# Theoretical Study of the One Self-Regulating Gene in the Modified Wagner Model

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## Abstract

**:**

## 1. Introduction

## 2. Studying the Sigmoid Function

#### 2.1. Introducing the Considered Sigmoid

- a continuous increasing function,
- the limit of ${f}_{a}$ as x approaches negative infinity is 0,
- the limit of ${f}_{a}$ as x approaches infinity is 1,
- ${f}_{a}\left(0\right)=a$ and ${f}_{a}^{\prime}\left(0\right)=1$,

#### 2.2. About $\lambda $ and $\mu $ Parameters

#### 2.3. Fixed Point of ${f}_{a}$

**Theorem**

**1.**

**Proof.**

- If $a\u2a7e{\displaystyle \frac{1}{{e}^{2}+1}}$, then ${H}^{\prime}\u2a7d0$ on $\mathbb{R}$, and we have $\underset{x\to -\infty}{lim}H\left(x\right)=\underset{x\to -\infty}{lim}1-x-\lambda x{e}^{-\mu x}}=+\infty $ and $\underset{x\to +\infty}{lim}H\left(x\right)}=-\infty $, which leads to the variations depicted in Figure 5.Consequently, if $a\u2a7e{\displaystyle \frac{1}{{e}^{2}+1}}$, then $H\left(x\right)=0$ has one unique solution, i.e., ${f}_{a}$ has one unique fixed point.
- If $a\u2a7d{\displaystyle \frac{1}{{e}^{2}+1}}$, then there exist two real numbers ${x}_{1}$ and ${x}_{2}$ such that the variation table of Figure 6 is satisfied for H.
- -
- $H\left({\displaystyle \frac{2}{\mu}}\right)=\left(\right)open="("\; close=")">1-{\displaystyle \frac{2}{\mu}}-{\displaystyle \frac{2\lambda}{\mu}}{e}^{-2}$. As $\mu \u2a7e4$, we deduce that $\frac{2}{\mu}}\u2a7d{\displaystyle \frac{1}{2}$. Additionally, ${e}^{-2}<{\displaystyle \frac{1}{4}}$ and $\frac{\lambda}{\mu}}={(1-a)}^{2}\in ]0,1[$, then $\frac{2\lambda}{\mu}}{e}^{-2}<{\displaystyle \frac{1}{2}$. As a conclusion, $H\left({\displaystyle \frac{2}{\mu}}\right)>0$.As $H\left({\displaystyle \frac{2}{\mu}}\right)>0$, H is increasing on $[{\displaystyle \frac{2}{\mu}},{x}_{2}]$, H decreases on $[{x}_{2},+\infty [$, and the limit of H of x as x approaches $+\infty $ equals $-\infty $, we can deduce that $H\left(x\right)=0$ has an unique solution on $\left(\right)$, on a point $\overline{x}>{x}_{2}$. In particular, ${f}_{a}$ has one unique fixed point in this interval.
- -
- We show that $H\left({x}_{1}\right)>0$, and so ${f}_{a}$ has no fixed point on $\left(\right)$.${H}^{\prime}\left({x}_{1}\right)=0$, then $\lambda {e}^{-\mu {x}_{1}}(\mu {x}_{1}-1)-1=0\u27fa\lambda {e}^{-\mu {x}_{1}}={\displaystyle \frac{1}{\mu {x}_{1}-1}}$. So: $\begin{array}{cc}H\left({x}_{1}\right)\hfill & =-\left(\right)open="("\; close=")">\lambda {x}_{1}{e}^{-\mu {x}_{1}}+{x}_{1}-1\hfill \end{array}$Furthermore, ${H}^{\prime}\left({\displaystyle \frac{1}{\mu}}\right)=\lambda {e}^{-1}(1-1)-1<0$, and ${H}^{\prime}$ is increasing on $\left(\right)$ (as ${H}^{\u2033}>0$ on this interval). As ${H}^{\prime}\left({x}_{1}\right)=0$, we can deduce that ${x}_{1}>{\displaystyle \frac{1}{\mu}}$. Consequently, $\mu {x}_{1}-1>0$. Let $j\left(x\right)=\mu {x}^{2}-\mu x+1$. Since $1+\lambda {e}^{-\mu {x}_{1}}>0$, thus $H\left({x}_{1}\right)$ is negative if and only if $j\left({x}_{1}\right)\u2a7e0$. We now investigate the sign of j.As $0<{\displaystyle \frac{1}{\mu}}<{x}_{1}<{\displaystyle \frac{2}{\mu}}\u2a7d{\displaystyle \frac{1}{2}}$, it is sufficient to study j on the interval $I=\left(\right)open="["\; close="]">0,{\displaystyle \frac{1}{2}}$.${j}^{\prime}\left(x\right)=\mu (2x-1)$, so j is strictly decreasing on I. The discriminant of the quadratic equation $j\left(x\right)=0$ being $\mu (\mu -4)\u2a7e0$, the latter has two solutions $\frac{\mu \pm \sqrt{\mu (\mu -4)}}{2\mu}$, which are equal when $\mu =4$ (i.e., when $a={\displaystyle \frac{1}{2}}$). Note that only $\frac{\mu -\sqrt{\mu (\mu -4)}}{2\mu}$ may belong to I, and that the latter is equal to $\frac{1}{2}}-\sqrt{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{\mu}}$ = $\frac{1}{2}}-\sqrt{{\displaystyle \frac{1}{4}}-a(1-a)$ = $\frac{1}{2}}-\sqrt{{\left(\right)}^{a}2$. We successively have$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}-\sqrt{{\left(\right)}^{a}2}\\ =& {\displaystyle \frac{{\displaystyle \frac{1}{4}}-{\left(\right)}^{a}2}{}{\displaystyle \frac{1}{2}}+\sqrt{{\left(\right)}^{a}2}}\end{array}$$If ${x}_{1}$ belongs to $\left(\right)$, then $j\left({x}_{1}\right)$ is positive and $H\left({x}_{1}\right)$ is negative.Let us show that ${x}_{1}\in \left(\right)open="["\; close="]">{\displaystyle \frac{\mu -\sqrt{\mu (\mu -4)}}{2\mu}},{\displaystyle \frac{1}{2}}$. As $a\u2a7d{\displaystyle \frac{1}{{e}^{2}+1}}$, we then have $a-{\displaystyle \frac{1}{2}}<0$, and so $\sqrt{{\left(\right)}^{a}2}$. Finally, $\frac{\mu -\sqrt{\mu (\mu -4)}}{2\mu}}=a$. As stated at the beginning of the proof, since ${x}_{1}$ is a root of H, ${x}_{1}$ is a fixed point for ${f}_{a}$ and thanks to (7).$a<{x}_{1}$, and so $j\left({x}_{1}\right)<0$, as j is decreasing with $j\left(a\right)=0$. To put it in a nutshell, $H\left({x}_{1}\right)>0$.

