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I have the linear system $y' = A(t)y$ where $A(t)$ is a continuous n-by-n matrix of period $\omega$. I want to show that if $-1$ is a multiplier of this system, then there is a solution of at least period $2\omega$. I am stuck on figuring out how to begin.

Here is the relevant material: enter image description here

So Step 1: What happens when $-1$ is a multiplier of the system? This means $-1$ is an eigenvalue of the nonsingular matrix $\exp(\omega R)$ where $R$ is a matrix. Here I do not know how to proceed.

KZ-Spectra
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    So, $-1$ is an eigenvalue of $\exp{(\omega R)}$. That means that there is also an eigenvector $v$ corresponding to this eigenvalue. Start with $v$ as an initial condition $y(0)$. What $y(\omega)$ equals? What $y(2\omega)$ equals? – Evgeny Dec 03 '18 at 12:00
  • Dear Nalt: Would you consider undeleting you post https://math.stackexchange.com/questions/3038615/showing-expta-leq-k-if-all-eigenvalues-have-real-part-negative-or-zero ? I was on the verge of posting what I think is a pretty detailed answer when--poof! The question was gone! Thank you. Cheers! – Robert Lewis Dec 14 '18 at 17:32
  • @RobertLewis Done. It was deleted because I found a similar post, but maybe it is okay. If a mod doesn't like it they'll handle it I suppose. – KZ-Spectra Dec 14 '18 at 17:35
  • Thank you! By the way, what was the "similar post" if you don't mind? – Robert Lewis Dec 14 '18 at 17:37
  • This: https://math.stackexchange.com/a/1374462/597047 – KZ-Spectra Dec 14 '18 at 17:39
  • @RobertLewis And by the way, I've just put up another question. If you find time, here it is https://math.stackexchange.com/q/3039700/597047 I look forward to your post! – KZ-Spectra Dec 14 '18 at 17:48
  • I'll try to get to your new question pretty soon--first, to finish up on the other one! Thanks! – Robert Lewis Dec 14 '18 at 17:52

1 Answers1

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The situation is that you know from the fundamental matrix solution $\Phi$ to your ODE system, $$ Φ'(t)=A(t)Φ(t),~~Φ(0)=I $$ that $Φ(ω)$ has an eigenvalue $-1$. Additionally you know from the periodicity of $A$ that $Φ(ω+t)=Φ(t)Φ(ω)$.

Let $v$ be an eigenvector, then $y(t)=Φ(t)v$ is a solution with initial value $y(0)=v$ and $$ y(ω)=Φ(ω)v=-v,~~y(2ω)=Φ(ω)^2v=v. $$ which means that the trajectory of that solution returns to the initial point.

Lutz Lehmann
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  • Why do we have $\Phi(\omega)^2$ instead of $\Phi(2\omega)$? I know we have $\Phi(2\omega + t) = \Phi(t)\Phi(2\omega)$, but don't see how this translates to having $\Phi(2\omega) = \Phi(\omega)^2$ – KZ-Spectra Dec 03 '18 at 16:06
  • You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct. – Lutz Lehmann Dec 03 '18 at 16:14
  • I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2\omega$". What you showed was there is a solution of period $2\omega$. So why is there the "at least" wording? – KZ-Spectra Dec 03 '18 at 16:20
  • This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=\cos(5t)$ has a period $2\pi$ with $y(0)=1$ and $y(\pi)=-1$, but the minimal period is $2\pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation. – Lutz Lehmann Dec 03 '18 at 16:28
  • Thank you. For my very first question I just had to write $\Phi(\omega+\omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot! – KZ-Spectra Dec 03 '18 at 16:34