Explain, for a Hamiltonian system, why an equilibrium point is a center for the

Question

Explain, for a Hamiltonian system, why an equilibrium point is a center for the non-linear Hamiltonian system if it is a center for the linearized system at the equilibrium point. Given a nonlinear system dx/dt = 1 - y^2 dy/dt = (1 + y^2)sin x. Is the system (l) - (2) Hamiltonian? If it is, please find a Hamiltonian function, and describe the solution curves in the phase plane. Now, define a new system by dividing 1 + y2, dx/dt = 1 - y^2/1 + y^2, dy/dt = sinx, Is the system (3) - (4) Hamiltonian? If it is, please find a Hamiltonian function, and describe the solution curves in the phase plane. Can you find solution curves of the system (l) - (2) in the phase plane? If yes, please describe them.

Explanation / Answer

Explain, for a Hamiltonian system, why an equilibrium point is a center for the

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---