# Boundedness and Global Attractivity of a Higher-Order Nonlinear Difference Equation

Authors: Xiu-Mei Jia; Wan-Tong Li
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We investigate the local stability, prime period-two
solutions, boundedness, invariant intervals, and global attractivity
of all positive solutions of the following difference equation:yn+1=(r+pyn+yn−k)/(qyn+yn−k),  n∈ℕ0,  where the parameters p,q,r∈(0,∞),k∈{1,2,3,…} and the initial conditions y−k,…,y0∈(0,∞). We show that the unique positive equilibrium of this equation is a global
attractor under certain conditions.


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## Body

## 1. Introduction and Preliminaries

Our aim in this paper is to study the dynamical behavior of the following rational difference equation(1.1)yn+1=r+pyn+yn-kqyn+yn-k,n∈ℕ0,
where p,q,r∈(0,∞), ℕ0∈{0,1,…}, k∈{1,2,3,…} and the initial conditions y-k,…,y0∈(0,∞).Whenk=1, (1.1) reduces to(1.2)yn+1=r+pyn+yn-1qyn+yn-1,n∈ℕ0.In [1] (see also [2]), the authors investigated the global convergence of solutions to (1.2) and they obtained the following result.Theorem 1.1.
Letp, q and r be positive numbers. Then every solution of (1.2) converges to the unique equilibrium or to a prime-two solution.The main purpose of this paper is to further consider the global attractivity of all positive solutions of (1.1). That is to say, we will prove that the unique positive equilibrium of (1.1) is a global attractor under certain conditions (see Theorem 4.10).For the general theory of difference equations, one can refer to the monographes [3] and [2]. For other related results on nonlinear difference equations, see, for example, [1–18].For the sake of convenience, we firstly present some definitions and known results which will be useful in the sequel.LetI be some interval of real numbers and let f:I×I→I be a continuously differentiable function. Then for initial conditions x-k,…,x0∈I, the difference equation(1.3)xn+1=f(xn,xn-k),n∈ℕ0
has a unique solution {xn}n=-k∞.A pointx̅ is called an equilibrium of (1.3) if(1.4)x̅=f(x̅,x̅).That is,xn=x̅ for n≥0 is a solution of (1.3), or equivalently, x̅ is a fixed point of f.An intervalJ⊆I is called an invariant interval of (1.3) if(1.5)x-k,…,x0∈J⇒xn∈J∀n∈ℕ0.That is, every solution of (1.3) with initial conditions in J remains in J.Let(1.6)P=∂f∂u(x̅,x̅),Q=∂f∂v(x̅,x̅)denote the partial derivatives off(u,v) evaluated at an equilibrium x̅ of (1.3). Then the linearized equation associated with (1.3) about the equilibrium x̅ is(1.7)zn+1=Pzn+Qzn-k,n=0,1,…,
and its characteristic equation is(1.8)λk+1-Pλk-Q=0.Lemma 1.2 (see [3]).
Assume thatP,Q∈R and k∈{1,2,…}. Then
(1.9)|P|+|Q|<1
is a sufficient condition for asymptotic stability of the difference equation (1.7). Suppose in addition that one of the following two cases holds:(i)
k odd and Q>0,(ii)
k even and PQ>0.Then (1.9) is also a necessary condition for the asymptotic stability of the difference equation (1.7).The following result will be useful in establishing the global attractivity character of the equilibrium of (1.1), and it is a reformulation of [2, 7].Lemma 1.3.
Suppose that a continuous functionf:[a,b]×[a,b]→[a,b] satisfies one of (i)–(iii):(i)
f(x,y) is nonincreasing in x,y, and
(1.10)∀(m,M)∈[a,b]×[a,b],(f(m,m)=M,f(M,M)=m)⇒m=M,(ii)
f(x,y) is nondecreasing in x and nonincreasing in y, and
(1.11)∀(m,M)∈[a,b]×[a,b],(f(m,M)=m,f(M,m)=M)⇒m=M,(iii)
f(x,y) is nonincreasing in x and nondecreasing in y, and
(1.12)∀(m,M)∈[a,b]×[a,b],(f(M,m)=m,f(m,M)=M)⇒m=M.(Note that fork odd this is equivalent to (1.3) having no prime period-two solution)Then (1.3) has a unique equilibrium in [a,b] and every solution with initial values in [a,b] converges to the equilibrium.This work is organized as follows. In Section2, the local stability and periodic character are discussed. In Section 3, the boundedness, invariant intervals of (1.1) are presented. Our main results are formulated and proved in Section 4, where the global attractivity of (1.1) is investigated.


## 2. Local Stability and Period-Two Solutions

The unique positive equilibrium of (1.1) is(2.1)y̅=(1+p)+(1+p)2+4r(1+q)2(1+q).The linearized equation associated with (1.1) about y̅ is(2.2)zn+1-(p-q)y̅-qr(q+1)[r+(p+1)y̅]zn+(p-q)y̅+r(q+1)[r+(p+1)y̅]zn-k=0,and its characteristic equation is(2.3)λk+1-(p-q)y̅-qr(q+1)[r+(p+1)y̅]λk+(p-q)y̅+r(q+1)[r+(p+1)y̅]=0.From this and Lemma1.2, we have the following result.Theorem 2.1.
Assume thatp,q,r∈(0,∞) and initial conditions y-k,…,y0∈(0,∞). Then the following stataments are true.(i)
If(2.4)(p-q)y̅-qr≥0,(p-3q-pq-1)y̅<2qr,
then the unique positive equilibrium y̅ of (1.1) is locally asymptotically stable;(ii)
If(2.5)(p-q)y̅-qr<0<(p-q)y̅+r,
then the unique positive equilibrium y̅ of (1.1) is locally asymptotically stable. In particular, if k is even, then the equilibrium y̅ is locally asymptotically stable if and only if (2.5) holds;(iii)
If(2.6)(p-q)y̅+r≤0,
then the unique positive equilibrium y̅ of (1.1) is locally asymptotically stable.In the following, we will consider the period-two solutions of (1.1).Let(2.7)…,ϕ,ψ,ϕ,ψ,…be a period-two solution of (1.1), where ϕ and ψ are two arbitrary positive real numbers.Ifk is even, then yn=yn-k, and ϕ and ψ satisfy the following system:(2.8)ϕ=r+pψ+ψqψ+ψ,ψ=r+pϕ+ϕqϕ+ϕ,then(ϕ-ψ)(p+1)=0, we have ϕ=ψ, which is a contradiction.Ifk is odd, then yn+1=yn-k, and ϕ and ψ satisfy the following system:(2.9)ϕ=r+pψ+ϕqψ+ϕ,ψ=r+pϕ+ψqϕ+ψ,thenϕ+ψ=1-p, ϕψ=p(1-p)/(q-1). By calculating, (1.1) has prime period-two solution if and only if(2.10)p<1,q>1,4r<(1-p)(q-1-pq-3p).From the above discussion, we have the following result.Theorem 2.2.
Equation (1.1) has a positive prime period-two solution
(2.11)…,ϕ,ψ,ϕ,ψ,…
if and only if
(2.12)kisodd,p<1,q>1,4r<(1-p)(q-pq-3p-1).
Furthermore, if (2.12) holds, then the prime period-two solution of (1.1) is “unique” and the values of ϕ and ψ are the positive roots of the quadratic equation
(2.13)t2-(1-p)t+r+p(1-p)q-1=0.


## 3. Boundedness and Invariant Intervals

In this section, we discuss the boundedness, invariant intervals of (1.1).


