For nonlinear difference equations of the form xn=F(n,xn−1,…,xn−m), it is usually difficult to find periodic solutions. In this paper, we consider a class of difference equations of
the form xn=anxn−1+bnf(xn−k), where {an},{bn} are periodic sequences and f is a nonlinear filtering function, and show how periodic solutions can be constructed. Several examples are also included to illustrate our results.

1. Introduction

There are good reasons to find “eventually periodic
solutions” of difference equations of the form xn=F(n,xn−1,xn−2,…,xn−m),n∈{0,1,2,…}. For instance, the well-known
logistic population model xn=λxn−1(L−xn−1),n∈{0,1,2,…} is of the above form, and the
study of the existence of its periodic solutions leads to chaotic solutions.
As another example in [1], Chen considers
the equation xn=xn−1+g(xn−k−1),n∈{0,1,2,…}, where k is a nonnegative integer, and g:R→R is a McCulloch-Pitts type
function g(ξ)={−1,ξ∈(σ,∞),1,ξ∈(−∞,σ], in which σ∈R is a constant which acts as a threshold. Chen
showed that all solutions of (1.3) are eventually periodic and pointed out that
such a result may lead to more complicated dynamical behavior of a more general
neural network. Recently, Zhu and Huang [2] discussed the periodic solutions of
the following difference equation: xn=axn−1+(1−a)f(xn−k),n∈{0,1,2,…}, where a∈(0,1),k is a positive integer, and f:R→R is a signal transmission function of the form
(1.9). In particular, they obtained the following
theorem.

Theorem A.

Let p,q∈{0,1,2,…}. If κ∈(ap+1,ap(1−ak−1)(1−ak+p−1))∩(1−aq+ap+q+k,1−aq+1(1−ak+p)1−a2k+p+q), then (1.5) has an eventually (2k+p+q)-periodic solution -periodic solution {xn}n=−k∞.

In this paper, we consider the following delay
difference equation: xn=anxn−1+bnf(xn−k),n∈{0,1,2,…}, where {an}n=0∞ and {bn}n=0∞ are positive ω-periodic sequences such that an+bn≤1 for n≥0.

The integer k is assumed to satisfy k=lω+1, for some nonnegative integer l. The function f can be chosen in a number of ways. Here, f is a filtering function of the
form f(x)={1,x∈(0,κ],0,x∈(−∞,0]∪(κ,∞), where the positive number κ can be regarded as a threshold term.
Therefore, if ω=1, then an=a,bn=b, and k=l+1 so that (1.7) reduces to xn=axn−1+bf(xn−l−1), which includes (1.5) as a special
case.

When l=0, we have xn=axn−1+bf(xn−1), which will also be included in
the following discussions.

Let Ω denote the set of real finite sequences of the
form {ϕ−k,ϕ−k+1,…,ϕ−1}. Given ϕ={ϕ−k,…,ϕ−1}∈Ω, if we let x−k=ϕ−k,…,x−1=ϕ−1, then we may compute x0,x1,… successively from (1.7) in a unique manner. Such
a sequence x={xn}n=−k∞ is called a solution of (1.7) determined by ϕ∈Ω. Recall that a positive integer η is a period of the sequence {xn}n=−k∞ if xη+n=xn for all n≥−k and that τ is the least period of {xn}n=−k∞ if τ is the least among all periods of {xn}n=−k∞. The sequence {xn}n=−k∞ is said to be τ-periodic if τ is the least period of {xn}n=−k∞. In case {xn}n=−k∞ is not periodic, it may happen that for some N≥−k, the subsequence {xn}n=N∞ is τ-periodic. Such a sequence is said to be
eventually τ-periodic. In other words, let us call {yj}j=−k∞ a translate of {xn}n=−k∞ if yj=xj+N+k for j∈{−k,−k+1,…}, where N is some integer greater than or equal to −k. Then, {xn}n=−k∞ is eventually τ-periodic if one of its translates is τ-periodic.

We will seek eventually periodic solutions
of (1.7). This is a rather difficult question since the existence question
depends on the sequences {an},{bn}, the “delay” k, and the control term κ.

