# Global Stability of a Rational Difference Equation

Authors: Guo-Mei Tang; Lin-Xia Hu; Gang Ma
Journal: Discrete Dynamics in Nature and Society
 (2010)

---

## Abstract

We consider the higher-order nonlinear difference equationxn+1=(p+qxn−k)/(1+xn+rxn−k),n=0,1,… with the parameters, and the initial conditions x−k,…,x0 are nonnegative real numbers. We investigate the periodic character, invariant intervals, and the global asymptotic stability of all positive solutions of the above-mentioned equation. In particular, our results solve the open problem introduced by Kulenović and Ladas in their monograph (see Kulenović and Ladas, 2002).


---

## Body

## 1. Introduction and Preliminaries

Our aim in this paper is to investigate the global behavior of solutions of the following nonlinear difference equation:(1.1)xn+1=p+qxn-k1+xn+rxn-k,n=0,1,…,
where the parameters p, q, r and the initial conditions x-k,…,x0 are nonnegative real numbers, k∈{1,2,…}.In 2002, Kulenović and Ladas [1] proposed the following open problem.Open Problem 1.
Assume thatp,q,r∈[0,∞) and k∈{2,3,…}. Investigate the global behavior of all positive solutions of (1.1).Consider the difference equation(1.2)xn+1=α+γxn-1A+Bxn+Cxn-1,n=0,1,…
with α,γ,A,B,C∈(0,∞), and the initial conditions x-1, x0 are nonnegative real numbers. Note that, the authors [1, 2] investigated this equation and studied (1.2).In this paper, we will consider the above open problem. Actually, we will investigate the global asymptotic stability and the invariant interval for all positive solutions of (1.1).For the global behavior of solutions of some related equations, see [3–9]. Other related results can be found in [10–19]. For the sake of convenience, we recall some definitions and theorems which will be useful in the sequel.Definition 1.1.
LetI be some interval of real numbers and let
(1.3)f:Im+1⟶I
be a continuously differential function. Then for every set of initial conditions y-k,…,y-1, y0∈I, the difference equation
(1.4)yn+1=f(yn,yn-1,…,yn-k),n=0,1,…,
has a unique solution {yn}n=-k∞.
A pointy¯ is called an equilibrium point of (1.4) if(1.5)y¯=f(y¯,y¯,…,y¯),
that is,
(1.6)yn=y¯forn≥0
is a solution of (1.4), or equivalently y¯ is a fixed point of f.Definition 1.2.
Lety¯ be an equilibrium point of (1.4). (i)
The equilibriumy¯ is called locally stable (or stable) if for every ɛ>0, there exists δ>0 such that for all y-k,…,y-1,y0∈I with ∑i=-ki=0|yi-y¯|<δ, we have |yn-y¯|<ɛ for all n≥k.(ii)
The equilibriumy¯ of (1.4) is called locally asymptotically stable (asymptotic stable) if it is locally stable, and if there exists γ>0 such that for all y-k,…,y-1,y0∈I with ∑i=-ki=0|yi-y¯|<γ, we have limn→∞yn=y¯.(iii)
The equilibriumy¯ of (1.4) is called a global attractor if for every y-k,…,y-1,y0∈I, we have limn→∞yn=y¯.(iv)
The equilibriumy¯ of (1.4) is globally asymptotically stable if it is locally stable and is a global attractor.(v)
The equilibriumy¯ of (1.4) is called unstable if it is not stable.(vi)
The equilibriumy¯ of (1.4) is called a source, or a repeller, if there exists r>0 such that for all y-k,…,y-1,y0∈I with ∑i=-ki=0|yi-y¯|<γ, there exists N≥1 such that |yN-y¯|≥r.An intervalJ⊆I is called an invariant interval for (1.4) if (1.7)y-k,…,y0∈J⟹yn∈J∀n>0.
That is, every solution of (1.4) with initial conditions in J remains in J.The linearized equation associated with (1.4) about the equilibrium y¯ is (1.8)yn+1=∑i=ok∂f∂ui(y¯,…,y¯)yn-i,n=0,1,….Its characteristic equation is(1.9)λk+1=∑i=ok∂f∂ui(y¯,…,y¯)λk-i.Theorem 1.3 (see [20]).
Assume thatf is a C1 function and let y¯ be an equilibrium of (1.4). Then the following statements are true: (i)
If all the roots of (1.9) lie in the open unit disk |λ|<1, then the equilibrium y¯ of (1.4) is asymptotically stable.(ii)
If at least one root of (1.9) has absolute value greater than one, then the equilibrium y¯ of (1.4) is unstable.Theorem 1.4 (see [20]).
Assume thatP,Q∈R and k∈{1,2,…}. Then
(1.10)|P|+|Q|<1
is a sufficient condition for the asymptotic stability of the difference equation
(1.11)yn+1=Pyn+Qyn-k,n=0,1,….Lemma 1.5 (see [4]).
Letp≥2 be a positive integer and assume that every positive solution of equation
(1.12)xn+1=α+βxn+γxn-kA+Bxn+Cnn-k,n=0,1,…
is periodic with period p. If C>0, then A=B=0.Lemma 1.6 (see [21]).
Consider the difference equation(1.13)yn+1=f(yn,yn-k),n=0,1,…,
where k∈{1,2,…}. Let I=[a,b] be some interval of real numbers and assume that
(1.14)f:[a,b]×[a,b]⟶[a,b]
is a continuous function satisfying the following properties: (a)
f(u,v) is nonincreasing in u and nondecreasing in v,(b)
if(m,M)∈[a,b]×[a,b] is a solution of the system
(1.15)m=f(M,m),M=f(m,M),
then m=M.Then (1.13) has a unique equilibrium y¯∈[a,b], and every solution of (1.13) converges to y¯.Lemma 1.7 (see [21]).
Consider the difference equation(1.16)yn+1=f(yn,yn-k),n=0,1,…,
where k∈{1,2,…}. Let I=[a,b] be some interval of real numbers, and assume that
(1.17)f:[a,b]×[a,b]⟶[a,b]
is a continuous function satisfying the following properties: (a)
f(u,v) is nonincreasing in each of its arguments,(b)
if(m,M)∈[a,b]×[a,b] is a solution of the system
(1.18)m=f(M,M),M=f(m,m),
then m=M.Then (1.16) has a unique equilibrium y¯∈[a,b], and every solution of (1.16) converges to y¯.


