# Existence of Periodic Solutions for a Class of Fourth-Order Difference Equation

Authors: Jia Wei; Xiaoling Han; Fumei Ye
Journal: Journal of Function Spaces
 (2022)

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## Abstract

We apply the continuation theorem of Mawhin to ensure that a fourth-order nonlinear difference equation of the formΔ4uk−2−akuαk+bkuβk=0 with periodic boundary conditions possesses at least one nontrivial positive solution, where Δuk=uk+1−uk is the forward difference operator, α>0,β>0 and α≠β, ak,bk are T-periodic functions and akbk>0. As applications, we give some examples to illustrate the application of these theorems.


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## Body

## 1. Introduction

In recent years, the theory of nonlinear difference equations has been widely used in the study of discrete models in the fields of economics, neural networks, ecology, etc. For the general background of difference equations, in particular, there are many authors who have discussed the existence and multiplicity of periodic solutions for discrete boundary value problems by exploiting various methods, including the method of upper and lower solutions, Leray-Schauder degree, fixed point theory, critical theory, and variational methods; see Bereanu and Mawhin [1], Cabada and Dimitrov [2], Graef et al. [3, 4], and Cai et al. [5–10] and the references therein.Letℕ+, Z, and ℝ denote the sets of all positive integers, integers, and real numbers, respectively. This paper considers the following fourth-order nonlinear difference equation:
(1)Δ4uk−2−akuαk+bkuβk=0,where α>0,β>0 and α≠β, ak,bk are T-periodic functions and Δuk=uk+1−uk is the forward difference operator.Equation (1) can be considered as a discrete analogue of a special case of the following fourth-order nonlinear differential equation:
(2)u′′′′−atuα+btuβ=0,t∈ℝ,which has been studied in [11, 12] when α=1,β=3. In [13], Yang and Han proved the existence of a periodic solution to equation (2) when α=n,β=n+2, where n is a positive integer.Whenak≡0,β=1, Peterson and Ridenhour [14] considered the disconjugacy of the following equation:
(3)Δ4uk−2+bkuk=0,k∈Z.In 2005, Cai et al. [5] studied the fourth-order nonlinear difference equation
(4)Δ4uk−2+fk,uk=0,k∈Z.By applying the linking theorem, they obtained some criteria for the existence and multiplicity of periodic solutions of equation (4).In fact, there is a big difference between the continuous case and the discrete case. For example, the basic ideas of calculus are not always applicable when studying difference equations, such as the embedding theorem. Therefore, we need to consider other methods to deal with the difference problem. The main tool used is the continuation theorem of Mawhin (see [15]).Motivated by the above works, the main aim of this paper is to investigate the existence of at least one positiveT-periodic solution of (1). In order to obtain the main results of (1), we assume that the coefficient functions ak and bk satisfy the following condition:F1: Suppose ak,bk are T−periodic functions and akbk>0 for all k∈Z. Furthermore, we assume that there exist positive constants a,A,b,B such that
(5)a=mink∈Zak,A=maxk∈Zak,b=mink∈Zbk,B=maxk∈Zbk.LetX be all real T-periodic sequences of the form u=ukk∈Z. Then, X is a Banach space under the norm u=maxk∈2,T+1Zuk.The main results in this paper are stated next: Theorems1 and 2.Theorem 1.
Let F1 hold, if α<β and the period T satisfies
(6)16≤T4≤8AR1α−1+BR1β−1,where R1=A/b1/β−α+ρ and ρ>0 small enough such that a/B1/β−α−ρ>0; then, equation (1) admits at least one positive T-periodic solution.Theorem 2.
Let F1 hold, if α>β and the period T satisfies
(7)16≤T4≤8AQ1α−1+BQ1β−1,where Q1=B/a1/α−β+τ and τ>0 small enough such that b/A1/α−β−τ>0; then, equation (1) admits at least one positive T-periodic solution.Theorem 3.
Supposeakbk≤0 and ak,bk are not identical to zero for all k∈Z; then, equation (1) has no positive solution.This paper is organized as follows: in Section2, we give some lemmas needed to prove the main results. Section 3 contains the proof of Theorem 1. Section 4 contains the proof of Theorem 2. Section 5 contains the proof of Theorem 3.


