# Multiple Solutions for Boundary Value Problems ofp-Laplacian Difference Equations Containing Both Advance and Retardation

Authors: Zhenguo Wang; Zhan Zhou
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

This paper concerns the existence of solutions for the Dirichlet boundary value problems ofp-Laplacian difference equations containing both advance and retardation depending on a parameter λ. Under some suitable assumptions, infinitely many solutions are obtained when λ lies in a given open interval. The approach is based on the critical point theory.


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## Body

## 1. Introduction

Letℤ and ℝ be the sets of integers and real numbers, respectively. For a,b∈ℤ, ℤa,b denotes the discrete interval a,a+1,…,b if a≤b.Assume thatN is a positive integer. In this paper, we consider the following boundary value problem of difference equation containing both advance and retardation:(1)−ΔϕpΔuk−1+2qkϕpuk=λfk,uk+1,uk,uk−1,k∈ℤ1,N,u0=uN+1=0,where λ is a positive real parameter, the forward difference operator is defined as Δuk=uk+1−uk. For every k∈ℤ1,N, qk≥0, fk,⋅:ℝ3⟶ℝ is a continuous function, ϕps is the p-Laplacian operator, that is, ϕps=sp−2s, and 1<p<+∞.As it is to know, difference equations have been widely used in various research fields such as computer science, economics, network, and control system. Relevant examples and mathematical models can be found in [1–3]. By means of critical point theory, many researchers devote themselves to the study of difference equations and achieve many excellent results, for example, the results on boundary value problems [4–16], periodic solutions [17–22], and homoclinic solutions [23–31] had been obtained.Difference equations containing both advance and retardation have many applications in physical and biological phenomena [32]. Yu et al. [23] obtained the existence of nontrivial homoclinic orbits for the following second-order difference equation containing both advance and retardation:(2)Luk−ωuk=fk,uk+T,uk,uk−T,k∈ℤ.The operatorL is a second-order difference operator given by(3)Luk=akuk+1+ak−1uk−1+bkuk,where ak and bk are real valued for each k∈ℤ and ω∈ℝ.Mei and Zhou [21] considered the existence of the periodic and subharmonic solutions of a 2nth order p-Laplacian difference equation containing both advances and retardations:(4)Δnrk−nφpΔnuk−n=−1nfk,uk+τ,…,uk+1,uk,uk−1,…,uk−τ,k∈ℤ.There are more about the results of the difference equations containing both advance and retardation which can be seen in [13, 19, 22, 24].Here, we are interested in using critical point theory to investigate infinitely many solutions for (1) containing both advance and retardation.In fact, there are some papers which studied the existence of infinitely many solutions for the boundary value problems of difference equations. Bonanno and Candito [33] in 2009 proved the existence of infinitely many solutions of the following discrete boundary value problem:(5)−ΔϕpΔuk−1+qkϕpuk=λfk,uk,k∈ℤ1,N,u0=uN+1=0.Recently, Zhou and Ling [10] studied the existence of infinitely many positive solutions for the following boundary value problem of the second order nonlinear difference equation with ϕc-Laplacian:(6)−ΔϕcΔuk−1=λfk,uk,k∈ℤ1,N,u0=uN+1=0.However, to the best of our knowledge, no similar results are obtained for problem (1) containing both advance and retardation. The main difficulty is caused by the advance and retardation. In this paper, we obtain some sufficient conditions to guarantee the existence of infinitely many solutions of (1). In fact, under some assumptions, we prove the existence of infinitely many solutions of equation (1) for each λ∈2+2Q/pB,2/pN+1p−1A in Theorem 1. Moreover, Theorem 2 guarantees the existence of infinity positive solutions for (1) by applying a strong maximum principle. Finally, we show that (1) possesses a sequence of distinct solutions which converges to zero for each λ∈2+2Q/pB∗,2p−11+q¯/pN+1p−1A∗ in Theorem 3.This paper is organized as follows. In Section2, some definitions and preliminaries on difference equations are collected. In Section 3, our main results are established. Finally, two examples are given to illustrate our main results.


