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A simple neutral multiplier and transverse phase make the shooting Jacobian invertible

Statement

At a phase-fixed periodic point $x_{*}$ with period $T$ , let $M = D_{x} ϕ_{T} (x_{*})$ , let $f_{*} = f (x_{*}) \neq 0$ , and let $c = D ψ (x_{*})$ for the affine phase $ψ (x) = c^{⊤} (x - x_{r e f})$ . The derivative of the phase-bordered shooting residual is $J = [\begin{array}{cc} M - I_{d} & f_{*} \\ c^{⊤} & 0 \end{array}] .$ It is invertible if and only if $1$ is an algebraically simple eigenvalue of $M$ and $c^{⊤} f_{*} \neq 0$ . In the simple case choose real left/right eigenvectors $w, f_{*}$ with $w^{⊤} f_{*} = 1$ , put $Q = I - f_{*} w^{⊤}$ and $W = \ker w^{⊤}$ , and let $K_{W} = ∥ ((M - I_{d}) ∣_{W})^{- 1} ∥_{2 \to 2}$ . All vector and matrix norms below are Euclidean and induced Euclidean norms, and the product norm on $(u, τ) \in R^{d} \times R$ is Euclidean. Then $∥ J^{- 1} ∥_{2 \to 2} \leq K_{W} ∥ Q ∥_{2} + ∥ w ∥_{2} + \frac{∥ f_{*} ∥_{2} (1 + ∥ c ∥_{2} K_{W} ∥ Q ∥_{2})}{∣ c^{⊤} f_{*} ∣} .$ The period column is the exact time-sensitivity $f_{*}$ .

Facts & Assumptions

Given: A nonconstant autonomous periodic orbit, the phase residual of The phase-bordered periodic-orbit BVP residual, its monodromy $M$ , and the affine phase covector $c$ .

[F1]

The phase-bordered residual is $G (x, T) = (ϕ_{T} (x) - x, ψ (x))$ , with $ψ (x) = c^{⊤} (x - x_{r e f})$ (The phase-bordered periodic-orbit BVP residual).

[F2]

The tangent vector $f_{*}$ is a right eigenvector of $M$ for eigenvalue $1$ (An autonomous periodic orbit has a neutral Floquet multiplier).

[F3]

For a simple eigenvalue, compatible left and right eigenvectors pair nontrivially and can be normalized by $w^{*} f_{*} = 1$ (For a simple eigenvalue, left and right eigenvectors pair nontrivially and may be normalized by $y^{*} x = 1$ , Compatible left and right eigenvectors for a simple eigenvalue).

[F4]

Eigenvalues are precisely the roots of the characteristic polynomial (For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ , For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix). Here algebraic multiplicity means the multiplicity of that root; over $C$ the characteristic polynomial factors into linear factors counted with multiplicity (A complex polynomial of degree $n$ has exactly $n$ roots counted with multiplicity) and a finite-dimensional operator has Jordan form (Jordan form over the base field exists exactly when the characteristic polynomial splits).

[F5]

The monodromy is the state derivative of the flow (Local periodic-orbit, Poincare-return, and Floquet data). For the $C^{2}$ field in [F1], the finite-regularity maximal-flow theorem supplies a $C^{2}$ flow with time derivative $\partial_{t} ϕ_{t} (x) = f (ϕ_{t} (x))$ (C¹ Euclidean maximal flows, variational dependence and the finite C² upgrade).

[F6]

The derivative of a composition is the composition of its derivatives (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F7]

For linear maps between finite-dimensional spaces of equal dimension, injectivity is equivalent to surjectivity (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Proof

1.1F1F5F6

By [F1] and [F5], differentiating in $x$ gives $M - I_{d}$ , differentiating in $T$ gives $f (ϕ_{T} (x_{*})) = f_{*}$ , and differentiating the phase row gives $(c^{⊤}, 0)$ . This is the displayed matrix by [F6].

