# The Existence of Multiple Periodic Solutions to a Class of Fourth-Order Difference Equations

Authors: Haoxin Wu; Zixian Tan; Xinfei Hu; Huafeng Xiao
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

In this paper, we study the multiplicity of periodic solutions for a class of fourth-order difference equations. We estimate a lower bound on the number of periodic solutions when the nonlinearity is asymptotic both at the origin and at infinity. Our results generalize the reference’ results.


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## Body

## 1. Introduction

On one hand, as we well know, many differential equations cannot be solved directly. In order to obtain approximate solutions iteratively by computer, we need to differentiate differential equations. For example, Gupta (cf. [1]) studied the bending of an elastic beam with simply-supported ends under an external force ex is described by the boundary-value problem(1)d4udx4=ex,0<x<1,u0=u1=u'′0=u'′1=0.If we discretize the first equation of (1), we can obtain the following fourth-order difference equation(2)△4xn−2=en.On the other hand, difference equations have been widely used as mathematical models to describe real-life situations in probability theory, matrix theory, electrical circuit analysis, combinatorial analysis, number theory, psychology and sociology, etc. For example, in 2015, Tariboon et al. (cf. [2]) introduced the quantum difference operators in orthogonal polynomials, basic hypergeometric functions, combinatorics, the calculus of variations, mechanics, and the theory of relativity. So it is worthwhile to explore this topic.Many scholars studied the qualitative properties of difference equations such as disconjugacy, stability, attractivity, oscillation, bifurcation, and boundary value problems (cf. [3, 4]). In 2003, Guo and Yu firstly built the variational structure for difference equations (cf. [5–7]). Then they converted the existence of periodic solutions of difference equations to that of critical points of the variational functional. Since then, the study of solutions to discrete dynamical systems has attached the attention of many researchers, for example, boundary value problems(cf. [8, 9]), periodic solutions, subharmonic solutions(cf. [5–7, 10–12]), positive solutions(cf. [9]), solutions with the minimal period(cf. [12, 13]), homoclinic orbits(cf. [14]), and heteroclinic orbits(cf. [15]).Focusing on fourth-order difference equations, the results are rare. To introduce the results, we give some notations. Letℕ,ℤ,ℝ be the set of all natural numbers, integers, and real numbers, respectively. For a,b∈ℤ, define Za=a,a+1,…,Za,b=a,a+1,…,b when a≤b. In 2005, Cai, Yu, and Guo (cf. [16]) studied the fourth-order difference equation(3)△2rn−2△2xn−2+fn,xn=0,n∈ℤ,where f∈Cℤ×ℝ,ℝ such that fn+T,z=fn,z for some positive integer T. When fn,⋅ was superlinear, using the linking theorem, some sufficient conditions were obtained for (3) to guarantee the existence and multiplicity of T-periodic solutions.In 2007, Zhao (cf. [17]) used the linking theorem to study the existence of solutions to the fourth-order difference equation(4)△2γn−2φp△2xn−2=fn,xn+2,xn+1,xn,xn−1,xn−2,n∈ℤ,xn∈ℝ.In his master’s thesis, he also gave some sufficient conditions to ensure the existence of homoclinic orbits of the following fourth-order difference equation(5)△4un−2−qnun+fn,un=0,n∈ℤ,un∈ℝ.In 2008, Sun and Zhou studied a high-dimensional fourth-order difference system(6)△2An−2△2xn−2−△Bn−1△xn−1+fn,xn=0,xn∈ℝN.Using the critical point theory, they (cf. [11, 18]) proved the existence and multiplicity of periodic solutions of (6) when f satisfies super-quadratic conditions, sub-quadratic conditions, and asymptotically quadratic conditions, respectively.In 2010, Cabada and Dimitrov studied the following fourth-order difference equation(7)uk+4+Muk=λgkfuk+ck,k∈0,1,2,…,T−1,uk∈ℝ,ui=uT+i,8=0,1,2,3.Using Krasnoselskii’s