Throughout this paper, empty sums are taken to be 0 and empty products to be 1. We will also need the following elementary
facts. If the real sequence {xn}n=−1∞ satisfies the recurrence
relation xn=anxn−1+bn,n∈{0,1,2,…}, then x0=a0x−1+b0,x1=a1x0+b1=a1(a0x−1+b0)+b1=a1a0x−1+a1b0+b1,x2=a2x1+b2=a2(a1a0x−1+a1b0+b1)+b2=a2a1a0x−1+a2a1b0+a2b1+b2, and by
induction, xn=α0,nx−1+α0,nα0,0b0+α0,nα0,1b1+⋯+α0,nα0,nbn=α0,n(x−1+b0α0,0+b1α0,1+⋯+bnα0,n), where α0,j=∏n=0jan,j∈{0,1,2,…}. Since {an} and {bn} are positive ω-periodic sequences, we see further
that α0,mω+i=(α0,ω−1)mα0,i,i∈{0,…,ω−1};m∈{0,1,2,…}, that ∑j=0mω+ibjα0,j=(b0α0,0+⋯+bω−1α0,ω−1)+(bωα0,ω+⋯+b2ω−1α0,2ω−1)+⋯+(b(m−1)ωα0,(m−1)ω+⋯+bmω−1α0,mω−1)+(bmωα0,mω+⋯+bmω+iα0,mω+i)=(b0α0,0+⋯+bω−1α0,ω−1){1+1α0,ω−1+⋯+1(α0,ω−1)m−1}+1(α0,ω−1)m{b0α0,0+⋯+biα0,i} for i∈{0,…,ω−1} and m∈{0,1,2,…}, and that xmω+i=α0,mω+i(x−1+∑j=0mω+ibjα0,j)=(α0,ω−1)mα0,ix−1+α0,ω−11−α0,ω−1m1−α0,ω−1α0,iβ0,ω−1+α0,iβ0,i for i∈{0,…,ω−1} and m∈{0,1,2,…}, where β0,j=∑k=0jbkα0,k,j∈{0,1,…,ω−1}.

2. Convergence of Solutions

The filtering function f will return 0 for inputs that fall below 0 or above the threshold constant κ.
For this reason, we will single out some subsets of Ω as follows: Ω−={{ϕ−k,…,ϕ−1}∈Ω∣ϕi≤0,−k≤i≤−1},Ω*={{ϕ−k,…,ϕ−1}∈Ω∣0<ϕi≤κ,−k≤i≤−1},Ω+={{ϕ−k,…,ϕ−1}∈Ω∣ϕi>κ,−k≤i≤−1}.

Let x={xn}n=−k∞ be the solution of (1.7) determined by ϕ∈Ω−. By (1.7), x0=a0x−1+b0f(x−k)=a0x−1≤0,x1=a1x0+b1f(x−k+1)=a1x0=a1a0x−1≤0. By induction, we may see
that xn=anan−1⋯a1a0x−1≤0,n∈{0,1,2,…}. Since 0≤limn→∞a0a1⋯an−1an≤limn→∞(max{a0,a1,…,aω−1})n+1=0, we see that limn→∞xn=0.

Next, let x={xn}n=−k∞ be the solution of (1.7) determined by ϕ∈Ω*. If κ≥1, then by (1.7), 0<x0=a0x−1+b0≤a0κ+b0=a0κ−a0+a0+b0≤a0(κ−1)+1≤κ,0<x1=a1x0+b1≤a1(a0κ+b0)+b1=a1κ+b1≤κ. By induction, we see
that 0<xn=anxn−1+bn≤anκ+bn≤κ,n∈{0,1,2,…}. By (1.7), we see
that xn=anxn−1+bnf(xn−k)=anxn−1+bn,n∈{0,1,2,…}. In view of (1.18), we see further
that limm→∞xmω+i=Ai,i∈{0,1,…,ω−1}, where Ai=α0,i(α0,ω−1β0,ω−11−α0,ω−1+β0,i),i∈{0,1,…,ω−1}.

Next, let x={xn}n=−k∞ be the solution of (1.7) determined by ϕ∈Ω+. Then, by (1.7), x0=a0x−1+b0f(x−k)=a0x−1, and by induction, xn=anan−1⋯a0x−1,n∈{0,1,2,…}. Although x−1>κ, since (2.4) holds, we see that {xn} is a strictly decreasing sequence tending to 0. Hence, there is a nonnegative integer j such that xj−1>κ but xj≤κ. Then, κ≥xj>xj+1>xj+2>⋯>xj+k−1. If we let ϕ={xj,xj+1,…,xj+k−1},
then ϕ∈Ω*. If κ≥1, then by what we have shown above, the solution {x˜n} of (1.7) determined by ϕ satisfies limm→∞x˜mω+i=Ai for i∈{0,1,…,ω−1}. By uniqueness, x˜n=xn+j+k for n≥0. In other words, the translate {x˜n} of the solution {xn}n=−k∞ satisfies limm→∞x˜mω+i=Ai for i∈{0,1,…,ω−1}.

We summarize the above discussions by means of the
following result.

Lemma 2.1.

A
solution
x={xn}n=−k∞ determined by ϕ∈Ω− will tend to 0; and if κ≥1, then a solution x={xn}n=−k∞ determined by ϕ∈Ω*∪Ω+ will satisfy (2.8) or one of its translates
will satisfy it.

