Fundamental matrix solution and Floquet theory

Question

Let the following matrix problem be given $$y'=M(t)y$$ where $y\in\mathbb{R}^n$ and $M(t+a)=M(t)$ for some $a>0$. So $M$ is periodic.
A fundamental matrix solution can be written as $X(t)=f(t)e^{At}$, where $f(t)$ is a $a$-periodic $n\times n$-matrix and $A$ a constant $n\times n$-matrix.

For $n=1$ and $n=2$:
1. How do I find $A$?
2. What must hold for $M(t)$ such that all solutions remain bounded when $t\rightarrow \pm \infty$?
3. What must hold for $M(t)$ such that all solutions are $a$-periodic?

Here $M(t)=m(t)\begin{pmatrix}a&b\\c&d\end{pmatrix}$ for $n=2$ with $m(t)$ again $a$-periodic and $a,b,c,d$ constants.

What I thought:
For $n=1$ you want to find $$X'=M(t)X$$ where $X(t)=f(t)e^{At}$. So this gives us $$f'(t)+Af(t)=M(t)f(t)$$ where $A$ is just a real number. How do I find it? And what must I do for the conditions in 2. and 3.?

For $n=2$ I would appreciate any hints.

Answer 1

These are several different questions that need extensive space to be answered in detail. I will only provide an outline of some key points.

The matrix $A$ can be directly calculated from the relation $$X(a)=f(a)e^{aA}=f(0)e^{aA}=e^{aA}$$ where the property $f(0)=\mathbb{I}$ has been used that follows from the fact that $X(0)=\mathbb{I}$. Thus, the general approach is to find $A$ from

$$A=\frac{1}{a}\ln(X(a))$$

but calculating $X(a)$ is a tedious task for most cases.

However, for $n=1$ the fundamendal "matrix" (scalar in this case) takes the simple form

$$X(t)=\exp\left(\int_0^t{M(s)ds}\right)\qquad\qquad \qquad(1)$$

This solution can be easily checked that satisfies the ODE $X'=M(t)X$. As a result $A$ in this case is given by

$$A=\frac{1}{a}\int_0^a{M(s)ds}\qquad\qquad \qquad(2)$$

Also, for a general dimension $n$, if $M(t)M(\tau)=M(\tau)M(t)$ for all $t,\tau$ (which is the case for your $n=2$ function $M(t)$) then (1), (2) also hold true.

2. For boundedness of the solutions as $t\rightarrow+\infty$ all eigenvalues of $e^{aA}$ should lie inside or on the unit circle. Note that the solution of $y'=M(t)y$ at times $t=ka$ is given by $$y(ka)= (e^{aA})^ky (0)$$ Thus, what is needed is $\|(e^{aA})^k\|$ to remain bounded which is the case if all eigenvalues of $e^{aA}$ should lie inside or on the unit circle (Edit: for repeated eigenvalues on the unit cycle, their eigenvectors need to also be linear independent). In fact if all eigenvalues of $e^{aA}$ lie within the unit circle then all solutions tend to zero as $t\rightarrow+\infty$. Reversing now time and going for $t\rightarrow -\infty$ you will need to change the conditions accordingly.

3. This property is valid for all initial conditions if and only if

$$e^{aA}=\mathbb{I}$$

Assume $y(t)$ a solution for some $y(0)$. Then $y(t)=f(t)e^{At}y(0)$ and $$y(t+a)=f(t+a)e^{A(t+a)}y(0)=f(t)e^{aA}e^{At}y(0)$$ We want $y(t+a)=y(t)$ or equivalently $$f(t)e^{aA}e^{At}y(0)=f(t)e^{At}y(0)$$ $f(t)$ is an invertible matrix which yields $$e^{aA}e^{At}y(0)=e^{At}y(0)$$ and for $t=0$ we have $$e^{aA}y(0)=y(0)$$ Since the above must hold true for all $y(0)$ we can select different initial conditions $y_i(0)$ such that $[y_1(0),y_2(0),\cdots,y_n(0)]=\mathbb{I}$ and $$e^{aA}=e^{aA}[y_1(0),y_2(0),\cdots,y_n(0)]=[e^{aA}y_1(0),e^{aA}y_2(0),\cdots,e^{aA}y_n(0)]=[y_1(0),y_2(0),\cdots,y_n(0)]=\mathbb{I}$$ It is easy to prove that this is also a sufficient condition.