Population Dynamics of Globally Coupled Degrade-and-Fire Oscillators

Abstract

This paper reports the analysis of a model of pulse-coupled oscillators with global inhibitory coupling, inspired by experiments on colonies of bacteria-embedded synthetic genetic circuits. Populations are represented by one-dimensional profiles and their time evolution is governed by a singular differential equation. We address the initial value problem and characterize the dynamics’ main features. In particular, we prove that all trajectory behaviors are asymptotically periodic, with asymptotic features only depending on the population cluster distribution and on the model parameters. A criterion is obtained for the existence of attracting periodic orbits, which reveals the existence of a sharp transition as the coupling parameter is increased. The transition separates a regime where any cluster distribution can be obtained in the limit of large times, to a situation where only trajectories with sufficiently large groups of synchronized oscillators perdure.

Appendices

Appendix 1: Discontinuous Dependence on Initial Conditions

Instantaneous resetting simplifies the analysis of  Eq. (1). However, it makes the proof of global existence of solutions rather delicate (and apparently unaccessible by standard approaches such as the Picard operator). It also implies that the solution dependence on initial profiles has discontinuities. In particular, this is the case for the first firing time function $T_{1}u$.

Indeed, there are examples of sequences $\{u_n\}$ of profiles that uniformly converge to a limit profile $u_\infty$ and for which we have $\lim \limits _{n\rightarrow +\infty }T_{1}u_n(x)\ne T_{1}u_\infty (x)$ for some $x\in (0,1]$.

To see this, let $\epsilon \leqslant 1$, let $u_\infty$ be a profile with a left plateau, i.e. $u_\infty (x)=u_\infty (x_{1})$ for all $x\in (0,x_{1}]$ ($x_{1}>0$), and let an approximating sequence be defined by

$$\begin{aligned} u_n(x)=\left\{ \begin{array}{lll} u_\infty (x)-\frac{1}{n}&{}\text {if}&{}0<x\leqslant \frac{x_{1}}{2}\\ u_\infty (x)&{}\text {if}&{}\frac{x_{1}}{2}<x\leqslant 1 \end{array}\right. \end{aligned}$$

We obviously have $T_{1}u_\infty (x)=T_{1}u_\infty (x_{1})$ for all $x\in (0,x_{1}]$ and direct calculations yield the following result:

$$\begin{aligned} \lim _{n\rightarrow +\infty }T_{1}u_n(x)=\left\{ \begin{array}{l} T_{1}u_\infty (x),\ \forall x\in (0,\frac{x_{1}}{2}]\\ T_{1}u_\infty (x)+\epsilon \eta \frac{x_{1}}{2}(T_{1}u_\infty (x_{1})-u_\infty (x_{1})+1),\ \ \forall x\in (\frac{x_{1}}{2},x_{1}] \end{array}\right. \end{aligned}$$

and the inequalities $0\leqslant u_\infty (x_{1})-T_{1}u_\infty (x_{1})<1$ imply $\lim \limits _{n\rightarrow +\infty }T_{1}u_n(x)> T_{1}u_\infty (x)$ for all $x\in (\frac{x_{1}}{2},x_{1}]$.

In addition, discontinuities may also result in the existence of attracting ghost orbits, depending on parameters. Ghost orbits are periodic cycles of profiles, viz. $\{u(x,t)\}_{(x,t)\in (0,1]\times \mathbb {R}^+}$ with $u(\cdot ,t+\tau +0)=u(\cdot ,t)$ for some $\tau >0$, which, while they do not satisfy Eq. (1), attract all trajectories in their neighborhood (uniform topology). As shown after Theorem 5.3, ghost orbits exist at bifurcation points in the parameter space, when a periodic orbit collapses.

Appendix 2: The Lower and Upper Traces of a Non-decreasing Function

Let $u:(0,1]\rightarrow (0,1]$ be a left continuous and non-decreasing function. Its lower trace $\underline{u}$ and respectively upper trace $\overline{u}$ are defined as follows:

$$\begin{aligned} \underline{u}(x)=\inf \left\{ y\in (0,1]\ :\ u(y)\geqslant u(x)\right\} ,\ \forall x\in (0,1], \end{aligned}$$

and

$$\begin{aligned} \overline{u}(x)=\sup \left\{ y\in (0,1]\ :\ u(y)\leqslant u(x)\right\} ,\ \forall x\in (0,1]. \end{aligned}$$

These functions satisfy the following basic properties.

• $0\leqslant \underline{u}(x)\leqslant x\leqslant \overline{u}(x)\leqslant 1$ for all $x\in (0,1]$.

• either $\underline{u}(x)< x$ or $x<\overline{u}(x)$ implies $u(y)=u(x)$ for all $y\in (\underline{u}(x),\overline{u}(x)]$.

• If $u$ is strictly increasing, then $\underline{u}(x)=\overline{u}(x)=x$ for all $x\in (0,1]$.⁷

• $\underline{u}\circ \underline{u}(x)=\underline{u}(x)$ iff $u$ is continuous at $x$.

• $\overline{u}\circ \overline{u}=\overline{u}$.

• Both functions $\underline{u}$ and $\overline{u}$ are left continuous and non-decreasing. (We prove the property for $\underline{u}$ here; the proof for $\overline{u}$ is similar and is left to the reader. Monotonicity is obvious and implies $\underline{u}(x-0)\leqslant \underline{u}(x)$. Left continuity is also evident in the case $\underline{u}(x)<x$. If, otherwise $\underline{u}(x)=x$, there must be a sequence $\{x_n\}_{n\in \mathbb {N}}$ such that $u(x_n)<u(x_{n+1})$ and $\lim \limits _{n\rightarrow +\infty } x_n=x$. The former condition implies $\underline{u}(x_n)>x_{n-1}$. Together with the latter, we obtain $\underline{u}(x-0)\geqslant x=\underline{u}(x)$ as desired.)

