# Global Asymptotic Stability of a Predator-Prey Model with Modified Leslie-Gower and Holling-Type II Schemes

Authors: Shengbin Yu
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We study the predator-prey model proposed by Aziz-Alaoui and Okiye (Appl. Math. Lett. 16 (2003) 1069–1075) First, the structure of equilibria and their linearized stability is investigated. Then, we provide two sufficient conditions on the global asymptotic stability of a positive equilibrium by employing the Fluctuation Lemma and Lyapunov direct method, respectively. The obtained results not only improve but also supplement existing ones.


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## Body

## 1. Introduction

One of the important interactions among species is the predator-prey relationship and it has been extensively studied because of its universal existence. There are many factors affecting the dynamics of predator-prey models. One of the familiar factors is the functional response, referring to the change in the density of prey attached per unit time per predator as the prey density changes. In the classical Lotka-Volterra model, the functional response is linear, which is valid first-order approximations of more general interaction. To build more realistic models, Holling [1] suggested three different kinds of functional responses, and Leslie and Gower [2] introduced the so-called Leslie-Gower functional response.Recently, Aziz-Alaoui and Daher Okiye [3] proposed and studied the following predator-prey model with modified Leslie-Gower and Holling-type II schemes,
(1.1)dxdt=(r1-b1x-a1yx+k1)x,dydt=(r2-a2yx+k2)y.
Here, all the parameters are positive, and we refer to Aziz-Alaoui and Daher Okiye [3] for their biological meanings. System (1.1) can be considered as a representation of an insect pest-spider food chain, nature abounds in systems which exemplify this model; see [3].Since then, system (1.1) and its nonautonomous versions have been studied by incorporating delay, impulses, harvesting, and so on (see, e.g., [4–11]). In spite of this extensive study, the dynamics of (1.1) is not fully understood and some existing results are not true. For example, the main result (Theorem  6 on global stability of a positive equilibrium) of Aziz-Alaoui and Daher Okiye [3] is not true as the condition (i) and condition (iii) cannot hold simultaneously. In fact, it follows from condition (i), (1/4a2b1)(a2r1(r1+4)+(r2+1)2(r1+b1k2))<r1k1/2a1, that 2a1a2r1<a2b1r1k1. On the other hand, condition (iii), 4(r1+b1k1)<a1, implies that a1>4b1k1. Then, one can have 8a2b1r1k1<a2b1r1k1, which is impossible. One purpose of this paper is to establish several sufficient conditions on the global asymptotic stability of a positive equilibrium.LetΩ0={(x,y):x≥0,y≥0}. As a result of biological meaning, we only consider solutions (x(t),y(t)) of (1.1) with (x(0),y(0))∈Ω0. Moreover, solutions (x(t),y(t)) of (1.1) with (x(0),y(0))∈Ω0 are called positive solutions. An equilibrium E*=(x*,y*) of (1.1) is called globally asymptotically stable if x(t)→x* and y(t)→y* as t→∞ for any positive solution (x(t),y(t)) of (1.1). System (1.1) is permanent if there exists 0<α<β such that, for any positive solution (x(t),y(t)) of (1.1),
(1.2)α≤min⁡{lim⁡inf⁡t→∞x(t),lim⁡inf⁡t→∞y(t)}≤max⁡{lim⁡sup⁡t→∞x(t),lim⁡sup⁡t→∞y(t)}≤β.
The remaining part of this paper is organized as follows. In Section 2, we discuss the structure of nonnegative equilibria to (1.1) and their linearized stability. This has not been done yet, and the results will motivate us to study global asymptotic stability of (1.1) in Section 3. The obtained results not only improve but also supplement existing ones.


