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Change ODE system to Polar to apply Poincare-Bendixson

  1. Dec 15, 2012 #1
    Question:
    Show that the system

    x'= x-y-x[x^2 + (3/2)y^2]
    y'= x+y -y[x^2 + (1/2)y^2]

    has at least one periodic orbit.


    I know that I need to apply Poincare-Bendixson Theorem. I can prove the first three points of it easily, but to create a trapping region, I believe that I need to switch this system to polar. I know that I need to make the substitutions y=r*sin(θ), dy= sin(θ)*dr + r*cos(θ) θ', x= r*cos(θ), dx= cos(θ)*dr - r*sin(θ)θ'. But when I do make the substitution, it makes the equations worse than before. Is there some other way to come up with a r min and r max for the trapping region? I just assumed it was a polar type question because its phase plane has a limit cycle on it.
     
  2. jcsd
  3. Dec 16, 2012 #2

    pasmith

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    Homework Helper

    If [itex]V(x,y) = \frac12(x^2 + y^2) = \frac12 r^2[/itex] then
    [tex]\dot V = x\dot x + y \dot y
    = x^2 + y^2 - x^4 - \frac{5}{2} x^2y^2 - \frac12 y^4[/tex]
    which after some rearrangement yields
    [tex]\dot V = x^2 + y^2 - \left(x^2 + \frac54y^2\right)^2 + \frac{17}{16}y^4[/tex]
    Looking at that, you can see that there exist [itex]0 < m < M[/itex] such that if [itex]0 < r < m[/itex] then [itex]\dot V > 0[/itex] (because the origin is unstable, so trajectories are locally away from it) and if [itex]r > M[/itex] then [itex]\dot V < 0[/itex] (because if [itex]|y|[/itex] is fixed then [itex]\dot V < 0[/itex] for [itex]|x|[/itex] sufficiently large), so that the trapping region is [itex]m \leq r \leq M[/itex].

    I don't think it's necessary to calculate [itex]m[/itex] and [itex]M[/itex]; it is enough to show that they exist.
     
    Last edited: Dec 16, 2012
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