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Existence and nonexistence of positive solutions to a class of nonlocal discrete Kirchhoff type equations

  • In this paper, we investigate the existence and nonexistence of positive solutions to a class of nonlocal partial difference equations via a variant version of the mountain pass theorem. The conditions in our obtained results release the classical (AR) condition in some sense.

    Citation: Yuhua Long. Existence and nonexistence of positive solutions to a class of nonlocal discrete Kirchhoff type equations[J]. AIMS Mathematics, 2023, 8(10): 24568-24589. doi: 10.3934/math.20231253

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  • In this paper, we investigate the existence and nonexistence of positive solutions to a class of nonlocal partial difference equations via a variant version of the mountain pass theorem. The conditions in our obtained results release the classical (AR) condition in some sense.



    Consider existence and nonexistence of positive solutions to the following type of nonlocal discrete Kirchhoff equation:

    [a+b(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)](Δ21x(i1,j)+Δ22x(i,j1))=f((i,j),x(i,j)),(i,j)[1,m]×[1,n], (1.1)

    subject to Dirichlet boundary conditions

    x(i,0)=x(i,n+1)=0,i[0,m+1],x(0,j)=x(m+1,j)=0,j[0,n+1], (1.2)

    where, given constants a,b>0 and m,n>0 are integers. For integers , let [,]={,+1,,} denote a discrete segment. Forward difference operators Δ1x(i,j)=x(i+1,j)x(i,j), Δ2x(i,j)=x(i,j+1)x(i,j) and Δ2x(i,j)=Δ(Δx(i,j)). R+ denotes the set of all nonnegative real numbers and the nonlinearity f((i,j),x) fulfills:

    (H1) f:[0,m+1]×[0,n+1]×RR+ is continuous in x. If x0, then f((i,j),x)0 for all (i,j)[0,m+1]×[0,n+1];

    (H2) for (i,j)[1,m]×[1,n], f((i,j),x)x3 is nondecreasing with respect to x0.

    Notice that (1.1) with Dirichlet boundary conditions (1.2) is usually taken in regard to the discrete analogue of the following Kirchhoff type problem:

    {(a+bΩ|u|2)u=f(x,u),inΩ,u=0,onΩ. (1.3)

    Owing to taking into account the effects of the changes in the length of a string during vibrations, (1.3) is an extension of the classical d'Alembert's wave equations [1]. Kirchhoff type equation (1.3) concerns not only the non-Newton mechanics, but also the physical laws of the universe, population dynamics models, the problem of plasma and so on. Consequently, it has captured keen research interest and there are many papers that have emerged. For example, Perera and Zhang [2] achieved a nontrivial solution by combining a critical group with the Yang index. Both in [3] and [4], the authors displayed results on multiple solutions including sign-changing solutions. Without the Ambrosetti-Rabinowitz condition, ground state solutions to the N-Kirchhoff equation was studied in [5]. For more interesting results, we refer the reader to [6,7] and references therein.

    It is well known that difference equation models are established in lots of areas, for instance, mechanical engineering, neural networks, biology, computer science and so on. For example, using difference equations, a two-patch SIR disease model was established in [8]. The authors studied the interaction between wild and sterile mosquitoes by a difference equation model in [9]. Because of wide applications, difference equations have been investigated extensively and many results have been achieved. Here we mention a few. Yu, Guo and Zuo [10] considered periodic solutions of second order self-adjoint difference equations, Zhou and Ling [11] presented results on positive solutions to a discrete two-point boundary value problem, and Kuang and Guo [12] dealt with heteroclinic solutions for p-Laplacian difference equations with a parameter and Nastasi, Tersian and Vetro [13] gave results on the existence of at least two non-zero homoclinic solutions without using Ambrosetti-Rabinowitz type-conditions.

    As pointed out in [14], partial difference equations, involving two or more discrete variables, have been used in recent investigations related to digital control systems, image processing, neural networks, population models and social behaviors. Recently, many authors turned their interest towards study of them. For example, Long and Zhang [15,16] achieved multiple solutions of second order partial difference equations via Morse theory. Meanwhile, results on periodic solutions of partial difference equations via critical point theorems were presented in [17,18,19].

    The nonlocal discrete Kirchhoff type equation (1.1), a basic nonlinear partial difference equation, not only contains bivariate sequences with two independent integer variables, but also involves the discrete Kirchhoff term

    b(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)(Δ21x(i1,j)+Δ22x(i,j1)).

    Thus, it is more difficult and interesting to study. Recently, based on critical point theory and variational methods, the authors [20] obtained the existence of at least three solutions. We move our attention to (1.1) and obtain some results. For example, we obtained sign-changing solutions in [21] and displayed results on infinitely many solutions in [22,23]. Also, in [24], we studied nontrivial solutions via Morse theory. Meanwhile, it is well known that positive solutions play an important role in research, there seems few results concerned with positive solutions of (1.1). Moreover, above mentioned results indicate that critical point theory is a strong candidate for study of (1.1). Consequently, in this paper, we manage to deal with the existence and nonexistence of positive solutions of (1.1) by employing variational methods together with a variant version of the mountain pass theorem, which can be found in [25].

    We arrange this paper as follows. In Section 2, we provide preliminaries and display our main results. We prove our main results at length in Section 3.

    Let the set of all bivariate sequences be denoted by

    S={x={x(i,j)}:x(i,j)R,(i,j)Z×Z}.