2.1F2F3F4F7step 1.1givenalgebra

Suppose $1$ is algebraically simple. Then [F4] implies that the complex eigenspace for eigenvalue $1$ is one-dimensional. By [F2] its spanning eigenvector $f_{*}$ is real, so the real kernel is $R f_{*}$ and rank-nullity gives ${rank}_{R} (M - I_{d}) = d - 1$ . Choose a real basis $v_{1}, \dots, v_{d - 1}$ of $range (M - I_{d})$ . The $d - 1$ real linear forms $w \mapsto v_{j}^{⊤} w$ are independent: if a linear combination vanishes for every $w$ , then the row $(\sum_{j} a_{j} v_{j})^{⊤}$ is zero, so $\sum_{j} a_{j} v_{j} = 0$ and basis independence forces every coefficient $a_{j}$ to vanish. Rank-nullity therefore gives a nonzero real $w_{0}$ with $v_{j}^{⊤} w_{0} = 0$ for every $j$ . Thus $w_{0}^{⊤} (M - I_{d}) = 0$ . It is a compatible left eigenvector with the real right eigenvector $f_{*}$ , so [F3], applied over $C$ , gives $w_{0}^{⊤} f_{*} \neq 0$ . This pairing is real; normalize $w = w_{0} / (w_{0}^{⊤} f_{*})$ , so $w$ is real, $w^{⊤} M = w^{⊤}$ , and $w^{⊤} f_{*} = 1$ . For $Q = I - f_{*} w^{⊤}$ and $W = \ker w^{⊤}$ , one has $Q f_{*} = 0$ , $Q ∣_{W} = I_{W}$ , and $Q M = M - f_{*} w^{⊤} M = M - f_{*} w^{⊤} = M - M f_{*} w^{⊤} = M Q$ . Hence $Q$ is the projection onto $W$ , $M$ preserves $W$ , and $M - I_{d}$ maps $W$ into $W$ . Its restriction to $W$ is injective: a kernel vector there would be an eigenvector independent of $f_{*}$ , contrary to simplicity. Since $W$ is finite dimensional, [F7] makes the restriction invertible. For $J (u, τ) = (r, s)$ , left multiplication of the first row by $w^{⊤}$ gives $τ = w^{⊤} r$ . Write $u = u_{W} + α f_{*}$ with $u_{W} \in W$ ; applying $Q$ to the first row and using its commutation with $M$ gives $u_{W} = ((M - I_{d}) ∣_{W})^{- 1} Q r$ , and the phase row gives $α = (s - c^{⊤} u_{W}) / (c^{⊤} f_{*})$ . Thus nonzero $c^{⊤} f_{*}$ gives a unique solution for every right-hand side, so $J$ is invertible. Moreover, $∣ τ ∣ \leq ∥ w ∥_{2} ∥ r ∥_{2}$ , $∥ u_{W} ∥_{2} \leq K_{W} ∥ Q ∥_{2} ∥ r ∥_{2}$ , and $∣ α ∣ \leq (∣ s ∣ + ∥ c ∥_{2} K_{W} ∥ Q ∥_{2} ∥ r ∥_{2}) / ∣ c^{⊤} f_{*} ∣$ . By the triangle inequality and $∥ r ∥_{2}, ∣ s ∣ \leq ∥ (r, s) ∥_{2}$ , these bounds yield the stated estimate.

2.2F2F4step 1.1algebraconstruct

If $c^{⊤} f_{*} = 0$ , then $J (f_{*}, 0) = 0$ , so $J$ is singular. Now suppose the algebraic multiplicity of $1$ exceeds one. If the complex eigenspace has dimension at least two, its real kernel also has dimension at least two: for the real matrix $M - I_{d}$ , its complex kernel consists exactly of $u + i v$ with $u$ and $v$ in the real kernel. Choose a real $v \in \ker (M - I_{d})$ independent of $f_{*}$ . If $c^{⊤} f_{*} = 0$ the preceding argument applies; otherwise $u = v - (c^{⊤} v) / (c^{⊤} f_{*}) f_{*}$ is nonzero, lies in $\ker (M - I_{d})$ , and satisfies $c^{⊤} u = 0$ , so $J (u, 0) = 0$ . If instead the eigenspace is one-dimensional, Jordan form [F4] supplies a generalized vector which, after rescaling, satisfies $(M - I_{d}) v = f_{*}$ over $C$ . Since $M - I_{d}$ and $f_{*}$ are real, $Re v$ is a real generalized vector satisfying the same equation; replace $v$ by this real vector. If $c^{⊤} f_{*} = 0$ , $J (f_{*}, 0) = 0$ ; if not, put $u = v - (c^{⊤} v) / (c^{⊤} f_{*}) f_{*}$ . Then $c^{⊤} u = 0$ and $(M - I_{d}) u = f_{*}$ , so $J (u, - 1) = 0$ . In every nonsimple case $J$ is singular.

3.1step 2.1step 2.2∎

Steps 2.1–2.2 prove both directions of the invertibility criterion.

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Sources

Y. A. Kuznetsov, Elements of Applied Bifurcation Theory, 2nd ed., Chapter 1, §1.5 (standard reference, not scraped)
E. J. Doedel and B. E. Oldeman, AUTO-07P full manual (2021) (standard reference, not scraped)