[Retracted] The Stability of Gauss Model Having One-Prey and Two-Predators

Abstract

The study of the dynamics of predator-prey interactions can be recognized as a major issue in mathematical biology. In the present paper, some Gauss predator-prey models in which three ecologically interacting species have been considered and the behavior of their solutions in the stability aspect have been investigated. The main aim of this paper is to consider the local and global stability properties of the equilibrium points for represented systems. Finally, stability of some examples of Gauss model with one prey and two predators is discussed.

1. Introduction

Gauss is one of the well-known scientists who studied in various area in mathematics such as mathematical biology and mathematical ecology. One of his famous models is predator-prey problem in which were obtained the fundamental results in order to be interpreted and analyzed by him. In 1934, Gauss introduced his standard model in mathematical biology. After two years, he and Smaragdov studied a generalization of the following model as a model for predator-prey interactions:𝑑𝑥(𝑥),𝑑𝑡=𝑎𝑥−𝑦𝑝𝑑𝑦(𝑥)𝑑𝑡=𝑦−𝛾+𝑐𝑝 (1.1) (see, for more details, [1]).

More general form of this model known as an intermediate model of predator-prey interactions is as follows:𝑑𝑥(𝑥)(𝑥),𝑑𝑡=𝑥𝑔−𝑦𝑝𝑑𝑦(𝑥.𝑑𝑡=𝑦−𝛾+𝑞) (1.2)

Since the predator-prey system is investigated and extended frequently, one of the more investigated extensions of predator-prey system is this model with two preys and one predator. One may see some analysis of predator-prey system having two preys and one predator in references such as [2–8]. Elabbasy and Lisena studied dynamics of periodic predator-prey system having two preys and one predator [2, 6]. Gakkhar and Singh investigated food web model consisting of two preys and one predator which was a harvested factor [3]. The predator-prey system with two preys and one predator consisting of effort rate harvested factor is studied in [4, 5, 7]. A dynamic behavior of the Holling-II system with two preys and one predator system and with impulsive effect concerning biological control is investigated in [8].

2. The Predator-Prey Gauss Model with One Prey and Two Predators

Let us consider a system of two predator species living in an ecosystem independently and each species baits the prey. The Gauss model with one prey and two predators may be written as follows:𝑑𝑥𝑑𝑡=𝑎₁𝑥−𝑦𝑝₁(𝑥)−𝑧𝑝₂(𝑥),𝑑𝑦𝑑𝑡=−𝛾₁𝑦+𝑐₁𝑦𝑝₁(𝑥),𝑑𝑧𝑑𝑡=−𝛾₂𝑧+𝑐₂𝑧𝑝₂(𝑥), (2.1) where 𝑥 is the density of prey species, and 𝑦 and 𝑧 are the densities of predators species. In this system, all of coefficients 𝑎₁,𝛾₁,𝛾₂,𝑐₁, and 𝑐₂ are positive and constant. Moreover, the prey enhances in absence of predators species and this increasing is limited by terms −𝑦𝑝₁(𝑥) and −𝑧𝑝₂(𝑥). In the absence of prey, density of predators populations decrease by exponential growth rate and the prey has positive efficiency on predator population to be a positive sign; the terms 𝑐₁𝑦𝑝₁(𝑥) and 𝑐₂𝑧𝑝₂(𝑥) prove this claim.

For example, consider two species fox and eagle living in an ecosystem and each of the two species baits of rabbit species. In addition to assumptions of Gauss’s model, assume that 𝑝₁(𝑥) and 𝑝₂(𝑥) have properties of 𝑝(𝑥) in the Gauss model (1.1).

In system (2.1), the following properties are held.(i)If the population density for one of the predators species is zero, then the system (2.1) converts to the system (1.1).(ii)If the population density of prey species is zero, then system (2.1) will be converted to system with two predator species that live in an ecosystem independently.(iii)Let the population density of two species be zero, then system (2.1) will be converted to an equation of growth rate.(iv)The solutions orbit of system (2.1) is located in the following set: int𝑅³₊=int𝑥_𝑖∣𝑥_𝑖⩾0,𝑖=1,2,3. (2.2)

The terms 𝑝₁(𝑥) and 𝑝₂(𝑥) have the properties described as follows:(i)𝑝₁(0)=0, 𝑝₁(𝑥) is continuous and differentiable for 𝑥≥0, 𝑑𝑝₁(𝑥)/𝑑𝑥>0,(ii)𝑝₂(0)=0, 𝑝₂(𝑥) is continuous and differentiable for 𝑥≥0, 𝑑𝑝₂(𝑥)/𝑑𝑥>0.