## 2. The Special Casepqr=0

If the parameterspqr=0, then (1.1) contains the following several equations. We now assume that all their parameters are positive(2.1)xn+1=qxn-k1+xn+rxn-k,n=0,1,…,(2.2)xn+1=p1+xn+rxn-k,n=0,1,…,(2.3)xn+1=0,n=0,1,…,(2.4)xn+1=p1+xn,n=0,1,…,(2.5)xn+1=qxn-k1+xn,n=0,1,…,(2.6)xn+1=p+qxn-k1+xn,n=0,1,….Equation (2.2) was studied in [19], where it is shown that the unique positive equilibrium is a global attractor. Equation (2.3) is trivial. Equation (2.4) is the Riccati equation [1]. Equation (2.5) can be reduces to (2.6), which was discussed in [22], and they showed that the unique positive equilibrium of (2.6) is globally asymptotically stable when q<1. So, here we only consider (2.1).Clearly,x¯=0 is always an equilibrium of (2.1) and when q>1, (2.1) also possesses the unique positive equilibrium x¯=(q-1)/(1+r).The linearized equation associated with (2.1) about the zero equilibrium is (2.7)zn+1-qzn-k=0.
The linearized equation associated with (2.1) about the positive equilibrium is (2.8)zn+1+x¯1+x¯+rx¯zn-q-rx¯1+x¯+rx¯zn-k=0.From this and by Theorem1.4, we have the following result.Theorem 2.1.
(a) Assume thatq<1. Then x¯=0 of (2.1) is locally asymptotically stable.
(b) Assume thatq>1 and r>1. Then the unique positive equilibrium x¯=(q-1)/(1+r) of (2.1) is locally asymptotically stable.Theorem 2.2.
Assume thatq>1 and r>1. Then the unique positive equilibrium x¯ of (2.1) is globally asymptotically stable.Proof.
By Theorem2.1, the positive equilibrium of (2.1) is locally asymptotically stable. It suffices to show that x¯ is a global attractor.
Let(2.9)f(x,y)=qy1+x+ryforx,y∈(0,∞),
then f(x,y) is nonincreasing in x and nondecreasing in y. So
(2.10)0≤f(x,y)≤qrforx,y∈(0,∞).
From
(2.11)qm1+M+rm=m,qM1+m+rM=M,
we have m=M.
Hence by Lemma1.6 the proof is complete.In the following sections we assume that all parameters in (1.1) are positive.