    For any x,yS, ı,ȷR, define ıx+ȷy={ıx(i,j)+ȷy(i,j)}. Then, S is a vector space. Define the subset X, an mn-dimensional Hilbert space, of S as

    X={xS:x(i,0)=x(i,n+1)=0fori[0,m+1],x(0,j)=x(m+1,j)=0forj[0,n+1]}.

    For any x,yX, endow with the inner product , on E as

    x,y=nj=1m+1i=1(Δ1x(i1,j)Δ1y(i1,j))+mi=1n+1j=1(Δ2x(i,j1)Δ2y(i,j1)), (2.1)

    which implies that the norm induced by (2.1) is

    x=x,x=(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)1/2,xX.

    For later use, we denote an mn-dimensional Hilbert space E, which is equipped with usual norm || and inner product (,), respectively. Then, E is isomorphic to X. Throughout this paper, xX is regarded as an extension of xE when it is necessary. In what follows, for 1s<+, let

    Ls{xS:xLs=(mi=1nj=1|x(i,j)|s)1s<+}

    and

    xL=sup(i,j)[1,m]×[1,n]|x(i,j)|<+.

    Then,

    xLsηsx,xX, (2.2)

    where ηs is the best constant for the embedded of X in Ls.

    For convenience, we give some notations. Denote the well-known discrete Laplacian acting on a function x(i,j):[0,1+m]×[0,1+n] by Ξx(i,j)=Δ21x(i1,j)+Δ22x(i,j1). From [26], we get that the distinct Dirichlet eigenvalues of the invertible operator Ξ on [1,m]×[1,n] can be expressed as 0<λ1<λ2λ3λmn. Specifically,

    λ1x22x2λmnx22. (2.3)

    Consider the following eigenvalue problem:

    {x2Ξ=μx3(i,j),[i,j][1,m]×[1,n]x(i,0)=x(i,n+1)=x(0,j)=x(m+1,j)=0,i[0,m+1],j[0,n+1]. (2.4)

    Denote the minimum eigenvalue and the maximum eigenvalue of (2.4) by μ1 and μmax, respectively. In the same manner as [22], we get that (2.4) has finitely many eigenvalues which all belong to [λ21,mnλ2mn]. Clearly, μ1>0.

    Write F((i,j),x)=x0f((i,j),τ)dτ. Consider the functional J:ER as the following:

    J(x)=a2(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)+b4(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)2mi=1nj=1F((i,j),x(i,j))=a2x2+b4x4mi=1nj=1F((i,j),x(i,j)), (2.5)

    then the continuity of f guarantees that JC1(E,R).

    Denote

    Φ(x)=a2(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)+b4(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)2

    and

    Ψ(x)=mi=1nj=1F((i,j),x(i,j)),

    then J(x)=Φ(x)Ψ(x). For each x,zE, we have

    Ψ(x),z=limτ0Ψ(x+τz)Ψ(x)τ=mi=1nj=1(f((i,j),x(i,j))z(i,j)). (2.6)

    Moreover, using Dirichlet boundary conditions, there holds

    Φ(x),z=limτ0Φ(x+τz)Φ(x)τ=a(nj=1m+1i=1(Δ1x(i1,j)Δ1z(i1,j))+mi=1n+1j=1(Δ2x(i,j1)Δ2z(i1,j)))+b(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)(nj=1m+1i=1(Δ1x(i1,j)Δ1z(i1,j))+mi=1n+1j=1(Δ2x(i,j1)Δ2z(i1,j)))=a(nj=1mi=1(Δ1x(i1,j)Δ1z(i1,j))nj=1Δ1x(i,j)z(m,j)+mi=1nj=1(Δ2x(i,j1)Δ2z(i1,j))mi=1Δ2x(i,j)z(i,n))+b(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)(nj=1mi=1(Δ1x(i1,j)Δ1z(i1,j))nj=1Δ1x(i,j)z(m,j)+mi=1nj=1(Δ2x(i,j1)Δ2z(i1,j))mi=1Δ2x(i,j)z(i,n))=[a+b(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)](nj=1mi=1(Δ1x(i1,j)Δ1z(i1,j))+mi=1nj=1(Δ2x(i,j1)Δ2z(i1,j))). (2.7)

    Recall the definition of , and joint (2.6) with (2.7), it follows that

    J(x),z=Ψ(x)Φ(x),z=(a+bx2)(nj=1mi=1(Δ1x(i1,j)Δ1z(i1,j))+mi=1nj=1(Δ2x(i,j1)Δ2z(i1,j)))mi=1nj=1(f((i,j),x(i,j))z(i,j)). (2.8)

    Accordingly, J(x),z=0 is equivalent to nj=1mi=1((Δ1x(i1,j)Δ1z(i1,j))+(Δ2x(i,j1)Δ2z(i1,j))(f((i,j),x(i,j))z(i,j)))=0. Since z is arbitrary, the critical point of J is just the solution of (1.1) with Dirichlet boundary conditions (1.2). Namely, to seek solutions of (1.1) with Dirichlet boundary conditions (1.2), it is equivalent to look for critical points of the functional J on E. Further, the assumption (H1) and the strong maximum principle guarantee that nontrivial critical points of J on E are actually positive solutions of (1.1) with Dirichlet boundary conditions (1.2).