☐

**Proposition**

**1.**

**Proof.**

**Theorem**

**2.**

**Proof.**

- The unique fixed-point can be found as follows: start with an arbitrary element ${x}_{0}$ in $\mathbb{R}$ and define a sequence ${\left({x}_{n}\right)}_{n\in \mathbb{N}}$ by ${x}_{n}={f}_{a}\left({x}_{n-1}\right)$, then ${x}_{n}\u27f6\overline{x}\left(a\right)$.
- For $a\u2a7e1/2$, as $|{f}_{a}^{\prime}\left(x\right)|<a$, we can deduce that ${f}_{a}$ is Lipschitz continuous, with a Lipschitz constant equal to a. As a well-known consequence, the convergence of the aforementioned sequence is at least geometric, with a common ratio of a. For $a<1/2$ the same conclusion holds but for a constant $\gamma <1$.

## 3. The 1-Dimensional Situation ($\mathit{n}=\mathbf{1}$)

#### 3.1. The Discrete Dynamical System under Consideration

#### 3.2. A Fundamental Case: $M=\left(1\right)$

- exists and is unique,
- is such that $0\u2a7da<\overline{x}\left(a\right)<1$,
- the convergence speed is geometric, of ratio equal to a ($\gamma $).

#### 3.3. General 1-D Case: $M=\left(m\right)$

#### 3.3.1. Fixed Points of ${g}_{a,m}$ When $m>0$

- If $a\u2a7e{\displaystyle \frac{1}{{e}^{2}+1}}$, then $\lambda {e}^{-2}-1$ is negative and then ${H}^{\prime}\u2a7d0$ over $\mathbb{R}$, and after the computation of the limits of ${H}_{m}$ as x approaches $\pm \infty $, we can deduce the table of variations depicted in Figure 12 (and which is independent of m).Consequently, if $a\u2a7e{\displaystyle \frac{1}{{e}^{2}+1}}$, then ${H}_{m}\left(x\right)=0$ has one unique solution, i.e., ${g}_{a,m}$ has one unique fixed point.
- If $a\u2a7d{\displaystyle \frac{1}{{e}^{2}+1}}$, then we obtain a curve similar to the fundamental case for ${H}_{m}$, see Figure 13.As previously, we remark that ${H}_{m}^{\prime}\left({x}_{1}\right)=0=\lambda {e}^{-\mu m{x}_{1}}(\mu m{x}_{1}-1)-1$, so $\lambda {e}^{-\mu m{x}_{1}}={\displaystyle \frac{1}{\mu m{x}_{1}-1}}$.As a consequence,$$\begin{array}{cc}{H}_{m}\left({x}_{1}\right)\hfill & =-\left(\right)open="("\; close=")">{\displaystyle \frac{{x}_{1}}{\mu m{x}_{1}-1}}+{x}_{1}-1\hfill \end{array}$$Again as previously, ${H}_{m}^{\prime}\left({\displaystyle \frac{1}{m\mu}}\right)=-1<0$, ${H}_{m}^{\prime}$ is increasing over $\left(\right)$, and ${H}_{m}^{\prime}\left({x}_{1}\right)=0$, so ${x}_{1}>{\displaystyle \frac{1}{\mu m}}$. Consequently, since $m>0$, then $\mu m{x}_{1}-1>0$, and ${H}_{m}\left({x}_{1}\right)$ has the opposite sign of ${j}_{m}\left({x}_{1}\right)$ where ${j}_{m}\left(x\right)=\mu m{x}^{2}-\mu mx+1$.Let us study the quadratic polynomial ${j}_{m}\left(x\right)$ on $\mathbb{R}$. Its discriminant $\Delta \left({j}_{m}\right)$ is equal to ${\mu}^{2}{m}^{2}-4\mu m=\mu m(\mu m-4)$, and it has the sign table described in Figure 14.
