Research article

Existence and multiplicity of nontrivial solutions to discrete elliptic Dirichlet problems


  • Received: 15 April 2022 Revised: 06 May 2022 Accepted: 09 May 2022 Published: 18 May 2022
  • In this paper, we study discrete elliptic Dirichlet problems. Applying a variational technique together with Morse theory, we establish several results on the existence and multiplicity of nontrivial solutions. Finally, two examples and numerical simulations are provided to illustrate our theoretical results.

    Citation: Yuhua Long, Huan Zhang. Existence and multiplicity of nontrivial solutions to discrete elliptic Dirichlet problems[J]. Electronic Research Archive, 2022, 30(7): 2681-2699. doi: 10.3934/era.2022137

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  • In this paper, we study discrete elliptic Dirichlet problems. Applying a variational technique together with Morse theory, we establish several results on the existence and multiplicity of nontrivial solutions. Finally, two examples and numerical simulations are provided to illustrate our theoretical results.



    Let N and Z be the sets of natural numbers and integers, respectively. For integers a, b, define Z(a,b):={a,a+1,...,b} with ab. Given the integers T1, T22, we write Ω:=Z(1,T1)×Z(1,T2). Consider the existence and multiplicity of nontrivial solutions to the following discrete elliptic problem:

    Δ21u(i1,j)+Δ22u(i,j1)+f((i,j),u(i,j))=0,(i,j)Ω, (1.1)

    with Dirichlet boundary conditions

    u(i,0)=u(i,T2+1)=0iZ(1,T1),u(0,j)=u(T1+1,j)=0jZ(1,T2), (1.2)

    where Δ1, Δ2 are the forward difference operators defined by Δ1u(i,j)=u(i+1,j)u(i,j), Δ2u(i,j)=u(i,j+1)u(i,j), and Δ2u(i,j)=Δ(Δu(i,j)). Here, f((i,j),u) is continuously differentiable with respect to u and f((i,j),0)=0.

    Advances in modern computing devices have made it increasingly convenient to determine the behavior of complex systems through simulations, contributing greatly to the increasing interest in discrete problems. As a result [1,2,3,4,5,6,7], difference equations have been widely investigated and numerous results have been obtained [8,9,10,11,12,13,14,15,16]. With the development of science and technology in modern society and the progress of mathematical research, the study of difference equations has gradually shifted to the study of partial difference equations. For example, [17,18,19] deal with discrete Kirchhoff-type problems, whereas [20,21,22] present several results on multiple solutions to partial difference equations.

    Equation (1.1) is a partial difference equation involving bivariate sequences with two independent integer variables over Ω. It is elementary, but illustrative of many problems that are of interest in various branches of science, such as chemical reactions, population dynamics with spatial migration, and even the computation and analysis of finite difference equations [23,24]. Therefore, Eq (1.1) has attracted considerable attention. For example, [24] shows that Eq (1.1) possesses at least two nontrivial solutions, while Zhang [25] studied Eq (1.1) using the extremum principle. Moreover, Eq (1.1) is regarded as a discrete analog of the partial differential equation

    2ux2+2uy2+f(x,y,u(x,y))=0,(x,y)Ω,

    which has been extensively studied. Consequently, investigating problem (1.1)–(1.2) is of practical significance.

    Morse theory is a very powerful tool for studying the existence of multiple solutions to both differential and difference equations having a variational structure [26,27,28,29,30]. Very recently, Long [18,19,20] studied partial difference equations via Morse theory and obtained rich results on the existence and multiplicity of nontrivial solutions. This encourages us to consider the existence and multiplicity of nontrivial solutions for problem (1.1)–(1.2) using Morse theory.

    The remainder of this paper is organized as follows. In Section 2, the variational structure and the corresponding functional are established according to (1.1)–(1.2). Moreover, we recall some related definitions and propositions that are beneficial to our results. Section 3 displays our main results and the corresponding proofs. Finally, two examples and numerical simulations are provided to illustrate our main results in Section 4.

    Let E be a T1T2-dimensional Euclidean space equipped with the usual inner product (,) and norm ||. Let

    S={u:Z(0,T1+1)×Z(0,T2+1)Rsuch thatu(i,0)=u(i,T2+1)=0,iZ(0,T1+1)andu(0,j)=u(T1+1,j)=0,jZ(0,T2+1)}.

