# Dynamics of a Predator-Prey System with Mixed Functional Responses

Authors: Hunki Baek; Dongseok Kim
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

A predator-prey system with two preys and one predator is considered. Especially, two different types of functional responses, Holling type and Beddington-DeAngelis type, are adopted. First, the boundedness of system is showed. Stabilities analysis of system is investigated via some properties about equilibrium points and stabilities of two subsystems without one of the preys of system. Also, persistence conditions of system are found out and some numerical examples are illustrated to substantiate our theoretical results.


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## Body

## 1. Introduction

Ecological systems are mainly characterized by the interaction between species and their surrounding natural environment ([1]). Among them, two-species continuous time ecological models with one predator and one prey have been studied for several functional responses such as Holling-Tanner type ([2–4]), Beddington-DeAngelis type ([5–7]), and ratio-dependent type ([8, 9]). However, it has been recognized that such two-species ecological models are not sufficient to explain various phenomena observed in nature ([10–13]). For this reason, in recent years, ecological models with three and more species have been investigated by many authors in [14–18]. Particularly, in this paper, we will deal with a three-species ecological system with two different preys and one predator.On the other hand, functional response between two species is known as the relationship between prey and predator. Most three-species systems in [5, 10, 11, 16, 18] have the same two functional responses. However, it is reasonable to consider two different functional responses since two preys in the system are different from each other. In fact, if one considers the handling time of the predator to capture the prey, one figures out that the predator has a Holling type-II functional response and if one thinks of the competitions of predators with one another to catch the prey, Beddington-DeAngelis type functional response could be suitable. Thus, in this paper, we consider the following system with two preys and one predator with mixed two functional responses:
(1)x1′(t)=x1(t)(a1-b1x1(t)-y(t)c+x1(t)),x2′(t)=x2(t)(a2-b2x2(t)-y(t)α+x2(t)+βy(t)),y′(t)=y(t)(-d+e1x1(t)c+x1(t)+e2x2(t)α+x2(t)+βy(t)),
where x1(t),x2(t), and y(t) represent the population density of two preys and the predator at time t, respectively. The constants ai(i=1,2) are called the intrinsic growth rates, bi(i=1,2) are the coefficients of intraspecific competition, c and α are the half-saturation constants, ei(i=1,2) are the per capita rate of predation of the predator, d is the death rate of the predator, and β scales the impact of the predator interference.The main purpose of this paper is to look into dynamical properties of system (1). In Section 2, the boundedness of system (1), which means the solution of system (1) initiating in the nonnegative octant is bounded, is studied. Stabilities analysis of system (1) is investigated via well-known properties about equilibrium points and the stabilities of two subsystems without one of the preys of system (1). Also, persistence conditions of the main system (1) are found out and some numerical examples are illustrated to substantiate our theoretical results in Section 4.