## 2. Preliminaries

LetE be a reflexive real Banach space and Iλ:E⟶ℝ be a function satisfying the following structure hypothesis:H1 Assume that λ is a real positive parameter. Iλ:=Φu−λΨu, ∀u∈E, where Φ,Ψ∈C1E,ℝ, Φ is coercive, that is, limu⟶∞Φu=+∞.Provided thatinfEΦ<r, write(7)φr=infv∈Φ−1−∞,rsupv∈Φ−1−∞,rΨv−Ψur−Φu,γ=liminfr⟶+∞φr,δ=liminfr⟶infEΦ+φr.Obviously,γ≥0 and δ≥0. When γ=0 or δ=0, we agree to read 1/γ or 1/δ as +∞.Lemma 1 (see [34]).
Assume that conditionH1 holds, and one has(a)
For everyr>infEΦ and every λ∈0,1/φr, the restriction of functional Iλ=Φu−λΨu to u∈Φ−1−∞,r admits a global minimum, which is a critical point (local minimum) of Iλ in E.(b)
Ifγ<+∞, then for each λ∈0,1/γ, the following alternative holds: b1Iλ possesses a global minimumb2 There is a sequence un of critical points (local minimum) of Iλ such that limn⟶+∞Φun=+∞.(c)
Ifδ<+∞, then for each λ∈0,1/δ, the following alternative holds:c1 There is a global minimum of Φ which is a local minimum of Iλc2 There is a sequence un of pairwise distinct critical points (local minimum) of Iλ, with limn⟶+∞Φun=infEΦ, which weakly converges to a global minimum of Φ.Consider theN-dimensional Banach space:(8)S=u:0,N+1⟶ℝsuch thatu0=uN+1=0,endowed with the norm(9)u=∑k=1N+1Δuk−1p+2qkukp1/p.We define another two norms onS as follows:(10)up=∑k=1N+1Δuk−1p1/p,u∞=maxk∈ℤ1,Nuk.According to Lemma 2.2 in [12], we have the following inequality:(11)u∞=maxk∈ℤ1,Nuk≤N+1p−1/p2up,for every u∈S.Lemma 2.
For everyu∈S, one has(12)u∞≤N+1p−1/p21+q¯1/pu,where q¯=minqk,k∈ℤ1,N.Proof.
Making use of (11), we obtain(13)u∞p+q¯u∞p≤12N+12p−1upp+∑k=1N+12qkukp≤N+1p−12pup,then u∞≤N+1p−1/p/21+q¯1/pu.
For everyu∈S, put(14)Φu=upp,Ψu=∑k=1NFk,uk+1,uk,Iλu=Φu−λΨu,where Fk,⋅ satisfies the following conditions: H2Fk,⋅∈C1ℝ2,ℝ, Fk,0,0=0, F0,.,.=0∀k∈ℤ1,NH3∂Fk−1,x2,x3/∂x2+∂Fk,x1,x2/∂x2=fk,x1,x2,x3, for k,x1,x2,x3∈ℤ1,N×ℝ3
Direct computation ensures thatIλ is a functional of class C1 on S with(15)∂Iλ∂uk=−ΔϕpΔuk−1+2qkϕpuk−λfk,uk+1,uk,uk−1,k∈ℤ1,N.
Then,u is a critical point of Iλ on S if and only if(16)−ΔϕpΔuk−1+2qkϕpuk=λfk,uk+1,uk,uk−1,k∈ℤ1,N.
That is, the functionIλ is just the variational framework of (1). For the reader’s convenience, we recall a consequence of strong maximum principle [8]. The strong maximum principle is used to obtain positive solutions to (1), that is, uk>0 for every k∈ℤ1,N.Lemma 3 (see [8]).
Fixu∈S such that(17)eitheruk>0or−ΔϕpΔuk−1+qkϕpuk≥0,∀k∈ℤ1,N.
Then, eitheru>0 or u≡0.Remark 1.
Assumefk,x1,x2,x3:ℝ3⟶ℝ is continuous for each k∈ℤ1,N and fk,x1,0,x3≥0 for all k,x1,x3∈ℤ1,N×ℝ2. Put(18)f∗k,x1,x2,x3=fk,x1,0,x3,ifx2≤0,fk,x1,x2,x3,ifx2>0,then f∗k,⋅∈Cℝ3,ℝ for each k∈ℤ1,N.
Consider the following boundary value problem:(19)−ΔϕpΔuk−1+2qkϕpuk=λf∗k,uk+1,uk,uk−1,k∈ℤ1,N,u0=uN+1=0.
From Lemma3, all solutions of (19) are either zero or positive; hence, they are also solutions for (1). When (19) possesses nontrivial solutions, (1) possesses positive solutions, independently of the sign of f.


