# Positive Solutions for a Class of Discrete Mixed Boundary Value Problems with thep,q-Laplacian Operator

Authors: Cuiping Li; Zhan Zhou
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we consider the existence of solutions for the discrete mixed boundary value problems involvingp,q-Laplacian operator. By using critical points theory, we obtain the existence of at least two positive solutions for the boundary value problem under appropriate assumptions on the nonlinearity.


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## Body

## 1. Introduction

In recent years, with the development of mechanical engineering, control system, computer science, and economics, the existence of solutions of difference equations has attracted wide attention (see [1–6]). For example, applying Ricceri variational principle to obtain the existence of multiple solutions [7–9], taking the invariant sets of descending flow to prove the existence of sign-changing solutions [10], making the linking theorem to get the existence and multiplicity of periodic solutions [11], and using critical point theory to obtain the existence of homoclinic solutions [12–15] and heteroclinic solutions [16].As we know, the fixed-point method and upper and lower solution techniques are important tools to solve the existence of solutions for boundary value problems (see [17, 18]). But recently, it is more common to use critical point theory to study Dirichlet boundary value problems (see [19–23]). More result on difference equations by using critical point theories can be referred to [24–27].In [28], D’Aguì et al. established the existence of at least two positive solutions for the following discrete Dirichlet boundary value problem:(1)−ΔϕpΔuk−1+qkϕpuk=λfk,uk,k∈ℤ1,N,u0=uN+1=0,where qk≥0 for all k∈1,2,…,N.Unlike this, D’Aguì et al. in [29] proved that there are at least two nonzero weak solutions for the following mixed boundary value problem:(2)−qxu′kp−2u′x′+sxukp−2ux=λfx,ux,x∈a,b,ua=u′b=0,where p>1,q,s∈L∞a,b with q0=essinfa,bq>0,s0=essinfa,bs≥0,f:a,b×ℝ⟶ℝ is an L1-Carathéodory function and λ is a real positive parameter.As a discrete analogy of the abovementioned problem, we consider the existence of positive solutions for the following discrete mixed boundary value problem:(3)−ΔϕpΔuk−1+skϕquk=λfk,uk,k∈ℤ1,N,u0=ΔuN=0,Dλfwhere ℤa,b denote the discrete interval a,a+1,…,b for any integers a and b with a<b, N be a positive integer, fk,u is continuous in u for each k∈ℤ1,N, Δuk=uk+1−uk is the forward difference operator, ϕr:ℝ⟶ℝ is the r-Laplacian given by ϕru=ur−2u with u∈ℝ, 1<q≤p<+∞, sk≥0 for all k∈ℤ1,N, and λ is a positive parameter.In this paper, under suitable assumptions on the nonlinearityf, we use the theory of two nonzero critical points (see [30]) to ensure that there are at least two nonzero solutions for problem Dλf. The two nonzero critical points theorem is an appropriate combination of local minimum theorem (see [31]) and classical Ambrosetti–Rabinowitz theorem (see [32]). An important hypothesis of mountain pass theorem is Palais–Smale condition. It satisfies the application of infinite dimensional space by requiring the condition that the nonlinear term is stronger than p-superlinearity at infinity. In order to obtain the existence of two nonzero solutions, we can assume the classical Ambrosetti–Rabinowitz condition and nonlinear algebraic condition (see (40) in Theorem 2) hold, that is, more widespread than the p-sublinearity at zero. Moreover, when we require that fk,0≥0 for all k∈ℤ1,N, we can use strong maximum principle to obtain the existence of positive solutions, which has been proved in Lemma 2.Lets∗=minsk:k∈ℤ1,N, a special case of our main result is stated as follows.Theorem 1.
Letf:ℝ⟶ℝ be a continuous function such that4(a)limt⟶0+fttp−1=+∞,4(b)limt⟶+∞fttp−1=+∞,then, for each λ∈0,1+s∗Np−1/pNpminsupc>0cq/maxξ≤c∫0ξftdt,supc>0cp/maxξ≤c∫0ξftdt, the problem(5)−ΔϕpΔuk−1+skϕquk=λfuk,k∈ℤ1,N,u0=ΔuN=0,admits at least two positive solutions.The structure of the article is as follows. In Section2, some basic definitions and properties are given. In Section 3, we give the main results. Under suitable hypothesis, Lemma 1 is used to obtain that the problem Dλf possesses at least two positive solutions. Finally, some examples are given to illustrate our main results.