Lemma 2.2.

If 0<κ<min{1,max{A0,A1,…,Aω−1}}, then for any solution {xn} of (1.7) determined by a ϕ∈Ω*∪Ω+, there exists an integer m∈{0,1,…} such that {xm−k,…,xm−1}∈Ω* and xm∈(κ,1).

Proof.

First
let {xn}n=−k∞ be the solution of (1.7) determined by a ϕ∈Ω*. If xn∈(0,κ] for all n∈{−k,−k+1,…}, then xn=anxn−1+bnf(xn−k)=anxn+bn,n∈{0,1,2,…}, so that by (1.18), we see
that limm→∞xmω+i=Ai,i∈{0,1,…,ω−1}. But, this is contrary to our
assumption that 0<κ<min{1,max{A0,A1,…,Aω−1}}. Hence, there is some nonnegative integer m such that xn∈(0,κ] for n∈{−k,−k+1,…,m−1} but xm∈(−∞,0]∪(κ,∞). Note that xm=amxm−1+bmf(xm−k)>0, which implies that xm∈(κ,∞). Moreover, since xm−1∈(0,κ]⊂(0,1), we then have xm=amxm−1+bm<am+bm≤1, so that xm∈(κ,1).

Next, let {xn}n=−k∞ be the solution of (1.7) determined by a ϕ∈Ω+. As seen in the discussions immediately
preceding Lemma 2.1, there is a nonnegative integer j such that {xj,xj+1,…,xj+k−1}∈Ω*. If xn∈(0,κ] for all n∈{j,j+1,…}, then as we have just explained, a translate {x˜n} of {xn} will satisfy limm→∞x˜mω+i=Ai,i∈{0,1,…,ω−1}. This is again a
contradiction. Hence, we may conclude our proof in a manner similar to the
above discussions. The proof is complete.

From the proof of Lemma 2.2, we see that if κ∈(0,min{1,max{A0,A1,…,Aω−1}}), then to study the limiting behavior of a
solution {xn}n=−k∞ determined by ϕ in Ω*∪Ω+, we may assume without loss of generality that ϕ∈Ω* and x0∈(κ,1). As an example, let us consider (1.11), where we
recall that a,b>0 and a+b≤1.

Example 2.3.

Let ab/(1−a2)≤κ<b/(1−a2). Then, (1.11) has a 2-periodic solution {xn}k=−1∞ with x−1∈(0,κ] and x0∈(κ,1). Indeed, let us choose x−1=ab/(1−a2) (and hence, x0=b/(1−a2)). Then, 0<x−1=ab1−a2≤κ,κ<x0=ax−1+b=b1−a2<1. Furthermore, x1=ax0=ab1−a2∈(0,κ],x2=ax1+b=a⋅ab1−a2+b=b1−a2=x0, so that x1=x3=x5=⋯ and x2=x4=x6=⋯ and x1≠x2.

3. Existence of Eventually Periodic Solutions

Recall that G[0](u)=u,G[1](u)=G(u),G[2](u)=(G∘G)(u)=G(G(u)),…,G[j](u)=G(G[j−1](u)) are the zeroth, first, second, and so forth
and the jth iterate of the function G(u). Also, recall the fact that if {un}n=0∞ is a sequence that satisfies un+1=G(un),n∈{0,1,2,…}, then {un} is a τ-periodic sequence if and only
if u0=G[τ](u0),u0≠G[j](u0),j=1,2,…,τ−1. For convenience,
denote αn=∏j=1naj,βn=∑j=1nbjαj,n∈{1,2,…}. Since αnβn+αn=a1⋯an+a2⋯anb1+a3⋯anb2+⋯+bn≤1, we see that αnβn1−αn≤1,n∈{1,2,…}.

Theorem 3.1.

Let k=lω+1,p=τω−1, and q=σω−1, where l,τ,σ∈{1,2,…,k−1}. Let I1(p)=[αωτ(αωl+(1−αωl)1−αωαωβω),αp(1−αωl)αωβω(1−αpαωl)(1−αω)),I2(p,q)=[M,αωτ+σ+l(1−αωl)+(1−αωσ)(1−αωτ+σ+2l)(1−αω)αωβω), where M=max{αnαωτ+l(αωl+1−αωl1−αωαωβω)+αnβn:n∈{0,1,…,q}}. If κ∈I1(p)∩I2(p,q) and 0<κ<min{αnβn1−αn:n∈{1,2,…,k−1}}, then (1.7) has an eventually (2k+p+q)-periodic solution {xn}n=−k∞ (which can be explicitly generated).

Proof.

