= Solution
At a positive spatially homogeneous <equilibrium point>, $v=u^2$ and $2+u-v=0$. Hence \b[$(u,v)=(2,4)$]. The reaction <Jacobian matrix> there is
$$
J=\begin{pmatrix}2&-2\\4a&-a\end{pmatrix},\qquad\operatorname{tr}J=2-a,\quad\det J=6a.
$$
The <two-dimensional Routh-Hurwitz stability criterion> gives \b[strict linear asymptotic stability for $a>2$], and instability for $0<a<2$.
For a spatial mode with $s=k^2$, the reaction-diffusion <linearization> is $J_s=J-\operatorname{diag}(s,ds)$. When $a>2$, its trace is negative for every $s\geq0$, while
$$
D(s)=\det J_s=ds^2+(a-2d)s+6a.
$$
A growing mode exists precisely when this quadratic is negative at some $s>0$. Its minimum must occur at positive $s$, and must be strictly negative, giving
$$
2d>a,\qquad(2d-a)^2>24ad.
$$
With $z=d/a$, the latter inequality becomes $4z^2-28z+1>0$. Selecting the root compatible with $2d>a$ yields the <Turing instability> threshold
$$
\boxed{d>\left(\frac72+2\sqrt3\right)a.}
$$
The unstable band is
$$
\frac{2d-a-\sqrt{(2d-a)^2-24ad}}{2d}<k^2<\frac{2d-a+\sqrt{(2d-a)^2-24ad}}{2d}.
$$
On a bounded domain a boundary condition must permit a mode in this band. Equality at the threshold gives a neutral mode, not positive growth.
At $a=2$, the homogeneous <linearization> has purely imaginary <eigenvalues>, so it is not strictly linearly asymptotically stable. If “stable” is read as nonlinear stability, the boundary deserves attention. Put $x=u-2$, $y=(v-4-x)/\sqrt3$, and $x=r\cos\theta$, $y=r\sin\theta$. The exact equations give $\dot r=r^2F(\theta)$, $\dot\theta=\omega+rG(\theta)$, where $\omega=2\sqrt3$,
$$
F=-\frac{\cos^2\theta\sin\theta}{\sqrt3}+\cos\theta\sin^2\theta,\qquad G=\frac{2\cos^3\theta}{\sqrt3}+\cos^2\theta\sin\theta+\sqrt3\cos\theta\sin^2\theta.
$$
Expanding $dr/d\theta$ through cubic order, $\int_0^{2\pi}F=0$ cancels the quadratic displacement. The correction from the quadratic change in $r$ also integrates to zero, because it is proportional to $(\int F)^2$. Since $\int_0^{2\pi}FG=\pi/(2\sqrt3)$, the <Poincaré return map> is
$$
r\longmapsto r-\frac{\pi}{24\sqrt3}r^3+O(r^4).
$$
This <cubic return-map stability of a weak focus> shows that $a=2$ is a weak attracting focus. The strict linear range used in the <Turing instability> calculation is $a>2$; the positive homogeneous equilibrium is also nonlinearly asymptotically stable at the boundary $a=2$. For that boundary, $\operatorname{tr}J_s=-(1+d)s<0$ for every nonzero spatial mode. The same strict <Turing instability> threshold and unstable band therefore also apply there, with the homogeneous mode understood through nonlinear stability.
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