How to calculate divergence of this function

Question

Could somebody explain the method used here to differentiate? I understand how to calculate divergence of simple functions, but the method above isn’t very clear. By the way, $x\in\mathbb R^3$, $a\in\mathbb R$.

Answer 1

\begin{align*}\nabla\cdot \left(\frac{x}{\|x\|^{2a}} \right)&=\sum_{i=1}^3\frac{\partial}{\partial x_i}\frac{x_i}{\|x\|^{2a}}\\ &=\sum_{i=1}^{3}\left[\frac{1}{\|x\|^{2a}}-2ax_i\|x\|^{-2a-1}\frac{x_i}{\|x\|} \right]\\ &=\sum_{i=1}^{3}\left[\frac{1}{\|x\|^{2a}} -2a\frac{x_i^2}{\|x\|^{2a+2}} \right]\\ &=\frac{3}{\|x\|^{2a}}-2a\frac{\|x\|^2}{\|x\|^{2a+2}}\\ &= \frac{3-2a}{\|x\|^{2a}} \end{align*}

Note that

$$\frac{\partial}{\partial x_1}\|x\|=\frac{\partial}{\partial x_1}\sqrt{x_1^2+x_2^2+x_3^2}=2x_1\frac{1}{2} (x_1^2+x_2^2+x_3^2)^{-\frac{1}{2}}=x_1\|x\|^{-1}=\frac{x_1}{\|x\|}$$

Answer 2

If $f:\Bbb R^3\to\Bbb R$ is a scalar function, and $\mathbf{g}:\Bbb R^3\to\Bbb R^3$ is a vector function, then we have $$\nabla.(f\mathbf{g})=f\nabla.\mathbf{g}+\mathbf{g}.\nabla f$$ (you can verify this simply by writing out both sides in full).

Here we have $f(x,y,z)=\dfrac{1}{||x||^{2a}}=(x^2+y^2+z^2)^{-a}$ and $\mathbf{g}(x,y,z)=(x,y,z)$.

So $\nabla.\mathbf{g}$ is just $3$, and I will let you work out $\nabla f$ for yourself.

It's true that the equation looks daunting when you first see it, but it's quite straightforward if you work it through.