From
the condition that l,τ,σ∈{1,2,…,k−1}, we have k−1≥ω.
By (3.5), we see that 1≥A0=α0,0(α0,ω−1β0,ω−11−α0,ω−1+β0,0)=11−αω(a0αωβ0,ω−1+b0−b0αω)=11−αω(a0αω(b0a0+b1a0a1+⋯+bω−1a0⋯aω−1)+b0−b0αω)=αωβω1−αω>κ. Hence, κ<max{A0,A1,…,Aω−1}. Thus, 0<κ<min{1,max{A0,A1,…,Aω−1}}. By Lemmas 2.1 and 2.2, we may look for our desired
eventually periodic solution {xn}n=−k∞ determined by ϕ∈Ω* such that x0∈(κ,1).

Define gn(u)=αnu+αnβnforn∈{0,1,2,…},hn(u)=anuforn∈{0,1,2,…}, and the mapping g by g(x)=(gq+1∘(hω∘⋯∘h1)[τ+l]∘gk−1)(x). We will show
that g(x)=αωτ+σ+l(αωlx+αωβω(1−αωl)1−αω)+αωβω(1−αωσ)1−αω, and that g maps D0=(κ,1) into D0 with a fixed point x*∈D0,
where x*=βωαωτ+σ+l+1(1−αωl)+βωαω(1−αωσ)(1−αωτ+σ+2l)(1−αω). The first assertion is easy to
show. Indeed, since gk−1(x)=αk−1x+αk−1βk−1,(hω∘⋯∘h1)[τ+l](x)=(aω⋯a1)τ+lx=αωτ+lx,gq+1(x)=αq+1x+αq+1βq+1, we see that ((hω∘⋯∘h1)[τ+l]∘gk−1)(x)=αωτ+l(αk−1x+αk−1βk−1),g(x)=αq+1αωτ+l(αk−1x+αk−1βk−1)+αq+1βq+1=αωτ+σ+l(αωlx+αωβω(1−αωl)1−αω)+αωβω(1−αωσ)1−αω.

We now show the second assertion. Note that the linear
maps gn and hn satisfy gmω(u)=αmωu+αmωβmω=αωmu+(1−αωm)1−αωαωβω,m∈{0,1,2,…},hmω∘hmω−1∘⋯∘h1(u)=αωmu,m∈{0,1,2,…}. Let gn(D0)=Dn for n∈{1,…,k−1}.
Since ϕ∈Ω* and x0∈D0, it is clear that the solution {xn} of (1.7) satisfies xn=gn(x0),n∈{1,…,k−1}. Moreover, it is easy to prove
that Dn=(gn(κ),gn(1)),n∈{1,…,k−1}. Indeed, we have κ<αnκ+αnβn=gn(κ)<αn+αnβn=gn(1)<αn+βn≤1,n∈{1,2,…,k−1}. That is, Dn⊂D0 holds for all n∈{0,…,k−1}. Let n1 be the largest integer such that xn>κ for n∈{0,1,…,n1+k−1}. Then, from (1.7), we can obtain xn+k−1=an+k−1⋯ak(αωlx0+αωβω(1−αωl)1−αω),n∈{1,2,…,n1+k}, which implies that xn+k−1∈Dn+k−1 for n∈{1,2,…,n1+k}, where Dn+k−1=an⋯a1gk−1(D0)=(αn(αωlκ+αωβω(1−αωl)1−αω),αn(αωl+αωβω(1−αωl)1−αω)). Since κ∈I1(p),
we have κ<αp(1−αωl)αωβω(1−αpαωl)(1−αω), that is, κ<αp(αωlκ+(1−αωl)αωβω(1−αω))<αp−1(αωlκ+(1−αωl)αωβω(1−αω))<⋯<αωlκ+(1−αωl)(1−αω)αωβω,αp(αωl+1−αωl1−αωαωβω)<αp−1(αωl+1−αωl1−αωαωβω)<⋯<αωl+(1−αωl)(1−αω)αωβω≤1, which shows that Dn+k−1⊂D0 for n∈{0,1,…,p}. Thus, n1≥p and xn+k−1∈Dn+k−1⊂(0,κ]forn∈{p+1,…,p+k}. In fact, from κ∈I1(p), we have xp+k=ap+kxp+k−1+bp+kf(xp)=ap+kxp+k−1=ap+kap+k−1xp+k−2=⋯=ap+k⋯ak(αωlx0+1−αωl1−αωαωβω)=αp+1(αωlx0+1−αωl1−αωαωβω)=αωτ(αωlx0+1−αωl1−αωαωβω)≤κ,xp+k+1=ap+k+1xp+k+bp+k+1f(xp+1)=ap+k+1xp+k<κ, and, by
induction, xp+2k−1=ap+2k−1xp+2k−2+bp+2k−1f(xp+k−1)=ap+2k+1xp+2k−2<κ. Then, it is easy to see that n1=p.

