Dynamics of a Model of Allelopathy and Bacteriocin with a Single Mutation

Abstract

In this paper we discuss a model of allelopathy and bacteriocin in the chemostat with a wild-type organism and a single mutant. Dynamical properties of this model show the basic competition between two microorganisms. A qualitative analysis about the boundary equilibrium, a state that microorganisms both vanish, is carried out. The existence and uniqueness of the interior equilibrium are proved by a technical reduction to the singularity of a matrix. Its dynamical properties are given by using the index theory of equilibria. We further discuss its bifurcations. Our results are demonstrated by numerical simulations.

1 Introduction

Antibiotic resistance among bacteria has become a worldwide public health threat (Neu [15], Levy and Marshall [13]). One of the limitations of using broad-spectrum antibiotics is that they kill almost any bacterial species not specifically resistant to the drug (Riley and Wertz [18]). Frequent use of these antibiotics result in an intensive selection pressure for the evolution of antibiotic resistance in both pathogen and commensal bacteria (Walker and Levy [23], Riley and Wertz [18]). Alternative methods of combating infection has been considered (Riley [17], Riley and Wertz [18]).

Allelopathy is the chemical inhibition of one species by another (Abell et al. [1]). Bacteriocin, a particular type of allelopathy, is a toxin produced by bacteria that is inhibit the growth of closely related species (Abell et al. [1]). Bacteriocins provide an alternative solution with their relatively narrow spectrum of killing activity. Special examples of bacteriocin productions involving more than two competing organisms, such as the bacteriocins of Escherichia coli and Klebsiella pneumonia, are given in Riley [17].

Various mathematical models have been proposed to examine the interaction between bacteriocin-producer and sensitive strains (Chao and Levin [5], Frank [7], Abell et al. [1], etc.). Recently, Abell et al. [1] proposed chemostat-type competition models with mutation and toxin production. Via numerical simulations, they showed how the coexistence of competitors depends upon the growth rates and toxin sensitivity.

A chemostat is a laboratory bio-reactor used to culture microorganisms (Smith and Waltman [20]). Competition for single and multiple resources, the evolution of resource acquisition, and competition among organisms have been investigated in ecology and biology using chemostats (Herbert et al. [8], Hsu et al. [9, 11], Monod [14], Novick and Szilard [16], Taylor and Williams [21]). Since 1950 (Monod [14], Novick and Szilard [16]), chemostat models have been studied. We refer to the monograph of Smith and Waltman [20], the surveys of Hsu and Waltman [10] and Ruan [19] and the references cited therein.

In this paper, we study the dynamics of the allelopathy model in the chemostat with a wild-type organism and a single mutation proposed by Abell et al. [1]. Let S(t) be the concentration of the nutrient at time t, x(t) be the density of the wild-type organism X at time t, and y(t) be the density of a mutant Y at time t, respectively. We consider the system

$$\{\begin{array}{c}S\prime (t)=D(S^0-S(t))-\frac{1}{\mathrm{\gamma }}{f}_1(S)x(t)-\frac{1}{\mathrm{\gamma }}{f}_2(S)y(t),\hfill \\ x\prime (t)=x(t)[(1-\alpha )f_1(S)-D]+\beta f_2(S)y(t),\hfill \\ y\prime (t)=y(t)[(1-\beta )f_2(S)-D]+\alpha f_1(S)x(t),\hfill \end{array}$$ (1.1)

where D is the dilution rate, γ is the yield constant, 0 ≤ α ≤ 1 is the rate at which X mutates, 0 ≤ β ≤ 1 is the rate at which Y reverts to X during reproduction, and f_i(S) = m_iS/(a_i + S) is MichaelisMenten-Monod function, in which m_i > 0 is the maximum growth rate and a_i > 0 is the Michaelis-Menten-Monod constant. It shows the competition between two microorganism X and Y. The single mutant of a parental species Y can revert to the wild-type X during reproduction.

After rescaling the variables and some parameters and using the same notation, the model can be written as follows (Abell et al. [1]):

$$\{\begin{array}{c}S\prime =1-S-f_1(s)x-f_2(S)y,\hfill \\ x\prime =x[(1-\alpha )f_1(S)-1]+\beta f_2(S)y,\hfill \\ y\prime =y[(1-\beta )f_2(S)-1]+\alpha f_1(S)x.\hfill \end{array}$$ (1.2)

Let Σ = 1 − S − x − y. Then, Σ′ = −Σ and system (1.2) can be replaced with

$$\{\begin{array}{c}\mathrm{\Sigma }\prime =-\mathrm{\Sigma },\hfill \\ x\prime =x[(1-\alpha )f_1(1-\mathrm{\Sigma }-x-y)-1]+\beta f_2(1-\mathrm{\Sigma }-x-y)y,\hfill \\ y\prime =y[(1-\beta )f_2(1-\mathrm{\Sigma }-x-y)-1]+\alpha f_1(1-\mathrm{\Sigma }-x-y)x.\hfill \end{array}$$ (1.3)

Since solutions of Σ′ = −Σ all tend to 0 as t → +∞, system (1.2) is asymptotic to the 2-dimensional system (Thieme [22], Smith and Waltman [20])

$$\{\begin{array}{c}x\prime =x[(1-\alpha )\frac{m_1(1-x-y)}{a_1+1-x-y}-1]+\beta \frac{m_2(1-x-y)}{a_2+1-x-y}y:=\tilde{f}(x,y),\hfill \\ y\prime =y[(1-\beta )\frac{m_2(1-x-y)}{a_2+1-x-y}-1]+\alpha \frac{m_1(1-x-y)}{a_1+1-x-y^{}}x:=\tilde{g}(x,y)\hfill \end{array}$$ (1.4)

in the region 𝒢 = {(x, y) : x ≥ 0, y ≥ 0, x + y ≤1}.

When α = β = 0, system (1.2) has been discussed in Hsu et al. [9]. Later, Hsu et al. [11] presented some results for the case β = 0 and 0 < α < 1, which is equivalent to the case α = 0 and 0 < β < 1. Therefore, we mainly consider the case 0 < α, β ≤ 1 in this paper.

Based on the study on system (1.2) in Abell et al. [1], we first discuss the properties of the boundary equilibrium E₀. We then consider the existence and uniqueness of the interior equilibrium E₁ by a singular matrix. Using the index of the equilibrium, we study the stability of E₁ by its Jacobian matrix.

The paper is organized as follows. In section 2 we consider the boundary equilibria. The existence of the interior equilibrium is addressed in section 3 and the properties of the interior equilibrium are given in section 4. Section 5 deals with possible bifurcations of the model. Numerical simulations and some remarks are given in section 6.

2 The Boundary Equilibria

To find the equalibria of system (1.4), we find zeros of the coupled equations f̃(x, y) = 0, g̃(x, y) = 0, i.e.,

$$\{\begin{array}{c}x[(1-\alpha )\frac{m_1(1-x-y)}{a_1+1-x-y}-1]+\beta \frac{m_2(1-x-y)}{a_2+1-x-y}y=0,\hfill \\ y[(1-\beta )\frac{m_2(1-x-y)}{a_2+1-x-y}-1]+\alpha \frac{m_1(1-x-y)}{a_1+1-x-y}x=0.\hfill \end{array}$$ (2.1)

We can see that E₀ = (0, 0) is the unique boundary equilibrium of system (1.4) when 0 < α ≤ 1 and 0 < β ≤ 1. In fact, when x = 0, (2.1) is equivalent to that

$$\begin{array}{ccc}\beta \frac{m_2(1-y)}{a_2+1-y}y=0\hfill & \text{and}\hfill & y[(1-\beta )\frac{m_2(1-y)}{a_2+1-y}-1]=0,\hfill \end{array}$$

implying that y = 0. Similarly, if y = 0, (2.1) exists if and only if x = 0. Furthermore, on x+y = 1, (2.1) is equivalent to that y = 0 and x = 0, which obviously do not exist on x + y = 1. So there is no other boundary equilibrium except E₀.

