# Existence of Monotone Solutions of  a  Difference Equation

Authors: Taixiang Sun; Hongjian Xi; Weizhen Quan
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

We consider the nonlinear difference equationxn+1=f(xn−k,xn−k+1,…,xn),   n=0,1,…,
where k∈{1,2,…} and the initial values x−k,x−k+1,…,x0∈(0,+∞). We give sufficient conditions under which this equation has monotone positive solutions which converge to the equilibrium, extending and including in this way some results of the literature.


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## Body

## 1. Introduction

In this paper, we study the existence of
monotone positive solutions which converge to the equilibrium of a nonlinear
difference  equation. Recently, there has been a lot of interest in studying
such solutions and the existence of some specific solutions (see [1–20]). In
[8], Karakostas and Stević studied the boundedness, global attractivity,
and oscillatory and asymptotic periodicity of the nonnegative solutions of the
difference equation(E1)xn+1=B+xn−ka0xn+⋯+ak−1xn−k+1+γ,n=0,1,…,where B≥0, γ>0, k∈{1,2,…},
and ai≥0 for every i∈{0,…,k−1} with ∑i=0k−1ai>0 and the initial conditions x−k,…,x0∈(0,+∞).
They proposed the following open problem.Open problem A.
Letγ=1,B=0,
and k≥2.
Is there a positive solution {xn} of (E1) such that xn→0 as n→∞?
In [5], Devault et al. studied the boundedness, global
stability, and periodic character of positive solutions of the difference
equation(E2)xn+1=p+xn−kxn,n=0,1,…,where k∈{1,2,…}, p∈(0,+∞), and the initial conditions x−k,…,x0∈(0,+∞).
They proposed the following Open problem B (which has been solved in [1, 18] by quite different methods).Open problem B.
Do there exist nonoscillatory solutions of
(E2)?
Recently, Stević [12] studied the following difference
equations:(E3)xn+1=p+xn−kα0xn+⋯+αk−1xn−k+1,n=0,1,…,(E4)xn+1=1+xn−kα0xn+⋯+αk−1xn−k+1,n=0,1,…,(E5)xn+1=α+xn−k1+α0xn+⋯+αk−1xn−k+1,n=0,1,…,where p>−1, α>0, k∈{1,2,…},
and αi≥0 for every i∈{0,…,k−1} with ∑i=0k−1αi=1 and the initial conditions x−k,…,x0∈(0,+∞).
He proved that (E3), (E4), and (E5) have positive solutions which decrease to
the equilibrium.The main theorem in this paper is motivated by the
above studies and [17]. In this paper, we consider the following nonlinear
difference equation:(1.1)xn+1=f(xn−k,xn−k+1,…,xn),n=0,1,…,where k∈{1,2,…},
the initial values x−k,x−k+1,…,x0∈(0,+∞),
and f∈C(Ek+1,E),
where C(Ek+1,E) denotes the set of all continuous maps from Ek+1 to E and E=(0,+∞) or E=[0,+∞).
Using arguments similar to ones developed in the proof of main theorem in [18],
we prove that under appropriate conditions (see (C1)–(C5) below) this difference equation has monotone
solutions converging to the equilibrium x¯.