### 3.1. Boundedness

Theorem 3.1.
All positive solutions of (1.1) are bounded.Proof.
Equation (1.1) can be written as
(3.1)yn+1=r+pyn+yn-kqyn+yn-k≥pyn+yn-kqyn+yn-k≥min{(p/q),1}(qyn+yn-k)qyn+yn-k=min{pq,1}
for all n≥0. We denote
(3.2)K=min{pq,1}.
Then
(3.3)yn+1=r+pyn+yn-kqyn+yn-k≤r+pyn+yn-k(q/2)K+(K/2)+(q/2)yn+(1/2)yn-k≤max{r,p,1}(1+yn+yn-k)min{(q/2)K+(K/2),(q/2),(1/2)}(1+yn+yn-k)=max{r,p,1}min{(q/2)K+(K/2),(q/2),(1/2)}
for all n>k. The proof is complete.Let{yn}n=-k∞ be a positive solution of (1.1). Then the following identities are easily established:(3.4)yn+1-1=(q-p)((r/(q-p))-yn)qyn+yn-k,n∈ℕ0,(3.5)yn+1-pq=((p-q)/q)((qr/(p-q))-yn-k)qyn+yn-k,n∈ℕ0,(3.6)yn+1-qrp-q=((p2-pq-q2r)/(p-q))(yn+(1/q))qyn+yn-k+((p-q-qr)/(p-q))(yn-k-(p/q))qyn+yn-k,n∈ℕ0,(3.7)yn+1-p+rq=r(1-yn)+((q-p-r)/q)yn-kqyn+yn-k,n∈ℕ0,(3.8)yn-yn+2(k+1)=(yn-(p/q))(q2yn+kyn+2k+1+qynyn+2k+1)qyn+2k+1(qyn+k+yn)+(r+pyn+k+yn)+pyn+k(yn-((p+qr)/p))+(yn2-yn-r)qyn+2k+1(qyn+k+yn)+(r+pyn+k+yn),n∈ℕ0.Whenq=p+r, the unique positive equilibrium of (1.1) is y̅=1, (3.4) becomes(3.9)yn+1-1=r(1-yn)qyn+yn-k,n∈ℕ0.Whenp=q(1+1+4r)/2, the unique positive equilibrium is y̅=p/q, (3.5) becomes(3.10)yn+1-pq=(p-q)/q((p/q)-yn-k)qyn+yn-k,n∈ℕ0,
and (3.8) becomes(3.11)yn-yn+2(k+1)=(yn-(p/q))(q2yn+kyn+2k+1+qynyn+2k+1+pyn+k+yn+(p-q)/q)qyn+2k+1(qyn+k+yn)+(r+pyn+k+yn),n∈ℕ0.Set(3.12)f(x,y)=r+px+yqx+y.Lemma 3.2.
Assume thatf(x,y) is defined in (3.12). Then the following statements are true:(i)
assumep<q. Then f(x,y) is strictly decreasing in x and increasing in y for x≥r/(q-p); and it is strictly decreasing in each of its arguments for x<r/(q-p);(ii)
assumep>q. Then f(x,y) is increasing in x and strictly decreasing in y for y≥qr/(p-q); and it is strictly decreasing in each of its arguments for y<qr/(p-q).Proof.
By calculating the partial derivatives of the functionf(x,y), we have
(3.13)fx′(x,y)=(p-q)y-qr(qx+y)2,fy′(x,y)=(q-p)x-r(qx+y)2,
from which these statements easily follow.


### 3.2. Invariant Interval

In this subsection, we discuss the invariant interval of (1.1).


#### 3.2.1. The Casep<q

Lemma 3.3.
Assume thatp<q, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>p/q for all n≥1;(ii)
If for someN≥0, yN>r/(q-p), then yN+1<1;(iii)
If for someN≥0, yN=r/(q-p), then yN+1=1;(iv)
If for someN≥0, yN<r/(q-p), then yN+1>1;(v)
Ifp<q<p+r, then (1.1) possesses an invariant interval [1,r/(q-p)] and y̅∈(1,r/(q-p));(vi)
Ifp+r<q<p+qr/p, then (1.1) possesses an invariant interval [r/(q-p),1] and y̅∈(r/(q-p),1);(vii)
Ifq≥p+qr/p, then (1.1) possesses an invariant interval [p/q,1] and y̅∈(p/q,1).Proof.
The proofs of (i)–(iv) are straightforward consequences of the identities (3.5) and (3.4). So we only prove (v)–(vii). By the condition (i) of Lemma 3.2, the function f(x,y) is strictly decreasing in x and increasing in y for x≥r/(q-p); and it is strictly decreasing in both arguments for x<r/(q-p).
(v) Using the decreasing character off, we obtain
(3.14)1=f(rq-p,rq-p)<f(x,y)<f(1,1)=r+p+1q+1<rq-p.
The inequalities
(3.15)1<rq-p,r+p+1q+1<rq-p
are equivalent to the inequality q<p+r.
On the other hand,y̅ is the unique positive root of quadratic equation
(3.16)(q+1)y2-(p+1)y-r=0.
Since
(3.17)(q+1)(rq-p)2-(p+1)rq-p-r=r(q+1)(p+r-q)(q-p)2>0,(q+1)-(p+1)-r=q-p-r<0,
then we have that y̅∈(1,r/(q-p)).
(vi) By using the monotonic character off, we obtain
(3.18)(q-p)(p+r)+rq2-pq+r=f(1,rq-p)≤f(x,y)≤f(rq-p,1)=1.
The inequalities
(3.19)(q-p)(p+r)+rq2-pq+r>rq-p,rq-p<1
follow from the inequality q>p+r.
On the other hand, similar to (v) it can be proved thaty̅∈(r/(q-p),1).
(vii) In this case note thatr/(q-p)≤p/q<1 holds, and using the monotonic character of f, we obtain
(3.20)pq<qr+pq+pq2+p=f(1,pq)≤f(x,y)≤f(pq,1)=qr+p2+qq(p+1)≤1.
Furthermore, similar to (v) it follows y̅∈(p/q,1). The proof is complete.Whenq=p+r, (3.9) implies that the following result holds.Lemma 3.4.
Assume thatq=p+r, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN>1, then yN+1<1;(ii)
If for someN≥0, yN=1, then yN+1=1;(iii)
If for someN≥0, yN<1, then yN+1>1.


