= Common-strength Hill repression criterion
{title2=$\max K_{m,n}>1$}
For $\dot x=\lambda/(1+y^m)-x$, $\dot y=\lambda/(1+x^n)-y$, the equilibrium slope product is $K=mn x^ny^m/[(1+x^n)(1+y^m)]$. 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 $0<m\le1<n$, define $Y(x)$ as the unique positive solution of $Y^{-1}+Y^{m-1}=x^{-1}+x^{n-1}$ and set $K_{m,n}(x)=mn x^nY(x)^m/[(1+x^n)(1+Y(x)^m)]$. All equilibria are parameterized by $\lambda=x(1+Y^m)$, whose <derivative> is $(1+Y^m)(1-K)/(1-mY^m/(1+Y^m))$. 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.
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