## 3. Local Stability and Period-Two Solutions

The equilibria of (1.1) are the solutions of the equation (3.1)x¯=p+qx¯1+x¯+rx¯.
So (1.1) possesses the unique positive equilibrium (3.2)x¯=q-1+(q-1)2+4p(r+1)2(r+1).
The linearized equation associated with (1.1) about the positive equilibrium is(3.3)zn+1+x¯1+x¯+rx¯zn-q-rx¯1+x¯+rx¯zn-k=0.By Theorem1.4, it is sufficient to show that in (3.4)|x¯1+x¯+rx¯|+|q-rx¯1+x¯+rx¯|<1.
Thus(3.5)|q-rx¯|<1+rx¯.Ifq-rx¯<0, then we have rx¯-q<1+rx¯, and it clearly holds.Ifq-rx¯≥0, then we have q-rx¯<1+rx¯, and
(3.6)q-1<2rx¯.Ifq≤1, the inequality (3.6) obviously holds. Suppose q>1, then we can get
(3.7)q-1<r(q-1)2+4p(r+1),(q-1)2<r2(q-1)2+4pr2(r+1),
from which it follows that(3.8)(r-1)(q-1)2+4pr2>0.So we have the following result.Theorem 3.1.
Assume that(3.9)eitherq≤1orq>1,(r-1)(q-1)2+4pr2>0.
Then the positive equilibrium x¯ of (1.1) is locally asymptotically stable.Theorem 3.2.
(a) Assume thatk is odd. Then (1.1) has a nonnegative prime period-two solution if and only if
(3.10)q>1,(r-1)(q-1)2+4pr2<0.
Further when (3.10) holds, the period-two solution is “unique” and the value of ϕ1 and ϕ2 are the positive roots of the quadratic equation
(3.11)t2-q-1rt+p1-r=0.
(b) Assume thatk is even. Equation (1.1) has no nonnegative prime period-two solution.Proof.
(a) Assume thatk is odd, then xn+1=xn-k. Let
(3.12)…,ϕ1,ϕ2,ϕ1,ϕ2,…
be a nonnegative prime period-two solution of (1.1). Then ϕ1, ϕ2 satisfy the following system:
(3.13)ϕ1=p+qϕ11+ϕ2+rϕ1,ϕ2=p+qϕ21+ϕ1+rϕ2.
Substituting the above two equations, we obtain(3.14)(ϕ1-ϕ2)(ϕ1+ϕ2-q-1r)=0.
Thus
(3.15)ϕ1+ϕ2=q-1r.
Adding them and using the above equations, we can get(3.16)ϕ1ϕ2=p1-r.
Clearly, in this case the discriminant of (3.11) is positive, that is(3.17)Δ=(r-1)(q-1)2+4pr2r2(r-1)>0,
and so ϕ1, ϕ2 are the positive roots of (3.11).
(b) Assume thatk is even, then xn=xn-k. If there exist distinctive nonnegative real number ϕ1 and ϕ2, such that (3.18)…,ϕ1,ϕ2,ϕ1,ϕ2,…
is a prime period-two solution of (1.1) and ϕ1, ϕ2 satisfy the following system:
(3.19)ϕ1=p+qϕ21+ϕ2+rϕ2,ϕ2=p+qϕ11+ϕ1+rϕ1,
which is equivalent to
(3.20)ϕ1+ϕ1ϕ2+rϕ1ϕ2=p+qϕ2,ϕ2+ϕ1ϕ2+rϕ1ϕ2=p+qϕ1.
Subtracting these two equation, we can get
(3.21)(ϕ1-ϕ2)(q+1)=0.
Then ϕ1=ϕ2. This contradicts the hypothesis that ϕ1≠ϕ2.
The proof is complete.Further, applying Lemma1.5, we have the following result about the period solutions.Theorem 3.3.
There is no integerp≥2 such that every positive solution of (1.1) is periodic with period p.


