}{-\,{2\vec{r} \over r^{3}}} $$