## 2. Nonnegative Equilibria and Their Linearized Stability

The Jacobian matrix of (1.1) is(2.1)J(x,y)=(r1-2b1x-a1k1y(x+k1)2-a1xx+k1a2y2(x+k2)2r2-2a2yx+k2).
An equilibrium E of (1.1) is (linearly) stable if the real parts of both eigenvalues of J(E) are negative and therefore a sufficient condition for stability is
(2.2)tr⁡(J(E))<0,det⁡⁡(J(E))>0.Obviously, (1.1) has three boundary equilibria, E0=(0,0), E1=(r1/b1,0), and E2=(0,r2k2/a2), whose Jacobian matrices are(2.3)(r100r2),(-r1-a1r1r1+b1k10r2),(r1-a1r2k2a2k10r22a2-r2),
respectively. As a direct consequence of (2.2), we have the following result.Proposition 2.1.
(i) BothE0 and E1 are unstable.
(ii)E2 is stable if a1r2k2>a2r1k1, while it is unstable if a1r2k2<a2r1k1.Besides the three boundary equilibria, (1.1) may have (componentwise) positive equilibria. Suppose that E^=(x^,y^) is such an equilibrium. Then,
(2.4)r1-b1x^-a1y^x^+k1=0,r2-a2y^x^+k2=0.
One can easily see that x^ satisfies
(2.5)a2b1x^2+Bx^+(a1r2k2-a2r1k1)=0,
where B≜a1r2-a2r1+a2b1k1. Moreover, for convenience, we denote Δ≜B2-4a2b1(a1r2k2-a2r1k1). Equation (2.5) can have at most two positive solutions, and hence (1.1) can have at most two positive equilibria. Precisely, we have the following three cases.Case 1.
Suppose one of the following conditions holds.(i)
a1r2k2<a2r1k1.(ii)
a1r2k2=a2r1k1 and B<0.(iii)
a1r2k2>a2r1k1, B<0, and Δ=0.
Then, (1.1) has a unique positive equilibrium E3,1=(x3,1,y3,1) with x3,1=(-B+Δ)/2a2b1 and y3,1=r2(x3,1+k2)/a2.Case 2.
Ifa1r2k2>a2r1k1, B<0, and Δ>0, then (1.1) has two positive equilibria E3,±=(x3,±,y3,±), where x3,±=(-B±Δ)/2a2b1 and y3,±=r2(x3,±+k2)/a2.Case 3.
If no condition in Case1 or Case 2 holds, then (1.1) has no positive equilibrium.For a positive equilibriumE^=(x^,y^), J(E^) can be simplified to(2.6)J(E^)=(x^(r1-b1k1-2b1x^)x^+k1-a1x^x^+k1r22a2-r2)
by using (2.4). By simple computation,  tr(J(E^))=(-2b1x^2+(r1-r2-b1k1)x^-k1r2)/(x^+k1), det(J(E^))=(r2x^(2a2b1x^+B))/a2(x^+k1).Then, one can easily see thatdet(J(E3,1))>0 for Case 1(i)-(ii), det(J(E3,1))=0 for Case 1(iii), det(J(E3,+))>0, and det(J(E3,-))<0. Therefore, we obtain the following.Proposition 2.2.
(i) The positive equilibriumE3,1 in Case 1(i)(ii) is stable if 2b1x3,12-(r1-r2-b1k1)x3,1+k1r2>0.
(ii) The positive equilibriumE3,- is unstable, while the positive equilibrium E3,+=(x3,+,y3,+) is stable if 2b1x3,+2-(r1-r2-b1k1)x3,++k1r2>0.Remark 2.3.
In [3, 7, 8], only existence of the positive equilibrium of (1.1) for Case 1(i) was considered, which is stable if either (a) r1≤r2 and k1≥k2 [3] or (b) a1r2k2<a2r1k1 and r1<b1k1 [7, 8]. Obviously, Proposition 2.2 greatly improves these results.Propositions2.1 and 2.2 naturally motivate us to seek sufficient conditions on global asymptotic stability of equilibrium to (1.1) and permanence of (1.1).Nindjin et al. [5] showed that if
(H1)a1r1r2+a1b1r2k2<a2b1r1k1,
then
(2.7)lim⁡sup⁡t→∞x(t)≤K≜r1b1,lim⁡sup⁡t→∞y(t)≤L≜r1r2+b1r2k2a2b1,(2.8)lim⁡inf⁡t→∞x(t)≥M≜r1k1-a1Lb1k1,lim⁡inf⁡t→∞y(t)≥N≜r2(M+k2)a2
for a positive solution (x(t),y(t)) of (1.1). Therefore, system (1.1) is permanent if (H1) holds. With the help of these bounds, it was shown that E2 is globally asymptotically stable if r1(k1+K)<a1N holds (see [5]).In the coming section, we present two results on the global asymptotic stability of a positive equilibrium, which not only supplement Theorem 7 of Nindjin et al. [5] but also improve it by including more situations.


