# Infinitely Many Solutions for Discrete Boundary Value Problems with thep,q-Laplacian Operator

Authors: Zhuomin Zhang; Zhan Zhou
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we consider the existence and multiplicity of solutions for a discrete Dirichlet boundary value problem involving thep,q-Laplacian. By using the critical point theory, we obtain the existence of infinitely many solutions under some suitable assumptions on the nonlinear term. Also, by our strong maximum principle, we can obtain the existence of infinitely many positive solutions.


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## Body

## 1. Introduction

LetN be a positive integer and denote with 1,N the discrete set 1,⋯,N. In this paper, we consider the existence of infinitely many solutions for the following discrete Dirichlet boundary value problem
(1)−Δpuj−1−Δquj−1+αjϕpuj+βjϕquj=λgj,uj,∀j∈1,N,u0=uN+1=0,whereΔruj≔ΔϕrΔuj is the discrete r-Laplacian, ϕru=ur−2u with u∈ℝ,Δuj=uj+1−uj is the forward difference operator, gj,·:ℝ⟶ℝ is continuous for each j∈1,N, 1<q≤p<+∞, λ is a positive parameter, and αj,βj≥0 for all j∈1,N.In the past decades, there has been tremendous interest in the study of difference equations, with the development of engineering, physics, economy, and so on (see [1–4]). Most results about the boundary value problems of difference equations are obtained by using the method of upper and lower solutions and fixed point methods (see [5–7]). In 2003, Guo and Yu [8] first applied the critical point theory to study the existence of periodic and subharmonic solutions for a second-order difference equation. Since then, the critical point theory has been employed to study difference equations, and many meaningful results have been obtained, concerning periodic solutions [9, 10], homoclinic solutions [11–13], heteroclinic solutions [14], and especially in boundary value problems [15–20]. For example, Candito and Giovannelli [21] established the existence of multiple solutions of the following problem
(2)−Δpuj−1=λfj,uj,j∈1,N,u0=uN+1=0.Later, Bonanno and Candito [22] established the existence of infinitely many solutions of the following problem
(3)−Δpuj−1+qkϕpuj=λfj,uj,j∈1,N,u0=uN+1=0,whereqj≥0 for all j∈1,N. Obviously, (2) is a special case (qj=0) of (3). After that, under different conditions, D’Aguì et al. [23] established the existence of at least two positive solutions of (3).In [24], Li and Zhou considered the following discrete mixed boundary value problem
(4)−Δpuj−1+sjϕquj=λfk,uk,j∈1,N,u0=ΔuN=0,wheresj≥0 for all j∈1,N. By using the critical point theory, the authors obtained the existence of at least two positive solutions for (4).The boundary value problems involving the sum of ap-Laplacian operator and of a q-Laplacian operator is more common, because this arises in the study of stationary solutions of reaction-diffusion systems (see [25]). For example, Mugnai and Papageorgiou [26] and Marano et al. [27] investigated the following Dirichlet problem
(5)−Δpu−μΔqu=fx,u,inΩ,u=0,on∂Ω,wheref:Ω×ℝ⟶ℝ satisfies Carathéodory’s conditions, and they obtained the existence of multiple solutions of (5).In [28], Nastasi et al. proved the existence of at least two positive solutions for problem (1). Compared with the discrete boundary value problem involving p-Laplacian operator, there are few results on the discrete boundary value problem with p,q-Laplacian operator except [28]. Inspired by the above results, we want to investigate the multiplicity of solutions for problem (1).In this paper, under suitable assumptions, we use the critical point theory obtained in [29] to establish the existence of infinitely many solutions for discrete p,q-Laplacian equations with Dirichlet type boundary conditions. Moreover, by our strong maximum principle, we can obtain the existence of infinitely many positive solutions of (1).The rest of this paper is organized as follows. In Section2, we recall the critical point theory and show some basic lemmas. In Section 3, our main results and proofs are presented. After that, we have two examples to explain our main results. We conclude our results in the last section.


