 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 isNegative trace and positive determinant are equivalent toThus 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.