# How to convert dynamical system to polar coordinates? [closed]

I have a dynamical system on the plane given by $$\dot{x}=-y+x\left(1-\sqrt{x^2+y^2}\right)\\ \\ \dot{y}=y+x\left(1-\sqrt{x^2+y^2}\right)$$

I want to convert this into polar coordinates as it will be easier for the question I am attempting to solve (it gives a hint to convert this into polar coordinates).

Problem is I don't know how to do this I know all the relations for polar coordinates to relate it to Cartesian coordinates for example I know the bit in the bracket would be $(1-r)$ but I don't know what $\dot{x}$ would and how to convert it into polar coordinates.

Any help?

• Answer this question: if you know $\dot{x}$ and $\dot{y}$ and also know how $r$ and $\phi$ related to $x$ and $y$, can you write $\dot{r}$ and $\dot{\phi}$ ? Apr 12, 2016 at 21:30
• What exactly you don't understand? Apr 12, 2016 at 21:38
• As you know, $r^2 = x^2+y^2$. What would happen if you calculate time derivative of left and right side of this equality? Apr 12, 2016 at 21:39
• Yes, that's it. If you don't remember chain rule for derivatives, it's the right time to revise it, because it's used everywhere in differential equations. Apr 12, 2016 at 21:58
• Having just done this calculation I think the original system was missing a bracket. The solution is much nicer if $$\dot{x}=\left(-y+x\right)\left(1-\sqrt{x^2+y^2}\right)\\ \\ \dot{y}=\left(y+x\right)\left(1-\sqrt{x^2+y^2}\right)$$ Sep 18, 2016 at 17:03

$1)$ From $x',y'$ to $r',\theta'$: $$r' = \left(\sqrt{x^2 +y^2} \right)' = \frac{(x^2 +y^2)'}{2 \sqrt{x^2 +y^2}}=\frac{xx' +yy'}{r}$$ and $$\theta' = \left(\arctan \frac{y}{x} \right)' = \frac{(y/x)'}{1+(y/x)^2} = \frac{y' x -x' y}{r^2}.$$
$2)$ From $r',\theta'$ to $x',y'$: $$x'= (r\cos\theta)'=r' \cos \theta -r \theta' \sin \theta$$ and $$y'= (r\sin\theta)'= r' \sin \theta + r \theta' \cos \theta .$$