Finding polar representation of $\dot{x} = -y+x(1-x^{2}-y^{2}), \dot{y} = x+y(1-x^{2}-y^{2}). $

Question

We have the system

$$\dot{x} = -y+x(1-x^{2}-y^{2})\\ \dot{y} = x+y(1-x^{2}-y^{2}). $$

Now this ends up as $$ \dot{r} = r(1-r^{2})\\ \dot{\theta} = 1 $$ in polar coordinates, but I do not see how.

I reckon we should use the standard transformation $x= r\cos{\theta}$ and $y = r \sin{\theta}$ but how one derive the latter representation? I see how $-x^{2}-y^{2} = - ( x^{2} + y^{2} ) = -r^{2}(\sin^{2}{\theta}+\cos^{2}{\theta}) = -r^{2}$, but that is just a small part of it I guess.

Answer 1

You have $$ r\dot r=x\dot x+y\dot y=r^2(1-r^2) $$ and for the angle $$ r^2\dot θ=x\dot y-y\dot x=r^2 $$ Now cancel the $r$'s on the left side ...