## 2. Boundedness of System (1)

First, let us consider the state spaceR+3={(x,y,z)T∣x≥0,y≥0,z≥0}. It is easy to see that the functions in the right-hand side of system (1) are continuous and have continuous partial derivatives on R+3. Moreover elementary calculations yield the fact that they are Lipschizian on R+3. Thus the solution of system (1) with nonnegative initial condition exists and is unique. In addition, the solution of system (1) initiating in the nonnegative octant is bounded as shown in the following theorem.Theorem 1.
The solution of system (1) initiating in R+3 is bounded for all t≥0.Proof.
Sincedxi(t)/dt≤xi(t)(ai-bix(t)),i=1,2, we have
(2)limsup⁡t→∞xi(t)≤aibi,i=1,2.
Define V(t)=e1x1(t)+e2x2(t)+y(t). Then
(3)dV(t)dt≤a1(a1+1)e1b1+a2(a2+1)e2b2-mV(t),
where m=min⁡{1,d}. So, by comparison theorem, we obtain that V(t)≤(M/m)+ke-mt for t≥0, where M=(a1(a1+1)e1/b1)+(a2(a2+1)e2/b2) and k is a constant of integration. Thus e1x1(t)+e2x2(t)+y(t)≤M/m for sufficiently large t, which means that all species are uniformly bounded for any initial value in R+3.It is easy to see that ifx1(0)>0, then x1(t)>0 for all t>0. The same is true for x2 and y components. Therefore, we conclude clearly that the first octant R+3 is an invariant domain of system (1).Now, we will discuss conditions that render certain species extinct. According to system (1), even if one of the preys is extinct, predator species could survive since the predator has two preys. However, the higher the death rate of the predator is, the higher the possibility of predator extinction is. Thus the following theorem indicates that if the death rate of the predator is less than a certain value depending on the growth rate of two preys, then the predator will not face extinction.Theorem 2.
A necessary condition for the predator speciesy to survive is
(4)d<a1e1a1+b1c+a2e2a2+αb2.Proof.
From the third equation of system (1), we get
(5)dy(t)dt=y(t)(-d+e1x1(t)c+x1(t)+e2x2(t)α+x2(t)+βy(t))≤y(t)(-d+e1x1(t)c+x1(t)+e2x2(t)α+x2(t)).
In the proof of Theorem 1, limsup⁡t→∞xi(t)≤ai/bi,i=1,2, is shown. Then
(6)dy(t)dt≤y(t)(-d+a1e1a1+b1c+a2e2a2+αb2)
and hence y(t)≤y(0)eAt, where A=-d+(a1e1/(a1+b1c))+(a2e2/(a2+αb2)). Thus if A<0, that is d>(a1e1/(a1+b1c))+(a2e2/(a2+αb2)), then lim⁡t→∞y(t)=0. Therefore d<(a1e1/(a1+b1c))+(a2e2/(a2+αb2)) is a necessary condition for the predator species y to survive.