#### 3.2.2. The Casep>q

Lemma 3.5.
Assume thatp>q, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>1 for all n≥1;(ii)
If for someN≥0, yN<qr/(p-q), then yN+k+1>p/q;(iii)
If for someN≥0, yN=qr/(p-q), then yN+k+1=p/q;(iv)
If for someN≥0, yN>qr/(p-q), then yN+k+1<p/q;(v)
Ifq<p<q(1+1+4r)/2, then (1.1) possesses an invariant interval [p/q,qr/(p-q)] and y̅∈(p/q,qr/(p-q));(vi)
Ifq(1+1+4r)/2<p<q+qr, then (1.1) possesses an invariant interval [qr/(p-q),p/q] and y̅∈(qr/(p-q),p/q);(vii)
Ifp≥q+qr, then (1.1) possesses an invariant interval [1,p/q] and y̅∈(1,p/q).Proof.
The proofs of (i)–(iv) are direct consequences of the identities (3.4) and (3.5). So we only give the proofs (v)–(vii). By Lemma 3.2 (ii), the function f(x,y) is increasing in x and strictly decreasing in y for y≥qr/(p-q); and it is strictly decreasing in each of its arguments for y<qr/(p-q).
(v) Using the decreasing character off, we obtain
(3.21)pq=f(qrp-q,qrp-q)≤f(x,y)≤f(pq,pq)=qr+p(p+1)p(q+1)≤qrp-q.
The inequalities
(3.22)qr+p(p+1)p(q+1)≤qrp-q,pq<qrp-q
are equivalent to the inequality p<q(1+1+4r)/2. That is, [p/q,qr/(p-q)] is an invariant interval of (1.1).
On the other hand, similar to Lemma3.3 (v), it can be proved that y̅∈(p/q,qr/(p-q)).
(vi) By using the monotonic character off, we obtain
(3.23)qrp-q≤(qr+p)(p-q)+pq2rq3r+p(p-q)=f(qrp-q,pq)≤f(x,y)≤f(pq,qrp-q)=pq.
The inequalities
(3.24)(qr+p)(p-q)+pq2rq3r+p(p-q)≥qrp-q,qrp-q≤pq
are equivalent to the inequality p>q(1+1+4r)/2.
On the other hand, similar to Lemma3.3 (v) it can be proved that y̅∈(qr/(p-q),p/q).
(vii) In this case note thatqr/(p-q)≤1<p/q holds. By the monotonic character of f, we have
(3.25)1<qr+pq+pq2+p=f(1,pq)≤f(x,y)≤f(pq,1)=qr+p2+qq(p+1)≤pq.
The inequalities
(3.26)qr+pq+pq2+p>1,qr+p2+qq(p+1)≤pq
are equivalent to the inequality p≥q+qr.
Furthermore, similar to Lemma3.3 (v), it follows y̅∈(1,p/q). The proof is complete.Lemma 3.6.
Assume thatp=q(1+1+4r)/2, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN<p/q, then yN+k+1>p/q;(ii)
If for someN≥0, yN=p/q, then yN+k+1=p/q;(iii)
If for someN≥0, yN>p/q, then yN+k+1<p/q;(iv)
If for someN≥0, yN>p/q, then yN>yN+2(k+1)>p/q;(v)
If for someN≥0, yN<p/q, then yN<yN+2(k+1)<p/q.Proof.
In this case, we have thatqr/(p-q)=p/q. These results follow from the identities (3.10) and (3.11) and the details are omitted.


## 3.1. Boundedness

Theorem 3.1.
All positive solutions of (1.1) are bounded.Proof.
Equation (1.1) can be written as
(3.1)yn+1=r+pyn+yn-kqyn+yn-k≥pyn+yn-kqyn+yn-k≥min{(p/q),1}(qyn+yn-k)qyn+yn-k=min{pq,1}
for all n≥0. We denote
(3.2)K=min{pq,1}.
Then
(3.3)yn+1=r+pyn+yn-kqyn+yn-k≤r+pyn+yn-k(q/2)K+(K/2)+(q/2)yn+(1/2)yn-k≤max{r,p,1}(1+yn+yn-k)min{(q/2)K+(K/2),(q/2),(1/2)}(1+yn+yn-k)=max{r,p,1}min{(q/2)K+(K/2),(q/2),(1/2)}
for all n>k. The proof is complete.Let{yn}n=-k∞ be a positive solution of (1.1). Then the following identities are easily established:(3.4)yn+1-1=(q-p)((r/(q-p))-yn)qyn+yn-k,n∈ℕ0,(3.5)yn+1-pq=((p-q)/q)((qr/(p-q))-yn-k)qyn+yn-k,n∈ℕ0,(3.6)yn+1-qrp-q=((p2-pq-q2r)/(p-q))(yn+(1/q))qyn+yn-k+((p-q-qr)/(p-q))(yn-k-(p/q))qyn+yn-k,n∈ℕ0,(3.7)yn+1-p+rq=r(1-yn)+((q-p-r)/q)yn-kqyn+yn-k,n∈ℕ0,(3.8)yn-yn+2(k+1)=(yn-(p/q))(q2yn+kyn+2k+1+qynyn+2k+1)qyn+2k+1(qyn+k+yn)+(r+pyn+k+yn)+pyn+k(yn-((p+qr)/p))+(yn2-yn-r)qyn+2k+1(qyn+k+yn)+(r+pyn+k+yn),n∈ℕ0.Whenq=p+r, the unique positive equilibrium of (1.1) is y̅=1, (3.4) becomes(3.9)yn+1-1=r(1-yn)qyn+yn-k,n∈ℕ0.Whenp=q(1+1+4r)/2, the unique positive equilibrium is y̅=p/q, (3.5) becomes(3.10)yn+1-pq=(p-q)/q((p/q)-yn-k)qyn+yn-k,n∈ℕ0,
and (3.8) becomes(3.11)yn-yn+2(k+1)=(yn-(p/q))(q2yn+kyn+2k+1+qynyn+2k+1+pyn+k+yn+(p-q)/q)qyn+2k+1(qyn+k+yn)+(r+pyn+k+yn),n∈ℕ0.Set(3.12)f(x,y)=r+px+yqx+y.Lemma 3.2.
Assume thatf(x,y) is defined in (3.12). Then the following statements are true:(i)
assumep<q. Then f(x,y) is strictly decreasing in x and increasing in y for x≥r/(q-p); and it is strictly decreasing in each of its arguments for x<r/(q-p);(ii)
assumep>q. Then f(x,y) is increasing in x and strictly decreasing in y for y≥qr/(p-q); and it is strictly decreasing in each of its arguments for y<qr/(p-q).Proof.
By calculating the partial derivatives of the functionf(x,y), we have
(3.13)fx′(x,y)=(p-q)y-qr(qx+y)2,fy′(x,y)=(q-p)x-r(qx+y)2,
from which these statements easily follow.


## 3.2. Invariant Interval

In this subsection, we discuss the invariant interval of (1.1).


### 3.2.1. The Casep<q

Lemma 3.3.
Assume thatp<q, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>p/q for all n≥1;(ii)
If for someN≥0, yN>r/(q-p), then yN+1<1;(iii)
If for someN≥0, yN=r/(q-p), then yN+1=1;(iv)
If for someN≥0, yN<r/(q-p), then yN+1>1;(v)
Ifp<q<p+r, then (1.1) possesses an invariant interval [1,r/(q-p)] and y̅∈(1,r/(q-p));(vi)
Ifp+r<q<p+qr/p, then (1.1) possesses an invariant interval [r/(q-p),1] and y̅∈(r/(q-p),1);(vii)
Ifq≥p+qr/p, then (1.1) possesses an invariant interval [p/q,1] and y̅∈(p/q,1).Proof.
The proofs of (i)–(iv) are straightforward consequences of the identities (3.5) and (3.4). So we only prove (v)–(vii). By the condition (i) of Lemma 3.2, the function f(x,y) is strictly decreasing in x and increasing in y for x≥r/(q-p); and it is strictly decreasing in both arguments for x<r/(q-p).
(v) Using the decreasing character off, we obtain
(3.14)1=f(rq-p,rq-p)<f(x,y)<f(1,1)=r+p+1q+1<rq-p.
The inequalities
(3.15)1<rq-p,r+p+1q+1<rq-p
are equivalent to the inequality q<p+r.
On the other hand,y̅ is the unique positive root of quadratic equation
(3.16)(q+1)y2-(p+1)y-r=0.
Since
(3.17)(q+1)(rq-p)2-(p+1)rq-p-r=r(q+1)(p+r-q)(q-p)2>0,(q+1)-(p+1)-r=q-p-r<0,
then we have that y̅∈(1,r/(q-p)).
(vi) By using the monotonic character off, we obtain
(3.18)(q-p)(p+r)+rq2-pq+r=f(1,rq-p)≤f(x,y)≤f(rq-p,1)=1.
The inequalities
(3.19)(q-p)(p+r)+rq2-pq+r>rq-p,rq-p<1
follow from the inequality q>p+r.
On the other hand, similar to (v) it can be proved thaty̅∈(r/(q-p),1).
(vii) In this case note thatr/(q-p)≤p/q<1 holds, and using the monotonic character of f, we obtain
(3.20)pq<qr+pq+pq2+p=f(1,pq)≤f(x,y)≤f(pq,1)=qr+p2+qq(p+1)≤1.
Furthermore, similar to (v) it follows y̅∈(p/q,1). The proof is complete.Whenq=p+r, (3.9) implies that the following result holds.Lemma 3.4.
Assume thatq=p+r, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN>1, then yN+1<1;(ii)
If for someN≥0, yN=1, then yN+1=1;(iii)
If for someN≥0, yN<1, then yN+1>1.