Taking n=p+k in (3.20), we have x2k+p−1=a2k+p−1⋯akgk−1(x0)=ak+p⋯a1gk−1(x0)=αωτ+l(αωlx0+αωβω(1−αωl)1−αω). Let n2 be the largest integer such that xn+2k+p−1∈(0,κ] for n∈{0,1,…,n2}. Then, it follows from (1.7) that xn+2k+p−1=∏j=2k+pn+2k+p−1ajx2k+p−1+∏j=2k+pn+2k+p−1aj∑j=2k+pn+2k+p−1bja2k+p⋯aj=αnx2k+p−1+αnβn=αnαωτ+l(αωlx0+1−αωl1−αωαωβω)+αnβn=gn(x2k+p−1) for n∈{1,2,…,n2+k}. This implies that xn+2k+p−1∈Dn+2k+p−1 for n∈{1,2,…,n2+k}, where Dn+2k+p−1=(gn(hω∘⋯∘h1)[τ+l]gk−1)(D0).

Substituting (3.21) with n1=p into (3.28), we have Dn+2k+p−1=(gn∘(hω∘⋯∘h1)[τ+l]gk−1(κ),gn(hω∘⋯∘h1)[τ+l]gk−1(1))=(αnαωτ+l(αωlκ+1−αωl1−αωαωβω)+αnβn,αnαωτ+l(αωl+1−αωl1−αωαωβω)+αnβn) for n∈{1,2,…,n2+k}. Since κ∈I2(p,q), we have αnαωτ+l(αωl+1−αωl1−αωαωβω)+αnβn≤κforn∈{0,1,…,q}. From (3.29), we further
have xn+2k+p−1∈Dn+2k+p−1⊂(0,κ]forn∈{0,1,…,q}. By (3.8), (3.24), (3.28), and (3.31) as
well as κ∈I2(p,q),
we have x2k+p+q=a2k+p+qx2k+p+q−1+b2k+p+q=a2k+p+q(αqαωτ+l(αωlx0+1−αωl1−αωαωβω)+αqβq)+b2k+p+q=aq+1(αqαωτ+l(αωlx0+1−αωl1−αωαωβω)+aq+1αqβq)+bq+1=αωτ+σ+l(αωlx0+1−αωl1−αωαωβω)+1−αωσ1−αωαωβω>αωτ+σ+l(αωlκ+1−αωl1−αωαωβω)+1−αωσ1−αωαωβω>κ,x2k+p+q+1=a2k+p+q+1x2k+p+q+b2k+p+q+1>a1κ+b1>κ,x2k+p+q+2=a2k+p+q+2x2k+p+q+1+b2k+p+q+2>a2(a1κ+b1)+b2>κ,⋮x2k+p+q+k−1=a2k+p+q+k−1x2k+p+q+k−2+b2k+p+q+k−1=∏j=2k+p+q+12k+p+q+k−1ajx2k+p+q+∏j=2k+p+q+12k+p+q+k−1aj∑j=2k+p+q+12k+p+q+k−1bja2k+p+q+1⋯aj=αk−1x2k+p+q+αk−1βk−1>αk−1κ+αk−1βk−1>κ.

Hence,xn+2k+p−1∈Dn+2k+p−1⊂D0forn∈{q+1,…,q+k}, which implies that n2=q. In particular, taking n=q+1 in (3.33) and (3.28), we have, respectively, x2k+p+q∈D2k+p+q⊂D0,x2k+p+q=g(x0)=αωτ+σ+l(αωlx0+αωβω(1−αωl)1−αω)+αωβω(1−αωσ)1−αω.

Since g is a linear map sending D0 into D0, then it is easy to see that it has a unique
fixed point x* in D0 which satisfies (3.13).