## 2. Preliminary Results

In this section, we introduce some notations and well-known results which will be used in the subsequent sections.Definition 4 (see [7], p. 12, B.1).
LetX,Y be real Banach spaces, L:DomL⊂X⟶Y be a linear mapping. The mapping L is said to be a Fredholm mapping of index zero if
(a)
ImL is closed in Y(b)
dimKerL=codimImL<+∞
IfL is a Fredholm mapping of index zero, then there exist continuous projectors P:X⟶X and Q:Y⟶Y such that
(8)ImP=KerL,KerQ=ImL=ImI−Q.
It follows that the restriction(9)LDomL∩KerP:I−PX⟶ImL
has an inverse which is denoted byKP.Definition 5 (see [7], pp. 12–13, B.1, B.2).
LetN:X⟶Y be a continuous mapping. If Ω is a bounded open subset of X, N is called L−compact on Ω¯ if QNΩ¯ is bounded and KPI−QN:Ω¯⟶X is compact. Since ImQ is isomorphic to KerL, there exists an isomorphism J:ImQ⟶KerL.Lemma 6 (Mawhin’s continuation theorem, see [7], Theorem IV.1).
LetL be a Fredholm mapping of index zero, Ω⊂X is an open bounded set, and let N be L−compact on Ω¯. Suppose
(1)
for eachλ∈0,1, Lu≠λNu for all u∈∂Ω∩DomL(2)
QNu≠0, for all u∈∂Ω∩KerL(3)
degJQN,Ω∩KerL,0≠0, where J:ImQ⟶KerL is an isomorphism
Then, the equationLu=Nu has at least one solution in Ω¯∩DomL.Define the operatorL:X⟶X by setting
(10)Lu=Δ4uk−2,u∈X.Direct calculation shows that(11)KerL=ℝandImL=v∑i=2T+1vi=0.Indeed, ifv=vkk∈Z∈ImL, then there is u=ukk∈Z∈X such that
(12)vk=Δ4uk−2,k∈Z.Thus,(13)∑i=2T+1vi=∑i=2T+1Δ4ui−2.Note thatukk∈Z∈X, it follows that
(14)Δ3ukk∈Z∈X.Furthermore, direct calculation shows that(15)∑i=2T+1Δ4ui−2=Δ3u1−Δ3u0+Δ3u2−Δ3u1+⋯+Δ3uT−Δ3uT−1=Δ3uT−Δ3u0=Δ3u0−Δ3u0=0.By virtue of the above facts, we have(16)∑i=2T+1vi=∑i=2T+1Δ4ui−2=0.SincedimX=T and L is a linear mapping, by the knowledge of linear algebra, we know that dimKerL⊕dimImL=dimX. It is easy to see that dimKerL=codimImL=1 and dimImL=T−1. It follows that ImL is closed in X. Therefore, the operator L is a Fredholm operator with index zero.Let us defineN:X⟶X by
(17)Nuk=akuαk−bkuβk.We defineP:X⟶KerL and Q:X⟶X as follows:
(18)Puk=Quk=1T∑i=2T+1ui.The operatorsP and Q are projections. Hence,
(19)ImP=KerL,KerQ=ImL.It follows thatLDomL∩KerP:I−PX⟶ImL has an inverse which is denoted by KP.In view of (17) and (18), for any u∈X, we can see that
(20)QNuk=1T∑i=2T+1aiuαi−biuβi,(21)I−QNuk=akuαk−bkuβk−1T∑i=2T+1aiuαi−biuβi.Since the Banach spaceX is finite dimensional, KP is linear. By virtue of the relations (20) and (21), we see that QN and KPI−QN are continuous on X. Hence, we know that if Ω is an open and bounded subset of X, then QNΩ¯ is bounded. It follows that
(22)KPI−QN:Ω¯⟶Xis compact. Therefore, the mappingN is L-compact on Ω¯ with any open and bounded subset Ω⊂X.Lemma 7 (see [16], Lemma 2.3).
Letukk∈Z be a real T-periodic sequence; then,
(23)max2≤i,j≤T+1ui−uj≤T316∑k=2T+1Δ4uk−2.