### 3.2.2. The Casep>q

Lemma 3.5.
Assume thatp>q, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>1 for all n≥1;(ii)
If for someN≥0, yN<qr/(p-q), then yN+k+1>p/q;(iii)
If for someN≥0, yN=qr/(p-q), then yN+k+1=p/q;(iv)
If for someN≥0, yN>qr/(p-q), then yN+k+1<p/q;(v)
Ifq<p<q(1+1+4r)/2, then (1.1) possesses an invariant interval [p/q,qr/(p-q)] and y̅∈(p/q,qr/(p-q));(vi)
Ifq(1+1+4r)/2<p<q+qr, then (1.1) possesses an invariant interval [qr/(p-q),p/q] and y̅∈(qr/(p-q),p/q);(vii)
Ifp≥q+qr, then (1.1) possesses an invariant interval [1,p/q] and y̅∈(1,p/q).Proof.
The proofs of (i)–(iv) are direct consequences of the identities (3.4) and (3.5). So we only give the proofs (v)–(vii). By Lemma 3.2 (ii), the function f(x,y) is increasing in x and strictly decreasing in y for y≥qr/(p-q); and it is strictly decreasing in each of its arguments for y<qr/(p-q).
(v) Using the decreasing character off, we obtain
(3.21)pq=f(qrp-q,qrp-q)≤f(x,y)≤f(pq,pq)=qr+p(p+1)p(q+1)≤qrp-q.
The inequalities
(3.22)qr+p(p+1)p(q+1)≤qrp-q,pq<qrp-q
are equivalent to the inequality p<q(1+1+4r)/2. That is, [p/q,qr/(p-q)] is an invariant interval of (1.1).
On the other hand, similar to Lemma3.3 (v), it can be proved that y̅∈(p/q,qr/(p-q)).
(vi) By using the monotonic character off, we obtain
(3.23)qrp-q≤(qr+p)(p-q)+pq2rq3r+p(p-q)=f(qrp-q,pq)≤f(x,y)≤f(pq,qrp-q)=pq.
The inequalities
(3.24)(qr+p)(p-q)+pq2rq3r+p(p-q)≥qrp-q,qrp-q≤pq
are equivalent to the inequality p>q(1+1+4r)/2.
On the other hand, similar to Lemma3.3 (v) it can be proved that y̅∈(qr/(p-q),p/q).
(vii) In this case note thatqr/(p-q)≤1<p/q holds. By the monotonic character of f, we have
(3.25)1<qr+pq+pq2+p=f(1,pq)≤f(x,y)≤f(pq,1)=qr+p2+qq(p+1)≤pq.
The inequalities
(3.26)qr+pq+pq2+p>1,qr+p2+qq(p+1)≤pq
are equivalent to the inequality p≥q+qr.
Furthermore, similar to Lemma3.3 (v), it follows y̅∈(1,p/q). The proof is complete.Lemma 3.6.
Assume thatp=q(1+1+4r)/2, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN<p/q, then yN+k+1>p/q;(ii)
If for someN≥0, yN=p/q, then yN+k+1=p/q;(iii)
If for someN≥0, yN>p/q, then yN+k+1<p/q;(iv)
If for someN≥0, yN>p/q, then yN>yN+2(k+1)>p/q;(v)
If for someN≥0, yN<p/q, then yN<yN+2(k+1)<p/q.Proof.
In this case, we have thatqr/(p-q)=p/q. These results follow from the identities (3.10) and (3.11) and the details are omitted.


## 3.2.1. The Casep<q

Lemma 3.3.
Assume thatp<q, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>p/q for all n≥1;(ii)
If for someN≥0, yN>r/(q-p), then yN+1<1;(iii)
If for someN≥0, yN=r/(q-p), then yN+1=1;(iv)
If for someN≥0, yN<r/(q-p), then yN+1>1;(v)
Ifp<q<p+r, then (1.1) possesses an invariant interval [1,r/(q-p)] and y̅∈(1,r/(q-p));(vi)
Ifp+r<q<p+qr/p, then (1.1) possesses an invariant interval [r/(q-p),1] and y̅∈(r/(q-p),1);(vii)
Ifq≥p+qr/p, then (1.1) possesses an invariant interval [p/q,1] and y̅∈(p/q,1).Proof.
The proofs of (i)–(iv) are straightforward consequences of the identities (3.5) and (3.4). So we only prove (v)–(vii). By the condition (i) of Lemma 3.2, the function f(x,y) is strictly decreasing in x and increasing in y for x≥r/(q-p); and it is strictly decreasing in both arguments for x<r/(q-p).
(v) Using the decreasing character off, we obtain
(3.14)1=f(rq-p,rq-p)<f(x,y)<f(1,1)=r+p+1q+1<rq-p.
The inequalities
(3.15)1<rq-p,r+p+1q+1<rq-p
are equivalent to the inequality q<p+r.
On the other hand,y̅ is the unique positive root of quadratic equation
(3.16)(q+1)y2-(p+1)y-r=0.
Since
(3.17)(q+1)(rq-p)2-(p+1)rq-p-r=r(q+1)(p+r-q)(q-p)2>0,(q+1)-(p+1)-r=q-p-r<0,
then we have that y̅∈(1,r/(q-p)).
(vi) By using the monotonic character off, we obtain
(3.18)(q-p)(p+r)+rq2-pq+r=f(1,rq-p)≤f(x,y)≤f(rq-p,1)=1.
The inequalities
(3.19)(q-p)(p+r)+rq2-pq+r>rq-p,rq-p<1
follow from the inequality q>p+r.
On the other hand, similar to (v) it can be proved thaty̅∈(r/(q-p),1).
(vii) In this case note thatr/(q-p)≤p/q<1 holds, and using the monotonic character of f, we obtain
(3.20)pq<qr+pq+pq2+p=f(1,pq)≤f(x,y)≤f(pq,1)=qr+p2+qq(p+1)≤1.
Furthermore, similar to (v) it follows y̅∈(p/q,1). The proof is complete.Whenq=p+r, (3.9) implies that the following result holds.Lemma 3.4.
Assume thatq=p+r, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN>1, then yN+1<1;(ii)
If for someN≥0, yN=1, then yN+1=1;(iii)
If for someN≥0, yN<1, then yN+1>1.


