= Solution
If $D=1$, both species have the same diffusion coefficient. For a perturbation proportional to $e^{\sigma t+ikx}$, diffusion replaces $J$ by $J-k^2I$, shifting both reaction eigenvalues to the left by $k^2$. Since the homogeneous equilibrium is already stable, every spatial mode remains stable.
For general $D$, put $q=k^2$. The mode matrix is
$$
A(q)=J-q\begin{pmatrix}D&0\\0&1\end{pmatrix}
=\begin{pmatrix}
\beta-1-Dq&\alpha^2\\
-\beta&-\alpha^2-q
\end{pmatrix}.
$$
Its trace is
$$
\operatorname{tr}A(q)=\beta-1-\alpha^2-(D+1)q<0,
$$
while its determinant is
$$
\Delta(q)=Dq^2+\bigl(D\alpha^2-(\beta-1)\bigr)q+\alpha^2.
$$
With negative trace, instability occurs exactly when $\Delta(q)<0$ for some $q>0$. The <two-species diffusion-driven instability criterion> says that this upward-opening quadratic has a negative minimum precisely when
$$
\beta-1-D\alpha^2>2\alpha\sqrt D.
$$
Equivalently, the condition is
$$
\boxed{\beta>(1+\alpha\sqrt D)^2}.
$$
This is the <Turing instability> region.
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