Both Folds Are Full Dynamical Saddle-Nodes

6. Saddle-nodes and complete modal cubics

6.1 Full dynamical nondegeneracy

A stationary turning point does not by itself establish a dynamical saddle-node. In the range of Theorem 5.1 fix $c$ and write the stationary curve as

$$R(s)=\sqrt{\tfrac{4\kappa}{3}se^s},\quad z(s)=\frac{|c|}{c}R(s)v_3,\quad m(s)=(-s,-s),\quad a(s)=|c|\sqrt{H_3(s)}.\tag{36}$$

In real coordinates $x(s)$ the stationary map and vector field are

$$Y(x,a)=\nabla F(x)-ah,\quad h=(\operatorname{Re}(v_3/c),\operatorname{Im}(v_3/c),0,0),\quad X(x,a)=-DY(x,a).\tag{37}$$

The order of $-D$ on the left matters. At a fold $a'=0$, differentiate $Y(x(s),a(s))=0$. Then $r=x'(s)$ spans $\ker H$ by Theorem 5.1, and

$$r^Th=R'(s)/|c|>0,\qquad d=r^TD^{-1}r>0.\tag{38}$$

Thus $l^T=r^TD^{-1}/d$ is a normalized left null vector of $A=-DH$, with $l^Tr=1$. The normalization uses the actual derivative $r$, not a unit vector. The load coefficient is

$$l^TX_a=\frac{r^Th}{d}>0.\tag{39}$$

Differentiate the stationary identity twice and multiply by $r^T$ to obtain

$$r^TY_{xx}[r,r]=(r^Th)a'',\qquad \tfrac12 l^TX_{xx}[r,r]=-\frac{(r^Th)a''}{2d}.\tag{40}$$

At a fold $a''=|c|H_3''/(2\sqrt{H_3})$ and

$$H_3''(s)=\tfrac{4\kappa}{3}\bigl[\mu^2e^s(s+2)+9e^{-s}(s-2)\bigr].\tag{41}$$

Equation (29) shows the sign is opposite to $Q'$. Thus $a''<0$ at $s_-$ and $a''>0$ at $s_+$, with neither zero. The one simple zero, seven decaying modes, and nonzero load and quadratic terms give both full saddle-nodes by the usual conditions (Kuznetsov, 2004). The quadratic coefficient is positive at the upper-input fold and negative at the lower one.

At $\mu=3\chi$, $s=\sqrt3$, the scalar folds meet. Strict convexity still gives a unique equilibrium at the critical input, but the Jacobian has a zero eigenvalue. Neither linear exponential stability nor a general positive local exponential estimate follows there. A cusp claim would require additional higher-order and unfolding conditions.

6.2 Independent all-coordinate blocks

Rotate the common equilibrium phase to make $z$ real. The three orthonormal node modes and the memories $(\delta m_1\pm\delta m_2)/\sqrt2$ give an orthogonal transform of the eight-dimensional source. The collective pair is

$$p=-\mu(\gamma\pm i\alpha).\tag{42}$$

For $\lambda=1,3$ the Hessian and mobility blocks, with a chosen memory sign, are

$$H_\lambda=\begin{pmatrix}k&0&\beta\\0&k&0\\\beta&0&h\end{pmatrix},\qquad D_\lambda=\begin{pmatrix}\gamma&-\alpha&0\\\alpha&\gamma&0\\0&0&\eta\end{pmatrix}.\tag{43}$$ $$k=\mu+\lambda w,\quad h=\kappa(1+s),\quad \beta^2=2\kappa\lambda sw,\quad w=e^{-s}.\tag{44}$$

For source mode $v_j$, $j\in\{1,3\}$, the coupling squared is $w^2R^2\sum_e(b_e^Tv_j)^2(b_e^Tv_\lambda)^2$. The sum is $j\lambda/2$ and $R^2=(4\kappa/j)se^s$, giving (44). Highest-mode drive pairs $\lambda=3$ with even memory and $\lambda=1$ with odd memory; low-mode drive reverses that assignment. Each block keeps radial and phase amplitude and one memory coordinate.

$$\begin{aligned} P_\lambda(p)&=(p^2+2\gamma kp+|c|^2k^2)(p+\eta h)-\eta\beta^2(\gamma p+|c|^2k)\\ &=p^3+A_2p^2+A_1p+A_0,\\ A_2&=2\gamma k+\eta h,\\ A_1&=|c|^2k^2+\eta\gamma(2hk-\beta^2),\\ A_0&=\eta|c|^2k(hk-\beta^2). \end{aligned}\tag{45}$$ $$\rho=k-\beta^2/h=\mu-\lambda Q(s).\tag{46}$$

Also $2hk-\beta^2=2\kappa[(1+s)\mu+\lambda w]>0$ and

$$A_2A_1-A_0=2\gamma|c|^2k^3+\eta|c|^2k\beta^2+ \eta\gamma(2\gamma k+\eta h)(2hk-\beta^2)>0.\tag{47}$$

Hence $A_2,A_1>0$. When $\rho>0$, vary $A_0$ from zero through $0<A_0<A_2A_1$ with $A_2,A_1$ held fixed. At zero there is a simple zero root and a stable quadratic. The zero moves left with derivative $-1/A_1$. A nonzero imaginary root would require $p=i\omega$, $\omega^2=A_1$, $A_0=A_2A_1$, excluded. All roots are strictly left. When $\rho=0$, zero is simple and the other two decay. When $\rho<0$, $P_\lambda(0)<0$ and $P_\lambda'(p)>0$ for $p\ge0$ give exactly one positive real root. The remaining roots have positive product and negative sum, hence negative real parts. This elementary argument is independent of the general inertia proof.

The two cubics and the collective pair count all eight eigenvalues with multiplicity. Numerical roots of the cubics and direct matrix spectra test different formulations with a shared numerical eigensolver backend.