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?

Thanks,

Fan

From:	Thomas Nordhaus
Subject:	Re: Question about global stablity?
Date:	Thu, 18 Nov 2004 23:42:52 +0100

"Fan Yang" 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
>

From:	Fan Yang
Subject:	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" wrote in message
news:qf8qp0t7aaiuhe8b20k371halimrfm0ajo@4ax.com...
> "Fan Yang" 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
> >
>

From:	James Meiss
Subject:	Re: Question about global stablity?
Date:	Thu, 18 Nov 2004 17:44:09 -0700

In article ,
"Fan Yang" wrote:

> 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?
>

No. For example

d/dt r= -r(1-r)
d/dt theta = f(r)

Has an asymptotically stable fixed point at the origin (r=0), and no
other fixed points (if f(r) is nonzer0 everywhere), but an unstable
periodic orbit at r=1, and infinity is an attractor.

--
Jim Meiss

From:	Fan Yang
Subject:	Re: Question about global stablity?
Date:	Thu, 18 Nov 2004 22:44:03 -0600

Dear James,

Thanks for your reply.

However, your example doen not satisfy my assumption. If all the
eigenvalues are less than -1, then there should be no periodic
orbit (r=1). Am I correct?

Thanks,

Fan

"James Meiss" wrote in message
news:jdm-9DE646.17440918112004@peabody.colorado.edu...
> In article ,
> "Fan Yang" wrote:
>
>> 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?
>>
>
> No. For example
>
> d/dt r= -r(1-r)
> d/dt theta = f(r)
>
> Has an asymptotically stable fixed point at the origin (r=0), and no
> other fixed points (if f(r) is nonzer0 everywhere), but an unstable
> periodic orbit at r=1, and infinity is an attractor.
>
> --
> Jim Meiss
>

From:	Thomas Nordhaus
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 08:10:27 +0100

"Fan Yang" schrieb:

>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.

Oops - Mixed it up with stability for maps. Your condition may be a
little too special, though. All smaller than some negative number may
be better.

>
>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
^^^
Actually it is 1. But it doesn't matter...

>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?

OK - lets have your first question first. What you mean ist: All fixed
points have eigenvalues < -1. Still, if there are more than one, they
can't be asymptotically stable by the same argument as in my first
post.

If there is only one fixed point there may still exist an (unstable)
limit cycle. This is the example of James Weiss. What problems do you
have with that counter example? Do you want to have all *limit sets*
to be locally asymptotically stable? Then you must extend your
concepts. It's not about eigenvalues anymore, but Lyapunov exponents
and you have to make the proper definitions.

Thomas

>
>Thanks again,
>
>Fan
>
>
>"Thomas Nordhaus" wrote in message
>news:qf8qp0t7aaiuhe8b20k371halimrfm0ajo@4ax.com...
>> "Fan Yang" 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
>> >
>>
>

From:	Thomas Nordhaus
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 08:18:24 +0100

Thomas Nordhaus schrieb:
>
>If there is only one fixed point there may still exist an (unstable)
>limit cycle. This is the example of James Weiss. What problems do you
>have with that counter example? Do you want to have all *limit sets*
>to be locally asymptotically stable? Then you must extend your
>concepts. It's not about eigenvalues anymore, but Lyapunov exponents
>and you have to make the proper definitions.

Mhmm. Never mind. I guess I misread your original question about the
Jacobian. So this is probably not a counter example, since for points
in the limit cycle the Jacobian will probably not satisfy your
conditions.

Thomas

From:	Thomas Nordhaus
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 08:32:50 +0100

"Fan Yang" 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?

Hi Fan - I thoroughly misunderstood your question, I guess. Sorry!
But the following idea might work to show your claim. I didn't check
all the technical arguments.

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.

