# Global Attractivity in a Discrete Mutualism Model with Infinite Deviating Arguments

Authors: Xiangdong Xie; Yalong Xue; Runxin Wu
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

A set of sufficient conditions is obtained for the global attractivity of the following two-species discrete mutualism model with infinite deviating arguments:x1(k+1)=x1kexp⁡r1(K1+α1∑s=0+∞J2sx2k-s)/(1+∑s=0+∞J2sx2k-s)-x1k and x2(k+1)=x2kexp⁡r2(K2+α2∑s=0+∞J1sx1k-s)/(1+∑s=0+∞J1sx1k-s)-x2k, where ri,Ki,αi, i=1,2, are all positive constants, ∑j=1+∞Ji(n)=1, and αi>Ki. Our results generalize the main result of Yang et al. (2014).


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## Body

## 1. Introduction

The aim of this paper is to investigate the stability property of the following two-species discrete mutualism model with infinite deviating arguments:(1)x1k+1=x1kexp⁡r1K1+α1∑s=0+∞J2sx2k-s1+∑s=0+∞J2sx2k-s-x1k,x2k+1=x2kexp⁡r2K2+α2∑s=0+∞J1sx1k-s1+∑s=0+∞J1sx1k-s-x2k,together with the initial conditions (2)xis=ϕis≥0,xi0>0,s=⋯,-k,-k+1,…,-2,-1,i=1,2.Li and Xu [1] studied the following two-species integrodifferential model of mutualism: (3)N1′t=r1tN1tK1t+α1t∫0∞J2sN2t-sds1+∫0∞J2sN2t-sds-N1t-σ1t,N2′t=r2tN2tK2t+α2t∫0∞J1sN1t-sds1+∫0∞J1sN1t-sds-N2t-σ2t.Under the assumption Ki(t),αi(t),i=1,2, are all positive periodic functions and αi>Ki,i=1,2, by applying the coincidence degree theory, they showed that system (3) admits at least one positive ω-periodic solution. Chen and You [2] argued that a general nonautonomous nonperiodic system is more appropriate, and for the general nonautonomous case, by using the differential inequality theory, they showed that the system is permanent. It brings to our attention that both [1, 2] did not consider the stability property of the system, and in [3], under the assumption ri,Ki,αi,i=1,2, are all positive constants, σi(t)≡0, we investigated the stability property of the system, and we showed that the system admits a unique globally attractive positive equilibrium. At the end of the paper, we pointed out “whether some parallel result could be established for the discrete type mutualism system is still unknown, we leave this for future investigation.”Previously, corresponding to system (3), Li and Yang [4] and Li [5] proposed the following two-species discrete model of mutualism with infinite deviating arguments: (4)x1k+1=x1kexp⁡r1kK1k+α1k∑s=0+∞J2sx2k-s1+∑s=0+∞J2sx2k-s-x1k-δ1k,x2k+1=x2kexp⁡r2kK2k+α2k∑s=0+∞J1sx1k-s1+∑s=0+∞J1sx1k-s-x2k-δ2k,where xi(k),i=1,2, is the density of mutualism species i at the kth generation and {ri(k)},{Ki(k)},{αi(k)},{Ji(k)}, and {δi(k)},i=1,2, are bounded nonnegative sequences such that (5)0<ril≤riu,0<ail≤aiu,0<Kil≤Kiu,0<δil≤δiu,∑j=1+∞Jin=1,αi>Ki.They showed that, under the above assumption, system (4) is permanent. Again, none of the papers [4, 5] considered the stability property of the system. To make an intensive study on this direction, in [6], we investigated the dynamic behaviors of the following autonomous mutualism system: (6)x1k+1=x1kexpr1K1+α1x2k1+x2k-x1k,x2k+1=x2kexp⁡r2K2+α2x1k1+x1k-x2k,where xi(k)(i=1,2) are the population density of the ith species at k-generation. We showed that if (H1)
ri,Ki,αi(i=1,2) are all positive constants and αi>Ki(i=1,2);(H2)
riαi≤1,(i=1,2)hold, system (6) admits a unique positive equilibrium (x1∗,x2∗), which is globally asymptotically stable. Our result shows that the dynamic behavior of the discrete type mutualism model is more complicated, and one could not expect to establish parallel result as that of continuous ones. Also, at the end of the paper, we pointed out “it seems interesting to incorporate the time delay to the system (6) and investigate the dynamic behaviors of the system, we leave this for future study.” However, to this day, we still did not study the correspondence topic on this area. For more background of system (3), (4), and (6) one could refer to [1–24] and the references cited therein. We mention here that, with σi(k)≠0,i=1,2, and all the coefficients being time-dependent, system (4) is a nonautonomous pure-delay system, and it is not an easy thing to investigate the stability property of the system. This motivated us to discuss the simple one, that is, the autonomous simple non-pure-delay system (1).Concerned with the stability property of system (1)-(2), we have the following result.Theorem 1.
Assume that (H1) and (H2) hold, and then system (1)-(2) admits a unique positive equilibrium (x1∗,x2∗), which is globally attractive.Remark 2.
Obviously, Theorem1 generalizes the main results of Yang et al. [6] to the infinite deviating arguments case. Theorem 1 can also be seen as the parallel result of the continuous one in [3]. Thus, we push on the study of the mutualism model.


