 Solution

Adding the two homogeneous reaction rates gives $a-u$, so the unique positive Brusselator equilibrium isThe reaction Jacobian matrix at that state isA real two-dimensional linear system has strict asymptotic stability exactly when $\tau <0$ and $\Delta >0$: the eigenvalues solve ${\lambda }^2-\tau \lambda +\Delta =0$. Real eigenvalues then have positive product and negative sum; a complex conjugate pair has real part $\tau /2$. Conversely, stable eigenvalues require those signs.

Since $a>0$, the Brusselator is homogeneously stable precisely forAt $b_H=1+a^2$, the linear eigenvalues are $\pm ia$, so strict decay is lost through a homogeneous Hopf bifurcation threshold. For larger $b$ the equilibrium is unstable. The source's matrix label $A$ is here written $J$ to distinguish it from the filament modulus in the preceding question.