My classmates and I are trying to figure out the following question: given the system $y'\left(t\right) = A\left(t\right)y\left(t\right)$, where $A\left(t\right) =
\left[ {\begin{array}{cc}
0 & 1 \\
\cos \left(t\right) & \sin \left(t\right) \\
\end{array} } \right]
$, find its characteristic exponents and determin e whether it has a periodic solution of period $2π$. So far, we have written it as $\left[ {\begin{array}{cc}
x \\
y \\
\end{array} } \right]' = \left[ {\begin{array}{cc}
0 & 1 \\
\cos \left(t\right) & \sin \left(t\right) \\
\end{array} } \right]\left[ {\begin{array}{cc}
x \\
y \\
\end{array} } \right]$, meaning $x' = y$ and $y' = x\cos \left(t\right) + y\sin \left(t\right)$. We can then write this as $x'' = x\cos \left(t\right) + x'\sin \left(t\right)$, so $x'' – x'\sin \left(t\right) – x\cos \left(t\right) = 0$. In another post, someone gave us a hint to solving this differential equation, and we came up with $x = ce^{-\cos \left(t\right)}$. However, from here we are lost; we understand the question involves u\sin g Floquet Theory, but it was not covered in great detail in class and the online resources we have found are difficult for us to understand because while the theory is described in a lot of higher-level detail, there aren't many worked through examples. Would someone be able to briefly explain Floquet Theory to us and how this question is relevant without providing a full answer so that we have a step in the right direction?
Thank you very much in advance.
Question Regarding Floquet Theory and Characteristic Exponents
ordinary differential equations
Best Answer
As @user539887 already noted, the solution of the system is $$ x(t)=c_1 e^{-\cos t}+c_2 e^{-\cos t} \int e^{\cos t}\, dt. $$ However, it will be more convenient if we choose $$ x(t)=c_1 e\cdot e^{-\cos t}+ c_2 e^{-\cos t} \int_0^{t} e^{\cos \tau}\, d\tau; $$ then we have $$ y(t)= c_1 e\cdot\sin t\cdot e^{-\cos t}+c_2\left( \sin t\cdot e^{-\cos t}\int_0^{t} e^{\cos \tau}\, d\tau+1 \right) $$ and $x(0)= c_1$, $y(0)=c_2$. Hence, the solution can be written as $$ \left(\begin{array}{c} x(t)\\y(t) \end{array}\right)= \Phi(t) \left(\begin{array}{c} x(0)\\y(0) \end{array}\right), $$ where $$ \Phi(t)=\left(\begin{array}{cc} e^{1-\cos t} & e^{-\cos t} \int_0^{t} e^{\cos \tau}\, d\tau\\ \sin t\cdot e^{1-\cos t} & \sin t\cdot e^{-\cos t}\int_0^{t} e^{\cos \tau}\, d\tau+1 \end{array}\right) $$ is the principal fundamental matrix of the system (i.e. the fundamental matrix that satisfies $\Phi(0)=I$).
Now we can calculate the Floquet multipliers. They are equal to the eigenvalues of the matrix $$ \Phi(2\pi)=\left(\begin{array}{cc} 1 & e^{-1} \int_0^{2\pi} e^{\cos \tau}\, d\tau\\ 0 & 1 \end{array}\right), $$ thus, $\lambda_1=\lambda_2= 1$ and the Floquet exponents are equal to zero.
Another way to solve this problem is to use the formula $$\prod_{i=1}^n\lambda_{i}=e^{\int_0^T {\mathrm trace}A(t)\,dt}.$$ For this system we have $$\tag{1} \lambda_1\lambda_2= e^{\int_0^T \sin t\,dt}=1. $$ Since $x(t)=e^{-\cos t}$, $y(t)=\sin t\, e^{-\cos t}$ is a periodic solution, there is a Floquet multiplier $\lambda_1=1$; the other one is given by (1).