## 2. Existence and Uniqueness of Positive Equilibrium

This section focuses on the existence and uniqueness of positive equilibrium of system (1). More precisely, we will prove the following result.Theorem 3.
Under the assumption of Theorem1, system (1)-(2) admits a unique positive equilibrium.Proof.
The positive equilibrium of system (1) satisfies (7)K1+α1x21+x2-x1=0,K2+α2x11+x1-x2=0,which is equivalent to (8)A1x12+A2x1+A3=0,B1x22+B2x2+B3=0,where (9)A1=α2+1,A2=-α1α2-K1+K2+1,A3=-K2α1-K1,B1=α1+1,B2=-α1α2+K1-K2+1,B3=-K1α2-K2.Now let us consider the function (10)F1x1=A1x12+A2x1+A3,and since A1>0,A3<0, it follows that F1(-∞)=+∞,F1(0)=0,F1(+∞)=+∞, and, hence, from the continuity of F1, there exist two points x1∗∗ and x1∗,x1∗∗<0<x1∗, such that F1(x1∗∗)=F1(x1∗)=0, and since F1(x1)=0 has at most two solutions, it means that F1(x1)=0 admits unique positive solution x1∗. Similarly, from B1>0,B3<0, one could prove F2(x2)=B1x22+B2x2+B3=0 admits unique positive solution x2∗. By simple computation, system (7) admits a unique positive solution E+(x1∗,x2∗), where (11)x1∗=-A2+A22-4A1A32A1,x2∗=-B2+B22-4B1B32B1.This ends the proof of Theorem 3.


