= Quadratic-Hill mutual repression switch
{title2=$\gamma>2$}
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> $-1\pm2/\gamma$, both negative; the symmetric state is then a <saddle point>. For $0<\gamma<2$ only the attracting symmetric state remains.
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