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Consider the linear periodic system in $\mathbb{R}^n$ \begin{equation} \begin{cases} \dot{x}(t) & = A(t)x(t),\\ x(0) & = x_0, \end{cases} \label{eq:Floquet} \tag{1} \end{equation} where $A(t)$ is a real $n\times n$ matrix function which is smooth in $t$ and periodic of period $T>0$. Floquet theory states that there exists at least 1 non-trivial solution $\chi(t)$ satisfying \begin{equation} \chi(t+T) = \mu\chi(t), \ \ t\in(-\infty, \infty), \label{eq:Floquet2} \tag{2} \end{equation} where $\mu$ is an eigenvalue of the Floquet matrix. $\mu$ is more well-known as a Floquet multiplier of the system. What is the necessary and sufficient conditions so that \eqref{eq:Floquet} has a non-trivial $T$-periodic solution? By non-trivial I meant a periodic solution with minimal period $T$.

It seems like according to \eqref{eq:Floquet2}, one would want to impose condition on the Floquet matrix such that it has eigenvalue $\mu=1$, but I know nothing about this, not to mention that this sounds like a very strong condition. Spefically, I am looking for conditions that stem from Floquet theory.

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    So, if you know that for any $x(0)$ there exists some special matrix $A$ that describes you what happens after time $T$, namely $x(T) = A x(0)$, and you want to find non-trivial periodic solutions with period that is commensurable to $T$, what kind of eigenvalues this matrix should have? – 2017-01-04
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    @Evgeny real and 1? Oh wait, complex I think....... ? – 2017-01-04
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    Close enough. If you know that there is a periodic solution with period $\frac{p}{q}T$, the trajectory which started from point $x(0)$ must return to itself after time $pT$. But at the same time it passes through the point $x(pT)$ which relationship with $x(0)$ is known. Do you see how you can answer your question now? – 2017-01-04
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    @Evgeny Apologise for the late reply, but from what I understand, your comment leads to the same conclusion as mine (see above). I can surely say that $x(t)=\Phi(t)x(0)$, where $\Phi(t)$ is a fundamental matrix of the system such that $\Phi(T)=I$. Now $x(T)=\Phi(T)x(0)$, where $\Phi(T)$ is the Floquet matrix (as far as how I defined it). So in order to have a non-trivial periodic solution, one requires $\Phi(T)$ to have unit eigenvalue? – 2017-01-05
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    @Evgeny I had a thought about this yesterday again, and I think it suffices to quote Floquet theorem, which is statement (1) above and mention that there exists a non-trivial $T$-periodic solution if $\Phi(T)$ has at least one eigenvalue $\mu=1$. If you can say anything more than this, I would be happy to listen about it. – 2017-01-05
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    My main concern here is that you are overlooking solutions that are periodic, but not with the period $T$, but $\frac{p}{q}T$. This is the case if the $T$ in your comments and the $T$ in your question are the same. Of course everything boils down to some matrix having additional unit eigenvalues, but you must consider this case. – 2017-01-05
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    @Evgeny The question only concerns about $T$-periodic solution, which I assumed $T$ is the minimal period. – 2017-01-05
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    Okay, then having (additional) unit eigenvalues is necessary and sufficient. By the way, could you cite the reference that you are using for Floquet theorem? – 2017-01-06
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    @Evgeny Note that one can give a counterexample to $\mu=1$ being necessary and sufficient: a periodic system may have constant nonzero solutions. – 2017-01-06
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    @JohnB I suspect that the counterexample is $A(t)$ being a constant matrix? So, what do you think is the condition then ? – 2017-01-06
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    @Evgeny I learned Floquet theory from a graduate course so I am not entirely sure where it's from, but the book we used is Chicone's ODEs with applications, see Chapter 2. I have the impression that people usually talked about monodromy matrix when they do Floquet theory. – 2017-01-06
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    @JohnB How can we get zero eigenvalue of monodromy matrix and hence Floquet matrix? My impression is that we can't. That's why we are not talking about this case – 2017-01-06
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    @Evgeny I have the following in mind. Let $A(t)$ be a block matrix with two blocks: $0$ and a $2\times2$ nonconstant matrix with period $T$. Assume the second block does not give rise to $T$-periodic solutions and so there are no $T$-periodic solutions (with minimal period $T$). Still one of the multipliers is $1$. – 2017-01-06
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    @JohnB Do I understand right that the main feature of your counter-example is that monodromy matrix has unit eigenvalue, but it corresponds to the solution that is not periodic in the strict sense (it is constant, so has no minimal period)? – 2017-01-07
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    @Evgeny Indeed, the OP does write "non-trivial". For the existence of $T$-periodic solutions (let alone minimal period, it could also be a multiple), the eigenvector of the constant part of the Floquet matrix must correspond to a nontrivial nonconstant part. – 2017-01-07
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    @JohnB 1) Well, I am pretty sure that here non-trivial means "other from identically zero solution". (Strictly speaking, $x(t) \equiv 0$ is not really a periodic solution. But it repeats itself, which is more important here than how "wiggly" it is.) And usual approach with the monodromy matrix is a bit insensitive to this detail. So, you are right, fixed and periodic points of monodromy matrix catch more than truly periodic solutions. Still I think that "being closed is more important than being wiggly". 2) Can you post as an answer how to distinguish truly periodic solutions from others? – 2017-01-07
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    @Evgeny I see, I understood "non-trivial" as periodic with minimal period $T$. Perhaps the OP can clarify this point? Sure, I can add an answer as soon as this point is clarified. – 2017-01-07
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    @JohnB Also I can't get rid of the impression that "two" Floquet theorems are being mixed here. "The first one" is for the initial system $\dot{x}(t) = A(t) x(t)$ and "the second one" that is obtained from linearization along periodic orbit of some non-autonomous system. This linearization always has extra unit eigenvalue because tangent vector to this periodic solution at the beginning and at the end of period is the same, hence it is an eigenvector. I can't find the general statement about periodic non-autonomous system that claims they always have unit eigenvalue. – 2017-01-07
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    Hi guys. Thanks for the much involved discussion. To be honest it was a question from my qualifying exam, and judging from the question, I would have to guess that non-trivial in this case means periodic solution with minimal period $T$. But I am happy to listen to answers for both interpretation of non-trivial solution. – 2017-01-07
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    @Evgeny I see your confusion. When I said Floquet theorem, I am referring to the first case. I am aware of the second case that you mentioned in your comment, that to me was THE motivation/origin of Floquet theory. In this question, you should just assume that you were given a linear periodic system that does not stem from linearising a non-autonomous nonlinear system about a limit cycle. – 2017-01-07
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    @JohnB I think I've guessed the way how we can distinguish constant periodic solutions from the non-constant :) – 2017-01-08
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    @CheeHan Actually I've confused myself... I've read your statement of Floquet theorem wrong way and thus didn't understood what version you've meant exactly. So, everything is clear now :) – 2017-01-08

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