# Dynamic Behavior of a Commensalism Model with Nonmonotonic Functional Response and Density-Dependent Birth Rates

Authors: Liang Zhao; Bin Qin; Xianbo Sun
Journal: Complexity
 (2018)

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## Abstract

In this paper, we propose and analyze a commensalism model with nonmonotonic functional response and density-dependent birth rates. The model can have at most four nonnegative equilibria. By applying the differential inequality theory, we show that each equilibrium can be globally attractive under suitable conditions. However, commensalism can be established only when resources for both species are large enough.


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## Body

## 1. Introduction

Commensalism is a long-term biological interaction in which members of one species gain benefits while those of the other species neither benefit nor are harmed. An example of it is that remora are specially adapted to attach themselves to larger fish that provide locomotion and food. In the last decades, commensalism has attracted the attention of many researchers ([1–16]). Complicated dynamics have been found in the study. For example, in [3], Lin considered the effects of partial closure and harvesting. Depending on the size of the harvesting area, species can go extinct, partially survive, or become permanent. He also showed in [4] that the final density of the species increases as the Allee effect increases. This is quite different from results for predator-prey system with Allee effect.Recently, Chen and Wu [5] proposed the following two species commensal symbiosis models with nonmonotonic functional response:(1)dxdt=xa1-b1x+c1yd1+y2,dydt=ya2-b2y,where a1,b1,c1,d1,a2,b2 are all positive constants. System (1) admits four nonnegative equilibria, A0(0,0), A1(a1/b1,0), A2(0,a2/b2), and A3(x∗,y∗), where (2)x∗=a1b22d1+a1a22+a2b2c1b1b22d1+a22,y∗=a2b2.The stability of the equilibria is summarized as follows (see Theorem 2.1 and 2.2 in [5] for more detail).Theorem A.
A 0 ( 0,0 ), A1(a1/b1,0), and A2(0,a2/b2) are unstable; A3(x∗,y∗) is globally asymptotically stable.When the interaction between the species is ignored, the growth for both species is described by traditional logistic equations. Indeed, without the presence ofy, the growth of the first species takes the form(3)dxdt=xa1-b1x,where a1 is the intrinsic growth rate and b1 is the density-dependent coefficient or the interspecific competition coefficient. However, in most situations, the intrinsic growth rate is not always constant. One model incorporating nonconstant intrinsic growth rate is the following density-dependent model:(4)dxdt=xa11a12+a13x-a14-b1x.For more details, see [6–8]. Combining this with (1), we propose the following commensalism model:(5)dxdt=xa11a12+a13x-a14-b1x+c1yd1+y2,dydt=ya21a22+a23y-a24-b2y,where aij (i=1,2,j=1,2,3,4) and b1,c1,d1, and b2 are all positive constants. Here x(t) and y(t) are the densities of the first and second species at time t, respectively. a11 and a21 stand for the total resources available per-unit-time for species x and y, respectively.The aim of this paper is to investigate the attractivity of equilibria of (5). The main tool is the differential inequality theory or comparison principle. To the best of our knowledge, this is the first time to use differential inequality in this direction for ecosystems. The rest of the paper is arranged as follows. In Section 2, we obtain the existence and global attractivity of equilibria of system (5). Section 3 is devoted to illustrating the feasibility of the main results through numeric simulations. We end this paper by a brief discussion.