## 3. Proof of Theorem1

Proof.
The content of Theorem1 is as follows: let F1 hold, if α<β and the period T satisfies
(24)16≤T4≤8AR1α−1+BR1β−1,where R1=A/b1/β−α+ρ and ρ>0 small enough such that a/B1/β−α−ρ>0; then, equation (1) admits at least one positive T-periodic solution. Now, we prove that the conclusion holds. We assume that α<β. From condition F1, we know that akbk>0, which include both positive and negative cases. So we need to classify the cases where both ak and bk are positive and both negative.Case 1.
If coefficient functionsak and bk are positive T-periodic functions, in view of F1, we have that 0<a≤ak≤A and 0<b≤bk≤B.
Let(25)Ω1=u∈XH1<uk<R1,which is an open set in X, where
(26)R1=R+ρ,R=Ab1/β−α,(27)H1=H−ρ,H=aB1/β−α,where ρ>0 small enough such that a/B1/β−α−ρ>0. Obviously, H1 and R1 are well defined.
By(28)α<β,0<a≤ak≤A,0<b≤bk≤B,
we obtain(29)0<H1<H≤akbk1/β−α≤R<R1,
uniformly fork∈Z.
Furthermore, it follows from (26), (27), and (29) that
(30)ak−bkH1β−α≥a−BH1β−α>0,ak−bkR1β−α≤A−bR1β−α<0.
Therefore,(31)ak−bkH1β−α>0,ak−bkR1β−α<0.
We prove that condition (1) of Lemma 6 holds. Let 0<λ<1 and u be such that
(32)Δ4uk−2−λakuαk+λbkuβk=0.
Summing from2 to T+1, we can see that
(33)∑i=2T+1Δ4ui−2−λaiuαi+λbiuβi=0.
Firstly, we claim that for eachλ∈0,1 and u∈∂Ω1∩DomL, Lu≠λNu. In fact, in view of (25), if u∈∂Ω1, then u=H1 or u=R1. We only prove the case of u=H1, similar to the proof of u=R1.
Whenu=H1, we can get that max2≤l,j≤T+1ul−uj<H1/2 or max2≤l,j≤T+1ul−uj≥H1/2 holds. If max2≤l,j≤T+1ul−uj<H1/2, thus H1/2<u≤H1. According to the above facts, we have
(34)0=∑i=2T+1λaiuαi−λbiuβi=λ∑i=2T+1uαiai−biuβ−αi≥λa−BH1β−α∑i=2T+1uαi>0.
This is a contradiction.
Ifmax2≤l,j≤T+1ul−uj≥H1/2, we see from Lemma 7 that
(35)0=∑i=2T+1Δ4ui−2−λaiuαi+λbiuβi≥∑i=2T+1Δ4ui−2−∑i=2T+1λaiuαi+λbiuβi>∑i=2T+1Δ4ui−2−∑i=2T+1aiuαi+biuβi≥16T3max2≤l,j≤T+1ul−uj−TAH1α−1+BH1β−1u≥8T3u−TAH1α−1+BH1β−1u=8T3−TAH1α−1+BH1β−1u≥0.
This is a contradiction.
Whenu=R1, we can get that max2≤l,j≤T+1ul−uj<R1/2 or max2≤l,j≤T+1ul−uj≥R1/2 holds. If max2≤l,j≤T+1ul−uj<R1/2, thus R1/2<u≤R1; further, we have
(36)0=∑i=2T+1λaiuαi−λbiuβi=λ∑i=2T+1uαiai−biuβ−αi≤λA−bR1β−α∑i=2T+1uαi<0.
This is a contradiction.
Ifmax2≤l,j≤T+1ul−uj≥R1/2, from Lemma 7, we have that