## 2. Preliminaries

LetX be a reflexive real Banach space and let Iλ:X⟶ℝ be a function satisfying the following structure hypothesis:(H)Iλu=Φu−λΨu for all u∈X, where Φ,Ψ:X⟶ℝ are two functions of class C1 on X with Φ coercive, i.e., lim∥u∥⟶∞Φu=+∞, and λ is a real positive parameterProvided thatinfXΦ<r, put
(6)φr=infu∈Φ−1−∞,rsupv∈Φ−1−∞,rΨv−Ψur−Φu,and
(7)γ=liminfr⟶+∞φr,δ=liminfr⟶infXΦ+φr.There is no doubt thatγ≥0 and δ≥0. When γ=0 (or δ=0), in the sequel, we agree to regard 1/γ (or 1/δ) as +∞.Now, we recall Theorem 2.1 of [29], which is our main tool for investigating problem (1).Lemma 1.
Assume that the condition (H) holds. We have
a For every r>infXΦ and every λ∈0,1/φr, the restriction of the functional Iλ=Φ−λΨ to Φ−1−∞,r admits a global minimum, which is a critical point (local minimum) of Iλ in X.
b If γ<+∞ then, for each λ∈0,1/γ, the following alternative holds: either
b1Iλ possesses a global minimum, or
b2 There is a sequence un of critical points (local minimum) of Iλ such that limn⟶+∞Φun=+∞
c If δ<+∞ then, for each λ∈0,1/δ, the following alternative holds: either
c1 There is a global minimum of Φ which is a local minimum of Iλ, or
c2 There is a sequence un of pairwise distinct critical points (local minima) of Iλ, with limn⟶+∞Φun=infXΦ, which weakly converges to a global minimum of ΦHere, we consider theN-dimensional Banach space
(8)Xd=u:0,N+1⟶ℝsuchthatu0=uN+1=0,and define the norm
(9)∥u∥r,h≔∑j=0NΔujr+∑j=1Nhjujr1/r,where h:1,N⟶ℝ, with hj≥0 for all j∈1,N, and r∈1,+∞. Then, let Xd be endowed with the norm ∥u∥=∥u∥p,α+∥u∥q,β. We denote the usual sup-norm by ∥u∥∞=maxj∈1,N∣uj∣, and then we consider the inequality (see ([30], Lemma 2.2)):
(10)u∞≤N+1r−1/r2ur,hforallu∈Xd.Lemma 2.
Leth=∑j=1Nhj. The following inequalities hold
(11)2N+1r−1/ru∞≤ur,h≤2rN+h1/ru∞.Proof.
The left-hand side of (11) follows by [30]. Consider the right-hand inequality,
(12)∥u∥r,hr=∑j=0NΔujr+∑j=1Nhjujr=Δu0r+ΔuNr+∑j=1N−1Δujr+∑j=1Nhjujr≤2∥u∥∞r+∑j=1N−12∥u∥∞r+∑j=1Nhj∥u∥∞r≤2rN+h∥u∥∞r.Put(13)A1u=1p∥u∥p,αp,A2u=1q∥u∥q,βqandΨu=∑j=1NGj,uj,forallu∈Xd,where the function G:1,N×ℝ⟶ℝ is given by Gj,t=∫0tgj,sds,forallt∈ℝ,j∈1,N.Clearly,A1,A2,Ψ∈C1Xd,ℝ and we have the following Gâteaux derivatives at the point u∈Xd:
