Strict Floquet cluster enclosures certify orbital asymptotic stability

By [F1]–[F3], exactly one multiplier of $M_{\ast }$ lies in $P_0$, and it is $1$ by [F2]. Thus the algebraic multiplicity of $1$ for $M_{\ast }$ is one. Every other multiplier lies in one of the polygons $P_j$, $j\ge 1$: their counts together with the count one in $P_0$ sum to $d$, while [F9] says a $d\times d$ characteristic polynomial has exactly $d$ roots counted with multiplicity. The remaining multipliers therefore all have modulus at most $1-\sigma <1$.

Nonconstancy and flow uniqueness [F6] imply $f(\gamma (0))\mathrm{\ne }0$, since otherwise the constant solution through $\gamma (0)$ would be the unique trajectory. Define $H=\{t\in \mathbb{R}:\gamma (t)=\gamma (0)\}$. Flow uniqueness makes $H$ an additive subgroup: a return to $\gamma (0)$ repeats the entire trajectory, so sums and differences of return times are return times. Continuity makes $H$ closed. One coordinate has nonzero derivative at zero and is strictly monotone on a sufficiently short interval; hence $H$ has no nonzero element there. Its positive infimum $S$ is therefore positive and belongs to $H$ by closedness. For any positive period $T$, division by $S$ gives $T=kS+r$ with integer $k\ge 1$ and $0\le r<S$; the subgroup property gives $r\in H$, so minimality forces $r=0$. Thus $T=kS$, where $S$ is the least positive period.

Let $M_S$ be the monodromy over $S$. By flow composition and the chain rule [F6], $M_{\ast }=M_S^k$. The phase tangent is a $1$ eigenvector of $M_S$ by [F2]. If $M_{S}v=\alpha v$ with ${\alpha }^k\mathrm{\ne }1$, then $M_{\ast }v={\alpha }^{k}v$ is a non-phase eigenvalue of $M_{\ast }$, so step 1.1 gives $\mathrm{\mid }\alpha {\mathrm{\mid }}^k<1$ and $\mathrm{\mid }\alpha \mathrm{\mid }<1$. If ${\alpha }^k=1$ and $\alpha \mathrm{\ne }1$, then $v$ and the phase tangent are independent eigenvectors of $M_{\ast }$ for eigenvalue $1$, contradicting its algebraic simplicity. If $\alpha =1$ and there were an eigenvector independent of the phase tangent, the same contradiction follows. If the eigenvalue $1$ of $M_S$ had a nontrivial Jordan chain, [F6] gives a generalized vector $v$ with $(M_S-I)v=f(\gamma (0))$ after rescaling. Then $(M_{\ast }-I)v=(M_S^k-I)v=kf(\gamma (0))\mathrm{\ne }0$, contradicting the algebraic simplicity of $1$ for $M_{\ast }$. Hence $1$ is simple for $M_S$ and every other eigenvalue of $M_S$ is strictly inside the unit circle.

The orbit with period $S$ is least-period by step 2.1. The certificate [F1] validates the phase-bordered shooting Jacobian for the possibly repeated period $T$, and that is the Jacobian used in the following nonsingularity check. For the exact real Jacobian $J$ at that root, [F7] gives $\mathrm{\parallel }I-CJD{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ with the recorded positive diagonal scale $D$. Regard $I-CJD$ as a complex matrix; [F8] makes $CJD$ invertible over $\mathbb{C}$ and hence over $\mathbb{R}$. The affine-chart transfer in [F7] then makes $J$ invertible; explicitly, $J^{-1}=D(CJD{)}^{-1}C$. Step 1.1 gives a simple phase multiplier, so the bordered criterion in [F8] implies $c^{\mathrm{\top }}f(\gamma (0))\mathrm{\ne }0$. The same affine phase is therefore transverse for the least-period orbit. Apply [F4] to identify the Poincare derivative eigenvalues with all non-phase eigenvalues of $M_S$; step 3.1 places them strictly inside the unit circle.

The certificate field is $C^2$ and its affine phase section is smooth; [F4]'s hypotheses therefore give a $C^1$ Poincare return map. Its derivative has no unit-circle multiplier, and all its eigenvalues are strictly inside by step 4.1. Shrink the section to a relatively compact patch whose closure meets the orbit only at $\gamma (0)$: transversality isolates that intersection along a short orbit arc, and the remaining compact arc stays away from $\gamma (0)$ by the least-period property. Apply the $C^1$ theorem [F5] on this patch to conclude orbital asymptotic stability.