## 3.2.2. The Casep>q

Lemma 3.5.
Assume thatp>q, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>1 for all n≥1;(ii)
If for someN≥0, yN<qr/(p-q), then yN+k+1>p/q;(iii)
If for someN≥0, yN=qr/(p-q), then yN+k+1=p/q;(iv)
If for someN≥0, yN>qr/(p-q), then yN+k+1<p/q;(v)
Ifq<p<q(1+1+4r)/2, then (1.1) possesses an invariant interval [p/q,qr/(p-q)] and y̅∈(p/q,qr/(p-q));(vi)
Ifq(1+1+4r)/2<p<q+qr, then (1.1) possesses an invariant interval [qr/(p-q),p/q] and y̅∈(qr/(p-q),p/q);(vii)
Ifp≥q+qr, then (1.1) possesses an invariant interval [1,p/q] and y̅∈(1,p/q).Proof.
The proofs of (i)–(iv) are direct consequences of the identities (3.4) and (3.5). So we only give the proofs (v)–(vii). By Lemma 3.2 (ii), the function f(x,y) is increasing in x and strictly decreasing in y for y≥qr/(p-q); and it is strictly decreasing in each of its arguments for y<qr/(p-q).
(v) Using the decreasing character off, we obtain
(3.21)pq=f(qrp-q,qrp-q)≤f(x,y)≤f(pq,pq)=qr+p(p+1)p(q+1)≤qrp-q.
The inequalities
(3.22)qr+p(p+1)p(q+1)≤qrp-q,pq<qrp-q
are equivalent to the inequality p<q(1+1+4r)/2. That is, [p/q,qr/(p-q)] is an invariant interval of (1.1).
On the other hand, similar to Lemma3.3 (v), it can be proved that y̅∈(p/q,qr/(p-q)).
(vi) By using the monotonic character off, we obtain
(3.23)qrp-q≤(qr+p)(p-q)+pq2rq3r+p(p-q)=f(qrp-q,pq)≤f(x,y)≤f(pq,qrp-q)=pq.
The inequalities
(3.24)(qr+p)(p-q)+pq2rq3r+p(p-q)≥qrp-q,qrp-q≤pq
are equivalent to the inequality p>q(1+1+4r)/2.
On the other hand, similar to Lemma3.3 (v) it can be proved that y̅∈(qr/(p-q),p/q).
(vii) In this case note thatqr/(p-q)≤1<p/q holds. By the monotonic character of f, we have
(3.25)1<qr+pq+pq2+p=f(1,pq)≤f(x,y)≤f(pq,1)=qr+p2+qq(p+1)≤pq.
The inequalities
(3.26)qr+pq+pq2+p>1,qr+p2+qq(p+1)≤pq
are equivalent to the inequality p≥q+qr.
Furthermore, similar to Lemma3.3 (v), it follows y̅∈(1,p/q). The proof is complete.Lemma 3.6.
Assume thatp=q(1+1+4r)/2, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN<p/q, then yN+k+1>p/q;(ii)
If for someN≥0, yN=p/q, then yN+k+1=p/q;(iii)
If for someN≥0, yN>p/q, then yN+k+1<p/q;(iv)
If for someN≥0, yN>p/q, then yN>yN+2(k+1)>p/q;(v)
If for someN≥0, yN<p/q, then yN<yN+2(k+1)<p/q.Proof.
In this case, we have thatqr/(p-q)=p/q. These results follow from the identities (3.10) and (3.11) and the details are omitted.


## 4. Global Attractivity

In this section, we discuss the global attractivity of the positive equilibrium of (1.1). We show that the unique positive equilibrium y̅ of (1.1) is a global attractor when p=q or p<1 and p<q≤pq+1+3p or q<p≤1.


### 4.1. The Casep=q

In this subsection, we discuss the behavior of positive solutions of (1.1) when p=q.Theorem 4.1.
Assume thatp=q holds, and {yn}n=-k∞ is a positive solution of (1.1). Then the unique positive equilibrium y̅ of (1.1) is a global attractor.Proof.
By the change of variables(4.1)yn=1+rp+1un,
Equation (1.1) reduces to the difference equation
(4.2)un+1=11+(pr/(p+1)2)un+(r/(p+1)2)un-k,n∈ℕ0.
The unique positive equilibrium u̅ of (4.2) is
(4.3)u̅=-(p+1)+(p+1)2+4r(p+1)2r.
Applying Lemma 1.3 in interval [0,1], then every positive solution of (1.1) converges to u̅. That is, u̅ is a global attractor. So, y̅ is a global attractor.


### 4.2. The Casep<q

In this subsection, we present global attractivity of (1.1) when p<q.The following result is straightforward consequence of the identity (3.7).Lemma 4.2.
Assume thatp<q holds, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
Suppose thatq<p+r. If for some N≥0, yN>1, then yN+1<(p+r)/q;(ii)
Suppose thatq>p+r. If for some N≥0, yN<1, then yN+1>(p+r)/q.Theorem 4.3.
Assume thatp<q, p<1 and q≤pq+1+3p hold. Let {yn}n=-k∞ be a positive solution of (1.1). Then the following statements hold true:(i)
Supposeq<p+r. If y0∈[1,r/(q-p)], then yn∈[1,r/(q-p)] for n≥1. Furthermore, every positive solution of (1.1) lies eventually in the interval [1,r/(q-p)].(ii)
Supposeq>p+r. If y0∈[r/(q-p),1], then yn∈[r/(q-p),1] for n≥1. Furthermore, every positive solution of (1.1) lies eventually in the interval [r/(q-p),1].Proof.
We only give the proof of (i), the proof of (ii) is similar and will be omitted. First, note that in this casep/q<1<(p+r)/q<r/(q-p) holds.
Ify0∈[1,r/(q-p)], then by Lemma 3.3 (iv), we have that y1>1, and by Lemma 4.2 (i), we obtain that y1<(p+r)/q<r/(q-p), which implies that y1∈[1,r/(q-p)], by induction, we have yn∈[1,r/(q-p)], for n≥1.
Now, to complete the proof it remains to show that wheny0∉[1,r/(q-p)], there exists N>0 such that yN∈[1,r/(q-p)].
Ify0∉[1,r/(q-p)], then we have the following two cases to be considered:
(a)
y0>r/(q-p);(b)
y0<1.
Case (a). From Lemma3.3 (ii), we see that y1<1. Thus, in the sequel, we only consider case (b).
Ify0<1, then by Lemma 3.3 (iv), we have y1>1, and from Lemma 4.2 (i), we have y2<(p+r)/q<r/(q-p). So y3>1 and y4<r/(q-p). By induction, there exists exactly one term greater than 1, which is followed by exactly one term less than r/(q-p), which is followed by exactly one term greater than 1, and so on. If for some N>0, 1≤yN≤r/(q-p), then the former assertion implies that the result is true.
So assume for the sake of contradiction, that for alln≥1, yn never enter the interval [1,r/(q-p)], then the sequence {yn}n=1∞ will oscillate relative to the interval [1,r/(q-p)] with semicycles of length one. Consider the subsequence {y2n}n=1∞ and {y2n+1}n=1∞ of solution {yn}n=-k∞, we have
(4.4)y2n<1,y2n+1>rq-pforn≥1.
Let
(4.5)L=limn→∞supy2n,l=limn→∞infy2n,M=limn→∞supy2n+1,m=limn→∞infy2n+1,
which in view of Theorem 3.1 exist as finite numbers, such that
(4.6)L≤r+pm+lqm+l,l≥r+pM+LqM+L,(4.7)M≤r+pl+mql+m,m≥r+pL+MqL+M.
From (4.6), we have q(Lm-lM)≤p(m-M)+(l-L), which implies that Lm-lM≤0. Also, from (4.7), we have q(lM-Lm)≤p(l-L)+(m-M), which implies that lM-Lm≤0. Thus lM-Lm=0 and L=l, M=m hold, from which it follows that limn→∞y2n and limn→∞y2n+1 exist.
Set
(4.8)limn→∞y2n=L,limn→∞y2n+1=M,
then L≤1, M≥r/(q-p) and L, M satisfies the system
(4.9)L=r+pM+LqM+L,M=r+pL+MqL+M,
which implies that L, M is a period-two solution of (1.1). Furthermore, in view of Theorem 2.2, (1.1) has no period-two solution when p<1 and q≤pq+1+3p hold. This is a contradiction, as desired. The proof is complete.Theorem 4.4.
Assume thatp<q, p<1 and q≤pq+1+3p hold. Then the unique positive equilibrium y̅ of (1.1) is a global attractor.Proof.
To complete the proof, there are four cases to be considered.
Case (i).q<p+r.
By Theorem4.3 (i), we know that all solutions of (1.1) lies eventually in the invariant interval [1,r/(q-p)]. Furthermore, the function f(x,y) is non-increasing in each of its arguments in the interval [1,r/(q-p)]. Thus, applying Lemma 1.3, every solution of (1.1) converges to y̅, that is, y̅ is a global attractor.
Case (ii).q=p+r.
In this case, the only positive equilibrium isy̅=1. In view of Lemma 3.4, we see that, after the first semicycle, the nontrivial solution oscillates about y̅ with semicycles of length one. Consider the subsequences {y2n}n=1∞ and {y2n+1}n=1∞ of any nontrivial solution {yn}n=-k∞ of (1.1). We have
(4.10)y2n<1,y2n+1>1,forn≥1,
or vice versa. Here, we may assume, without loss of generality, that y2n<1 and y2n+1>1, for n≥1.
Let
(4.11)L=limn→∞supy2n,l=limn→∞infy2n,M=limn→∞supy2n+1,m=limn→∞infy2n+1,
which, in view of Theorem 3.1, exist. Then as the same argument in Theorem 4.3, we can see that limn→∞y2n and limn→∞y2n+1 exist.
Set
(4.12)limn→∞y2n=L,limn→∞y2n+1=M,
then L≤1, M≥1. If L≠M, then, also as the same argument in Theorem 4.3, we can see that L, M is a period-two solution of (1.1), which contradicts Theorem 2.2. Thus L=M, from which it follows that limn→∞yn=1, which implies that y̅=1 is a global attractor.
Case (iii).p+r<q<p+(q/p)r.
By Theorem4.3 (ii), we know that all solutions of (1.1) lies eventually in the invariant interval [r/(q-p),1]. Furthermore, the function f(x,y) decreases in x and increases in y in the interval [r/(q-p),1]. Thus, applying Lemma 1.3, every solution of converges to y̅, that is, y̅ is a global attractor.
Case (iv).q≥p+(q/p)r.
In this case, we note thatr/(q-p)≤p/q<1 holds. From Theorem 4.3 (ii) and Lemma 3.3 (i), we know that all solutions of (1.1) eventually enter the invariant interval [p/q,1]. Hence, by using the same argument in (iii), y̅ is a global attractor. The proof is complete.