fixed point theorem, the authors (cf. [19]) obtained some sufficient conditions to guarantee the existence and nonexistence of periodic solutions.In 2011, Zhang studied the fourth-order difference equation with boundary value problems as follows(8)△2pn−2△xn−2=λfn,xn,n∈Z2,T,xn∈ℝ,x1=xT+1=0,△2x0=△2xT=0.Using the linking theorem, the author (cf. [20]) proved the existence and multiplicity of periodic solutions to (8) with super-quadratic and sub-quadratic nonlinearity, respectively.In 2020, Han and Ye (cf. [21]) studied the fourth-order nonlinear difference equation with periodic boundary conditions of the form(9)△4uk−2−akukα+bkukβ=0,uk∈ℝ,they applied the continuation theorem of Mawhin to ensure the existence of a nontrivial positive periodic solution. Also, they gave some sufficient conditions to ensure that (9) has no positive periodic solutions.There are a few conclusions about the estimation of the number of periodic solutions in the literature. In 2006, Guo and Yu (cf. [10]) studied the second-order difference equation(10)△2xn−1+fxn0,xn∈ℝ,when f is super-linear, using the ZP geometrical index theory, the authors proved that (10) exists at least p−1 distinct Zp-orbits of solutions with the minimal period p. They also gave an example to illustrate that p−1 is the best lower bound.In 2008, He and Chen investigated the multiplicity of periodic solutions for the following non-autonomous equation:(11)Δ2xn−1−xn+qnfn,xn=0,n∈ℤ,xn∈ℝ.Using the critical point theory and Clark theorem, they (cf. [22]) gave some lower bound of periodic solutions to (11).In 2013, Graef, Kong, and Wang studied the discrete fourth-order periodic boundary value problem(12)△4xn−2−△pn−1△xn−1+qnxn=fn,xn,n∈Z1,N,xn∈ℝ,△ix−1=△ixN−1,i=0,1,2,3.Using a variant of the Liusternik-Schnirelman theory, the authors (cf. [23]) obtained some sufficient conditions to guarantee the existence of multiple periodic solutions. Later, Dhar and Kong (cf. [24]) generalized the results to the following difference equation(13)△2rn−2△2xn−2−△pn−1△xn−1δ+qnxnσ=λfn,xn+μgn,xnn∈Z1,T,xn∈ℝ,△ix−1=△ixT−1,i=0,1,2,3.More results from this direction, we refer to [25].In this paper, we will study a class of fourth-order difference system when the nonlinearity is asymptotical both at the origin and at infinity. The rest of the paper is divided into three parts. In Section2, we will study a class of a low dimensional fourth-order difference equation and estimate the lower bound of the number of periodic solutions. In Section 3, we will study a class of a high dimensional fourth-order difference system and find the lower bound of the number of periodic solutions. In Section 4, we will give three examples to illustrate our results.


## 2. A Low-Dimensional Fourth-Order Difference Equation

Consider the nonlinear fourth-order difference equation(14)Δ4xn−2+fn,xn=0,n∈ℤ,xn∈ℝ,where △ is the forward difference operator defined by △xn=xn+1−xn, △2xn=△△xn.Assume that:(F1)f∈Cℤ×ℝ,ℝ, and fn,z is odd with respect to variable z.(F2) There existT∈ℕ, T≥4 such that, fn+T,z=fn,z, for all n,z∈ℤ×ℝ.(F3) There exist two functionsαi:ℤ1,T⟶ℝi=0,∞ such thatfn,z=α0nz+oz as z⟶0 for all n∈Z1,T;fn,z=α∞nz+oz as z⟶∞ for all n∈Z1,T.Defineα0=minα0n,n∈Z1,T,α∞=minα∞n,n∈Z1,T.LetE be the space of sequences x=⋯,x−n,…,x0,x1,…,xn,⋯=xnn=−∞+∞, i.e.(15)E=x|x=xnn=−∞∞,xn∈ℝ,n∈ℤ.For anyx,y∈E,a,b∈ℝ, define(16)ax+by≔axn+bynn=−∞+∞,then E is a vector space. Define the subspace ET of E as follows(17)ET=x∈E|xn+T=xn,∀n∈ℤ.Define the norm⋅ and inner product ⋅,⋅ on ET by(18)x2=∑n=1Txn2,x,y∑n=1Txn⋅yn,∀x,y∈ET.Define a linear mapψ:ET⟶ℝT by(19)ψx=x1,x2,…,xTτ,where τ denotes the transpose of a vector. Then ψ, defined in (19), is a linear homeomorphism with x=ψx and ET,⋅,⋅ is a Hilbert space, which can be identified with ℝT.The functionalJ on the Hilbert space ET is(20)Jx=12∑n=1TΔ2xn−12−∑n=1TFn,xn,where Fn,z=∫0zfn,sds.Arguing similar as reference [6], we can prove the following result.Lemma 1.
