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Pointwise comparisons used in kernel stability

PDF p. 17: Proof of Lemma 2.8, Cases 1–3

Statement

Assume countable choice. Let G=(V,E) be a finite undirected graph with nonempty vertex set, self-loops allowed, and symmetric binary adjacency matrix A. Let θ be a stable state, meaning a local minimum on the phase torus of the energy EG(θ)=12∑x,y∈VAx,y(1−cos⁡(θx−θy)), and choose representatives θx∈(−π,π]. Define K(α,β)=sin⁡(∣α∣−min⁡{∣β∣,π/2}). For each y, let ry=∑xAx,ycos⁡(θx−θy).

For every x,y∈V: if 0≤θy≤π/2, then K(θx,θy)≥sin⁡(θx−θy); if −π/2≤θy≤0, then K(θx,θy)≥−sin⁡(θx−θy). If ∣θy∣≥π/2, then K(θx,θy)=−cos⁡θx and ∑xAx,yK(θx,θy)=−ryRe⁡(e−iθy)≥0.

Facts & Assumptions

[F1]

The kernel has branches K(α,β)=sin⁡(∣α∣−∣β∣) for ∣β∣≤π/2 and K(α,β)=−cos⁡α for ∣β∣≥π/2 (Kernel for stability amplification).

[F2]

Sine is nonnegative on [0,π]. Sine is odd, cosine is even, and cos⁡t=sin⁡(π/2−t) (Angular order and elementary inverse-trigonometric bounds, Parity and the Pythagorean identity for sine and cosine, Quarter-turn values and shifts by pi/2 and pi).

[F3]

For real u,v, ei(u−v)=eiue−iv=cos⁡(u−v)+isin⁡(u−v), and real parts distribute over finite sums (Phasors, trigonometric identities and periodic derivatives).

[F4]

Under countable choice, for the graph and stable state in the Statement, ry≥0 and ∑xAx,yeiθx=ryeiθy (The Axiom of Countable Choice (AC_ω), Second-order necessary conditions for stable states). The graph, phase, energy and stable-state conventions are those of Adjacency, degrees, graph Laplacian, and spectral notation, Finite phase tori, local lifts and the flat metric, Homogeneous Kuramoto energy and Equilibria, stable states, synchronized states, and spurious minima.

[F5]

For all real u,v, sin⁡(u+v)=sin⁡ucos⁡v+cos⁡usin⁡v (The addition formulas for sine and cosine).

Proof

Given: The graph, stable state, representatives and kernel in the Statement, under countable choice.

Goal: Prove the two pointwise inequalities and the branch identity and nonnegative sum in the third case.

1.1F1F2F5algebra

Write a=θx and b=θy. If 0≤b≤π/2 and a≥0, [F1] gives K(a,b)=sin⁡(a−b). If instead a<0, put t=∣a∣∈[0,π]. By [F1], parity and the sine addition formula in [F5], K(a,b)−sin⁡(a−b)=sin⁡(t−b)+sin⁡(t+b)=2sin⁡tcos⁡b≥0. Here sin⁡t≥0 by [F2], and cos⁡b=sin⁡(π/2−b)≥0 because π/2−b∈[0,π]. This proves the first comparison.

2.1F1F2F5step 1.1algebra

If −π/2≤b≤0 and a≤0, [F1] and oddness give K(a,b)=−sin⁡(a−b). If a>0, put s=∣b∣∈[0,π/2]. By [F5], K(a,b)+sin⁡(a−b)=sin⁡(a−s)+sin⁡(a+s)=2sin⁡acos⁡s≥0, by the same sign argument as in step 1.1. This proves the second comparison, including the endpoints.

3.1F1F2F3F4algebra∎

If ∣b∣≥π/2, [F1] gives K(a,b)=−cos⁡a. Taking the real part of the phasor identity in [F4] and using [F3] yields ∑xAx,ycos⁡θx=rycos⁡θy. Thus the kernel sum is −rycos⁡θy=−ryRe⁡(e−iθy). Since ∣θy∣∈[π/2,π], cosine is nonpositive there: by [F2], cos⁡∣θy∣=−sin⁡(∣θy∣−π/2)≤0, and cosine is even. Together with ry≥0, this proves the nonnegativity and completes the proof.

Source notes

The three comparisons are attributed to Expander graphs are globally synchronizing, PDF page 17, proof of Lemma 2.8. The proof above uses the stated mathematical suppliers.

Depends on

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