= Solution
The species $u$ is the <activator>, since increasing $u$ increases its own production $u^2/v$; $v$ is the <inhibitor>, since increasing $v$ reduces that production. The unique positive uniform <steady state> solves $v=u^2$ and $u^2/v=bu$, giving $\boxed{u_*=1/b,\ v_*=1/b^2}$.
The reaction <Jacobian matrix> at this state is
$$
J=\begin{pmatrix}b&-b^2\\2/b&-1\end{pmatrix},\qquad \operatorname{tr}J=b-1,\qquad\det J=b.
$$
Its <eigenvalues> solve $\lambda^2+(1-b)\lambda+b=0$. Since $b>0$, both have negative real part precisely for $0<b<1$. This proves stability of the spatially uniform reaction kinetics.
A perturbation proportional to $e^{\lambda t+ikx}$ replaces $J$ by $J-k^2\operatorname{diag}(1,d)$. Put $q=k^2$. Its trace is $b-1-(1+d)q<0$ in the kinetically stable range, and its determinant is
$$
D(q)=dq^2+(1-bd)q+b.
$$
A growing spatial mode exists exactly when this determinant is negative for some $q>0$. Its minimum occurs at $q=(bd-1)/(2d)$, which is positive only if $bd>1$, and the minimum is negative precisely if $(bd-1)^2>4bd$. Together these conditions reduce to the <Turing threshold of the quadratic activator-inhibitor model>,
$$
\boxed{0<b<1,\qquad bd>3+2\sqrt2.}
$$
The unstable band is
$$
\frac{bd-1-\sqrt{(bd-1)^2-4bd}}{2d}<k^2<\frac{bd-1+\sqrt{(bd-1)^2-4bd}}{2d}.
$$
At the <Turing instability> threshold the band shrinks to its double root, yielding $\boxed{k_c^2=(1+\sqrt2)/d}$. On a bounded spatial interval, an allowed <wavenumber> must additionally lie in the unstable band; the displayed parameter condition assumes the continuous spatial spectrum.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-3-turing-region.png]
{title=Diffusion-driven instability above the threshold curve for b between zero and one}
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