Assume thatF satisfies (F1)–(F3). Then J is continuously differentiable on ET. Critical points of J are periodic solutions to equation (14).
Due to the identification ofET with ℝT, we write x∈ET as x=x1,x2,…,xTτ. For convenience, Jx is written as(21)Jx=12xτAx−∑n=1TFn,xn,where(22)A=6−410…1−4−46−41…011−46−4…00⋮⋮⋮⋮…⋮⋮1000…6−4−4100…−46T×T.
In order to compute the eigenvalues ofA, we introduce matrix B as follows(23)B=010…00001…00⋮⋮⋮…⋮⋮000…01100…00T×T.
By direct computation, the eigenvalues ofB can be given by(24)λBk=e2kπi/T,k=0,1,2,…,T−1.
SinceA=−4B+B2+BT−2−4BT−1+6BT, then(25)λAk=−4λBk+λBk2+λBkT−2−4λBkT−1+6,=42sinα−sin2α2sinα+sin2α,=16sin2α1−cosα1+cosα=16sin4α,where α=kπ/T. Obviously, 0 is an eigenvalue of A with eigenvector η0=1/T1,1,…,1. Other eigenvalues of A are positive. When T is odd, for any k=1,2,…,T−1/2, λAk is the eigenvalue of A with multiplicity 2 and λAk=λAT−k and the maximal eigenvalue is 16cos4π/2T. When T is even, for any k=1,2,…,T/2−1, λk is the eigenvalue of A with multiplicity 2. Furthermore, λAk=λAT−k,λT/2=16 is the maximal eigenvalue of A and the eigenvector corresponding to 16 is ηe=1/T1,−1,1,…,−1. For any k,1≤k≤⌊T−1/2⌋, we denote(26)ηk1=1Tcos2kπT,cos2kπ⋅2T,…,cos2kπ⋅TT;,and(27)ηk2=1Tsin2kπT,sin2kπ⋅2T,…,sin2kπ⋅TT.
It is easy to see thatηk1 and ηk2 are the eigenvalues of A corresponding to λAk. Define λmin=minλAk|k=0,1,2,…,T−1 and λmax=maxλAk|k=0,1,2,…,T−1. By straightforward computation, one obtains(28)λmin=0,λmax=16cos4π2T,if T is odd,16,if T is even.
By (F3), integrating both side of the two equalities, we get(29)Fn,xn=12α0nxn2+oxn2,asxn⟶0,Fn,xn=12α∞nxn2+oxn2,asxn⟶∞.
Fori=0,∞, set(30)rin,xn=Fn,xn−12αinxn2,Rin,x=∑n=1Trin,xn.Proposition 1.
Rin,x=ox2,∇Rin,x=ox,i=0,∞.
This proposition is obvious.
