= Solution
Adding the two homogeneous reaction rates gives $a-u$, so the unique positive <Brusselator> equilibrium is
$$
\boxed{u_*=a,\qquad v_*=b/a.}
$$
The reaction <Jacobian matrix> at that state is
$$
J=\begin{pmatrix}b-1&a^2\\-b&-a^2\end{pmatrix},\qquad
\tau=\operatorname{tr}J=b-1-a^2,\qquad\Delta=\det J=a^2.
$$
A 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 for
$$
\boxed{b<1+a^2.}
$$
At $b_H=1+a^2$, 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.
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