Writing a system of ODEs in polar coordinates

Question

I have this system of equations:

$$\dot{x}=x-y-x(x^2+y^2)+\frac{xy}{\sqrt{x^2+y^2}} \\ \dot{y}=x+y-y(x^2+y^2)-\frac{x^2}{\sqrt{x^2+y^2}}$$

How can I get this in polar coordinates ? I know that $r^2=x^2+y^2$, but how can I find $\dot{r}$ or $\dot{\theta}$ ?

Answer 1

When we use polar co-ordinates, we let $x=r\cos\theta$ and $y=r\sin \theta$. We can differentiate each of these equations to find $\dot{x}$ and $\dot{y}$ (and this will help you solve your question). If you want to find $\dot{r}$ or $\dot{\theta}$, just differentiate the equations $\tan\theta=y/x$ and $r^{2}=x^{2}+y^{2}$ implicitly.

For example, $$\dot{x}=\frac{d}{dt}(r\cos\theta)=\frac{dr}{dt}\cos\theta-r\sin\theta\frac{d\theta}{dt}=\dot{r}\cos\theta-r\dot{\theta}\sin\theta$$

Answer 2

Setting $$ x=r\cos\theta,\ y=r\sin\theta, $$ we have \begin{eqnarray} \dot{r}\cos\theta-r\dot{\theta}\sin\theta&=&\dot{x}=r(\cos\theta-\sin\theta)-r^3\cos\theta+r\cos\theta\sin\theta,\\ \dot{r}\sin\theta+r\dot{\theta}\cos\theta&=&\dot{y}=r(\cos\theta+\sin\theta)-r^3\sin\theta-r\cos^2\theta. \end{eqnarray} It follows that $$ \dot{r}=r-r^3,\quad \dot{\theta}=1-\cos\theta. $$