Prove all solutions to $ x^{\prime}=A(t) x, \quad x(0)=x_0 $ are $T$-periodic.

Question

Suppose $A(-t)=-A(t)$ and $A(t+T)=A(t)$. Prove all solutions to $$ x^{\prime}=A(t) x, \quad x(0)=x_0 $$ are $2T$-periodic.

My try

$t_0=0$ here. $A$ is an $n \times n$ matrix.

Method 1:

We know that the solution is $x(t)=\Pi(t,t_0)x_0$ where $\Pi(t,t_0)$ is the principal matrix(normalized fundamental matrix).

Then $$x(t+T)=\Pi(t+T,t_0)x_0=\Pi(t,t_0)x_0=x(t).$$ [We used $A(t+T)=A(t)$ is proving $\Pi(t+T,t_0+T)=\Pi(t,t_0)$]

Method 2:

If we want to use $A(t+T)=A(t)$ directly then

$$x(t)=exp\left(\int_{t_0}^{t}A(s)ds\right)$$ then $$x(t+T)=exp\left(\int_{t_0+T}^{t+T}A(s)ds\right)=exp\left(\int_{t_0}^{t}A(u+T)du\right)=exp\left(\int_{t_0}^{t}A(u)du\right)=x(t)$$.

There are some mistakes in my proof as well because my calculation implies that $A$ is T periodic will imply Floquet multiplier is 1. Where did we use $A(-t)=A(t)$?

Answer 1

Let us denote $f(t)=\Pi(t,0)$

Claim $f(-t)=f(t)$

We have $f'(-t)=A(-t)f(t)=A(t)f(t)$ and $f(-t)|_{t=0}=f(0)=Id$ then by uniqueness $f(-t)=f(t)$. Here $'=\frac{d}{dx}$.

Now consider $f(t-2T)$ then $f'(t-2T)=A(t-2T)f(t-2T)=A(t)f(t-2T)$ and at $t=T$ we have $f(t-2T)|_{t=T}=f(-T)=f(T)=f(t)|_{t=T}$.

Hence, $f(t-2T)=f(t)$ and the solution is $x=f(t)x_0$ which is $2T$ periodic.