Polar coordinates transformation of $\dot{x} = -y + x(1-x^2-y^2)$ and $\dot{y} = x + y(1-x^2-y^2)$

Question

We have $$\begin{cases}\dot{x} = -y + x(1-x^2-y^2)\\ \dot{y} = x + y(1-x^2-y^2) \end{cases}$$ By taking $x=r\cos\theta$, $y=r\sin\theta$, I am able to get to the expression $$\dot{r} = r(1-r^2) + \frac{r\sin\theta(\dot{\theta}+1)}{\cos\theta}.$$ I can't seem to see where I have gone wrong as I am supposed to deduce that $\dot{r} = r(1-r^2)$ and $\dot{\theta} = 1$

Can someone explain where I am going wrong.

Answer 1

It is not clear how you got your expression. Anyway, in order to obtain $\dot{r} = r(1-r^2)$, you may multiply the first equation by $2x$ and the second one by $2y$. Aftersumming them, we get $$2x\dot{x} +2y\dot{y} = -2xy + 2x^2(1-x^2-y^2)+ 2xy + 2y^2(1-x^2-y^2)$$ Now, the left-hand side is the derivative of $r^2=x^2+y^2$ with respect to $t$ that is $2r\dot{r}$.

Can you take it from here?

Moreover from $\dot{x} = -y + x(1-x^2-y^2)$ and $\dot{r} = r(1-r^2)$, we have that $$\dot{r}\cos(\theta)-r\sin(\theta)\dot{\theta}=-r\sin(\theta)+\cos(\theta)\dot{r}$$ which implies $r\sin(\theta)(\dot{\theta}-1)=0$.

Answer 2

Just some algebra... After substitution we have system $$ \begin{align} \dot{r}\cos\theta-r\sin\theta(\dot{\theta}-1)&=r\cos\theta(1-r^2)\tag{1}\\ \dot{r}\sin\theta+r\cos\theta(\dot{\theta}-1)&=r\sin\theta(1-r^2)\tag{2}\\ \end{align} $$ then $$ \begin{align} \cos\theta\times(1)+\sin\theta\times(2)&\Longrightarrow\dot r=r(1-r^2)\\ \sin\theta\times(1)-\cos\theta\times(2)&\Longrightarrow r(\dot \theta-1)=0. \end{align} $$