Denote bydiaga1,a2,…,aT the T×T diagonal matrix with diagonal elements a1,a2,…,aT. Set Bi=A−diagαi1,αi2, ⋯,αiT, i.e.,(31)Bi=6−αi1−410…1−4−46−αi2−41…011−46−αi3−4…00⋮⋮⋮⋮…⋮⋮1000…6−αiT−1−4−4100…−46−αiTT×T,where i=0,∞. Then (21) can be rewritten as(32)Jx=12xτBix−Rin,x,i=0,∞.
To apply critical point theory to prove our main results, we shall state some basic notations and lemmas, which can be found in reference [26].
LetH be a real Hilbert space, f be a real-valued C1 function on H. The norm and inner product of H are given by ⋅H and ⋅,⋅H, respectively. We shall assume that f0=0 and that f satisfies the following Palais-Smale type condition (for short, (PS)-condition):(a)
If a sequenceyk⊂H is such that fyk<0, fyk is bounded below, and f′yk⟶0, then yk contains a convergent subsequence.For verification of Condition (A) in particular cases, it will be convenient to verify the following two conditions, which taken together are equivalent to the condition(A):(b)
For everyk<0 there exists R>0 and a>0 such that f′xH≥a for x such that k<fx<0 and xH>R;(c)
If a bounded sequenceyk⊂H is such that fyk<0,fyk is bounded below, and f′yk⟶0, then yk contains a convergent subsequence.
A functionq:H⟶ℝ is called a quadratic form if there exists a symmetric bilinear form β:H×H⟶ℝ such that qx=βx,x. A quadratic form q is continuous if and only if there is a constant K such that qx≤KxH2. In this case, there is a continuous symmetric linear operator L from H to the space dual to H such that qx=1/2Lx,xH and q is differentiable to all orders, the first and second derivatives being given by q′x=Lx, q'′x=L⋅,⋅H. We shall say that q is regular if LxH≥αxH for some α>0. The index iq is defined to be the maximal dimension of a subspace of H on which q is negative.Lemma 2.
LetHbe a real Hilbert space andJbe real valued,C1functional onH. Suppose further(1)
J0=0 and f is an even, i.e., J−x=Jx for all x∈H.(2)
J satisfies Condition (C).(3)
J is bounded below on bounded sets.(4)
Jx=q1x+oxH2, as x⟶0, where q1 is a quadratic form of index k.(5)
Jx=q2x+r2x, where q2 is a continuous regular quadratic form of finite index l<k and wherer2x=oxH2andr2′x=oxHasx⟶∞.Then for each integer m such that l<m≤k, there is at least one antipodal pair xm,−xm of critical points of J such that fxm=cmJ≤0.Lemma 3.
Assume that (F1)–(F3) hold. Then J satisfies Conditions (1)–(3) of Lemma 2.Proof.
Obviously,J0=0. Since f is odd in z, then J is even in z. Thus (1) of Lemma 2 holds.
Letyk⊂ET be a bounded sequence such that fyk<0,fyk is bounded below and f′yk⟶0. Since ET is finite-dimensional, then yk contains a convergence subsequence. Thus J satisfies Condition (C).
LetA⊂ET be a bounded set. Since the minimal eigenvalue of A is 0, then 1/2xτAx≥0. Since f is continuous, so is F, which implies that Fn,z is bounded for n,z∈ℤ1,T×A. Subsequently, J is bounded below on bounded sets. Thus (1) of Lemma 2 holds.
Now let’s state and prove the main results.Theorem 1.
Suppose thatf satisfies (F1)–(F3). If there exist 1≤l≤k≤⌊T/2⌋ such that λAl−1<α∞<λAl≤λAk−1<α0<λAk, then (14) possesses at least 2k−l pairs of nonzero distinct T-periodic solutions.Proof.
According to Lemmas1 and 2, we need to check conditions (4)-(5) of Lemma 2.
