Differential Equation: Periodicity of a circle with zero radius in polar coordinates

Question

I am given the following diff. equation in polar coordinates:

$$\dfrac{dr}{dt} = r(1 + a~\cos \theta - r^2) \\ \dfrac{d \theta}{dt} = 1$$

where $a$ is a positive number and is less than $1$.

I am asked to show the circle $r = 0$ is a periodic orbit with period $2 \pi$.

I am not sure if we can have the notion of a minimal period when $r = 0$, or can we?

Answer 1

the circle $r = 0$ is a periodic orbit with period $2 \pi$.

This is a crazy assertion, on at least two related counts. First the "circle $r=0$" is not a circle in the usual meaning of the term but a single point, namely, the origin. (Recall that, in polar coordinates, the origin is represented by $r=0$ and every $\theta$.) Second, the origin is a fixed point, that is, when the system starts from $r=0$ it stays there. This is not what is called a periodic orbit.

One says that a dynamics $(x(t))$ has a periodic orbit starting from $x(0)$ if $T=\inf\{t\gt0\mid x(t)=x(0)\}$ is finite and positive, then $T$ is the period of the orbit passing by $x(0)$. In the present case:

For every $r_0\gt0$, the circle of equation $r = r_0$ (that is, the circle with center $(0,0)$ and radius $r_0$) is a periodic orbit with period $2 \pi$.