How to differentiate the numerator of this vector field $\frac{\vec r}{|r|^3}$?

Question

I was studying a nice solution of how to apply the divergence theorem for a vector field with a singularity at the origin.

However, the solution doesn't give any concrete computations and just makes the claim that $\large\frac {\vec r}{|r|^3}$ is divergence-free; I am guessing that $\vec r$ is just a radius / position vector $(x,y,z)$.

So, I would like to check this, but I am slightly confused about how to differentiate the numerator.

The denominator is clearly $(x^2 + y^2 + z^2)^{\frac{3}{2}}$, so I know how to take derivatives of this term, w.r.t. either $x$, $y$, or $z$ - just have to be careful when using the chain rule to account for all the factors.

What about the numerator? Would the derivative of it, w.r.t., say, $x$, be ... $r_x$ just be... $1$? So, for simplicity just looking at $div \vec r$ gives $r_x + r_y + r_z$ = 1 + 1 + 1 = 3.

Am I ok with my computations?

Thanks,

Answer 1

$u=(x,y,z)$ and $f(x,y,z)=(x/\|u\|^3,y/\|u\|^3,z/\|u\|^3)$

$\partial_x(x/\|u\|^3)= 1/\|u\|^3-3 x^2/\|u\|^{5}$

$\partial_y(y/\|u\|^3)= 1/\|u\|^3-3 y^2/\|u\|^{5}$

$\partial_z(z/\|u\|^3)= 1/\|u\|^3-3 z^2/\|u\|^{5}$

$\partial_x(x/\|u\|^3)+\partial_y(y/\|u\|^3)+\partial_z(z/\|u\|^3)= 3/\|u\|^3-3(x^2+y^2+z^2)/\|u\|^{5}= 3/\|u\|^3-3(\|u\|^3)=0$

Answer 2

Consider the function $f(x,y,z)= -(x^2+y^2+z^2)^{-1/2}$. Then its gradient is your vector field. Write $F= (f_x,f_y,f_z)$, so that the divergence reads

$${\rm div}\, F= f_{xx}+f_{yy}+f_{zz}$$

i.e. you want to show that $F$ has null Laplacian, i.e. that $f$ is harmonic. Note that $f$ is symmetric in $x,y,z$, so it suffices to compute only $f_{xx}$. This is $$\partial_x ({x}{(x^2+y^2+z^2)^{-3/2}})=(x^2+y^2+z^2)^{-3/2}-3x^2(x^2+y^2+z^2)^{-5/2}$$

Summing gives the desired equality. More generally, $f(\vec x)=\lVert \vec x\rVert^{2-n}$ is always harmonic (where $n$ is the number of variables).