In order to determine the qualitative properties of E₀, we calculate the Jacobian of system (1.4) at E₀, i.e.,

$$J(E_0)=\left[\begin{array}{cc}\hfill \frac{(1-\alpha )m_1}{a_1+1}-1\hfill & \hfill \frac{\beta m_2}{a_2+1}\hfill \\ \hfill \frac{\alpha m_1}{a_1+1}\hfill & \hfill \frac{(1-\beta )m_2}{a_2+1}-1\hfill \end{array}\right]$$

and see that eigenvalues are zeros of the polynomial P(λ) ≔ λ² − T₀λ + D₀, where

$$\begin{array}{c}{T}_0:=\text{tr}J(E_0)=\frac{(1-\alpha )m_1}{a_1+1}+\frac{(1-\beta )m_2}{a_2+1}-2,\hfill \\ D_0:=\text{det}J(E_0)=\frac{(1-\alpha -\beta )m_1{m}_2}{(a_1+1)(a_2+1)}-\frac{(1-\alpha )m_1}{a_1+1}-\frac{(1-\beta )m_2}{a_2+1}+1.\hfill \end{array}$$

Since the discriminant

$${\mathrm{\Delta }}_0≔T_0^2-4D_0={(\frac{(1-\alpha )m_1}{a_1+1}-\frac{(1-\beta )m_2}{a_2+1})}^2+4\frac{\alpha \beta m_1{m}_2}{(a_1+1)(a_2+1)}>0,$$ (2.2)

It follows that E₀ is neither a focus nor of the center type.

Theorem 1 (i) E₀ is a saddle if D₀ < 0. Moreover, the trajectories starting from 𝒢 all go far away from E₀, (ii) E₀ is a stable node if D₀ > 0 and T₀ < 0. (iii) E₀ is an unstable node if D₀ > 0 and T₀ > 0. (iv) If D₀ = 0, then the equilibrium E₀ is a saddle-node and E₀ is stable in 𝒢.

Proof. In order to prove (i), we note that the stable manifold is tangent to the eigenvector (x_s, y_s) at E₀ and the unstable manifold is tangent to the eigenvector (x_u, y_u) at E₀, where x_s, y_s, x_u and y_u satisfy

$$\begin{array}{c}\frac{1}{2}[\frac{(1-\alpha )m_1}{a_1+1}-4-\frac{(1-\beta )m_2}{a_2+1}-\sqrt{{(\frac{(1-\alpha )m_1}{a_1+1}-\frac{(1-\beta )m_2}{a_2+1})}^2+4\frac{\alpha m_1\beta m_2}{(a_1+1)(a_2+1)}]}{x}_u+\frac{\beta m_2}{a_2+1}{y}_u=0,\hfill \\ \frac{1}{2}[\frac{(1-\alpha )m_1}{a_1+1}-4-\frac{(1-\beta )m_2}{a_2+1}+\sqrt{{(\frac{(1-\alpha )m_1}{a_1+1}-\frac{(1-\beta )m_2}{a_2+1})}^2+4\frac{\alpha m_1\beta m_2}{(a_1+1)(a_2+1)}]}{x}_s+\frac{\beta m_2}{a_2+1}{y}_s=0,\hfill \end{array}$$

which means that x_uy_u > 0 and x_sy_s < 0. Therefore, the intersection of the stable manifold and 𝒢 is empty, while the intersection of the stable manifold and 𝒢 is nonempty. This proves (i).

When D₀ > 0, we assure that T₀ ≠ 0; otherwise, Δ₀ = −4D₀ < 0, a contradiction to (2.2). Thus we only need to discuss the case of T₀ < 0 and the case of T₀ > 0. In the two cases the assumption D₀ > 0 implies results (ii) and (iii) respectively.

In order to discuss the case (iv), we need to determine the qualitative properties in the degenerate case. Applying a time-scaling transformation t = τ (a₁ + 1 − x − y)(a₂+1−x−y), we can reduce system (1.4) orbital-equivalently to the polynomial differential system

$$\{\begin{array}{c}x\prime =f(x,y),\hfill \\ y\prime =g(x,y),\hfill \end{array}$$ (2.3)

where

$$\begin{array}{c}\begin{array}{c}f(x,y)≔[(1-\alpha )m_1-1-a_1](a_2+1)x+\beta m_2(a_1+1)y+\{a_1+(a_2+2)[1-(1-\alpha )m_1]\}{x}^2+[(a_2+2)(1-(1-\alpha )m_1)+a_1-\beta m_2(a_1+2)]\mathit{\text{xy}}-\beta m_2(a_1+2)y^2+[(1-\alpha )m_1-1]x^3+[\beta m_2-2+2m_1(1-\alpha )]x^{2}y+[2\beta m_2-1+(1-\alpha )m_1]{\mathit{\text{xy}}}^2+\beta m_2{y}^3,\hfill \\ g(x,y)≔\alpha m_1(a_2+1)x+[(1-\beta )m_2-a_2-1](a_1+1)y-m_1\alpha (a_2+2)x^2+[(1-(1-\beta )m_2)(a_1+2)+a_2-\alpha m_1(a_2+2)]\mathit{\text{xy}}+\{a_2+(a_1+2)[1-(1-\beta )m_2]\}{y}^2+\alpha m_1{x}^3+[2m_1\alpha -1+m_2(1-\beta )]x^{2}y+[\alpha m_1-2+2(1-\beta )m_2]{\mathit{\text{xy}}}^2+[(1-\beta )m_2-1]y^3.\hfill \end{array}\hfill \end{array}$$

If D₀ = 0 then

$$\{(1-\alpha )m_1-(a_1+1)\}\{(1-\beta )m_2-(a_2+1)\}=\alpha \beta m_1{m}_2.$$ (2.4)

Assume that T₀ ≥ 0. Then

$$(a_2+1)[(1-\alpha )m_1-(a_1+1)]\ge -(a_1+1)[(1-\beta )m_2-(a_2+1)].$$

Substituting (2.4) in it, we get

$$(a_2+1){((1-\alpha )m_1-(a_1+1))}^2\le -\alpha \beta m_1{m}_2(a_1+1),$$

which holds if and only if its both sides vanish. This contradicts to the inequality βm₂αm₁(a₁ + 1) > 0. In fact, β > 0, α > 0, m₁ > 0, m₂ > 0, and a₁ > 0. It concludes that T₀ < 0. Therefore, only one of eigenvalues of the system vanishes but the other is equal to T₀ < 0.

Furthermore, with a transformation

$$\{\begin{array}{c}\begin{array}{c}u=-\alpha m_1(a_2+1)x+((1-\alpha )m_1-a_1-1)(a_2+1)y,\hfill \\ \upsilon =-\alpha m_1(a_2+1)x-((1-\beta )m_2-a_2-1)(a_1+1)y,\hfill \end{array}\hfill \end{array}$$

we change system (2.3) into the form

$$\{\begin{array}{cc}\dot{u}\hfill & =U(u,\upsilon ),\hfill \\ \dot{\upsilon }\hfill & =\mu \upsilon +V(u,\upsilon ),\hfill \end{array}$$ (2.5)

where μ = (1 − α)m₁(a₂ + 1) + (1 − β)m₂(a₁ + 1) − 2(a₁ + 1)(a₂ + 1) and U, V are O(|u|² + |υ|²) and shown in Appendix A. By the Implicit Function Theorem, there is a unique function υ = ϕ(u) such that ϕ(0) = 0 and V (u, ϕ(u)) = 0. Actually, we can solve from μυ + V (u, v) = 0 that

$$\varphi (u)=\alpha m_1\{(a_2+1)[(a_2+2)(1-(1-\alpha )m_1)+a_1]-[(1-\beta )m_2-(a_2+1)](a_1+1)(a_2+2)\}{u}^2/[(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-2(a_1+1)(a_2+1)+O({|u|}^3),$$

Substituting υ = ϕ (u) in the first equation of (2.5), we get

$$U(u,\varphi (u))=a_1\alpha m_1{(a_2+1)}^2{u}^2+O({|u|}^3),$$

which implies that E₀ is a saddle-node of system (2.5). i.e., E₀ is a saddle-node of system (1.4).