3. Local Stability

We use the linearisation method to study the stability of the system (2.1). By this means, we calculate the Jacobian matrix, which may be found as follows:𝐽|_{(𝑥,𝑦,𝑧)}=𝑎₁−𝑦𝑑𝑝₁(𝑥)𝑑𝑥−𝑧𝑑𝑝₂(𝑥)𝑑𝑥−𝑝₁(𝑥)−𝑝₂(𝑥)𝑐₁𝑦𝑑𝑝₁(𝑥)𝑑𝑥−𝛾₁+𝑐₁𝑝₁(𝑥)0𝑐₂𝑧𝑑𝑝₂(𝑥)𝑑𝑥0−𝛾₂+𝑐₂𝑝₂(𝑥). (3.1)

Now, let (𝑥,𝑦,𝑧) be the equilibrium point of system (2.1). Then𝐴=𝐽|_{(𝑥,𝑦,𝑧)}=𝑎₁−𝑦𝑑𝑝₁(𝑥)−𝑑𝑥𝑧𝑑𝑝₂(𝑥)𝑑𝑥−𝑝₁(𝑥)−𝑝₂(𝑥)𝑐₁𝑦𝑑𝑝₁(𝑥)𝑑𝑥−𝛾₁+𝑐₁𝑝₁(𝑥)0𝑐₂𝑧𝑑𝑝₂(𝑥)𝑑𝑥0−𝛾₂+𝑐₂𝑝₂(𝑥). (3.2)

And so, if tr𝐴<0 and det𝐴>0, therefore, the system (2.1) is locally asymptotically stable at the equilibrium point (𝑥,𝑦,𝑧).

Let𝐴₁=𝑎₁−𝑦𝑑𝑝₁(𝑥)−𝑑𝑥𝑧𝑑𝑝₂(𝑥),𝐴𝑑𝑥₂=𝑐₁𝑦𝑑𝑝₁(𝑥),𝐴𝑑𝑥₃=−𝛾₁+𝑐₁𝑝₁(𝑥),𝐴₄=𝑐₂𝑧𝑑𝑝₂(𝑥),𝐴𝑑𝑥₅=−𝛾₂+𝑐₂𝑝₂(𝑥). (3.3) Then,𝐴=𝐽|_{(𝑥,𝑦,𝑧)}=𝐴₁−𝑝₂(𝑥)−𝑝₁(𝑥)𝐴₂𝐴₃0𝐴₄0𝐴₅. (3.4) Therefore,det𝐴=𝐴₁𝐴₃𝐴₅+𝑝₂(𝑥)𝐴₂𝐴₅+𝑝₁(𝑥)𝐴₃𝐴₄,tr𝐴=𝐴₁+𝐴₃+𝐴₅. (3.5) Hence, the system (2.1) is locally asymptotically stable if 𝐴₁𝐴₃𝐴₅+𝑝₂(𝑥)𝐴₂𝐴₅+𝑝₁(𝑥)𝐴₃𝐴₄>0 and 𝐴₁+𝐴₃+𝐴₅<0.

So, the following proposition is proved.

Proposition 3.1. Let 𝐴₁𝐴₃𝐴₅+𝑝₂(𝑥)𝐴₂𝐴₅+𝑝₁(𝑥)𝐴₃𝐴₄>0 and 𝐴₁+𝐴₃+𝐴₅<0. The system (2.1) is locally asymptotically stable at its equilibrium point (𝑥,𝑦,𝑧), whenever it exists.

4. Global Stability

In this section, we will prove the global stability of the system (2.1) by constructing a suitable Lyapunov function.

Theorem 4.1. The system (2.1) is globally asymptotically stable at equilibrium point (𝑥,𝑦,𝑧), where 𝑥<𝑥, 𝑦>𝑦 and 𝑧>𝑧.

