= Solution
The homogeneous <fixed point> must satisfy $v=u^2$. The singular point $u=v=0$ is outside the domain of the reaction term, so substitution in the other equation gives the unique regular <spatially homogeneous equilibrium>
$$
\boxed{u_*=1+r,\qquad v_*=(1+r)^2.}
$$
This is the <basally forced quadratic activator-inhibitor model>. The <linearization> of its reaction terms gives the <Jacobian matrix>
$$
J=\begin{pmatrix}\dfrac{r-1}{r+1}&-\dfrac{r}{(1+r)^2}\\2q(1+r)&-q\end{pmatrix}.
$$
Put $a=(r-1)/(r+1)$, so $-1<a<1$. Its <trace> and <determinant> are $a-q$ and $q$, and its <eigenvalues> are
$$
\lambda_\pm=\frac{a-q\pm\sqrt{(a-q)^2-4q}}2.
$$
Since $q>0$, both <eigenvalues> have negative real part exactly when $q>a$. Hence
$$
\boxed{q>\frac{r-1}{r+1}:\text{ linearly asymptotically stable};\qquad 0<q<\frac{r-1}{r+1}:\text{ unstable}.}
$$
For $r\leq1$ the equilibrium is always strictly linearly stable. For $r>1$, equality $q=a$ is an oscillatory threshold with $\lambda_\pm=\pm i\sqrt a$; the nonlinear qualification at this threshold is proved below.
Now insert a perturbation proportional to $e^{\lambda t+ikx}$. The <diffusion> terms subtract $k^2\operatorname{diag}(1,p)$ from $J$. With $s=k^2$ the mode <trace> and <determinant> are
$$
\tau(s)=a-q-(1+p)s,\qquad \Delta(s)=ps^2+(q-pa)s+q.
$$
For a strictly stable homogeneous state, $q>a$ makes $\tau(s)<0$ for all $s\geq0$. A nonzero <Fourier mode> can therefore grow only if $\Delta(s)<0$, in which case its two real <eigenvalues> have opposite signs. The <determinant> parabola has its minimum at $s_*=(pa-q)/(2p)$, and that minimum is $q-(pa-q)^2/(4p)$. The <two-species Turing criterion> is consequently derived here as
$$
\boxed{q>a,\qquad pa>q,\qquad (pa-q)^2>4pq.}
$$
These force $a>0$, hence $r>1$. Equivalently, solving $pa-q>2\sqrt{pq}$ for $\sqrt p$ gives
$$
\boxed{r>1,\qquad q>a,\qquad p>p_c=\frac{q}{(\sqrt{1+a}-1)^2}.}
$$
Thus the <inhibitor in a reaction-diffusion system> must diffuse sufficiently fast. Inside this region, the growing <wavenumbers> obey
$$
s_-<k^2<s_+,\qquad s_\pm=\frac{pa-q\pm\sqrt{(pa-q)^2-4pq}}{2p}.
$$
At onset, the minimum touches zero and the two <determinant> roots coalesce. Therefore
$$
\boxed{k_c^2=\frac{pa-q}{2p}=\sqrt{\frac qp},\qquad \lambda_c=\frac{2\pi}{k_c}=2\pi\left(\frac pq\right)^{1/4}.}
$$
Here $p$ is evaluated on the onset surface. These are the <Turing threshold of a basally forced quadratic activator-inhibitor model> and its continuum critical <wavelength>. On a finite or domain with <periodic boundary conditions> an admissible nonzero spatial mode must additionally fall in the stated band. Equality in the <diffusion> criterion is marginal onset, while $q=a$ is the distinct homogeneous oscillatory threshold.
For completeness, the <weak focus at the Hopf threshold of a basal activator-inhibitor model> resolves the nonlinear homogeneous stability when $q=a>0$. Let $h=(1+a)/2$, $\beta=(1-a)/2$, $\omega=\sqrt a$, and set $u=(1+r)(1+x)$, $v=(1+r)^2(1+y)$, $X=x$, $Y=(hy-a x)/\omega$. Expanding the rational reaction term to cubic order gives
$$
\dot X=-\omega Y+F_2+F_3+O(4),\qquad\dot Y=\omega X+G_2+G_3+O(4),
$$
where
$$
F_2=\frac{(\beta X-\omega Y)^2}{h},\quad F_3=-\frac{(aX+\omega Y)(\beta X-\omega Y)^2}{h^2},\quad G_2=\omega(hX^2-F_2),\quad G_3=-\omega F_3.
$$
Write $X=\varrho\cos\theta$, $Y=\varrho\sin\theta$ and evaluate these <homogeneous polynomials> on the <unit circle>. If $A=\cos\theta F_2+\sin\theta G_2$, $B=\cos\theta F_3+\sin\theta G_3$, and $C=\cos\theta G_2-\sin\theta F_2$, division of the radial and angular equations gives
$$
\frac{d\varrho}{d\theta}=\frac A\omega\varrho^2+\left(\frac B\omega-\frac{AC}{\omega^2}\right)\varrho^3+O(\varrho^4).
$$
The angular integral of $A$ is zero because it changes sign under $\theta\mapsto\theta+\pi$. Iterating its quadratic term once over a full cycle contributes the square of this zero integral, so the first nonzero <return map> term is cubic. Using $\langle\cos^4\theta\rangle=3/8$, $\langle\cos^2\theta\sin^2\theta\rangle=1/8$ and the corresponding sixth moments gives $\langle B\rangle=a/4$ and $\langle AC\rangle/\omega=(a+1)/8$. The <Poincaré map> is
$$
\varrho\longmapsto\varrho+\frac{2\pi}{\omega}\frac{a-1}{8}\varrho^3+O(\varrho^4).
$$
Because $0<a<1$, sufficiently small positive radii decrease on every return. The equilibrium at $q=a$ is therefore \b[nonlinearly asymptotically stable, with weak rather than exponential attraction], despite its neutral linear <eigenvalues>. The strict inequalities above describe the ordinary <diffusion>-driven instability from a strictly linearly stable homogeneous state.
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