## 3. Stability Analysis of System (1)

In order to study stabilities of equilibria of system (1), we first take into account a subsystem of system (1) when the second prey (x2) is absent as follows:
(7)x1′(t)=x1(t)(a1-b1x1(t))-x1(t)y(t)c+x1(t),y′(t)=y(t)(-d+e1x1(t)c+x1(t)).Kolmogorov's theorem in [19] assumes the existence of either a stable equilibrium point or a stable limit cycle behavior in the positive quadrant of phase space of a two-dimensional (2D) dynamical system, provided certain conditions are satisfied.In fact, it is easy to see that the subsystem (7) is a Kolmogorov system under the following condition:
(8)0<cde1-d<a1b1.
For this reason, from now on, we assume that subsystem (7) satisfies condition (8). By applying the local stability analysis ([20]) to Kolmogorov system (7) we have the following results.(1)
The equilibrium pointE10=(0,0) always exists and is a saddle point.(2)
The equilibrium pointE11=(a1/b1,0) always exists and is a saddle point under condition (8).(3)
The positive equilibrium pointE12=(x^,y^) exists, where
(9)x^=cde1-d,y^=(a1-b1x^)(c+x^),and it is a locally asymptotically stable point if the following condition holds:
(10)d>e1(a1-b1c)a1+b1c.
Moreover, if the condition d<e1(a1-b1c)/(a1+b1c) holds, the solution of subsystem (7) approaches to a stable limit cycle even though the system is not a Kolmogorov system.Secondly, we focus on another subsystem of system (1) when the first prey (x1) is absent as follows:
(11)x2′(t)=x2(t)(a2-b2x2(t))-x2(t)y(t)α+x2(t)+βy(t),y′(t)=y(t)(-d+e2x2(t)α+x2(t)+βy(t)).
Subsystem (11) is a Kolmogorov system if the following condition is satisfied:
(12)0<dαe2-d<a2b2.
Simple calculation yields that there exist at most three nonnegative equilibrium points of subsystem (11). Moreover, the stability of such equilibrium points can be studied by applying the local stability analysis to subsystem (11) as the previous case. Thus we summarize results about local stability as follows.(1)
The equilibrium pointE20=(0,0) always exists and is a saddle point.(2)
The equilibrium pointE21=(a2/b2,0) always exists and is also a saddle point under condition (12).(3)
The positive equilibrium pointE22=(x-,y-) exists, where
(13)x-=(a2β+d-e2)+(βa2+d-e2)2+4b2αdβ2b2β,y-=(e2-d)x--dαdβ.In [15], the authors have investigated the local stability of the equilibrium point E22.Theorem 3 (see [15]).
The positive equilibrium pointE22=(x-,y-) of Kolmogorov system (11) is locally asymptotically stable if one of the following sets of conditions is satisfied:(1)
e2β≥1,(2)
e2β<1 and Δ12-4Δ2≤0,(3)
e2β<1 and Δ12-4Δ2>0, with 0<x-≤R1 or R2≤x-<1.
However, the solution of subsystem (11) approaches to a stable limit cycle for R1<x-<R2. Here Δ1=d(1-e2β)(d-e2)/b2e22β, Δ2=αd2(1-e2β)/b2e22β, R1=(1/2)(-Δ1-Δ12-4Δ2), and R2=(1/2)(-Δ1+Δ12-4Δ2).Now, we turn our concerns on system (1) to investigate the existence and local stability of the equilibrium points of the system. In fact, there are at most seven nonnegative equilibrium points of system (1). The existence conditions of them are mentioned in the following lemma.Lemma 4.
(1) The trivial equilibrium point E0=(0,0,0) and one-prey equilibrium points E1=(a1/b1,0,0) and E2=(0,a2/b2,0) always exist.
(2) Two-species equilibrium points E3=(a1/b1,a2/b2,0), E4=(x^,0,y^), and E5=(0,x-,y-) exist in the interior of positive quadrant of x1x2, x1y, and x2y planes, respectively, if the Kolmogorov conditions 0<cd/(e1-d)<a1/b1 and 0<dα/(e2-d)<a2/b2 hold, where x^,y^ and x-,y- are given in (9) and (13), respectively.
(3) The positive equilibrium point E6=(x1*,x2*,y*) exists in the interior of the first octant if
(14)e2<d<e1,c(d-e2)e1+e2-d<x1*<min⁡{cde1-d,a1b1},
where
(15)x2*=((d-e1)x1*+cd)(α+β(a1-b1x1*)(c+x1*))(e1+e2-d)x1*+c(e2-d),y*=(a1-b1x1*)(c+x1*)
and x1* satisfies the following equation:
(16)A5x1*5+A4x1*4+A3x1*3+A2x1*2+A1x1*+A0=0.
Here,
(17)A0=b2c(α+a1βc)2de2-a2c(α+a1βc)e2(-d+e2)+a1c2(-d+e2)2,A1=2b2βc(a1-b1c)(α+a1βc)de2+b2(α+a1βc)2(d-e1)e2-a2βc(a1-b1c)e2(-d+e2)-b1c2(-d+e2)2-a2(α+a1βc)e2(-d+e1+e2)+2a1c(-d+e2)(-d+e1+e2),A2=b2β2c(a1-b1c)2de2-2b1b2βc(α+a1βc)de2+2b2β(a1-b1c)(α+a1βc)(d-e1)e2+a2b1βce2(-d+e2)-a2β(a1-b1c)e2(-d+e1+e2)-2b1c(-d+e2)(-d+e1+e2)+a1(-d+e1+e2)2,A3=-2b1b2β2c(a1-b1c)de2+b2β2(a1-b1c)2(d-e1)e2-2b1b2β(α+a1βc)(d-e1)e2+(-d+e1+e2)(a2b1βe2-b1(-d+e1+e2)),A4=b1b2(-2a1d+3b1cd+2a1e1-2b1ce1)e2β2,A5=b12b2(d-e1)e2β2.Proof.
We only consider the existence of the positive equilibrium pointE6. It is easy to see that the equilibrium point E6=(x1*,x2*,y*) exists in the interior of the first octant if and only if there exists a positive solution to the following algebraic nonlinear simultaneous equations:
(18)f1(x,y,z)=a1-b1x1(t)-y(t)c+x1(t)=0,f2(x,y,z)=a2-b2x2(t)-y(t)α+x2(t)+βy(t)=0,f3(x,y,z)=-d+e1x1(t)c+x1(t)+e2x2(t)α+x2(t)+βy(t)=0.
From the first and third equations in (18) we can have
(19)x2*=((d-c)x1*+dc)(α+β(a1-b1x1*)(c+x1*))(e1+e2-d)x1*+c(e2-d),y*=(a1-b1x1*)(c+x1*).
By using (19) in the second equation of (18) and by elementary calculation we can obtain the following equation:
(20)A5x1*5+A4x1*4+A3x1*3+A2x1*2+A1x1*+A0=0.
Since the degree of equation (20) is 5, it has at least one real root x1*. Moreover if condition (14) is satisfied then all values of x1*,x2*, and y* are positive.It is worth noting that since predator dies out in the absence of all preys the equilibrium point(0,0,zc) with zc>0 does not exist.In order to investigate stabilities of the equilibrium points, we need to consider the variational matrixV(x1,x2,y) of system (1). Thus we get the following matrix:
(21)V(x1,x2,y)=(v11v12v13v21v22v23v31v32v33),
where
(22)v11=a1-2b1x1-cy(c+x1)2,v12=0,v13=-x1c+x1,v21=0,v22=a2-2b2x2-y(α+βy)(α+x2+βy)2,v23=-x2(α+x2)(α+x2+βy)2,v31=ce1y(c+x1)2,v32=e2y(α+βy)(α+x2+βy)2,v33=-d+e1x1c+x1+e2x2(α+x2)(α+x2+βy)2.
By using the variational matrix (21), local stabilities of system (1) near the equilibrium points are studied in the following theorems.Theorem 5.
(1) The trivial equilibrium point E0=(0,0,0) is a hyperbolic saddle point. In fact, near E0 both prey populations are increasing while the predator population is decreasing. And the equilibrium points E1=(a1/b1,0,0) and E2=(0,a2/b2,0) are also hyperbolic saddle points.
(2) The equilibrium point E3=(a1/b1,a2/b2,0) is always unstable; actually, a saddle point with locally stable manifold in the x1x2 plane and with local unstable manifold in the y declines if Kolmogorov conditions (8) and (12) hold.
(3) The equilibrium point E4=(x^,0,y^) is stable if (1-a2β)y^>a2α and is unstable if (1-a2β)y^<a2α.
(4) Assume that hypotheses of Theorem 3 hold; then the equilibrium point E5=(0,x-,y-) is stable if y->a1c and is unstable if y-<a1c.Proof.
(1) The eigenvalues of the matrixV(0,0,0) are a1,a2, and -d and their eigenvectors are (1,0,0),(0,1,0), and (0,0,-1). Furthermore, the eigenvalues of the matricies V(a1/b1,0,0) and V(0,a2/b2,0) are -a1,a2,(a1e1/(b1c+a1))-d and a1,-a2,(a2e2/(b2α+a2))-d, respectively. Thus the equilibrium points E0,E1, and E2 are hyperbolic saddle.
(2) The eigenvalues of the matrixV(a1/b1,a2/b2,0) are -a1,-a2, and (a1e1/(b1c+a1))+(a2e2/(b2α+a2))-d. Therefore, since Kolmogorov conditions (8) and (12) are satisfied, the sign of (a1e1/(b1c+a1))+(a2e2/(b2α+a2))-d is always positive. Thus the point E3 is unstable.
(3) Now, consider the equilibrium pointE4. The point E4=(x^,0,y^) has the same stability as E12 in the interior of positive coordinate plane x1y. Furthermore, since the equilibrium point E12 is always stable under condition (8), the local stability of the point E4 depends on the sign of the eigenvalue a2-(y^/(α+βy^)) of the x2-direction.