Proof. Let us consider a suitable Lyapunov function 𝑣𝑥,𝑦,𝑧=𝑥+ℎ𝑦+𝑘𝑧, (4.1) where ℎ=1/𝑐₁ and 𝑘=1/𝑐₂. Obviously 𝑣 is a positive definite. Now take the derivative from the last Lyapunov system with respect to the time 𝑡. So we have 𝑑𝑣=𝑑𝑡𝑑𝑥𝑑𝑡+ℎ𝑑𝑦𝑑𝑡+𝑘𝑑𝑧.𝑑𝑡 (4.2) By substituting 𝑑𝑥/𝑑𝑡, 𝑑𝑦/𝑑𝑡 and 𝑑𝑧/𝑑𝑡 in the system (2.1), we obtain the said derivative as follows: 𝑑𝑣=𝑎𝑑𝑡¹𝑥−𝑦𝑝₁(𝑥)−𝑧𝑝₂(𝑥)+ℎ−𝛾₁𝑦+𝑐₁𝑦𝑝₁(𝑥)+𝑘−𝛾₂𝑧+𝑐₂𝑧𝑝₂(𝑥). (4.3)
Since (𝑥,𝑦,𝑧) is an equilibrium point for system (2.1), so we can add the zero equations to 𝑑𝑣/𝑑𝑡 as follows: 𝑑𝑣=𝑎𝑑𝑡¹𝑥−𝑦𝑝₁(𝑥)−𝑧𝑝₂(𝑥)+ℎ−𝛾₁𝑦+𝑐₁𝑦𝑝₁(𝑥)+𝑘−𝛾₂𝑧+𝑐₂𝑧𝑝₂(𝑥=𝑎₁)𝑥−𝑦𝑝₁(𝑥)−𝑧𝑝₂(𝑥)−𝑎₁𝑥−𝑦𝑝₁(𝑥)−𝑧𝑝₂(𝑥)+ℎ−𝛾₁𝑦+𝑐₁𝑦𝑝₁(𝑥)−−𝛾₁𝑦+𝑐₁𝑦𝑝₁(𝑥)+𝑘−𝛾₂𝑧+𝑐₂𝑧𝑝₂(𝑥)−−𝛾₂𝑧+𝑐₂𝑧𝑝₂(𝑥=𝑎₁)(𝑥−𝑥)−𝑦𝑝₁(𝑥)+𝑦𝑝₁(𝑥)−𝑧𝑝₂(𝑥)+𝑧𝑝₂(𝑥)+ℎ−𝛾₁𝑦−𝑦+𝑐₁𝑦𝑝₁(𝑥)−𝑐₁𝑦𝑝₁(𝑥)+𝑘−𝛾₂(𝑧−𝑧)+𝑐₂𝑧𝑝₂(𝑥)−𝑐₂𝑧𝑝₂(𝑥), (4.4) and by putting ℎ=1/𝑐₁ and 𝑘=1/𝑐₂, we find out 𝑑𝑣𝑑𝑡=𝑎₁(𝑥−𝑥)−𝛾₁𝑐₁𝑦−𝑦−𝛾₂𝑐₂(𝑧−𝑧). (4.5) Therefore, 𝑑𝑣/𝑑𝑡<0 if 𝑥<𝑥, 𝑦>𝑦 and 𝑧>𝑧. This completes the proof.

5. Analysis of Examples

In this section, we present some examples of Gauss’s model and analyze the stability of them.

5.1. Analysis of Example 1

Consider the following system:𝑑𝑥𝑎𝑑𝑡=𝑥₁−𝑎₂𝑦−𝑎₃𝑧,𝑑𝑦𝑑𝑡=𝑦−𝑏₁+𝑏₂𝑥,𝑑𝑧𝑑𝑡=𝑧−𝑐₁+𝑐₂𝑥. (5.1)

