Global attraction and nonlinear gain

4 Sharp curvature and global attraction

Lemma 4.1 (Reduced edge curvature). For $d\in\mathbb C,r=|d|,s=W_0(r^2/(2\kappa)),w=e^{-s}$, the two real Hessian eigenvalues of $\kappa(s+s^2/2)$ are

$$\lambda_{\rm tan}=w,\qquad\lambda_{\rm rad}=w\frac{1-s}{1+s}.\tag{18}$$

Both are one at the origin by continuity. The sharp scalar lower bound is $-\chi$, where

$$\chi=(2-\sqrt3)e^{-\sqrt3}=0.0474058943568\ldots.\tag{19}$$

Proof. From $se^s=r^2/(2\kappa)$, $ds/dr=2s/[r(1+s)]$ for $r>0$. Differentiate the edge gradient $wd$: its tangential eigenvalue is $w$ and its radial eigenvalue $w+r\,dw/dr$ is (18). For $s\le1$ both are nonnegative; for $s>1$,

$$\frac d{ds}\left[e^{-s}\frac{s-1}{s+1}\right]=e^{-s}\frac{3-s^2}{(s+1)^2}.\tag{20}$$

The magnitude peaks at $s=\sqrt3$ with value $\chi$. Every $s\ge0$ is attained for some $r$, proving scalar sharpness without assuming simultaneous worst directions on all graph edges.

Pull back the edge Hessians through $z\mapsto b_e^\top z$:

$$\nabla^2 V_{\rm eff}\ge\mu I-\chi L_R\ge\delta I,\qquad L_R=\operatorname{diag}(L,L),\quad\delta=\mu-\chi\|L\|.\tag{21}$$

The spectral norm includes all multiplicities and collective modes.

Theorem 4.2 (Unique sustained state and global attraction). Under the positivity hypotheses and $\delta>0$, every constant complex $f$ has one globally asymptotically stable equilibrium, with $\|z_*\|\le\|\ell\|/\mu$, $m_*\le0$.

Proof. The coercive $V_{\rm eff}$ is strongly convex by (21), so its unique critical point is the unique minimizer $z_*$. The strictly convex memory slices restore $m_*$. At equilibrium $K(m_*)z_*=\ell$, giving the norm bound, and (15) gives the memory sign.

Along any bounded trajectory $V$ converges to a finite limit. Write $g=\gamma\|Kz-\ell\|^2+\eta\|F_m\|^2$. An omega-limit point with $g>0$ would have a ball where $g\ge\epsilon>0$. Bounded speed on a compact neighborhood forces a positive residence time in that ball after each visit to a smaller concentric ball, unless the trajectory never leaves, in which case $V\to-\infty$. Infinitely many disjoint residence intervals likewise force unbounded loss. Therefore all omega-limit points have $g=0$; compactness and the unique equilibrium then imply convergence of the entire trajectory.

Moreover $V(z,m)\ge V_{\rm eff}(z)\ge V_{\rm eff}(z_*)=V(z_*,m_*)$, with equality only there. On a small sphere around the equilibrium the excess has a positive minimum. Initial states in a smaller sublevel cannot cross it by (10), proving Lyapunov stability. This direct finite-dimensional invariance argument [LaSalle 1976] gives global convergence, without a uniform global exponential or intertrajectory contraction estimate. Failure of the strict sufficient condition alone establishes neither instability nor multistability.

5 Complete local stability and a parameter bound

At equilibrium $q_*=(\Re z_*,\Im z_*)^\top$, set

$$H=\begin{pmatrix}K_R&B\\B^\top&H_m\end{pmatrix},\quad B_{\cdot e}=e^{m_{*,e}}\operatorname{diag}(B_e,B_e)q_*,\quad H_m=\operatorname{diag}\big(\kappa+\tfrac12 e^{m_{*,e}}|b_e^\top z_*|^2\big).\tag{22}$$

Differentiating stationary memory gives $dm/dq=-H_m^{-1}B^\top$, hence

$$S=K_R-BH_m^{-1}B^\top=\nabla^2 V_{\rm eff}(z_*)\ge\delta I.\tag{23}$$

Consequently $H>0$. The constant-mobility Jacobian is

$$A_{\rm full}=-DH,\qquad D=\operatorname{diag}(G,\eta I),\qquad\tfrac12(D+D^\top)=\operatorname{diag}(\gamma I,\eta I).\tag{24}$$

Partial amplitude curvature alone cannot establish this joint-state result.

Proposition 5.1 (Linear rate and nonlinear local stability). With $\sigma=\min\{\gamma,\eta\}$, define

$$\theta_0=\frac{\|\ell\|\sqrt{\sum_e\|b_e\|^4}}{\kappa\mu},\qquad h_0=\frac{\min\{\delta,\kappa\}}{(1+\theta_0)^2}>0.\tag{25}$$

Then $H\ge h_0 I$, and for $\|v\|_H=(v^\top Hv)^{1/2}$,

$$\|v(t)\|_H\le e^{-\sigma h_0 t}\|v(0)\|_H,\qquad\max_{\lambda\in\operatorname{spec}(A_{\rm full})}\Re\lambda\le-\sigma h_0.\tag{26}$$

Every positive rate strictly below $\sigma h_0$ holds for nonlinear evolution in a sufficiently small neighborhood in the equilibrium Hessian norm.