(14)A1′u,v=∑j=0NϕpΔujΔvj+∑j=1Nαjϕpujvj,(15)A2′u,v=∑j=0NϕqΔujΔvj+∑j=1Nβjϕqujvj,(16)Ψ′u,v=∑j=1Ngj,ujvj,for all v∈Xd. Now, for r∈1,+∞,
(17)∑j=0NϕrΔujΔvj=∑j=0NϕrΔujvj+1−ϕrΔujvj=∑j=1NϕrΔuj−1vj−∑j=1NϕrΔujvj=−∑j=1NΔϕrΔuj−1vj.If we plug this result back into the calculation of Gâteaux derivatives above, then(18)A1′u,v=∑j=1N−ΔϕpΔuj−1+αjϕpujvj,(19)A2′u,v=∑j=1N−ΔϕqΔuj−1+βjϕqujvj,for all u,v∈Xd. Let
(20)Φu=A1u+A2u.Consider the functionalIλ:Xd⟶ℝ given as
(21)Iλu=Φu−λΨu,forallu∈Xd.We have(22)Iλ′u,v=∑j=1N−Δpuj−1−Δquj−1+αjϕpuj+βjϕquj−λgj,ujvj,for all u,v∈Xd. Thus, u∈Xd is a solution of problem (1) if and only if u is a critical point of Iλ.Lemma 3.
Fixu∈Xd such that either
(23)uj>0or−Δpuj−1−Δquj−1+αjϕpuj+βjϕquj≥0,for all j∈1,N. Then, either u>0 in 1,N or u≡0.Proof.
Fixu∈Xd\0 and Z=j∈1,N:uj≤0. If Z=∅, then, u>0. Now, if minZ=1, we can get
(24)−Δpu0−Δqu0+α1ϕpu1+β1ϕqu1≥0,which implies that
(25)ΔϕpΔu0+ΔϕqΔu0≤α1ϕpu1+β1ϕqu1≤0.
Thus,(26)ϕpΔu1+ϕqΔu1≤ϕpΔu0+ϕqΔu0.
Sinceϕp and ϕq are both strictly increasing, we have Δu1≤Δu0, which implies u2−u1≤u1−0≤0. It follows that u2≤0, then ΔϕpΔu1+ΔϕqΔu1≤α2ϕpu2+β2ϕqu2≤0. An easy induction gives
(27)0=uN+1≤uN≤⋯≤u1≤0.
That isu≡0, and this is absurd. Next, we assume that minZ=z∈2,N,
(28)ΔϕpΔuz−1+ΔϕqΔuz−1≤αzϕpuz+βzϕquz≤0.
Due to the monotonicity ofϕp and ϕq, Δuz≤Δuz−1, which means uz+1−uz≤uz−uz−1. Because uz−1>0, we have uz+1<uz≤0. By repeating this argument, it is easy to see
(29)0=uN+1<uN<⋯<uz≤0,which leads to a contradiction.Now, consider the functionG+:1,N×ℝ⟶ℝ given as
(30)G+j,t=∫0tgj,s+ds,forallt∈ℝ,j∈1,N,where s+=maxs,0. Now, we define Iλ+u=Φu−λΨ+u,forallu∈Xd, where Ψ+u=∑j=1NG+j,uj. Similarly, the critical points of Iλ+ are the solutions of the following problem
(31)−Δpuj−1−Δquj−1+αjϕpuj+βjϕquj=λgj,u+j,∀j∈1,N,u0=uN+1=0.Lemma 4.
Ifgj,0≥0 for all j∈1,N, then each nonzero critical point of Iλ+ is a positive solution of (1).Proof.
We note that each positive solutionu∈Xd of (31) is a positive solution of (1). By an application of Lemma 3, we conclude that u>0. It follows that the nonzero solutions of (31) are positive and hence are positive solutions of (1).