(4) Similar to the case of the pointE4, the point E5=(0,x-,y-) has the same stability behavior as E22 in the interior of positive coordinate plane x2y. Thus, since the point E22 is locally stable, if one of the conditions of Theorem 3 is satisfied, then the point E5 is locally stable or unstable along the x1-direction according to the sign of the eigenvalue a1-(y-/c) of the x1-direction.Example 6.
In this example we simulate system (1) numerically by using Runge-Kutta method of order 4 to substantiate Theorem 5 when the parameters are as follows:
(23)a1=0.8,b1=1,a2=0.7,b2=1,c=0.5,d=0.2,e1=0.8,e2=0.9,α=1.3,β=0.5.
Then it follows from Theorem 5 (4) that E5=(0,x-,y-)=(0,0.4516,0.5609) is stable since y->a1c=0.4. Figure 1 illustrates the phase portrait of system (1) and time series for x1(t),x2(t), and y(t) when initial condition is (0.2,0.2,0.2).Figure 1
(a) A phase portrait of system (1) with initial condition (0.2,0.2,0.2) and (b)–(d) time series for x1(t),x2(t), and y(t), respectively, when a1=0.8,b1=1,a2=0.7,b2=1,c=0.5,d=0.2,e1=0.8,e2=0.9,α=1.3, and β=0.5.In order to discuss the stability of the equilibrium pointE6=(x1*,x2*,y*), let V*=(vij*) be the variational matrix at E6. Then it follows from (21) that V* can be written as follows:(24)V*=(-b1x1*+x1*y*(c+x1*)20-x1*c+x1*0-b2x2*+x2*y*(α+x2*+βy*)2-x2*(α+x2*)(α+x2*+βy*)2ce1y*(c+x1*)2e2y*(α+βy*)(α+x2+βy*)2-e2βx2*y*(α+x2*+βy*)2).Thus the characteristic equation of the matrixV* is obtained as λ3+B1λ2+B2λ+B3=0, where B1=-(v11*+v22*+v33*),B2=v11*v22*+v11*v33*+v22*v33*-v13*v31*-v23*v32*,B3=(v23*v32*-v22*v33*)v11*+v22*v13*v31*,v11*=-b1x1*+(x1*y*/(c+x1*)2),v12*=0,v13*=-x1*/(c+x1*),v21*=0,v22*=-b2x2*+(x2*y*/(α+x2*+βy*)2),v23*=-x2*(α+x2*)/(α+x2*+βy*)2,v31*=ce1y*/(c+x1*)2,v32*=e2y*(α+βy*)/(α+x2+βy*)2, and v33*=-e2βx2*y*/(α+x2*+βy*)2.From the Routh-Hurwitz criterion ([20]), we know that E6=(x1*,x2*,y*) is locally asymptotically stable if and only if B1, B3, and B1B2-B3 are positive. It is not easy to find the conditions B1>0 and B3>0 and B1B2-B3>0. However, we give a sufficient condition to guarantee the local stability of the equilibrium point E6=(x1*,x2*,y*) in the following theorem.Theorem 7.
Suppose that the positive equilibrium pointE6 exists in the interior of the positive octant. Then E6 is locally asymptotically stable if
(25)y*<b1(c+x1*)2,(26)y*<b2(α+x1*+βy*)2.Proof.
It is from elementary calculation thatB1>0, B3>0, and B1B2-B3>0 under conditions (25) and (26).Example 8.
In order to substantiate Theorem7, we set the parameters as follows:
(27)a1=0.8,b1=1,a2=0.7,b2=1,c=0.8,d=0.5,e1=0.9,e2=0.3,α=1.3,β=0.5.
The point E6=(x1*,x2*,y*)=(0.667992,0.603106,0.193787) is locally stable since y*<b1(c+x1*)2=2.155 and y*<b2(α+x1*+βy*)2=4.2638. The phase portrait of system (1) and time series for x1(t),x2(t), and y(t) are shown in Figure 2.
On the other hand, if one of the conditions (25) and (26) is not satisfied, the positive equilibrium point E6 could not be stable. In order to illustrate an example, we take the parameters as follows:
(28)a1=0.8,b1=0.5,a2=0.7,b2=1,c=0.2,d=0.5,e1=0.9,e2=0.3,α=1.3,β=0.5.
Then we have the point E6=(0.171044,0.567449,0.265103). Since y*>b1(c+x1*)2=0.068836 and y*<b2(α+x1*+βy*)2=2.57152, the point E6 does not satisfy condition (25) and moreover Figure 3 exhibits numerically that E6 is unstable. As shown in Figure 3 even if the positive point becomes an unstable point a stable limit cycle could occur.Figure 2
(a) A phase portrait of system (1) with initial condition (0.2,0.2,0.2) and (b)–(d) time series for x1(t),x2(t), and y(t), respectively, when a1=0.8,b1=1,a2=0.7,b2=1,c=0.8,d=0.5,e1=0.9,e2=0.3,α=1.3, and β=0.5.Figure 3
(a) A phase portrait of system (1) with initial condition (0.171,0.567,0.265) and (b)–(d) time series for x1(t),x2(t), and y(t), respectively, when a1=0.8,b1=0.5,a2=0.7,b2=1,c=0.2,d=0.5,e1=0.9,e2=0.3,α=1.3, and β=0.5.


