The following equations,
\begin{equation} \begin{split} \dot{F} &= -C + \frac{F}{\sqrt{F^2 + C^2}}\left(\alpha -\left(\frac{F}{a}\right)^2 - \left(\frac{C}{b}\right)^2\right),\\ \dot{C} &= F + \frac{C}{\sqrt{F^2+ C^2}}\left(\alpha -\left(\frac{F}{a}\right)^2 - \left(\frac{C}{b}\right)^2\right), \label{eq:LiCy} \end{split} \end{equation}
illustrates a elliptic limit cycle.
$F = F(t)\\ C = C(t) \\ a \rightarrow semi-major axis \\ b \rightarrow semi-minor axis \\ \alpha \, \text{is a scaling constant, and is normally represented as 1 in the equation of an ellipse}$
We would like to represent the limit cycle through the functions $\dot{r}(t)$ and $\dot{\theta}(t)$. Where the solution is found by solving the following equations,
$\dot{r}(t)\, \text{from} \\ x\dot{x} + y\dot{y}$
and
$\dot{\theta}(t) \, \text{from}\\ x\dot{y} - y \dot{x}$
The example we tried to follow, gave the following polar coordinates for the circle,
$x(t) = r(t)\cos(\theta(t))\quad \text{and} \quad y(t)=r(t)\sin(\theta(t))$ where $r(t) = \sqrt{x^2 + y^2}$ and we found $\dot{r}= \alpha -r^2$ and $\dot{\theta}(t)=1$.
Therefore we tried to implement new polar coordinates for an ellipse, where $x(t) = a\cdot r(t)\cos(\theta(t))\quad \text{and} \quad y(t)=b\cdot r(t)\sin(\theta(t))$
which did not work out as we had hoped. We are wondering where it went wrong? And how we could illustrate this elliptic limit cycle by using the given equations?
In polar coordinates $r \cos(\theta) = F$, $r \sin(\theta) = C$, the equations become $$ \eqalign{\dot{\theta} &= 1\cr \dot{r} &= 1 + \dfrac{(a^2-1) b^2 \cos^2(\theta) - a^2}{a^2 b^2} r^2\cr}$$ Modulo a change of origin, we can take $\theta = t$. The equation for $r$ then has an integrating factor $$ \mu(t) = \exp\left(-{\frac { \left( {a}^{2}{b}^{2}-2\,{a}^{2}-{b}^{2} \right) t}{2{a} ^{2}{b}^{2}}}-{\frac { \left( a^2-1 \right) \sin \left( 2\,t \right) }{4{a}^{2}}} \right)$$ and we find that $ \mu(t) r(t) - \int \mu(t) \; dt$ is constant. Thus (since $\mu(0)=1$)
$$ \mu(t) r(t) - \int_0^t \mu(s)\; ds = r(0) $$
In order to have a periodic solution with period $2\pi$, what we need is $$ (\mu(2\pi) -1) r(0) = \int_0^{2\pi} \mu(s)\; ds $$
If $r^*$ is the value of $r(0)$ that makes this true, we have $\mu(2\pi) (r(2\pi) - r^*) = r(0) - r^*$. Thus the periodic solution will be unstable if $\mu(2\pi) < 1$, i.e. $a^2 b^2 - 2 a^2 - b^2 > 0$, and stable if $a^2 b^2 - 2 a^2 - b^2 < 0$.