Moreover, since a₁αm₁(a₂ + 1)² > 0, the two hyperbolic sectors lie on the left but the parabolic sector lies on the right. So E₀ is stable in 𝒢.

3 Existence of Interior Equilibria

It is much more difficult to determine the interior equilibria because much more complex computation is needed. In this section we consider (1.4) in the interior of the region 𝒢, denoted by int𝒢.

Theorem 2 System (1.4) has at most one interior equilibrium. Furthermore, system (1.4) has exactly one interior equilibrium E₁ = (x₁, y₁) if and only if E₀ is unstable in 𝒢, where

$$\begin{array}{cc}{x}_1=\frac{\beta m_2{z}_1(a_1+z_1)(1-z_1)}{(a_1+z_1)\beta m_2{z}_1-(a_2+z_1)(K_1{z}_1-a_1)},\hfill & y_1=\frac{-(a_2+z_1)(K_1{z}_1-a_1)(1-z_1)}{(a_1+z_1)\beta m_2{z}_1-(a_2+z_1)(K_1{z}_1-a_1)},\hfill \\ z_1=\{\begin{array}{cc}\frac{-(a_2{K}_1+a_1{K}_2)+\sqrt{\mathrm{\Delta }}}{2(\alpha \beta m_1{m}_2-K_1{K}_2)},\hfill & \alpha \beta m_1{m}_2\ne K_1{K}_2,\hfill \\ \frac{a_1{a}_2}{a_2{K}_1+a_1{K}_2},\hfill & \alpha \beta m_1{m}_2=K_1{K}_2,\hfill \end{array}\hfill & \hfill \\ \mathrm{\Delta }={(a_2{K}_1-a_1{K}_2)}^2+4\alpha \beta m_1{m}_2{a}_1{a}_2,\hfill & \hfill \\ K_1=(1-\alpha )m_1-1,\hfill & K_2=(1-\beta )m_2-1.\hfill \end{array}$$

Proof. The system (2.1) of determining equilibria is equivalent to the system

$$\{\begin{array}{c}x[(1-\alpha )m_1(1-x-y)-a_1-1+x+y](a_2+1-x-y)+\beta m_2(1-x-y)(a_1+1-x-y)y=0,\hfill \\ y[(1-\beta )m_2(1-x-y)-a_2-1+x+y](a_1+1-x-y)+\alpha m_1(1-x-y)(a_2+1-x-y)x=0,\hfill \end{array}$$

which can be rewritten as

$$\left[\begin{array}{cc}\hfill (K_{1}z-a_1)(a_2+z)\hfill & \hfill \beta m_{2}z(a_1+z)\hfill \\ \hfill \alpha m_{1}z(a_2+z)\hfill & \hfill (K_{2}z-a_2)(a_1+z)\hfill \end{array}\right]\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right],$$ (3.1)

where z ≔ 1 − x − y. The system has an interior equilibrium if and only if there is a pair of nonzero x and y such that (3.1) holds. It follows that the determinant of coefficient matrix of (3.1) is equal to zero, i.e.,

$$\mathrm{\Upsilon }(z)≔(\alpha \beta m_1{m}_2-K_1{K}_2)z^2+(a_2{K}_1+a_1{K}_2)z-a_1{a}_2=0.$$ (3.2)

Since the region int𝒢 requires 0 < z < 1, we need to discuss positive roots of the quadratic equation (3.2).

Case 1: αβm₁m₂ − K₁K₂ > 0. In this case the quadratic equation (3.2) surely has a positive root, which must be

$$z_1=\frac{-(a_2{K}_1+a_1{K}_2)+\sqrt{\mathrm{\Delta }}}{2(\alpha \beta m_1{m}_2-K_1{K}_2)},$$ (3.3)

because

$$\mathrm{\Delta }≔{(a_2{K}_1-a_1{K}_2)}^2+4\alpha \beta m_1{m}_2{a}_1{a}_2>0$$ (3.4)

and −a₁a₂ < 0. Thus, the quadratic equation (3.2) has at most one zero in the interval (0, 1) and system (1.4) has at most one interior equilibrium. Furthermore, with z₁ given by (3.3) we can determine the coordinates x₁, y₁ of the corresponding equilibrium. Actually, from the first equation in (3.1), where z is replaced by z₁, we get

$$\{(a_2+z_1)(K_1{z}_1-a_1)\}{x}_1=-\{(a_1+z_1)\beta m_2{z}_1\}{y}_1.$$

Let

$$\xi =\frac{x_1}{(a_1+z)\beta m_2{z}_1}.$$ (3.5)

It implies that

$$y_1=-(a_2+z_1)(K_1{z}_1-a_1)\xi .$$ (3.6)

Noting that z₁ = 1 − x₁ − y₁, from (3.5) and (3.6) we obtain

$$1-z_1=[(a_1+z_1)\beta m_2{z}_1-(a_2+z_1)(K_1{z}_1-a_1)]\xi ,$$

which implies that

$$\xi =\frac{1-z_1}{(a_1+z_1)\beta m_2{z}_1-(a_2+z_1)(K_1{z}_1-a_1)}.$$

Thus, by (3.5) and (3.6) we uniquely obtain

$$\{\begin{array}{cc}{x}_1\hfill & =\frac{\beta m_2{z}_1(a_1+z_1)(1-z_1)}{(a_1+z_1)\beta m_2{z}_1-(a_2+z_1)(K_1{z}_1-a_1)},\hfill \\ y_1\hfill & =\frac{-(a_2+z_1)(K_1{z}_1-a_1)(1-z_1)}{(a_1+z_1)\beta m_2{z}_1-(a_2+z_1)(K_1{z}_1-a_1)}.\hfill \end{array}$$ (3.7)

Case 2: αβm₁m₂ − K₁K₂ ≤ 0. In this case we have K₁K₂ > 0, implying that a₂K₁ + a₁K₂, the coefficient of the first degree term in ϒ (z), does not vanish. Since equation (3.2) has no positive roots when a₂K₁ + a₁K₂ < 0, we only need to discuss in the case that a₂K₁ + a₁K₂ > 0. When a₂K₁ + a₁K₂ > 0, we obviously have

$$K_1>0,K_2>0.$$ (3.8)

In the circumstance that αβm₁m₂ − K₁K₂ = 0, equation (3.2) reduces to a linear equation and has a unique root z₁ = a₁a₂/(a₂K₁ + a₁K₂). It lies in (0, 1) if and only if a₂K₁ + a₁K₂ − a₁a₂ > 0. Such a z₁ determines the coordinates x₁, y₁ by the same (3.7) as in the discussion in Case 1. In the other circumstance, i.e., αβm₁m₂ − K₁K₂ < 0, the quadratic equation (3.2) has two positive roots

$$z_1=\frac{-(a_2{K}_1+a_1{K}_2)+\sqrt{\mathrm{\Delta }}}{2(\alpha \beta m_1{m}_2-K_1{K}_2)},z_2=\frac{-(a_2{K}_1+a_1{K}_2)-\sqrt{\mathrm{\Delta }}}{2(\alpha \beta m_1{m}_2-K_1{K}_2)},$$

where Δ are defined in (3.4). Clearly, z₁ < z₂. We claim that system (1.4) has at most one interior equilibrium, the same point E₁ = (x₁, y₁) as in Case 1, where x₁ and y₁ are given in (3.7). It suffices to prove that the equilibrium E₂ = (x₂, y₂), determined by z₂, locates outside the region int𝒢. For an indirect proof, assume that x₂ > 0, y₂ > 0. It is clear that 0 < z₂ ≔ 1 − (x₂+y₂) < 1. Since x₂, y₂ have the same expressions as x₁, y₁ in (3.7) where z₁ is replaced with z₂ and the numerator of x₂, i.e., βm₂z₂(a₁ +z₂)(1 − z₂), is positive, we see that the common denominator (a₁+z₂)βm₂z₂− (a₂+z₂)(K₁z₂− a₁) is also positive. It follows from the numerator of y₂, i.e., − (a₂ +z₂)(K₁z₂ −a₁)(1 − z₂), that K₁z₂ − a₁ < 0. It is equivalent to say

$$-K_1(a_2{K}_1+a_1{K}_2)-2a_1(\alpha \beta m_1{m}_2-K_1{K}_2)>K_1\sqrt{\mathrm{\Delta }}.$$