In our context, the traces provide information about the group structure of a population at time $t$: $\underline{u}(x,t)=\overline{u}(x,t)=x$ means that cell $x$ is isolated, while $\underline{u}(x,t)<\overline{u}(x,t)$ means that all cells $y\in (\underline{u}(x,t),\overline{u}(x,t)]$ belong to the same group.

The properties of the lower trace above imply that this function can be entirely determined by its plateaus; namely by considering the following decomposition:

$$\begin{aligned} (0,1]=\mathcal{C}_<\cup \mathcal{C}_=, \end{aligned}$$

where

$$\begin{aligned} \mathcal{C}_<=\left\{ x\in (0,1]\ : \underline{u}(x)<x\right\} \quad \text {and}\quad \mathcal{C}_==\left\{ x\in (0,1]\ : \underline{u}(x)=x\right\} , \end{aligned}$$

the second item above imposes the existence of a countable (possibly empty) set $\mathcal{D}$ such that $\mathcal{C}_<=\bigcup \limits _{i\in \mathcal{D}}(x_i^{-},x_i^+]$ where $x_i^-<x_i^+\leqslant x_{i+1}^-$ for all $i$. (Notice that $\mathcal{C}_=$ is empty when $u$ (or $\underline{u}$) is a step function.) In other words, every countable (possibly empty) collection of pairwise disjoint semi-open intervals in $(0,1]$ uniquely defines a lower trace function.

The upper trace function depends only on the lower trace, i.e. $\overline{u}=\overline{u}\circ \underline{u}$ (and vice-versa, we have $\underline{u}=\underline{u}\circ \overline{u}$). One can prove this fact using the sets $\mathcal{C}_<$ and $\mathcal{C}_=$ and the analogous decomposition for the upper trace. However, for our purpose, it is more convenient to use the following characterization:

$$\begin{aligned} \overline{u}(x)=\inf \left\{ \underline{u}(y)\ :\ x<\underline{u}(y)\right\} ,\ \forall x\in (0,1], \end{aligned}$$

(13)

with the convention that $\inf \emptyset =1$ in this expression. To prove this relation, notice first that we must have $\overline{u}(x)\leqslant \inf \left\{ \underline{u}(y)\ :\ x<\underline{u}(y)\right\}$. Indeed, otherwise there existed $y$ such that $x<\underline{u}(y)$ and $\underline{u}(y)<\overline{u}(x)$. Using that the former inequality is equivalent to $\overline{u}(x)<y$, it results that we must have

$$\begin{aligned} \underline{u}(y)<\overline{u}(x)<y, \end{aligned}$$

(14)

which is clearly incompatible with the definition of the traces. Secondly, still by using contradiction, assume that $\overline{u}(x)<\inf \left\{ \underline{u}(y)\ :\ x<\underline{u}(y)\right\}$. This implies the existence of $z$ such that $\overline{u}(x)<z<\inf \left\{ \underline{u}(y)\ :\ x<\underline{u}(y)\right\}$. However, the first inequality here implies $u(x)<u(z)$ and then $x<\underline{u}(z)$ which contradicts the second inequality.

Similar arguments prove the following relation:

$$\begin{aligned} \overline{u}(x)=\underline{u}(\overline{u}(x)+0)=\inf \{\underline{u}(y)\ :\ \overline{u}(x)<y\},\ \forall x\in (0,\underline{u}(1)). \end{aligned}$$

(15)

Indeed, as before, we must have $\overline{u}(x)\leqslant \inf \{\underline{u}(y)\ :\ \overline{u}(x)<y\}$ because the converse would yield to the double inequality (14) otherwise. Now if there existed $z$ such that

$$\begin{aligned} \overline{u}(x)<z< \inf \{\underline{u}(y)\ :\ \overline{u}(x)<y\}, \end{aligned}$$

the right inequality would imply $z<\underline{u}(y)\leqslant y$ for all $\overline{u}(x)<y$, hence $z\leqslant \overline{u}(x)$ holds, which contradicts the left inequality.

In the main text, we also refer to the following relation:

$$\begin{aligned} \int _0^1\underline{u}(x)dx+\int _0^1\overline{u}(x)dx=1. \end{aligned}$$

(16)

In order to prove this relation, consider again the decomposition $\mathcal{C}_<\cup \mathcal{C}_=$ with $\mathcal{C}_<=\bigcup \limits _{i\in \mathcal{D}}(x_i^{-},x_i^+]$. A moment’s reflection yields

$$\begin{aligned} \int _0^1\underline{u}(x)dx= & {} \sum _{i\in \mathcal{D}}x_i^-(x_i^+-x_i^-)+\int _\mathcal{C_=}xdx\quad \text {and}\quad \int _0^1\overline{u}(x)dx\\= & {} \sum _{i\in \mathcal{D}}x_i^+(x_i^+-x_i^-)+\int _\mathcal{C_=}xdx, \end{aligned}$$

and then

$$\begin{aligned} \int _0^1\underline{u}(x)dx+\int _0^1\overline{u}(x)dx=\sum _{i\in \mathcal{D}}(x_i^+)^2-(x_i^-)^2+2\int _\mathcal{C_=}xdx. \end{aligned}$$

Equation (16) then directly follows from the fact that

$$\begin{aligned} (x_i^+)^2-(x_i^-)^2=2\int _{x_i^{-}}^{x_i^+}xdx. \end{aligned}$$