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Poincarè stabillity!

  1. Feb 16, 2012 #1
    Hey Guys!

    I'm having MAJOR difficulty with a problem regarding stability of a DE. The problem goes as following:

    Find the equation of the phase paths of x˙=1+x^2, y˙=−2xy. It is obvious from the phase diagram that y=0 is Poincaré stable. Show that for the path y=0, all paths which start in (x+1)^2+y^2=δ^2 subsequently remain in a circle of radius δ[1+(1+δ)^2] centered on y=0

    Finding phase paths is easy, and is expressed by:

    y=C/(1+x^2)

    I now see the phase paths all converges to y=0, and therefor is poincarè stabile, but the last part of the exercise is causing me alot of problems...

    To show that all paths which starts in (x+1)^2+y^2=δ^2, i have tried finding the tangent between the circle and the phase paths, which will give me the highest y-value at x=0. Since the phase paths are symetric around x=0, and maximun value at the point. But I only get complex solutions, and they are no good...

    Any tips on how to go from here?
     
  2. jcsd
  3. Feb 20, 2012 #2
    Do you have the result that any trajectory that passes through the point (-1,δ) reaches it's max at (0,2δ)? This is easy to show and is essentially just as good a result for qualitative purposes. Finding the trajectory that intersects the circle at exactly one point (I think this is want you tried to do), combined with the result above should give your result but finding that intersection gives an ugly equation.
     
  4. Feb 20, 2012 #3
    Thanks for replying on this subject, Alan2, I do apreciate it!

    That's actually the first result I got, but i realized that it would not give the max value of any trajectory starting fra the small circle. My logic told me that the trajectory I was looking for did not start in (-1,δ), but in a point where the tangent av the circle was equal to the tangent of the trajectory. That's when i started looking for the trajectory that would intersect the circle in just one point, and trace that trajectory to x=0 and compute it's value. Then I would compare this to the known radius of the bigger circle
    How is the trajectory from the point (-1,δ) as good a result as any? I dont see the logic in this...
    Thanks again for any help!
     
  5. Feb 20, 2012 #4
    Yes you are correct. There is some value of x<-1 where the trajectory intersects the circle in exactly one point, that trajectory is tangent to the circle, and it achieves a larger max at x=0 than does the trajectory that passes through the circle at x=-1. I wrote out the equations and I wouldn't want to solve it. It just looks to me as if it's not worth it. From a strictly qualitative point of view, I would just want to show that the trajectories can't escape. So I might say if the point passes through this unknown point then clearly it also passes through (-1,a), where a is larger than delta, and thus attains its max of 2a at x=0. I guess I would judge that it's not worth my time to find the exact coordinates of that unknown point if all that I really wanted to show was that the trajectories are stable.
     
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