= Solution
An even number of negative weights gives $\Gamma>0$ when all three connections are nonzero. Put $r=\Gamma^{1/3}$. The three cube roots give
$$
\boxed{\lambda_0=\frac{r-1}\tau,\qquad\lambda_\pm=\frac{-1-r/2\pm i\sqrt3r/2}\tau}.
$$
The complex pair always has negative real part, and the real <eigenvalue> is negative exactly when $r<1$. Therefore \b[strict linear asymptotic stability requires $\Gamma<1$]. The origin is unstable when $\Gamma>1$. A zero connection gives $\Gamma=0$ and all eigenvalues $-1/\tau$, also stable, though the matrix need not be diagonalizable.
At $\Gamma=1$, linearization has a simple zero eigenvalue, so a strict negative-real-part test alone does not settle nonlinear stability. For completeness, take a nonzero real eigenvector $c$ of the weighted loop matrix $M=\beta W$, satisfying $Mc=c$, and set $x_i=c_i u_i$. The local equations are
$$
\tau\dot u_i=-u_i+u_{i-1}-\tfrac13\beta^2c_{i-1}^2u_{i-1}^3+O(u_{i-1}^5).
$$
The center direction has all $u_i=a$. Averaging the three equations on the center manifold gives $\tau\dot a=-(\beta^2\sum_i c_i^2/9)a^3+O(a^5)$. Its leading coefficient is negative, while the other modes decay exponentially. Hence the exact boundary origin is locally asymptotically stable for the specified hyperbolic-tangent nonlinearity, with algebraic rather than exponential decay. This qualification distinguishes actual nonlinear stability from the usual strict Jacobian criterion.
Back to article page