## 2. Main Result

In this section, we assume thatf satisfies the following conditions.(C1) f∈C(Ek+1,E) and f(z0,z1,…,zk) is increasing in z0 (i.e., f(a,z1,…,zk)>f(b,z1,…,zk) if a>b), where E=(0,+∞) or E=[0,+∞) and k≥1 is an integer.(C2)
Equation (1.1) has the unique nonnegative equilibrium, denoted by x¯.(C3) A={(z0,z1,…,zk):z0≥z1≥⋯≥zk≥f(z0,z1,…,zk)≥x¯} is an unbounded connected (closed) set.Now we formulate and prove the main result of this
paper.Theorem 2.1.
Letf satisfy (C1)–(C3).
If there exists g∈C([x¯,+∞)k+1,[x¯,+∞)) such that the following two conditions hold:
(C4)A⊂B={(z0,z1,…,zk):g(z0,z1,…,zk)≥z0≥z1≥⋯≥zk≥x¯},
(C5)zk=f(g(z0,z1,…,zk),z0,z1,…,zk−1) for any (z0,z1,…,zk)∈[x¯,+∞)k+1, then (1.1) has a
monotone positive solution which converges to the equilibrium x¯.Proof.
DefineF:A→B by(2.1)F(z0,z1,…,zk)=(u0,u1,…,uk)≡(z1,z2,…,zk,f(z0,z1,…,zk)),for all (z0,z1,…,zk)∈A.
Claim 1.
F is well defined.
Proof of Claim1.
From (2.1) and the definition of A,
we have(2.2)ui=zi+1,fori∈{0,1,…,k−1},uk=f(z0,z1,…,zk)≥x¯.It follows from (2.2) and (C5) that(2.3)f(z0,u0,…,uk−1)=uk=f(g(u0,u1,…,uk),u0,u1,…,uk−1),which with (C1) implies(2.4)g(u0,u1,…,uk)=z0≥u0≥⋯≥uk≥x¯.Thus, (u0,u1,…,uk)∈B.
Claim 1 is proved.
Claim 2.
F is a bijection from A to B.
Proof of Claim2.
Letz=(z0,z1,…,zk) and y=(y0,y1,…,yk)∈A with z≠y.
If zi≠yi for some i∈{1,…,k},
then F(y)≠F(z).
If z0≠y0 and zi=yi for every i∈{1,…,k},
then from (C1) we have(2.5)f(z0,z1,…,zk)≠f(y0,z1,…,zk),which also implies F(y)≠F(z).
On the other hand, for anyu=(u0,u1,…,uk)∈B,
we have(2.6)g(u0,u1,…,uk)≥u0≥u1≥⋯≥uk≥x¯.Choose(2.7)z=(z0,z1,…,zk)≡(g(u0,u1,…,uk),u0,u1,…,uk−1).It follows from (2.6), (2.7), and (C5) that(2.8)zk=uk−1≥uk=f(g(u0,u1,…,uk),u0,u1,…,uk−1)=f(z0,z1,…,zk)≥x¯,which implies z∈A. From (2.1) and (C5), we obtain(2.9)F(z)=(z1,…,zk,f(z0,z1,…,zk))=(u0,…,uk−1,f(g(u0,u1,…,uk),u0,u1,…,uk−1))=(u0,u1,…,uk)=u.Claim 2 is proved.
Furthermore, sinceF−1(u0,u1,…,uk)=(g(u0,u1,…,uk),u0,u1,…,uk−1) is continuous, F is a homeomorphism from A to B.
SinceA⊂B and F is a homeomorphism from A onto B,
it follows that F−1(A)⊂F−1(B)=A. By induction, we have(2.10)x¯=(x¯,x¯,…,x¯)∈F−n(A)⊂F−n+1(A)for every positive integer n.
Because A is an unbounded connected closed set, we know
that F−n(A) is an unbounded connected closed set for every
positive integer n.
Let(2.11)S=⋂i=0∞F−i(A).
Claim 3.
S is an unbounded connected set.
Proof of Claim3.
Indeed, ifS is a bounded connected closed set, then there
exists β>0 such that S⊂B(x¯,β)≡{x∈Ek+1:d(x¯,x)<β}.
Since F−n(A) is an unbounded connected closed set for every
positive integer n,
it follows that Kn=[{x:d(x¯,x)≤2β}−B(x¯,β)]∩F−n(A)≠∅ and Kn is a bounded closed set. Let xn∈Kn,
then there exist the positive integers n1<n2<⋯<nk<⋯ and a point v∈{x:d(x¯,x)≤2β}−B(x¯,β) such that limk→∞xnk=v.
Notice that v∉S.
On the other hand, for every positive integer n,
there exists N such that xnk∈F−n(A) if nk>N,
which implies v∈F−n(A).
Thus v∈S,
which is a contradiction. Claim 3 is proved.
Now suppose that{xn}n=−k∞ is a positive solution of (1.1) with (x−k,…,x0)∈S−x¯;
we can show that for all positive integer n, (2.12)Fn(x−k,…,x0)=(xn−k,xn−k+1,…,xn)∈A.Thus, {xn}n=−k∞ is a monotone positive solution. Let(2.13)lim⁡n→∞xn=a,then(2.14)a=f(a,a,…,a)≥x¯.It follows from (C2) that a=x¯.
Thus, {xn}n=−k∞ is a nontrivial monotone positive solution
which converges to x¯.
Theorem 2.1 is proved.Remark 2.2.
From the proof of Theorem2.1, we can conclude
that (1.1) has infinitely many monotone positive solutions which converge to the
equilibrium x¯.Remark 2.3.
In
[21], Stević gave another proof of Claim 3 of Theorem 2.1 for
the case of equation xn=xn−k/(1+xn−1+⋯+xn−(k−1)).