## 2. Preliminaries

In this section, we recall some definitions, notations, and properties. Consider theN-dimensional Banach space:(6)S=u:ℤ0,N+1⟶ℝ:u0=ΔuN=0,and define the norm(7)u=∑k=1N+1Δuk−1p1/p,and u∞=maxuk:k∈ℤ1,N is another norm in S.Proposition 1.
The following inequality holds:(8)u∞≤maxpNp−11+s∗Np−11/qupp+∑k=1Nskukqq1/q,pNp−11+s∗Np−11/pupp+∑k=1Nskukqq1/p.Proof.
Letu∈S, then there exist k∗∈ℤ1,N such that uk∗=maxuk:k∈ℤ1,N.
Since(9)uk∗=∑k=1k∗Δuk−1≤∑k=1k∗11−1/p∑k=1k∗Δuk−1p1/p≤N1−1/pu,then(10)u∞p≤Np−1up.
Ifu∞>1, then(11)1+s∗Np−1u∞qp≤u∞p+Np−1∑k=1Nskukqp≤Np−1pup+∑k=1Nskukq≤Np−1pup+Np−1q∑k=1Nskukq,that is,(12)u∞≤pNp−11+s∗Np−11/qupp+∑k=1Nskukqq1/q.
Ifu∞≤1, then(13)1+s∗Np−1u∞pp≤u∞p+Np−1∑k=1Nskukqp≤Np−1pup+Np−1q∑k=1Nskukq,that is,(14)u∞≤pNp−11+s∗Np−11/pupp+∑k=1Nskukqq1/p.
In summary, we have(15)u∞≤maxpNp−11+s∗Np−11/qupp+∑k=1Nskukqq1/q,pNp−11+s∗Np−11/pupp+∑k=1Nskukqq1/p.
Put(16)Fk,t≔∫0tfk,ξdξ,∀k,t∈ℤ1,N×ℝ,and consider the function Jλ:S⟶ℝ for all λ>0 by(17)Jλ=Φ−λΨ,where(18)Φ≔Φ1+Φ2,Φ1u=upp,Φ2u=∑k=1Nskukqq,Ψu≔∑k=1NFk,uk.
It is clear thatΦ1,Φ2,Ψ∈C1S,ℝ and their Gâteaux derivatives at the point u∈S are given by(19)Φ1′uv=∑k=1N+1ϕpΔuk−1vk=−∑k=1N+1ϕpΔuk−1vk−1=∑k=1NϕpΔuk−1vk−∑k=0NϕpΔukvk=−∑k=1NΔϕpΔuk−1vk,Φ2′uv=∑k=1Nskϕqukvk,Ψ′uv=∑k=1Nfk,ukvk,for all u,v∈S. So, we have(20)Jλ′uv=∑k=1N−ΔϕpΔuk−1+skϕquk−λfk,ukvk.
Hence, a critical pointu of Jλ is a solution of problem Dλf.
Now, we recall a definition and a two nonzero critical points theorem for the reader’s convenience.Definition 1.
LetX be a real Banach space; we say that a Gâteaux differentiable function Jλ:X⟶ℝ satisfies the PS-condition, if any sequence unn∈N⊆X such that(i)
Jλun⟶c∈ℝ, as n⟶+∞(ii)
Jλ′un⟶0, as n⟶+∞, has a convergent subsequenceLemma 1.
LetX be a real Banach space and Φ,Ψ∈C1S,ℝ such that infXΦ=Φ0=Ψ0=0. Assume that there are r∈ℝ and ω∈X, with 0<Φω<r, such that(21)supu∈Φ−1−∞,rΨur<ΨωΦω,and for each(22)λ∈Λ=ΦωΨω,rsupu∈Φ−1−∞,rΨu,the functional Jλ=Φ−λΨ satisfies the PS-condition and it is unbounded from below.Then, for eachλ∈Λ, the functional Jλ admits at least two nonzero critical points uλ,1,uλ,2 such that Jλuλ,1<0<Jλuλ,2.In order to obtain the positive solution of problemDλf, we establish the following strong maximum principle.Lemma 2.
Fixu∈S such that either(23)uk>0or−ΔϕpΔuk−1+skϕquk≥0,for all k∈ℤ1,N. Then, either u>0 in ℤ1,N or u≡0.Proof.
Letj∈ℤ1,N such that(24)uj=minuk:k∈ℤ1,N.
Ifuj>0, then it is easy know that u>0 in ℤ1,N.
Ifuj≤0, then by (23), we have(25)−ΔϕpΔuj−1≥−sjϕquj≥0,that is,(26)ϕpΔuj≤ϕpΔuj−1.
Sinceϕpu is increasing in u, we have(27)Δuj≤Δuj−1.
By the definition ofuj, we know that(28)Δuj≥0,Δuj−1≤0.
