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Classifaction of equilibrium points for a Hamiltonian System

  1. May 19, 2016 #1
    1. The problem statement, all variables and given/known data
    For the system:
    [tex]
    \frac{dx}{dt}=x\cos{xy}
    \: \:
    \frac{dy}{dt}=-y\cos{xy}[/tex]
    (a) is Hamiltonian with the function:
    [tex]
    H(x,y)=\sin{xy}[/tex]
    (b) Sketch the level sets of H, and
    (c) sketch the phase portrait of the system. Include a description of all equilibrium points and any saddle connections.
    2. Relevant equations


    3. The attempt at a solution
    [tex]\frac{\partial H}{\partial y}=y\cos{xy}=-g \\
    \frac{\partial H}{\partial x}=x\cos{xy}=f[/tex]
    So the function is Hamiltonian. I see that the equilibrium points are (0,0) and (±√π/2,±√π/2) by inspection. The problem I have is that the second set of equilibria have complex roots, but I don't see any of that behavior when I graph the phase portrait with pplane.
     
  2. jcsd
  3. May 19, 2016 #2

    Mark44

    Staff: Mentor

    The equation ##\cos(xy) = 0## has many solutions of the form ##xy = (2k + 1)\frac{\pi}2##, with k an integer. The point you show is just one equilibrium point.
    How are you getting this (the complex roots)?
     
  4. May 19, 2016 #3
    I didn't approximate so I may have made a mistake typing it into my calculator. I'll approximate and see if I get something else. Maybe that's the problem.
     
  5. May 20, 2016 #4

    Mark44

    Staff: Mentor

    I used plain old algebra, so I don't see how using approximations would be helpful.
     
  6. May 20, 2016 #5
    I get a Jacobian matrix of:
    [tex]
    \begin{bmatrix}
    \cos{xy}-y\sin{xy} & -x^2\sin{xy} \\
    xy\sin{xy} & xy\sin{xy}-\cos{xy}
    \end{bmatrix}
    [/tex]
    So then my matrix for that second point becomes:
    [tex]
    \begin{bmatrix}
    -\sqrt{\frac{\pi}{2}} & -\frac{\pi}{2} \\
    \frac{\pi}{2} & \frac{\pi}{2}
    \end{bmatrix}
    [/tex]
    Then using the trace and determinant I get:
    [tex]
    \lambda^2-\Bigg(\frac{\pi}{2}-\sqrt{\frac{\pi}{2}}\Bigg)\lambda-\Bigg(\Big(\frac{\pi}{2}\Big)^\frac{3}{2}-\Big(\frac{\pi}{2}\Big)^2\Bigg)[/tex]
    So then the quadradic would be:
    [tex]
    \frac{\Bigg(\frac{\pi}{2}-\sqrt{\frac{\pi}{2}}\Bigg)\pm\sqrt{\Bigg(\frac{\pi}{2}-\sqrt{\frac{\pi}{2}}\Bigg)^2+4\Bigg(\big(\frac{\pi}{2}\big)^\frac{3}{2}-\big(\frac{\pi}{2}\big)^2\Bigg)}}{2}[/tex]

    That is why I wanted to approximate and when I do, I still get a spiral, but this time a source. I used 1.51 as an approximation for π/2 and 1.25 for√π/2 then my approximation for b in the quadradic was 0.32 and c was 0.50. I hope I'm not completely missing something here.
     
    Last edited: May 20, 2016
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