## 4. Boundedness and Invariant Interval

In this section, we will investigate the boundedness and invariant interval of (1.1).Theorem 4.1.
Every solution of (1.1) is bounded from above and from below by positive constants.Proof.
Let{xn}n=-k∞ be a solution of (1.1). Clearly, if the solution is bounded from above by a constant M, then
(4.1)xn+1≥p1+(r+1)M∀n≥-k,
and so it is also bounded from below. Now for the sake of contradiction assume that the solution is not bounded from above. Then there exists a subsequences {xnm+1}m=0∞ such that
(4.2)lim⁡m→∞nm=∞,lim⁡m→∞xnm+1=∞,
and also
(4.3)xnm+1=max⁡{xn:n≤nm}form≥0.
From (1.1) we see that
(4.4)xn+1<p+qxn-kforn≥0,
and so
(4.5)lim⁡m→∞xnm+1=lim⁡m→∞xnm-k=∞.
Hence, for sufficiently large m,
(4.6)0≤xnm+1-xnm-k=p+qxnm-k1+xnm+rxnm-k-xnm-k=p+[(q-1)-xnm-rxnm-k]xnm-k1+xnm+rxnm-k<0,
which is a contradiction. The proof is complete.Let(4.7)f(u,v)=p+qv1+u+rv.
Then the following statements are true.Lemma 4.2.
(a) Assume thatpr≤q. Then f(u,v) is decreasing in u for each v and increasing in v for each u.
(b) Assume thatpr>q. Then f(u,v) is decreasing in u for each v and decreasing in v for u∈[0,(pr-q)/q], and increasing in v for u∈[(pr-q)/q,∞).Proof.
The proofs of (a) and (b) are simple and will be omitted.Lemma 4.3.
Equation (1.1) possesses the following invariant intervals: (a)
[0,q/r], when pr≤q;(b)
[(pr-q)/q,q/r], when q<pr<q+q2/r;(c)
[0,p], when pr=q+q2/r;(d)
[q/r,(pr-q)/q], when q+q2/r<pr<pq+q;(e)
[q/r,p], when pr≥pq+q.Proof.
(a) Setg(x)=(p+qx)/(1+rx), so g(x) is increasing for x and g(q/r)≤q/r. Using (1.1) we see that when x-k,…,x-1,x0∈[0,q/r], then
(4.8)x1=p+qx-k1+x0+rx-k≤p+qx-k1+rx-k≤g(qr)≤qr.
The proof follows by induction.
(b) Using the monotonic character of the functionf(u,v) which is described by Lemma 4.2(b) and the condition that q<pr<q+q2/r, when x-k,…,x-1,x0∈[(pr-q)/q,q/r], we can get (4.9)pr-qq≤f(qr,pr-qq)≤x1=p+qx-k1+x0+rx-k=f(xo,x-k)≤f(pr-qq,qr)=qr.
The proof follows by induction.
(c) Seth(x)=(p+qx)/(1+p+rx), g(x)=(p+qx)/(1+rx), then h(x) is increasing for x and g(x) is decreasing for x when pr=q+q2/r, we see that when x-k,…,x-1,x0∈[0,p], then(4.10)x1=p+qx-k1+x0+rx-k≥p+qx-k1+p+rx-k≥h(0)=p1+p>0,x1=p+qx-k1+x0+rx-k≤p+qx-k1+rx-k≤g(0)=p.
The proof follows by induction.
(d) Using the monotonic character of the functionf(u,v) which is described by Lemma 4.2(b) and the condition q+q2/r<pr<pq+q, when x-k,…,x-1,x0∈[q/r,(pr-q)/q], we obtain(4.11)qr=f(pr-qq,pr-qq)≤x1=p+qx-k1+x0+rx-k=f(xo,x-k)≤f(qr,qr)=pr+q2r+(r+1)q<pr-qq.
The proof follows by induction.
(e) In view of the conditionpr≥pq+q, we can get (pr-q)/q>p>q/r. By using the monotonic character of the function f(u,v) which is described by Lemma 4.2(b) and the condition pr≥pq+q, when x-k,…,x-1,x0∈[q/r,p], we have (4.12)qr≤p+qp1+p+pr=f(p,p)≤x1=p+qx-k1+x0+rx-k=f(xo,x-k)≤f(qr,qr)=pr+q2r+q+qr<p.
The proof follows by induction.
The proof is complete.