By Proposition1 and (32), Jx=1/2xτB0x−R0n,x and R0n,x=ox2 as x⟶0. If λAk−1<α0<λAk, then there are 2k−1 eigenvalues of A smaller than α0, which are λA0,λA1,…,λAk−1, λAT−k+1,…,λAT. Define by E2k−1 the negative eigenspace of B0. Then dimE2k−1=2k−1. Hence the index of B0 is 2k−1 when x⟶0.
By Proposition1 and (32), Jx=1/2xτB∞x−R∞n,x and R∞n,x=ox2,∇R∞n,x=ox as x⟶∞. If λAl−1<α∞<λAl, then there are 2l−1 eigenvalues of A smaller than α∞, which are λA0,λA1,…, λAl−1, λAT−l+1,…,λAT. Define by E2l−1 the negative eigenspace of B∞. Then dimE2l−1=2l−1. Hence index of B∞ is 2l−1 when x⟶∞. Using Lemma 2, J has 2k−l pairs difference nonzero critical points. Thus (14) has 2k−l pairs of nonzero distinct T-periodic solutions.Corollary 1.
Suppose thatf satisfies (F1)–(F3). If there exist 1≤l<⌊T/2⌋ such that λAl−1<α∞<λAl≤λmax<α0, then (14) possesses at least T−2l+1 pairs of nonzero distinct T-periodic solutions.Proof.
Whenα0>λmax, the negative space of J is ET as x⟶0. Then the index k of B0 is T as x⟶0. The index of B∞ is 2l−1 when x⟶∞. Applying Theorem 1, (14) has T−2l+1 pairs of nonzero distinct T-periodic solutions.Corollary 2.
Suppose thatf satisfies (F1)–(F3). If α∞<0≤λAk−1<α0<λAk, then (14) possesses at least 2k−1 non-zero distinct solutions.Proof.
Ifα∞<0, then Jx≥0 for all x∈ET. Thus index l of B∞ is 0 when x⟶∞. The index of B0 is 2k−1 when x⟶0. Applying Theorem 1, (14) possesses 2k−1 pairs nonzero distinct T-periodic solutions.Corollary 3.
Suppose thatf satisfies (F1)–(F3). If α∞<0<λmax<α0, then (14) possesses at least T pairs of nonzero distinct T-periodic solutions.Proof.
Whenα0>λmax, the negative space of J is ET as x⟶0. Then the index k of A0 is T. When α∞<0, then Jx≥0 for all x∈ET. Thus index l of A∞ is 0 when x⟶0. Applying Theorem 1, (14) possesses T pairs of nonzero distinct T-periodic solutions.


## 3. A High-Dimensional Fourth-Order Difference System

In this section, we consider the nonlinear fourth-order difference system(33)Δ4xn−2+gn,xn=0,n∈ℤ,xn∈ℝm,m∈ℕ.Assume that:(G1)g∈Cℤ×ℝm,ℝm, and gn,z is odd with respect to variable z.(G2) There existG∈C1ℤ×ℝm,ℝ and T≥2 such that ∇zGn,z=gn,z and Gn+T,z=Gn,z, for all n,z∈ℤ×ℝm.(G3) There exist two matrix functionsAi:Z1,T⟶GLm,ℝi=0,∞ such thatgn,z=A0nz+oz as z⟶0 for all n∈Z1,T;gn,z=A∞nz+oz as z⟶∞ for all n∈Z1,T,where GLm,ℝ denotes the general linear group of m×m real matrices.LetH be the space of vector sequences x=⋯,x−n,…,x0,x1,…,xn,⋯=xnn=−∞+∞, i.e.(34)H=x|x=xnn=−∞∞,xn=xn1,xn2,…,xnmτ∈ℝm,n∈ℤ..For anyx,y∈H,a,b∈ℝ, define(35)ax+by≔axn+bynn=−∞∞,then H is a vector space. Define the subspace HT of H as follows(36)HT=x∈H|xn+T=xn,∀n∈ℤ.Define the norm⋅ and inner product ⋅,⋅ on HT given by(37)x2=∑n=1Txn2,x,y=∑n=1Txn,yn,∀x,y∈HT,where ⋅ and ⋅,⋅ denote the usual norm and inner product in ℝm. Define a linear map ψ:HT⟶ℝmT by(38)ψx=x1τ,x2τ,…,xTττ,then ψ, defined in (38), is a linear homeomorphism with x=ψx and HT,⋅,⋅ is a Hilbert space, which can be identified with ℝmT.The functionalJ, defined on the Hilbert space HT, is(39)Jx=12∑n=1TΔ2xn−12−∑n=1TGn,xn.Arguing similar as reference [6], we can prove the following result.Lemma 4.
