conversion of ode system to polar

Question

$x' = -y + x(1 - x^2 - y^2)$

$y' = x + y(1 - x^2 - y^2)$

I am trying to learn how to transform a dynamical system into its polar coordinate version. I already got

$r' = r(1 - r^2)$

However, I am not getting

$\theta' = 1$

I started with

$yx' = -y^2 + xy (1 - x^2 - y^2)$

$xy' = x^2 + xy (1 - x^2 - y^2)$

Then

$xy' - yx' = x^2 + y^2$

$(y/x)' = r^2$

$(\tan \theta)' = r^2$

$\theta' \sec^2 \theta = r^2$

$\theta' = r^2 \cos^2 \theta$

$\theta' = x^2 \neq 1?$

Answer 1

In dealing dynamical systems I prefer not to lose physical dimensions.. helps to give a dimension check at each stage:

$x'/\omega = -y + x(1 - x^2/a^2 - y^2/a^2)$

$y'/\omega = x + y(1 - x^2/a^2 - y^2/a^2)$

You can always set $ \omega=1, a=1 $ for differentiation. Start with $r^2= x^2+y^2$ etc..

$r' = r(1 - r^2)$ is alright but more beneficial with physical dimensions

Primed on time $x'$ is linear velocity, and angular $ \dfrac{d\theta}{dt}= \omega.$

Proceeding same way as you did for differentials,

$$ \dfrac{r'}{\omega }= \dfrac{r(a^2-r^2)}{a^2}$$

$$ \dfrac{d\theta}{dt}= \dfrac{\omega \cdot r}{a}$$

Answer 2

$$ r(t)= \sqrt{x^2(t)+y^2(t)} $$ $$ r'(t)= \frac{1}{r}(xx'+yy')=r(1-r^2) $$ $$ \theta(t) = \arctan(\frac{y(t)}{x(t)}) $$ $$ \theta'(t)= \frac{x^2(t)}{x^2(t)+y^2(t)} \frac{y'(t)x(t)-x'(t)y(t)}{x^2(t)}=1 $$

Answer 3

This line is correct: $$xy' - yx' = x^2 + y^2$$ Then we have : $$\dfrac {xy' - yx' }{ x^2 + y^2}=1$$ $$(\arctan \frac y x )'=1$$ $$\theta '=1$$ Since we have: $$\theta ' = (\arctan \frac y x )'= \dfrac {xy' - yx' }{ x^2 + y^2}$$