= Saturating positive feedback
<Saturating positive feedback> limits the reinforcing rate as activity increases. Here is an explicitly synthetic <ordinary differential equation> for normalized nonnegative input $u$, response $x$, and return strength $\beta$:
$$
\frac{dx}{dt}=f(x;u,\beta)=u+\frac{\beta x^2}{1+x^2}-x,\qquad u\geq0,\quad\beta\geq0.
$$
The return is an increasing, nonproportional function of $x$, while removal is proportional to $x$. Time and activities are <dimensionless variables>. At $x=0$ the <vector field> points into the nonnegative domain, and $dx/dt\leq u+\beta-x$ prevents unbounded growth there. This makes the toy useful for asking about a bounded switch; it supplies no molecular mechanism for the chosen rate law.
At an <equilibrium point>, $f=0$. For a simple <equilibrium point>, the sign of the <derivative>
$$
f_x=\frac{2\beta x}{(1+x^2)^2}-1
$$
determines local <asymptotic stability>. If $\beta=1$, the maximum reinforcing slope is $3\sqrt3/8<1$, so $f$ decreases strictly: every $u\geq0$ has one <asymptotically stable> <equilibrium point> despite <positive feedback>.
Three <equilibrium points> become possible only if the reinforcing slope exceeds removal's slope somewhere. Its maximum is $3\sqrt3\beta/8$, at $x=1/\sqrt3$. Consequently this toy requires
$$
\beta>\frac{8}{3\sqrt3}\simeq1.53960
$$
for two distinct folds. This is a necessary threshold for a <bistability> interval in this toy, not a universal biochemical number. At a fold, both $f=0$ and $f_x=0$, giving
$$
(1+x^2)^2=2\beta x,\qquad u=x-\frac{\beta x^2}{1+x^2}.
$$
For $\beta=1.8$, the folds are at $(x,u)\simeq(0.350003,0.153563)$ and $(0.887240,0.094405)$. Between these input values there are two <asymptotically stable> <equilibrium points> separated by an unstable one. At $u=0.12$, the three activities are approximately $0.171331$, $0.621230$ and $1.127439$; the low and high ones are stable.
A sufficiently slow increasing sweep follows the low branch to $u_{\mathrm{on}}\simeq0.153563$ and then jumps upward. A decreasing sweep follows the high branch to $u_{\mathrm{off}}\simeq0.094405$ and then jumps downward: <hysteresis>. These are limiting quasistatic thresholds, with relaxation between input changes and negligible perturbations that could cause earlier switching. Finite sweep rates and fluctuations can alter observed thresholds. A switchlike response with only one stable <equilibrium point> does not have this equilibrium <hysteresis> mechanism.
Back to article page
