Deriving Bistability In An Autoactivation Model

From positive feedback to a fixed-point calculation

Positive feedback can make a signaling response persist after the input that initiated it has changed. To establish whether a particular pathway model can support two stable signaling levels, however, it is not enough to say that the pathway has positive feedback. We must derive its steady states, check that they are physically possible, and determine their stability.

Consider an idealized pathway in which the active species $x$ promotes its own production. Activation saturates as $x$ increases, basal production continues even when $x=0$, and the active species is removed by a first-order process. These assumptions give a simple model that exposes the mathematical conditions for bistability. They do not imply that every biochemical autoactivation pathway has this exact kinetic form.

Model and assumptions: Let $x(t)\geq 0$ be the concentration of active pathway component, measured in micromolar. Its production consists of a constant basal rate $a>0$ and a self-activation rate with maximum $V>0$, activation scale $K>0$, and Hill exponent 2. Removal is first order, with rate constant $\delta>0$. The well-mixed, deterministic model is $$ \frac{dx}{dt}=a+V\frac{x^2}{K^2+x^2}-\delta x. $$ The parameters $a$ and $V$ have units of concentration per time; $K$ has units of concentration; and $\delta$ has units of inverse time. The model assumes constant parameters, no explicit delay, and no additional feedback or state variables.

The squared Hill term is a saturating, cooperative activation function: it is near zero at low $x$, increases as $x$ rises, and approaches one at high $x$. Basal production ensures that the vector field at $x=0$ is positive. Degradation eventually dominates at sufficiently large $x$, because production is bounded while $\delta x$ keeps increasing.

Dimensionless steady-state equation

Use the concentration scale $K$ and time scale $1/\delta$. Define $y=x/K$, $\tau=\delta t$, $\alpha=V/(\delta K)$, and $\beta=a/(\delta K)$. Both $\alpha$ and $\beta$ are dimensionless and positive. Substitution into the model gives

\frac{dy}{d\tau}=F(y)=\beta+\alpha\frac{y^2}{1+y^2}-y.

A fixed point is a value $y^*\geq 0$ for which $F(y^*)=0$. Multiplication by $1+y^2$, which is strictly positive for all real $y$, gives an equivalent polynomial equation:

P(y)=y^3-(\alpha+\beta)y^2+y-\beta=0.

This is a cubic, so it has at most three real roots. But that fact alone does not establish three biologically meaningful steady states: roots must be real and positive. A useful form of the same equation is $F(y)=-P(y)/(1+y^2)$. Since $F(0)=\beta>0$ and $F(y)$ tends to negative values for large $y$, at least one positive fixed point exists.

Definition: In this one-dimensional model, a fixed point $y^*$ is locally asymptotically stable if a small displacement from it produces a vector field that moves the state back toward $y^*$. If $F'(y^*)<0$, the fixed point is locally asymptotically stable; if $F'(y^*)>0$, it is unstable. When $F'(y^*)=0$, this derivative test is inconclusive and the vector field must be examined more closely.

When can there be three fixed points?

The slope of the vector field is

F'(y)=\frac{2\alpha y}{(1+y^2)^2}-1.

The first term is the slope contributed by self-activation; the $-1$ is the slope contributed by linear removal in dimensionless time. For $F$ to turn upward over some range, the activation slope must exceed one somewhere. The function $2y/(1+y^2)^2$, for $y\geq0$, has its maximum at $y=1/\sqrt{3}$. Differentiating it gives $2(1-3y^2)/(1+y^2)^3$, which is zero at that point, positive below it, and negative above it. Its maximum value is $3\sqrt{3}/8$. Therefore $F'$ can be positive only if

\alpha>\frac{8}{3\sqrt{3}}\approx 1.5396.

