# Find equilibrium solutions and stability when the system is in polar coordinates

For the autonomous system in polar coordinates given by

$$\dot{r}=r-r^2$$

$$\dot{\theta} = \sin ^2 \theta$$

what are all equilibrium points of the system and how do we determine their stability?

I know that $$(0,0), (1,0), \text{and} (-1,0)$$ are equilibrium points. Is there way to get back to Cartesian coordinates and then finding equilibrium points? How do we find the stability of these equilibrium points?

And for every initial condition $$x_0\in \Bbb{R}^2$$, what is the $$\omega -$$limit set of the orbit starting at $$x_0$$? I know that for $$r=1$$, we have $$\dot{r}=0$$, so every orbit eventually gets attracted to the $$r=1$$ circle so what would be the $$\omega -$$limit set?

I can only speak for the first part, but converting to cartesian coordinates might makes things messy. Rather, the velocity of the point can be expressed in terms of two components, tangential and radial

$$v = \begin{bmatrix}v_r \\ v_t \end{bmatrix} = \begin{bmatrix}\dot{r} \\ r\dot{\theta} \end{bmatrix} = \begin{bmatrix}r(1-r) \\ r\sin^2{\theta} \end{bmatrix}$$

Now ,the points which have zero velocity are

$$\begin{bmatrix}r(1-r) \\ r\sin^2{\theta} \end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix} \implies \begin{bmatrix}r^* \\ \theta^*\end{bmatrix} = \begin{bmatrix} 0,1 \\ 0, \pi\end{bmatrix}$$

This would correspond to the points $$(0,0), (1,0), (-1,0)$$

Can you visualise what will happen based on radial and tangential velocities to see what happens at each point? I'm guessing that none of them would be stable (because tangential velocity is always in positive rotation sense, so for positive angular displacements will be unstable and negative be stable - hence neutral stability)

• Yes that's what I thought too. Since $\dot{\theta}>0$, so it will be unstable. But I am not sure. Thanks Dhanvi.
– John
Commented Aug 11, 2020 at 13:39