Find the index of the equilibrium points of the system (Question on solution)

Question

I have the following system: $$\dot{x} = 2xy$$ $$\dot{y} = 3x^2-y^2$$

I have the following solution: The system has one equilibrium point at the origin. Let the curve $\Gamma$ surrounding the origin be the ellipse $x = \cos\theta$, $y = \sqrt{3}\sin\theta$ for $0\leq\theta\leq 2\pi$. Then $$\dot{x} = 2xy = \sqrt{3}\sin2\theta$$ $$\dot{y} = 3\cos2\theta$$

Then on $\Gamma$ $$\tan\phi = \sqrt{3}\tan[\frac{\pi}{2}-2\theta].$$ As $\theta$ increases from $0$ to $2\pi$, $\phi$ decreases $0$ to $-4\pi$. Hence the index is 2.

My question about this solution is how do I see that as $\theta$ increases from $0$ to $2\pi$, $\phi$ decreases $0$ to $-4\pi$? I feel like this is a typo because at $\theta = 0$ isn't $\phi = \frac{\pi}{2}$?

Any help and comments would be greatly appreciated. Thank you.

Answer 1

(this is not an answer)

Are you sure you know, what are you doing? You seem to think about the ellipse as about the solution, but it is just an assisting construct to further analysis of the field.

Answer 2

I'm 7 years late, here it is anyways. The following picture allows you to visualize why the index of the fixed point should be 2 (edit: the index is -2!).

Answer 3

To add some formulas to the plots:

If one sets $$z=\sqrt3x+iy,$$ then $$\dot z=i\bar z^2,$$ which also allows to directly read off the index $-2$. The negative sign is due to the conjugation switching the orientation of any rotation around the origin, the value $2$ due to the square. The constant factor $i$ has no influence on the index.