= Solution
A positive <spatially homogeneous equilibrium> satisfies $v=c+u$ and $au=bv$, giving
$$
\boxed{u_*=\frac{bc}{a-b},\qquad v_*=\frac{ac}{a-b}.}
$$
It exists iff $a>b$. The reaction <Jacobian matrix> is
$$
J=u_*\begin{pmatrix}1&-1\\a^2/b&-a\end{pmatrix},\qquad
\operatorname{tr}J=u_*(1-a),\quad\det J=u_*^2\frac{a(a-b)}b.
$$
Thus <linear stability analysis> gives strict homogeneous stability precisely when
$$
\boxed{a>1,\qquad0<b<a.}
$$
The boundaries are not included: at $a=1$ the <linearization> has purely imaginary exponents, and at $a=b$ the positive equilibrium ceases to exist.
For a mode $\cos(kx)$, put $q=k^2$. The perturbation matrix is $J-\operatorname{diag}(q,dq)$. In the homogeneously stable region its <trace> is still negative, so a growing mode exists iff its <determinant> is negative:
$$
D(q)=dq^2+u_*(a-d)q+u_*^2\frac{a(a-b)}b<0.
$$
This upward quadratic is negative for some $q>0$ precisely when
$$
\boxed{d>a>1,\qquad\frac{4da^2}{(d+a)^2}<b<a.}
$$
Such a <Turing instability> region is nonempty iff $d>1$. The unstable band is
$$
\frac{u_*}{2d}\left[d-a-\sqrt{(d-a)^2-\frac{4da(a-b)}b}\right]<k^2<
\frac{u_*}{2d}\left[d-a+\sqrt{(d-a)^2-\frac{4da(a-b)}b}\right].
$$
For a finite domain only the admissible boundary-condition <wavenumbers> in this band grow. The region shown below describes a continuous <wavenumber> <spectrum>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4-turing.png]
{title=Homogeneous stability and diffusion-driven instability for d=4}
At onset the minimum of $D$ is zero and $q_c=u_*(d-a)/(2d)$. On the threshold $b=4da^2/(d+a)^2$, substitution gives $u_*=4dac/(d-a)^2$. Hence
$$
\boxed{k_c=\sqrt{\frac{2ac}{d-a}}.}
$$
This last expression is the threshold <wavenumber>; away from onset the fastest-growing mode need not coincide with the minimum of the <determinant>.
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