Consider matrix ODE: $$\dot{\phi}(t)=A(t)\phi(t)$$ Where $A(t)$ is a given periodic matrix with period $T$, i.e. $A(x)=A(x+T)$, for all $x\in \mathbb{R}$. Then the solution $\phi(t)$ satisfies the identity: $$\phi(t+T)=\phi(t)\phi^{-1}(0)\phi(T)$$ This could be explained as $\phi^{-1}\phi(T)=\int_{0}^T\phi(t)$. Now we consider solving the equation: $e^{TB}=\phi^{-1}(0)\phi(T)$. At least formally, it could be solved: $$B=\frac{1}{T}\log\left(\frac{\phi(T)}{\phi(0)}\right)$$ (Unfortunately, $\log$…