### 4.3. The Casep>q

In this subsection, we discuss the global behavior of (1.1) when p>q.The following three results are the direct consequences of equations (3.4), (3.5), (3.6), and (3.8).Lemma 4.5.
Assume thatq<p<q(1+1+4r)/2, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN<p/q, then yN<yN+2(k+1)<qr/(p-q);(ii)
If for someN≥0, yN>qr/(p-q), then p/q<yN+2(k+1)<yN;(iii)
If for someN≥0, p/q≤yN≤qr/(p-q), then p/q≤yN+2(k+1)≤qr/(p-q).Lemma 4.6.
Assume thatq(1+1+4r)/2<p<q+qr, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN<qr/(p-q), then yN<yN+2(k+1)<p/q;(ii)
If for someN≥0, yN>p/q, then qr/(p-q)<yN+2(k+1)<yN;(iii)
If for someN≥0, qr/(p-q)≤yN≤p/q, then qr/(p-q)≤yN+2(k+1)≤p/q.Lemma 4.7.
Assume thatp≥q+qr, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>1 for n≥1;(ii)
If for someN≥0, yN>p/q, then 1<yN+2(k+1)<p/q and yN+2(k+1)<yN;(iii)
If for someN≥0, 1<yN≤p/q, then 1<yN+2(k+1)≤p/q.Theorem 4.8.
Assume thatp>q holds, and let {yn}n=-k∞ be a positive solution of (1.1). Then the following statements hold true:(i)
Ifp<q(1+1+4r)/2, then every positive solution of (1.1) lies eventually in the interval [p/q,qr/(p-q)].(ii)
Ifq(1+1+4r)/2<p<q+qr, then every positive solution of (1.1) lies eventually in the interval [qr/(p-q),p/q].(iii)
Ifp≥q+qr, then every positive solution of (1.1) lies eventually in the interval [1,p/q].Proof.
We only give the proof of (i), the proofs of (ii) and (iii) are similar and will be omitted.
Whenq<p<q(1+1+4r)/2, recall that from Lemma 3.5, [p/q,qr/(p-q)] is an invariant interval and so it follows that every solution of (1.1) with k+1 consecutive values in [p/q,qr/(p-q)], lies eventually in this interval. If the solution is not eventually in [p/q,qr/(p-q)], there are three cases to be considered.
Case (i). If for someN≥0, yN>qr/(p-q), then there are two cases to be considered. If yN+2(k+1)n≥qr/(p-q) for every n∈N, then by Lemma 4.5, we have
(4.13)yN+2(k+1)(n-1)>yN+2(k+1)n>pq,
hence, the subsequence {yN+2(k+1)n} is strictly monotonically decreasing convergent and its limit S satisfies S≥qr/(p-q). Taking limit on both sides of (3.8), we obtain a contradiction. If for some n0, yN+2(k+1)n0<qr/(p-q), then by Lemma 4.5 we obtain that {yN+2(k+1)n} is eventually in the interval [p/q,qr/(p-q)].
Case (ii). If for someN≥0, yN<p/q, then there are two cases to be considered. If yN+2(k+1)n<p/q for every n∈N, then by Lemma 4.5 we obtain
(4.14)yN+2(k+1)(n-1)<yN+2(k+1)n<qrp-q,
which implies that the subsequence {yN+2(k+1)n} is convergent. Then as the same argument in case (i), obtain a contradiction. If for some n0, yN+2(k+1)n0>p/q, then by Lemma 4.5 we have that {yN+2(k+1)n} is eventually in the interval [p/q,qr/(p-q)].
Case (iii). If for someN≥0,p/q≤yN≤qr/(p-q), then by Lemma 4.5 it follows that p/q≤yN+2(k+1)n≤qr/(p-q). Assume that there is a subsequence {yN0+2(k+1)n} such that yN0+2(k+1)n≥qr/(p-q), or yN0+2(k+1)n≤p/q, for every n∈N. Then its limit S satisfies S≥qr/(p-q), or S≤p/q. Taking limit on both sides of (3.8), obtain a contradiction. Hence, for all N∈{1,2,…,2(k+1)} the subsequences {yN+2(k+1)n} are eventually in the interval [p/q,qr/(p-q)].Theorem 4.9.
Assume thatp>q and p≤1 hold. Then the unique positive equilibrium y̅ of (1.1) is a global attractor.Proof.
The proof will be accomplished by considering the following four cases.
Case (i).p<q(1+1+4r)/2.
By part (i) of Theorem4.8, we know that all positive solutions of (1.1) lie eventually in the invariant interval [p/q,qr/(p-q)]. Furthermore, the function f(x,y) is nonincreasing in each of its arguments in the interval [p/q,qr/(p-q)]. Thus, applying Lemma 1.3, every solution of (1.1) converges to y̅, that is, y̅ is a global attractor.
Case (ii).p=q(1+1+4r)/2.
In this case, the only positive equilibrium of (1.1) is y̅=p/q. From Lemma 3.6 and (3.11), we know that each of the 2(k+1) subsequences
(4.15){y2(k+1)n+i}n=0∞fori=1,2,…,2(k+1)
of any solution {yn}n=-k∞ of (1.1) is either identically equal to p/q or strictly monotonically convergent and its limit is greater than zero. Set
(4.16)Li=limn→∞y2(k+1)n+ifori=1,2,…,2(k+1).
Then, clearly,
(4.17)…,L1,L2,…,L2(k+1),…
is a period solution of (1.1) with period 2(k+1). By applying (3.11) to the solution (4.17) and using the fact Li>0 for i=1,2,…,2(k+1), we see that
(4.18)Li=pqfori=1,2,…,2(k+1),
and so
(4.19)limn→∞yn=pq,
which implies that y̅=p/q is a global attractor.
Case (iii).q(1+1+4r)/2<p<q+qr.
By Theorem4.8 (ii), all positive solutions of (1.1) eventually enter the invariant interval [qr/(p-q),p/q]. Furthermore, the function f(x,y) increases in x and decreases in y in the interval [qr/(p-q),p/q]. Thus, applying Lemma 1.3 and assumption p≤1, every solution of (1.1) converges to y̅. So, y̅ is a global attractor.
Case (iv).p≥q+qr.
In this case, we note thatqr/(p-q)≤1<p/q holds. In view of Theorem 4.8 (iii), we obtain that all solutions of (1.1) eventually enter the invariant interval [1,p/q]. Furthermore, the function f(x,y) increases in x and decreases in y in the interval [1,p/q]. Then using the same argument in case (iii), every solution of (1.1) converges to y̅. Thus the equilibrium y̅ is a global attractor. The proof is complete.Finally, we summarize our results and obtain the following theorem, which shows thaty̅ is a global attractor in three cases.Theorem 4.10.
The unique positive equilibriumy̅ of (1.1) is a global attractor, when one of the following three cases holds:(i)
p=q;(ii)
p<1 and p<q≤pq+1+3p;(iii)
q<p≤1.