This threshold is necessary for three distinct fixed points, but it is not sufficient. When the inequality holds, $F'$ has two positive zeros. Call them $y_-$ and $y_+$, with $y_-<y_+$. The vector field decreases up to $y_-$, increases between the two zeros, and decreases after $y_+$. Thus $y_-$ is a local minimum and $y_+$ a local maximum.

Three distinct positive fixed points occur exactly when the minimum is below zero and the maximum is above zero:

F(y_-)<0 \quad\text{and}\quad F(y_+)>0.

These inequalities ensure a crossing between zero and the minimum, another between the minimum and maximum, and a third after the maximum. Conversely, if either inequality fails, this three-crossing pattern is absent. Equality corresponds to a point where the curve touches zero at an extremum rather than crossing it, so it does not give three distinct fixed points.

Stability follows from the direction of the flow

When there are three distinct positive fixed points, write them in increasing order as $y_1<y_2<y_3$. At $y=0$, $F$ is positive; at sufficiently large $y$, it is negative. With three crossings, the sign of $F$ therefore alternates: positive, negative, positive, negative as $y$ increases across successive roots. At $y_1$, trajectories on either side move toward the root, so $y_1$ is stable. At $y_2$, nearby trajectories move away, so $y_2$ is unstable. At $y_3$, they move toward the root, making it stable.

The same conclusion follows from the derivative test. Since $F=-P/(1+y^2)$, at a root $y_i$ we have $F'(y_i)=-P'(y_i)/(1+y_i^2)$. If the three roots are distinct, $P'(y_1)>0$, $P'(y_2)<0$, and $P'(y_3)>0$, giving negative, positive, and negative slopes for $F$, respectively.

Worked Example: Three Positive Steady States

Suppose $K=1$ micromolar and $\delta=0.1$ per hour. Choose $\alpha=101/48\approx2.1042$ and $\beta=19/240\approx0.07917$. In dimensional terms, $V=\alpha\delta K\approx0.2104$ micromolar per hour and $a=\beta\delta K\approx0.007917$ micromolar per hour. These are hypothetical values.

The polynomial is $P(y)=y^3-(131/60)y^2+y-19/240$, because $\alpha+\beta=131/60$. It factors as $$ P(y)=(y-0.1)(y-0.5)(y-19/12). $$ Thus the three fixed points are $y_1=0.1$, $y_2=0.5$, and $y_3=19/12\approx1.5833$. Since $x=Ky$, the corresponding concentrations are $0.100$, $0.500$, and $1.583$ micromolar.

To check stability numerically, at a root $r$, $F'(r)=-P'(r)/(1+r^2)$. For the first root, $P'(0.1)=(0.1-0.5)(0.1-19/12)\approx0.5933$, so $F'(0.1)\approx-0.5875$. At the middle root, $P'(0.5)=(0.5-0.1)(0.5-19/12)\approx-0.4333$, so $F'(0.5)\approx0.3467$. At the third root, $P'(19/12)=(19/12-0.1)(19/12-0.5)\approx1.607$, so $F'(19/12)\approx-0.4582$. The outer fixed points are stable and the middle one is unstable.

Worked Example: Feedback Too Weak for Three Fixed Points

Let $\alpha=1$ and $\beta=0.1$. The maximum possible value of the activation contribution to $F'$ is $3\sqrt{3}\alpha/8\approx0.6495$, which is less than one. Therefore $F'(y)<0$ for all $y\geq0$: the vector field decreases continuously, and it can cross zero only once.

Solving $y^3-1.1y^2+y-0.1=0$ gives one positive root, $y\approx0.1125$. For example, substituting $0.1125$ into the polynomial gives approximately $0.00000195$, consistent with that rounded root. The root is stable because $F'$ is negative there. This model has one positive steady state, not bistability.

Worked Example: A Feedback Threshold Is Not Sufficient

Let $\alpha=2$ and $\beta=0.5$. Since $2>8/(3\sqrt{3})$, the slope condition permits a local minimum and maximum. It does not, by itself, guarantee three roots.