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- If $m\in \left(\right)open="]"\; close="[">0,{\displaystyle \frac{4}{\mu}}$, then ${j}_{m}\left({x}_{1}\right)=\mu m{x}_{1}^{2}-\mu m{x}_{1}+1=\mu m{\left(\right)}^{{x}_{1}}2$. As $0<m<{\displaystyle \frac{4}{\mu}}$, we can conclude that ${j}_{m}\left({x}_{1}\right)>0$. So ${H}_{m}\left({x}_{1}\right)<0$. For the same reasons, ${H}_{m}\left({x}_{2}\right)$ is negative and ${H}_{m}$ has thus only one root, which belongs to $\left(\right)$. Thus ${g}_{m,a}$ has only one fixed point in this interval.
- -
- If $m={\displaystyle \frac{4}{\mu}}$, then$${H}_{\frac{4}{\mu}}\left({x}_{1}\right)={\displaystyle \frac{-{(2{x}_{1}-1)}^{2}}{\left(\right)open="("\; close=")">1+\lambda {e}^{-4{x}_{1}}}}$$
- -
- If $m>{\displaystyle \frac{4}{\mu}}$, then $\Delta \left({j}_{m}\right)>0$. So ${j}_{m}$ is positive outside its two roots and negative otherwise. $\frac{1}{2}}\pm \sqrt{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{m\mu}}$. The largest one is outside $I=\left(\right)open="["\; close="]">0,{\displaystyle \frac{1}{2}}$. Let us first focus on ${z}_{1}={\displaystyle \frac{1}{2}}-\sqrt{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{m\mu}}}={\displaystyle \frac{{\displaystyle \frac{1}{\mu m}}}{{\displaystyle \frac{1}{2}}+\sqrt{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{\mu m}}}}}$. Since $0\u2a7d{\displaystyle \frac{1}{\mu m}}\u2a7d{\displaystyle \frac{1}{4}}$ then $\frac{1}{2}}\u2a7d{\displaystyle \frac{1}{2}}+\sqrt{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{\mu m}}}\u2a7d1$ and thus $\frac{1}{\mu m}}\u2a7d{z}_{1}\u2a7d{\displaystyle \frac{2}{\mu m}$. From $\frac{1}{\mu m}}\u2a7d{z}_{1}\u2a7d{\displaystyle \frac{1}{2}$, one can thus deduce that $\frac{2}{\mu m}}\u2a7d{z}_{2}\u2a7d1$. Thus, $\frac{2}{\mu m}$ is in $]{z}_{1},{z}_{2}[$ and ${j}_{m}\left({\displaystyle \frac{2}{\mu m}}\right)<0$. In other words,$${H}_{m}\left({x}_{2}\right)>{H}_{m}\left({\displaystyle \frac{2}{\mu m}}\right)>0.$$
- ∗
- If $\mu m$ is large (i.e., $\frac{1}{\mu m}$ is close to 0), ${z}_{1}$ is close to $\frac{1}{\mu m}$. The left root of H, ${x}_{1}$ would be s.t. ${x}_{1}\ge {z}_{1}$ and ${j}_{m}\left({x}_{1}\right)<0$. In such a case ${H}_{m}\left({x}_{1}\right)>0$ and there is only one fixed point in $]{x}_{2},\infty [$.
- ∗
- If $\mu m$ is close to 4, ${z}_{1}$ is close to $\frac{2}{\mu m}$. The left root of H, ${x}_{1}$ would be s.t. ${x}_{1}\le {z}_{1}$ and ${j}_{m}\left({x}_{1}\right)>0$. In such a case ${H}_{m}\left({x}_{1}\right)<0$ and there is one fixed point in $\left(\right)$, one in $\left(\right)$ and one in $]{x}_{2},\infty [$.
- ∗
- if ${x}_{1}={z}_{1}$, ${j}_{m}\left({x}_{1}\right)=0$ and so ${H}_{m}\left({x}_{1}\right)=0$. There is one fixed point ${x}_{1}$ in $\left(\right)$ and one in $]{x}_{2},\infty [$.