Taking the square of both sides, we have

$$4a_1\alpha \beta m_1{m}_2{K}_1(a_2{K}_1-a_1{K}_2)+4a_1^2{\alpha }^2{\beta }^2{m}_1^2{m}_2^2>4K_1^2\alpha \beta m_1{m}_2{a}_1{a}_2,$$

which implies that αβm₁m₂ − K₁K₂ > 0, a contradiction to the assumption that αβm₁m₂ − K₁K₂ < 0.

In summary, in all cases we considered above, system (1.4) has at most one equilibrium in the region int𝒢. Now we further prove that system (1.4) has at least one equilibrium in this open region when E₀ is unstable. Construct a closed curve Γ with line segments AB ≔ {(x, y) : x + y = 1, 0 ≤ x, y ≤ 1}, BB₀ ≔ {(x, y) : y = 0, 0 < ε ≤ x ≤ 1}, B₀A₀ ≔ {(x, y) : x + y = ε, 0 ≤ x, y ≤ ε} and A₀A ≔ {(x, y) : x = 0, 0 < ε ≤ y ≤ 1}. Restricted to AB, system (1.4) reduces to the form dx/dt = −x, dy/dt = −y, implying that both x and y decrease and trajectories staring from AB all enter𝒢. Restricted to A₀A, system (1.4) implies that dx/dt = βm₂(1 − y)y/(a₂ + 1 − y) > 0, i.e., x(t) increases, and therefore trajectories staring from A₀A all enter𝒢. Similarly, trajectories staring from BB₀ all enter𝒢. Moreover, all trajectories starting from B₀A₀ go away from E₀ as ε is chosen sufficiently small when E₀ is unstable. This proves that all trajectories starting from Γ enter the region surrounded by Γ. Therefore, the winding number of the vector field (1.4) along the curve Γ is equal to 1. As indicated in [2, p.313] and [24, Chpater 3], there is at least one equilibrium in the interior of the region bounded by the curve Γ. As a consequence, system (1.4) has exactly one interior equilibrium E₁ when E₀ is unstable.

Furthermore, we claim that no equilibrium exists in int𝒢 when E₀ is stable. In this case, by Theorem 1, we have T₀ < 0 and D₀ ≥ 0, which respectively implies that $(\frac{(1-\alpha )m_1}{a_1+1}-1)+(\frac{(1-\beta )m_2}{a_2+1}-1)<0\text{and}(\frac{(1-\alpha )m_1}{a_1+1}-1)(\frac{(1-\beta )m_2}{a_2+1}-1)\ge \frac{\alpha \beta m_1{m}_2}{(a_1+1)(a_2+1)}>0$. It means that $\frac{(1-\alpha )m_1}{a_1+1}<1\text{and}\frac{(1-\beta )m_2}{a_2+1}<1$. Hence,

$$K_1<a_1,K_2<a_2,$$ (3.9)

$$\mathrm{\Upsilon }(1)=\alpha \beta m_1{m}_2-K_1{K}_2+a_1{K}_2+a_2{K}_1-a_1{a}_2\le 0,$$ (3.10)

where ϒ is defined in (3.2). In the case that αβm₁m₂ − K₁K₂ > 0 the quadratic function ϒ is convex and has no zeros in (0, 1) by (3.10) and the fact

$$\mathrm{\Upsilon }(0)=-a_1{a}_2<0.$$ (3.11)

In the case that αβm₁m₂ − K₁K₂ = 0 the function ϒ, linking two non-positive ϒ(0) and ϒ(1) linearly, also has no zeros in (0, 1). In the case that αβm₁m₂ − K₁K₂ < 0 the quadratic function ϒ is concave. If it has a positive zero then, as discussed in the above Case 2, (3.8) holds, i.e., K₁ > 0 and K₂ > 0. By (3.9), 0 < K₁K₂ < a₁a₂. It follows that

$$\sigma ≔\frac{-a_1{a}_2}{\alpha \beta m_1{m}_2-K_1{K}_2}>1.$$ (3.12)

If ϒ(1) < 0 then, by (3.11) and (3.4), the function ϒ has either no or two zeros in (0, 1), but (3.12), where we note that σ is equal to the product of two zeros, implies that at least one zero of ϒ lies outside [−1, 1]. This proves that ϒ has no zeros in (0, 1). If ϒ(1) = 0 then one zero of ϒ is z₁ = 1 and the other zero is z₂ = σ > 1 by (3.12). It also implies that ϒ has no zeros in (0, 1). Consequently, system (1.4) has no interior equilibria.

Summarizing the above discussion we see that system (1.4) has exactly one interior equilibrium if and only if E₀ is unstable.

The proof of Theorem 2 also gives a computable condition for the existence of the interior equilibrium, i.e., system (1.4) has exactly one interior equilibrium if and only if

$$0<z_1<1,K_1{z}_1-a_1<0,$$ (3.13)

where z₁,K₁,K₂ are defined in Theorem 2.

4 Properties of the Interior Equilibrium

By Theorem 2, we only need to discuss the unique interior equilibrium E₁ in the region int𝒢 when the boundary equilibrium E₀ is unstable.

Theorem 3 E₁ is asymptotically stable if it exists. Moreover, E₁ is a stable node if

$${\delta }_1≔m_2{z}_1^2{(a_1+z_1)}^4{(a_2+z_1)}^2{\beta }^2+c_1\beta +c_0\ge 0,$$ (4.1)

where

$$\begin{array}{cc}{c}_1=\hfill & 2{(a_1+z_1)}^2(a_2+z_1)m_2{z}_1[m_1{z}_1(1+\alpha )(a_1+z_1){(a_2+z_1)}^2-z_1{(a_1+z_1)}^2(a_2+z_1)-m_1{a}_1{x}_1{(a_2+z_1)}^2-2m_1{a}_2{y}_1{z}_1(a_1+z_1)(a_2+z_1)+m_2{a}_2{y}_1{(a_1+z_1)}^2],\hfill \\ c_2=\hfill & \begin{array}{c}{(1-\alpha )}^2{m}_1{z}_1^2{(a_1+z_1)}^2{(a_2+z_1)}^4-2(1-\alpha )m_1{m}_2{z}_1^2{(a_1+z_1)}^3{(a_2+z_1)}^3+m_2{z}_1^2{(a_1+z_1)}^4{(a_2+z_1)}^2-2(1-\alpha )m_1^2{a}_1{x}_1{z}_1(a_1+z_1){(a_2+z_1)}^2+2(1-2\alpha )m_1{m}_2{a}_1{x}_1{z}_1{(a_1+z_1)}^2{(a_2+z_1)}^3+2(1-\alpha )m_1{m}_2{a}_2{y}_1{z}_1{(a_1+z_1)}^3{(a_2+z_1)}^2-2m_2^2{a}_2{y}_1{z}_1{(a_1+z_1)}^4(a_2+z_1)+m_1^2{a}_1^2{x}_1^2{(a_2+z_1)}^4+2m_1{m}_2{a}_1{a}_2{x}_1{x}_2{(a_1+z_1)}^2{(a_2+z_1)}^2+m_2^2{a}_2^2{y}_1^2{(a_1+z_1)}^4.\hfill \end{array}\hfill \end{array}$$

and x₁, y₁, z₁ are defined by parameters α, β, a₁, a₂, m₁, m₂ as in (3.7) and (3.3).