## 3. Main Results

Put(20)Q=∑k=1Nqk,κ=11+QN+1p−1,B=limsupξ⟶+∞∑k=1N−1Fk,ξ,ξ+FN,0,ξξp.Now, we consider the suitable oscillating behavior ofFk,x1,x2 when x1p+x2p goes to +∞. We have the following theorem.Theorem 1.
Assume thatH2andH3 are satisfied and there exist two real sequences anandcn, with limn⟶+∞cn=+∞, such that(21)anp<2p−21+q¯cnp1+QN+1p−1,for alln∈ℤ,(22)A=liminfn⟶+∞∑k=1Nmaxx1p+x2p≤cnpFk,x1,x2−∑k=1N−1Fk,an,an−FN,0,an2p−21+q¯cnp−1+QN+1p−1anp<κB.Then, problem (1) admits an unbounded sequence of solutions for each λ∈2+2Q/pB,2/pN+1p−1A.Proof.
Our aim is to apply Lemma1(b) to prove our conclusion. Fix λ∈2+2Q/pB,2/pN+1p−1A, which clearly H1 holds. Our conclusion needs to provide that γ<+∞. Put rn=2p−11+q¯cnp/pN+1p−1 for all n∈Z. Owing to (12), if u≤prn1/p, then uk+1p+ukp≤cnp for every k∈ℤ1,N, and we obtain(23)φrn=infu≤prn1/psupu≤prn1/p∑k=1NFk,uk+1,uk−∑k=1NFk,uk+1,ukrn−up/p≤infu≤prn1/p∑k=1Nmaxuk+1p+ukp≤cnpFk,uk+1,uk−∑k=1NFk,uk+1,uk2p−11+q¯cnp/pN+1p−1−up/p.
Letgnk=an for every k∈ℤ1,N, and gn0=gnN+1=0. Clearly, gn⊂S and gnp=21+Qanp. From (21), we obtain gn≤prn1/p, and one has(24)φrn≤pN+1p−12∑x1p+x2p≤cnpNFk,x1,x2−∑k=1N−1Fk,an,an−FN,0,an2p−21+q¯cnp−1+QN+1p−1anp.
Hence,γ≤liminfn⟶∞φrn≤pN+1p−1A/2<+∞.
Now, we verify thatIλ is unbounded from below. By (20), let dn be a positive real sequence with limn⟶∞dn=+∞ and(25)B=limn⟶+∞∑k=1N−1Fk,dn,dn+FN,0,dndnp.
First, we assume thatB=+∞, and then we can fix M>2+2Q/pλ, and there exists νM∈ℤ such that(26)∑k=1N−1Fk,dn,dn+FN,0,dn≥Mdnp,∀n>νM.
We take a sequencesn in S such that snk=dn for every k∈ℤ1,N, sn0=snN+1=0, and one has(27)Iλsn=Φsn−λΨsn=snpp−λ∑k=1N−1Fk,dn,dn+FN,0,dn<2+2Qp−λMdnp,for all n>νM. That is, limn⟶+∞Iλsn=−∞.
Next, we assume thatB<+∞. Since λ>2+2Q/pB, from (25), we can fix ε>0 such that ε<B−2+2Q/pλ, arguing as before, there is a νε∈ℤ such that(28)∑k=1N−1Fk,dn,dn+FN,0,dn≥B−εdnp,∀n>νε.
By choosingsn in S as above, we obtain that(29)Iλsn<2+2Qp−λB−εdnp,for all n>νε. Hence, one has limn⟶+∞Iλsn=−∞.
We have verified all assumptions of Lemma1(b); then, there is a sequence un of critical points (local minima) of Iλ such that limn⟶+∞Φun=+∞.
According to Theorem1, it is easy to obtain the following corollary.Corollary 1.
LetH2andH3 be satisfied, and assume that(30)A=liminfξ⟶+∞∑k=1Nmaxx1p+x2p≤ξpFk,x1,x2ξp<2p−21+q¯κB.Then, problem (1) admits an unbounded sequence of solutions for each λ∈2+2Q/pB,2p−11+q¯/pN+1p−1A.Proof.
Letcn be a real sequence with limn⟶∞cn=+∞ such that(31)A=limn⟶+∞∑k=1Nmaxx1p+x2p≤cnpFk,x1,x2cnp.
We takean=0 for all n∈ℤ in (21); combining with H2, (30) and (31), we can apply Theorem 1 to reach the conclusion.