## 5. Semicycles Analysis

Let{xn}n=-k∞ be a positive solution of (1.1). Then we have the following equations:(5.1)xn+1-qr=qr(pr-q)/q-xn1+xn+rxn-k,forn≥0,(5.2)xn+1-p=-pxn+(pr-q)xn-k1+xn+rxn-k,forn≥0,(5.3)xn+1-pr-qq=(1/q)(q+q2/r-pr)+((pr-q)/q)(q/r-xn)+(r/q)(q+q2/r-pr)xn-k1+xn+rxn-k,forn≥0,(5.4)xn+1-x¯=r(x¯-q/r)(x¯-xn-k)+x¯(x¯-xn)1+xn+rxn-k,forn≥0,(5.5)xn+2-xn=rxn+1-k(q/r-xn)(rxn-k+xn+1)+(q+r)xn-k(pr/(q+r)-xn)+(p-xn-xn2)(1+xn+rxn-k)(1+rxn+1-k)+p+qxn-k,forn≥0.Ifpr=q+q2/r, then the unique positive equilibrium is x¯=q/r, and (5.1) and (5.5) change into(5.6)xn+1-qr=qrq/r-xn1+xn+rxn-k,forn≥0,(5.7)xn+2-xn=(q/r-xn)(r2xn+1-kxn-k+rxnxn+1-k+rxn+1-k+(pr2/q)xn-k+xn+q/r+1)(1+xn+rxn-k)(1+rxn+1-k)+p+qxn-k,forn≥0.The following lemma is straightforward consequences of identities (5.1)–(5.7).Lemma 5.1.
Assume thatpr≤q and let {xn}n=-k∞ be a solution of (1.1). Then the following statements are true: (i)
xn≤q/r for all n≥1;(ii)
if for someN≥0, xN-k≤x¯ and xN≥x¯, then xN+1≤x¯;(iii)
if for someN≥0, xN-k>x¯ and xN<x¯, then xN+1>x¯;(iv)
0≤x¯≤q/r.Lemma 5.2.
Assume thatq<pr<q+q2/r and let {xn}n=-k∞ be a solution of (1.1). Then the following statements are true: (i)
if for someN≥0, xN<q/r, then xN+1>(pr-q)/q;(ii)
if for someN≥0, xN<(pr-q)/q, then xN+1>q/r;(iii)
if for someN≥0, xN=(pr-q)/q, then xN+1=q/r;(iv)
if for someN≥0, xN>(pr-q)/q, then xN+1<q/r;(v)
if for someN≥0, (pr-q)/q≤xN≤q/r, then (pr-q)/q≤xn≤q/r for n≥N;(vi)
if for someN≥0, xN-k≤x¯, and xN≥x¯, then xN+1≤x¯;(vii)
if for someN≥0, xN-k>x¯, and xN<x¯, then xN+1>x¯;(viii)
if for someN≥0, xN<(pr-q)/q, then xN+2>xN;(ix)
if for someN≥0, xN>q/r, then xN+2<xN;(x)
(pr-q)/q≤x¯≤q/r.Lemma 5.3.
Assume thatpr=q+q2/r and let {xn}n=-k∞ be a solution of (1.1). Then the following statements are true: (i)
if for someN≥0, xN>q/r, then xN+1<q/r;(ii)
if for someN≥0, xN=q/r, then xN+1=q/r;(iii)
if for someN≥0, xN<q/r, then xN+1>q/r;(iv)
if for someN≥0, xN≥x¯, then xN+1≤x¯;(v)
if for someN≥0, xN<x¯, then xN+1>x¯;(vi)
if for someN≥0, xN>q/r, then xN+2<xN;(vii)
if for someN≥0, xN<q/r, then xN+2>xN;(viii)
x¯=q/r.Lemma 5.4.
Assume thatq+q2/r<pr<pq+q and let {xn}n=-k∞ be a solution of (1.1). Then the following statements are true: (i)