Thomas

>
>Thanks,
>
>Fan
>

From:	James Meiss
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 09:11:15 -0700

In article ,
Thomas Nordhaus wrote:

> "Fan Yang" 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?
>
> Hi Fan - I thoroughly misunderstood your question, I guess. Sorry!
> But the following idea might work to show your claim. I didn't check
> all the technical arguments.
>
> 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.

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.

--
Jim Meiss

From:	Thomas Nordhaus
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 18:21:17 +0100

James Meiss schrieb:

>In article ,
> Thomas Nordhaus 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.

From:	James Meiss
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 11:50:49 -0700

In article ,
Thomas Nordhaus wrote:

> James Meiss schrieb:
>
> >In article ,
> > Thomas Nordhaus 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.

Okay, though it is a bit messy to write here. Let A(t) be the 2x2 matrix

A = [[-1+acos^2(t), 1-acos(t)sin(t)],[-1-a cos(t)sin(t), -1+a sin^2(t)]]

It is easy to see that both eigenvalues of A(t) are negative for a<2.
However, there are two explicit, independent solutions:

x(t) = [ cos(t), sin(t)] exp((a-1)t)
y(t) = [ sin(t), cos(t)] exp(-t)

The x solution is unbounded when 1 < a. Thus for 1< a< 2 both local
eigenvalues are negative, but the solution is unbounded, hence the
system is unstable.

Amazing, isn't it!

--
Jim Meiss

From:	Thomas Nordhaus
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 20:55:21 +0100

James Meiss schrieb:

>In article ,
> Thomas Nordhaus wrote:

>>
>> 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.
>
>Okay, though it is a bit messy to write here. Let A(t) be the 2x2 matrix
>
>A = [[-1+acos^2(t), 1-acos(t)sin(t)],[-1-a cos(t)sin(t), -1+a sin^2(t)]]
>
>It is easy to see that both eigenvalues of A(t) are negative for a<2.
>However, there are two explicit, independent solutions:
>
>x(t) = [ cos(t), sin(t)] exp((a-1)t)
>y(t) = [ sin(t), cos(t)] exp(-t)
>
>The x solution is unbounded when 1 < a. Thus for 1< a< 2 both local
>eigenvalues are negative, but the solution is unbounded, hence the
>system is unstable.
>
>Amazing, isn't it!

Right...
Shoot! I knew that example (it is on page 121 in Hale's book on ODE)
but I thought we're talking about Equations of the form x' = f(x). So
this *not* a counterexample, but it shows what problems one might
have. So - where to go from here? Trivial, profound or
counter-example?

Thomas

From:	James Meiss
Subject:	Re: Question about global stablity?
Date:	Fri, 19 Nov 2004 16:12:53 -0700

In article ,
Thomas Nordhaus wrote:

> James Meiss schrieb:
>
> >In article ,
> > Thomas Nordhaus wrote:
>
> >>
> >> 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.
> >
> >Okay, though it is a bit messy to write here. Let A(t) be the 2x2 matrix
> >
> >A = [[-1+acos^2(t), 1-acos(t)sin(t)],[-1-a cos(t)sin(t), -1+a sin^2(t)]]
> >
> >It is easy to see that both eigenvalues of A(t) are negative for a<2.
> >However, there are two explicit, independent solutions:
> >
> >x(t) = [ cos(t), sin(t)] exp((a-1)t)
> >y(t) = [ sin(t), cos(t)] exp(-t)
> >
> >The x solution is unbounded when 1 < a. Thus for 1< a< 2 both local
> >eigenvalues are negative, but the solution is unbounded, hence the
> >system is unstable.
> >
> >Amazing, isn't it!
>
> Right...
> Shoot! I knew that example (it is on page 121 in Hale's book on ODE)
> but I thought we're talking about Equations of the form x' = f(x). So
> this *not* a counterexample, but it shows what problems one might
> have. So - where to go from here? Trivial, profound or
> counter-example?
>
> Thomas

I suppose one could try to construct a nonlinear system for which this
is a linearization...though I don't know how to do that. It would be fun
to try.

--
Jim Meiss