## 3. Main Results

Let(32)α=∑j=1Nαj,β=∑j=1Nβj,L∞j=liminft⟶+∞Gj,ttpandL∞=minj∈1,NL∞j.The main results are as follows.Theorem 5.
Assume thatL∞>0, and there are two real sequences an and bn, with limn⟶+∞an=+∞, such that
(33)bn<min2anα+21/pN+1p−1/p,2anβ+21/qN+1q−1/q,foreveryn∈ℕ,(34)A∞≔liminfn⟶+∞∑j=1Nmax∣t∣≤anGj,t−∑j=1NGj,bn2anp/pN+1p−1+2anq/qN+1q−1−2+α/pbnp−2+β/qbnq<qL∞2p+2qN+α+β.
Then for eachλ∈2p+2qN+α+β/qL∞,1/A∞, problem (1) admits an unbounded sequence of solutions.Proof.
Fixλin2p+2qN+α+β/qL∞,1/A∞, then, we can take the real Banach space Xd as defined in Section 2, and the definitions of Φ,Ψ,Iλ are the same as before. We will prove Theorem 5 by applying Lemma 1 part (b) to function Iλ. Since (H) is trivial to prove, it suffices to prove γ<+∞ and Iλ turns out to be unbounded from below. To this end, let
(35)ρn≔2anppN+1p−1andσn≔2anqqN+1q−1,foreveryn∈ℕ.
Since, owing to (10), if ∥u∥p,α≤pρn1/p then ∥u∥∞≤an, and if ∥u∥q,β≤qσn1/q then ∥u∥∞≤an. So, let rn=ρn+σn. From Φu≤rn, we have ∥u∥∞≤an.
We obtain(36)φrn≤infΦu≤rn∑j=1Nmax∣t∣≤anGj,t−∑j=1NGj,ujrn−Φu.
Then, we definewj such that wnj=bn for every j∈1,N, wn0=wnN+1=0. Clearly wnj∈Xd and Φwn<rn owing to (33). One has
(37)φrn≤∑j=1Nmax∣t∣≤anGj,t−∑j=1NGj,c2anp/pN+1p−1+2anq/qN+1q−1−2+α/pbnp−2+β/qbnq.
Therefore,γ≤liminfn⟶+∞φrn≤A∞<+∞. It remains to show that Iλ is unbounded from below.
Letun⊂Xd be a sequence with unj≥1 for j∈1,N such that limn⟶∞∥un∥=+∞. Because L∞>0, fix L such that L∞>L>2p+2qN+α+β/qλ, and we deduce that there is δj>0 such that Gj,t>Ltp for all t>δj. Moreover, since Gj,t is a continuous function, there exists a constant Cj≥0 such that Gj,t≥Ltp−Cj for all t∈0,δj. Thus, Gj,t≥Ltp−Cj for all t≥0 and j∈1,N. It follows that
(38)Ψun=∑j=1NGj,unj≥∑j=1NLunjp−Cj≥L∥un∥∞p−C,foralln∈ℕ,where C=∑j=1NCj. Since ∥un∥∞≥1, one has
(39)Iλun=∥un∥p,αpp+∥un∥q,βqq−λ∑j=1NGj,unj≤2pN+αp∥un∥∞p+2qN+βq∥un∥∞q−λL∥un∥∞p+λC≤2p+2qN+α+βq−λL∥un∥∞p+λC.
As2p+2qN+α+β/q−λL<0, it is obvious that limn⟶+∞Iλun=−∞. Hence, Iλ is unbounded from below and the proof is complete.Let(40)B∞=limsupt⟶+∞∑j=1NGj,ttp.The following theorem can be obtained if we change some of the conditions.Theorem 6.
Assume that there are two real sequencesan and bn, with limn⟶+∞an=+∞, such that (33) holds and
(41)A∞≔liminfn⟶+∞∑j=1Nmax∣t∣≤anGj,t−∑j=1NGj,bn2anp/pN+1p−1+2anq/qN+1q−1−2+α/pbnp−2+β/qbnq<B∞4+α+β.
Then, for eachλ∈4+α+β/qB∞,1/A∞, problem (1) admits an unbounded sequence of solutions.Proof.
The first half of the argument is analogous to that in Theorem5, and put Φ,Ψ,Iλ,rn as above. So, we have γ≤liminfn⟶+∞φrn≤A∞<+∞.
Our task now is to verify thatIλ is unbounded from below. First, we assume that B∞=+∞. Fix M such that B∞>M>4+α+β/qλ, and let tn be a sequence with tn≥1 and limn⟶+∞tn=+∞, such that
(42)∑j=1NGj,tn>Mtnp,foralln∈ℕ.
Taking the sequencexn in Xd defined by xnj=tn for every j∈1,N, xn0=xnN+1=0, we have
(43)Iλxn=∥xn∥p,αpp+∥xn∥q,βqq−λ∑j=1NGj,xnj=2+αptnp+2+βqtnq−λ∑j=1NGj,tn<2+αptnp+2+βqtnq−λMtnp<α+β+4q−λMtnp.
It is easy to seelimn⟶+∞Iλxn=−∞.
Then, we assume thatB∞<+∞ and fix ε>0 such that ε<B∞−4+α+β/qλ. Let tn be a sequence with tn≥1, such that limn⟶+∞tn=+∞ and