## 3. Proof of Theorem1

Now we state several lemmas which will be useful in the proof of Theorem1.Lemma 4 (see [25]).
Letf(u)=uexp⁡(α-βu), where α and β are positive constants, and then f(u) is nondecreasing for u∈(0,1/β].Lemma 5 (see [25]).
Assume that sequence{u(k)} satisfies(12)uk+1=ukexp⁡α-βuk,k=1,2,…,where α and β are positive constants and u(0)>0. Then(i)
Ifα<2, then limk→+∞⁡u(k)=α/β.(ii)
Ifα≤1, then uk≤1/β,k=2,3,….Lemma 6 (see [26]).
Suppose that functionsf,g:Z+×[0,∞)→[0,∞) satisfy f(k,x)≤g(k,x)(f(k,x)≥g(k,x)) for k∈Z+ and x∈[0,∞) and g(k,x) is nondecreasing with respect to x. If {x(k)} and {u(k)} are the nonnegative solutions of the following difference equations:(13)xk+1=fk,xk,uk+1=gk,uk,respectively, and x0≤u(0)(x(0)≥u(0)), then(14)xk≤ukxk≥uk,∀k≥0.Lemma 7 (see [27]).
Letx:Z→R be nonnegative bounded sequences, and let H:N→R be nonnegative sequences such that ∑n=0∞H(n)=1. Then (15)liminfn→+∞⁡xn≤liminfn→+∞⁡∑s=-∞nHn-sxs≤limsupn→+∞⁡∑s=-∞nHn-sxs≤limsupn→+∞⁡xn.Lemma 8.
Letgi(x)=(Ki+αix)/(1+x),i=1,2, assume that αi>Ki, and then gi(x) are the strictly increasing function of x.Proof.
Since(16)gi′x=-Ki-αi1+x2>0,i=1,2,the conclusion of Lemma 8 immediately follows.Now we are in the position to prove the main result of this paper.Proof of Theorem1.
Let(x1(k),x2(k)) be arbitrary solution of system (1) with initial condition (2). Denote (17)Ui=limsupk→+∞⁡xik,Vi=liminfk→+∞⁡xik,i=1,2.We claim that U1=V1=x1∗ and U2=V2=x2∗.
From the first equation of system (1), we obtain(18)x1k+1=x1kexp⁡r1K1+α1∑s=0+∞J2sx2k-s1+∑s=0+∞J2sx2k-s-x1k≤x1kexp⁡r1α1+α1∑s=0+∞J2sx2k-s1+∑s=0+∞J2sx2k-s-x1k≤x1kexp⁡r1α1-r1x1k,k=0,1,2,…,considering the auxiliary equation as follows: (19)uk+1=ukexp⁡r1α1-r1uk,k=0,1,2,….Because of 0<r1α1≤1, according to (ii) of Lemma 5, we can obtain u(k)≤1/r1 for all k≥2, where u(k) is arbitrary positive solution of (18) with initial value u(0)>0. From Lemma 4, fu=uexp⁡(r1α1-r1u) is nondecreasing for u∈(0,1/r1]. According to Lemma 6 we can obtain x1(k)≤u(k) for all k≥2, where u(k) is the solution of (19) with the initial value u(2)=x1(2). According to (i) of Lemma 5, we can obtain (20)U1=limsupk→+∞⁡x1k≤limk→+∞⁡uk=α1.From (20) and Lemma 7 we have (21)limsupk→+∞⁡∑s=0+∞J1sx1k-s=limsupn→+∞⁡∑s=-∞kJ1k-sx1s≤limsupk→+∞⁡x1k≤α1.From the second equation of system (1), we obtain (22)x2k+1≤x2kexp⁡r2α2-r2x2k,k=0,1,2,….Similar to the above analysis, we have (23)U2=limsupk→+∞⁡x2k≤α2.From (23) and Lemma 7 we have (24)limsupk→+∞⁡∑s=0+∞J2sx2k-s=limsupn→+∞⁡∑s=-∞kJ2k-sx2s=limsupk→+∞⁡x2k≤α2.For ε>0 enough small, without loss of generality, we may assume that ε<(1/2)min⁡{K1,K2}, and it follows from (20)–(24) that there is an integer k1>2 such that, for all k>k1, (25)x1k<α1+ε=defM1x1,x2k<α2+ε=defM1x2,(26)∑s=0+∞J1sx1k-s<α1+ε=defM1x1,∑s=0+∞J2sx2k-s<α2+ε=defM1x2.For k≥k1, according to the first equation of system (1) we can obtain (27)x1k+1=x1kexp⁡r1K1+α1∑s=0+∞J2sx2k-s1+∑s=0+∞J2sx2k-s-x1k≥x1kexp⁡r1K1+K1∑s=0+∞J2sx2k-s1+∑s=0+∞J2sx2k-s-x1k≥x1kexp⁡r1K1-r1x1k,considering the auxiliary equation as follows: (28)uk+1=ukexp⁡r1K1-r1uk.According to (ii) of Lemma 5, we can obtain u(k)≤1/r1 for all k≥k2, where u(k) is arbitrary positive solution of (28) with initial value u(k2)>0. From Lemma 4, f(u)=uexp⁡(r1K1-r1u) is nondecreasing for u∈(0,1/r1]. According to Lemma 6 we can obtain x1(k)≥u(k) for all k≥2, where u(k) is the solution of (28) with the initial value u(k2)=x1(k2). According to (i) of Lemma 5, we have (29)V1=liminfk→+∞⁡x1k≥limk→+∞⁡uk=K1.From (29) and Lemma 7 