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?
Best Answer
Hamiltonian systems can only have eigenvalues with either zero-real part (centers) or a pair of positive and negative real parts (saddles). Because Hamiltonian systems are volume-preserving in phase space, there can never be a Lyapunov-type function near a fixed point of Hamiltonian system. In other words, there cannot be any sink or source type fixed points in Hamiltonian systems.