    Define the inner product , on S as

    u,v=T1+1i=1T2j=11u(i1,j)1v(i1,j)+T1i=1T2+1j=12u(i,j1)2v(i,j1),u,vS,

    and let the induced norm be

    u=u,u=(T1+1i=1T2j=1|1u(i1,j)|2+T1i=1T2+1j=1|2u(i,j1)|2)12,uS.

    Then, S is a Hilbert space and is isomorphic to E. Here and hereafter, we take uS as an extension of uE when necessary.

    Consider the functional I:SR in the following form:

    I(u)=12T1+1i=1T2j=1|Δ1u(i1,j)|2+12T1i=1T2+1j=1|Δ2u(i,j1)|2T1i=1T2j=1F((i,j),u(i,j))=12u2T1i=1T2j=1F((i,j),u(i,j)),uS, (2.1)

    where F((i,j),u)=u0f((i,j),τ)dτ. Note that f((i,j),u) is continuously differentiable with respect to u. It is clear that IC2(S,R) and solutions of the problem (1.1)–(1.2) are precisely the critical points of I(u). Moreover, for any u,vS, using the Dirichlet boundary conditions gives

    I(u),v=T2j=1T1+1i=1(Δ1u(i1,j)Δ1v(i1,j))+T1i=1T2+1j=1(Δ2u(i,j1)Δ2v(i1,j))T1i=1T2j=1(f((i,j),u(i,j))v(i,j))=T1i=1T2j=1{Δ21u(i1,j)+Δ22u(i,j1)+f((i,j),u(i,j))}v(i,j). (2.2)

    Let Ξ be the discrete Laplacian, which is defined by Ξu(i,j)=Δ21u(i1,j)+Δ22u(i,j1). According to the conclusion of [31], we know that Ξ is invertible and the distinct Dirichlet eigenvalues of Ξ on [1,T1]×[1,T2] can be denoted by 0<λ1<λ2λ3λT1T2. Let ϕk=(ϕk(1),ϕk(2),,ϕk(T1T2))tr, k[1,T1T2], where each ϕk is an eigenvector corresponding to the eigenvalue λk. Let W=span{ϕ1,...,ϕk1}, W0=span{ϕk}, W+=(WW0). Then, S can be expressed as

    S=WW+W0.

    For later use, we define another norm as u2=(T1i=1T2j=1|u(i,j)|2)12. Then, for any uS, we have that

    λ1u22u2λT1T2u22.

    Next, we recall some preliminaries with respect to Morse theory.

    We say that the functional I satisfies the Cerami condition ((C) for short) if any sequence {un}S satisfying I(un)c, (1+un)I(un)0 as n has a convergent subsequence. Note that if (C) is satisfied, then the deformation condition ((D) for short) is satisfied [32].

    Definition 2.1. [28,33] Let u0 be an isolated critical group of I with I(u0)=cR, and U be a neighborhood of u0. The group

    Cq(I,u0):=Hq(IcU,IcUu0),qZ

    is called the q-th critical group of I at u0. Let κ={uS|I(u)=0}. For all aR, each critical point of I is greater than a and IC2(S,R) satisfies (D). Then, the group

    Cq(I,):=Hq(S,Ia),qZ

    is called the q-th critical group of I at infinity.

    To calculate the critical group at infinity, we need the following auxiliary proposition.

    Proposition 2.1. [34]Suppose that S is a Hilbert space, {ItC2(S,R)|t[0,1]}. It and tIt are locally continuous. If I0 and I1 satisfy (C), and there exist aR and δ>0 such that

    It(u)a(1+u)It(u)δ,t[0,1],

    then

    Cq(I0,)=Cq(I1,),qZ. (2.3)

    In particular, if there is some R>0 such that

    inft[0,1],u>R(1+u)It(u)>0 (2.4)

    and

    inft[0,1],uR(1+u)It(u)>, (2.5)

    then Eq (2.3) is satisfied.

    The following three propositions are important in obtaining some nonzero critical points.

    Proposition 2.2. [26]Let S be a real Hilbert space, IC2(S,R). Suppose that u0 is the isolated critical point of I with limited Morse index μ(u0) and null dimension ν(u0). I(u0) is a Fredholm operator. Moreover, if u0 is the local minimizer of I, then

    Cq(I,u0)δq,0Z,q=0,1,2,.

    Proposition 2.3. [34]Assume that IC2(S,R) with S=S+S and 0 is the isolated critical point of I. If I has a local linking structure at 0 with k=dimS<, then

    Cq(I,0)δq,kZ,k=μ0ork=μ0+ν0.

