I am stuck on an attempted calculation in which I aim to find the Floquet Exponents of the Meissner equation (ie the Hill equation where the parametric driving is a square wave).
$\ddot{y} + (1 + r f(t) ) y = 0$
where $f(t) = 1$ for $0<t\leq T/2$, and $f(t) = -1$ for $T/2<t \leq T$.
Furthermore $f(t+T)=f(t)$.
Since the pieces are exactly solvable we write $$ \dot Y = \frac{d}{dt} \begin{pmatrix} \dot y \\ y \end{pmatrix} = \begin{pmatrix} 0 & -(1+r) \\ 1 & r \end{pmatrix} \begin{pmatrix} \dot y \\ y \end{pmatrix} = M_+ Y $$ for $0<t\leq T/2$. And $$ \dot Y = \frac{d}{dt} \begin{pmatrix} \dot y \\ y \end{pmatrix} = \begin{pmatrix} 0 & -(1-r) \\ 1 & r \end{pmatrix} \begin{pmatrix} \dot y \\ y \end{pmatrix} = M_- Y $$ for $T/2<t\leq T$. The result is that $$ Y(T) = e^{T M_-/2} e^{T M_+/2} Y(0) = B Y(0) $$ I believe the Floquet multipliers are then the solutions to the characteristic equation
$$ |B - \rho I | = 0 $$
This equation is found to be
$$ \rho^2 - 2 \rho\left(\cos(\sqrt{1-r} \,T/2 ) \cos(\sqrt{1+r} \,T/2) -\frac{\sin(\sqrt{1-r}\, T/2) \sin(\sqrt{1+r}\, T/2)}{\sqrt{1-r^2}} \right) + 1 = 0 $$
I cannot see a way to massage this into the very simple form given on the wikipedia page, which states that the floquet exponents for $T=2$ are the roots of the equation $$ \lambda^2 - 2 \lambda \cos \sqrt{r} \cosh \sqrt{r} + 1 = 0 $$ Any pointers will be much appreciated.