## 4.1. The Casep=q

In this subsection, we discuss the behavior of positive solutions of (1.1) when p=q.Theorem 4.1.
Assume thatp=q holds, and {yn}n=-k∞ is a positive solution of (1.1). Then the unique positive equilibrium y̅ of (1.1) is a global attractor.Proof.
By the change of variables(4.1)yn=1+rp+1un,
Equation (1.1) reduces to the difference equation
(4.2)un+1=11+(pr/(p+1)2)un+(r/(p+1)2)un-k,n∈ℕ0.
The unique positive equilibrium u̅ of (4.2) is
(4.3)u̅=-(p+1)+(p+1)2+4r(p+1)2r.
Applying Lemma 1.3 in interval [0,1], then every positive solution of (1.1) converges to u̅. That is, u̅ is a global attractor. So, y̅ is a global attractor.


## 4.2. The Casep<q

In this subsection, we present global attractivity of (1.1) when p<q.The following result is straightforward consequence of the identity (3.7).Lemma 4.2.
Assume thatp<q holds, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
Suppose thatq<p+r. If for some N≥0, yN>1, then yN+1<(p+r)/q;(ii)
Suppose thatq>p+r. If for some N≥0, yN<1, then yN+1>(p+r)/q.Theorem 4.3.
Assume thatp<q, p<1 and q≤pq+1+3p hold. Let {yn}n=-k∞ be a positive solution of (1.1). Then the following statements hold true:(i)
Supposeq<p+r. If y0∈[1,r/(q-p)], then yn∈[1,r/(q-p)] for n≥1. Furthermore, every positive solution of (1.1) lies eventually in the interval [1,r/(q-p)].(ii)
Supposeq>p+r. If y0∈[r/(q-p),1], then yn∈[r/(q-p),1] for n≥1. Furthermore, every positive solution of (1.1) lies eventually in the interval [r/(q-p),1].Proof.
We only give the proof of (i), the proof of (ii) is similar and will be omitted. First, note that in this casep/q<1<(p+r)/q<r/(q-p) holds.
Ify0∈[1,r/(q-p)], then by Lemma 3.3 (iv), we have that y1>1, and by Lemma 4.2 (i), we obtain that y1<(p+r)/q<r/(q-p), which implies that y1∈[1,r/(q-p)], by induction, we have yn∈[1,r/(q-p)], for n≥1.
Now, to complete the proof it remains to show that wheny0∉[1,r/(q-p)], there exists N>0 such that yN∈[1,r/(q-p)].
Ify0∉[1,r/(q-p)], then we have the following two cases to be considered:
(a)
y0>r/(q-p);(b)
y0<1.
Case (a). From Lemma3.3 (ii), we see that y1<1. Thus, in the sequel, we only consider case (b).
Ify0<1, then by Lemma 3.3 (iv), we have y1>1, and from Lemma 4.2 (i), we have y2<(p+r)/q<r/(q-p). So y3>1 and y4<r/(q-p). By induction, there exists exactly one term greater than 1, which is followed by exactly one term less than r/(q-p), which is followed by exactly one term greater than 1, and so on. If for some N>0, 1≤yN≤r/(q-p), then the former assertion implies that the result is true.
So assume for the sake of contradiction, that for alln≥1, yn never enter the interval [1,r/(q-p)], then the sequence {yn}n=1∞ will oscillate relative to the interval [1,r/(q-p)] with semicycles of length one. Consider the subsequence {y2n}n=1∞ and {y2n+1}n=1∞ of solution {yn}n=-k∞, we have
(4.4)y2n<1,y2n+1>rq-pforn≥1.
Let
(4.5)L=limn→∞supy2n,l=limn→∞infy2n,M=limn→∞supy2n+1,m=limn→∞infy2n+1,
which in view of Theorem 3.1 exist as finite numbers, such that
(4.6)L≤r+pm+lqm+l,l≥r+pM+LqM+L,(4.7)M≤r+pl+mql+m,m≥r+pL+MqL+M.
From (4.6), we have q(Lm-lM)≤p(m-M)+(l-L), which implies that Lm-lM≤0. Also, from (4.7), we have q(lM-Lm)≤p(l-L)+(m-M), which implies that lM-Lm≤0. Thus lM-Lm=0 and L=l, M=m hold, from which it follows that limn→∞y2n and limn→∞y2n+1 exist.
Set
(4.8)limn→∞y2n=L,limn→∞y2n+1=M,
then L≤1, M≥r/(q-p) and L, M satisfies the system
(4.9)L=r+pM+LqM+L,M=r+pL+MqL+M,
which implies that L, M is a period-two solution of (1.1). Furthermore, in view of Theorem 2.2, (1.1) has no period-two solution when p<1 and q≤pq+1+3p hold. This is a contradiction, as desired. The proof is complete.Theorem 4.4.
Assume thatp<q, p<1 and q≤pq+1+3p hold. Then the unique positive equilibrium y̅ of (1.1) is a global attractor.Proof.
To complete the proof, there are four cases to be considered.
Case (i).q<p+r.
By Theorem4.3 (i), we know that all solutions of (1.1) lies eventually in the invariant interval [1,r/(q-p)]. Furthermore, the function f(x,y) is non-increasing in each of its arguments in the interval [1,r/(q-p)]. Thus, applying Lemma 1.3, every solution of (1.1) converges to y̅, that is, y̅ is a global attractor.
Case (ii).q=p+r.
In this case, the only positive equilibrium isy̅=1. In view of Lemma 3.4, we see that, after the first semicycle, the nontrivial solution oscillates about y̅ with semicycles of length one. Consider the subsequences {y2n}n=1∞ and {y2n+1}n=1∞ of any nontrivial solution {yn}n=-k∞ of (1.1). We have
(4.10)y2n<1,y2n+1>1,forn≥1,
or vice versa. Here, we may assume, without loss of generality, that y2n<1 and y2n+1>1, for n≥1.
Let
(4.11)L=limn→∞supy2n,l=limn→∞infy2n,M=limn→∞supy2n+1,m=limn→∞infy2n+1,
which, in view of Theorem 3.1, exist. Then as the same argument in Theorem 4.3, we can see that limn→∞y2n and limn→∞y2n+1 exist.
Set
(4.12)limn→∞y2n=L,limn→∞y2n+1=M,
then L≤1, M≥1. If L≠M, then, also as the same argument in Theorem 4.3, we can see that L, M is a period-two solution of (1.1), which contradicts Theorem 2.2. Thus L=M, from which it follows that limn→∞yn=1, which implies that y̅=1 is a global attractor.
Case (iii).p+r<q<p+(q/p)r.
By Theorem4.3 (ii), we know that all solutions of (1.1) lies eventually in the invariant interval [r/(q-p),1]. Furthermore, the function f(x,y) decreases in x and increases in y in the interval [r/(q-p),1]. Thus, applying Lemma 1.3, every solution of converges to y̅, that is, y̅ is a global attractor.
Case (iv).q≥p+(q/p)r.
In this case, we note thatr/(q-p)≤p/q<1 holds. From Theorem 4.3 (ii) and Lemma 3.3 (i), we know that all solutions of (1.1) eventually enter the invariant interval [p/q,1]. Hence, by using the same argument in (iii), y̅ is a global attractor. The proof is complete.


