Equilibrium points in Dyson's Toy Cell Model

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Equilibrium points in Dyson's Toy Cell Model<br>When real analysis meets biology<br>CasualPhysicsEnjoyer<br>Jul 16, 2026

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Thanks to David Pfau for the tutelage, ideas and discussion. All mistakes are mine.<br>In my last piece on Dyson's toy model of a cell, we modelled a cell as a Markov chain where each monomer is one of three possible states - active, inactive, and empty. In that post, ψ(x) is meant to be the probability factor that increases when the number of active monomers increases.<br>I explained that in Dyson's model, the fraction of active monomers settles in equilibrium points where<br>\(x = \dfrac{\psi(x)}{q + n + \psi(x)}\)

where x is the fraction of monomers that are active, and q and n are the (positive) statistical weights of the empty and inactive states respectively, and ψ(x) is the probability factor that determines the likelihood of an empty monomer becoming active.<br>In this post I wanted to think out loud on some of the math around this equilibrium condition. Dyson claims that ψ(x) should take the form:<br>\(\psi(x) = b^{x}\)

It's a little unclear to me why this should be the case. In a different post I'll clarify why this makes sense. But for now, let's take ψ(x) as given. Dyson then rewrites it as<br>\(x = \frac{\psi(x)}{q + n + \psi(x)} = \frac{b^{x}}{q + n + b^{x}} = \frac{1}{1 + a\,b^{-x}}\)

through the substitutions ψ(x) = bˣ and a = q + n, giving<br>\(\varphi(x) = \left(1 + a\,b^{-x}\right)^{-1}\)

Both parameters are positive. For ψ(x) = bˣ to increase with the active fraction — the autocatalytic assumption — we need b > 1. And a = q + n > 0, since q and n are positive.<br>And this reduces the problem to figuring out the roots of φ(x) = x — the points where φ crosses the diagonal.

Now, we have some questions.<br>Is it always true that we have three equilibrium points?

Could we have two or one equilibrium points, or zero?

If x starts near any of these equilibrium points, how can we tell if one is stable or not?

What are the sets of a and b that correspond to the different numbers of equilibrium points?

What is the derivative of φ(x) at the equilibrium points?

Can we bound the size of the derivatives at each equilibrium point?

Let's first say a few things about this function. It's bounded between 0 and 1. It's continuous. And it's differentiable infinitely many times. It's also monotonic.<br>To get the number of equilibrium points in the theory of life system, we want to find the number of intersections that this function makes with the y = x line. In other words, we want to find the number of roots that satisfies the equation<br>\(\varphi(x) = x\)

This is a tricky problem and should not be taken for granted! It's not guaranteed that we have three equilibrium points all the time.<br>a = 20, b = 400

By changing the parameters, we can squish the fixed points around. For example, we can make the central and right fixed point go quite far to the right:

But by just playing around with the parameters, we find that not only can we get three equilibrium points, we can also get two. If we set the following parameters to a = 9, b = 91, then we get

We can also get one equilibrium point<br>a = 9, b = 130

Can we get zero equilibrium points? To do this, we would need to show that the following equation has no roots:<br>\(\varphi(x) - x = 0 \quad\Longleftrightarrow\quad \frac{1}{1 + a\,b^{-x}} - x = 0\)

We can answer whether this is true with the the intermediate value theorem. The theorem says that if g is continuous on [0, 1] and g(0) and g(1) have opposite signs, then there is at least one c ∈ (0, 1) with g(c) = 0. This is a diagram of this result is:

Apply it to g(x) = φ(x) - x, which looks like

But note that<br>\(g(0) = \varphi(0) = \frac{1}{1+a} > 0,\)

and also that<br>\(g(1) = \varphi(1) - 1 = \frac{1}{1 + a/b} - 1 = -\frac{a/b}{1 + a/b}

and that g is continuous, since φ is.<br>Since g(0) > 0 and g(1) never get zero roots — there is always at least one equilibrium point.<br>How many roots as a function of *a* and *b*?

Indeed, it's possible that φ(x) has between one and three roots. The function φ(x) is a function of a and b. So the next question is what values of a and b correspond to how many equilibrium points there are in the system?<br>Dyson puts a diagram of how many roots there are based on the values of a and b here.

Region tour

Here is a tour of what the equilibrium points look like as we sweep across values a and b. The plot on the left shows the φ(x) curve intersecting with y, and the plots on the right show where we are for a given value of a and b. Dyson labels the region where we only have an alive state as the Garden of Eden. And similarly we have a region where we only have one dead state.<br>a = 5

a = 9 , b ∈ [76, 91])

a = 16 (wider, b ∈ [174, 1078])

a = 30 (very tall band, b ∈ [396, 1.8×10⁵])

So the next question is - what values of a and b give three equilibrium points? What is the functional form of those black...