    Throughout this paper, we denote a universal constant by c unless specified otherwise. To seek critical points of the functional (2.5), we recall the concept of the Cerami condition at level c ((C)c for short), which is a weak version of the Palais-Smale condition ((PS) for short) and introduced by Cerami [27], as well as a variant version of the mountain pass theorem, which plays an important role in proofs of our main results.

    Definition 2.1. Let J(x)C1(E,R). If any sequence {xκ}E satisfying

    {J(xκ)}cand(1+xκ)J(xκ)0,asκ+

    possesses a convergent subsequence in E, then J satisfies (C)c. For all cR, if J(x) satisfies (C)c, then J(x) is called satisfying the (C).

    Proposition 2.1. [25] Let JC1(E,R). Assume that

    max{J(0),J(x1)}α<βinfx=ρJ(x)

    for some α<β, ρ>0 and x1E with x1>β. Then there is a sequence {xκ} of E satisfying

    J(xκ)cβ>0and(1+xκ)J(xκ)0,asκ, (2.9)

    where

    c=infγΓmax0τ1J(γ(τ))andΓ={γC([0,1],E):γ(0)=0,γ(1)=x1}.

    Further, if (C)c is satisfied, then J has a critical value c.

    Assume that

    (H3) for any (i,j)[1,m]×[1,n],

    limx0f((i,j),x)ax=p(i,j),limx+f((i,j),x)x3=q(i,j)0,

    where 0p(i,j),q(i,j)+ and pL<λ1.

    Remark 2.1. The assumption (H3) means that the nonlinearity f possesses asymptotic behavior at zero and infinity. Usually, the asymptotically 4-linear condition

    limx0f(i,j),xax=λ,limx+f(i,j),xbx3=μ, (2.10)

    or the following classic 4-superlinear condition of Ambrosetti and Rabinowitz (AR)

    ν>4:νF((i,j),x)xf((i,j),x),|x|large, (2.11)

    is crucial to certify the mountain pass geometry and prove the boundedness of Cerami or Palais-Smale sequences in E. Clearly, our assumption (H3) is weaker than (2.10) and indicates that (2.11) does not hold any more. Further, q(i,j)+ in (H3) indicates that f is 4-superlinear at infinity, which is weaker than (2.11).

    Set

    Λ=inf{x4:xX,mi=1nj=1q(i,j)x4(i,j)=1}. (2.12)

    Remark 2.2. By (2.12), we have that Λ is positive.

    First, Λ is attainable. Let a minimizing sequence of Λ be denoted by {xl}E, then {xl} is bounded and satisfies mi=1nj=1q(i,j)x4l(i,j)=1. Choose a subsequence of {xl}, still denoted by {xl}. Then there exists ˉx1E such that xlˉx1. Hence,

    mi=1nj=1q(i,j)x4l(i,j)mi=1nj=1q(i,j)ˉx14(i,j),asl+,

    and

    mi=1nj=1q(i,j)ˉx14(i,j)=1.

    Therefore,

    Λ(nj=1m+1i=1|Δ1ˉx1(i1,j)|2+mi=1n+1j=1|Δ2ˉx1(i,j1)|2)2Λ,

    which leads to

    Λ=(nj=1m+1i=1|Δ1ˉx1(i1,j)|2+mi=1n+1j=1|Δ2ˉx1(i,j1)|2)2=ˉx14.

    Namely, Λ is attainable.

    Further, ˉx1(i,j)>0 for all (i,j)[1,m]×[1,n]. In fact, if qL<+ and q(i,j)()0, then Λ>0 and there exists ˉx1E such that ˉx14=1 and mi=1nj=1q(i,j)ˉx14(i,j)=1. Moreover, ˉx1(i,j)>0 for all (i,j)[1,m]×[1,n]. Furthermore, assume ˉx1(i,j)0 on (i,j)[1,m]×[1,n]. Otherwise, we can replace ˉx1 by |ˉx1|. So, the strong maximum principle implies that ˉx1(i,j)>0 for all (i,j)[1,m]×[1,n].

    Now we display our main results as following:

    Theorem 2.1. Assume f((i,j),x) satisfies (H1)(H3). Then:

    (ⅰ) If Λ1b, there is no any positive solution of (1.1) with Dirichlet boundary conditions (1.2);

    (ⅱ) If Λ<1b, there is at least one positive solution of (1.1) with Dirichlet boundary conditions (1.2).

    Corollary 2.3. Assume (H1)(H3) hold with q(i,j)>0. Then:

    (ⅰ) If bμ1, (1.1) with Dirichlet boundary conditions (1.2) admits no any positive solution;

    (ⅱ) If bμ1<<+, (1.1) with Dirichlet boundary conditions (1.2) possesses at least one positive solution.

    Proof. Note that q(i,j)>0 ensures that Λ=μ1. Then, Theorem 2.2 guarantees conclusions in Corollary 2.3 are true.

    Remark 2.3. Owing to (ⅱ) of Corollary 2.3, (1.1) with Dirichlet boundary conditions (1.2) admits at least one positive solution if q(i,j)bμı, ı2.

    Theorem 2.4. Assume q(i,j) and f((i,j),x) satisfies (H1)(H3). If 4<k<6 such that

    limx+f((i,j),x)xk1=0,(i,j)[1,m]×[1,n].

    Then, there exists at least one positive solution for (1.1) with Dirichlet boundary conditions (1.2).

    Remark 2.4. It is necessary to point out that it is not difficult to find many functions satisfying our conditions in Theorems 2.2 and 2.4, but (2.11) is not satisfied.

