$x' = y + x (\alpha - x^2 - y^2)$
$y' = -x + y (1 - x^2 - y^2)$
I believe that I have figured out its polar coordinate form.
$r' = r \left[ (\alpha - 1) \cos^2 \theta + 1 - r^2 \right]$
$\theta' = -1 - (\alpha - 1) \sin \theta \cos \theta$
I am trying to calculate the divergence.
$F_r = r' = r \left[ (\alpha - 1) \cos^2 \theta + 1 - r^2 \right]$
$F_\theta = \theta' = -1 - (\alpha - 1) \sin \theta \cos \theta$
$F_z = 0$
I should be using the cylindrical version of the divergence method, no?
$ \text{div} F = \frac{1}{r} \frac{d}{dr} \left( r F_r \right) + \frac{1}{r} \frac{d F_\theta}{d \theta} + \frac{d F_z}{dz}$
When I substitute everything in and work out the algebra, I get
$ r F_r = (\alpha - 1) r^2 \cos^2 \theta + r^2 - r^4$
$\frac{d}{dr}(r F_r) = 2 (\alpha - 1) r \cos^2 \theta + 2r - 4r^3$
$\frac{1}{r} \frac{d}{dr}(r F_r) = 2 (\alpha - 1) \cos^2 \theta + 2 - 4r^2$
and
$F_\theta = -1 - (\alpha - 1) \sin \theta \cos \theta$
$\frac{d F_\theta}{d \theta} = -(\alpha - 1) \left( \cos^2 \theta - \sin^2 \theta \right)$
$\frac{1}{r} \frac{d F_\theta}{d \theta} = -\frac{\alpha - 1}{r} \left( \cos^2 \theta - \sin^2 \theta \right)$
so
$ \text{div} F = 2 (\alpha - 1) \cos^2 \theta + 2 - 4 r^2 - \frac{\alpha - 1}{r} \left( \cos^2 \theta - \sin^2 \theta \right)$
But I was checking the book and it seems that I should be getting
$\text{div} F = \alpha + 1 - 4 r^2$
I checked and tried many times but it doesn't seem to work out. Where did I go wrong? Maybe it is my derivation?
$\text{div} F$">
$\theta'$">
$r'$">