## 2. The Main Result

We first consider the existence of equilibria of (5). An equilibrium of (5) satisfies the equilibrium equations,(6)xa11a12+a13x-a14-b1x+c1yd1+y2=0,(7)ya21a22+a23y-a24-b2y=0.If a21≤a22a24 then (7) only has the unique nonnegative solution y=0 while if a21>a22a24, then, besides y=0, (7) also has a unique positive solution y∗=(-(a24a23+a22b2)+(a24a23+a22b2)2+4b2a23(a21-a22a24))/2b2a23. Substituting y=0 into (6), we see that if a11≤a12a14, then x=0 is the only nonnegative solution while if a11>a12a14, besides x=0, (6) also has a unique positive solution x∗=(-(a14a13+a12b1)+(a14a13+a12b1)2+4b1a13(a11-a12a14))/2b1a13. Similarly, substituting y=y∗ into (6), we can get that x=0 is the only nonnegative solution if a11≤a12a14∗ while, besides x=0, (6) also has a unique positive solution x∗∗=(-(a14∗a13+a12b1)+(a14∗a13+a12b1)2+4b1a13(a11-a12a14∗))/2b1a13 if a11>a12a14∗, where a14∗=a14-c1y∗/(d1+(y∗)2). In summary, we have obtained the following result.Proposition 1.
The following statements on equilibria of (5) are valid. (i)
Ifa11≤a12a14 and a21≤a22a24 then there is only the trivial equilibrium A0=(0,0).(ii)
Ifa11>a12a14 and a21≤a22a24 then, besides A0, there is also the nontrivial boundary equilibrium A1=(x∗,0).(iii)
Ifa21>a22a24 and a11≤a12a14∗ then there are only the two equilibria A0 and A2(0,y∗).(iv)
Ifa21>a22a24 and a12a14∗<a11≤a12a14 then there are only three equilibria A0, A2, and A3(x∗∗,y∗).(v)
Ifa21>a22a24 and a11>a12a14 then there are only four equilibria A0, A1, A2, and A3.Before analyzing the stability of the equilibria of (5), we first consider the dynamic behavior of the following equation: (8)dydt=ya21a22+a23y-a24-b2ywith y(0)=y0∈[0,+∞). Clearly, every such solution of (8) is nonnegative.Lemma 2.
The following statements on (8) hold. (i)
Ifa21>a22a24 then the unique positive equilibrium y∗ is globally attractive in (0,+∞).(ii)
Ifa21≤a22a24 then the equilibrium y=0 is globally attractive in [0,+∞).Proof.
Denote(9)Fy=a21a22+a23y-a24-b2y.Note that(10)Fy=-b2a23y2-a24a23+a22b2y+a21-a22a24a22+a23y.
(i) Whena21>a22a24, it is easy to see that F only has the unique positive zero y∗. Observe that yF(y)>0 for y∈(0,y∗) and yF(y)<0 for y>y∗. It follows easily that limt→+∞y(t)=y∗ if y0>0; that is, y∗ is globally attractive in (0,+∞).
(ii) Whena21≤a22a24, clearly F can not have positive zero and hence y=0 is the only equilibrium. As yF(y)<0 for y>0, we obtain limt→+∞y(t)=0. This completes the proof.Now we are ready to study the attractivity of the equilibria of (5).Theorem 3.
(i) Assume thata21≤a22a24 and a11<a12a14. Then A0 is globally attractive in [0,+∞)×[0,+∞).
(ii) Suppose thata21≤a22a24 and a11>a12a14. Then A1 is globally attractive in (0,+∞)×[0,+∞).
(iii) Assume thata21>a22a24 and a11<a12a14∗. Then A2 is globally attractive in [0,+∞)×(0,+∞).
(iv) Assume thata21>a22a24 and a11>a12a14∗. Then the unique positive equilibrium A3 is globally attractive in (0,+∞)×(0,+∞).Proof.
First, assume thata21≤a22a24. By Lemma 2 (ii), we have limt→+∞y(t)=0 for y0∈[0,+∞).
(i) Asa11<a12a14, we can choose ε>0 small enough so that a11<a12(a14-c1ε/d1). For this ε, there exists T1≥0 such that y(t)<ε for t≥T1. This, together with the first equation of (5), gives(11)dxdt≤xa11a12+a13x-a14-b1x+c1εd1fort≥T1.It follows from the choice of ε, Lemma 2 (ii), and the comparison principle that limt→+∞x(t)=0. Therefore, limt→+∞(x(t),y(t))=A0; that is, A0 is globally attractive in [0,+∞)×[0,+∞).
(ii) On the one hand, for anyε>0 such that a14(ε)=a14-c1ε/d1>0, there exists Tε≥0 such that y(t)<ε for t≥T1. Thus it follows from the first equation of (5) that(12)dxdt≤xa11a12+a13x-a14ε-b1xfort≥Tε.Note that a11>a12a14>a12a14(ε). By comparison principle and Lemma 2 (i), (13)limsupt→+∞⁡xt≤x∗ε=-a14εa13+a12b1+a14εa13+a12b12+4b1a13a11-a12a14ε2b1a13.Letting ε→0+ gives limsupt→+∞x(t)≤x∗. On the other hand, note that(14)dxdt≥xa11a12+a13x-a14-b1x.Again, by comparison principle and Lemma 2 (i), we have liminft→+∞x(t)≥x∗. It follows that limt→+∞x(t)=x∗. In summary, limt→+∞(x(t),y(t))=A1; namely, A1 is globally attractive in (0,+∞)×[0,+∞).
Now suppose thata21>a22a24. Then limt→+∞y(t)=y∗ for y0>0 by Lemma 2 (i).
(iii) Sincea11<a12a14∗, we choose ε>0 sufficiently small so that a11<a12a14∗(ε), where a14∗(ε)=a14-c1(y∗+ε)/(d1+(y∗-ε)2). For this ε, there exists T^ε≥0 such that(15)y∗-ε<yt<y∗+εfort≥T^ε.This, combined with the first equation of (5), gives(16)dxdt≤xa11a12+a13x-a14∗ε-b1xfort≥T^ε.Then limt→+∞x(t)=0 by the choice of ε, Lemma 2 (ii), and the comparison principle. Thus we have shown limt→+∞(x(t),y(t))=A2; that is, A2 is globally attractive in [0,+∞)×(0,+∞).
(iv) This timea11>a12a14∗. For any ε>0 such that a^14∗=a14-c1(y∗+ε)/(d1+(y∗-ε)2)>0 and a11>a12a~14∗(ε), there exists T~ε≥0 such that(17)y∗-ε<yt<y∗+εfort≥T~ε,where a~14∗(ε)=a14-c1(y∗-ε)/(d1+(y∗+ε)2). Again, employing the first equation of (5), we have(18)xa11a12+a13x-a~14∗ε-b1x≤dxdt≤xa11a12+a13x-a^14∗ε-b1xfor t≥T~ε. Note that a11>a12a~14∗(ε)>a12a^14∗(ε). Applying Lemma 2 (i) and comparison principle again, we have(19)x~∗ε≤liminft→+∞⁡xt≤limsupt→+∞⁡xt≤y^∗ε,where (20)x~∗∗ε=-a~14∗εa13+a12b1+a~14∗εa13+a12b12+4b1a13a11-a12a~14∗ε2b1a13,x^∗∗ε=-a^14∗εa13+a12b1+a^14∗εa13+a12b12+4b1a13a11-a12a^14∗ε2b1a13.Letting ε→0+, we get x∗∗≤liminft→+∞x(t)≤limsupt→+∞x(t)≤x∗∗ and hence limt→+∞x(t)=x∗∗. It follows that limt→+∞(x(t),y(t))=A3 and so A3 is globally attractive in (0,+∞)×(0,+∞). This completes the proof.