    Proposition 2.4. [35]Let IC2(S,R) satisfy (D). Then,

    (J1) if Cq(I,)0 holds for some q, then I possesses a critical point u such that Cq(I,u)0;

    (J2) if 0 is the isolated critical point of I and there exists some q such that Cq(I,)Cq(I,0), then I has a nonzero critical point.

    In our proofs, we also require the following Mountain Pass Lemma.

    Proposition 2.5. [33]Let S be a real Banach space and IC1(S,R) satisfy the Palais–Smale condition ((PS) for short). Further, if I(0)=0 and

    (J3) there exist constants ρ, a>0 such thatI|Bρa,

    (J4) there is some eSBρ such that I(e)0.

    Then, I possesses a critical value ca given by

    c=infhΓsupx[0,1]I(h(x)),

    where

    Γ={hC([0,1],S)|h(0)=0,h(1)=e}.

    In this section, we state our main results and provide detailed proofs. First, the following assumptions are required:

    (f1) There exists k2 such that

    λklim inf|u|f((i,j),u)ulim sup|u|f((i,j),u)uλk+1,(i,j)Ω.

    (f2) There exists a subsequence {u(1)n,k}span{ϕk} such that u(1)n,kun1 as un. Then, there exist δ1, N1>0 such that

    T1i=1T2j=1(f((i,j),un(i,j))λkun(i,j))u(1)n,k(i,j)δ1,nN1,(i,j)Ω.

    (f3) There exists a subsequence {u(1)n,k+1}span{ϕk+1} such that u(1)n,k+1un1 as un. Then, there exist δ2, N2>0 such that

    T1i=1T2j=1(λk+1un(i,j)f((i,j),un(i,j)))u(1)n,k+1(i,j)δ2,nN2,(i,j)Ω.

    We are now in a position to state our main results.

    Theorem 3.1. Let (f1), (f2), (f3) hold. Moreover, for all (i,j)Ω,

    (V1)F((i,j),0)<λ1,

    (V2)F((i,j),u)>0 for all uR

    are satisfied. Then, problem (1.1)–(1.2) possesses at least three nontrivial solutions, among which one is positive and one is negative.

    Consider the following sign condition:

    (F0±) There exists δ>0 such that

    ±(2F((i,j),u)λmu2)0,|u(i,j)|δ,(i,j)Ω.

    Then, we can state the following theorem.

    Theorem 3.2. Suppose (f1), (f2), (f3) are satisfied. For all (i,j)Ω, let

    (V3)f((i,j),0)=λm,

    (V4) there exists u00 such that f((i,j),u0)=0.

    Then, problem (1.1)–(1.2) possesses at least four nontrivial solutions if one of the following conditions is met:

    (i)(F0+) with 2mk,     (ii)(F0) with 2<mk+1.

    Theorem 3.3. Assume (f1), (f2), (f3), and (V3) are true. Further, if k=1 and either

    (iii)(F0+) with m1,     (iv)(F0) with m2,

    then problem (1.1)–(1.2) admits at least two nontrivial solutions.

    To prove our results, (C) is necessary. First, we present a detailed proof to show that I satisfies (C).

    Lemma 3.1. Let (f1), (f2), and (f3) hold. Then, I satisfies (C).

    Proof. Suppose that {un}S and there exists a constant c such that

    I(un)c,(1+un)I(un)0,asn.

    Because S is a T1T2-dimensional space, it suffices to show that {un} is bounded. Otherwise, we can assume that

    un,asn.

    Denote ¯un=unun. Then, ¯un=1, which means that {¯un} has some subsequences. Without loss of generality, we set the subsequence to be the sequence. Moreover, there exists ¯uS with ¯u=1 such that

    ¯un¯u,asn.

    For any φS, we obtain

    I(un),φun=¯un,φT1i=1T2j=1(f((i,j),un(i,j))un,φ(i,j)).

    From (f1), there exist bλk+1 and N>0 such that

    |f((i,j),un(i,j))|b(1+|un(i,j)|),uS,n>N,(i,j)Ω.

    Set b1=bun such that, for n>N, we have

    |f((i,j),un(i,j))|unb1(1+|¯un(i,j)|),uS,(i,j)Ω. (3.1)

    If nN, then because f((i,j),) is continuous in , we have

    |f((i,j),un(i,j))|unmax{|f((i,j),un(i,j))|un},uS,(i,j)Ω. (3.2)

    Therefore, Eqs (3.1) and (3.2) ensure that {f((i,j),un)un} is bounded. Consequently, {f((i,j),un)un} has a convergent subsequence. We still denote this by {f((i,j),un)un}. Using (f1) once more, we can assume that there exists some p satisfying λkpλk+1 such that

    f((i,j),un)unp¯u,asn.