(37)0=∑i=2T+1Δ4ui−2−λaiuαi+λbiuβi≥∑i=2T+1Δ4ui−2−∑i=2T+1λaiuαi+λbiuβi>∑i=2T+1Δ4ui−2−∑i=2T+1aiuαi+biuβi≥16T3max2≤l,j≤T+1ul−uj−TAR1α−1+BR1β−1u≥8T3u−TAR1α−1+BR1β−1u=8T3−TAR1α−1+BR1β−1u≥0,
but this is a contradiction. Hence, foru∈∂Ω1 and λ∈0,1, we have
(38)Δ4uk−2−λakuαk+λbkuβk≠0.
Therefore, we verify that condition (1) of Lemma 6 holds for Ω1.
Ifu∈∂Ω1∩KerL, then u=H1k∈Z or u=R1k∈Z. By virtue of (31), we conclude that
(39)QNuk=1T∑i=2T+1uαiai−biuβ−αi≠0.
Hence,QNu≠0 for each u∈∂Ω1∩KerL.
Next let us considerH1+R1/2, the arithmetic mean of H1 and R1. We define G:X×ℝ⟶X as follows, for all μ∈0,1,
(40)Gu,μ=−1−μu−H1+R12+μ1T∑i=2T+1uαiai−biuβ−αi.
Clearly, we find that(41)Gu,μ≠0,∀u∈∂Ω1∩KerL.
By using the homotopy invariance theorem, it is easy to see that(42)degQN,Ω1∩KerL,0=degGu,1,Ω1∩KerL,0=degGu,0,Ω1∩KerL,0=−1≠0.
Therefore, conditions (1)–(3) of Lemma 6 hold for Ω1.
Furthermore, according to the above reasoning, we deduce that (1) has at least one positive solution in Ω¯1.Case 2.
If the coefficient functionsak,bk are negative T-periodic functions, in view of F1, we have that −A≤ak≤−a<0 and −B≤bk≤−b<0.
Leta~k=−ak,b~k=−bk; then, we see that
(43)0<a≤a~k≤A,0<b≤b~k≤B.
It is obvious that (1) is equivalent to the equation
(44)Δ4uk−2+a~kuαk−b~kuβk=0.
Let0<λ<1 and u be such that
(45)Δ4uk−2+λa~kuαk−λb~kuβk=0.
Summing from2 to T+1, we can see that
(46)∑i=2T+1Δ4ui−2+λa~iuαi−λb~iuβi=0.
Firstly, we claim that for eachλ∈0,1 and u∈∂Ω1∩DomL, Lu≠λNu. In fact, in view of (25), if u∈∂Ω1, then u=H1 or u=R1. We only prove the case of u=H1, similar to the proof of u=R1.
Whenu=H1, we can get that
(47)max2≤l,j≤T+1ul−uj<H12or
(48)max2≤l,j≤T+1ul−uj≥H12holds.
Ifmax2≤l,j≤T+1ul−uj<H1/2, thus
(49)H12<u≤H1.
Further, we have(50)0=∑i=2T+1λa~iuαi−λb~iuβi=λ∑i=2T+1uαia~i−b~iuβ−αi≥λa−BH1β−α∑i=2T+1uαi>0.
This is a contradiction.
Ifmax2≤l,j≤T+1ul−uj≥H1/2, we see from Lemma 7 that
(51)0=∑i=2T+1Δ4ui−2−λa~iuαi+λb~iuβi≥∑i=2T+1Δ4ui−2−∑i=2T+1λa~iuαi+λb~iuβi>∑i=2T+1Δ4ui−2−∑i=2T+1a~iuαi+b~iuβi≥16T3max2≤l,j≤T+1ul−uj−TAH1α−1+BH1β−1u≥8T3u−TAH1α−1+BH1β−1u=8T3−TAH1α−1+BH1β−1u≥0,
but this is a contradiction. Hence, for allu∈∂Ω1 and λ∈0,1, we have
(52)Δ4uk−2+λa~kuαk−λb~kuβk≠0.
Therefore, we verify that condition (1) of Lemma 6 holds for Ω1.
The remaining proof is similar to the proof of Case1, and so we omit it. Furthermore, according to the above reasoning, we deduce that (44) has at least one positive solution in Ω¯1.