From the argument of Remark1, we have the following theorem and corollary.Theorem 2.
Iffk,x1,0,x3≥0 for every k,x1,x3∈ℤ1,N×ℝ2 and the hypotheses of Theorem 1 hold. Then, problem (1) admits an unbounded sequence of positive solutions for each λ∈2+2Q/pB,2/pN+1p−1A.Corollary 2.
Iffk,x1,0,x3≥0 for every k,x1,x3∈ℤ1,N×ℝ2 and the hypotheses of Corollary 1 are satisfied, then problem (1) admits an unbounded sequence of positive solutions for each λ∈2+2Q/pB,2p−11+q¯/pN+1p−1A.Next, we consider the oscillating behavior ofFk,x1,x2 when x1p+x2p goes to 0. We obtain the following theorem.Theorem 3.
Assume thatH2andH3 are satisfied and(32)A∗<2p−21+q¯κB∗,where A∗=liminfξ⟶0+∑k=1Nmaxx1p+x2p≤ξpFk,x1,x2/ξp and B∗=limsupξ⟶0+∑k=1N−1Fk,ξ,ξ+FN,0,ξ/ξp. Then, problem (1) possesses a sequence of pairwise distinct solutions, which converges to zero for each λ∈2+2Q/pB∗,2p−11+q¯/pN+1p−1A∗.Proof.
We will check the partc of Lemma 1. Fix λ∈2+2Q/pB∗,2p−11+q¯/pN+1p−1A∗, let cn be a positive real sequence such that limn⟶∞cn=0 and(33)A∗=limn⟶+∞∑k=1Nmaxx1p+x2p≤cnpFk,x1,x2cnp.
As before, we letrn=2p−11+q¯cnp/pN+1p−1 for each n∈Z. In view of (12), note that u≤prn1/p implies uk+1p+ukp≤cnp for every k∈ℤ1,N, by the definition of φ, we have(34)φrn=infu≤prn1/psupu≤prn1/p∑k=1NFk,uk+1,uk−∑k=1NFk,uk+1,ukrn−up/p≤pN+1p−12p−11+q¯∑k=1Nmaxx1p+x2p≤cnpFk,x1,x2cnp,then δ≤liminfn⟶∞φrn≤pN+1p−1A∗/2p−11+q¯<+∞.
Clearly, 0 is a global minimum ofΦ in S, and Iλ0=0 by H2.
Moreover, we can verify that 0 is not a local minimum ofIλ. Given a positive real sequence dn with limn⟶+∞dn=0 such that(35)B∗=limn⟶+∞∑k=1N−1Fk,dn,dn+FN,0,dndnp,if B∗=+∞, fix M>2+2Q/pλ, and there exists νM∈ℤ such that(36)∑k=1N−1Fk,dn,dn+FN,0,dn>Mdnp,∀n>νM.
Letvn⊂S be a sequence satisfying vnk=dn for every k∈ℤ1,N, vn0=vnN+1=0. Since vn=2+2Q1/pdn⟶0 as n⟶∞, we have(37)Iλvn=Φvn−λΨvn=vnpp−λ∑k=1N−1Fk,dn,dn+FN,0,dn<2+2Qp−λMdnp<0,for all n>νM. Thus, 0 is not a local minimum of Iλ.
IfB∗<+∞, since λ>2+2Q/pB∗, we can also find a positive real sequence dn with limn⟶+∞dn=0 such that (35) holds. Fix 0<ε<B∗−2+2Q/pλ, and there exists νε∈ℤ, such that(38)∑k=1N−1Fk,dn,dn+FN,0,dn>B∗−εdnp,∀n>νε.
Arguing as before, there exist a sequencevn⊂S, such that(39)Iλvn<2+2Qp−λB∗−εdnp<0,for all n>νε. Obviously, 0 is not a local minimum of Iλ, and c2 of Lemma 1(c) is true.
A similar result to Theorem2 is obtained as follows.Theorem 4.
Iffk,x1,0,x3≥0 for every k,x1,x3∈ℤ1,N×ℝ2 and the hypotheses of Theorem 3 hold, then problem (1) admits a sequence of pairwise distinct positive solutions, which converges to zero for each λ∈2+2Q/pB∗,2p−11+q¯/pN+1p−1A∗.


