Let's say we have linear system $\frac{dx}{dt} = A(t)x(t)$, where $A(t)$ is a periodic matrix, $x(t)$ a vector, and the associated fundamental matrix is $\Phi(t)$ such that $\Phi(0)=Identity$ . What can we say about the fundamental matrix of the shifted system $\frac{dx}{dt} = A(t+s)x(t)$ where $s$ is not equal to the period of $A$? How do the Lypanov exponents of the two systems relate?
thanks
For a linear system $$\dot x(t)=A(t)x(t)$$ the solution is given by $x(t)=\Phi(t)x(0)$, where $$\Phi(t)=\exp({\int_0^tA(\tau)d\tau})$$ So for the shifted system, $$\frac{dx}{dt} = A(t+s)x(t)$$ the associated fundamental matrix is $$\begin{align}\Phi(t)&=\exp({\int_0^tA(\tau+s)d\tau})=\exp({\int_s^{t+s}A(v)dv})\\ &=\exp({-\int_0^s A(v)dv+\int_0^{t+s}A(v)dv})= \Phi(s)^{-1}\Phi(t+s)\end{align}$$ So the solution of the system is given by $$x(t)=\Phi(s)^{-1}\Phi(t+s)x(0)$$ and then you can relate the Lyapunov exponents of the two systems.