= Solution
Put $y=k^2\geq0$. At the threshold of a <Turing instability>, the maximum of
$$
\sigma(y)=-Dy-r+\rho\frac{y}{1+y}
$$
touches zero. Hence
$$
\sigma(y_c)=0,
\qquad
\frac{d\sigma}{dy}(y_c)
=-D+\frac{\rho}{(1+y_c)^2}=0.
$$
The second equation gives $\rho=D(1+y_c)^2$. Substitution in the first gives
$$
0=-Dy_c-r+Dy_c(1+y_c)=Dy_c^2-r.
$$
Therefore
$$
y_c=\sqrt{\frac rD},
\qquad
\boxed{k_c=\left(\frac rD\right)^{1/4}},
$$
and
$$
\boxed{
\rho_c(r)=D\left(1+\sqrt{\frac rD}\right)^2
=(\sqrt r+\sqrt D)^2.}
$$
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