we can obtain (30)liminfk→+∞⁡∑s=0+∞J1sx1k-s≥liminfk→+∞⁡x1k≥K1.From the second equation of system (1), we obtain (31)x2k+1≥x2kexp⁡r2K2-r2x2k.Similar to the analysis of (27)–(30), we have (32)V2=liminfk→+∞⁡x2k≥K2,liminfk→+∞⁡∑s=0+∞J2sx2k-s≥K2.Then, for the above ε>0, there is an integer k2>k1 such that, for all k>k2, (33)x1k>K1-ε=defm1x1,x2k>K2-ε=defm1x2;∑s=0+∞J1sx1k-s>K1-ε=defm1x1,∑s=0+∞J2sx2k-s>K2-ε=defm1x2.Noting that, from Lemma 8, gi(x)=(Ki+αix)/(1+x) (αi>Ki) is a strictly increasing function, then, from the first and second equations of system (1) and (26), we have (34)x1k+1=x1kexpr1K1+α1∑s=0+∞J2sx2k-s1+∑s=0+∞J2sx2k-s-x1k≤x1kexp⁡r1K1+α1M1x21+M1x2-x1k,x2k+1=x2kexp⁡r2K2+α1∑s=0+∞J1sx1k-s1+∑s=0+∞J1sx1k-s-x2k≤x2kexp⁡r2K2+α2M1x11+M1x1-x2k.From (34), similarly to the analysis of (18)–(24), we can finally obtain(35)limsupk→+∞⁡∑s=0+∞J1sx1k-s≤limsupk→+∞⁡x1k≤K1+α1M1x21+M1x2,limsupk→+∞⁡∑s=0+∞J2sx2k-s≤limsupk→+∞⁡x2k≤K2+α2M1x11+M1x1.For the above ε>0, it follows from (35) that there exists an integer k3>k2 such that, for all k>k3, (36)x1k<K1+α1M1x21+M1x2+ε2=defM2x1,x2k<K2+α2M1x11+M1x1+ε2=defM2x2;∑s=0+∞J1sx1k-s<K1+α1M1x21+M1x2+ε2=defM2x1,∑s=0+∞J2sx2k-s<K2+α2M1x11+M1x1+ε2=defM2x2.It then follows from (25), (26), and (36) that (37)M2xi<M1xi,i=1,2.For k≥k3, from the strictly increasing function gi(x)=(Ki+αix)/(1+x),αi>Ki,i=1,2, and (33), we can obtain (38)x1k+1≥x1kexp⁡r1K1+α1m1x21+m1x2-x1k,x2k+1≥x2kexp⁡r2K2+α2m1x11+m1x1-x2k.From (38), similar to the analysis of (27)–(32), we can obtain (39)V1=liminfk→+∞⁡x1k≥K1+α1m1x21+m1x2,liminfk→+∞⁡∑s=0+∞J1sx1k-s≥K1+α1m1x21+m1x2,V2=liminfk→+∞⁡x2k≥K2+α2m1x11+m1x1,liminfk→+∞⁡∑s=0+∞J1sx1k-s≥K2+α2m1x11+m1x1.For the above ε>0, it follows from (39) that there is an integer k4>k3 such that, for all k>k4, (40)x1k>K1+α1m1x21+m1x2-ε2=defm2x1,x2k>K2+α2m1x11+m1x1-ε2=defm2x2,∑s=0+∞J1sx1k-s>K1+α1m1x21+m1x2-ε2=defm2x1,∑s=0+∞J2sx2k-s>K2+α2m1x11+m1x1-ε2=defm2x2.Noting that (41)K1+α1m1x21+m1x2>K1,K2+α2m1x11+m1x1>K2,i=1,2.Then from (33) and (40) we have (42)m2xi>m1xi,i=1,2.Continuing the above steps, we can get four sequences {Mkx1}, {Mkx2}, {mkx1}, and {mkx2} such that (43)Mkx1=K1+α1Mk-1x21+Mk-1x2+εk,Mkx2=K2+α2Mk-1x11+Mk-1x1+εk;mkx1=K1+α1mk-1x21+mk-1x2-εk,mkx2=K2+α2mk-1x11+mk-1x1-εk.Clearly, we have (44)mkx1<V1≤U1<Mkx1,mkx2<V2≤U2<Mkx2,k=0,1,2,….
Now, we will prove{Mkxi}(i=1,2) is monotonically decreasing and {mkxi}(i=1,2) is monotonically increasing by means of inductive method.
First of all, from (37) and (42) we have M2xi<M1xi,m2xi>m1xi(i=1,2). For k≥2, we assume that Mkxi<Mk-1xi and mkxi>mk-1xi,i=1,2, holds, and then from the strictly increasing of function gi(x)=(Ki+αix)/(1+x),i=1,2, it immediately follows that (45)Mk+1x1=K1+α1Mkx21+Mkx2+εk+1<K1+α1Mk-1x21+Mk-1x2+εk=Mkx1;Mk+1x2=K2+α2Mkx11+Mkx1+εk+1<K2+α2Mk-1x11+Mk-1x1+εk=Mkx2,mk+1x1=K1+α1mkx21+mkx2-εk+1>K1+α1mk-1x21+mk-1x2-εk=Mkx1;mk+1x2=K2+α2mkx11+mkx1-εk+1>K2+α2mk-1x11+mk-1x1-εk=Mkx2.Equations of (45) show that {Mkxi}(i=1,2) is monotonically decreasing and {mkxi}(i=1,2) is monotonically increasing. Consequently, limk→+∞{Mkxi} and limk→+∞{mkxi}(i=1,2) both exist. Let(46)limk→+∞⁡Mkxi=X¯i,limk→+∞⁡mkxi=X_i,i=1,2.From (43), we have (47)X¯1=K1+α1X¯21+X¯2;X¯2=K2+α2X¯11+X¯1;X_1=K1+α1X_21+X_2;X_2=K2+α2X_11+X_1;Here, (47) shows that (X¯1,X¯2) and (X_1,X_2) are all solutions of system (7). However, system (7) has unique positive solution (x1∗,x2∗). Therefore (48)Ui=Vi=limk→+∞⁡xik=xi∗,i=1,2;that is, E+(x1∗,x2∗) is globally attractive. The proof of the theorem is completed.