Proof. Qualitative properties of E₁ are determined by the signs of the trace T₁, the determinant D₁ and the discriminant Δ₁ of the Jacobian matrix of system (1.4) at E₁. Simple computation shows that

$$\begin{array}{ccc}{T}_1\hfill & =\hfill & \frac{(K_1{z}_1-a_1)(a_1+z_1)-m_1{a}_1{x}_1}{{(a_1+z_1)}^2}+\frac{(K_2{z}_1-a_2)(a_2+z_1)-m_2{a}_2{y}_1}{{(a_2+z_1)}^2},\hfill \\ D_1\hfill & =\hfill & 1-\frac{(1-\alpha )m_1{z}_1}{a_1+z_1}-\frac{(1-\beta )m_2{z}_1}{a_2+z_1}+\frac{(1-\alpha -\beta )m_1{m}_2{z}_1^2}{(a_1+z_1)(a_2+z_1)}+\frac{m_1{a}_1{x}_1}{{(a_1+z_1)}^2}+\frac{m_2{a}_2{y}_1}{{(a_2+z_1)}^2}-\frac{(1-\alpha -\beta )m_1{m}_2{a}_2{z}_1{y}_1}{(a_1+z_1){(a_2+z_1)}^2}-\frac{(1-\alpha -\beta )m_1{m}_2{a}_1{z}_1{x}_1}{{(a_1+z_1)}^2(a_2+z_1)},\hfill \\ {\Delta }_1\hfill & =\hfill & \frac{{(1-\alpha )}^2{m}_1{z}_1^2}{{(a_1+z_1)}^2}+\frac{2m_1{m}_2{z}_1^2(\alpha (1+\beta )-(1-\beta ))}{(a_1+z_1)(a_2+z_1)}+\frac{{(1-\beta )}^2{m}_2{z}_1^2}{{(a_2+z_1)}^2}-\frac{2(1-\alpha )m_1^2{a}_1{x}_1{z}_1}{{(a_1+z_1)}^3}+\frac{2m_1{m}_2{a}_1{x}_1{z}_1(1-2\alpha -\beta )}{{(a_1+z_1)}^2(a_2+z_1)}+\frac{2m_1{m}_2{a}_2{y}_1{z}_1(1-\alpha -2\beta )}{(a_1+z_1){(a_2+z_1)}^2}-\frac{2(1-\beta )m_2^2{a}_2{y}_1{z}_1}{{(a_2+z_1)}^3}+\frac{m_1^2{a}_1^2{x}_1^2}{{(a_1+z_1)}^4}+\frac{2m_1{m}_2{a}_1{a}_2{x}_1{x}_2}{{(a_1+z_1)}^2{(a_2+z_1)}^2}+\frac{m_2^2{a}_2^2{y}_1^2}{{(a_2+z_1)}^4}\hfill \end{array}$$

From the first equation in (3.1) we see that K₂z₁ −a₂ < −2m₁z₁(a₂ +z₁)x₁/{(a₁ + z₁)y₁} < 0 since a₁, a₂,m₁, K₂, x₁, y₁, z₁ are all positive. Similarly K₂z₁ − a₂ < 0 by the second equation in (3.1). It follows that T₁ < 0. We further claim that

$$D_1\ge 0.$$

In fact, if D₁ < 0, i.e., E₁ is a saddle, then the index of E₁ is equal to −1. Consider the closed curve Γ composed in the proof of Theorem 2. The fact that all trajectories starting from Γ enter the region surrounded by Γ implies that the winding number of the vector field (1.4) along the curve Γ is equal to 1. This makes a contradiction to the Theorem on the sum of indices ([2, p.313], [24, Chpater 3]) because of the uniqueness of the interior equilibrium (given in Theorem 2).

In what follows we discuss the case D₁ > 0 and the case D₁ = 0 separately.

In the case D₁ > 0, E₁ is either a stable node or a stable focus because T₁ < 0. Furthermore, E₁ is a node if the inequality Δ₁ ≥ 0 holds. This inequality can be simplified as the condition (4.1) because a₁ + z₁ > 0 and a₂ + z₁ > 0.

In the case D₁ = 0 the equilibrium E₁ is degenerate. Clearly, ${\mathrm{\Delta }}_1=T_1^2>0$ in this case, i.e., condition (4.1) is satisfied naturally. Since T₁ < 0, the two eigenvalues at E₁ are λ₁ = 0 and λ₂ = T₁ < 0. This enables us to diagonalize the linear part of the cubic system (2.3) so that the system is transformed into the form

$$\{\begin{array}{c}\frac{\mathit{\text{dx}}}{\mathit{\text{dt}}}=G_2(x,y)+G_3(x,y),\hfill \\ \frac{\mathit{\text{dy}}}{\mathit{\text{dt}}}=-y+H_2(x,y)+H_3(x,y),\hfill \end{array}$$ (4.2)

where G_j, H_j are homogeneous polynomials of degree j (j = 2, 3) and the interior equilibrium E₁ is translated to the origin. By Theorem 7.1 in [24, Chapter 2], the origin of (4.2) is either a stable node or a saddle or a saddle-node. However, by Bendixson’s formula (see [3], [24, Chapter 3])

$$J=1+\frac{e-h}{2},$$

where J is the index of the equilibrium, h is the number of hyperbolic sectors near the equilibrium and e is the number of elliptic sectors, we get either J = −1 when E₁ is a saddle or J = 0 when E₁ is a saddle-node, both of which contradict to the Theorem on the Sum of Indices ([2, p.313], [24, Chapter 3]) because the winding number of the vector field (1.4) along the curve Γ is equal to 1, as proved in the paragraph above (3.10). This implies that E₁ is a stable node.

5 Bifurcations

It is indicated in sections 3 and 4 that system (1.4) has either exact one equilibrium in 𝒢 when D₀ > 0 and T₀ < 0 or exact two equilibria in 𝒢 in the other case. The following theorem displays the mechanism for the new one to arise. Let T₀₁ ≔ T₀(a₁ + 1)(a₂ + 1), D₀₁ ≔ D₀(a₁ + 1)²(a₂ + 1)², and

$$\begin{array}{cc}{\mu }_0\hfill & ≔[3(1-\alpha )m_1(a_2+1)-4(a_1+1)(a_2+1)+(1-\beta )m_2(a_1+1)-\sqrt{T_{01}^2-4D_{01}}][2(a_1+a_2+2)-(1-\beta )m_2(3a_1+5)-3(1-\alpha )m_1(a_2+1)+4(a_1+1)(a_2+1)+\sqrt{T_{01}^2-4D_{01}}]+4\alpha \beta m_1{m}_2(a_1+2)(a_2+1)-2[a_1+(a_2+2)(1-(1-\alpha )m_1)][3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_2+1)-4(a_1+1)(a_2+1)+\sqrt{T_{01}^2-4D_{01}}],\hfill \\ {\nu }_{10}\hfill & ≔(3-\alpha )m_1(a_2+1)-4(1+a_1)(1+a_2)+(1-\beta )m_2(a_1+1)+\sqrt{T_{01}^2-4D_{01}}.\hfill \end{array}$$

where T₀ and D₀ are defined as in Theorem 1.

