# Coordinate transformation of ODEs

I have the following system of ODEs:

$x' = \alpha x - y - x(x^2 + y^2), y' = x + \alpha y - y(x^2+y^2)$,

where $\alpha \in \mathbb{R}$ is a free parameter.

I have to make a coordinate transformation using polar coordinates, $x = r \cos \theta, y = r \sin \theta$.

I recognize that the $(x^2 + y^2) = r^2$. I am having difficulty in substituting the derivatives. I get that:

$x'(t) = r'(t) \cos (\theta (t))-r(t) \theta '(t) \sin (\theta (t))$,

and

$y'(t) =r'(t) \sin (\theta (t))+r(t) \theta '(t) \cos (\theta (t))$

The problem as you can see is that this autonomous system of ODEs that originally just had an $x'(t)$ and $y'(t)$, now has a $r'(t)$ and $\theta'(t)$ in one ODE. I am unable to write down a system of ODEs as:

$r'(t) = ...$ and $\theta'(t) =...$

• Multiply the first equation by $\cos ( \theta)$ and the second euation by $\sin ( \theta)$ and add them to elimnate $\theta$'. As similar trick can be used to elimante $r'$. – Donald Splutterwit Apr 30 '18 at 0:00
• Hi @DonaldSplutterwit that was very helpful. Thanks! – Thomas Moore Apr 30 '18 at 2:42

## 1 Answer

First

$$\dot x \equiv \cos(\phi)\dot r -r\sin(\phi)\dot\phi\\ \dot y \equiv \sin(\phi)\dot r+r\cos(\phi)\dot \phi$$

Second

$$\alpha x - y - x (x^2 + y^2) \equiv -r (-\alpha \cos(\phi) + r^2\cos(\phi) + \sin(\phi))\\ x + \alpha y - y (x^2 + y^2)\equiv r(\cos(\phi) + (\alpha - r^2) \sin(\phi))$$

and now solving for $\dot r, \dot\phi$

$$\cos(\phi)\dot r -r\sin(\phi)\dot\phi = -r (-\alpha \cos(\phi) + r^2\cos(\phi) + \sin(\phi))\\ \sin(\phi)\dot r+r\cos(\phi)\dot \phi = r(\cos(\phi) + (\alpha - r^2) \sin(\phi))$$

we get finally

$$\dot r = \alpha r - r^3\\ \dot\phi = 1$$

• Hello. Thanks for your answer! – Thomas Moore Apr 30 '18 at 2:42