if for someN≥0, xN>q/r, then xN+1<(pr-q)/q;(ii)
if for someN≥0, xN<(pr-q)/q, then xN+1>q/r;(iii)
if for someN≥0, xN=(pr-q)/q, then xN+1=q/r;(iv)
if for someN≥0, xN>(pr-q)/q, then xN+1<q/r;(v)
if for someN≥0, q/r≤xN≤(pr-q)/q, then q/r≤xn≤(pr-q)/q for n≥N;(vi)
if for someN≥0, xN-k≤x¯ and xN≤x¯, then xN+1≥x¯;(vii)
if for someN≥0, xN-k>x¯ and xN>x¯, then xN+1<x¯;(viii)
if for someN≥0, xN<(pr-q)/q, then xN+1>q/r;(ix)
if for someN≥0, xN<q/r, then xN+2>xN;(x)
if for someN≥0, xN>(pr-q)/q, then xN+2<xN;(xi)
q/r≤x¯≤(pr-q)/q.Lemma 5.5.
Assume thatpr≥pq+q and let {xn}n=-k∞ be a solution of (1.1). Then the following statements are true: (i)
xn≤p for all n≥1;(ii)
if for someN≥0, xN≤p, then xN+1≥q/r;(iii)
if for someN≥0, q/r≤xN≤p, then q/r≤xn≤p for n≥N;(iv)
if for someN≥0, xN-k≤x¯ and xN≤x¯, then xN+1≥x¯;(v)
if for someN≥0, xN-k>x¯ and xN>x¯, then xN+1<x¯;(vi)
if for someN≥0, xN<q/r, then xN+2>xN;(vii)
if for someN≥0, xN>p, then xN+2<xN;(viii)
q/r≤x¯≤p.The following results are consequences of Lemmas5.1–5.5.Theorem 5.6.
Let{xn}n=-k∞ be a nontrivial solution of (1.1) and x¯ is the unique positive equilibrium point of (1.1). Then the following statements are true. (a)
Assume thatpr≤q. Then except possibly for the first semicycle, every oscillatory solution of (1.1) has semicycles of length at most k.(b)
Assume thatq<pr<q+q2/r. Then, except possibly for the first semicycle, every oscillatory solution of (1.1) which lies in the invariant interval [(pr-q)/q,q/r] has semicycles of length at most k.(c)
Assume thatpr=q+q2/r. Then after the first semicycle, every oscillatory solution of (1.1) about the equilibrium point x¯ with semicycle of length one.(d)
Assume thatq+q2/r<pr<pq+q. Then, except possibly for the first semicycle, every oscillatory solution of (1.1) which lies in the invariant interval [q/r,(pr-q)/q] has semicycles that is either of length at least k-1, or of length at most k+1.(e)
Assume thatpr≥pq+q. Then, except possibly for the first semicycle, every oscillatory solution of (1.1) which lies in the invariant interval [q/r,p] has semicycles that is either of length at least k-1, or of length at most k+1.