    Example 2.1. Given M>0, for any (i,j)[1,m]×[1,n], set

    f((i,j),x)={0,<x0;x3,0<xM;Mx3,M<x<+.

    By simple calculation, we have that f((i,j),x) satisfies (H1)(H3) with p(i,j)0, q(i,j)M and

    F((i,j),x)=14Mx4120M5,x>M.

    If (2.11) is met, then there exists ν>4 such that

    Mx4(ν4M420x4)Mx4,for largex,

    which means that ν4. And it is a contradiction.

    Example 2.2. For any (i,j)[1,m]×[1,n], set

    f((i,j),x)={0,<x0;x3lnx,0<x<+; (2.13)

    (2.13) means that p(i,j)0, q(i,j)+ and (H1)(H3) are satisfied. Meanwhile,

    F((i,j),x)=14x4lnx116x4+116e4,x>e,

    which indicates that

    x4lnx(ν4ν16lnx)+νe416x4lnx,for largex

    holds for ν>4 impossible. Subsequently, (2.11) is not satisfied.

    Theorem 2.5. Suppose (H1) holds and q(i,j)=+. Moreover:

    (H4) There exist some positive constant C and 4<k<6 satisfying

    |f((i,j),x)|C(|x|k1+1),(i,j)[1,m]×[1,n],xR;

    (H5) Let G((i,j),x)=xf((i,j),x)4F((i,j),x), there exists θ1 such that

    θG((i,j),x)G((i,j),ωx),(i,j)[1,m]×[1,n],xRand0ω1;

    (H6) There exists δ>0 such that F((i,j),x)a2λ1x2 holds for |x|<δ.

    Then, problem (1.1) with Dirichlet boundary conditions (1.2) admits at least one positive solution.

    Remark 2.5. In some sense, Theorem 2.5 extends Theorem 2.2 in two ways. On the one hand, (H5) is equivalent to (H2) when θ=1 and gives some general when θ>1. For example, set

    f((i,j),x)=4x3ln(1+x4)+2sinx,(i,j)[1,m]×[1,n],

    direct computation yields that f satisfies (H5) but does not satisfy (H2). On the other hand, (H6) is weaker than (H3). For example, set

    f((i,j),x)={0,<x0;4x3ln(1+x4)+2xlnx,x>0.

    Then, f satisfies both (H5) and (H6), but neither (H2) nor (H3).

    Proof of Theorem 2.2. We give the proof of (ⅰ) by contradiction. As to the proof of (ⅱ), we complete it by Proposition 2.1 in 2 steps: First, we are to verify that there exists a sequence {xκ}κNE such that (2.9) in Proposition 2.1 is true. Second, we show the functional J satisfies the (C)c in E. Since E is finite dimensional, we only need to prove {xκ} is bounded.

    (ⅰ) Suppose that xE is positive and solves (1.1) with Dirichlet boundary conditions (1.2), by (2.8), it follows that

    (a+bx2)x2=mi=1nj=1(f((i,j),x(i,j))x(i,j)).

    Together with (H1)(H3), we obtain

    bx4<mi=1nj=1(f((i,j),x(i,j))x(i,j))mi=1nj=1(q(i,j)x4(i,j)),

    which implies that Λ<1b. This is a contradiction and Theorem 2.2(ⅰ) is verified.

    (ⅱ) We are to complete the proof by applying Proposition 2.1. Thus we begin the proof with showing that there exist ρ,β>0 such that J(x)β for xE with x=ρ, and J(τx1), as τ+. Indeed, thanks to (H1) and (H3), for any ϵ>0, there exists ˆM=ˆM(ϵ)>0 such that

    F((i,j),x)12a(pL+ϵ)x2+ˆMx4,(i,j)[1,m]×[1,n],xR. (3.1)

    Choosing a suitable ϵ>0 such that (pL+ϵ)<λ1. Combining (3.1) with (2.2), (2.3) and (2.5), it follows that

    J(x)=a2x2+b4x4mi=1nj=1F((i,j),x(i,j))a2x2+b4x412a(pL+ϵ)mi=1nj=1x2(i,j)ˆMmi=1nj=1x4(i,j)a2(1(pL+ϵ)λ1)x2+b4x4ˆMη44x4.

    Therefore, we can select small ρ>0 such that

    J(x)a4(1(pL+ϵ)λ1)ρ2β>0,xEwithx=ρ.

    Since Λ<1b, we have

    limτ+J(τˉx1(i,j))τ4=limτ+[aˉx122τ2+b4ˉx14mi=1nj=1F((i,j),τˉx1(i,j))τ4]=b4ˉx14mi=1nj=1limτ+F((i,j),τˉx1(i,j))τ4ˉx41(i,j)ˉx41(i,j)=b4ˉx1414mi=1nj=1q(i,j)ˉx41(i,j)=b4Λ14<0,

    which implies that J(τˉx1) as τ+ for all (i,j)[1,m]×[1,n]. Therefore, there exists τ0>0 large enough such that

    J(τ0ˉx1)<0andmax{J(0),J(τ0ˉx1)}0<βinfx=ρJ(x).

    Define

    Γ={γC([0,1],E):γ(0)=0,γ(1)=τ0ˉx1},

    and

    c=infγΓmax0ϖ1J(γ(ϖ)).