Proof. Since $m_*\le0$, $\|B_{\cdot e}\|\le\|b_e\|^2\|z_*\|$, hence $\|H_m^{-1}B^\top\|\le\kappa^{-1}\|B\|\le\theta_0$. Completing the square gives

$$(\xi,\nu)^\top H(\xi,\nu)=\xi^\top S\xi+(\nu+H_m^{-1}B^\top\xi)^\top H_m(\nu+H_m^{-1}B^\top\xi).\tag{27}$$

The inverse shear has norm at most $1+\theta_0$. Combine (23) and $H_m\ge\kappa I$ to get $H\ge h_0I$. For $E_v=v^\top Hv/2$, the linearized system gives

$$\dot E_v=-(Hv)^\top\operatorname{diag}(\gamma I,\eta I)(Hv)\le-\sigma h_0 v^\top Hv=-2\sigma h_0E_v.\tag{28}$$

Integrate to obtain the norm and spectral bounds. The nonlinear vector field differs by $O(\|v\|^2)$, adding $O(\|v\|^3)$ to $\dot E_v$. Bound it by $2\epsilon E_v$ in a small invariant sublevel for any $0<\epsilon<\sigma h_0$.

The exact linear estimate, local nonlinear rate and numerical eigenvalues remain distinct. A fixed nonzero input fixes phase; at zero input positive $\mu,\kappa$ also remove zero decay rates despite the graph collective mode. For the path, $\|L\|=3,\mu=2/5$. Since $\sqrt3>5/3,\ 2-\sqrt3<1/3$ and the cubic Taylor polynomial for the exponential at $5/3$ exceeds four, $\chi<1/12,\delta>3/20$. With unit input $|c|=\sqrt{9/5}>1$, and $\sqrt{\sum_e\|b_e\|^4}=\sqrt8<3$, giving $\theta_0<15/2,\ h_0>3/1445$, and

$$\sigma h_0>9/7225.\tag{29}$$

This is conservative: (25) evaluated numerically gives rate $\sigma h_0\approx0.00393374$, while the complete numerical spectral abscissa is about $-0.24$.

6 What the drive imposes and feedback changes

6.1 Collective control and symmetric contrast gain

With $b_e^\top\mathbf1=0$, $\bar z=n^{-1}\mathbf1^\top z,\ \bar f=n^{-1}\mathbf1^\top f$ satisfy

$$\bar z(t)=\frac{\bar f}{c\mu}+\left(\bar z(0)-\frac{\bar f}{c\mu}\right)e^{-c\mu t}.\tag{30}$$

A matched fixed carrier has the same mean. Uniform input and initial amplitude with zero memory keep all contrasts and memories zero. Thus sustained collective amplitude controls for excitation without adaptive edge organization.

For the path take

$$v=(1,0,-1)^\top/\sqrt2,\qquad f=av,\quad a>0.\tag{31}$$

Signed reflection (swap endpoints and reverse sign of amplitude) preserves this preparation. Starting from zero gives the invariant sector $z=uv$ and equal memories. At equilibrium $m_1=m_2=-s$; $Lv=v$ and $|b_e^\top z|^2=|u|^2/2$ imply

$$\frac{a^2}{|c|^2}=4\kappa s e^s(\mu+e^{-s})^2,\qquad u=\frac{a/c}{\mu+e^{-s}}.\tag{32}$$

For $a>0$, $s>0$. The right side starts at zero and diverges; its derivative has the sign of $\mu+e^{-s}(1-s)/(1+s)$, positive for $\mu>\chi$ by Lemma 4.1. Thus it specifies exactly one $s$ there. The fixed carrier gives $u_0=(a/c)/(\mu+1)$, and

$$G=\frac{|u|}{|u_0|}=\frac{\mu+1}{\mu+e^{-s}}>1.\tag{33}$$

The drive specifies the spatial mode and relative phase. Feedback produces nonzero equal memories, persistent softening and excess amplitude. For the worked parameters $s=0.07303438894\ldots,\ G=1.0529729143\ldots$, about 5.30% gain. This is a fixed-preparation prediction. A one-port input retains mean and contrast, and can sustain unequal memories as shown below, without implying every asymmetric input does.

6.2 Equal stationary power and continuing work

At equilibrium $Kz_*=\ell,\ F_m=0$, so

$$P_*=\Re(\ell^*c\ell)=\gamma\|\ell\|^2=\frac{\gamma\|f\|^2}{\gamma^2+\alpha^2}.\tag{34}$$

The same identity holds for the fixed carrier. Equal-norm inputs have equal stationary power in this contract while differing in amplitude, memory, energy and response.

Corollary 6.1 (Work required for sustained organization). Under Theorem 4.2 with $f\ne0$,

$$\int_0^T P_{\rm in}(t)\,dt=P_*T+o(T),\qquad P_*>0.\tag{35}$$

Dissipation has the same leading growth, while stored energy converges.

Proof. Full convergence gives $P_{\rm in}\to P_*,F\to F_*$. Split the integral at a time after which the difference in power is below any $\epsilon>0$. The early finite integral divided by $T$ vanishes; the remaining average error is at most $\epsilon$. Integrate (8) and use bounded energy change to transfer the leading term to dissipation.

An unreplenished finite work supply cannot maintain the nonzero state forever. Constant forcing represents powered dissipation rather than autonomous oscillation sustained indefinitely by initial energy.