## 4. Examples

Finally, we give two examples to illustrate our results.Example 1.
Consider the boundary value problem (1) with(40)fk,x1,x2,x3=1+ε+cosεln1+x2p+x3p−εsinεln1+x2p+x3ppx2p−2x2+1+ε+cosεln1+x1p+x2p−εsinεln1+x1p+x2ppx2p−2x2,for k∈ℤ1,N. Put(41)Fk,x1,x2=1+x1p+x2p1+ε+cosεln1+|x1|p+|x2|p−2−ε.
Obviously,Fk,x1,x2 satisfies conditions H2 and H3. It is easy to see that(42)maxx1p+x2p≤ξpFk,x1,x2=1+ξp1+ε+cosεln1+ξp−2−ε.
Clearly,(43)A=liminfξ⟶+∞∑k=1Nmaxx1p+x2p≤ξpFk,x1,x2ξp=liminfξ⟶+∞∑k=1N1+ξp1+ε+cosεln1+ξp−2−εξp=Nliminfξ⟶+∞1+ξp1+ε+cosεln1+ξp−2−εξp=εN,and(44)B=limsupξ⟶+∞∑k=1N−1Fk,ξ,ξ+FN,0,ξξp=limsupξ⟶+∞2N−11+ε+2N−1cosεln1+2ξp+cosεln1+ξp≥2N−1ε+2≥2+ε.
Letε be sufficiently small, then 2+2Q/pB≤2+2Q/2+εp≤2p−11+q¯/pN+1p−1εN. Hence, for each λ∈2+2Q/pB,2p−11+q¯/pN+1p−1εN, (1) admits an unbounded sequence of solutions by Corollary 1. Moreover, we have fk,x1,0,x3=0 for all k,x1,x3∈ℤ1,N×ℝ2; according to Corollary 2, (1) admits an unbounded sequence of positive solutions.Example 2.
Consider the boundary value problem (1) with(45)fk,x1,x2,x3=sinεlnx2p+x3p+εcosεlnx2p+x3p+sinεlnx1p+x2p+εcosεlnx1p+x2p+2+2εpx2p−2x2,if x12+x22+x32≠0 and fk,0,0,0=0 for k∈ℤ1,N. Let(46)Fk,x1,x2=x1p+x2psinεlnx1p+x2p+1+ε,ifx12+x22≠0,0,ifx12+x22=0,then H3 holds. Clearly, H2 holds and(47)maxx1p+x2p≤ξpFk,x1,x2=ξpsinεlnξp+1+ε.
By computation, we obtain(48)A∗=liminfξ⟶0+∑k=1Nmaxx1p+x2p≤ξpFk,x1,x2ξp=liminfξ⟶0+∑k=1Nξpsinεlnξp+1+εξp=Nliminfξ⟶0+sinεlnξp+1+ε=εN,B∗=limsupξ⟶0+∑k=1N−1Fk,ξ,ξ+FN,0,ξξp=limsupξ⟶0+2N−11+ε+2N−1sinεln2ξp+sinεlnξp≥2N−1ε+2≥2+ε.
Obviously, we have2+2Q/pB∗≤2+2Q/2+εp≤2p−11+q¯/pN+1p−1εN, when ε be sufficiently small. On the contrary, fk,x1,0,x3=0 for all k,x1,x3∈ℤ1,N×ℝ2.
The computations ensure that all the assumptions of Theorem4 are satisfied. Then, for each λ∈2+2Q/pB∗,2p−11+q¯/pN+1p−1εN, (1) admits a sequence of pairwise distinct positive solutions, which converges to zero.



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