Assume thatG satisfies (G1)–(G3). Then J is continuously differentiable on HT. Critical points of J are periodic solutions to (33).
Due to the identification ofHT with ℝmT, we write x∈HT as x=x1τ,x2τ,…,xTττ. For convenience, Jx is written as(40)Jx=12xτDx−∑n=1TGn,xn,where(41)D=6I−4II0…I−4I−4I6I−4II…0II−4I6I−4I…00⋮⋮⋮⋮…⋮⋮I000…6I−4I−4II00…−4I6ImT×mT,where I is the m×m identity matrix. By (G2) and (G3), we have(42)Gn,xn=12xnτA0nxn+oxn2,asxn⟶0,(43)Gn,xn=12xnτA∞nxn+oxn2,asxn⟶∞.
Fori=0,∞, set(44)rin,xn=Gn,xn−12xnτAinxn,Rin,x=∑n=1Trin,xn.Proposition 2.
Rin,x=ox2,i=0,∞.Proof.
Ifx⟶0, then xn⟶0 for all n∈Z1,T. It follows from (42) and (44) that r0n,xn=oxn2. For any ϵ>0, there exists δ>0 such that r0n,xn≤ϵxn2 for xn≤δ. If x≤δ, then xn≤δ for all n∈Z1,T. Thus(45)R0n,x=∑n=1Tr0n,xn≤∑n=1Tεxn2=εx2.
The arbitrary ofϵ implies that R0n,x=ox2. The other conclusion can be proved similarly.
Fori=0,∞, define Di=D−D¯i, where the matrices D¯i is given as follows(46)D¯i=Ai100…00Ai20…000Ai3…0⋮⋮⋮…⋮000…AiTmT×mT,
(40) can be rewritten as(47)Jx=12xτDix−Rin,x,i=0,∞.
HT can be decomposed as follows(48)HT=Hi+⊕Hi0⊕Hi−,i=0,∞,where Hi+ (respectively Hi−, Hi0) denote the subspaces of HT on which Di is positive definite (respectively negative definite, null), i=0,∞. Set(49)ni=dimHi0,i=0,∞,where dimA denotes the dimension of the space A.Definition 1.
A indexI of D1 and D2 is defined as follows(50)ID1,D2=dimM+D−D1∩M−D−D2−codimM+D−D1+M−D−D2,where M+D−Di (respectively M−D−Di,M0D−Di denotes the subspace of HT on which D−Di is positive definite (respectively negative definite, null).
In order to apply critical point theory to prove our main results, we need the following lemmas.Lemma 5.
[27] Let c0<c∞<0 be two constants. Let H be a real Hilbert space and f∈C1H,ℝ be a functional which satisfies the following assumption:(1)
fx=1/2Lx,x+Φx, where L is a bounded selfadjoint operator. Φ∈C1H,ℝ is a functional whose Frechet derivative Φ′∈CH,H is a compact operator.(2)
Φ−x=Φx.(3)
Any sequencexk⊂H such that fxk⟶c∈c0,c∞ and f′xk⟶0 as n⟶+∞ has a convergent subsequence.(4)
There are two closed subspaces ofH, H+ and H−, and a constant ρ>0 such thatThen the number of pairs of nontrivial critical points of f is greater or equal to dimH+∩H−−codimH++H−. Moreover, the corresponding critical values belong to c0,c∞.
