**5 - Applications**

We consider the following system

It is easy to verify that the system is Hamiltonian and that

The associated Jacobian matrix is

First case :

The fixed points are

We have:

and so A_0 is a center. Moreover,

A_1, A_2 and A_3 have the same eigenvalues and are saddles.

Second case :

We have now the system

The only fixed point is A_0=(0,0). The matrix M is now identically zero and A_0 is a degererate saddle. The Hamiltonian in this case is written

where

Exercice: Try to do the corresponding program.