Theorem 4 If ν₁₀μ₀ ≠ 0, system (1.4) experiences a transcritical bifurcation at E₀ when D₀ = 0. More concretely, for sufficiently small D₀, a stable equilibrium appears in the first quadrant when ν₁₀μ₀D₀ > 0 and an unstable equilibrium appears in the third quadrant when ν₁₀μ₀D₀ < 0.

Proof. As shown in Theorem 1, E₀ is a saddle-node as D₀ = 0. Applying the transformation

$$\{\begin{array}{c}u=-\alpha m_1(a_2+1)x+\{[(1-\alpha )m_1-a_1-1](a_2+1)-\frac{1}{2}(T_{01}+\sqrt{T_{01}^2+4D_{01}})\}y,\hfill \\ \upsilon =-\alpha m_1(a_2+1)x+\{[(1-\alpha )m_1-a_1-1](a_2+1)-\frac{1}{2}(T_{01}-\sqrt{T_{01}^2-4D_{01}})\}y\hfill \end{array}$$ (5.1)

to diagonalize the linear part of system of (2.3), we change system (2.3) into the form

$$\left[\begin{array}{c}\hfill u\prime \hfill \\ \hfill \upsilon \prime \hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \frac{1}{2}(T_{01}+\sqrt{T_{01}^2-4D_{01})}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{2}(T_{01}-\sqrt{T_{01}^2-4D_{01})}\hfill \end{array}\right]\left[\begin{array}{c}\hfill u\hfill \\ \hfill \upsilon \hfill \end{array}\right]+\left[\begin{array}{c}\hfill q_1(u,\upsilon )\hfill \\ q_2(u,\upsilon )\hfill \end{array}\right],$$ (5.2)

where q₁(u, υ) and q₂(u,υ) are composed of those terms of degree 2 or 3. With a rescaling $d\tau =(T_{01}-\sqrt{T_{01}^2-4D_{01})}\mathit{\text{ds}}$, the system can be reduced to the following suspended system of (5.2)

$$\{\begin{array}{c}u\prime =2D_{01}u+(T_{01}-\sqrt{T_{01}^2-4D_{01}})q_1(u,\upsilon ),\hfill \\ \upsilon \prime =\frac{1}{2}{(T_{01}-\sqrt{T_{01}^2-4D_{01}})}^2\upsilon +(T_{01}-\sqrt{T_{01}^2-4D_{01}})q_2(u,\upsilon ),\hfill \\ D_{01}^{\prime }=0.\hfill \end{array}$$ (5.3)

By the center manifold theory (see [4]), as D₀ = 0 system (5.3) has a 2-dimensional center manifold 𝒲^c : υ = W(u,D₀₁) near E₀. In order to obtain the second order approximation of function W, let

$$W(u,D_{01})≔\phi (u,D_{01})+O(|u,D_{01}{|}^3),$$ (5.4)

where φ(u,D₀₁) ≔ ν₁u² + ν₂uυ + ν₃υ², and let

$$(𝒩\phi )(u,D_{01})≔\phi \prime (u,D_{01})\{2D_{01}u+(T_{01}-\sqrt{T_{01}^2-4D_{01}})q_1(u,\phi (u,D_{01}))\}-\frac{1}{2}{(T_{01}-\sqrt{T_{01}^2-4D_{01}})}^2\phi (u,D_{01})-(T_{01}-\sqrt{T_{01}^2-4D_{01}})q_2(u,\phi (u,D_{01})).$$

By Theorem 3 in [4, Chapter 1], from the requirement (𝒩φ)(u,D₀₁) = O(|u,D₀₁|³) we can solve ν₂ = ν₃ = 0 and

$${\nu }_1=\left(\frac{{\nu }_{10}}{4\alpha m_1(a_2+1)(T_{01}^2-4D_{01})}\right)\{[3(1-\alpha )m_1(a_2+1)-4(a_1+1)(a_2+1)+(1-\beta )m_2(a_1+1)+\sqrt{T_{01}^2-4D_{01}}][2(a_1+a_2+2)-(1-\beta )m_2(3a_1+5)-3(1-\alpha )m_1(a_2+1)+4(a_1+1)(a_2+1)+\sqrt{T_{01}^2-4D_{01}}]+4\alpha \beta m_1{m}_2(a_1+2)(a_2+1)-2[a_1+(a_2+2)(1-(1-\alpha )m_1)][3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_2+1)-4(a_1+1)(a_2+1)+\sqrt{T_{01}^2-4D_{01}}]\}.$$

Thus we can substitute (5.4) in (5.3) and obtain the equation

$$u\prime =2D_{01}u+\mu u^2+{\mu }_1{u}^3+O{(|u,D_{01}|)}^4,$$ (5.5)

the restricted system on 𝒲^c, where

$$\mu ≔\frac{{\nu }_{10}{\mu }_0}{4\alpha m_1(a_2+1)(T_{01}^2-4D_{01})},$$ (5.6)

μ₀ is defined in the beginning of this section and μ₁ is expressed in Appendix A. Clearly, the expression (5.5) shows that a transcritical bifurcation ([6, p.201]) occurs at E₀ as D₀₁ varies through the bifurcation value D₀₁ = 0 when μ ≠ 0. Since D₀₁ has the same zeros with D₀, the transcritical bifurcation occurs at E₀ as D₀ varies through the bifurcation value D₀ = 0. More concretely, when ν₁₀μ₀D₀ < 0 the origin O is stable and the other equilibrium A₋₁ appears on the negative u-axis but is unstable; when D₀ = 0 the two equilibria coincide at O; when ν₁₀μ₀D₀ > 0 the origin O remains an equilibrium but is unstable while a stable equilibrium A₁ arises on the positive u-axis.

Finally, the transformation (5.1) gives the correspondence between A₁, A₋₁ and the equilibria B₁, B₋₁ of system (2.3), which lie in the first quadrant and the third one respectively. Note that the third quadrant is of no practical interests.

This proof shows that equilibrium B₁ (actually the same as the interior equilibrium E₁) arises from a transcritical bifurcation as D₀₁ varies through the bifurcation value D₀ = 0. This explains how E₁ is produced. We ignore B₋₁ since it does not appear in the first quadrant.

In Theorem 4, the required nondegenerate condition ν₁₀μ₀ ≠ 0 appears in the expression (5.6) of μ. If this condition is violated, i.e., ν₁₀μ₀ = 0, system (5.5) turns into the form

$$u\prime =2D_{01}u+{\mu }_1{u}^3+O{(|u,D_{01}|)}^4$$ (5.7)

and the coefficient μ₁ becomes a decisive quantity. Unfortunately, μ₁ = 0. If μ₁ ≠ 0, a pitchfork bifurcation ([6, p.201]) occurs at E₀ as D₀₁ passing through the bifurcation value D₀₁ = 0. It means that the origin is the unique equilibrium as D₀₁μ₁ > 0, implying as known in section 3 that the function ϒ defined in (3.2) has no real zeros, i.e.,

$$\begin{array}{cc}\alpha \beta m_1{m}_2>K_1{K}_2,\hfill & \mathrm{\Delta }={(a_2{K}_1-a_1{K}_2)}^2+4\alpha \beta m_1{m}_2{a}_1{a}_2<0.\hfill \end{array}$$ (5.8)

This is an obvious contradiction because all parameters are positive. For the same reason we assure that the first nonzero coefficient in the expansion (5.7) appears in an even term. The corresponding bifurcations can be discussed similarly to the proof of Theorem 4.

Note that no bifurcations occur at the interior equilibrium E₁ although its degeneracy for D₁ = 0 is shown in the proof of Theorem 3. In fact, when D₁ = 0 the asymptotically stable equilibrium E₁ does not coincide with E₀; otherwise, their coincidence gives a saddle-node, which contradicts to the stability of E₁ as shown in Theorem 4. Moreover, ignoring E₀, there also does not occur a bifurcation at E₁, which lies in the interior of 𝒢 for D₁ ≥ 0; otherwise, the only possibility is the pitchfork bifurcation as shown in Theorem 4, which produces three equilibria in the first quadrant as D₁ > 0 and makes a contradiction to the uniqueness of interior equilibria (given in Theorem 2).

