 Common-strength Hill repression criterion

For $\dot{x}=\lambda /(1+y^m)-x$, $\dot{y}=\lambda /(1+x^n)-y$, the equilibrium slope product is $K=mn{x}^n{y}^m/[(1+x^n)(1+y^m)]$. A value $K>1$ gives a saddle separating attracting alternatives. Both exponents greater than one allow this at sufficiently large common $\lambda$. If both are at most one it is impossible. If $0<m\le 1<n$, define $Y(x)$ as the unique positive solution of $Y^{-1}+Y^{m-1}=x^{-1}+x^{n-1}$ and set $K_{m,n}(x)=mn{x}^{n}Y(x{)}^m/[(1+x^n)(1+Y(x{)}^m)]$. All equilibria are parameterized by $\lambda =x(1+Y^m)$, whose derivative is $(1+Y^m)(1-K)/(1-m{Y}^m/(1+Y^m))$. Thus multistability for some common strength occurs exactly when ${\max }_{x>0}{K}_{m,n}(x)>1$. For $m=1$ the maximum is $(n-1{)}^2[(n+1)/(n-1){]}^{(n+1)/n}/(4n)$. Consequently $mn>1$ alone is not sufficient when the two synthesis strengths are required to coincide.