Hamiltonian Equation

May 2010
1
0
Just wondering if you have
dx/dt = y^3 - y
dy/dt = -x

then you can find the Hamiltonian which is H(x,y) = (x^2)/2 +(y^4)/2 -(y^2)/2

My question is what does the hamiltonian say about the nature of the steady states i know that (0,0) is a saddle and (0,1) is a centre

My second question is how can you describe the nature of the critical points of H(x,y). Im assuming it might help to draw a phase plane

thanks
 
May 2010
43
1
I believe the Hamiltonian is actually \(\displaystyle x^2/2+y^2/2-y4/4\). The steady states (i.e. critical points) occur at (x,y) = (0,0), (0,1), and (0,-1). Isn't (0,0) the center and (0,1) and (0,-1) the saddles?

\(\displaystyle H = x^2/2 + V(y), with V(y) =y^2/2-y^4/4\)

The potential function, \(\displaystyle V(y)\) has the following properties:

\(\displaystyle V'(y) = 2y -y^3\)
\(\displaystyle V''(y) = 2 - 3y^2\)

Therefore V(0) is a minimum, and hence a stable point in the system, and V(1) and V(-1) are maxima in the potential function, and hence unstable.

Not sure this is getting at your question...