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An autonomous periodic orbit has a neutral Floquet multiplier

Statement

Let f be an autonomous C1 vector field, let γ(t)=ϕt(x∗) be a nonconstant periodic solution of period T>0, and let M=DxϕT(x∗) be its monodromy matrix. Then

Mf(x∗)=f(x∗).

In particular, 1 is a Floquet multiplier.

Facts & Assumptions

Given: The autonomous vector field, its C1 local flow on a common tube through time T, and the nonconstant periodic solution γ(T)=x∗.

[F1]

The variational derivative DxϕT(x∗) exists on a common compact tube (Initial-data sensitivity of an ODE flow on a common compact tube).

[F2]

A C1 autonomous field has a unique maximal C1 flow with ∂tϕt(x)=f(ϕt(x)) (C¹ Euclidean maximal flows, variational dependence and the finite C² upgrade). Uniqueness gives ϕT(ϕs(x∗))=ϕT+s(x∗) wherever both sides are defined: as functions of the elapsed time, both sides solve the same IVP from ϕs(x∗).

[F4]

A monodromy matrix is the period-time variational fundamental matrix, and its eigenvalues are the Floquet multipliers (Local periodic-orbit, Poincare-return, and Floquet data).

Proof

1.1givenF2

Nonconstancy implies f(x∗)≠0: otherwise the constant curve through x∗ is an IVP solution, and uniqueness in [F2] forces γ to be that constant curve.

2.1F1F2F3F4step 1.1∎

Differentiate the flow identity in [F2] with respect to s at s=0. The chain rule [F3] gives DxϕT(x∗)f(x∗) on the left, while the time derivative on the right is f(ϕT(x∗))=f(x∗). By [F1] the left derivative is the monodromy action, hence Mf(x∗)=f(x∗). Since f(x∗)≠0 by step 1.1, this is an eigenvector with eigenvalue 1, which is a Floquet multiplier by [F4].

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