    According to Proposition 2.1, it yields that cβ>0 and there exists a sequence {xκ}κNE such that

    J(xκ)c,(1+xκ)J(xκ)0,asκ. (3.2)

    Let {xκ}E. Our next task is to prove that {xκ} has a convergent subsequence, also written by {xκ}. Since E is an mn-dimensional Hilbert space, it suffices to show the boundedness of {xκ}. Arguing indirectly, suppose that xκ+ as κ+. Write yκ=xκxκ, which follows that yκ=1. Therefore, {yκ} possesses a subsequence, still denoted by {yκ}, satisfying yκ(i,j)y(i,j) as κ for all (i,j)[1,m]×[1,n].

    We assume y=0 and denote Ω1{(i,j):(i,j)[1,m]×[1,n]such thatx(i,j)>0} and y+=max{y,0}. In view of (H1) and (H3), there exists ˜M>0 such that

    f((i,j),x)x3˜M,(i,j)Ω1. (3.3)

    Owing to (H1), (2.8), (3.2) and (3.3), we have

    b=limκ+mi=1nj=1f((i,j),xκ((i,j))xκ((i,j)xκ4=limκ+mi=1nj=1f((i,j),xκ((i,j))x3κ(i,j)/y3κ(i,j)yκ(i,j)=limκ+mi=1nj=1f((i,j),xκ((i,j))x3κ(i,j)y4κ(i,j)=limκ+(i,j)Ω1f((i,j),xκ((i,j))x3κ(i,j)(y+κ(i,j))4˜Mlimκ+mi=1nj=1(y+κ(i,j))4=0.

    Obviously, it is impossible.

    We assume y0 and set

    pκ(i,j)={0,ifxκ(i,j)0;f((i,j),xκ(i,j))x3κ(i,j),ifxκ(i,j)>0.

    We obtain that 0pκ(i,j)˜M for all (i,j)[1,m]×[1,n]. Consequently, we can assume that there exists a function h(i,j) such that

    pκ(i,j)h(i,j),asκ+.

    Now, for any zE, we have

    mi=1nj=1pκ(i,j)y3κ(i,j)z(i,j)=mi=1nj=1pκ(i,j)(y+κ(i,j))3z(i,j)mi=1nj=1h(i,j)(y+(i,j))3z(i,j). (3.4)

    Furthermore, recall yκ=xκxκ, for all zE, (2.8), (3.2) and (H1) induce that

    (J(xκ),z)xκ3=a+bxκ2xκ3xκ,zmi=1nj=1(f((i,j),xκ(i,j))xκ3,z(i,j))=a+bxκ2xκ2yκ,zmi=1nj=1(f((i,j),x+κ(i,j))xκ3,z(i,j))0,asκ. (3.5)

    Thus,

    mi=1nj=1(f((i,j),x+κ(i,j))xκ3,z(i,j))by,z. (3.6)

    Combining (3.4) with (3.6), we obtain that

    limκ+[mi=1nj=1(f((i,j),x+κ(i,j))xκ3,z(i,j))mi=1nj=1pκ(i,j)(y+κ(i,j))3z(i,j)]=by,zmi=1nj=1h(i,j)(y+(i,j))3z(i,j)=0,zE.

    By (H3), it yields that

    by,zmi=1nj=1q(i,j)(y+(i,j))3z(i,j)=0. (3.7)

    Set z=y, then y2=0 and yy+0. Consider (3.7) with the boundary conditions (1.2), we have

    {b(Δ21y(i1,j)+Δ22y(i,j1))=q(i,j)(y+(i,j))3,(i,j)[1,m]×[1,n],y(i,0)=y(i,n+1)=0,y(0,j)=y(m+1,j)=0,i[0,m+1],j[0,n+1],

    and the maximum principle implies y=y+>0. Hence,

    by,zmi=1nj=1q(i,j)y3(i,j)z(i,j)=0,zE.

    Let z=yκy. Note that y=1, (3.5) gives

    y2y,z=1bmi=1nj=1q(i,j)y3(i,j)z(i,j)=0,zE,

    which contradicts Λ<1b. Therefore, {xκ} is bounded in E and J satisfies (C)c.

    Consequently, Proposition 2.1 ensures that x is a nontrivial critical point of J, which means that there exists at least one positive solution for (1.1) with Dirichlet boundary conditions (1.2). Thus, we have verified Theorem 2.2.

    Proof of Theorem 2.4. Applying Proposition 2.1, we finish the proof of Theorem 2.4 by 3 steps.

    Step 1. We show that there exist constants ρ,β>0 such that J(x)β for xE with x=ρ.

    Set 0<ϵ<λ1pL. Since limx+f((i,j),x)xk1=0 holds for all (i,j)[1,m]×[1,n], by (H1) and (H3), there exists constant α>0 such that

    |f((i,j),x)|a(pL+ϵ)|x|+a|x|k1,(i,j)[1,m]×[1,n],

    which induces that

    F((i,j),x)a(pL+ϵ)2x2+ak|x|k,(i,j)[1,m]×[1,n]. (3.8)

    Then, (2.2), (2.3) and (3.8) yield that

    J(x)=a2x2+b4x4mi=1nj=1F((i,j),x(i,j))a2x2+b4x412a(pL+ϵ)mi=1nj=1x2(i,j)akmi=1nj=1xk(i,j)a2(1(pL+ϵ)λ1)x2+b4x4akηkkxk.

    Note that 4<k<6, there exist constants ρ,β>0 such that

    J(x)a4(1(pL+ϵ)λ1)ρ2β>0

    for xE with x=ρ.