In the above system, all coefficients 𝑎₁,𝑎₂,𝑎₃,𝑏₁,𝑏₂,𝑐₁, and 𝑐₂ are positive constants. In the system (5.1), efficiency of the predator species on the preys species and also efficiency of the preys species on the predator species are linear. The points (0,0,0),(𝑏₁/𝑏₂,𝑎₁/𝑎₂,0), and (𝑐₁/𝑐₂,0,𝑎₁/𝑎₃) are equilibrium points of the system (5.1) which may be analyzed by using the Jacobian matrix. We first calculate the Jacobian matrix of system (5.1) as follows:𝐽|_{(𝑥,𝑦,𝑧)}=𝑎₁−𝑎₂𝑦−𝑎₃𝑧−𝑎₂𝑥−𝑎₃𝑥𝑏₂𝑦−𝑏₁+𝑏₂𝑐𝑥0₃𝑧0−𝑐₁+𝑐₂𝑥. (5.2) So𝐽|_(0,0,0)=𝑎₁000−𝑏₁000−𝑐₁. (5.3) Thus, the origin is a saddle point for the system (5.1).

The Jacobian matrix for the above system at the equilibrium point (𝑏₁/𝑏₂,𝑎₁/𝑎₂,0) is given by𝐴₁=𝐽|_{(𝑏1/𝑏2,𝑎1/𝑎2,0)}=𝑎0−²𝑏₁𝑏₂−𝑎₃𝑏₁𝑏₂𝑏₂𝑎₁𝑎₂𝑏0000¹𝑐₂𝑏₂. (5.4) Thus,det𝐴₁=𝑏−𝜆𝐼¹𝑐₂𝑏₂−𝜆𝜆²+𝑎₁𝑏₁. (5.5) Hence, the equilibrium point (𝑏₁/𝑏₂,𝑎₁/𝑎₂,0) is hyperbolic point for the system (5.1).

The Jacobian matrix at the equilibrium point (𝑐₁/𝑐₂,0,𝑎₁/𝑎₃) is obtained by𝐴₂=𝐽|_{(𝑐1/𝑐2,0,𝑎1/𝑎3)}=0𝑎₂𝑐₁𝑐₂−𝑎₃𝑐₁𝑐₂0−𝑏₁+𝑏₂𝑐₁𝑐₂0𝑎₁𝑐₃𝑎₃00. (5.6) and sodet𝐴₂−𝜆𝐼=−𝑏₁+𝑏₂𝑐₁𝑐₂−𝜆𝜆²+𝑎₁𝑐₁. (5.7) Thus, the equilibrium point (𝑐₁/𝑐₂,0,𝑎₁/𝑎₃) is hyperbolic point for the system (5.1).

In fact, we worked out the following proposition.

Proposition 5.1. The following statements are true for the system (5.1).
(𝑖) The equilibrium point (0,0,0) is a saddle point.
(𝑖𝑖) The equilibrium points (𝑏₁/𝑏₂,𝑎₁/𝑎₂,0) and (𝑐₁/𝑐₂,0,𝑎₁/𝑎₃) are hyperbolic points.

5.2. Analysis of Example 2

In the second example, we consider that there are two predator species which live in an ecosystem independently, but their food is the same, that is, one species prey that has interaction between members species whose mathematical model is as follows:𝑑𝑥𝑎𝑑𝑡=𝑥₁−𝑎₂𝑥−𝑎₃𝑦−𝑎₄𝑧,𝑑𝑦𝑦𝑑𝑡=𝑦1−𝑣₁𝑥,𝑑𝑧𝑧𝑑𝑡=𝑧1−𝑣₂𝑥. (5.8)

It is clear that 𝑥≠0.

Moreover, the equilibrium points of system (5.8) are given by𝑎₁𝑎₂,𝑎,0,0¹𝑎₂+𝑎₄𝑣₂𝑣,0,²𝑎₁𝑎₂+𝑎₄𝑣₂,𝑎₁𝑎₂+𝑎₃𝑣₁,𝑎₁𝑣₁𝑎₂+𝑎₃𝑣₁,0. (5.9) Also, its Jacobian matrix is as follows:𝐽|_{(𝑥,𝑦,𝑧)}=𝑎₁−2𝑎₂𝑥−𝑎₃𝑦−𝑎₄𝑧−𝑎₃𝑥−𝑎₄𝑥𝑦²𝑣₁𝑥²1−2𝑦𝑣₁𝑥0𝑧²𝑣₂𝑥²𝑧01−2𝑣₂𝑥. (5.10)

Now by substituting the equilibrium point (𝑎₁/𝑎₂,0,0) and simplifying, the Jacobian matrix becomes𝐽|_{(𝑎1/𝑎2,0,0)}=−𝑎₁−𝑎₁𝑎₃𝑎₂−𝑎₁𝑎₄𝑎₂010001. (5.11)

The eigenvalues of the above matrix are 𝜆₁=𝜆₂=0 and 𝜆₃=−𝑎₁. Thus, the system (5.8) at equilibrium point (𝑎₁/𝑎₂,0,0) is stable.