By combining (27) with (28), we get uj+1=uj=uj−1. If j−1=0, we have uj=uj−1=0. Otherwise, j−1∈ℤ1,N, replacing j−1 by j, we know uj−2=uj−1. Continuing in this way, we have uj=uj−1=⋯=u0=0. Similarly, we have uj=uj+1=⋯=uN+1. Thus, uk=u0=0 and ∀k∈ℤ1,N.
Now, put(29)F+k,t=∫0tfk,ξ+dξ,∀k,t∈ℤ1,N×ℝ,where ξ+=maxξ,0.
DefineJλ+=Φ1+Φ2−λΨ+ and Ψ+u:=∑k=1NF+k,uk. Standard arguments show that Jλ+∈C1S,R and the critical points of Jλ+ are precisely the solutions of the following problem:(30)−ΔϕpΔuk−1+skϕquk=λfk,u+k,k∈ℤ1,N,u0=ΔuN=0.Dλf+Lemma 3.
Iffk,0≥0 for each k∈ℤ1,N, any nonzero critical point of the functional Jλ+ is a positive solution of problem Dλf.Proof.
Since a critical point ofJλ+ is a solution of problem Dλf+, the conclusion follows by the discrete maximum principle ([33], Proposition 1).
Next, we suppose thatfk,0≥0 and fk,x=fk,0 for all x≤0 and for all k∈ℤ1,N. Put(31)L∞≔mink∈ℤ1,Nliminft⟶+∞Fk,ttp,s˜=∑k=1Nsk,we have the following result.Lemma 4.
IfL∞>0, then Jλ satisfies PS-condition and it is unbounded from below for all λ∈2pN+s˜−2p−1/qL∞,+∞.Proof.
Letλ>2pN+s˜−2p−1/qL∞. We consider a sequence unn∈N⊆S such that Jλun⟶c∈ℝ and Jλ′un⟶0, as n⟶+∞. Let un+=maxun,0 and un−=max−un,0 for all n∈N. We first prove that un− is bounded. On one hand, we have(32)Δun−k−1p≤−ϕpΔunk−1Δun−k−1,sk∣un−k∣q=−sk∣unk∣q−2unkun−k,for all k∈ℤ1,N. So,(33)un−p=∑k=1N+1Δun−k−1p≤−∑k=1N+1ϕpΔunk−1un−k+∑k=1N=1ϕpΔunk−1un−k−1=−∑k=1NϕpΔunk−1un−k+∑k=1NϕpΔunkun−k=∑k=1NΔϕpΔunk−1un−k=−Φ1′unun−,∑k=1Nskun−kq=−∑k=1Nskϕqunkun−k=−Φ1′unun−.
On the other hand, we assume that(34)fk,u=fk,u,ifu>0,fk,0,ifu≤0,for each k∈ℤ1,N, then(35)Ψ′unun−=∑k=1Nfk,unkun−k≥0.
Therefore,(36)un−p≤un−p+∑k=1Nskukq≤−Φ1′unun−−Φ2′unun−+λΨ′unun−=−Jλ′unun−,for all n∈N, which leads to un−p−1⟶0 as n⟶+∞. So, we have un−⟶0 as n⟶+∞. It means that there exists an M>0 such that un−≤M. From (10) we know that un−∞≤N1−1/pM=γ, for all k∈ℤ1,N.
Next, we suppose that the sequenceun is unbounded, that is, un+ is unbounded.
AsL∞>0, we know that there exists an l∈R such that L∞>l>2pN+s˜−2p−1/λq. From the definition of L∞, there is δk>0 such that Fk,t>ltp for all t>δk. Furthermore, since Fk,t is a continuous function, there exists a constant Ck≥0 such that Fk,t≥ltp−Ck with t∈−γ,δk. Thus, Fk,t≥ltp−Ck for all s≥−γ and k∈ℤ1,N. We can obtain that(37)∑k=1NFk,unk≥∑k=1Nlunkp−C≥lun∞p−C,for all k∈ℤ1,N, where C=∑k=1NCk, that is,(38)Ψun≥lun∞p−C.
Hence, for allun such that un∞≥1, we conclude that(39)Jλun=∑k=1N+1Δunk−1pp+∑k=1Nskunkqq−λΨun≤2p−1p∑k=1Nunkp+∑k=1Nunk−1p+1q∑k=1Nsk∣unk∣q−λΨun≤2p−12N−1pun∞p+s˜qun∞q−λΨun≤2pN+s˜−2p−1q−λlun∞p+λC.
Since2pN+s˜−2p−1/q−λl<0, we can get limn⟶+∞Jλun=−∞ and this is absurd. Hence, Jλ satisfies PS-condition.
Letun be such that un− is bounded and un+ is unbounded. From the proof above we can see that Jλ is unbounded from below.


