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So I'm curious how to actually find Floquet multipliers given some differential equation $y'=A(t)y$. It's (usually) easy to find the fundamental matrix $\Phi(t)$ by other methods. And I feel like I'm missing something dumb, but what can I do from there to get the eigenvalues for the matrix $C$?


For a more concrete example at hand, consider $y'=(a+b\cos t)y$. It is easy to see that $y=(C_0e^{at})e^{b\sin t}$ and the period of the periodic matrix is $p=2\pi$. So now I have: $$\Phi(t+p)C=\Phi(t)\implies(C_0e^{A(t+2\pi)})e^{B\sin{(t+2\pi)}}C=(C_0e^{At})e^{B\sin{t}}$$ How do I get $C$, or more specifically, the eigenvalues of C once I have this setup? I'm pretty sure I did something incorrectly.

EDIT: Sorry, figured it out with anon's help. I completely mixed up my notions for computing the fundamental matrix.

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First do you mean $A$ and $B$ to be $a$ and $b$ ? Second, doesn't it follow that $C = \mathrm{e}^{-2 \pi a}$ ? –  Sasha Sep 29 '11 at 5:46

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It seems to me your only issue is that you did your calculation wrong. As you know, if $\Phi(t)$ is a fundamental matrix and $\omega$ its fundamental period then the eigenvalues of $\Phi(t+\omega)^{-1}\Phi(t)$ are the multipliers (Floquet's theorem tells us this is invariant with $t$; in fact one of my texts simply defines it with $t=0$ as an arbitrary choice). If $A(t)=a+b\cos t$ then $\Phi(t)=\exp(at+b\sin t)$ - I'll assume a one-dimensional setting for convenience - so $$C=\exp\left(-a(t+2\pi)-b\sin(t+2\pi)\right)\exp\left(at+b\sin t\right)=\exp(-2\pi a).$$

We're only working in one dimension so $e^{-2\pi a}$ is your Floquet multiplier.

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Wow, I completely mixed up my understanding of what a fundamental matrix was, and that's what led to my mistake. Thanks for the help! –  Dustin Tran Sep 29 '11 at 5:47

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