## 3. Numeric Simulations

In this section, we provide numeric simulations to illustrate the four situations in Theorem3.Example 4.
Leta11=3, a12=2, a13=2, a14=3, b1=3, c1=2, d1=3, a21=1, a22=2, a23=3, a24=1, and b2=2. Then (5) becomes(21)dxdt=x32+2x-2-3x+2y3+y2,dydt=y12+3y-1-2y.Clearly, a21=1<2=a22a24 and a11=3<6=a12a14. By Theorem 3(i), the boundary equilibrium A0(0,0) is globally attractive. Figure 1 strongly supports it.Figure 1
Solutions(x(t),y(t)) of system (21) with the initial conditions (x(0),y(0))=(1,0.3), (0.4,2), and (0.7,1.1).Example 5.
Consider(22)dxdt=x31+x-2-x+2y3+y2,dydt=y41+y-1-y.Corresponding to (5), a11=3,a12=1,a13=1,a14=2,b1=1,c1=2,d1=3,a21=4,a22=1,a23=1,a24=1,b2=1. Obviously, a11>a12a14>a12a14∗ and a21>a22a24. It follows from Theorem 3 (iv) that A3=(0.5,1) is globally attractive (see Figure 2).Figure 2
Solutions(x(t),y(t)) of system (22), with the initial conditions (x(0),y(0))=(1,0.3), (0.4,2), and (0.7,1.1).Example 6.
Consider(23)dxdt=x31+x-2-x+2y3+y2,dydt=y12+3y-1-2y;that is, we take a11=3,a12=1,a13=1,a14=2,b1=1,c1=2,d1=3,a21=1,a22=2,a23=3,a24=1,b2=2 in (5). This time, a11>a12a14 and a21<a22a24. Therefore, A1 is globally attractive by Theorem 3 (ii), which is illustrated by Figure 3.Figure 3
Solutions(x(t),y(t)) of system (23) with the initial conditions (x(0),y(0))=(1,0.3), (0.4,2), and (0.7,1.1).Example 7.
Finally, leta11=3,a12=2,a13=2,a14=3,b1=3,c1=2,d1=3,a21=4,a22=1,a23=1,a24=1,b2=1 in (5); that is, consider(24)dxdt=x32+2x-2-3x+2y3+y2,dydt=y41+y-1-y.Note that a21>a22a24. We can calculate that y∗=1 and a14∗=5/2. Then we see that a11=3<5=a12a14∗. Thus it follows from Theorem 3 (iii) that A2 is globally attractive (see Figure 4).Figure 4
Solutions(x(t),y(t)) of system (24) with the initial conditions (x(0),y(0))=(1,2),(0.4,1.5), and (2.5,0.8).


## 4. Discussion

In this paper, inspired by the work in [17–19], we proposed a commensalism model under the assumption that the intrinsic growth rates of both species are density-dependent. The model can have at most four equilibria. For the first time, differential inequality has been applied to obtain the global attractivity of equilibria of such ecosystem models. Depending on the availability of resources, each of the possible equilibria can be globally attractive. This implies that density-dependent birth rates play an important role in the dynamics. Though the dynamics can be complicated, from the point view of commensalism, commensalism can be established only when resources for both species are large enough (see Theorem 3 (iv)). Hence, these results agree with those of Chen and Wu [5] (see Theorem A in Introduction).As we know, delay always exists in many biological processes. We will leave the effect of delay on the dynamics for future study.



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