= Solution
For $q=k^2$, the mode <matrix> is $J-q\operatorname{diag}(D_1,D_2)$. Its trace remains negative. Its <determinant> is
$$
D_1D_2q^2-\frac{D_2\beta(1-\lambda\beta)}{\alpha-\beta}q
+\frac{\beta^2}{\alpha-\beta},
\qquad \lambda=\frac{D_1}{D_2}.
$$
It is negative for some $q>0$ exactly when
$$
1-\lambda\beta>0,
\qquad
(1-\lambda\beta)^2>4\lambda(\alpha-\beta).
$$
Together with homogeneous stability, the spatial-instability region is
$$
\beta>1,\qquad 0<\lambda<\frac1\beta,\qquad
\beta<\alpha<\beta+\frac{(1-\lambda\beta)^2}{4\lambda}.
$$
For fixed $\beta$, this is the region below the last curve, above $\alpha=\beta$, and left of $\lambda=1/\beta$.
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