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.
Best Answer
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!).