Sharp Curvature and Complete Dynamical Stability

3. Stationary reduction and endpoint uniqueness

3.1 Exact memory elimination and curvature

At fixed $z$ the tilt is strictly convex in each memory:

$$(F_{mm})_{ee}=\kappa+\tfrac12 e^{m_e}|b_e^Tz|^2\ge\kappa.\tag{10}$$

Its minimizing value is $m_e=-s_e$, with

$$s_ee^{s_e}=\frac{|b_e^Tz|^2}{2\kappa},\qquad s_e=W_0\!\left(\frac{|b_e^Tz|^2}{2\kappa}\right)\ge0.\tag{11}$$

Only the real nonnegative principal Lambert branch is used (Corless et al., 1996). Substitution gives

$$V_{\rm eff}(z)=\tfrac\mu2\lVert z\rVert^2+\sum_e\kappa(s_e+\tfrac12s_e^2)-\operatorname{Re}(\ell^*z).\tag{12}$$

Critical points lift bijectively by (11) to complete stationary points. The elimination is stationary. The evolution of memory away from stationarity remains in the source.

For the edge coordinate $d=b_e^Tz\in\mathbb R^2$ the gradient is $wd$, with $w=e^{-s}$. The tangential curvature is $w$, and the radial curvature follows by $ds/d|d|=2s/[(1+s)|d|]$, extended continuously to $d=0$:

$$w\frac{1-s}{1+s}=-Q(s),\qquad Q(s)=e^{-s}\frac{s-1}{s+1}.\tag{13}$$ $$Q'(s)=e^{-s}\frac{3-s^2}{(1+s)^2},\qquad \chi=\max_{s\ge0}Q(s)=(2-\sqrt3)e^{-\sqrt3}.\tag{14}$$

The unique maximizer is $\sqrt3$. In real amplitude coordinates,

$$\nabla^2V_{\rm eff}\ge\mu I-\chi L_R,\qquad L_R=\operatorname{diag}(L,L).\tag{15}$$

Theorem 3.1 (Closed sufficient uniqueness condition). For a finite real incidence network and positive $\kappa,\gamma,\alpha,\eta$, if

$$\mu\ge\chi\lVert L\rVert,\qquad\mu>0,\tag{16}$$

then every constant complex input gives one globally asymptotically stable complete equilibrium. At equality $V_{\rm eff}$ is strictly convex and coercive; uniform strong convexity and a general positive exponential rate do not follow.

Proof. Strict inequality gives strong convexity by (15). At equality, with $L\ne0$, restrict to any nonconstant line $z(t)=z_0+t\xi$. Set $d_e(t)=b_e^Tz(t)$, $\delta_e=b_e^T\xi$ and let $\phi_e$ be the corresponding edge potential. Its line second derivative is

$$\mu\lVert\xi\rVert^2-\chi\sum_e|b_e^T\xi|^2+ \sum_e\bigl(d^2\phi_e(d_e(t))[\delta_e,\delta_e]+\chi|\delta_e|^2\bigr).\tag{17}$$

The first group and every edge remainder are nonnegative. If the first group vanishes, some $\delta_e\ne0$ because $\mu>0$ and $\xi\ne0$. Its remainder vanishes only at $s_e=\sqrt3$ with a radial $\delta_e$. The former requires

$$|d_e(0)+t\delta_e|^2=2\kappa\sqrt3 e^{\sqrt3}.\tag{18}$$

This nonconstant quadratic equality holds at most twice. The line derivative is strictly increasing on every interval, which proves strict convexity. If $L=0$, the positive amplitude quadratic suffices. Bound (9) remains coercive after minimizing in $m$. The unique critical point is the strict global minimum of $V_{\rm eff}$; it lifts to the unique strict global minimum of $V$ because the minimizing memory is unique. Proposition 2.1 supplies attraction. Compact sublevels and positive definiteness of $V-V_*$ supply Lyapunov stability. $\square$

The path endpoint below can have a zero Hessian direction. Failure of (16) does not determine uniqueness at a specified input. For example, $f=0$ gives only $(0,0)$ for any $\mu>0$.

4. Full stability from stationary curvature

4.1 Return the eliminated memory

At a stationary point in coordinates $(q,m)$, the complete Hessian is

$$H=\begin{pmatrix}K_R&B\\B^T&H_m\end{pmatrix},\quad B_{:e}=e^{m_e}\operatorname{diag}(B_e,B_e)q,\quad H_m=\operatorname{diag}_e(\kappa+\tfrac12 e^{m_e}|b_e^Tz|^2).\tag{19}$$ $$S=K_R-BH_m^{-1}B^T=\nabla^2V_{\rm eff}.\tag{20}$$ $$(\xi,\nu)^TH(\xi,\nu)=\xi^TS\xi+ (\nu+H_m^{-1}B^T\xi)^TH_m(\nu+H_m^{-1}B^T\xi).\tag{21}$$

Thus $H$ has the negative and zero counts of $S$ plus $p$ additional positive directions. Positivity of the clamped $K$ alone cannot classify the complete system.

Lemma 4.1 (Mobility–inertia correspondence). For real symmetric $H$ and real $D$ with $\operatorname{Sym}D>0$, the open right- and left-half-plane counts of $A=-DH$ equal the negative and positive inertia of $H$, with algebraic multiplicities over $\mathbb C$. Zero is semisimple of multiplicity $\dim\ker H$. No nonzero imaginary eigenvalue exists.

Proof. $D$ is invertible and

$$\operatorname{Sym}(D^{-1})=D^{-T}(\operatorname{Sym}D)D^{-1}>0.\tag{22}$$

If $Av=i\omega v$, $v\ne0$, $\omega\in\mathbb R\setminus\{0\}$, then $Hv\ne0$ and

$$(Hv)^*D(Hv)=-i\omega v^*Hv.\tag{23}$$

The left has strictly positive real part and the right has zero real part, since $H$ is Hermitian. Also $\ker A=\ker H$. A zero Jordan chain $Aw=v_0\ne0$ with $Hv_0=0$ would require

$$0=v_0^*Hw=-v_0^*D^{-1}v_0,\tag{24}$$

contradicting (22). Now homotope $D_t=(1-t)I+tD$ for $0\le t\le1$. The symmetric part remains positive. The imaginary axis is excluded apart from zero, whose fixed algebraic multiplicity prevents any additional eigenvalue reaching it. Continuity of characteristic roots preserves both open-half-plane counts from $-H$ at $t=0$. $\square$

For (4), the full linearization is $-DH$. Positive $H_*$ gives a Hurwitz Jacobian and local nonlinear exponential stability. With $\sigma=\min(\gamma,\eta)$, $h_*=\lambda_{\min}(H_*)$ and $E=v^TH_*v/2$ for the perturbation $v$,

$$\dot E\le-2\sigma h_*E+O(E^{3/2}).\tag{25}$$

For every $0<r<\sigma h_*$ a sufficiently small sublevel is invariant, with $\lVert v(t)\rVert_{H_*}\le e^{-rt}\lVert v(0)\rVert_{H_*}$. The linear estimate has rate $\sigma h_*$. No positive global curvature bound is needed.

4.2 Varying mobility at fixed state

With $H$ fixed, changes in mobilities preserving $\operatorname{Sym}D>0$ can affect transient behavior, eigenvalues and rates without changing the index, stabilizing a negative Hessian direction, or crossing through Hopf. Holding the stationary state fixed as $c$ changes requires $f=c\ell$ with fixed $\ell$. Holding physical $f$ fixed instead changes $\ell=f/c$ and generally the equilibrium and $H$.