Next, we assert that there is a ϕ*∈Ω* such that the solution {xn} determined by ϕ* satisfies x0=x*, and that {xn} is a periodic solution of (1.7) with minimal
period 2k+p+q. To see this, we choose ϕ−1=(x*−b0)/a0 and arbitrary ϕ−2,…,ϕ−k∈(0,κ]. Then, clearly, the solution {xn} of (1.7) determined by ϕ−k,…,ϕ−1 will satisfy x0=x*. Furthermore, we may show that x−1=ϕ−1∈(0,κ]. Indeed, from αωτ+σ+l+αω>αωσ+αωτ+σ+2l+1, we have αωτ+σ+l(1−αωl)+(1−αωσ)>1−αω−αωτ+σ+2l+αωτ+σ+2l+1,αωτ+σ+l(1−αωl)+(1−αωσ)1−αωτ+σ+2l>(1−αω)=αωβω(1−αω)αωβω>b0(1−αω)αωβω, hence, x*=βωαωτ+σ+l+1(1−αωl)+βωαω(1−αωσ)(1−αωτ+σ+2l)(1−αω)>b0. Thus, ϕ−1=(x*−b0)/a0>0. Next, from 0≤αωτ+σ+2l−αωτ+σ+3l, we get αωτ+l−1≤αωτ+2l+αωτ+l(1−αωl)−1−αω2τ+σ+4l−αω2τ+σ+3l(1−αωl)+αωτ+σ+2l, so that αωτ+l−11−αωτ+σ+2l≤αωτ+2l+αωτ+l(1−αωl)−1,1+αωτ+σ+l−αωσ1−αωτ+σ+2l≤αωσ+τ+2l+αωσ+τ+l(1−αωl)+1−αωσ,αωτ+σ+l+1(1−αωl)+αω(1−αωσ)1−αωτ+σ+2l≤αωσ+τ+2l+1+αωσ+τ+l+1(1−αωl)+αω(1−αωσ),αωτ+σ+l+1(1−αωl)+αω(1−αωσ)(1−αωτ+σ+2l)(1−αω)≤αωσ+τ+2l+11−αω+αωσ+τ+l+1(1−αωl)1−αω+αω(1−αωσ)1−αω. On the other hand, by (3.5), we
have αω1−αω≤1βω, so that αωτ+σ+l+1βω(1−αωl)+αωβω(1−αωσ)(1−αωτ+σ+2l)(1−αω)≤αωσ+τ+2l+αωσ+τ+l+1βω(1−αωl)1−αω+αωβω(1−αωσ)1−αω=αωσ+τ+l(αωl+1−αωl1−αωαωβω)+αωβω(1−αωσ)1−αω=αωσ+τ+l(αωl+1−αωl1−αωαωβω)+ασωβσω=a0αqαωτ+l(αωl+1−αωl1−αωαωβω)+a0αqβq+b0. In view of our assumption that κ∈I2(p,q),
we may now see that x−1≤κ.

In view of the above discussions, we see that 0<xn≤κ for n∈{−k,…,−1},xn>κ for n∈{0,…,p+k−1}, and 0<xn≤κ for n∈{p+k,…,p+2k+q−1}. Since x* is the unique fixed point of g(x) in D0, we have g(x*)=x*,g[2](x*)=x*,…,g[n](x*)=x*, and so forth, and hence, x2k+p+q=g(x*)=x*,x2k+p+q+1=a2k+p+q+1x2k+p+q+b2k+p+q+1f(xk+p+q+1)=a1x*+b1=x1>κ,⋮xp+2k+q+k−1=ap+2k+q+k−1xp+2k+q+k−2+bp+2k+q+k−1f(xp+q+2k−1)=ak−1xp+2k+q+k−2+bk−1=ak−1xk−2+bk−1=xk−1>κ,xp+2k+q+k=ap+2k+q+kxp+2k+q+k−1+bp+2k+q+kf(xp+q+2k)=akxk−1=xk>κ,xp+2k+q+k+1=ap+2k+q+k+1xp+2k+q+k+bp+2k+q+k+1f(xp+q+2k+1)=ap+2k+q+k+1xp+2k+q+k=ak+1xk=xk+1>κ,⋮xp+2k+q+k+p−1=ap+2k+q+k+p−1xp+2k+q+k+p−2+bp+2k+q+k+p−1f(xp+q+2k+p−1)=ap+2k+q+k+p−1xp+2k+q+k+p−2=ak+p−1xk+p−2=xk+p−1>κ,xp+2k+q+k+p=azp+2k+q+k+pxp+2k+q+k+p−1+bp+2k+q+k+pf(xp+q+2k+p)=ap+kxp+k−1+bp+kf(xp)=ak+pxk+p−1=xk+p≤κ,⋮x2k+p+q+2k+p+q−1=a2k+p+q−1x2k+p+q+2k+p+q−2+bp+2k+q−1f(x2k+p+q+k+p+q−1)=a2k+p+q−1x2k+p+q−2+b2k+p+q−1=x2k+p+q−1≤κ,x2(2k+p+q)=g[2](x*)=g(x2k+p+q)=g(x*)=x*, and so forth.
Thus, xn>κforn∈{0,…,p+k−1},0<xn≤κforn∈{p+k,…,p+2k+q−1},xn>κforn∈{p+2k+q,…,p+2k+q+p+k−1},0<xn≤κforn∈{p+2k+q+p+k,…,2(p+2k+q)−1}, and so forth.

