= Solution
A homogeneous <equilibrium> has $v_0=u_0^2$. Besides $u_0=0$, a nonzero <equilibrium> satisfies $r=u_0^2-u_0$, giving $u_\pm=(1\pm\sqrt{1+4r})/2$. Concentrations are nonnegative, so only nonnegative roots are physically admissible. At a positive <equilibrium>, the reaction <Jacobian matrix> is
$$
J=\begin{pmatrix}u_0&-u_0\\2su_0&-s\end{pmatrix},\qquad \operatorname{tr}J=u_0-s,\qquad\det J=su_0(2u_0-1).
$$
Negative <trace> and positive <determinant> are equivalent to
$$
\boxed{\tfrac12<u_0<s.}
$$
Thus a positive homogeneous stable branch exists precisely when $s>1/2$ and $-1/4<r<s^2-s$, on the upper root $u_+$. The lower positive branch has negative <determinant> and is unstable. At $r=-1/4$ the branches meet and have a zero <eigenvalue>; at $u_0=s$ the strict linear stability condition also fails.
The trivial <equilibrium> $(0,0)$ has <eigenvalues> $r,-s$, so it is asymptotically stable for $r<0$. It exists for all $r$, independently of the nonzero branches. If negative $u_0$ were allowed mathematically, the negative branch for $r>0$ would also be homogeneously stable, but it does not represent a chemical concentration.
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