Existence of periodic orbit of the ODE system $\dot{r}=r-r^3 \cos^2(\theta),\,\dot{\theta}= 1$

Question

Consider the system of ODEs (in polar coordinates): $$\dot{r}=r-r^3 \cos^2(\theta)$$ $$\dot{\theta}= 1$$

If we take $r_1 = \frac{1}{4}$ then $\dot{r}> 0$, and if we take $r_2 = \frac{3}{2}$ then $\dot{r}<0$, so $S:=\{r\in \mathbb{R}:r_1 \le r\le r_2\}$ should be a positively invariant set. Thus there should be a periodic orbit in $S$. But either I'm mistaken or not, and if not then what does the orbit look like?

Answer 1

Since $\theta' = 1$, we may as well treat $\theta$ as the independent variable. Changing (dependent) variables to $u := r^{-2}$ and rearranging gives $$u' + 2 u = 2 \cos^2 \theta,$$ which is a linear, first-order ODE and can hence be solved by elementary techniques (including two integrations). The involved integrals can be evaluated in closed form and this leads to the solutions $$u = \tfrac{1}{4}(2 + \sin 2 \theta + \cos 2 \theta + C e^{-2 \theta}) .$$ So, $u$ (and thus $r$) is periodic iff $C = 0$. Substituting shows that this holds only for the initial condition $u(0) = \tfrac{3}{4}$, or equivalently, $r(0) = u(0)^{-1 / 2} = \frac{2}{\sqrt{3}}$, which agrees with Semiclassical's comment.

This plot shows indicates the behavior for the three qualitative types of solutions; the blue plot is the unique periodic solution.

This plot was generated using the following Maple code:

dsolve(diff(r(theta), theta) = r(theta) - (r(theta)^3) * cos(theta)^2, r(theta));
op(2, %[2]);
[seq(eval(op(2, %), _C1 = C), C in {-3, 0, 3})];
plots:-polarplot(%, axiscoordinates = polar, coordinateview=[0..2, (-2 * Pi)..(2 * Pi)]);