# Determining the stability of an equilibrium point of a system of non-linear odes.

I am considering the following system of odes: $$x' = \sin(y) \\ y' = \cos(x).$$ I have calculated a Hamiltonian for the system: $$H(x,y) = -\sin(x) -\cos(y).$$ The equilibrium points are obviously given by: $$\sin(y)=0\Rightarrow y_{eq}=n\pi, n\in\mathbb{Z} \\ \cos(x)=0\Rightarrow x_{eq}=(2m+1)\pi/2, m\in\mathbb{Z}$$ I would now like to investigate the stability of these equilibria. Linearising the system around $\vec{x}_{el}$ yields the Jacobian matrix: $$\left(\begin{matrix} 0 & (-1)^n \\ -(-1)^m & 0 \end{matrix}\right),$$ the eigenvalues are $\lambda=\pm i$ for $(m+n)$ even and $\lambda = \pm 1$ if $(m+n)$ odd. So in the latter case the equilibrium is unstable. I am now left with determining the stability of the former case.

I am a bit confused how to do this. I know that if I can find a Lyapunov energy function, this might tell me something, but I forget how to exactly use that. Furthermore, I have a feeling that a Lyapunov energy function is just the Hamiltonian expanded around an equilibrium point. Is that correct? I expanded the Hamiltonian for $m=1$ and $n=0$, which yields in the second order: $$H(x,y)\approx -(x-3\pi/2)^2 + y^2,$$ but I don't know what I could do with this or if it is even helpful at all. Can you guys help me?

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In a Hamiltonian system, the trajectories are on curves $H(x,y) =$constant. For your system, those curves look like this:
As you can see, some of the equilibrium points are centres (stable but not asymptotically stable). These are the ones that are maxima or minima of $H(x,y)$, where $H(x,y) = 2$ or $-2$. A strict local maximum/minimum of $H(x,y)$ is always a centre, the trajectories near it being closed curves around the equilibrium point. The others are saddle points, where the eigenvalues (as you found) are $\pm 1$.