    Hence,

    21¯u(i1,j)+22¯u(i,j1)+p¯u=0,(i,j)Ω,

    which means that ¯u is the nontrivial solution of

    21u(i1,j)+22u(i,j1)+pu=0,(i,j)Ω

    with boundary conditions

    ¯u(i,0)=¯u(i,T2+1)=0,iZ(1,T1),¯u(0,j)=¯u(T1+1,j)=0,jZ(1,T2).

    Together with the maximum principle and unique continuation property, this implies that pλk or pλk+1 for λkpλk+1.

    If pλk, then ¯uW0 and

    u(1)n,kun1,n.

    In fact, if ¯uW0, then p¯u=0, which leads to ¯u=0. In view of ¯u=1, this is a contradiction. Therefore, as n, we have that

    T1i=1T2j=1(f((i,j),un(i,j))λkun(i,j))u(1)n,k(i,j)=I(un),u(1)n,kunI(un)0. (3.3)

    Obviously, Eq (3.3) is inconsistent with (f2).

    If pλk+1, then ¯uspan{ϕk+1} and

    u(1)n,k+1un1,n.

    Thus,

    T1i=1T2j=1(λk+1un(i,j)f((i,j),un(i,j)))u(1)n,k+1(i,j)=I(un),u(1)n,k+1unI(un)0

    as n, which contradicts (f3). Therefore, {un} is bounded.

    To calculate critical groups at infinity, we have the following lemma.

    Lemma 3.2. Let μ=dim(W0W). If (f1), (f2), and (f3) are satisfied, then Cq(I,)δq,μZ.

    Proof. First, for t[0,1], let It:SR be given as

    It(u)=12T1+1i=1T2j=1|1u(i1,j)|2+12T1i=1T2+1j=1|2u(i,j1)|21t4λkT1i=1T2j=1|u(i,j)|21t4λk+1T1i=1T2j=1|u(i,j)|2tT1i=1T2j=1F((i,j),u(i,j))=12u21t4(λk+λk+1)u22tT1i=1T2j=1F((i,j),u(i,j)).

    In the following, we prove that Eq (2.4) in Proposition 2.1 is true. Otherwise, there exists {un}S, tn[0,1] such that

    un,(1+un)It(un)0,n.

    Denote ¯un=unun. Then, ¯un=1 and there exists ¯uS satisfying ¯u=1 such that ¯un¯u as n. From Lemma 3.1, we know that

    f((i,j),un)unp¯u,n.

    Suppose that tnt0, whereby is easy to show that ¯u is a solution subject to

    {21¯u(i1,j)+22¯u(i,j1)ξ(t0)¯u=0,(i,j)Ω,¯u(i,0)=¯u(i,T2+1)=0iZ(1,T1),¯u(0,j)=¯u(T1+1,j)=0jZ(1,T2),

    where ξ(t0)=1t02λk+1t02λk+1+t0p and λkξ(t0)λk+1. By the maximum principle and unique continuation property, we find that

    ξ(t0)λkorξ(t0)λk+1.

    Furthermore, t0=1 means that tn1 as n. Therefore,

    u(1)n,kun1oru(1)n,k+1un1.

    Note that (f2) and (f3) are valid, and so

    T1i=1T2j=1(f((i,j),un(i,j))λkun(i,j))u(1)n,k(i,j)δ1>0,n>N1

    or

    T1i=1T2j=1(λk+1un(i,j)f((i,j),un(i,j)))u(1)n,k+1(i,j)δ2>0,n>N2.

    Considering λkλk+1, we have

    T1i=1T2j=1(f((i,j),un)λkun(i,j))u(1)n,k(i,j)+1tn2tn(λkλk+1)u(1)n,k20,n,

    which implies that

    T1i=1T2j=1(f((i,j),un)λkun(i,j))u(1)n,k(i,j)+o(1)=1tn2tn(λkλk+1)u(1)n,k20,n.

    Therefore,

    lim supnT1i=1T2j=1(f((i,j),un)λkun(i,j))u(1)n,k(i,j)0

    or

    lim supnT1i=1T2j=1(λk+1un(i,j)f((i,j),un))u(1)n,k+1(i,j)0,

    which guarantees that Eq (2.4) is satisfied.