## 4. Proof of Theorem2

Proof.
The content of Theorem2 is as follows: let F1 hold, if α>β and the period T satisfies
(53)16≤T4≤8AQ1α−1+BQ1β−1,where Q1=B/a1/α−β+τ and τ>0 small enough such that b/A1/α−β−τ>0; then, equation (1) admits at least one positive T-periodic solution.Now, we prove that the conclusion holds. Similarly, in the case of α>β, we need to discuss the case where the coefficient functions ak and bk are both positive and negative, respectively.Case 1.
If coefficient functionsak and bk are positive T-periodic functions, we have that 0<a≤ak≤A and 0<b≤bk≤B.
Let(54)Ω2=u∈XP1<uk<Q1,which is an open set in X, where
(55)Q1=Q+τ,Q=Ba1/α−β,(56)P1=P−τ,P=bA1/α−β,
whereτ>0 small enough such that b/A1/α−β−τ>0. Obviously, P1 and Q1 are well defined.
Byα>β, 0<a≤ak≤A, and 0<b≤bk≤B, we obtain
(57)0<P1<P≤akbk1/α−β≤Q<Q1,
uniformly fork∈Z.
By virtue of (55) and (56), we obtain
(58)akP1α−β−bk≤AP1α−β−b<0,akQ1α−β−bk≥aQ1α−β−B>0.
Therefore,(59)akP1α−β−bk<0,akQ1α−β−bk>0,
uniformly fork∈Z.
The remaining proof is similar to the proof of Theorem1, and so we omit it. Furthermore, we deduce that (1) has at least one positive T-periodic solution in Ω¯2.Case 2.
If the coefficient functionsak,bk are negative T-periodic functions, we have that −A≤ak≤−a<0 and −B≤bk≤−b<0.
Leta~k=−ak,b~k=−bk. Then, we can see that
(60)0<a≤a~k≤A,0<b≤b~k≤B.
It is obvious that(61)−a~kP1α−β+b~k>0,−a~kQ1α−β+b~k<0,
uniformly fork∈Z.
The remaining proof is similar to the proof of Theorem1, and so we omit it. Furthermore, we conclude that (1) has at least one positive T-periodic solution in Ω¯2.


## 5. Proof of Theorem3

Proof.
The content of Theorem3 is as follows: suppose akbk≤0 and ak,bk are not identical to zero for all k∈Z; then, equation (1) has no positive solution.
Summing equation (1) from 2 to T+1, we obtain that
(62)∑i=2T+1Δ4ui−2−aiuαi+biuβi=0.
In view of(63)DomL=uu∈X,Δuk+T=Δuk,Δ2uk+T=Δ2uk,Δ3uk+T=Δ3uk.
Hence,(64)∑i=2T+1−aiuαi+biuβi=0.
Ifak>0 and bk≤0, it follows from (64) that (1) does not have any positive solution. Other cases are similar.


## 6. Example

Example 1.
The difference equation(65)Δ4uk−2−1100sin2kπT+150uk+1200cos2kπT+1100u3k=0,
is one of the form (1), where a=1/100, A=3/100, b=1/100, B=3/200, α=1, and β=3. Let ρ=6/6, we obtain
(66)16<T4≤320031+62.
Therefore, we can prove that (65) has at least one positive T-periodic solution in Ω¯1, where
(67)Ω1=u∈X26<uk<63+66.Example 2.
The difference equation(68)Δ4uk−2+15000cos2kπT+12500u3k−12000sin2kπT+1500uk=0,
is one of the form (1), where −a=−1/5000, −A=−3/5000, −b=−3/2000, −B=−1/400, α=3, and β=1. Let τ=10/4, we obtain
(69)16<T4≤6400083+125.
Therefore, we can prove that (68) has at least one positive T-periodic solution in Ω¯2, where
(70)Ω2=u∈X104<uk<11104.



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