By induction, we may see that xn>κ for n∈{m(p+2k+q),…,m(p+2k+q)+p+k−1},0<xn≤κ for n∈{m(p+2k+q)+p+k,…,(m+1)(p+2k+q)−1}, where m∈{0,1,2,…}, and xn(2k+p+q)=x*,xn(2k+p+q)+1=x1,…,xn(2k+p+q)+2k+p+q−1=x2k+p+q−1. This shows that {xn} is an eventually periodic solution of (1.7),
whose minimal period is 2k+p+q.
The proof is complete.

We remark that in the above result, l cannot be 0. We may, however, show the following by similar
considerations.

Theorem 3.2.

Let k=1,I1=[αω3βω(1−αω3),α2ω−1αωβω1−αω3),I2=[M,α2ω−1αωβω1−αω3), where M=max{αnαω3βω1−αω3+αnβn:n∈{0,1,…,ω−1}}. If κ∈I1∩I2 and 0<κ<min{αnβn1−αn:n∈{1,2,…,ω}}, then (1.7) has an eventually 3ω-periodic solution {xn}n=−k∞ (which can be generated explicitly).

Proof.

Similar
to the proof of the Theorem 3.1, set (3.10) and define the mapping g by g(x)=gω∘(hω∘hω−1∘⋯∘h1)[2](x). We may show that g(x)=αω(αω2x)+αωβω, and that g maps D0=(κ,1) into D0 with a unique fixed point x*∈D0, where x*=αωβω1−αω3. Let us choose x−1=1a0(αωβω1−αω3−b0). By αω>αω(1−αω3)>b0αωβωαω(1−αω3)=b0βω(1−αω3), we have αωβω(1−αω3)>b0, and hence, x−1=1a0(αωβω1−αω3−b0)>0. Since αω3(b0+αωβω−1)=αω4βω, then αωβω1−αω3−b0=αω4βω1−αω3+αωβω−1, and hence, x−1=1a0(αωβω1−αω3−b0)=αω−1(αω3βω1−αω3)+αω−1βω−1≤κ,x0=αωβω1−αω3>κ,x1=a1x0=a1αωβω1−αω3>κ,⋮x2ω−1=a2ω−1⋯a1αωβω1−αω3>κ,x2ω=αω3βω1−αω3≤κ,⋮x3ω−1=αω−1αω3βω1−αω3+αω−1βω−1≤κ,x3ω=αω4βω1−αω3+αωβω=x0,⋮ so that {xn} is an eventually 3ω-periodic solution of the system (1.7).

4. Examples and Remarks

Let {an},{bn} be 2-periodic sequences, k=3,p=1,q=1, and I1=[α2(α2+α2β2),a1α2β21−a1α2),I2=[α22(α2+α2β2),α23+11−α24α2β2)∩[α1α22(α2+α2β2)+α1β1,α23+11−α24α2β2). Suppose κ∈I1∩I2 and 0<κ<min{α11−α1β1,α21−α2β2}. Consider the following
“delay” difference equation: xn=anxn−1+bnf(xn−3),n∈{0,1,2,…}. We can check that (4.3) has an
eventually 8-periodic solution {xn}n=−3∞ with x0∈(κ,1).

In fact, as in the proof of Theorem 3.1,
let x*=α24β2+α2β21−α24,ϕ−1=x*−b0a0, and ϕ−2,ϕ−3 be arbitrary numbers in (0,κ]. Then, as shown in the proof of Theorem 3.1, the
solution of (4.3) determined by ϕ−3,ϕ−2,ϕ−1 satisfies x−1=ϕ−1∈(0,κ] and x0=x*.

Since κ∈I2,
we have x0>κ. On the other hand, by (3.3), and (α23+1)/(1+α22)(1+α2)<1, hence κ<x0<1,x1=a1x0+b1f(x−2)=a1x0+b1>a1κ+b1>κ,x2=a2x1+b2f(x−1)=a2x1+b2=a2(a1x0+b1)+b2=a1a2x0+a2b1+b2=α2x0+α2β2>α2κ+α2β2>κ,x3=a3x2+b3f(x0)=a3x2>a1(α2κ+α2β2)>κ,x4=a4x3+b4f(x1)=a4x3=a2a1(α2x0+α2β2)=α2(α2x0+α2β2)<α2(α2+α2β2)≤κ,x5=a5x4+b5f(x2)=a5α2(α2x0+α2β2)=a1α2(α2x0+α2β2)<α2(α2+α2β2)≤κ,x6=a6x5+b5f(x3)=a6x5=a2a1α2(α2x0+α2β2)=α22(α2x0+α2β2)<α22(α2+α2β2)≤κ,x7=a7x6+b7f(x4)=a7α22(α2x0+α2β2)+b7=α1α22(α2x0+α2β2)+b1=α1α22(α2x0+α2β2)+α1β1<α1α22(α2+α2β2)+α1β1≤κ,x8=a8x7+b8f(x5)=a8x7+b8=a8(α1α22(α2x0+α2β2)+α1β1)+b8=α23(α2x0+α2β2)+α2β2=α23(α2α24β2+α2β21−α24+α2β2)+α2β2=α24β2+α2β21−α24=x0,x9=a9x8+b9f(x6)=a9x8+b9=a9x0+b9=a1x0+b1=x1,x10=a10x9+b10f(x7)=a10x9+b10=a2x9+b2=a2x1+b2=x2,x11=a11x10+b11f(x8)=a11x10=a9x2=x3,⋮ so that {xn} is an eventually 8-periodic solution of the system (4.3).