## 3. Global Asymptotic Stability of a Positive Equilibrium

The first result is established by employing the Fluctuation Lemma, and we refer to [12–16] for details.Theorem 3.1.
In addition to(H1), further suppose that
(H2)2a2b1M+(a2b1k1-a2r1-a1r2)>0,
where M is defined in (2.8). Then, system (1.1) has a unique positive equilibrium which is globally asymptotically stable.Proof.
Obviously,(H1) implies a1r2k2<a2r1k1, that is, condition (i) of Case 1 holds. Thus, (1.1) has a unique positive equilibrium. Let (x(t),y(t)) be any positive solution of (1.1). By the results at the end of Section 2,  x¯≜limsupt→∞x(t)≥x_≜liminft→∞x(t)≥M,  y¯≜limsupt→∞y(t)≥y_≜liminft→∞y(t)>0.
We claimx¯=x_. Otherwise, x¯>x_. According to the Fluctuation lemma, there exist sequences ξn→∞, ηn→∞, τn→∞, and σn→∞ as n→∞ such that x˙(ξn)→0, x˙(ηn)→0, x(ξn)→x¯, x(ηn)→x_, y˙(τn)→0, y˙(σn)→0, y(τn)→y¯, and y(σn)→y_ as n→∞. First, from the second equation of (1.1),
(3.1)y˙(τn)≤(r2-a2y(τn)sup⁡t≥τn⁡x(t)+k2)y(τn),y˙(τn)≥(r2-a2y(τn)inf⁡t≥τn⁡x(t)+k2)y(τn).
Letting n→∞, we obtain that 0≤(r2-a2y¯/(x¯+k2))y¯ and 0≥(r2-a2y¯/(x_+k2))y¯. Hence,
(3.2)r2(x_+k2)a2≤y¯≤r2(x¯+k2)a2.
Similar arguments as above also produce
(3.3)r2(x_+k2)a2≤y_≤r2(x¯+k2)a2.
Second, from the first equation of (1.1),
(3.4)x˙(ξn)=(r1-b1x(ξn)-a1y(ξn)x(ξn)+k1)x(ξn).
Equation (3.4) implies  x˙(ξn)≤(r1-b1x(ξn)-(a1inft≥ξny(t)/(x(ξn)+k1)))x(ξn).
Taking limit asn→∞, one obtains 0≤(r1-b1x¯-a1y_/(x¯+k1))x¯. This, combined with (3.3), gives us 0≤(r1-b1x¯-a1r2(x_+k2)/a2(x¯+k1))x¯. It follows that
(3.5)(a2r1-a2b1k1)x¯-a2b1x¯2+a2r1k1≥a1r2(x_+k2).
Similarly, one can show that
(3.6)(a2r1-a2b1k1)x_-a2b1x_2+a2r1k1≤a1r2(x¯+k2).
Multiplying (3.5) by -1 and adding it to (3.6), we have
(3.7)a2b1(x¯2-x_2)+(a2b1k1-a2r1-a1r2)(x¯-x_)≤0.
Due to x¯>x_, one gets a2b1(x¯+x_)+(a2b1k1-a2r1-a1r2)≤0 which contradicts (H2). Therefore, x¯=x_, and the claim is proved.
The claim implies thatlimt→∞x(t) exists and we denote it by x*. Then, it follows from (3.2) and (3.3) that limt→∞y(t) exists and limt→∞y(t)≜y*=r2(x*+k2)/a2>0. Letting n→∞ in (3.4) gives us r1-b1x*-a1r2(x*+k2)/a2(x*+k1)=0. Then, one can see that (x*,y*) satisfies (2.4), that is, (x*,y*) is a positive equilibrium of (1.1). This completes the proof as the positive equilibrium is unique.Theorem 3.2.
Suppose that (1.1) has a unique positive equilibrium E*=(x*,y*). Further assume that
(H3)(2k2+k12k1k2)a1L+a12<b1(x*+k1),L<k2,
where L is defined in (2.7). Then, E* is globally asymptotically stable.Proof.
Let(x(t),y(t)) be any positive solution of (1.1). From (H3), we can choose an ɛ>0 such that
(3.8)(2k2+k12k1k2)a1(L+ɛ)+a12<b1(x*+k1),L+ɛ<k2.
Moreover, it follows from (2.7) that there exists T>0 such that
(3.9)0<y(t)≤L+ɛfort≥T.
According to the proof of Theorem 6 in [3], let
(3.10)V(x,y)=(x*+k1)(x-x*-x*ln⁡⁡(xx*))+a1(x*+k2)a2(y-y*-y*ln⁡⁡(yy*)).
Then, by the positivity of x, (3.8), and (3.9),
(3.11)dVdt=(-b1(x*+k1)+a1yx+k1)(x-x*)2-a1(y-y*)2+(-a1+a1yx+k2)(x-x*)(y-y*)≤(-b1(x*+k1)+a1yk1)(x-x*)2-a1(y-y*)2+(a1+a1yx+k2)(x-x*)2+(y-y*)22≤(-b1(x*+k1)+a1yk1)(x-x*)2-a1(y-y*)2+(a1+a1yk2)(x-x*)2+(y-y*)22=(-b1(x*+k1)+a1yk1+a1y2k2+a12)(x-x*)2+(a1y2k2-a12)(y-y*)2≤(-b1(x*+k1)+(2k2+k12k1k2)a1(L+ɛ)+a12)(x-x*)2+a12(L+ɛk2-1)(y-y*)2<0.
Therefore, E*(x*,y*) is globally asymptotically stable, and this completes the proof.



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