Solving $F'(y)=0$ gives a local minimum near $y_-=0.2956$ and a local maximum at $y_+=1$. At the minimum, $$ F(0.2956)\approx0.5+\frac{2(0.2956)^2}{1+(0.2956)^2}-0.2956\approx0.365. $$ The minimum is still above zero, and $F(1)=0.5$ is also above zero. The curve remains positive through both turning points and eventually decreases through zero only once. The cubic has one positive root, approximately $y=2.142$. This example shows why the turning-point heights must also be checked.

Derivation, diagnostics, and limitations

The stability classification can also be justified by perturbing a fixed point. Let $y=y^*+\varepsilon$, where $\varepsilon$ is a small displacement. A first-order Taylor expansion gives $d\varepsilon/d\tau\approx F'(y^*)\varepsilon$, with solution $\varepsilon(\tau)\approx\varepsilon(0)e^{F'(y^*)\tau}$. If $F'(y^*)<0$, the displacement decays; if it is positive, it grows. This argument explains the local derivative test and also its limitation: when $F'(y^*)=0$, the first-order approximation does not decide the behavior.

In practice, the derivative and sign pattern provide useful checks on an algebraic or numerical calculation. Confirm that all reported roots are real and positive, substitute each root into $F(y)$ to check its residual, and verify that the slopes or nearby vector-field signs match the proposed stability labels. A solver’s output alone is not evidence of bistability; it may include negative roots, nearly repeated roots, or numerical approximations with substantial residuals.

The Hill exponent matters. In this model, if the exponent is changed from 2 to 1, then $F'(y)=\alpha/(1+y)^2-1$, a decreasing function of $y$. The vector field can first rise and then fall, but it starts above zero and eventually becomes negative, so it has only one positive crossing. The squared activation term is what allows the slope to fall below, then rise above, and then fall below the degradation slope. Other model forms may have different conditions; the threshold derived here is specific to this stated equation.

A graphical sign analysis is often the clearest way to see the number and stability of fixed points. The cubic polynomial is useful for exact factoring or numerical root finding, but solving a general cubic formula is usually less informative than checking the extrema of $F$. Neither approach replaces a biological assessment of whether the assumed production and degradation mechanisms are appropriate. Moreover, a scalar deterministic model says nothing by itself about stochastic switching, parameter uncertainty, or how a pathway moves between stable states under changing input.

A common mistake is to infer bistability from strong feedback alone, or from $\alpha>8/(3\sqrt{3})$ alone. That inequality only permits the vector field to have a local minimum and maximum; three fixed points require those extrema to lie on opposite sides of zero. Another error is to call all three roots stable. In the three-root case, the middle fixed point is unstable, separating the local basins of attraction of the two stable outer states.

Key takeaway: For this saturating Hill-exponent-2 autoactivation model, three distinct positive fixed points require both sufficiently strong activation and turning points that straddle zero. When they exist, the lower and upper fixed points are stable and the middle one is unstable.

Check Your Understanding

Answer each question, then compare your reasoning with the brief solutions below.

For the stated model, what are the units of $a$, $V$, $K$, and $\delta$?
Why is $\alpha>8/(3\sqrt{3})$ necessary but not sufficient for three distinct fixed points?
If three distinct positive roots are $y_1<y_2<y_3$, which are locally stable?
What should be checked before interpreting three numerical polynomial roots as three pathway steady states?

Solutions: (1) $a$ and $V$ are concentration per time, $K$ is concentration, and $\delta$ is inverse time. (2) The inequality permits a region where $F'$ is positive and therefore two turning points; the local minimum must still be below zero and the local maximum above zero. (3) $y_1$ and $y_3$ are stable, while $y_2$ is unstable, as follows from the alternating sign of $F$ across the three crossings. (4) Check that the roots are real and positive, have small residuals in the fixed-point equation, and have stability classifications consistent with $F'$ or nearby signs of $F$.