# How can I determine $P(t)$ and $B$?

Please consider the following theorem:

Theorem (Floquet): Consider the equation $$\dot{x} = A(t)x$$ with $$A(t)$$ a continuous $$T$$-periodic $$n\times n$$-matrix. Each fundamental matrix $$\Phi(t)$$ of the equation $$\dot{x} = A(t)x$$ can be written as the product of two $$n\times n$$-matrices $$\Phi(t) = P(t)e^{Bt}$$ with $$P(t)$$ $$T$$-periodic and $$B$$ a constant $$n\times n$$-matrix.

I want to solve the following exercise:

Consider the equation $$\dot{x} = A(t)x$$ with $$A(t)$$ a smooth $$T$$-periodic $$n\times n$$-matrix, $$x\in\mathbb{R}^n$$, $$f(t)$$ a smooth scalar $$T$$-periodic function. Consider the case when $$n = 1, A(t) = f(t)$$. Determine $$P(t)$$ and $$B$$ in the Floquet theorem.

What I've tried: If $$n = 1, A(t) = f(t)$$ the fundamental matrix $$\Phi(t) = x(t)$$ is a scalar function and the solution to the equation $$\dot{x} = f(t)x$$. According to the Floquet theoren we can write $$\Phi(t) = x(t) = P(t)e^{Bt} = p(t)e^{bt}$$ where $$p(t)$$ is a $$T$$-periodic scalar function and $$b$$ a scalar. If I substitute this into the original equation I get the following $$\dot{x} = f(t)p(t)e^{bt}$$ Unfortunately I don't really know how to proceed from here..

Question: How should I solve this exercise?

Given that $$x(t)=p(t)e^{bt}$$ we get by substituting into the original equation $$\frac{dx}{dt}=f(t)x$$ that $$p(t)$$ must satisfy:

$$\dot{p}+(b-f(t))p=0\Rightarrow e^{\int^{t} (b-f(t'))dt'}p(t)=C\Rightarrow p(t)=Ce^{-bt}e^{\int^{t}f(t')dt'}$$

and thus the most general solution to the problem is:

$$x(t)=Ce^{\int^t f(t')dt'}=x(t_0)\exp\int_{t_0}^{t}f(t')dt'$$

for some arbitrary time $$t_0$$. Note that there is some arbitrariness in the way we can define b in this simple problem. To finish we should choose it's value such that $$p(t)$$ is periodic. We have that

$$\frac{d}{dt}\ln\frac{p(t+T)}{p(t)}=\frac{d}{dt}\int_t^{t+T}(f(t')-b)dt'=f(t+T)-f(t)=0$$

and therefore for every t we get that:

$$p(t+T)=p(t)\exp\int_{t_0}^{t_0+T}(f(t')-b)dt', A\in\mathbb{R}^+$$

However from mean value theorem we are guaranteed that there exists a $$\xi\in(0,T)$$ (the mean value of f in that interval) s.t:

$$\frac{1}{T}\int_{t_0}^{t_0+T}f(t')dt'=f(\xi)$$

If we pick $$b=f(\xi)$$ we get a periodic $$p(t)$$ and the Floquet theorem requirements are fulfilled.