Examples

Divergence of vector fields

Divergence of vector fields

Consider the vector field $\mathbf{F}=(x,y)$. This is a radial field where each vector has length $r$. $\nabla \cdot \mathbf{F}=+2$ everywhere.

Now consider the vector field $\mathbf{F}=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})=\frac{1}{r^2}(x,y)$. We have $\nabla \cdot \mathbf{F}=0$ except at $(0,0)$, where it is infinity (undefined).

Divergence in polar coordinates

Divergence in polar coordinates

Let $\mathbf{F}=F_r{\mathbf{e}}_r+F_{\theta }{\mathbf{e}}_{\theta }$ give a vector field in polar coordinates. Here $F_r$ and $F_{\theta }$ are scalar coordinate functions, ${\mathbf{e}}_r=(\frac{x}{r},\frac{y}{r})$ is the unit vector in the radial direction, and ${\mathbf{e}}_{\theta }=(\frac{-y}{r},\frac{x}{r})$ is the unit vector in the $+\theta$ direction.
Problem: Find a formula for the divergence $\nabla \cdot \mathbf{F}$ in polar coordinates.

Solution: Write $\nabla \cdot \mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}$ in terms of polar coordinates. We have in Cartesian coordinates:

$$\mathbf{F}=\left(x\frac{F_r}{r}-y\frac{F_{\theta }}{r},y\frac{F_r}{r}+x\frac{F_{\theta }}{r}\right)$$

so:

$$\begin{array}{rl}\nabla \cdot \mathbf{F}& =\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}\\ & =\frac{F_r}{r}+x\frac{\partial }{\partial x}\left(\frac{F_r}{r}\right)-y\frac{\partial }{\partial x}\left(\frac{F_{\theta }}{r}\right)\\ & +\frac{F_r}{r}+y\frac{\partial }{\partial y}\left(\frac{F_r}{r}\right)+x\frac{\partial }{\partial y}\left(\frac{F_{\theta }}{r}\right).\end{array}$$

The chain rule gives us formulas to convert derivatives:

$$\begin{array}{rl}\frac{\partial f}{\partial x}& =\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta }\frac{\partial \theta }{\partial x}\\ & =\frac{x}{r}\frac{\partial f}{\partial r}+\frac{-y}{r^2}\frac{\partial f}{\partial \theta }\\ \frac{\partial f}{\partial y}& =\frac{\partial f}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial f}{\partial \theta }\frac{\partial \theta }{\partial y}\\ & =\frac{y}{r}\frac{\partial f}{\partial r}+\frac{x}{r^2}\frac{\partial f}{\partial \theta }.\end{array}$$

So we obtain, after some cancellation and rearranging:

$$\begin{array}{rl}\nabla \cdot \mathbf{F}& =2\frac{F_r}{r}+\frac{x^2+y^2}{r}\frac{\partial }{\partial r}\left(\frac{F_r}{r}\right)+\frac{y^2+x^2}{r^2}\frac{\partial }{\partial \theta }\left(\frac{F_{\theta }}{r}\right)\\ & =2\frac{F_r}{r}+r\frac{\partial }{\partial r}\left(\frac{F_r}{r}\right)+\frac{1}{r}\frac{\partial }{\partial \theta }{F}_{\theta }\\ & =\frac{1}{r}\frac{\partial }{\partial r}(r{F}_r)+\frac{1}{r}\frac{\partial }{\partial \theta }{F}_{\theta }.\end{array}$$