# Obtaining the polar form of the planar system?

Consider the two variables $x_{1},x_{2}$ which are functions of time $t$, we refer $\dot{x_{1}}$ as the time derivative of $x_{1}$.

Let $\alpha$ be a constant parameter, and consider the following system of equations:

\begin{align*} \dot{x_{1}} &= \alpha x_{1} - x_{2} - x_{1}(x_{1}^2 + x_{2}^2)\\ \dot{x_{2}} &= x_{1} + \alpha x_{2} - x_{2}(x_{1}^2 + x_{2}^2). \end{align*}

Now we transform the above system to polar $(\rho,\theta)$ coordinates. I think after the transformation the above equations lead to the following:

\begin{align*} \dot{\rho} &= \rho(\alpha - \rho^2)\\ \dot{\theta} &= 1. \end{align*}

Where $x_{1} = \rho \cos(\theta)$ and $x_{2} = \rho \sin(\theta)$. If I substitute then I am thinking about the partial derivatives involved and I think which variable I should differentiate with respect to the time? I can observe that the $x_{1}^2 +x_{2}^2 = 1$ which could simplify a lot but how should I deal with $\dot{x_{1}}, \dot{x_{2}}$ which can help me in getting the polar form of the above system?

I don't understand your last paragraph (notice for instance $x_1^2+x_2^2 = \rho^2$, not $1$) or how you got your second set of equations. To transform your differential equations, just apply the product and chain rules: for instance the first equation becomes $$\dot\rho \cos(\theta) - \rho \sin(\theta)\dot\theta = \alpha \rho \cos(\theta) - \rho \sin(\theta) - \rho^3 \cos(\theta),$$ and you will get something similar for the second equation. You can then apply standard variable elimination techniques to isolate equations for $\dot\rho$ and $\dot\theta$ and try to simplify them. Give it a try and let me know if you need more help.