Also, the Jacobian matrix at the equilibrium point (𝑎₁/(𝑎₂+𝑎₄𝑣₂),0,𝑣₂𝑎₁/(𝑎₂+𝑎₄𝑣₂)) is given by𝐽|_{(𝑎1/(𝑎2+𝑎4𝑣2),0,𝑣2𝑎1/(𝑎2+𝑎4𝑣2))}=−𝑎₁𝑎₂𝑎₂+𝑎₄𝑣₂−𝑎₁𝑎₃𝑎₂+𝑎₄𝑣₂−𝑎₁𝑎₄𝑎₂+𝑎₄𝑣₂𝑣010₂01. (5.12) The eigenvalues of the above matrix are 𝜆₁=1,𝜆₂=−1 and 𝜆₃=𝑎₁𝑎₂/(𝑎₂+𝑎₄𝑣₂), and so the equilibrium point (𝑎₁/𝑎₂,0,0) is a saddle point for the system (5.8).

Finally, the last equilibrium point is given by𝑎₁𝑎₂+𝑎₃𝑣₁,𝑎₁𝑣₁𝑎₂+𝑎₃𝑣₁,0, (5.13) where its Jacobian matrix is as follows:𝐴=𝐽|_{(𝑎1/(𝑎2+𝑎3𝑣1),𝑎1𝑣1/(𝑎2+𝑎3𝑣1),0)}=−𝑎₁𝑎₂𝑎₂+𝑎₃𝑣₁−𝑎₁𝑎₃𝑎₂+𝑎₃𝑣₁−𝑎₁𝑎₄𝑎₂+𝑎₃𝑣₂𝑣₁−10001. (5.14) and so𝑎tr𝐴=−¹𝑎₂𝑎₂+𝑎₃𝑣₁𝑎<0,det𝐴=¹𝑎₂𝑎₂+𝑎₃𝑣₁+𝑎₁𝑎₃𝑣₁𝑎₂+𝑎₃𝑣₁>0. (5.15) Therefore, the system (5.8) is locally asymptotically stable at the above equilibrium point.

Therefore, we can summarize the above facts in the following proposition.

Proposition 5.2. For the system (5.8), the following statements are held.(i)It has three equilibrium points which are as follows: 𝑎₁𝑎₂,𝑎,0,0¹𝑎₂+𝑎₄𝑣₂𝑣,0,²𝑎₁𝑎₂+𝑎₄𝑣₂,𝑎₁𝑎₂+𝑎₃𝑣₁,𝑎₁𝑣₁𝑎₂+𝑎₃𝑣₁,0. (5.16)(ii)The said system is stable at the equilibrium point (𝑎₁/𝑎₂,0,0).(iii)The point (𝑎₁/(𝑎₂+𝑎₄𝑣₂),0,𝑣₂𝑎₁/(𝑎₂+𝑎₄𝑣₂)) is saddle point for the above system.(iv)The point (𝑎₁/(𝑎₂+𝑎₃𝑣₁),𝑎₁𝑣₁/(𝑎₂+𝑎₃𝑣₁),0) is locally asymptotically stable for the said system.

6. Conclusion

By adding some inequalities, one may make the Gauss system having one prey and two predators asymptotically stable globally. Furthermore, it can be guessed that the generalization of the Gauss system with existing 𝑛-preys and 𝑚-predators is globally asymptotically stable provided the following statements hold.(i)The density of each prey is less than the corresponding component in related equilibrium point.(ii)The density of each predator is greater than the corresponding component in related equilibrium point. Moreover, one may determine the local stability for some particular model of the Gauss system having one prey and two predators such as (5.1) and (5.8) by using the linearisation method.