(a)
fx>c0 for x∈H+,(b)
fx<c∞<0 for x∈H−∩SρSρ=x∈H|x=ρ.Lemma 6.
Assume that (G1)–(G3) hold. Then J satisfies conditions (1), (2) and (4) of Lemma 5.Proof.
By Lemma4, J∈C1HT,ℝ and Φ∈C1. Observe that Lx,x=xτDx and Φx=∑n=1TGn,xn. Since Φ′x=g1,x1τ,…,gn,xnττ, which is a continuous function defined on a finite-dimensional space, thus Φ′x is compact. Since D is a symmetry matrix, thus L is a bounded self-adjoint operator. Since g is odd in z, then Φ is even in z. Hence (1) and (2) of Lemma 5 hold.
SetH+=H∞+, H−=H0−. There exist positive constants α,β such that(51)D0x,x≤−αx2,∀x∈H−,D∞x,x≥βx2,∀x∈H+.
It follows that, for anyx∈H−,(52)Jx≤−α2x2+ox2,asx⟶0.
Then there exist positive constantsρ and γ such that(53)Jx<−α4ρ2<0,∀x∈H−∩Sρ.
Settingc∞=−α/4ρ2, (4) (b) is satisfied.
By (G3), there exists R0>0 such that(54)R∞n,x≤β4x2,∀x>R0.
SinceR∞n,x is continuous with respect to n,x, R∞n,x is bounded for n∈Z1,T,x≤R0, whose boundedness is denoted by M. Then M is finite. Thus(55)R∞n,x≤β4x2+M,then for every x∈H+,(56)Jx≥β4x2−M,thus J is bounded from below on H+. Setting c0=infx∈H+Jx−ω with ω>0 such that c0<c∞, then (4) (a) is satisfied. □
Now let’s state and prove the main results.Theorem 2.
Suppose thatf satisfies (G1)-(G3). If n0=n∞=0, then (3.1) possesses at least ID¯0,D¯∞ pairs of nonzero distinct solutions.Proof.
We need to check Conditions (3) of Lemma5. Let xk be a sequence such that J′xk⟶0, Jxk⟶c, where c∈ℝ such that c<0. Now we show that xk is bounded. Suppose to the opposite, we can choose xk⟶+∞ as k⟶+∞. Clearly, xk can be written as xk=xk++xk−∈H∞+⊕H∞−. On one hand, since(57)0≤limsupk⟶∞J′xk,xk+−xk−xk2≤limsupk⟶∞J′xkxkxk2=0,then(58)limsupk⟶∞J′xk,xk+−xk−xk2=0.
On the other hand,(59)J′xk,xk+−xk−=xτDx−∇R∞xk,xk+−xk−.
Obversing thatlimk⟶∞∇R∞xk,xk+−xk−/xk2=0, then(60)J′xk,xk+−xk−≥minα′,β′−o1x2,where α′,β′ satisfies xτDx≥α′x2 for x∈H∞+ and xτDx≤−β′x2 for x∈H∞−. Thus(61)liminfk⟶∞J′xk,xk+−xk−xk2≥minα′,β′>0,which contradicts with (58).
Applying Lemma5, (33) possesses at least dimH+∩H−−codimH++H−=ID¯∞,D¯0 pairs of nonzero solutions if ID¯∞,D¯0>0.
IfID¯∞,D¯0<0, then ID¯0,D¯∞>0. In this case, we replace J by −J and let H+=H0− and H−=H∞+. It is easy to check that Conditions (1)–(4) of Lemma 5 are satisfied. Applying Lemma 5, (33) possesses at least ID¯0,D¯∞ pairs of nonzero T-periodic solutions.
In conclusion, (33) has at least ID¯0,D¯∞ pairs of nonzero solutions.Theorem 3.
Suppose thatg satisfies (G1)–(G3). Assume more,(G4)∇R∞n,z is bounded.G5±R∞n,x⟶±∞ as x⟶+∞, uniformly for n∈Z1,T.then (33) possesses at least ID¯∞,D¯0 pairs of nonzero distinct T-periodic solutions.Proof.
