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A network describing how gene products affect the production of other gene products. Positive and negative regulatory interactions can generate multistability, oscillations, and noise-sensitive transitions.

// Target: mathematical-biology.bigb

Assume the decreasing functions are continuously differentiable so their local gains are defined. In the positive quadrant, the nullclines are $x=f(y)$ and $y=g(x)$, both decreasing. The phase plane arrows point right when $x<f(y)$ and left when $x>f(y)$; they point up when $y<g(x)$ and down when $y>g(x)$. Both axes point into the quadrant. Since $f,g$ are bounded, sufficiently large $x$ or $y$ decreases, giving a bounded forward-invariant rectangle containing all equilibria and eventually containing each trajectory. Thus mutual repression can have one attracting intersection or several intersections separating different basins of attraction.

An equilibrium obeys $x_{\ast }=f(y_{\ast })$, $y_{\ast }=g(x_{\ast })$. Its Jacobian matrix and eigenvalues areThe product is nonnegative. Hence an intersection is a stable equilibrium with two negative real eigenvalues when $f^{\prime }{g}^{\prime }<1$, a saddle equilibrium when $f^{\prime }{g}^{\prime }>1$, and nonhyperbolic when $f^{\prime }{g}^{\prime }=1$. This is the mutual repression stability criterion. To see why an intervening saddle normally accompanies multistability, eliminate $y$ and set $F(x)=f(g(x))-x$. Bounded positivity gives $F(0)>0$ and $F(x)<0$ for large $x$, whileTwo downward, stable crossings require an upward crossing between them. For simple roots that middle crossing has $f^{\prime }{g}^{\prime }>1$ and is a saddle equilibrium. Its stable manifold separates the attracting states. Thus robust multistability requires the loop gain to exceed one at an intervening equilibrium, with the stable equilibria themselves having gain below one. A tangency at gain one is the threshold where equilibria can merge; degenerate crossings require higher-order analysis. A gain exceeding one somewhere is necessary for multiple intersections, but by itself does not prove that a particular pair of nullclines has multiple stable intersections.

Define the logarithmic repression sensitivitiesBoth are negative. At equilibrium $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$, so their product is exactly $f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })$. The condition can therefore be expressed asEquivalently, the product of the magnitudes of the two negative logarithmic gains must exceed one at that saddle equilibrium. The divergence of the vector field is identically $-2$, so the Bendixson-Dulac criterion excludes periodic orbits in the positive quadrant. The relevant stable long-time states in this bounded phase plane are thus equilibria, rather than attracting limit cycles. The following sketches illustrate the single-attractor and two-attractor possibilities; the saddle in the second panel lies on the diagonal, with its stable diagonal separating the two basins.

For this equal-strength power-law mutual repression, the logarithmic gains and their product at equilibrium areThe mutual repression stability criterion immediately excludes multistability if $mn\le 1$, since $P<mn$ at every finite positive equilibrium. The common prefactor $\lambda$ gives a stronger exclusion: if either exponent is at most one, multistability is impossible for every $\lambda$, including some cases with $mn>1$.

To prove the additional cases, suppose $m\le 1<n$. At equilibrium the two expressions for $\lambda$ implyPut $a=x/(1+x)$ and $b=y/(1+y)$. Because $m\le 1$, the left side is at least $a$, so $a\le b(1-b{)}^{n-1}$. It follows thatThe maximum in the second inequality occurs at $b=2/(n+1)$. Its strict bound follows from $4n/(n+1{)}^2\le 1$, $m\le 1$, and $((n-1)/(n+1){)}^{n-1}<1$. Interchanging the components proves the case $n\le 1<m$; the case with both exponents at most one was already covered by $mn\le 1$. Every root of $F(x)=f(g(x))-x$ therefore has $F^{\prime }<0$. Since $F$ begins positive and ends negative, there is exactly one root: two downward crossings would require another crossing with nonnegative derivative between them. The unique equilibrium is stable.

For completeness, neither exponent can be relaxed above one in this exclusion. If $m>1$ and $n>1$, take $\lambda$ sufficiently large. On $x\in [\lambda /2,\lambda ]$, $g(x)=O({\lambda }^{1-m})\to 0$, so $f(g(x))=\lambda (1+o(1))$ maps that interval into itself. It has a fixed point withIt is a stable equilibrium. By exchanging the components, there is a second stable fixed point with $y\sim \lambda$ and $x\sim {\lambda }^{1-n}$. These two regions are disjoint for large $\lambda$, proving bistability for some production strength. Hence the full answer isFor example, $m=1/2,n=3$ has $mn>1$ but still a unique stable equilibrium for all $\lambda$. Thus the product-only necessary condition for strong feedback is not sufficient when both production prefactors are constrained to be equal.