    It is easy to show that Eq (2.5) is satisfied and I0, I1 satisfy (C). In fact, because

    I0(u)=12u214(λk+λk+1)u22,uS,

    we have that

    21¯u(i1,j)+22¯u(i,j1)+12(λk+λk+1)¯u2=0,(i,j)Ω,

    which is impossible for ¯u=1. Consequently, I0 satisfies (C). Moreover,

    tIt=14(λk+λk+1)u2T1i=1T2j=1F((i,j),u(i,j))

    is continuous. According to Proposition 2.1, we have that

    Cq(I,)=Cq(I1,)Cq(I0,),qZ.

    Note that u=0 is a unique nondegenerate critical point of I0 with μ0=dim(W0W). Thus,

    Cq(I0,)Cq(I0,0)δq,μZ,μ=μ0=dim(W0W).

    Further, we have that Cq(I,)δq,μZ.

    To gain some mountain pass-type critical points by applying the cut-technique, we verify the following compactness conditions.

    Lemma 3.3. Let

    f+((i,j),u)={f((i,j),u),u0,0,u<0, (3.4)

    such that

    λklim infu+f+((i,j),u)ulim supu+f+((i,j),u)uλk+1.

    Then, the functional I+:SR defined by

    I+(u)=12u2T1i=1T2j=1F+((i,j),u(i,j))

    satisfies (PS), where F+((i,j),u)=u0f+((i,j),τ)dτ.

    Proof. Let {un}S be a (PS)c sequence, that is,

    I+(un)c,I+(un)0,n.

    Similar to Lemma 3.1, assume that {un} is unbounded. Then, we have

    |un(i,j)|,n,(i,j)Ω. (3.5)

    Denote vn=unun, so that vn=1. As S is a T1T2-dimensional Hilbert space, there exists vS satisfying v=1 such that vnv as n. Hence, for any φS, it holds that

    I+(un),φun=vn,φT1i=1T2j=1(f+((i,j),un(i,j))un,φ(i,j)). (3.6)

    For any (i,j)Ω, write v+(i,j)=max{v(i,j),0}. Then, there exists α satisfying λkαλk+1 such that

    limnf+((i,j),un(i,j))un=limnf+((i,j),un(i,j))un(i,j)vn(i,j)=αv+(i,j).

    Together with Eq (3.6), this yields

    T1i=1T2j=1{Δ21v(i1,j)+Δ22v(i,j1)+αv+(i,j)}φ(i,j)=0,n,

    which implies that v is the nontrivial solution of

    Δ21v(i1,j)+Δ22v(i,j1)+αv+(i,j)=0,(i,j)Ω (3.7)

    with boundary conditions

    v(i,0)=v(i,T2+1)=0,iZ(1,T1),v(0,j)=v(T1+1,j)=0,jZ(1,T2). (3.8)

    Denote v(i0,j0):=min{v(i,j)|(i,j)Ω}. We aim to show that v(i0,j0)>0. Otherwise, αv+(i,j)=0 and

    Δ21v(i01,j0)+Δ22v(i0,j01)=0,(i,j)Ω,

    which means that Δ1v(i01,j0)=Δ2v(i0,j01)=0. Moreover, v(i01,j0)=v(i0,j0)=v(i0,j01). Therefore, v0 for all (i,j)Ω. Additionally, v is the nontrivial solution of Eqs (3.7)–(3.8), which implies that v(i0,j0)>0, and so v(i,j)>0. Recall that λkαλk+1, we so we have that {un} is bounded.

    In the same manner as Lemma 3.3, we can state the following.

    Lemma 3.4. If

    λklim infuf((i,j),u)ulim supuf((i,j),u)uλk+1,

    where

    f((i,j),u)={f((i,j),u),u0,0,u>0, (3.9)

    then the functional I:SR defined by

    I(u)=12u2T1i=1T2j=1F((i,j),u(i,j))

    satisfies (PS), where F((i,j),u)=u0f((i,j),τ)dτ.

    Lemma 3.5. If all conditions of Theorem 3.1 are fulfilled, then I+ has a critical point u+>0 such that Cq(I+,u+)δq,1Z and I has a critical point u<0 such that Cq(I,u)δq,1Z.

