Re: Question about global stablity?

From: Fan Yang (yang_at_cae.wisc.edu)
Date: 11/18/04

  • Next message: Fan Yang: "Re: Question about global stablity?" 
    Date: Thu, 18 Nov 2004 17:41:12 -0600

    Dear Thoms,

    Thanks a lot for your reply. :-)

    First, I do mean all the eigenvalues are real and < -1. It implies
    the local asymptotical stabality because we are in continuous
    time ODE formulation.

    Right, you gave a good example. However, the eigenvalue
    of x=1/2 is 1/2. It doesn't satisfy our assumption that the Jacobian
    matrix has all the eigenvalues less than -1 for all x. Btw, we can
    also make the assumption that there exists only ONE fixed point.
    Under this condition, can we infer the global A.S. of this fixed
    point?

    Thanks again,

    Fan

    "Thomas Nordhaus" <thnord2002@yahoo.de> wrote in message
    news:qf8qp0t7aaiuhe8b20k371halimrfm0ajo@4ax.com...
    > "Fan Yang" <yang@cae.wisc.edu> schrieb:
    >
    > >hi,
    > >
    > >For ODE system dx/dt = f(x), x in R^n. The Jacobian matrix
    > >Df(x) is diagonizable and has all the eigenvalues less
    > >than -1 for all x in R^n. Of course, it is locally asymptotically
    > >stable. Can we say that the fixed point is also globally
    > >asymptotically stable?
    >
    > Hmm. So all eigenvalues are real and less than -1? That would imply
    > instability. I guess you mean >-1 and absolute value < 1.
    >
    > No, you can't imply global a.s. If there is a second fixed point q,
    > the first one can't be globally asymptotically stable. It is very easy
    > to construct such a case. Take x' = x(x-1/2). That has a fixed point 0
    > whis is locally a.s. with eigenvalue -1/2 and it also has a fixed
    > point x = 1/2. The solution x(t) = 1/2 for all t will surely not
    > converge towards 0.
    >
    > Thomas
    >
    > >
    > >Thanks,
    > >
    > >Fan
    > >
    >

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