= Solution
For a small <Fourier mode> $u=Ue^{ikx+\sigma t}$, $v=Ve^{ikx+\sigma t}$, the inhibitor equation gives $V=U/(1+k^2)$. The <linearization> of the cubic reaction at zero is $-rU$. Thus the <fast-inhibitor cubic activator growth rate> is
$$
\boxed{\sigma(k)=-r-Dk^2+\frac{\rho k^2}{1+k^2}.}
$$
It is an <even function> of $k$, starts at $\sigma(0)=-r<0$, and tends to $-\infty$ as $|k|\to\infty$. In terms of $q=k^2$, its derivative is $-D+\rho/(1+q)^2$. If $\rho>D$, the maxima occur at $k=\pm\sqrt{\sqrt{\rho/D}-1}$, with
$$
\boxed{\sigma_{\max}=-r+(\sqrt\rho-\sqrt D)^2.}
$$
If $\rho\leq D$, the maximum is $-r$ at $k=0$ and all modes decay. For $\rho>D$, the zero state is unstable when $(\sqrt\rho-\sqrt D)^2>r$, and the stability boundary is
$$
\boxed{\rho=(\sqrt r+\sqrt D)^2,}
$$
where this curve lies in the allowed parameter region.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-3-growth-rates.png]
{title=Growth-rate curves and the zero-state stability boundary of the fast-inhibitor model}
Putting $\widetilde u=1-u$, $\widetilde v=1-v$ and $\widetilde r=1-r$ negates both original equations and restores precisely the same form, because $\widetilde u(\widetilde u-\widetilde r)(\widetilde u-1)=-u(u-r)(u-1)$. Therefore the state $u=v=1$ has the corresponding boundary $\boxed{\rho=(\sqrt{1-r}+\sqrt D)^2}$.
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