Quadratic-Hill mutual repression switch ($\gamma >2$)

For $\dot{Y}=-Y+\gamma /(1+Z^2)$ and $\dot{Z}=-Z+\gamma /(1+Y^2)$, subtraction of the equilibrium conditions gives $(Y-Z)(1-YZ)=0$. The symmetric state solves $x^3+x=\gamma$. Two asymmetric positive states exist precisely for $\gamma >2$: $Y_{\pm }=(\gamma \pm \sqrt{{\gamma }^2-4})/2$, $Z_{\pm }=Y_{\mp }$. Their Jacobian matrix has eigenvalues $-1\pm 2/\gamma$, both negative; the symmetric state is then a saddle point. For $0<\gamma <2$ only the attracting symmetric state remains.