Letxk be a sequence such that J′xk⟶0, Jxk⟶c, where c<0. Now we show that xk is bounded. Suppose to the contrary, we can choose xk⟶+∞ as k⟶+∞. Clearly, xk=xk++xk0+xk−∈H∞+⊕H∞0⊕H∞−. Since J′xk⟶0, we get(62)J′xk,xk+=xkτD∞xk++∇R∞n,xnk,xnk+|≤xk+.
By (G4), there exists c1>0 such that ∇R∞n,xnk≤c1. Then the above inequality implies that(63)βxk+2≤xkτD∞xk+≤xk++c1xk+,thus xk+ is bounded. Arguing similarly, we prove xk− is bounded. Thus, there exists c2>0 such that xτD∞x≤c2. Subsequently, we have(64)c−o1≤Jxk=12xkτD∞xk−R∞n,xk≤c2−R∞n,xk.
ThereforeR∞n,xk is bounded from above. G5+ implies that xk is bounded. If G5− holds, we can also prove that xk is bounded by modifying (3.11). Thus, xk is bounded and it contains a convergent subsequence.
Applying Lemma5, (33) possesses at least ID¯∞,D¯0 pairs of nonzero T-periodic solutions.


## 4. Examples

In this section, we will give three examples to illustrate our results. Consider the following difference equation(65)Δ2xn−1+fn,xn=0,where(66)fn,z=14z−sin2nπ15z3,whenz<1,1087−107z−67+17zsin2nπ15,when1≤z≤8,12z−sin2nπ15z1/3,whenz>8.It is easy to check thatf satisfies (F1)–(F3). Applying Theorem 1, (65) has 4-pairs nonzero distinct 15-periodic solutions since 1/2<16sin43π/15=16sin412π/15<16sin44π/15=16sin411π/15<16sin45π/15=16sin410π/15<16sin46π/15=16sin49π/15<14.Next, we will consider the following fourth-order difference system(67)Δ4xn−2+gn,xn=0,n∈ℤ,xn∈ℝ3,where g is continuous and gn+4,z=gn,z for all n∈ℤ and z∈ℝ. Let ρ1≤1 and ρ2 large enough,(68)gx=A0x+x12,x22,x32τ,x∈Bρ1,A∞x+sinx1,sinx2,sinx3τ,=x∈ℝ3∖Bρ2..Set(69)A0=13−1021000019,A∞=1300−20003.Compute the eigenvalues ofD0 and D∞. The space E8 is split into(70)E8=E0+⊕E0−=E∞+⊕E∞−,where E0+, E0− and E∞+, E∞− are denoted by positive, negative eigenspaces of D0 and D∞ respectively. Computing directly, we obtain(71)ID∞,D0=dimM+D−D∞∩M−D−D0−dimM−D−D∞∩M+D−D0=14.Applying Theorem2, (67) possesses 14 pairs of 8-periodic solutions.Consider fourth-order difference system (67), where g is give below(72)gx=A0x+x13,x23,x33,x∈Bρ1,A∞x+x1−1/3,x2−1/3,x3−1/3τ,x∉Bρ2..Set(73)A0=13−1021000019,A∞=400040004.Compute the eigenvalue ofD0 and D∞. The space E8 is split into(74)E8=E0+⊕E00⊕E0−=E∞+⊕E∞0⊕E∞−,where E0+, E0−, E00 and E∞+, E∞−, E∞0 are denoted by positive, negative eigenspaces, and null space of D0 and D∞, respectively. Computing directly, we obtain(75)ID∞,D0=dimM+D−D∞∩M−D−D0,−dimM−D−D∞⊕M0D−D∞∩M+D−D0⊕M0D−D0=5.Applying Theorem 3, (67) possesses 5 pairs of 8-periodic solutions.



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