## 3. Example and Some Remarks

In this section, we will give an application of Theorem2.1 and some remarks.Example 3.1.
Consider the equation(3.1)xn+1=p+a+xn−kb+∑i=0k−1aixn−i,n=0,1,…,where k∈{1,2,…} and ai≥0 for every i∈{0,…,k−1} with s=∑i=0k−1ai>0 and the initial  conditions x−k,…,x0∈(0,+∞).
If a,b,p∈[0,+∞) satisfy one of
the following conditions:
(i)
a=0,(ii)
b/s≥a>0,
then (3.1) has monotone positive solutions which
converge to the unique nonnegative equilibrium.Proof.
LetE=(0,+∞) if b=0 and let E=[0,+∞) if b>0.
Define f∈C(Ek+1,E) by(3.2)f(z0,z1,…,zk)=p+a+z0b+∑i=0k−1aizk−i,for all (z0,z1,…,zk)∈Ek+1.
Then, (3.1) has the unique nonnegative equilibrium(3.3)x¯=1+ps−b+(1+ps−b)2+4s(pb+a)2s≥p.
LetA={(z0,z1,…,zk):z0≥z1≥⋯≥zk≥f(z0,z1,…,zk)≥x¯} and define(3.4)g(z0,z1,…,zk)=(zk−p)(b+∑i=0k−1aizk−i−1)−a,for all (z0,z1,…,zk)∈[x¯,+∞)k+1;
then(3.5)g(z0,z1,…,zk)≥(x¯−p)(b+sx¯)−a=a+x¯b+sx¯(b+sx¯)−a=x¯.Thus g∈C([x¯,+∞)k+1,[x¯,+∞)).
It is easy to check that the conditions(C1),(C2), and (C5) hold. Now we show that A is an unbounded connected set.
It follows from conditions (i) and (ii) thatb≥as;
then(3.6)f(x,x,…,x)=p+a+xb+xs=p+1s−b−ass(b+sx),F(x)=x−f(x,x,…,x)=sx2−(ps−b+1)x−pb−ab+sxare increasing in x in [x¯,+∞).
Thus c≥c≥⋯≥c≥f(c,c,…,c)≥x¯ for any c≥x¯,
which implies that (c,…,c)∈A and A is unbounded.
Let(z0,z1,…,zk)∈A and Ai={(z0,…,z0,tz0+(1−t)zi,zi+1,…,zk):0≤t≤1} for 0≤i≤k;
then Ai is a connected set. Since(3.7)z0≥z1≥⋯≥zk≥f(z0,z1,…,zk)≥x¯,f(x,x,…,x)=p+a+xb+xsare
increasing in x,
we know that(3.8)z0≥⋯≥z0≥tz0+(1−t)zi≥zi+1≥⋯≥zk≥f(z0,z1,…,zk)≥f(z0,…,z0,tz0+(1−t)zi,zi+1,…,zk)≥f(z0,…,z0)≥x¯,from which it follows that Ai⊂A.
Again since (z0,…,z0,zi+1,…,zk)∈Ai∩Ai+1 for any 0≤i≤k−1, ∪i=0kAi∪{(c,c,…,c):c≥x¯} is a connected subset of A and (z0,z1,…,zk)∈A0,
which implies that A is an unbounded connected set. Thus, the
condition (C3) holds.
On the other hand, letz=(z0,z1,…,zk)∈A,
then(3.9)z0≥z1≥⋯≥zk≥f(z0,z1,…,zk)=p+a+z0b+∑i=0k−1aizk−i≥x¯.It follows from (3.9) that(3.10)g(z0,z1,…,zk)=(zk−p)(b+∑i=0k−1aizk−i−1)−a≥a+z0b+∑i=0k−1aizk−i(b+∑i=0k−1aizk−i)−a=z0,which implies z∈B={(z0,z1,…,zk):g(z0,z1,…,zk)≥z0≥z1≥⋯≥zk≥x¯}.
Thus, condition (C4) holds.
By Theorem2.1, we know that (3.1) has monotone positive
solutions which converge to the unique nonnegative equilibrium x¯=[1+ps−b+(1+ps−b)2+4s(pb+a)]/2s.Remark 3.2.
(i)
Leta=b=0 and a0=1>a1=⋯=ak−1=0,
then (3.1) reduces to (E2).
(ii) Takea=p=0 and b=1 in Example 3.1; then we have solved Open problem A.Example 3.3.
Consider the following equations:(3.11)xn+1=f(xn−k,xn−k+1,…,xn)=xn−k3xn,n=0,1,…,(3.12)xn+1=f(xn−k,xn−k+1,…,xn)=xn21+∑i=0k−1aixn−i−1,n=0,1,…,(3.13)xn+1=f(xn−k,xn−k+1,…,xn)=xn−k1/2+∑i=0k−1aixn−i,n=0,1,…,where k∈{1,2,…} and ai≥0 for every i∈{0,…,k−1} with ∑i=0k−1ai=1 and the initial conditions x−k,…,x0∈(0,+∞).
Then
(i)  equation
(3.11) satisfies conditions (C1) and (C2), but A={(z0,z1,…,zk):z0≥z1≥⋯≥zk≥f(z0,z1,…,zk)≥x¯}={(x¯,…,x¯}) since z0≥z1≥⋯≥zk≥z03/zk≥x¯=1 implies z0=z1=⋯=zk=1;
thus condition (C3) does not hold;
(ii)  using
arguments similar to ones developed in the proof of Example3.1, it is easy to
check that conditions (C2) and (C3) hold, but f(z0,z1,…,zk) is decreasing in z0,
which implies that condition (C1) does not hold for (3.12);
(iii)  equation
(3.13) satisfies condition (C1) and has two nonnegative equilibria: x¯1=0 and x¯2=1/2,
which implies that condition (C2) does not hold; using arguments similar to
ones developed in the proof of Example 3.1, it is easy to check that A={(z0,z1,…,zk):z0≥z1≥⋯≥zk≥f(z0,z1,…,zk)≥x¯2} is an unbounded connected set.Remark 3.4.
From Example3.3, we see that all the conditions (C1),(C2), and (C3) are necessary, in the sense that no pair of
such conditions implies the remaining condition.Remark 3.5.
Ifk=0 and the conditions (C1)–(C3) are satisfied, then automatically the
difference equation xn+1=f(xn) has monotone positive solutions converging to x¯.



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