    Step 2. We claim that J(tψ1) as t+, where ψ1 is the eigenfunction corresponding to μ1.

    Write Ω{(i,j):(i,j)[1,m]×[1,n]}. Then, there exists some α>0 such that

    min(i,j)¯Ω0ψ1(x)α>0,

    where Ω0{(i,j):(i,j)Ωsuch thatψ1(x(i,j))>0}. Obviously, Ω0 and Ω0¯Ω0⊂⊂Ω. Hence, tψ1(x)+ as t+ in ¯Ω0. In view of (H1) and (H2), it follows that

    04F((i,j),x)f((i,j),x)x,(i,j)Ω,x0,

    and F((i,j),x)x4 is nondecreasing in x>0. Since q(i,j)+, we get

    F((i,j),tψ1)t4ψ41F((i,j),tα)t4α4+ast+,(i,j)Ω0.

    Hence, for any K>0, there exists T>0 such that

    F((i,j),tα)t4α4K>0,tT,(i,j)Ω0.

    Therefore, for tT, choose K>0 large enough such that

    J(tψ1)t4=a2ψ12t2+b4ψ14mi=1nj=1F((i,j),tψ1(i,j))t4a2ψ12t2+b4ψ14(i,j)Ω0F((i,j),tψ1(i,j))t4ψ41(i,j)ψ41(i,j)a2T2ψ12+b4ψ14K(i,j)Ω0ψ41(i,j)a2T2ψ12+b4ψ14Kα4Ω0<0,

    where Ω0 denotes the number of (i,j) and (i,j)Ω0. Therefore, J(tψ1)+ as t+.

    Step 3. Let

    Γ={γC([0,1],E):γ(0)=0,γ(1)=t0ψ1}andc=infγΓmax0ϖ1J(γ(ϖ)).

    Proposition 2.1 means that cβ>0 and there exists a sequence {xκ}κNE such that

    J(xκ)c,(1+xκ)J(xκ)0,asκ+.

    In the following, we are to show that the sequence {xκ}κNE possesses a convergent subsequence, still denoted by {xκ}. In fact, owing to the finite dimension of space E, it is necessary to verify that {xκ} is bounded. Or else, we suppose xκ+. Let ϱ=2[(ca)12+(cb)14]. Set

    tκ=ϱxκ,~yκ=tκxκ=ϱxκxκ. (3.9)

    Then, ~yκ=ϱ and {~yκ}, {~yκ+} are bounded in E, which imply that there exists ˜yE such that

    ~yκ(i,j)˜y(i,j),{~yκ+}(i,j)˜y+(i,j),for(i,j)Ω.

    Denote Ω1={(i,j):(i,j)Ω0andy+(i,j)>0}. Thus (3.9) implies that x+κ(i,j)+ in Ω1. Therefore, by q(i,j)+, for any ˇM>0 and κ large enough, we have

    f((i,j),x+κ)(x+κ)3ˇM,(i,j)Ω1.

    Notice that J(xκ)c, (2.8), (3.9) and (H1) lead to

    (J(xκ),xκ)xκ4=axκ2+bxκ4xκ4mi=1nj=1(f((i,j),x+κ(i,j))xκ(i,j)xκ4)=a+bxκ2xκ2mi=1nj=1(f((i,j),x+κ(i,j))ϱ4xκ(i,j)3(~yκ(i,j))4)0,asκ.

    Hence, there holds

    b=limκ+mi=1nj=1(f((i,j),x+κ(i,j))ϱ4xκ(i,j)3(~yκ(i,j))4)=1ϱ4limκ+mi=1nj=1(f((i,j),x+κ(i,j))x+κ(i,j)3(~y+κ(i,j))4)1ϱ4limκ+(i,j)Ω1(f((i,j),x+κ(i,j))x+κ(i,j)3(~y+κ(i,j))4)ˇMϱ4limκ+(i,j)Ω1(~y+κ(i,j))4ˇMϱ4(i,j)Ω1(~y+(i,j))4. (3.10)

    Evidently, (3.10) means Ω1. Otherwise, for large ˇM, (3.10) is impossible. Thus, ~y+(i,j)=0 for all (i,j)[1,m]×[1,n]. Moreover, by (3.8), we get

    limκ+mi=1nj=1F((i,j),y+κ(i,j))=0.

    Recall ϱ=2[(ca)12+(cb)14]. Therefore,

    limκ+J(~yκ)=limκ+(a2~yκ2+b4~yκ4mi=1nj=1F((i,j),~yκ(i,j)))=aϱ22+bϱ44=c+aϱ22+b((ca)2+4(ca)32(cb)14+6ca(cb)12+4(ca)12(cb)34)=c+aϱ22+L, (3.11)

    where L=b((ca)2+4(ca)32(cb)14+6ca(cb)12+4(ca)12(cb)34)>0.

