Find scalar function so that vector field has no divergence

Question

Given is the vector field $\vec{w} = f \left( |\vec{x}| \right)\vec{x} $.

How do I find the scalar function $f$ so that $\vec{\nabla} \cdot \vec{w} = 0$ ??

Answer 1

The argument of $f$ is the magnitude of $\vec{x}$ but $\vec{x}=\vec{x}(x,y,z)$, so the argument of $f$ is able to change each $x,y,z$ independently. Therefore, let $f\left(\left|\vec{x}\right|\right)=g(x,y,z)=X(x)Y(y)Z(z)$ such that $\nabla\cdot g(x,y,z)\vec{x}=0$

Then

$$x\frac{\partial g}{dx}+y\frac{\partial g}{\partial y}+z\frac{\partial g}{\partial z}+3g(x,y,z)=0$$

$$xX'YZ+yXY'Z+zXYZ'=-3XYZ$$

$$x\frac{X'}{X}+y\frac{Y'}{Y}+z\frac{Z'}{Z}=-3$$

$$y\frac{Y'}{Y}+z\frac{Z'}{Z}=-x\frac{X'}{X}-3$$

This is possible if each side is constant

\begin{cases} -x\frac{X'}{X}-3=\eta \\ y\frac{Y'}{Y}=\eta-z\frac{Z'}{Z} \end{cases}

By the same argument, the second equation is possible if it is constant

$$y\frac{Y'}{Y}=\eta-z\frac{Z'}{Z}=\xi$$

So

\begin{cases} -x\frac{X'}{X}-3=\eta \\ y\frac{Y'}{Y}=\xi \\ \eta-z\frac{Z'}{Z}=\xi \end{cases}

Resulting in first order ordinary differential equations

\begin{cases} X'+\frac{3+\eta}{x}X=0 \\ Y'-\frac{\xi}{y}Y=0 \\ Z'+\frac{\xi-\eta}{z}Z=0 \end{cases}

Which have solutions

\begin{cases} X(x)=C_1x^{-(3+\eta)} \\ Y(y)=C_2y^\xi \\ Z(z)=C_3z^{\eta-\xi} \end{cases}

$$\boxed{\therefore g(x,y,z)=\tilde{C}x^{-(3+\eta)}y^{\xi}z^{\eta-\xi}}$$

For any constants $\eta$, $\xi$, and $\tilde{C}$ where $\tilde{C}=C_1C_2C_3$