If $e^{Ty}$ is a Floquet multiplier, then there is a solution of the form $x(t)=e^{yt}p(t)$

Question

question:

Let $A(t)$ be a continuous, non-constant, real matrix-valued, $T>0$ periodic function $A(t+T)=A(t)$. For the system $$\dot x =A(t)x$$ we want to show that if $e^{Ty}$ is a Floquet multiplier, then there is a solution of the form $x(t)=e^{yt}p(t)$.

Here it is picture wise...

I am a bit stuck on this. Here is what I have done so far:

I know that the solution to our ode looks like $$x(t)=X(t)x_0$$

I know that given $X(t)$ is a fundamental matrix (fm), we know $X(t+T)$ is a fm as well. Then, we know that there exists a matrix M (monodromy matrix) such that $$X(t+T)=X(t)M$$

Then by the existence of matrix log, we have that there is a matrix B so that $$M=e^{BT}$$

Then, let $p(t)=X(t)e^{-Bt}$.

$X(T+t)=X(t+T)e^{-B(t+T)}=X(t+T)e^{-Bt}e^{-BT}=X(t)Me^{B(t+T)}=X(t)e^{-Bt}=p(t)$. So we have $$X(t)=p(t)e^{Bt}$$

Plugging that into the solution we get $$x(t)=p(t)e^{Bt}x_0$$ and so for $x_0=1$ we get $$x(t)=p(t)e^{Bt}$$ But I am confused how to use the fact that $e^{Ty}$ are floqeut multipliers and how to get them involved in the solution.

I know that the floquet multipliers are the eigenvalues of $X(T)$.

Answer 1

Let me state a theorem first.

Theorem 1: Suppose $A(\cdot)$ is a $T$-periodic continuous matrix-valued function. The non-autonomous linear system $\dot{x} = A(t)x$ has at least one non-trivial solution $X(t)$ satisfying $X(t+T) = \mu X(t), t\in(-\infty,\infty)$, where $\mu$ is a Floquet multiplier.

Let us denote the system $\dot{x} = A(t)x$ by (#). Given $e^{T\gamma}$ is a Floquet multiplier (now if I understand you correctly, then $\gamma\in\mathbb{C}$ should be the corresponding Floquet exponent), Theorem 1 asserts that there exists a non-trivial solution $X(t)$ of (#) such that $$ X(t + T) = e^{T\gamma}X(t). $$ In particular, \begin{align*} X(t) & = X(t+T)e^{-T\gamma} \\ & = \Big[e^{\gamma t}e^{-\gamma t}\Big]X(t+T)e^{-T\gamma } \\ & = e^{\gamma t}\Big(X(t+T)e^{-\gamma (t+T)}\Big) \\ & = e^{\gamma t}p(t+T) \\ & = e^{\gamma t}p(t) \end{align*} where $p(t) = X(t)e^{-\gamma t}$ is $T$-periodic.