(44)B∞+εtnp>∑j=1NGj,tn>B∞−εtnp,∀n∈ℕ.
Let the sequencexn in Xd be the same as the case where B∞=+∞, such that
(45)Iλxn<4+α+βq−λB∞−εbnp,which implies that limn⟶+∞Iλxn=−∞.
So, in both cases,Iλ is unbounded from below, which completes the proof of Theorem 6.
Let(46)B0≔limsupt⟶0+∑j=1NGj,ttq.
Applying part (c) of Lemma1, we get the following theorem.Theorem 7.
Assume that there exist two real sequencescnanddn, with limn⟶+∞dn=0, such that
(47)cn<min2dnα+21/pN+1p−1/p,2dnβ+21/qN+1q−1/q,foreveryn∈ℕ,(48)A0≔liminfn⟶+∞∑j=1Nmax∣t∣≤dnGj,t−∑j=1NGj,cn2dnp/pN+1p−1+2dnq/qN+1q−1−2+α/pcnp−2+β/qcnq<B04+α+β.
Then, for eachλ∈4+α+β/qB0,1/A0, problem (1) admits a sequence of nonzero solutions which converges to zero.Proof.
Fixλ in 4+α+β/qB0,1/A0, and we can take the real Banach space Xd and functional Φ,Ψ,Iλ as defined in Section 2. Our aim is to apply Lemma 1 part (c) to function Iλ. To this end, let
(49)ρn≔2dnppN+1p−1andσn≔2dnqqN+1q−1,foreveryn∈ℕ.
Owing to (10), if ∥u∥p,α≤pρn1/p then ∥u∥∞≤dn, and if ∥u∥q,β≤qσn1/q then ∥u∥∞≤dn. So, let rn=ρn+σn. It follows that if Φu≤rn, then ∥u∥∞≤dn. We obtain
(50)φrn≤infΦu≤rn∑j=1Nmax∣t∣≤dnGj,t−∑j=1NGj,ujrn−∥u∥p,αp/p−∥u∥q,βq/q.
Now, for eachn∈ℕ, let vnj be defined by vnj=cn for every j∈1,N, vn0=vnN+1=0. Clearly vnj∈Xd, and Φvn≤rn from (47). We have
(51)φrn≤∑j=1Nmaxt≤dnGj,t−∑j=1NGj,cn2dnp/pN+1p−1+2dnq/qN+1q−1−2+α/pcnp−2+β/qcnq.
Hence,δ≤liminfn⟶+∞φrn≤A0<+∞ follows.
In fact,infXdΦ=0, so our task now is to verify that the 0 is not a local minimum of Iλ. First, assume that B0=+∞. Fix M such that B0>M>4+α+β/qλ, and let sn be a sequence of positive numbers, with sn≤1 and limn⟶+∞sn=0, such that
(52)∑j=1NGj,sn>Msnq,foralln∈ℕ.
Thus, taking the sequenceyn in Xd, let ynj=sn for every j∈1,N, yn0=ynN+1=0. Some tedious manipulation yields
(53)Iλyn<4+α+βq−λMsnq,which implies that Iλyn<0.
Then, we assume thatB0<+∞ and fix ε>0 such that ε<B0−4+α+β/qλ. Let sn be a sequence of positive numbers, with sn≤1, such that limn⟶+∞sn=0 and
(54)B0+εsnq>∑j=1NGj,sn>B0−εsnq,∀n∈ℕ.
Choosing the sameyn in Xd as the case B0=+∞, one has
(55)Iλyn<4+α+βq−λB0−εsnq.
That isIλyn<0. Since 0 is the global minimum of Φ, in both cases, u=0 is not a local minimum of Iλ and the proof is complete.By setting(56)A∗≔liminfn⟶+∞∑j=1Nmax∣t∣≤anGj,t2anp/pN+1p−1+2anq/qN+1q−1,A¯∞≔liminft⟶+∞∑j=1Nmax∣ξ∣≤tGj,ξtq+tp,we get the following consequences.Corollary 8.
Assume that(57)A¯∞<2qpN+1p−14+α+βB∞.
Then, for eachλ∈4+α+β/qB∞,2q/pN+1p−1A¯∞, problem (1) admits an unbounded sequence of solutions.Proof.
Letan be a sequence of positive numbers with limn⟶∞an=+∞, such that
(58)A¯∞=liminfn⟶+∞∑j=1Nmax∣ξ∣≤anGj,ξanq+anp.
After simple scaling and calculation, we have(59)A∗≤pN+1p−12qA¯∞.
Takingbn=0 for each n∈ℕ, from Theorem 6, the conclusion follows.Ifgj,0 satisfies the nonnegative condition, we have the following conclusion.Corollary 9.
Assume thatgj,0≥0 for all j∈1,N, and
(60)A¯∞<2qpN+1p−14+α+βB∞.
Then, for eachλ∈4+α+β/qB∞,2q/pN+1p−1A¯∞, problem (1) admits an unbounded sequence of positive solutions.Proof.
Let(61)g+j,t=gj,t,ift>0,gj,0,ift≤0.
Sincegj,0≥0,
