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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

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

$$ |B - \rho I | = 0 $$

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.

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    Are you sure your solution of the characteristic equation is correct? Because I see that B is expressed with exponentials of $M_{-}$ and $M_{+}$ or so, given their look I would expext some exp() terms to pop up at some point. I've pushed the calculus to the exponentials of the two above matrices and there's a exp(r) in both of them; i would not expect it to disapear at the end, but maybe i'm wrong.. – 2017-02-05
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    The exponentials lead to the $\sin$ and $\cos$. The piecewise solutions are both oscilliatory so with frequencies $\omega = \sqrt{1 \pm r}$ so I think one would expect it to be a function of $\sin(\sqrt{1 \pm r} T/2)$ and $\cos(\sqrt{1 \pm r} T/2)$. More specifically $\exp \frac{T}{2} \begin{pmatrix} 0 & -\omega^2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \cos(\omega t/2) & - \omega \sin(\omega t/2) \\ \sin(\omega t/2)/\omega & \cos(\omega t/2) \end{pmatrix}$. Personally I cannot see how $e^{r}$ would appear. – 2017-02-05
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    I must do the complete calculus then, i'll tell you what i get – 2017-02-05
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    Excellent - let me know what you find. – 2017-02-05

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