[Math] Phase portrait of ODE in polar coordinates

dynamical systemsordinary differential equationspolar coordinatesreal-analysisstability-in-odes

Given the system of ODEs in polar coordinates, $$r' = r(1-r^2)(4-r^2)$$ $$\theta'=2-r^2,$$

one can determine its equilibrium points and limit cycles as follows:

$\gamma_1:= \begin{cases} r = 0,\\ \theta = 2t\end{cases}$, $\gamma_2:= \begin{cases} r = 1,\\ \theta = t\end{cases}$, $\gamma_3:= \begin{cases} r = 2,\\ \theta = -2t\end{cases}$.

$\gamma_1$ corresponds to $(0,0)$ in the $xy$-plane, and $\gamma_2$ and $\gamma_3$ correspond to circles.

Now I need to sketch the phase portrait of this system and, based on this sketch, determine the stability of the equilibrium points and limit cycles.

Does one need to solve this system explicitly in order to sketch the phase portrait, or is there a neater way to do it, without solving the system?

I've also tried with solving for the ODE in terms of $\frac{d\theta}{dt}$, but it doesn't appear to be an equation which is easy to plot either.

Also, do you think that I've found all the possible limit cycles, or maybe missed something? I'm new to this kind of analysis.

Best Answer

Hints: I would have made this a comment if I could.

Here is the polar plot.

enter image description here

Here is the transformed system, if you want to validate the rest of your analysis using other approaches.

$$x' = x \left(-x^2-y^2+1\right) \left(-x^2-y^2+4\right)-\frac{y \left(-x^2-y^2+2\right)}{\sqrt{x^2+y^2}} \\ y' = y \left(-x^2-y^2+1\right) \left(-x^2-y^2+4\right) + \frac{x \left(-x^2-y^2+2\right)}{\sqrt{x^2+y^2}}$$

Update Reduced the plot area to see further detail in the areas of interest.

enter image description here

Related Question