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Equilibrium solutions in double well potential duffing oscillator

  1. Mar 18, 2013 #1
    1. The problem statement, all variables and given/known data

    I am trying to show that for a duffing oscillator described by
    x''+2g x'+ax+bx^3=0
    with a<0, b>0
    the equilibria at [itex]x=+- \sqrt{-a/b}[/itex] are stable

    2. Relevant equations
    I used coupled equations, and the characteristic equation of a linear system
    3. The attempt at a solution

    Coupled equations, x and x' related by
    [itex]d/dt(x,x')=[[0,1],[-a-bx^2, -2g]](x,x')[/itex]

    Setting x=+-[itex]\sqrt{-a/b}[/itex] gives the characteristic is

    [itex]\lambda^2+2g \lambda+a+b(-a/b)=0[/itex]
    But this can't be right since, i know from literature that the eigens are
    [itex]\lambda 1=-g + \sqrt{g^2+2a}[/itex]
    [itex]\lambda 2=-g- \sqrt{-g^2+2a}[/itex]
    and this gives eigens 0 and -2g
     
    Last edited: Mar 18, 2013
  2. jcsd
  3. Mar 19, 2013 #2
    After you have reduced the equation to the system and found the equilibrium point, you have to linearize the system about the equilibrium point, and then find the eigenvalue of the linear system.
     
  4. Mar 19, 2013 #3

    TSny

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    Gold Member

    Since no one has chimed in yet, I'll throw in my 2 cents (and it might not be worth even that). I believe that one way to examine the behavior near an equilibrium point is to "linearize" the equation in the neighborhood of the equilibrium point. So, you could let z = x-xo where xo is an equilibrium point. Rewrite the differential equation in terms of z but keep only terms linear in z. Then find the characteristic equation for this linearized differential equation.

    [Edit: Wow, I did not see voko's response. Sorry.]
     
  5. Apr 11, 2013 #4
    I'd done all the linearising stuff - I found my problem; I needed to use a Jacobean matrix, instead of just a system
     
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