    To get a contradiction, first of all, we prove that if (H1) and (H3) hold and a sequence {xκ}E satisfying

    J(xκ)xκ0,asκ+,

    then {xκ} possesses a subsequence, denoted by {xκ} once more, such that

    J(txκ)at22xκ2+1+t44κ+J(xκ),t>0,κ1. (3.12)

    Since J(xκ)xκ0 as κ+, for a subsequence {xκ}, we may assume that

    1κ<J(xκ)xκ=(a+bxκ2)xκ2mi=1nj=1f((i,j),xκ(i,j))xκ(i,j)<1κ,κ1. (3.13)

    Hence, for any t>0 and positive integer κ, by (3.13), it follows that

    J(txκ)=at22xκ2+bt44xκ4mi=1nj=1F((i,j),txκ(i,j))at22xκ2+t44[(a+b)xκ2]xκ2mi=1nj=1F((i,j),txκ(i,j))at22xκ2+t44(1κ+mi=1nj=1f((i,j),xκ(i,j))xκ(i,j))mi=1nj=1F((i,j),txκ(i,j))=at22xκ2+t44κ+mi=1nj=1(t44f((i,j),xκ(i,j))xκ(i,j)F((i,j),txκ(i,j))). (3.14)

    For any fixed (i,j)[1,m]×[1,n] and κ1, set

    h(t)=t44f((i,j),xκ)xκF((i,j),txκ).

    By (H1) and (H2), direct computation yields that

    dh(t)dt=t3f((i,j),xκ)xκf((i,j),txκ)xκ=t3xκ(f((i,j),xκ)f((i,j),txκ)t3),

    and

    dh(t)dt0,0<t1;dh(t)dt0,t1.

    Then,

    h(t)h(1)=14f((i,j),xκ)xκF((i,j),xκ),t>0.

    Therefore, (3.14) implies that

    J(txκ)at22xκ2+t44κ+mi=1nj=1(14f((i,j),xκ(i,j))xκ(i,j)F((i,j),xκ(i,j))). (3.15)

    Moreover, by (3.13), we have

    J(xκ)=a2xκ2+b4xκ4mi=1nj=1F((i,j),xκ(i,j))=14(2a+bxκ2)xκ2mi=1nj=1F((i,j),xκ(i,j))14(a+bxκ2)xκ2mi=1nj=1F((i,j),xκ(i,j))14mi=1(f((i,j),xκ(i,j))xκ(i,j)1κ)mi=1nj=1F((i,j),xκ(i,j)),

    which ensures that

    mi=1(14f((i,j),xκ(i,j))xκ(i,j)F((i,j),xκ(i,j)))14κ+J(xκ). (3.16)

    As a result, (3.15) and (3.16) guarantee that (3.12) holds.

    In view of (3.12), we have

    limκ+J(~yκ)=limκ+J(tκxκ)limκ+(at2κ2xκ2+1+t4κ4κ+J(xκ))aϱ22+limκ+(1+t4κ4κ+J(xκ))=c+aϱ22. (3.17)

    Evidently, (3.11) contradicts (3.17). Subsequently, {xκ} is bounded in E. Thus, all conditions of Proposition 2.1 are verified and the proof of Theorem 2.4 is completed.

    Proof of Theorem 2.5. In order to complete the proof by Proposition 2.1, what is first to do is to prove that if (H1), (H4) and (H5) are satisfied and q(i,j)+, then J(x) satisfies (C)c. To this end, we assume that {xκ}κNE is the (C)c sequence, that is, for cR,

    J(xκ)c,(1+xκ)J(xκ)0,asκ+. (3.18)

    Then, for κ large enough, (2.5) and (2.8) yield that

    1+cJ(xκ)14(J(xκ),xκ)=a4xκ2+mi=1(14f((i,j),xκ(i,j))xκ(i,j)F((i,j),xκ(i,j))). (3.19)

    Since E is an mn-dimensional Hilbert space, it is sufficient to show that {xκ} possesses a bounded subsequence, still denoted by {xκ}. Or else, we may assume that xκ+ as κ+. Set ^yκ=xκxκ, then ^yκ=1 and {yκ} is bounded. Thus, there exists ˆyE such that ^yκ(i,j)ˆy(i,j) holds for all (i,j)[1,m]×[1,n].

    We suppose that ˆy0. Because xκ+, we have |xκ|+ as κ+. For q(i,j)+, it follows that

    limκ+f((i,j),x+κ(i,j))(x+κ(i,j))3=+,(i,j)[1,m]×[1,n]. (3.20)

    Meanwhile, (2.8) and (H1) induce that

    J(xκ),xκ=axκ2+bxκ4mi=1nj=114(f((i,j),xκ(i,j))xκ(i,j))=xκ4(axκ2+bmi=1nj=1(f((i,j),x+κ(i,j))(x+κ(i,j))3^yκ4(i,j))).

    Together with (3.18), it follows that

    J(xκ),xκxκ4=axκ2+bmi=1nj=1f((i,j),x+κ(i,j))(x+κ(i,j))3^yκ4(i,j)0,asκ+.

    Hence,

    blim infκ+mi=1nj=1f((i,j),xκ(i,j))xκ(i,j)3^yκ4(i,j)=+,

    which is a contradiction.

    We suppose that ˆy=0. Let {κ} be a sequence of real numbers such that J(κxκ)=max[0,1]J(xκ). For any integer s>0, set ^yκs=(8sb)1/4^yκ. By (H4), we have

    |F((i,j),x)|c|x|k+˜c|x|.

