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Math Help - Limit cycle in a system

  1. #1
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    Limit cycle in a system

    Consider the two-dimensional system x'=Ax-r^2x where r=\parallel x\parallel and A is a 2x2 constant real matrix with complex eigenvalues \alpha\pm i\omega. Prove that there exists at least one limit cycle for \alpha>0 and that there are none for \alpha<0.
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  2. #2
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    Let's deal with say \alpha<0. Then the trace of the matrix is \tau=2\alpha<0 (unstable!)

    To find equilibria, we solve Ax-r^2x=0\Rightarrow x=0. Since A has distinct eigenvalues, this is the only one.

    By the Bendixon criterion, if {\rm div}(Ax-r^2x) maintains its sign in the vicinity of the equilibrium, then there can be no limit circle there. We calculate

    {\rm div}(Ax-r^2x)=\frac{\partial}{\partial x_1}(a_{11}x_1+a_{12}x_2-r^2x_1)+\ldots=\tau-4r^2,

    so for \varrho>0, we can apply Bendixon's criterion in \{x:||x||\leq \varrho\}.
    Last edited by Rebesques; August 22nd 2007 at 06:35 AM. Reason: msn chatting
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    Re: Limit cycle in a system

    I have a similar problem. I'm stuck on the first part, for alpha greater than 0. I first tried a level sets approach. That is to say, I attempted to find an annulus such that the dot product of the normal vector field <x, y> with the vector field given by the differential equation was positive on the inner circle and negative on the outer circle. I failed miserably. The professor suggests that I don't use that approach, and instead try to use "what you know about linear algebra". However, I'm getting nowhere quite fast. In short I'm stumped.

    Anyone know how to tackle the OP problem for alpha > 0???
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  4. #4
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    Re: Limit cycle in a system

    I was looking at the trace-determinant plane for the matrix (A - (r^2)I) and have found that a "center" will exist if r equals the square root of alpha. That said, I've only done this when the parameter the matrix depends on is independent of the input vectors. In this case r is the norm of the input vector x. That said, it seems like this will only imply the existence of a periodic solution if the periodic solution is a circle, that way r would remain constant and the matrix (A - (r^2)I) will also remain constant in its description of the derivative. Is this on the right track or is it all wrong?
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