Floquet's Theorem/Proof 2
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Theorem
Let $\mathbf A \left({t}\right)$ be a continuous matrix function with period $T$.
Let $\Phi \left({t}\right)$ be a fundamental matrix of the Floquet system $\mathbf x' = \mathbf A \left({t}\right) \mathbf x$.
Then $\Phi \left({t + T}\right)$ is also a fundamental matrix.
Moreover, there exists:
- A nonsingular, continuously differentiable matrix function $\mathbf P \left({t}\right)$ with period $T$
- A constant (possibly complex) matrix $\mathbf B$ such that:
- $\Phi \left({t}\right) = \mathbf P \left({t}\right) e^{\mathbf Bt}$
Proof
Let $\map S t = \map \Phi {t + T} {\map \Phi T}^{-1}$.
Then:
\(\ds \map {\frac \d {\d t} } {\map S t}\) | \(=\) | \(\ds \map {\Phi'} {t + T} {\map \Phi T}^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\mathbf A} {t + T} \map \Phi {t + T} {\map \Phi T}^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\mathbf A} t \map S t\) |
So $\map S t$ is a fundamental matrix and:
- $\map S 0 = Id$
Then:
- $\map S t = \map \Phi t$
which means that:
- $\map \Phi {t + T} = \map \Phi t \map \Phi T$
Hence by the existence of the matrix logarithm, there exists a matrix $\mathbf B$ such that:
- $\map \Phi T = e^{\mathbf B T}$
Defining $\map {\mathbf P} t = \map \Phi t e^{-\mathbf B t}$, it follows that:
\(\ds \map {\mathbf P} {t + T}\) | \(=\) | \(\ds \map \Phi {t + T} e^{-\mathbf B t - \mathbf B T}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \Phi t \map \Phi T e^{-\mathbf B T} e^{-\mathbf B t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \Phi t e^{-\mathbf B t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\mathbf P} t\) |
and hence $\map {\mathbf P} t$ is a periodic real function with period $T$.
As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds.
$\blacksquare$
Source of Name
This entry was named for Achille Marie Gaston Floquet.