## 4. Examples

In this section we shall give an example to illustrate the feasibility of the main result.Example 1.
Consider the following example:(49)x1k+1=x1kexp⁡30.2+0.3∑s=0∞e-1/ee-sx2n-s1+∑s=0∞e-1/ee-sx2n-s-x1k,x2k+1=x2kexp⁡0.50.5+1.5∑s=0∞e-1/ee-sx1n-s1+∑s=0∞e-1/ee-sx1n-s-x2k.Corresponding to system (1), we have r1=3,K1=0.2,α1=0.3,r2=0.5,K2=0.5,α2=1.5, and hence (50)αi>Ki,i=1,2.Also, (51)r1α1=0.9<1,r2α2=0.75<1.Hence, all the conditions of Theorem 1 hold, and it follows from Theorem 1 that system (49) admits a unique globally attractive positive equilibrium. Figure 1 supports this assertion.Figure 1
Dynamic behaviors of the solution(x1(n),x2(n)) of system (49), with the initial conditions (x1(s),x2(s))=(0.8,0.4), (0.5,0.5), and (1.5,1.5), s=⋯,-n,-n+1,…,-1,0, respectively.Example 2.
Consider the following example:(52)x1k+1=x1kexp⁡50.2+0.3∑s=0∞e-1/ee-sx2n-s1+∑s=0∞e-1/ee-sx2n-s-x1k,x2k+1=x2kexp⁡10.5+1.5∑s=0∞e-1/ee-sx1n-s1+∑s=0∞e-1/ee-sx1n-s-x2k.Corresponding to system (1), we have r1=5,K1=0.2,α1=0.3,r2=1,K2=0.5,α2=1.5, and, obviously, (53)αi>Ki,i=1,2.However, (54)r1α1=1.5>1,r2α2=1.5>1.Hence, condition (H2) in Theorem 1 could not be satisfied, and Theorem 1 could not be applied to this example. However, numeric simulation (Figure 2) also shows that system (52) admits a unique globally attractive positive equilibrium.Figure 2
Dynamic behaviors of the solution(x1(n),x2(n)) of system (52), with the initial conditions (x1(s),x2(s))=(0.8,0.4), (0.5,0.5), and (1.5,1.5), s=⋯,-n,-n+1,…,-1,0, respectively.


## 5. Discussion

In [6], Yang et al. proposed system (6); under the assumption αi>Ki,i=1,2, they showed that if riαi≤1, then the mutualism model admits a unique globally asymptotically stable positive equilibrium.In this paper, we try to incorporate the infinite deviating arguments, and, by developing the analysis technique of Yang et al. [6] and using the difference inequality of Chen [7], we also obtain the sufficient conditions which ensure the global attractivity of the positive equilibrium. Example 1 shows the feasibility of our main result.Since condition (H2) is the most important restriction on the coefficients of the system, one interesting issue is whether the result of Theorem 3 could hold if (H2) is not satisfied. Example 2 shows that our result (Theorem 3) still have room to improve. We leave this for future investigation.



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