    Proof. We prove the case of I+; the proof of I is similar and is omitted for brevity. With the aid of Proposition 2.5, we need only prove that I+ satisfies (J3), (J4). According to (V1), there exist ρ, ρ1>0 such that F((i,j),0)<ρ1<λ1 and

    F((i,j),u)12ρu2,|u(i,j)|ρ.

    Then, for any (i,j)Ω and uS with uλ1ρ1, we have

    I+(u)=12u2T1i=1T2j=1F+((i,j),u(i,j))=12u2uUF((i,j),u(i,j))12u212ρ1uU|u(i,j)|212u212ρ1T1i=1T2j=1|u(i,j)|212u212ρ1λ1u2>0,

    with U={(i,j)Ω|u(i,j)0}, which ensures (J3) is valid.

    Using (f1), there exist γ>λk1(λ1), b2R such that

    F((i,j),u)γ2u2+b2,uR.

    For t>ρ, choose espan{ϕ1}. Then, we have

    I+(te)=12te2T1i=1T2j=1F((i,j),te(i,j))12te2γte222b2T1T2=12te2γte22λ1b2T1T2=t22(1γλ1)e2b2T1T20.

    Therefore, (J4) is satisfied.

    By the Mountain Pass Lemma, I+ possesses a critical point u+0. Moreover, there exists a sequence {u+n} such that I+(u+n)I+(u+) as n. For any φS, we have

    I+(u+n),φ=u+n,φT1i=1T2j=1(f+((i,j),u+n(i,j)),φ(i,j)).

    Letting n,

    limnf+((i,j),u+n(i,j))=limnf+((i,j),u+n(i,j))u+n(i,j)u+n(i,j)=αu+(i,j),(i,j)Ω,

    which leads to

    Δ21u+(i1,j)+Δ22u+(i,j1)+αu+(i,j)=0

    with λkαλk+1 and u+(i,j):=max{u(i,j),0}. Similar to the proof of Lemma 3.3, u+ is positive.

    We now calculate Cq(I+,u+). Recalling Eq (2.1), we find that

    I(u+)v,v0,vS

    and there exists some v00 such that

    I(u+)v0,v=0,vS,

    which implies that v0 is a solution of

    {Δ21v0(i1,j)+Δ22v0(i,j1)F((i,j),u+)v0(i,j)=0,(i,j)Ω,v0(i,0)=v0(i,T2+1)=0,iZ(1,T1),v0(0,j)=v0(T1+1,j)=0,jZ(1,T2).

    Combining this with (V2), it follows that

    {Δ21v(i1,j)+Δ22v(i,j1)λF((i,j),u+)v(i,j)=0,(i,j)Ω,v(i,0)=v(i,T2+1)=0,iZ(1,T1),v(0,j)=v(T1+1,j)=0,jZ(1,T2)

    admits an eigenvalue λ=1 such that dimker(I(u+))=1. Hence, we conclude that Cq(I,u+)Cq(I+,u+)δq,1Z. This completes the proof.

    It is time for us to give the detailed proof of Theorem 3.1 using Proposition 2.4.

    Proof of Theorem 3.1 Given I(0)=0 and (V1), then

    I(0)u,u(1F((i,j),0)λ1)u2,(i,j)Ω,

    which implies that 0 is the local minimizer of I. From Proposition 2.2, we have

    Cq(I,0)δq,0Z,qZ. (3.10)

    Furthermore, Lemma 3.2 ensures that

    Cq(I,)δq,μZ,qZ.

    Then, according to Proposition 2.4, there exists some critical point u10 of I such that

    Cμ(I,u1)0. (3.11)

    Consequently, we conclude that u+, u and u1 are nontrivial critical points of I with u+>0 and u<0. The proof of Theorem 3.1 is achieved.

    Before verifying Theorem 3.2 using Proposition 2.3, we need the following lemma about local linking.

    Lemma 3.6. Let (V3) and (F0+) (or (F0)) hold. Then, I has a local linking at 0 with respect to

    S=SS+,

    where S=span{ϕ1,ϕ2,...,ϕm} (or S=span{ϕ1,ϕ2,...,ϕm1}).

    Proof. Suppose that (F0+) is satisfied. Then, there exists δ>0 such that |u(i,j)|δ, uδλT1T2, and

    F((i,j),u)12λmu2.

    For uS with 0<uδλT1T2, we have

    I(u)=12u2T1i=1T2j=1F((i,j),u(i,j))12u212λmu2212(1λmλm)u2=0. (3.12)

    For uS+ with 0<u<δλT1T2, we obtain

    I(u)12u212λmλm+1u2=12(1λmλm+1)u2>0. (3.13)

    Obviously, I(0)=0. Combining this with Eqs (3.12) and (3.13), it is clear that I has a local linking at 0.

