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Science Forum Index » Nonlinear Science Forum » Hopf bifurcation stability
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| Konstantin Dubrovinsky |
 Posted: Thu May 22, 2003 6:36 pm |
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Hello!
I have a question about Hopf bifurcation.
As far as I understand, its stability is determined by Floquet theory. That
is to say, all eigenvalues of fundamental solution matrix should be
negative. But in "Elementary Stability and Bifurcation Theory" (Gerard Iooss
and Daniel D. Joseph) it is stated that one eigenvalue is always zero (or is
it???). I find it very confusing. Could you please clarify my
misunderstanding, or, perhaps, suggest some literature on this subject.
Thank you in advance! |
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| Konstantin Dubrovinsky |
| Posted: Sat May 24, 2003 6:55 pm |
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Thank you very much for your reply!!!
I was really glad to read it!
So, as far as I understand, one of the Floquet multipliers is always 0. If
the limit cycle is represented as [x(t), y(t)], then this zero multiplier is
going to correspond to perturbation epsilon*[dx/dt,dy/dt], where epsilon is
"small". That is to say, if we take a perturbation along the periodic
solution, we still get a periodic solution, while if we perturb "in some
other direction", the perturbation will either grow exponentially, or
decrease exponentially, depending on the sing of the corresponding Floquet
multiplier.
Could you please tell me if I am right?
Thank you again. |
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| Louis M. Pecora |
| Posted: Sun May 25, 2003 8:49 am |
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In article <3ece574f$1@puffinus.its.uu.se>, Konstantin Dubrovinsky
<kodo7115@student.uu.se> wrote:
Quote: Hello!
I have a question about Hopf bifurcation.
As far as I understand, its stability is determined by Floquet theory. That
is to say, all eigenvalues of fundamental solution matrix should be
negative. But in "Elementary Stability and Bifurcation Theory" (Gerard Iooss
and Daniel D. Joseph) it is stated that one eigenvalue is always zero (or is
it???). I find it very confusing. Could you please clarify my
misunderstanding, or, perhaps, suggest some literature on this subject.
Thank you in advance!
It depends on the "orbit" you are examining. For a fixed point all the
exponents must be negative for stability. For other trajectories there
will always be a zero exponent (maybe more than one, but at least one)
along the trajectory (I am assuming you are looking at autonomous
ODEs).
For a periodic orbit (limit cycle) you examine the Floquet multipliers.
They should give the stability away from the orbit, i.e. if you perturb
the orbit, will the perturbation damp out or grow?
--
Lou Pecora
- My views are my own. |
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| Louis M. Pecora |
| Posted: Sun May 25, 2003 1:46 pm |
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Guest
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In article <3ed0fed4@puffinus.its.uu.se>, Konstantin Dubrovinsky
<kodo7115@student.uu.se> wrote:
Quote: So, as far as I understand, one of the Floquet multipliers is always 0.
Technically, the multiplier is 1 the Floquet exponent is 0.
Quote: If
the limit cycle is represented as [x(t), y(t)], then this zero multiplier is
going to correspond to perturbation epsilon*[dx/dt,dy/dt], where epsilon is
"small". That is to say, if we take a perturbation along the periodic
solution, we still get a periodic solution, while if we perturb "in some
other direction", the perturbation will either grow exponentially, or
decrease exponentially, depending on the sing of the corresponding Floquet
multiplier.
Could you please tell me if I am right?
Yes, that looks right.
--
Lou Pecora
- My views are my own. |
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