 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 giveThe complex pair always has negative real part, and the real eigenvalue is negative exactly when $r<1$. Therefore 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 areThe 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.