6 Numerical Simulations and Remarks

We simulate orbits of the system (1.4) to demonstrate our Theorems 2 and 3. We consider the following cases (S1), (S2) and (S3):

	Parameter values (m1, a1, α), (m2, a2, β)	Equilibrium Ei	Eigenvalues λ1, λ2 of J(Ei)	Properties
(S1)	(5, 0.7, 0.1), (1, 0.4, 0.1)	E0 = (0, 0)	λ1 = 1.657, λ2 = −0.368	saddle
		E1 = (0.692, 0.109)	λ1 = −3.117, λ2 = −0.712	stable node
(S2)	(3.1, 0.7, 0.2), (2.4, 0.4, 0.1)	E0 = (0, 0)	λ1 = 0.754, λ2 = 0.247	unstable node
		E1 = (0.200, 0.499)	λ1 = −0.327, λ2 = −1.418	stable node
(S3)	(0.2, 0.4, 0.2), (0.3, 1, 0.1)	E0 = (0, 0)	λ1 = −0.852, λ2 = −0.899	stable node

The phase portraits in the cases (S1), (S2) and (S3) are plotted separately in Figure 1, showing that the wild-type organism X and the mutant Y coexist in both (S1) and (S2), E₀ is a saddle in case (S1) but an unstable node in case (S2), and both X and Y finally go to extinction in case (S3) while E₀ is a stable node.

Figure 1.

Phase portraits in cases (S1), (S2) and (S3).

When a₁ > 0, a₂ > 0 are small enough and m₁ > 0, m₂ > 0 are large enough, we see that D₀ < 0 as α is small and D₀ > 0 as α is large and β is small, implying that the condition D₀ = 0 in Theorems 1 and 4 is reasonable. However, it is still difficult to prove either D₁ ≠ 0 or D₁ = 0 for some parameters. Fixing a₁ = 0.7, a₂ = 0.4, m₁ = 5 and m₂ = 1 for instance, consider the function

$$D_1(\beta ):=1-\frac{5(1-\alpha )z_1}{0.7+z_1}-\frac{(1-\beta )z_1}{0.4+z_1}+\frac{5(1-\alpha -\beta )z_1^2}{(0.7+z_1)(0.4+z_1)}+\frac{3.5x_1}{{(0.7+z_1)}^2}+\frac{0.4y_1}{{(0.4+z_1)}^2}-\frac{2(1-\alpha -\beta )z_1{y}_1}{(0.7+z_1){(0.4+z_1)}^2}-\frac{3.5(1-\alpha -\beta )z_1{x}_1}{{(0.7+z_1)}^2(0.4+z_1)},$$

parametrized by α, where x₁, y₁, z₁ are shown in Appendix B. It is shown in Figure 2 that D₁(β) has no zeros when α = 0.1, 0.4, 0.9. It seems natural to assert that, in contrast to the statement of Theorem 3, E₁ is a stable focus if δ₁ < 0, but it is hard to find a choice of parameters for the simulation of focus at E₁. It suggests a conjecture that E₁ is a node.

Figure 2.

The graph of D₁(β) when α = 0.1 (left), 0.4 (middle), and 0.9 (right), separately.

Furthermore, with the same parameters a₁ = 0.7, a₂ = 0.4, m₁ = 5 and m₂ = 1, we calculate the function δ₁(β) = ς₂β² + ς₁β + ς₀, where ς₀, ς₁, ς₂ are shown in Appendix B. Its graphs, plotted in Figure 3 for the three choices α = 0.1, 0.4 and 0.9, show that δ₁ has no zeros in (0, 1). However, it is still difficult to rule out the possibility of the inequality δ₁ < 0.

Figure 3.

The graph of δ₁(β) when α = 0.1 (left), 0.4 (middle), and 0.9 (right), separately.

We have studied the dynamics of an allelopathy model with a single mutant. It was shown that the model has a unique stable interior equilibrium under certain conditions, which differs from the typical dynamics of chemostat models. It will be interesting to study the dynamics of the general model (11) in Abell et al. [1] with multiple mutants. We leave this for future consideration.

Appendix A: Expressions of U(u, υ), V (u,υ)

$$\begin{array}{cc}U(u,\upsilon ):\hfill & =a_1\alpha m_1{(a_2+1)}^2{u}^2+\{-\alpha m_1(a_2+1)[(1-(1-\alpha )m_1)(a_2+2)+a_1-\beta m_2(a_1+2)]+[(1-\alpha )m_1-1-a_1](a_2+1)[-\alpha m_1(a_2+2)+(1-(1-\beta )m_2)(a_1+2)+a_2]\}u\upsilon +(a_2+1)\{\alpha \beta m_1{m}_2(a_1+2)+((1-\alpha )m_1-1-a_1)[(1-(1-\beta )m_2)(a_1+2)+a_2]\}{\upsilon }^2-\alpha a_1{m}_1(a_2+1)u^3+(a_2+1)\{-\alpha m_1(-2+2(1-\alpha )m_1+\beta m_2)+((1-\alpha )m_1-1-a_1)[(1-\beta )m_2-1+2\alpha m_1]\}{u}^2\upsilon +(a_2+1)\{\alpha \beta m_1{m}_2(a_1+2))+((1-\alpha )m_1-1-a_1)[(1-(1-\beta )m_2)(a_1+2)+a_2]\}u{\upsilon }^2+\{-\alpha m_1(a_2+1)\beta m_2+((1-\alpha )m_1-1-a_1)(a_2+1)((1-\beta )m_2-1)\}{\upsilon }^3,\hfill \\ V(u,\upsilon ):\hfill & =\alpha m_1\{-(a_2+1)[(1-(1-\alpha )m_1)(a_2+2)+a_1]+[(1-\beta )m_2-a_2-1](a_1+1)(a_2+2)\}{u}^2+\{-\alpha m_1(a_2+1)[(1-(1-\alpha )m_1)(a_2+2)+a_1-\beta (a_1+2)]-[(1-\beta )m_2-a_2-1](a_1+1)(-\alpha m_1(a_2+2)+(1-(1-\beta )m_2)(a_1+2)+a_2)\}u\upsilon +\{\alpha \beta m_1{m}_2(a_2+1)(a_1+2)-[(1-\beta )m_2-a_2-1](a_1+1)[(1-(1-\beta )m_2)(a_1+2)+a_2]\}{\upsilon }^2+\{-\alpha m_1(a_2+1)(-1+(1-\alpha )m_1)-[(1-\beta )m_2-a_2-1](a_1+1)\alpha m_1\}{u}^3+\{-\alpha m_1(a_2+1)(-2+2(1-\alpha )m_1+\beta m_2)-[(1-\beta )m_2-a_2-1](a_1+1)[(1-\beta )m_2-1+2\alpha m_1]\}{u}^2\upsilon +\{-\alpha m_1(a_2+1)(-1+(1-\alpha )m_1+2\beta m_2)-[(1-\beta )m_2-a_2-1](a_1+1)(2(1-\beta )m_2-2+\alpha m_1)\}u{\upsilon }^2+\{-\alpha m_1(a_2+1)\beta m_2-[(1-\beta )m_2-a_2-1](a_1+1)[(1-\beta )m_2-1]\}{\upsilon }^3.\hfill \\ {\mu }_1\hfill & =\frac{1}{\alpha m_1{(a_2+1)}^2[{((1-\alpha )m_1(a_2+1)-(1-\beta )m_2(a_1+1))}^2+4\alpha \beta m_1{m}_2(a_1+1)(a_2+1)]}·\{\frac{((1-\beta )m_2-a_2-1)(a_1+1)}{2((1-\alpha )m_1-a_1-1-\alpha m_1)}\{-2{\alpha }^2\beta m_1^2{m}_2(a_1+2){(a_2+1)}^2+\frac{1}{2}(a_1+(a_2+2)(1-(1-\alpha )m_1))(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)-\mathrm{\Theta })(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta })+\alpha m_1(a_2+1)((1-(1-\alpha )m_1)(a_2+2)+a_1-\beta m_2(a_1+2))(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)+4(a_1+1)(a_2+1))-\frac{1}{2}(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)-\mathrm{\Theta })[-2(a_2+(1-(1-\beta )m_2)(a_1+2))\alpha m_1(a_2+1)+\frac{1}{2}(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)-\mathrm{\Theta })(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta })-\frac{1}{2}((1-(1-\beta )m_2)(a_1+2)+a_2-\alpha m_1(a_2+2))(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)+4(a_1+1)(a_2+1))]\}+\frac{1}{\alpha m_1\mathrm{\Theta }}[\frac{1}{8}((1-\alpha )m_1-1){(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta })}^3+\frac{1}{4}\alpha m_1(a_2+1)(\beta m_2-2+2(1-\alpha )m_1){(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta })}^2+\frac{1}{2}{\alpha }^2{m}_1^2{(a_2+1)}^2(2\beta m_2-1+(1-\alpha )m_1)(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta })+{\alpha }^3\beta m_1^3{m}_2{(a_2+1)}^3-\frac{1}{2(a_2+1)}(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)-\mathrm{\Theta })(\frac{1}{8}{(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta })}^3+\frac{1}{4}(2\alpha m_1-1+(1-\beta )m_2)(a_2+1){(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta })}^2+\frac{1}{2}\alpha m_1(\alpha m_1-2+2(1-\beta )m_2)(3(1-\alpha )m_1(a_2+1)+(1-\beta )m_2(a_1+1)-4(1+a_1)(1+a_2)+\mathrm{\Theta }){(a_2+1)}^2+((1-\beta )m_2-1){\alpha }^2{m}_1^2{(a_2+1)}^3)]\},\hfill \end{array}$$

