How to transform polar dynamical set of equation to cartesian?

Question

I have been fiddling around trying to understand coordinate transformations in dynamical systems. I know the standard way to convert the system from cartesian coordinates to polar, but am having trouble reversing it.

I know the standard transformation equations: $$ r\cos(\theta)=x,\quad r\sin(\theta)=y,\quad r^2 = x^2+y^2,\quad \theta = \arctan\left(\frac{y}{x}\right), $$

$$ \frac{\partial x}{\partial r} =\cos(\theta),\quad \frac{\partial x}{\partial \theta} =-r\sin(\theta),\quad \frac{\partial y}{\partial r} =\sin(\theta),\quad \frac{\partial y}{\partial \theta} =r\cos(\theta), $$

$$ \frac{\partial r}{\partial x} =\frac{x}{r}=\cos(\theta),\quad \frac{\partial \theta}{\partial x} =\frac{-y}{x^2+y^2}=\frac{-\sin(\theta)}{r}, $$

$$ \frac{\partial r}{\partial y} =\sin(\theta),\quad \frac{\partial \theta}{\partial y} =\frac{x}{x^2+y^2}=\frac{\cos(\theta)}{r} $$

If we have an example system of equations: \begin{aligned} \dot{r} &= m r - r^3, \\ \dot{\theta} &= w + v r^2 \end{aligned} where $w, v, m$ are arbitrary constants.

What are the steps I follow to make the appropriate transformation to $\dot{x}, \dot{y}$?

Answer 1

Use $\dot{x}=\frac{\dot{r}x}{r}-y\dot{\theta}$ (proof is an exercise) together with $r^2=x^2+y^2$. The treatment of $\dot{y}$ is similar.

Answer 2

I don't really know why you want to do this, since these seem conveniently represented by the polar coordinates, you can convert as follows $$ \frac{d}{dt}\left(\sqrt{x(t)^2+y(t)^2} \right)=m\sqrt{x(t)^2+y(t)^2}-(x(t)^2+y(t)^2)^{3/2}\\ \frac{d}{dt}\arctan\left(\frac{y(t)}{x(t)}\right)=w+v(x(t)^2+y(t)^2) $$

Answer 3

$\begin{cases} \dot x=\dot r\cos(\theta)-r\dot\theta\sin(\theta)=(mr-r^3)\cos(\theta)-r(w+vr^2)\sin(\theta)\\ \dot y=\dot r\sin(\theta)+r\dot\theta\cos(\theta)=(mr-r^3)\sin(\theta)+r(w+vr^2)\cos(\theta)\\ \end{cases}$

$\begin{cases} \dot x=x(m-x^2-y^2)-y(w+vx^2+vy^2)\\ \dot y=y(m-x^2-y^2)+x(w+vx^2+vy^2) \end{cases}$