## 3. Main Results

The main results of this paper are as follows.Theorem 2.
Letf:ℤ1,N×ℝ⟶ℝ be a continuous function satisfying fk,0≥0 for all k∈ℤ1,N. If there are two constants c and d with d<c such that(40)pNp−11+s∗Np−1∑k=1Nmaxξ∣≤cFk,ξmax1cq,1cp<min∑k=1NFk,ddpp−1+dqq−1s˜,qL∞2pN+s˜−2p−1.Then, for eachλ∈Λ1 with(41)Λ1≔maxdpp−1+dqq−1s˜∑k=1NFk,d,2pN+s˜−2p−1qL∞,1+s∗Np−1/pNp−1mincq,cp∑k=1Nmaxξ∣≤cFk,ξ,the problem Dλf admits at least two positive solutions.Proof.
PutΦ,Ψ as in (18). It is clear that infXΦ=Φ0=Ψ0=0. According to Lemma 3, we know that a nonzero critical point in S of the functional Jλ+ is precisely a positive solution of problem Dλf. Next, we just need to prove condition (21) of Lemma 1.
We observe thatL∞>0 from (40) and Λ1 is nondegenerate. Fix λ∈Λ1, Lemma 4 ensures that Jλ satisfies PS-condition for all λ>2pN+s˜−2p−1/qL∞ and it is unbounded from below. We let u∈Φ−1−∞,r, that is, up/p+∑k=1Nskukq/q≤r. Put(42)r=1+s∗Np−1pNp−1mincq,cp.
Ifr=1+s∗Np−1/pNp−1cq, it means that c≥1. According to (8), we obtain(43)uk≤u∞≤maxpNp−11+s∗Np−11/qr1/q,pNp−11+s∗Np−11/pr1/p=maxc,cq/p=c.
Ifr=1+s∗Np−1/pNp−1cp, we know 0<c<1, then(44)uk≤u∞≤maxpNp−11+s∗Np−11/qr1/q,pNp−11+s∗Np−11/pr1/p=maxcp/q,c=c.
To sum up, we know thatuk≤c for all k∈ℤ1,N. Furthermore, we have(45)Ψu=∑k=1NFk,uk≤∑k=1Nmaxξ≤cFk,ξ,for all u∈S with Φu≤r. Hence,(46)supu∈Φ−1−∞,rΨur≤pNp−11+s∗Np−1∑k=1Nmaxξ≤cFk,ξmax1cq,1cp.
Now, letωk=d for all k∈ℤ1,N and ω0=ΔωN=0. Clearly, ω∈S. It is easy to account that Φω=dpp−1+dqq−1s˜, then(47)ΨωΦω=∑k=1NFk,ddpp−1+dqq−1s˜.
Consequently, from (46), (47), and assumption (40), we can obtain(48)supu∈Φ−1−∞,rΨur<ΨωΦω.
Moreover, because0<d<c and from (40), we obtain(49)0<dpp−1+dqq−1s˜<1+s∗Np−1pNp−1mincq,cp,that is mean that 0<Φω<r.
Hence, the problemDλf admits at least two positive solutions by Lemma 1 and Lemma 3 for all λ∈Λ1.Remark 1.
Iffk,t is a nonnegative function and there are two positive constants c,d with d<c such that(50)pNp−11+s∗Np−1max∑k=1NFk,ccq,∑k=1NFk,ccp<min∑k=1NFk,ddpp−1+dqq−1s˜,qL∞2pN+s˜−2p−1,then the result of Theorem 2 is also valid for each λ∈Λ2 with(51)Λ2≔maxdpp−1+dqq−1s˜∑k=1NFk,d,2pN+s˜−2p−1qL∞,1+s∗Np−1pNp−1mincq∑k=1NFk,c,cp∑k=1NFk,c.
There are some consequences of Theorem2 as follows.Corollary 1.
Letg:ℝ⟶[0,+∞ be a continuous function such that fk,t=αkgt, where αk>0 for all k∈ℤ1,N. Put A=∑k=1Nαk, Gt=∫0tgξdξ for all t∈ℝ and L∞∗:=mink∈1,Nαkliminft⟶+∞Gt/tp>0.
If there existsc>d>0 such that(52)pNp−11+s∗Np−1AGcmax1cq,1cp<minAGddpp−1+dqq−1s˜,qL∞∗2pN+s˜−2p−1,then the problem Dλf has at least two positive solutions for each λ∈Λ3 with(53)Λ3≔maxdpp−1+dqq−1s˜AGd,2pN+s˜−2p−1qL∞∗,1+s∗Np−1pNp−1mincq,cpAGc.Proof.
