# Average Conditions for the Permanence of a Bounded Discrete Predator-Prey System

Authors: Yong-Hong Fan; Lin-Lin Wang
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

Average conditions are obtained for the permanence of a discrete bounded system with Holling type II functional responseu(n+1)=u(n)exp{a(n)-b(n)u(n)-c(n)v(n)/(u(n)+m(n)v(n))},
v(n+1)=v(n)exp{-d(n)+e(n)u(n)/(u(n)+m(n)v(n))}. The method involves the application of estimates of uniform upper and lower bounds of solutions. When these results are applied to some special delay population models with multiple delays, some new results are obtained and some known results are generalized.


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## Body

## 1. Introduction

In this paper, we will study the permanence of the following discrete system:(1)u(n+1)=u(n)exp{a(n)-b(n)u(n)-c(n)v(n)u(n)+m(n)v(n)},v(n+1)=v(n)exp{-d(n)+e(n)u(n)u(n)+m(n)v(n)},
where the sequences a(n), b(n), c(n), d(n), e(n), m(n) are all assumed to be bounded and b(n), c(n), e(n), m(n) are all positive for n∈ℤ.If all the coefficients of the previous system (1) are periodic sequences with period ω, in [1], the authors obtained the following.Theorem 1 (see [1]).
Assume that(2)a¯>(cm)¯,e¯>d¯,
hold; then the periodic system (1) is permanent.In [2], by a standard comparison argument, they proved the following.Theorem 2.
Assume that(3)aL>cMmL,eL>dM,
hold; then the bounded system (1) is permanent.In the previous two theorems, we used the denotation as follows. For a bounded sequenceg(n), we define
(4)gM=sup{g(k)∣k∈ℤ},gL=inf{g(k)∣k∈ℤ}.And for a given periodic sequence with periodω, its average value is defined as
(5)f¯=1ω∑i=0ω-1‍f(i).Throughout this paper, we always assume thatbL>0,  cL>0,  eL>0,  mL>0.If all the coefficients of system (1) are periodic sequences with period ω, then it is a special form of the bounded coefficients of system (1), but from Theorem 2, we cannot obtain Theorem 1; that is to say, there is a gap between Theorems 1 and 2. In this paper, we attempt to fill in this gap.In order to illustrate our main results, similar to the corresponding definitions of the bounded continuous function in [3], we first introduce some notations.For a bounded sequencef:ℤ→ℝ, we define the lower average of f by
(6)AL(f)=limr→∞inft-s≥r1t-s∑k=stf(k).Some remarks:(a)
For a bounded sequencef, define the upper average AM(f) of f by replacing inf with sup in (6).(b)
Iff is ω-periodic, then
(7)AL(f)=AM(f)=f¯.(c)
The following inequalities hold true:(8)fL≤AL(f)≤AM(f)≤fM.(d)
For anyα,β∈R, the lower average satisfies
(9)AL(αf+βg)=αAL(f)+βAL(g).Proof.
We only prove that (b) hold; (c) and (d) can be proved similarly as that in [3]. Setting t-s=nω+αn, where αn∈[0,ω-1], in the following, we assume that n is sufficiently large; then
(10)1t-s∑k=st‍f(k)=1nω+αn∑k=ss+nω-1‍f(k)+1nω+αn∑k=s+nωt‍f(k)=nnω+αn∑k=0ω-1‍f(k)+(αn+1)fMnω+αn,
from the previous equality, we have
(11)limn→∞1t-s∑k=st‍f(k)=1ω∑k=0ω-1‍f(k);
therefore
(12)AL(f)=AM(f)=1ω∑k=0ω-1‍f(k)=f¯,
which completes the proof.During the study of the permanence for the bounded system, in view of the property (b), one can usually use the lower average or upper average instead of the sup and inf values. And we call the condition obtained by using the method of lower average or upper average as “average conditions.” For the permanence results with “average conditions," one can refer to [4–7], and so forth.For the permanence of system (1), we have the following.Theorem 3.
Assume that(13)AL(a(n)-c(n)m(n))>0,AL(e(n)-d(n))>0,AL(d)>0;
then the bounded system (1) is permanent.Obviously, Theorem3 includes both Theorems 1 and 2. Therefore, this theorem is a bridge that combines the bounded system and the periodic system.


