Floquet's Theorem

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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 1

We assume the two hypotheses of the theorem.


We have that:

\(\ds \map {\frac \d {\d t} } {\map \Phi {t + T} }\) \(=\) \(\ds \map {\Phi'} {t + T}\)
\(\ds \) \(=\) \(\ds \map {\mathbf A} {t + T} \map \Phi {t + T}\)
\(\ds \) \(=\) \(\ds \map {\mathbf A} t \map \Phi {t + T}\)

So the first implication of the theorem holds, that is: that $\map \Phi {t + T}$ is a fundamental matrix.


Because $\map \Phi t$ and $\map \Phi {t + T}$ are both fundamental matrices, there must exist some matrix $\mathbf C$ such that:

$\map \Phi {t + T} = \map \Phi t \mathbf C$

Hence by the existence of the matrix logarithm, there exists a matrix $\mathbf B$ such that:

$\mathbf C = e^{\mathbf BT}$


Defining $\map {\mathbf P} t = \map \Phi t e^{-\mathbf B t}$, it follows that:

\(\ds \map {\mathbf P} {t + T}\) \(=\) \(\ds \map \Phi {t + T} e^{-\mathbf B t - \mathbf B T}\)
\(\ds \) \(=\) \(\ds \map \Phi t C e^{-\mathbf B T} e^{-\mathbf B t}\)
\(\ds \) \(=\) \(\ds \map \Phi t e^{-\mathbf B t}\)
\(\ds \) \(=\) \(\ds \map {\mathbf P} t\)

and hence $\map {\mathbf P} t$ is periodic with period $T$.

As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds.

$\blacksquare$


Proof 2

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 function with period $T$.

As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds.

$\blacksquare$


Also see


Source of Name

This entry was named for Achille Marie Gaston Floquet.