## 6. Global Stability

In this section, we will investigate global stability of the positive equilibrium pointx¯ of (1.1).Theorem 6.1.
Assume that (3.9) holds and let {xn}n=-k∞ be a positive solution of (1.1). Then every solution of (1.1) eventually enters the invariant interval (a)
[0,q/r] if pr≤q;(b)
[(pr-q)/q,q/r] if q<pr<q+q2/r;(c)
[q/r,(pr-q)/q] if q+q2/r<pr<pq+q;(d)
[q/r,p] if pr≥pq+q.Proof.
(a) The proof is a direct consequence of Lemma5.1.
(b) From Lemma5.2(v) we know that if there exist an integer N such that xN∈[(pr-q)/q,q/r], then xn∈[(pr-q)/q,q/r] for n≥N and the result follows. Now assume for the sake of contradiction that all terms of {xn} never enter the invariant interval [(pr-q)/q,q/r] for n≥0. Notice that Lemma 5.2(ii) implies that xn+1>q/r for xn<(pr-q)/q. Further using Lemma 5.2(viii) and (ix), we obtain that the subsequence {x2n}n=0∞ and {x2n+1}n=0∞ are both monotonous. If one of them is decreasing, then it is bounded above by (pr-q)/q, and if one of them is decreasing, then it is bounded below by q/r. Thus limn→∞x2n and limn→∞x2n+1 exist. Set(6.1)lim⁡n→∞x2n=L,lim⁡n→∞x2n+1=M,
then L≤(pr-q)/q and M≥q/r, or vice versa. From which it follows that
(6.2)…,L,M,L,M,…
is a period-two solution of (1.1), which is a contradiction, since when (2.3) holds, (1.1) has no period-two solution.
(c) The proof is similar to (b), so will be omitted.
(d) In view of Lemma5.5(i) and (iii), we know that xn≤p for all n≥1 and [q/r,p] is an invariant interval of (1.1). If there exist an integer N such that xN∈[q/r,p], then xn∈[q/r,p] for n≥N, from which it follows that the result is true. Now assume for the sake of contradiction that terms of {xn} never enter the invariant interval [q/r,p], then they all lie in the interval [0,q/r]. Noticing that x1≤q/r and pr≥pq+q hold, we get(6.3)x2-x1=p-x1+rx1-k(q/r-x1)-x121+x1+rx1-k≥p-q/r-(q/r)21+x1+rx1-k>0,
from which it follows by induction that the sequence {xn} is increasing in the interval [0,q/r]. Hence, limn→∞xn exists and limn→∞xn≤q/r, which is a contradiction because (1.1) has no equilibrium point in the interval [0,q/r].
The proof is complete.Theorem 6.2.
Assume that (3.9) holds. Then the positive equilibrium x¯ is a global attractor of (1.1).Proof.
The proof is finished by considering the following five cases.Case 1 (whenpr≤q).
By Lemma4.3(a) and Theorem 6.1(a), we know that (1.1) possesses an invariant interval [0,q/r] and every solution of (1.1) eventually enters the interval [0,q/r]. Further, it is easy to see that f(u,v) decreases in u and increases in v in [0,q/r].Finally observe that when (3.9) holds, the only solution of the system
(6.4)p+qm1+M+rm=m,p+qM1+m+rM=M,
