= Solution
At a positive <spatially homogeneous equilibrium>,
$$
uv^2=a,
\qquad
v-uv^2=0,
$$
so
$$
(u_*,v_*)=\left(\frac1a,a\right).
$$
The <Jacobian matrix> of the reaction terms there is
$$
J=\begin{pmatrix}a^2&2\\-a^2&-1\end{pmatrix},
\qquad
\operatorname{tr}J=a^2-1,
\qquad
\det J=a^2.
$$
The <linear stability analysis> of a two-dimensional system requires negative trace and positive determinant. Since $a>0$, the homogeneous equilibrium is stable exactly when
$$
\boxed{0<a<1}.
$$
For a spatial <Fourier mode> proportional to $\cos(kx)$, put $q=k^2$. The linearized matrix is
$$
J_q=J-q\begin{pmatrix}D&0\\0&1\end{pmatrix}.
$$
Its trace is still negative when $0<a<1$, while
$$
\det J_q=Dq^2+(D-a^2)q+a^2.
$$
A <Turing instability> occurs for some $q>0$ exactly when this quadratic has a negative minimum. This requires
$$
D<a^2,
\qquad
(a^2-D)^2>4Da^2.
$$
Writing $d=\sqrt D$, the second inequality reduces to $d<a(\sqrt2-1)$. Hence, for continuously available wave numbers, the condition is
$$
\boxed{D<a^2(\sqrt2-1)^2=a^2(3-2\sqrt2).}
$$
On a bounded spatial domain, one additionally requires an allowed wave number $k$ for which $\det J_{k^2}<0$.
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