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Dx/dt = y, dy/dt = 2x - 4x

  1. Oct 17, 2009 #1
    Consider the system:
    dx/dt = y, dy/dt = 2x - 4x3 - y.

    I know that the Hamiltonian H(x,y) = y2/2 - x2 + x4 + y2/2 = y2 - x2 + x4. But how do I show that H is a Lyapunov function for this system. Please help.
     
  2. jcsd
  3. Oct 17, 2009 #2
    Re: Proof

    Is it:
    d/dt H(x(t),y(t)) = d/dt(y2 - x2 + x4) = y dy/dt + dx/dt(-2x + 4x3) = y(2x - 4x3 - y) + y(-2x + 4x3) = 2xy - 4x3y - y2 - 2xy + 4x3y = -y2 < 0. Since dH/dt < 0, this is a Lyapunov function.
     
  4. Oct 18, 2009 #3
    Re: Proof

    Also show that H is always positive for nonzero x,y. Then you are done.
     
  5. Oct 18, 2009 #4
    Re: Proof

    Cool.
    Cheers.
     
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