where $\mathrm{\Theta }≔\sqrt{{((1-\alpha )m_1(a_2+1)-(1-\beta )m_2(a_1+1))}^2+4\alpha \beta m_1{m}_2(a_1+1)(a_2+1)}$.

Appendix B: Expressions of x₁, y₁, z₁, ς₀, ς₁, ς ₂

$$\begin{array}{cc}\hfill z_1:& =\frac{1}{\beta }(-0.2+0.25\alpha +0.0875\beta +0.0125\varpi ),\hfill \\ \hfill x_1:& =-0.00625(20\alpha -73\beta +\varpi -16)(20\alpha +63\beta +\varpi -16)(20\alpha +7\beta +\varpi -16)/(64\varpi -1024+3840\alpha -160\alpha \varpi +832\beta -80\beta \varpi -4800{\alpha }^2+100{\alpha }^2\varpi -3760\alpha \beta +135\alpha \beta \varpi +35{\beta }^2\varpi +2000{\alpha }^3+3400{\alpha }^2\beta +1645\alpha {\beta }^2+245{\beta }^3),\hfill \\ \hfill y_1:& =\frac{1}{\beta }(25.6\varpi -409.6+2048\alpha -96\alpha \varpi -793.6\beta +38.4\beta \varpi -3840{\alpha }^2+120{\alpha }^2\varpi +2080\alpha \beta -68\alpha \beta \varpi +755.2{\beta }^2-64{\beta }^2\varpi +3200{\alpha }^3-50{\alpha }^3\varpi -1480{\alpha }^2\beta +25{\alpha }^2\beta \varpi -2798\alpha {\beta }^2+94.875\alpha {\beta }^2\varpi +448{\beta }^3-1000{\alpha }^4+150{\alpha }^3\beta +2072.5{\alpha }^2{\beta }^2+664.125\alpha {\beta }^3)/(64\varpi -1024+3840\alpha -160\alpha \varpi +832\beta -80\beta \varpi -4800{\alpha }^2+100{\alpha }^2\varpi -3760\alpha \beta +135\alpha \beta \varpi +35{\beta }^2\varpi +2000{\alpha }^3+3400{\alpha }^2\beta +1645\alpha {\beta }^2+245{\beta }^3),\hfill \\ \hfill {\varsigma }_0& =25z_1^8{\alpha }^2-40z_1^8\alpha +16z_1^8+75z_1^7{\alpha }^2-117z_1^7\alpha +45.6z_1^7+21z_1^6{x}_1\alpha -28z_1^6{x}_1+3.2z_1^6{y}_1-4z_1^6{y}_1\alpha +92.25z_1^6{\alpha }^2-139.8z_1^6\alpha +52.89z_1^6-62.3z_1^5{x}_1+44.1z_1^5{x}_1\alpha +9.04z_1^5{y}_1-11.6z_1^5{y}_1\alpha +59.6z_1^5{\alpha }^2-87.41z_1^5\alpha +31.982z_1^5+12.25z_1^4{x}_1^2+2.8z_1^4{x}_1{y}_1-54.25z_1^4{x}_1+35.7z_1^4{x}_1\alpha +0.16z_1^4{y}_1^2+9.992z_1^4{y}_1-13.24z_1^4{y}_1\alpha +21.36z_1^4{\alpha }^2-30.204z_1^4\alpha +10.6521z_1^4+19.6z_1^3{x}_1^2-23.212z_1^3{x}_1+6.16z_1^3{x}_1{y}_1+13.944z_1^3{x}_1\alpha +0.448z_1^3{y}_1^2+5.3816z_1^3{y}_1-7.42z_1^3{y}_1\alpha +4.032z_1^3{\alpha }^2-5.4768z_1^3\alpha +1.8564z_1^3+0.132496z_1^2+0.3136z_1^2{\alpha }^2-4.8944z_1^2{x}_1+11.76z_1^2{x}_1^2-0.40768z_1^2\alpha +4.956z_1^2{x}_1{y}_1+2.6208z_1^2{x}_1\alpha +0.4704z_1^2{y}_1^2+1.40728z_1^2{y}_1-2.0384z_1^2{y}_1\alpha +3.136z_1{x}_1^2+0.18816z_1{x}_1\alpha -0.40768z_1{x}_1+0.142688z_1{y}_1+1.7248z_1{x}_1{y}_1-0.21952z_1{y}_1\alpha +0.21952z_1{y}_1^2+0.3136x_1^2+0.219520x_1{y}_1+0.038416y_1^2,\hfill \\ \hfill {\varsigma }_1:& =10z_1^8\alpha +8z_1^8+33z_1^7\alpha +25.8z_1^7+44.7z_1^6\alpha -7z_1^6{x}_1-7.2z_1^6{y}_1+34.02z_1^6-18.2z_1^5{x}_1-20.64z_1^5{y}_1+31.79z_1^5\alpha +23.446z_1^5+12.516z_1^4\alpha -18.55z_1^4{x}_1-23.232z_1^4{y}_1+8.8998z_1^4+2.5872z_1^3\alpha +1.764z_1^3-9.268z_1^3{x}_1-12.8016z_1^3{y}_1-2.2736z_1^2{x}_1+0.21952z_1^2\alpha -3.44568z_1^2{y}_1+0.142688z_1^2-0.21952x_1{z}_1-0.362208y_1{z}_1,\hfill \\ \hfill {\varsigma }_2:& =z_1^8+3.6z_1^7+5.34z_1^6+4.172z_1^5+1.8081z_1^4+0.41160z_1^3+0.038416z_1^2,\hfill \end{array}$$

where $\mathrm{\varpi }≔\sqrt{256-640\alpha +224\beta +400{\alpha }^2+280\alpha \beta +49{\beta }^2}$.