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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

    (3.8)

    Denote . We aim to show that . Otherwise, and

    which means that . Moreover, . Therefore, for all . Additionally, is the nontrivial solution of Eqs (3.7)–(3.8), which implies that , and so . Recall that , we so we have that is bounded.

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

    Lemma 3.4. If

    where

    (3.9)

    then the functional defined by

    satisfies , where .

    Lemma 3.5. If all conditions of Theorem 3.1 are fulfilled, then has a critical point such that and has a critical point such that .

    Proof. We prove the case of ; the proof of is similar and is omitted for brevity. With the aid of Proposition 2.5, we need only prove that satisfies , . According to , there exist , such that and

    Then, for any and with , we have

    with , which ensures is valid.

    Using , there exist , such that

    For , choose . Then, we have

    Therefore, is satisfied.

    By the Mountain Pass Lemma, possesses a critical point . Moreover, there exists a sequence such that as . For any , we have

    Letting ,

    which leads to

    with and . Similar to the proof of Lemma 3.3, is positive.

    We now calculate . Recalling Eq (2.1), we find that

    and there exists some such that

    which implies that is a solution of

    Combining this with , it follows that

    admits an eigenvalue such that . Hence, we conclude that . 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 and , then

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

    (3.10)

    Furthermore, Lemma 3.2 ensures that

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

    (3.11)

    Consequently, we conclude that , and are nontrivial critical points of with and . 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 and (or ) hold. Then, has a local linking at 0 with respect to

    where (or ).

    Proof. Suppose that is satisfied. Then, there exists such that , , and

    For with , we have

    (3.12)

    For with , we obtain

    (3.13)

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

    Proof of Theorem 3.2 Denote , . Let be true. Then, 0 is degenerate. Taking account of Proposition 2.3, we find that

    In contrast, Lemma 3.2 gives . Therefore, Proposition 2.4 guarantees the existence of such that

    Moreover, implies that . Thus, .

    From , there exists such that . Without loss of generality, we can assume that . In the sequence, we intend to obtain the local minimizer of . Define

    (3.14)

    and let

    where . Then, is coercive and continuous. Therefore, there exists a minimizer of . From the maximum principle, we deduce that or for all . Moreover, means that 0 is not a minimizer. Consequently, is a local minimizer of and , .

    Denote , where

    The corresponding functional is then given by

    If is a nonzero critical point of , then is a critical point of with

    Define

    (3.15)

    and construct the corresponding functional as

    with . Then, is a local minimizer of for satisfying , which leads to being a local minimizer of . Thus, is fulfilled. Applying yields

    which ensures that is satisfied. By Lemma 2.5, possesses a critical point . Furthermore, possesses a critical point with , . As a result, is a critical point of satisfying , . Similarly, is a critical point of satisfying . Therefore, , , are four nontrivial solutions of and , are positive.

    For the case , repeating the above steps shows that has four nontrivial solutions, among which there are two negative solutions. Therefore, 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 , . Because , and there exists some critical point such that . Moreover, is the critical point of satisfying . From , we conclude that . If , the Morse equality implies that

    (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 , , and consider

    (4.1)

    with the boundary value conditions of Eq (1.2).

    Because , it follows that

    It is not difficult to verify that , , and

    which means that , and are satisfied. If as , then we obtain

    Therefore, is satisfied. Similarly, 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 , , and consider

    (4.2)

    with the boundary value conditions of Eq (1.2).

    Denote

    Then, and there exists such that , which means that is satisfied.

    A direct computation yields

    and

    and so and is valid.

    As , if , then

    Therefore, is satisfied. Similar to , we can show that is satisfied.

    In the following, we verify and . If we write

    then is positive-definite and the eigenvalues of are

    Let , . Then, , which means that is satisfied. Further, there exists such that, when , the following holds for any :

    In fact, for any , we can choose , and then for . This means that and . Thus, 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: , , , and .

    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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