    Proof of Theorem 3.2 Denote μ0=dimspan{ϕ1,...,ϕm1}, ν0=dimspan{ϕm}. Let (V3) be true. Then, 0 is degenerate. Taking account of Proposition 2.3, we find that

    Cq(I,0)δq,μ0+ν0Z,qZ.

    In contrast, Lemma 3.2 gives Cq(I,)δq,μZ. Therefore, Proposition 2.4 guarantees the existence of u such that

    Cμ(I,u)0.

    Moreover, mk implies that μ0+ν0μ. Thus, u0.

    From (V4), there exists u00 such that f((i,j),u0)=0. Without loss of generality, we can assume that u0>0. In the sequence, we intend to obtain the local minimizer of I. Define

    ˜f((i,j),u)={0,u<0,f((i,j),u),u[0,u0],0,u>u0, (3.14)

    and let

    ˜I(u)=12u2T1i=1T2j=1˜F((i,j),u(i,j)),uS,

    where ˜F((i,j),u)=u0˜f((i,j),τ)dτ. Then, ˜I is coercive and continuous. Therefore, there exists a minimizer ˜u0 of ˜I. From the maximum principle, we deduce that ˜u0=0 or 0<˜u0(i,j)<u0 for all (i,j)Ω. Moreover, (V3) means that 0 is not a minimizer. Consequently, ˜u00 is a local minimizer of I and Cq(I,˜u0)δq,0Z, qZ.

    Denote ˆF((i,j),u)=u0ˆf((i,j),τ)dτ, where

    ˆf((i,j),v)=f((i,j),v+˜u0)f((i,j),˜u0),(i,j)Ω,vS.

    The corresponding functional is then given by

    ˆI(v)=12v2T1i=1T2j=1ˆF((i,j),v(i,j)),vS.

    If v is a nonzero critical point of ˆI, then ˜u0+v is a critical point of I with

    Cq(ˆI,v)=Cq(I,˜u0+v).

    Define

    ˆf+((i,j),v)={ˆf((i,j),v),v0,0,v<0, (3.15)

    and construct the corresponding functional as

    ˆI+(v)=12v2T1i=1T2j=1ˆF+((i,j),v(i,j)),vS,

    with ˆF+((i,j),u)=u0ˆf+((i,j),τ)dτ. Then, ˜u0 is a local minimizer of I for ˆI+ satisfying (PS), which leads to v=0 being a local minimizer of ˆI+. Thus, (J3) is fulfilled. Applying (f1) yields

    ˆI+(te)0,t+,

    which ensures that (J4) is satisfied. By Lemma 2.5, ˆI+ possesses a critical point v+>0. Furthermore, ˆI possesses a critical point v+ with Cq(ˆI,v+)δq,1Z, qZ. As a result, u+=˜u0+v+ is a critical point of I satisfying Cq(I,u+)δq,1Z, qZ. Similarly, u<˜u0 is a critical point of I satisfying Cq(I,u)δq,1Z. Therefore, u, ˜u0, u± are four nontrivial solutions of I and ˜u0, u+ are positive.

    For the case u0<0, repeating the above steps shows that I has four nontrivial solutions, among which there are two negative solutions. Therefore, I admits four nontrivial solutions and the proof is finished.

    Proof of Theorem 3.3 Similar to the proof of Theorem 3.2, we find that Cq(I,0)δq,μ0+ν0Z, qZ. Because k=1, Cq(I,)δq,1Z and there exists some critical point ¯u such that C1(I,¯u)0. Moreover, u is the critical point of I satisfying Cq(I,¯u)δq,1Z. From m1, we conclude that ¯u0. If κ={0,¯u}, the Morse equality implies that

    (1)μ0+ν0+(1)1=(1)1. (3.16)

    Of course, Eq (3.16) is impossible. Therefore, problem (1.1)–(1.2) possesses at least two nontrivial solutions.

    Finally, we present two examples to verify the feasibility of our results.

    Example 4.1. Take T1=3, T2=2, and consider

    Δ21u(i1,j)+Δ22u(i,j1)+(λ12λk+λk+12)u(i,j)1+[u(i,j)]2+λk+λk+12u(i,j)=0 (4.1)

    with the boundary value conditions of Eq (1.2).

