I have the following ODE:
$\ddot{x}-(1-x^{2}-\dot{x}^{2})\dot{x}+x=0$
Now I want to convert this to polar coordinates so that I have an equation for $\dot{\theta}$ and $\dot{r}$. The first thing I have tried is reducing the order to first order by introducing:
$u=x$ and $v=\dot{x}$
But this then leads to a very complicated formula due to the $\dot{x}^{2}$ term. So I am not sure how to then simplfiy that to get the polar ODE's I am looking for. Would appreciate some help with how to simplify such a complex system or perhaps another approach to making turning this ODE into polar coordinates.
Thanks a lot!
Best Answer
Following your first step: \begin{align} \dot u & = v \\ \dot v & = (1 - u^2 - v^2)v - u \end{align} Let $u = r\cos \theta$ and $v = r \sin \theta$, then note that $r^2 = u^2 + v^2$ so $r \dot r = u \dot u + v \dot v$ and $\tan \theta = v/u$ and so $\sec^2 \theta \dot \theta = \frac{u \dot v - v \dot u }{u^2}$. I will do the $r$ equation, you can plug in to find one for $\dot \theta$: $$ \dot r = (1-r^2)r \sin^2 \theta $$