Re: Question about global stablity?

From: Thomas Nordhaus (thnord2002_at_yahoo.de)
Date: 11/19/04 

Date: Fri, 19 Nov 2004 18:21:17 +0100

 James Meiss <jdm@NOSPAM.invalid> schrieb:

>In article <iv7rp095r4v7kj7q2joo3210al9ba3kicp@4ax.com>,
> Thomas Nordhaus <thnord2002@yahoo.de> wrote:

>>
>> Let Phi(x) be the corresponding flow map after one unit of time. Then
>> your condition implies that all eigenvalues of Phi have modulus < K
>> where 0 <= K < 1. So Banach's fixed point theorem should show that Phi
>> is a global uniform contraction which has exactly one fixed point
>> which is globally asymptotically stable, then.
>
>I also didn't read the question closely enough, obviously.
>
>However, I'm not sure your idea is correct. Just because the Jacobian,
>Df(x) has negative eigenvalues for all x, doesn't mean that the time one
>map will. I think a counter example is the famous example of
>Marcus-Yamabe (in all the ode books, such as Perko) is one where a
>linear system is everywhere "locally" stable, but the linear flow is not.
 ^^^^^^

Sorry, I don't know this counter-example. Could you state it? For
*linear* equations x' = Ax with all eigenvalues in the left half plane
the linear flow e^(tA) is a contraction and 0 is globally
asymptotically stable.

The argument with the time-1 map isn't very clean, I agree. But for
linear contracting flows one can always find a T>0 such that
|Phi(T,x)| < |x|/2, for example.

Thomas

>
>There is the relation
> det(Phi(t)) = exp ( int ( tr ( Df(x(s)) ), s=0..t)
>where Phi(t) is the linearized time-t map along the orbit from x(0) to
>x(t).
>
>So if tr(Df(x)) < 0, say, then you can guarantee that det(Phi(t)) < 1.
>But does that mean its eigenvalues are all smaller than one? No.

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