## 2. Preliminaries

In order to prove Theorem3, we need some lemmas below. The first lemma could be found in [8].Lemma 4 (see [8, Corollary 2.5]).
Letu(n) be a positive solution of the following inequality:
(14)u(n+1)≤u(n)exp{a1(n)-b1(n)u(n)},n∈ℤ,u(0)>0,
if b1L>0 and a1M>0; then
(15)limsupn→∞u(n)≤min{a1Mb1Lexp{a1M},1b1Lexp{a1M-1}}.We should point out that whenb1L=0, the conclusion of the previous lemma is not true. That is, b1L>0 is a necessary condition. We give an example to illustrate it.Example 5.
Consider the following inequality:(16)u(n+1)≤u(n)exp{12+1n-1nu(n)},n∈ℤ,u(0)>0.
Obviously, u(n)=n is a solution of it, but limsupn→∞u(n)=+∞.Lemma 6.
Letu(n) be a solution of the following inequality:
(17)u(n+1)≥u(n)exp{a2(n)-b2(n)u(n)},n∈ℤ,u(0)>0,
and bounded above; if b2L>0 and
(18)AL(a2)>0,
then there exists some positive constant L such that
(19)liminfn→∞u(n)≥L.To prove this lemma, we give two claims in what follows. First, by using mathematical induction, we can easily obtain the following.Claim 1.
Ifu(n) is a solution of (17), then
(20)u(n)>0foranyn∈ℤ.In what follows, we use contradiction to prove the lemma.Claim 2.
Assume thatu(k) is a solution of (17) and bounded above by a positive constant M; if (19) does not hold, then there exist positive integer sequences {sn} and {tn} such that
(21)0≤sn<tn,tn-sn≥n+1,u(sn)≥u(0)n,u(k)≤u(0)nforsn<k≤tn.Proof of the claim.
Notice that(22)a2(n)-b2(n)u(n)≥a2L-b2MM≥-γforanyn∈ℤ,
where γ>0 is a constant.
If (19) does not hold, then from Claim 1, liminfn→∞u(n)=0, thus, for any positive integer n≥1, there exist tn>0 such that
(23)u(tn)<u(0)nexp{-nγ}.
In addition, there exists a numbersn such that 0≤sn<tn, u(sn)≥u(0)/n and u(k)≤u(0)/n for sn<k≤tn. In the following, we only need to prove that tn-sn≥n+1. From the first equation of (17), we have
(24)u(0)n≤u(sn)≤u(tn)exp{-∑k=sntn-1‍(a2(k)-b2(k)u(k))}≤u(tn)exp{γ(tn-sn-1)}≤u(0)nexp{-nγ}exp{γ(tn-sn-1)},
which implies that tn-sn≥n+1. This completes the proof of Claim 2.Proof of Lemma6.
From the first equation of (17), we have
(25)a2(k)≤lnu(k+1)u(k)+b2(k)u(k),
by Claim 2, we obtain that if (19) does not hold, then for any n≥1, we have
(26)∑k=sntn-1‍a2(k)≤∑k=sntn-1‍lnu(k+1)u(k)+∑k=sntn-1‍b2(k)u(k),
which implies that
(27)1tn-sn-1∑k=sntn-1‍a2(k)≤1tn-sn-1lnu(tn)u(sn)+b2(sn)u(sn)tn-sn-1+tn-sn-2tn-sn-1u(0)b2Mn<-nγtn-sn-1+b2(sn)u(sn)tn-sn-1+tn-sn-2tn-sn-1u(0)b2Mn.
Notice that
(28)limn→∞b2(sn)u(sn)tn-sn-1=0,limn→∞tn-sn-2tn-sn-1u(0)b2Mn=0;
thus, by (27), we have
(29)limn→∞1tn-sn-1∑k=sntn-1‍a2(k)≤0.
This is in contradiction to (18); the proof is complete.From Lemmas4 and 6, we have the following.Theorem 7.
Letu(n) be a solution of the following inequality:
(30)u(n)exp{a2(n)-b2(n)u(n)}≤u(n+1)≤u(n)exp{a1(n)-b1(n)u(n)},n∈ℤ,u(0)>0;
if
(31)b1L>0,b2L>0,AL(a2)>0,
then there exist some positive constants L and M such that
(32)L≤liminft→∞u(t)≤limsupt→∞u(t)≤M.From Theorem7, we can easily obtain the following.Corollary 8.
Letu(n) be a solution of the following inequality:
(33)u(n)exp{a2(n)-b2(n)u(n-k)}≤u(n+1)≤u(n)exp{a1(n)-b1(n)u(n-k)},
for any n∈ℤ, u(i)>0, -k+1≤i≤0. If
(34)b1L>0,b2L>0,AL(a2)>0;
then the conclusion of Theorem 7 also holds true, where k is a positive integer.


## 3. Permanence

In this section, we give some applications of Theorem7. First we use it to prove Theorem 3.Proof of Theorem3.
From the first equation of (1), we have
(35)u(n)exp{a(n)-c(n)m(n)-b(n)u(n)}≤u(n+1)≤u(n)exp{a(n)-b(n)u(n)};
by Theorem 7 and the condition (13), we can obtain that there must exist some positive constants L1 and M1 such that
(36)L1≤liminfn→∞u(n)≤limsupn→∞u(n)≤M1,
for any solution (u(n),v(n)) of (1) with positive initial conditions u(0)>0 and v(0)>0.
From the second equation of (1), we have
(37)v(n+1)=v(n)exp{e(n)-d(n)-e(n)m(n)v(n)u(n)+m(n)v(n)}≥v(t)[e(n)-d(n)-e(n)m(n)v(n)L1];
by Theorem 7 and condition (13), we can obtain that there exists a positive constant L2 such that
(38)liminfn→∞v(n)≥L2.
Sety(n)=1/v(n); then from the second equation of (1), we can obtain
(39)y(n+1)=y(n)exp{d(n)-e(n)u(n)u(n)y(n)+m(n)y(n)}≥y(n)exp{d(n)-e(n)M1m(n)y(n)},
for sufficiently large t; by Theorem 7 and (13), we have
(40)limsupn→∞v(n)≤M2.
By (36), (38), and (40), we complete the proof.Through some similar analysis as in [9], we have the following.Corollary 9.
Assume that any positive solution of the periodic equationu(n+1)=f(n,u(n))(f(n+ω,u)=f(n,ω),∀n∈ℤ) satisfies
(41)u(n)exp{a2(n)-b2(n)u(n)}≤u(n+1)≤u(n)exp{a1(n)-b1(n)u(n)},n∈ℤ,u(0)>0,
where ai(n),bi(n) (i=1,2) are all ω-periodic sequences; if
(42)b1L>0,b2L>0,a2¯>0,
then the periodic equation u(n+1)=f(n,u(n)) has at least one ω-periodic positive solution.We should point out that the previous corollary can be generalized to then-dimensional situation. As a direct application of the previous corollary, we have the following.Theorem 10.
Assume that(43)a¯>(cm)¯,e¯>d¯>0,
hold; then the periodic system (1) (the coefficients of the system (1) are all periodic sequences with a common period ω) has at least one ω-periodic positive solution.This theorem generalized Theorem  3.1 in [10].



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