## 4. Persistence of System (1)

The term persistence is given to systems in which strict solutions do not approach the boundary of the nonnegative cones as time goes to infinity. Therefore, for the continuous biological system, survival of all interacting species and the persistence are equivalent. In the following theorem, we find out some persistence conditions of system (1).Theorem 9.
Suppose that system (1) has no nontrivial periodic solutions in the boundary planes and satisfies the hypothesis of Theorem 3 and condition e1(a1-b1c)/(a1+b1c)<d<a1e1/(a1+b1c) holds. Then the necessary conditions for the persistence of system (1) are
(29)λ-1=a1-y-c≥0,(30)λ^2=a2-y^α+βy^≥0,
and the sufficient conditions for the persistence of system (1) are
(31)λ-1=a1-y-c>0,(32)λ^2=a2-y^α+βy^>0.Proof.
Note that the boundedness of system (1) is shown in Theorem 1 and E12 is locally stable under Kolmogrov condition (12). Since E12 and E22 are locally stable by the assumptions, the signs of the eigenvalues λ^2 and λ-1 determine the stability of the equilibrium points E4=(x^,0,y^) and E5=(0,x-,y-). In fact, if there are no nontrivial periodic solutions in the x2y plane and (29) does not hold (i.e., λ-1<0) then there is an orbit in the positive cone, which approaches to E5. Hence, condition (29) is one of the necessary conditions for the persistence. Similarly, we obtain the other necessary condition (30) for the persistence of system (1) by applying the same method as mentioned above to the equilibrium point E4.
Now, we will use the abstract theorem of Freedman and Waltman [19] to figure out sufficient conditions for the persistence of system (1). In order to do this, consider the growth functions f1,f2, and f3 in (18) of system (1). Then it is shown that the following four conditions are satisfied.(1)
Clearly, we have∂fi/∂y<0,∂f3/∂xi>0,i=1,2.(2)
Each prey population grows up to its carrying capacity in the absence of predators; that is,f1(0,0,0)=a1>0 and f2(0,0,0)=a2>0 and (∂fi/∂xi)(x1,x2,0)=bi<0(i=1,2) and f1(a1/b1,0,0)=0=f2(0,a2/b2,0). Furthermore, the predator population dies out in the absence of preys; that is, consider f3(0,0,0)=-d<0.(3)
∂f1/∂x2=0 and ∂f2/∂x1=0. There exists exactly one point E3=(b1/a1,b2/a2,0) satisfying fi(b1/a1,b2/a2,0)=0, i=1,2.(4)
In the absence of each prey species the predator can survive on the other prey. This is always true under the Kolmogrov conditions (8) and (12). There exist uniquely E4=(x^,0,y^) and E5=(0,x-,y-) satisfying f1(x^,0,y^)=f3(x^,0,y^)=f2(0,x-,y-)=f3(0,x-,y-)=0. According to Kolmogrov conditions (8) and (12), we can get that f3(a1/b1,0,0)>0 and f3(0,a2/b2,0)>0, respectively.(5)
It follows from (8), (12), (31), and (32) that the inequalities f3(a1/b1,a2/b2,0)>0, f1(0,x-,y-)>0, and f2(x^,0,y^)>0 hold.
Therefore, by Freedman and Waltman theorem ([19]), system (1) persists under the hypotheses.Example 10.
Now let the parameters be as follows:(33)a1=0.8,b1=0.5,a2=1.0,b2=1.0,c=0.8,d=0.199,e1=0.3,e2=0.5,α=1.0,β=0.7.
Then it follows from [15, 21] that subsystems (7) and (11) have no periodic solutions in the boundary planes. Furthermore, it is not difficult to see that system (1) satisfies conditions (2) of Theorem 3 and the other hypotheses in Theorem 9. Thus, all species in system (1) can coexist as time goes away. Figure 4 shows a phase portrait and time series of all species of system (1) with initial condition (0.7,0.3,0.7).Figure 4
(a) A phase portrait of system (1) with initial condition (0.7,0.3,0.7) and (b)–(d) time series for x1(t),x2(t), and y(t), respectively, when a1=0.8,b1=0.5,a2=1.0,b2=1.0,c=0.8,d=0.199,e1=0.3,e2=0.5,α=1.0, and β=0.7.Theorem 11.