    Note that ^yκs0 as κ+ and F(,x) is continuous in x, we achieve

    limκ+F((i,j),^yκs)=0,(i,j)[1,m]×[1,n]. (3.21)

    Since xκ+ as κ+, we obtain 0(8sb)1/4xκ1 is true as κ large enough. Together with the definitions of J(κxκ) and κ, it yields that

    J(κxκ)J((8sb)1/4xκxκ)=J(^yκs)=a2^yκs2+b4^yκs4mi=1nj=1F((i,j),^yκs(i,j))2smi=1nj=1F((i,j),^yκs(i,j)). (3.22)

    Consider s>0 is arbitrary, (3.21) and (3.22) imply that

    J(κxκ)+,κ+. (3.23)

    For 0κ1, (H5) means that there exists θ1 such that θG((i,j),xκ)G((i,j),κxκ). Notice that J(0)=0 and J(xκ)c, then 0<κ<1 for κ large enough. Therefore,

    J(κxκ),κxκ=aκxκ2+bκxκ4mi=1nj=1κxκf((i,j),κxκ(i,j))=κdJ(xκ)d|=κ=0,

    that is,

    aκxκ2+bκxκ4=mi=1nj=1κxκf((i,j),κxκ(i,j)). (3.24)

    Combining (3.23) with (3.24), it follows that

    a4xκ2+14mi=1nj=1G((i,j),xκ(i,j))a4θκxκ2+14θmi=1nj=1G((i,j),κxκ(i,j))=1θ[a4κxκ2+mi=1nj=1(14κxκf((i,j),κxκ(i,j))F((i,j),κxκ(i,j)))]=1θ[a2κxκ2+b4κxκ4mi=1nj=1F((i,j),κxκ(i,j))]=1θJ(κxκ)+,asκ+.

    Namely, as κ+, there has

    a4κxκ2+mi=1nj=1(14κxκf((i,j),κxκ(i,j))F((i,j),κxκ(i,j)))+,

    which contradicts (3.19). Thus, {xκ} is bounded. Therefore, J(x) satisfies (C)c.

    Next, we claim that there exist some ρ,β>0 such that J(x)β for all xE with x=ρ.

    In fact, (H5) and (H6) imply that there exists C1>0 such that

    F((i,j),x)a2λ1x2+C1|x|k,[i,j][1,m]×[1,n],xR. (3.25)

    Thus, by (2.2), (2.3), (2.5) and (3.25), it follows that

    J(x)=a2x2+b4x4mi=1nj=1F((i,j),x(i,j))a2x2+b4x4aλ12mi=1nj=1x2(i,j)C1mi=1nj=1|x(i,j)|kb4x4C1ηkkxk, (3.26)

    where k>4. Given a small ρ>0, (3.26) means that

    J(x)βb4ρ4C1ηkkρk>0,xEandx=ρ.

    Last, we show there exists ˘xE with ˘x>ρ such that J(˘x)<0. Since q(i,j)+, that is, limx+f((i,j),x)x3+, for any ε>0, there exists ˘M>0 such that f((i,j),x)x31ε holds for all x>˘M and (i,j)[0,1+m]×[0,n+1]. Set c(ε)=˘M3ε, then,

    f((i,j),x)1εx3c(ε),x0,(i,j)[0,1+m]×[0,n+1].

    Therefore, for all x0, (i,j)[0,1+m]×[0,n+1] and 0ω1, we have

    f((i,j),ωx)1εω3x3c(ε).

    Then,

    f((i,j),ωx)x1εω3x4c(ε)x,x0. (3.27)

    Integrating both sides of (3.27) on [0, 1] with respect to ω, we have

    F((i,j),x)14εx4c(ε)x,x0,

    which ensures that

    F((i,j),tφ1)14εt4φ41c(ε)tφ1, (3.28)

    where φ1 is the normal eigenfunction corresponding to λ1. Dividing by t4, (3.28) means that

    F((i,j),tφ1)t414εφ41c(ε)φ1t3.

    Thus,

    mi=1nj=1F((i,j),tφ1)t4mi=1nj=1(14εφ41c(ε)φ1t3). (3.29)

    Let t+, (3.29) leads to

    lim inft+mi=1nj=1F((i,j),tφ1)t4mi=1nj=114εφ41. (3.30)

    Notice that ε>0 is arbitrary and let ε0, then (3.30) indicates that

    lim inft+mi=1nj=1F((i,j),tφ1)t4=+.

    Therefore,

    J(tφ1)t4=aφ122t2+bφ124mi=1nj=1F((i,j),tφ1)t4,ast+.

    Therefore, there exists t0 large enough such that J(˘x)<0, where ˘x=t0φ1.

    Define

    Γ={γC([0,1],E):γ(0)=0,γ(1)=˘x},c=infγΓmax0ϖ1J(γ(ϖ)),

    then cβ>0. Hence, Proposition 2.1 guarantees that J has at least a nontrivial critical point. Therefore, the proof of Theorem 2.5 is completed.

    Due to their wide applications, partial difference equations have been studied extensively. We all know that the discrete Kirchhoff term

    b(nj=1m+1i=1|Δ1x(i1,j)|2+mi=1n+1j=1|Δ2x(i,j1)|2)(Δ21x(i1,j)+Δ22x(i,j1))

    makes it not only more difficult but also more interesting to study. In this paper, we investigate the existence and nonexistence of positive solutions to a class of partial difference equations which involve the discrete Kirchhoff term. First, we established the corresponding variational functional on a suitable variational function space. Then, we obtained a series of results on the existence and nonexistence of positive solutions via a variant version of the mountain pass theorem. The conditions in our obtained results release the classical (AR) condition.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The author wishes to thank the handling editor and the referees for their valuable comments and suggestions.

    The author declares no competing interest in this paper.



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