    Because f((i,j),u)=(λ12λk+λk+12)u(i,j)1+[u(i,j)]2+λk+λk+12u(i,j), it follows that

    F((i,j),u)=λk+λk+12u4+(32(λk+λk+1)λ12)u2+λ12>0.

    It is not difficult to verify that f((i,j),0)=0, F((i,j),0)=λ12<λ1, and

    λklim|u|f((i,j),u)u=λk+λk+12λk+1,

    which means that (f1), (V1) and (V2) are satisfied. If u(1)n,kun1 as un, then we obtain

    3i=12j=1((λ12λk+λk+12)un(i,j)1+[un(i,j)]2+λk+λk+12un(i,j)λkun(i,j))u(1)n,k(i,j)=3i=12j=1((λ12λk+λk+12)un(i,j)1+[un(i,j)]2+λk+1λk2un(i,j))u(1)n,k(i,j)=3i=12j=1[λ12λk+λk+121+1[u(1)n,k(i,j)]2+λk+1λk2[u(1)n,k(i,j)]2]+.

    Therefore, (f2) is satisfied. Similarly, (f3) is valid. Therefore, Theorem 3.1 guarantees that problem (4.1)–(1.2) admits at least three nontrivial solutions, of which one is positive and one is negative.

    Example 4.2. Take T1=3, T2=2, and consider

    Δ21u(i1,j)+Δ22u(i,j1)+2(λmλk+λk+12)u2u2+λk+λk+12u=0 (4.2)

    with the boundary value conditions of Eq (1.2).

    Denote

    f((i,j),u)=2(λmλk+λk+12)u2u2+λk+λk+12u.

    Then, f((i,j),0)=0 and there exists u=±4λmλk+λk+10 such that f((i,j),u)=0, which means that (V4) is satisfied.

    A direct computation yields

    F((i,j),u)=(λk+λk+12λm)ln(2u2)+λk+λk+14u2

    and

    f((i,j),u)=2(λmλk+λk+12)(2+u2)(2u2)2+λk+λk+12,

    and so f((i,j),0)=λm and (V3) is valid.

    As un, if u(1)n,kun1, then

    3i=12j=1(2(λmλk+λk+12)un(i,j)2[un(i,j)]2+λk+λk+12un(i,j)λkun(i,j))u(1)n,k(i,j)=3i=12j=1(2(λmλk+λk+12)un(i,j)1+[un(i,j)]2+λk+1λk2un(i,j))u(1)n,k(i,j)=3i=12j=1[2λm(λk+λk+1)1+1[u(1)n,k(i,j)]2+λk+1λk2[u(1)n,k(i,j)]2]+.

    Therefore, (f2) is satisfied. Similar to (f2), we can show that (f3) is satisfied.

    In the following, we verify (f1) and (F0+). If we write

    A=(410100141010014001100410010141001014),

    then A is positive-definite and the eigenvalues of A are

    λ1=32,λ2=3,λ3=52,λ4=3+2,λ5=5,λ6=5+2.

    Let m=2, k=3. Then, 52lim inf|u|f((i,j),u)u=lim sup|u|f((i,j),u)u=43+2, which means that (f1) is satisfied. Further, there exists δ>0 such that, when |u(i,j)|δ, the following holds for any (i,j)Z(1,3)×Z(1,2):

    2F((i,j),u)3u2=2ln(2u2)+u20.

    In fact, for any (i,j)Ω, we can choose δ=1>0, and then 0<[u(i,j)]21 for 0<|u(i,j)|1. This means that ln(2u2)0 and 2F((i,j),u)3u20. Thus, (F0+) holds and all conditions of Theorem 3.2 are satisfied. Consequently, Theorem 3.2 ensures that (4.2)–(1.2) admits at least four nontrivial solutions.

    More clearly, using Matlab, we find that problem (4.2)–(1.2) has 36 nontrivial solutions. Some examples of these solutions are as follows: (0.5061,0.6548,0.5061,0.5061,0.6548,0.5061), (0.5061,0.6548,0.5061,0.5061,0.6548,0.5061), (1.9858,2.1015,0.6084,1.9858,2.1015,0.6084), and (1.2137,4.7492,1.2137,13.9612,1.5032,13.9612).

    This paper is supported by the National Natural Science Foundation of China (NSFC) (Grant No. 11971126). The authors wish to thank the handling editor and the anonymous referees for their valuable comments and suggestions.

    All authors declare no conflicts of interest in this paper.



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