is m=M. Further, Lemma 5.1 implies that (1.1) has a unique equilibrium x¯∈[0,q/r]. Thus, in view of Lemma 1.6, every solution of (1.1) converges to x¯. So the unique positive equilibrium x¯ is a global attractor of (1.1).Case 2 (whenq<pr<q+q2/r).
By Lemma4.3(b) and Theorem 6.1(b), we know that (1.1) possesses an invariant interval [(pr-q)/q,q/r] and every solution of (1.1) eventually enters the interval [(pr-q)/q,q/r]. Further, it is easy to see that f(u,v) decreases in u and increases in v in [(pr-q)/q,q/r]. Then using the same argument in Case 1, (1.1) has a unique equilibrium x¯∈[(pr-q)/q,q/r] and every solution of (1.1) converges to x¯. So the unique positive equilibrium x¯ is a global attractor of (1.1).Case 3 (whenpr=q+q2/r).
In view of part (c) of Theorem5.6, we know that, after the first semicycle, the nontrivial solution oscillates about x¯ with semicycles of length one. Considering the subsequences {x2n}n=0∞ and {x2n+1}n=0∞, we have
(6.5)x2n>qr,x2n+1<qrforn≥0,
or
(6.6)x2n<qr,x2n+1>qrforn≥0.
Let us consider Case 1. Case 2 can be handled in a similar way. In view of Theorem 4.1 and Lemma 5.3, we know that {xn}n=-k∞ is bounded and xn+1=q/r if xn=q/r and
(6.7)x1<x3<x5<⋯<qr<⋯<x4<x2<x0.
So limn→∞x2n and limn→∞x2n+1 exist.Let
(6.8)L=lim⁡n→∞x2n,M=lim⁡n→∞x2n+1,
then L≥q/r,M≤q/r. If L≠M, then L,M is a period-two solution of (1.1). Furthermore, the condition (3.9) holds. This contradicts Theorem 3.2. Thus L=M and limn→∞xn=q/r. So x¯=q/r is a global attractor of (1.1).Case 4 (whenq+q2/r<pr<pq+q).
By Theorem6.1(c), we know that every solution of (1.1) eventually enters the interval [q/r,(pr-q)/q]. Furthermore, it is easy to see that the function f(u,v) decreases in each of its arguments in the interval [q/r,(pr-q)/q]. Let m,M∈[q/r,(pr-q)/q] is a solution of the system
(6.9)p+qm1+m+rm=M,p+qM1+M+rM=m,
that is, the solution of the system
(6.10)p+qm=M+mM+rmM,p+qM=m+mM+rmM.
Then (m-M)(q+1)=0, which implies that m=M. Employing Lemma 1.7, we see that (1.1) has a unique equilibrium x¯∈[q/r,(pr-q)/q] and every solution of (1.1) converges to x¯. Thus the unique positive equilibrium x¯ is a global attractor of (1.1).Case 5.
Whenpr≥pq+q. By Theorem 6.1(d), we know that every solution of (1.1) eventually enters the interval [q/r,p]. Further, it is clear to see that the function f(u,v) decreases in each of its arguments in the interval [q/r,p]. Then, using the same argument as in Case 4, (1.1) has a unique equilibrium x¯∈[q/r,p] and every solution of (1.1) converges to x¯. Thus the unique positive equilibrium x¯ is a global attractor of (1.1).
The proof is complete.In view of Theorems3.1 and 6.2, we have the following result, which solves Open Problem 1 when conditions (3.9) holds.Theorem 6.3.
Assumed that (3.9) holds. Then the positive equilibrium of (1.1) is globally asymptotically stable.



---

*Source: 432379-2011-01-20.xml*