(62)max0≤s≤t∫0sg+j,ξdξ=max0≤s≤t∫0sgj,ξdξ,for all t≥0. From Corollary 8, we know that problem (1) with g replaced by g+ admits an unbounded sequence of solutions for each λ∈4+α+β/qB∞,2q/pN+1p−1A¯∞. Then, all these solutions are positive solutions of problem (1) by Lemma 4.
Let(63)A¯0≔liminft⟶0+∑j=1Nmax∣ξ∣≤tGj,ξtq+tp.
Arguing as in the proof of Corollary8 and taking cn=0 for each n∈1,N, by Theorem 7, we have the following corollary.Corollary 10.
Assume that(64)A¯0<2qpN+1p−14+α+βB0.
Then, for eachλ∈4+α+β/qB0,2q/pN+1p−1A¯0, problem (1) admits a sequence of nonzero solutions which converges to zero.Arguing as in Corollary9, we have the following result.Corollary 11.
Assume thatgj,0≥0 for all j∈1,N, and
(65)A¯0<2qpN+1p−14+α+βB0.
Then, for eachλ∈4+α+β/qB0,2q/pN+1p−1A¯0, problem (1) admits a sequence of positive solutions which converges to zero.Finally, we give two easy examples to illustrate our results.Example 1.
Letα=β=0,q=2,p=3,(66)gj,x=gx=3x2sin12ln∣x∣+12x2cos12ln∣x∣+258x2,ifx≠0,0,ifx=0.
for eachj∈1,N. Then,
(67)liminft⟶+∞maxξ≤t∫0ξ3x2sinlnx/2+x2coslnx/2/2+25x2/8dxt2+t3=liminft⟶+∞t3sinlnt/2+25t3/24t2+t3=124,and
(68)limsupt⟶+∞∫0t3x2sinlnx/2+x2coslnx/2/2+25x2/8dxt3=limsupt⟶+∞t3sinlnt/2+25t3/24t3=4924.
By choosingN=3, we have
(69)2qpN+1p−14+α+β=148.
From the above calculation, we obtain(70)A¯∞=liminft⟶+∞∑j=13maxξ≤t∫0ξx23sinlnx/2+coslnx/2/2+25/8dxt2+t3=18,(71)B∞=limsupt⟶+∞∑j=13∫0tx23sinlnx/2+coslnx/2/2+25/8dxt3=498.
It is clear thatA¯∞<2qB∞/pN+1p−14+α+β, by Corollary 9, the problem
(72)−∣Δuj∣+1Δuj+∣Δuj−1∣+1Δuj−1=12guj,∀j∈1,3,u0=u4=0,admits an unbounded sequence of positive solutions.Example 2.
Letq=2,p>2 and
(73)gj,x=gx=x2+2ε+2cosεln∣x∣−εsinεln∣x∣,ifx≠0,0,ifx=0,
for eachj∈1,N. Then,
(74)Gj,x=Gx=∫0xgsds=x21+ε+cosεlnx,for x>0. Since gx≥0 for x≥0, Gx is increasing. We have
(75)A¯0=liminft⟶0+∑j=1Nmax0≤ξ≤tGj,ξtq+tp=Nliminft⟶0+t21+ε+cosεlntt2+tp=Nε,(76)B0=limsupt⟶0+∑j=1NGj,ttq=Nlimsupt⟶0+t21+ε+cosεlntt2=N2+ε.
Letε be a sufficiently small constant, such that
(77)Nε<2qpN+1p−14+α+βN2+ε.
Then, by Corollary11, for each λ∈4+α+β/qB0,2q/pN+1p−1A¯0, problem (1) admits a sequence of positive solutions which converges to zero.


## 4. Conclusions

In this paper, we consider a discrete Dirichlet boundary value problem involving thep,q-Laplacian. Unlike the existing result in [28], which is the existence of at least two positive solutions, we consider the existence of infinitely many solutions for problem (1) for the first time. In fact, by using Theorem 2.1 of [29], we show that problem (1) admits a sequence of pairwise distinct solutions under some appropriate assumptions on the nonlinear term near at infinity and at the origin. Moreover, we prove the existence of infinitely many positive solutions through our strong maximum principle. It seems that we can use the method in this paper to study other similar problems, such as the existence and multiplicity of solutions for difference equations with different boundary value conditions. This will be left as our future work.



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*Source: 1980285-2021-09-27.xml*