Consider the functionf:ℤ1,N×ℝ⟶ℝ is given as(54)fk,ξ=αkgξ,∀k∈ℤ1,N,ξ∈ℝ,so that(55)∑k=1Nmaxξ≤cFk,ξ=AGc,∑k=1NFk,d=AGd.
Then, the conclusion can be obtained by Theorem2.Corollary 2.
Assumef be a continuous function with fk,0≥0 and(56)limsupt⟶0+Fk,ttp=+∞,(57)limt⟶+∞Fk,ttp=+∞,for all k∈ℤ0,N. Put λ∗=1+s∗Np−1/pNp−1minsupc>0cq/∑k=1Nmaxξ∣≤cFk,ξ,supc>0cp/∑k=1Nmaxξ∣≤cFk,ξ. Then, for each λ∈0,λ∗, the problem Dλf admits at least two positive solutions.Proof.
We know thatL∞=+∞ from (57). Fix λ∈0,λ∗, and then there exists c>0 such that(58)λ<1+s∗Np−1pNp−1minsupc>0cq∑k=1Nmaxξ≤cFk,ξ,supc>0cp∑k=1Nmaxξ≤cFk,ξ.
From (56), we can also obtain(59)limsupt⟶0+∑k=1NFk,ttp=+∞,and then there exists d∈0,c such that ∑k=1NFk,d/dpp−1+dqq−1s˜>1/λ. Therefore, Theorem 2 ensures the conclusion.Remark 2.
Iffk,t is a nonnegative function for all k,t∈ℤ1,N×0,+∞ As long as condition (56) holds for at least one k∈ℤ1,N, then Corollary 2 ensures that the solutions are obtained for each(60)λ∈0,1+s∗Np−1pNp−1minsupc>0cq∑k=1NFk,c,supc>0cp∑k=1NFk,c.Remark 3.
Whenfk,t=ft for all k∈ℤ1,N, Theorem 1 can be ensured by Corollary 2. Obviously, condition (4(a)) implies f0≥0. Specially, if f is nonnegative, we only need condition (4(a)) to get the corresponding solution for each(61)λ∈0,1+s∗Np−1pNpminsupc>0cqFc,supc>0cpFc.Example 1.
Letp=4,q=2,N=3,sk=12,andfk,t=et.
Put(62)xc=c2ec−1.
Then,(63)x′c=c2−cec−2ec−12.
Letzc=2−cec−2, then z′c=1−cec. So, zc is increasing in c∈0,1 and decreasing in c∈1,+∞. Since z0=0 and z+∞=−∞, there exists an unique c1∈1,+∞ such that zc1=0. Thus, xc in increasing in c∈0,c1 and decreasing in c∈c1,+∞. This means that supc>0xc=xc1. In fact, c1≈1.5936.
Similarly, putyc=c4/ec−1, we can show that there exists a unique c2∈3,+∞ such that supc>0yc=yc2. In fact, c2≈3.9207.
Since(64)yc2=c24ec2−1>c14ec1−1=c12xc1>xc1,then(65)1+s∗Np−1pNpminsupc>0xc,supc>0yc=1+s∗Np−1pNpxc1≈0.6496.
Therefore, for eachλ∈0,0.6496, the problem(66)−Δϕ4Δuk−1+12ϕ2uk=λeuk,k=1,2,3,u0=Δu3=0,admits at least two positive solutions.Example 2.
LetN=3,p=3,andq=2 and f be a function as follows:(67)fk,t=0,ift<0,kt3,if0≤t≤1,kt3+225t2−225,ift>1.
From Remark1, we can choose c=1 and d=0.02. Easy calculation shows that(68)pNp−11+s∗Np−1∑k=1NFk,cmax1cq,1cp=pNp−11+s∗Np−1∑k=13∫01kt3dt≈1.3693,∑k=1NFk,ddpp−1+dqq−1s˜=∑k=13∫00.02kt3dtdpp−1+dqq−1s˜≈3.1386,qL∞2pN+s˜−2p−1=166limt⟶+∞kt3+225t2−225t2≈3.4091,which satisfy condition (50). Thus, for each λ∈0.3186,0.7303, the problem(69)−Δϕ3Δuk−1+8ϕ2uk=λfk,uk,k=1,2,3,u0=Δu3=0,admits at least two positive solutions.



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