Let us examine Hill's equation $\ddot x+Q(t)x=0$, where $Q$ is piecewise continuous and with a period $T$. Let $\mu_{1,2}$ be the multiplicators. Let $\lambda$ be the characteristic exponent. How can I prove the following
Theorem:
a) $\mu_{1}\not=\mu_{2}\Rightarrow$ Hill's equation has two linearly independent solutions $f_1(t)=e^{\lambda t}p_1(t)$ and $f_2(t)=e^{\lambda t}p_2(t)$, where $p_i$ are $T$-periodic.
b) $\mu_{1}=\mu_{2}\Rightarrow$ Hill's equation has a nontrivial periodic solution $p(t)$ with period $T$ if $\mu_{1}=\mu_{2}=1$ and period $2T$ if $\mu_{1}=\mu_{2}=-1$. If y(t) is а solution linearly independent with $p$, then $y(t+T)=\mu_{1}y(t)+\theta p(t)$, where $\theta$ is a constant.
Furthermore, if $x_1$ and $x_1$ are two solutions such that
$$x_1(0)=1, \dot x_1(0)=0$$ $$x_2(0)=0, \dot x_2(1)=0$$ then $\theta=0\Leftrightarrow (x_1(T)+\dot x_2(T)=\pm 2) \land (x_2(T)=0) \land (\dot x_1(T)=0)$
This is a supplementary theorem from my notes, left without proof. Is there an online source that proves the result?
Edit:
p.s. The theorem is attributed to Floquet.