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Show that the system has no closed orbits by finding a Lyapunov

  1. Feb 21, 2013 #1
    Show that the system has no closed orbits by finding a Lyapunov ....

    1. The problem statement, all variables and given/known data

    I'm at the point in the problem where I need constants a and b satisfying

    ax2(y-x3) + by2(-x-y3) < 0
    and ax2+bx2 > 0

    for all (x,y)≠(0,0).

    2. Relevant equations

    Just in case you're wondering, this is to satisfy the V(x,y)=ax2+by2 > 0 and ΔV(x,y)•<y-x3, -x-y3> < 0 so I can apply that one theorem to my problem.

    3. The attempt at a solution

    Well, it seems reasonable to choose a,b>0 to ensure ax2+bx2 > 0, but I'm having trouble figuring out how to make ax2(y-x3) + by2(-x-y3) < 0 simultaneously.
     
  2. jcsd
  3. Feb 21, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    Re: Show that the system has no closed orbits by finding a Lyapunov ..

    For x=y, this can be simplified to ##(a-b)x^3 - (a+b)x^5 <0 ##, which cannot be true both for positive and negative small x.
     
  4. Feb 21, 2013 #3

    pasmith

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    Homework Helper

    Re: Show that the system has no closed orbits by finding a Lyapunov ..

    You won't be able to do that: on the assumption that [itex]V = ax^2 + by^2[/itex] and [itex]\dot V = ax^2(y - x^3) - by^2(x + y^3)[/itex] your system is
    [tex]
    \dot x = \textstyle\frac12 x(y - x^3) \\
    \dot y = -\textstyle\frac12 y(x + y^3).
    [/tex]
    The fixed point at the origin is some kind of non-hyperbolic saddle: x = 0 and y = 0 are invariant. On x = 0, [itex]\dot y < 0[/itex] for [itex]y \neq 0[/itex] and on y = 0, [itex]\dot x < 0[/itex] for [itex]x \neq 0[/itex]. Thus there are points arbitrarily close to the origin where motion is unambiguously away from the origin, and at these points [itex]\dot V > 0[/itex].

    However, you now know that no periodic orbit can cross the coordinate axes, so the origin cannot be inside a periodic orbit. Since the origin is the only fixed point (any other fixed point must satisfy [itex]x^8 = -1[/itex]), there can be no periodic orbits (because in a 2D system every periodic orbit must enclose at least one fixed point).
     
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