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How to prove that any nondegenerate critical point of a 2D Hamilotonian system is either a saddle or a center ? By definition a critical point of an autonomous system is nondegenerate if the Jacobian evaluated at this point is non-zero. Also system has a saddle if the eigenvalues of the corresponding Jacobian matrix has a positive and a negative real part; and a center if it has purely imaginary eigen values.

Any help is appreciated.

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Let the Hamiltonian equations be $$ \dot{x} = \partial_p H , \ \ \dot{p}=-\partial_x H$$ If $(x_0,p_0)$ is a critical point then the linearized equations are $$ \delta \dot{x}= \partial_{px} H \; \delta x + \partial_{pp} H \; \delta p, \ \ \ \delta \dot{p}= \partial_{xx} H \; \delta x + \partial_{xp} H \; \delta p $$ In other words, $$ \left( \begin{matrix} \delta \dot{x} \\ \delta \dot{p} \end{matrix} \right) = \left( \begin{matrix} a & b \\ c & -a \end{matrix} \right) \left( \begin{matrix} \delta {x} \\ \delta {p} \end{matrix} \right) $$ The characteristic polynomial is $\lambda^2 - (a^2+bc)$ from which the conclusion follows.

H. H. Rugh
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  • @ H. H. Rugh, thank you. Will it be the same proof it it's been asked to use the Liouville's theorem ? Which states that, for a Hamiltonian system, the phase flow $(p(t),q(t))$ preserves the area, $\mathcal{S}=\iint_{S}dp~dq.$ –  Sep 04 '16 at 19:44
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    Yes, provided that you know that for a flow that preserves area the Jacobian has zero trace (the $a$ and $-a$ in the above matrix). – H. H. Rugh Sep 04 '16 at 19:46