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Poincare Bendixson's Theorem

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    System in polar coordinates
    [itex]\dot{r}[/itex] = 2r - r3(2 + sin([itex]\theta[/itex])),
    [itex]\dot{\theta}[/itex] = 3 - r2

    Use a trapping region to show there is at least one periodic orbit?


    2. Relevant equations

    By using Poincare Bendixson's Theorem

    3. The attempt at a solution

    I am struggling to know where to start.. Do I start by considering
    g([itex]\theta[/itex]) = 2 + sin([itex]\theta[/itex]). Any help would be greatly appreciated?
     
  2. jcsd
  3. Apr 9, 2012 #2

    Dick

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    Your g(θ) is always between 3 and 1. Try and find a small value r1 such that r' on the circle r=r1 is always positive. Now find a larger value of r2 such that r' on the circle r=r2 is always negative. Doesn't that make the region between the two circles a trapping region? Now think about fixed points.
     
  4. Apr 10, 2012 #3
    Okay so does this mean

    1≤g(θ)≤3 So you get

    [itex]\dot{r}[/itex] = 2r - r3g(θ)
    [itex]\dot{r}[/itex] > 2r - 3r3 > 0 and so 2/3 > r2

    [itex]\dot{r}[/itex] < 2r - r3 < 0 and so r2 > 2

    and √(2/3) < r < √2
     
  5. Apr 10, 2012 #4

    Dick

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    No, you don't really have to solve for anything. Pick a small value of r like r=1/2. Can you show r'>0 if r=1/2?
     
  6. Apr 10, 2012 #5
    So if I have to give bounds for the periodic orbit I can pick any value for r(small) provided r'>0 and then choose another value (this time larger) and show r'<0? How would I then determine its stability characteristics?
     
  7. Apr 10, 2012 #6

    Dick

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    If you can find such values then you have a trapped region between the two circles. Can you have a fixed point in between? Try and figure out why not. Where can a fixed point be? Then use Poincare Bendixson clearly, yes?
     
    Last edited: Apr 10, 2012
  8. Apr 10, 2012 #7
    There can't be any fixed points in trapping region. There can be a fixed point at the origin? So to determing the stability I determine the fixed point at the origin?
     
  9. Apr 10, 2012 #8

    Dick

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    Sure there is a fixed point at the origin. But that doesn't matter since it's not in your trapping region. Then what does your theorem tell you?
     
  10. Apr 10, 2012 #9
    So the flow enters the trapping region about the origin which has no critical points which means it is a stable periodic orbit in the trapping region..
     
  11. Apr 10, 2012 #10

    Dick

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    Not quite. It tells you there IS a periodic orbit in the trapping region. What's the exact statement of the theorem you are using?
     
  12. Apr 10, 2012 #11
    Suppose [itex]\alpha[/itex](x0) enters and does not leave some closed and bounded domain D that contains no critical points. This means that [itex]\phi[/itex](x0, t) [itex]\in[/itex] D for all t≥[itex]\tau[/itex], for some [itex]\tau[/itex]≥0. Then there is at least one periodic orbit in D and this orbit is in the [itex]\omega[/itex]-limit set of x0.

    What does this mean in terms of stability? I have no idea how to determine the stability?
     
  13. Apr 10, 2012 #12

    Dick

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    I'm maybe overextending myself here, but didn't the original problem ask you to show that there is a periodic orbit? I don't think it said anything about stability?
     
  14. Apr 10, 2012 #13
    Yes it did, but I didn't realise there was a second part that asks to determine its stability characteristics..
     
  15. Apr 10, 2012 #14

    Dick

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    Ok, dynamical systems aren't really something I know well, so I might have to recuse myself here. Sorry.
     
  16. Apr 10, 2012 #15
    No problem thanks so much for your help!
     
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