#### 3.3.2. Fixed Points of ${g}_{a,m}$ When $m<0$

**Lemma**

**1.**

**Proof.**

**Theorem**

**3.**

**Proof.**

## 4. Conclusions and Future Work

## Author Contributions

## Conflicts of Interest

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**Figure 1.**Graph of the sigmoid function ${f}_{a}\left(x\right)={\displaystyle \frac{1}{1+\lambda {e}^{-\mu x}}}$ with $\lambda ={\displaystyle \frac{1-a}{a}}>0$, $\mu ={\displaystyle \frac{1}{a(1-a)}}$ and $a\in ]0,1[$.

**Figure 8.**Graph of numerical solution of equation $\overline{x}={\displaystyle \frac{1}{1+\lambda \left(a\right){e}^{-\mu \left(a\right)\overline{x}}}}$ when a is varying in $]0,1[$.

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**MDPI and ACS Style**

Guyeux, C.; Couchot, J.-F.; Le Rouzic, A.; Bahi, J.M.; Marangio, L.
Theoretical Study of the One Self-Regulating Gene in the Modified Wagner Model. *Mathematics* **2018**, *6*, 58.
https://doi.org/10.3390/math6040058

**AMA Style**

Guyeux C, Couchot J-F, Le Rouzic A, Bahi JM, Marangio L.
Theoretical Study of the One Self-Regulating Gene in the Modified Wagner Model. *Mathematics*. 2018; 6(4):58.
https://doi.org/10.3390/math6040058

**Chicago/Turabian Style**

Guyeux, Christophe, Jean-François Couchot, Arnaud Le Rouzic, Jacques M. Bahi, and Luigi Marangio.
2018. "Theoretical Study of the One Self-Regulating Gene in the Modified Wagner Model" *Mathematics* 6, no. 4: 58.
https://doi.org/10.3390/math6040058