= Solution
For a spatially homogeneous equilibrium of the <Brusselator>, the reaction terms obey
$$
0=\alpha-(\beta+1)u+u^2v,
\qquad
0=\beta u-u^2v.
$$
Positivity gives $uv=\beta$, and substitution into the first equation gives
$$
(u_*,v_*)=\left(\alpha,\frac\beta\alpha\right).
$$
The reaction Jacobian there is
$$
J=
\begin{pmatrix}
\beta-1&\alpha^2\\
-\beta&-\alpha^2
\end{pmatrix},
$$
so
$$
\operatorname{tr}J=\beta-1-\alpha^2<0,
\qquad
\det J=\alpha^2>0.
$$
The <linear stability analysis> therefore makes the equilibrium asymptotically stable. It is a stable node when
$$
(\beta-1-\alpha^2)^2\geq4\alpha^2
$$
and a stable focus when the reverse strict inequality holds. Since the upper-right entry of $J$ is positive and the lower-left entry is negative, focus trajectories rotate clockwise. Thus the local phase portrait consists of trajectories approaching the fixed point, either directly as a node or while spiralling clockwise as a focus.
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