## 4.3. The Casep>q

In this subsection, we discuss the global behavior of (1.1) when p>q.The following three results are the direct consequences of equations (3.4), (3.5), (3.6), and (3.8).Lemma 4.5.
Assume thatq<p<q(1+1+4r)/2, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN<p/q, then yN<yN+2(k+1)<qr/(p-q);(ii)
If for someN≥0, yN>qr/(p-q), then p/q<yN+2(k+1)<yN;(iii)
If for someN≥0, p/q≤yN≤qr/(p-q), then p/q≤yN+2(k+1)≤qr/(p-q).Lemma 4.6.
Assume thatq(1+1+4r)/2<p<q+qr, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
If for someN≥0, yN<qr/(p-q), then yN<yN+2(k+1)<p/q;(ii)
If for someN≥0, yN>p/q, then qr/(p-q)<yN+2(k+1)<yN;(iii)
If for someN≥0, qr/(p-q)≤yN≤p/q, then qr/(p-q)≤yN+2(k+1)≤p/q.Lemma 4.7.
Assume thatp≥q+qr, and {yn}n=-k∞ is a positive solution of (1.1). Then the following statements are true:(i)
yn>1 for n≥1;(ii)
If for someN≥0, yN>p/q, then 1<yN+2(k+1)<p/q and yN+2(k+1)<yN;(iii)
If for someN≥0, 1<yN≤p/q, then 1<yN+2(k+1)≤p/q.Theorem 4.8.
Assume thatp>q holds, and let {yn}n=-k∞ be a positive solution of (1.1). Then the following statements hold true:(i)
Ifp<q(1+1+4r)/2, then every positive solution of (1.1) lies eventually in the interval [p/q,qr/(p-q)].(ii)
Ifq(1+1+4r)/2<p<q+qr, then every positive solution of (1.1) lies eventually in the interval [qr/(p-q),p/q].(iii)
Ifp≥q+qr, then every positive solution of (1.1) lies eventually in the interval [1,p/q].Proof.
We only give the proof of (i), the proofs of (ii) and (iii) are similar and will be omitted.
Whenq<p<q(1+1+4r)/2, recall that from Lemma 3.5, [p/q,qr/(p-q)] is an invariant interval and so it follows that every solution of (1.1) with k+1 consecutive values in [p/q,qr/(p-q)], lies eventually in this interval. If the solution is not eventually in [p/q,qr/(p-q)], there are three cases to be considered.
Case (i). If for someN≥0, yN>qr/(p-q), then there are two cases to be considered. If yN+2(k+1)n≥qr/(p-q) for every n∈N, then by Lemma 4.5, we have
(4.13)yN+2(k+1)(n-1)>yN+2(k+1)n>pq,
hence, the subsequence {yN+2(k+1)n} is strictly monotonically decreasing convergent and its limit S satisfies S≥qr/(p-q). Taking limit on both sides of (3.8), we obtain a contradiction. If for some n0, yN+2(k+1)n0<qr/(p-q), then by Lemma 4.5 we obtain that {yN+2(k+1)n} is eventually in the interval [p/q,qr/(p-q)].
Case (ii). If for someN≥0, yN<p/q, then there are two cases to be considered. If yN+2(k+1)n<p/q for every n∈N, then by Lemma 4.5 we obtain
(4.14)yN+2(k+1)(n-1)<yN+2(k+1)n<qrp-q,
which implies that the subsequence {yN+2(k+1)n} is convergent. Then as the same argument in case (i), obtain a contradiction. If for some n0, yN+2(k+1)n0>p/q, then by Lemma 4.5 we have that {yN+2(k+1)n} is eventually in the interval [p/q,qr/(p-q)].
Case (iii). If for someN≥0,p/q≤yN≤qr/(p-q), then by Lemma 4.5 it follows that p/q≤yN+2(k+1)n≤qr/(p-q). Assume that there is a subsequence {yN0+2(k+1)n} such that yN0+2(k+1)n≥qr/(p-q), or yN0+2(k+1)n≤p/q, for every n∈N. Then its limit S satisfies S≥qr/(p-q), or S≤p/q. Taking limit on both sides of (3.8), obtain a contradiction. Hence, for all N∈{1,2,…,2(k+1)} the subsequences {yN+2(k+1)n} are eventually in the interval [p/q,qr/(p-q)].Theorem 4.9.
Assume thatp>q and p≤1 hold. Then the unique positive equilibrium y̅ of (1.1) is a global attractor.Proof.
The proof will be accomplished by considering the following four cases.
Case (i).p<q(1+1+4r)/2.
By part (i) of Theorem4.8, we know that all positive solutions of (1.1) lie eventually in the invariant interval [p/q,qr/(p-q)]. Furthermore, the function f(x,y) is nonincreasing in each of its arguments in the interval [p/q,qr/(p-q)]. Thus, applying Lemma 1.3, every solution of (1.1) converges to y̅, that is, y̅ is a global attractor.
Case (ii).p=q(1+1+4r)/2.
In this case, the only positive equilibrium of (1.1) is y̅=p/q. From Lemma 3.6 and (3.11), we know that each of the 2(k+1) subsequences
(4.15){y2(k+1)n+i}n=0∞fori=1,2,…,2(k+1)
of any solution {yn}n=-k∞ of (1.1) is either identically equal to p/q or strictly monotonically convergent and its limit is greater than zero. Set
(4.16)Li=limn→∞y2(k+1)n+ifori=1,2,…,2(k+1).
Then, clearly,
(4.17)…,L1,L2,…,L2(k+1),…
is a period solution of (1.1) with period 2(k+1). By applying (3.11) to the solution (4.17) and using the fact Li>0 for i=1,2,…,2(k+1), we see that
(4.18)Li=pqfori=1,2,…,2(k+1),
and so
(4.19)limn→∞yn=pq,
which implies that y̅=p/q is a global attractor.
Case (iii).q(1+1+4r)/2<p<q+qr.
By Theorem4.8 (ii), all positive solutions of (1.1) eventually enter the invariant interval [qr/(p-q),p/q]. Furthermore, the function f(x,y) increases in x and decreases in y in the interval [qr/(p-q),p/q]. Thus, applying Lemma 1.3 and assumption p≤1, every solution of (1.1) converges to y̅. So, y̅ is a global attractor.
Case (iv).p≥q+qr.
In this case, we note thatqr/(p-q)≤1<p/q holds. In view of Theorem 4.8 (iii), we obtain that all solutions of (1.1) eventually enter the invariant interval [1,p/q]. Furthermore, the function f(x,y) increases in x and decreases in y in the interval [1,p/q]. Then using the same argument in case (iii), every solution of (1.1) converges to y̅. Thus the equilibrium y̅ is a global attractor. The proof is complete.Finally, we summarize our results and obtain the following theorem, which shows thaty̅ is a global attractor in three cases.Theorem 4.10.
The unique positive equilibriumy̅ of (1.1) is a global attractor, when one of the following three cases holds:(i)
p=q;(ii)
p<1 and p<q≤pq+1+3p;(iii)
q<p≤1.



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