Suppose that conditions (31) and (32) are satisfied and system (1) has a finite number of limit cycles in the x1y plane or in the x2y plane. Then system (1) persists if
(34)∫0T‍(a1-v2(t)c)dt>0,(35)∫0T‍(a2-v1(t)α+βv1(t))dt>0
hold for each of the limit cycles (0,u2(t),v2(t)) and (u1(t),0,v1(t)) in the x2y plane and in the x1y plane, respectively, if it exists. Here, T means the period of the limit cycle.Proof.
First, consider a limit cycle(0,u2(t),v2(t)) in x2y plane. Then the variational matrix about the limit cycle (0,u2(t),v2(t)) can be written as(36)V1=(a1-v2(t)c000a2-2b2u(t)-v2(t)(α+βv2(t))(α+u2(t)+βv2(t))2-u2(t)(α+u2(t))(α+u2(t)+βv2(t))2e1v2(t)c-e2v2(t)(α+βv2(t))(α+u2(t)+βv2(t))2-d+u2(t)(α+u2(t))(α+u2(t)+βv2(t))2).Now, let (x1,x2,y) be a solution of system (1) with positive initial condition (z1,z2,z3) sufficiently close to the limit cycle. It is easily obtained from the variational matrix V1 that ∂x1/∂z1 is a solution of system dx1/dt=(a1-(v2(t)/c))x1 with x1(0)=1. Thus, we obtain
(37)∂x1∂z1(t,z1,z2,z3)=exp⁡(∫0t‍(a1-v2(s)c)ds).
From Taylor expansion, we get
(38)x1(t,z1,z2,z3)-x1(t,0,z2,z3)≈exp⁡(∫0t‍(a1-v2(s)c)ds)z1.
Thus x1 increases or decreases as the value of ∫0t‍(a1-(v2(s)/c))ds is positive or negative, respectively. Since E5 and the limit cycle (0,u2(t),v2(t)) are the only possible limit in the x2y plane of trajectories with positive initial condition, the trajectories go away from the x2y plane if conditions (31) and (34) are satisfied.
Similar argument can be applied to each limit cycle(u1(t),0,v1(t)) to obtain the fact that the trajectories go away from the x1y plane if conditions (32) and (35) hold by considering the variational matrix V2 about the limit cycle (u1(t),0,v1(t)) as follows:(39)V2=(a1-2b1u1-cv1(t)(c+u1)20-u1c+u10a2-v1(t)α+βv1(t)0ce1v1(t)(c+u1)2e2v1(t)α+βv1(t)-d+e1u1(t)c+u1(t)).Therefore the proof is complete.


## 5. Conclusions and Remarks

In this paper, we have considered a predator-prey system with two preys and one predator with two different types of functional responses, Holling type and Beddington-DeAngelis type. Until now, many researches for two-prey and one-predator systems have dealt with the same functional responses to describe the relationship between prey and predator even if the two preys are different from each other. Thus in this research we adopted two different types of functional responses to model the relationship between two different preys and predator. We investigated stabilities of system about equilibrium points by virtue of stabilities of two subsystems without one of the preys of system. Also, we found out conditions that guarantee that the system is persistent. In addition, some numerical examples are illustrated to substantiate our theoretical results.Generally speaking, if food is abundant the predators do not interfere with each other to get it; otherwise there is intense competition between predators to get food. In order to describe such kind of phenomenon, we use the mixed functional responses. Thus, due to the mixed functional responses, one can see from Theorem5 that the value β of the impact of the predator interference to catch the prey x2 has an effect on the extinction of another prey x1. Thus even though ecological systems have two different functional responses, they have a variety of dynamical behaviors.



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*Source: 536019-2014-09-14.xml*