Next, let an≡a and bn≡1−a in (1.7). We have αn1−αnβn=an1−an(ba+ba2+⋯+ban)=1,αωτ(αωl+αωβω(1−αωl)1−αω)=ap+1(αωl+1−αωl)=ap+1,αpαωβω(1−αωl)(1−αpαωl)(1−αω)=ap(1−ak−1)1−ap+k−1,αiαωτ+l(αωl+αωβω(1−αωl)1−αω)+αiβi=ap+k+i+1−ai≤ap+q+k+1−aq,i∈{0,…,q−1},αωτ+σ+l(1−αωl)+(1−αωσ)(1−αωτ+σ+2l)(1−αω)αωβω=1−aq+1(1−ak+p)1−ap+q+2k. Hence, I1(p)=(ap+1,ap(1−ak−1)(1−ak+p−1)),I2(p,q)⊃(1−aq+ap+q+k,1−aq+1(1−ak+p)1−a2k+p+q). Form the above, we can see that
Theorem A is just a special case of Theorem 3.1, hence Theorem 3.1 is an extension
of Theorem A.

Further, if k=1 in (1.7), then the intervals I1 and I2 in Theorem 3.2 are,
respectively, I1=I2=[a21+a+a2,a1+a+a2).

Corollary 4.1.

Let an≡a,bn≡1−a, and k=1. If κ∈(0,1)∩[a21+a+a2,a1+a+a2), then (1.7) has an eventually 3-periodic solution {xn}n=−k∞ (which can be generated explicitly).

As our final remark, note that under the conditions of
Theorems 3.1 or 3.2 if {xn'} is an arbitrary solution of (1.7) with x−k',…,x−2',x−1'∈(0,κ] such that x0'∈(κ,1), then in view of the proofs of Theorems 3.1 or
3.2, limj→∞g[j](x0')=x*=x0. This shows, by means of the
continuity properties of the maps gn and hn, that limn→∞|xn'−xn|=0. Note that the requirement x−k',…,x−2',x−1'∈(0,κ] with x0'∈(κ,1) is the same as requiring x−1'=1a0(x0'−b0)∈(1a0(κ−b0),1a0(1−b0))∩(0,κ]. In other words, let {xn'} be a solution determined by ϕ−k,…,ϕ−1∈(0,κ] such that ϕ−1∈(1a0(κ−b0),1a0(1−b0))∩(0,κ], then {xn'} will be “attracted” to the periodic
solution {xn} in the proofs of Theorems 3.1 or 3.2. We
remark that (1−b0)/a0>0. Thus, if 1a0(κ−b0)≤κ,then the above intersection is
nonempty. And, if κ−b0≤0,1a0(1−b0)>κ, then (1a0(κ−b0),1a0(1−b0))∩(0,κ]=(0,κ]. Since a0 and b0 can be chosen in arbitrary manners in Theorems
3.1 and 3.2, such additional conditions can easily be achieved once κ is determined.

We may illustrate the above discussions by the following
example. Let k=1 and an=1/2=bn for all n∈{0,1,2,…}. According to Corollary 4.1, if κ∈(0,1)∩[17,27)=[17,27),then the solution {xn} of (1.7) determined by x0=x* in (3.50), that is, x−1=1/7, is eventually 3-periodic. Furthermore, let {xn'} be the solution determined by x−1'=ϕ−1. If ϕ−1≤0, then by Lemma 2.1, limn→∞xn'=0. If ϕ−1∈(2κ−1,1)∩(0,κ]=(0,κ], then the solution {xn'} will satisfy limn→∞|xn'−xn|=0. If ϕ−1>κ, then by Lemma 2.2, a translate {yn} of {xn'} will satisfy limn→∞|yn−xn|=0.

Acknowledgment

Project supported by the National Natural Science
Foundation of China (10661011).

ChenY.All solutions of a class of difference equations are truncated periodicZhuH.HuangL.Asymptotic behavior of solutions for a class of delay difference equation