# Three Nontrivial Solutions for Second-Order Partial Difference Equation via Morse Theory

Authors: Yuhua Long; Huan Zhang
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In the present paper, we consider a second-order nonlinear partial difference equation with Dirichlet boundary conditions. Applying variational method together with the Morse theory, we establish a criterion to obtain at least three nontrivial solutions. An example is also elaborated to demonstrate our main result.


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## Body

## 1. Introduction

Letℕ and ℤ are natural number set and integer set, respectively. Denote ℤa,b≔a,a+1,⋯,b for any a, b∈ℤ and a≤b. Δ1 and Δ2 are the forward difference operators defined by Δ1ui,j=ui+1,j−ui,j and Δ2ui,j=ui,j+1−ui,j. Given integers T1, T2≥2, write Ω≔ℤ1,T1×ℤ1,T2, we deal with self-adjoint partial difference equation of the form
(1)Δ1pi−1,jΔ1ui−1,j+Δ2ri,j−1Δ2ui,j−1+qi,jui,j=−fi,j,ui,j,i,j∈Ω,subject to Dirichlet boundary conditions
(2)ui,0=ui,T2+1=0i∈ℤ1,T1,u0,j=uT1+1,j=0j∈ℤ1,T2.Here,fi,j,u:Ω×ℝ⟶ℝ is differentiable in u and there exists a function Fi,j,u such that Fi,j,u=∫0ufi,j,τdτ for each i,j∈Ω. Further, for all i,j∈Ω, throughout this paper, we assume thatf0fi,j,0=0,ppi,j>0, ri,j>0, qi,j≤0 and qi,j≡0.As usual, ifu satisfies (1.1)-(1.2), we say u is a solution of (1) and (2). According to f0, it is easy to see that (1) and (2) admit a trivial solution u=0. Meanwhile, what we care about is the existence and multiplicity of nontrivial solutions of (1)-(2).Consider (1), a partial difference equation, involving functions with two discrete variables, it can be used to many investigations related to image processing, population models, and digital control systems [1]. Due to the rapid development of modern digital computing devices, more and more important information about the behavior of complex systems can be revealed by simulations by modern digital computing devices in a simple way, which contributes greatly to the increasing interest in discrete problems and they are investigated in many literatures, for example, [2–8].Letri,j−1≡0 in (1), it becomes an ordinary difference equation, namely,
(3)ΔptΔut−1+qtut=ft,ut,t∈ℤ.(3) has captured many interests and has been studied extensively. Here, mention a few and [9] considered the existence of periodic solutions via critical theory. Ma and Guo [10] discussed homoclinic orbits. [11] got sign-changing solutions, whereas, when ri,j−1≠0, it seems that there are rare literatures.Moreover, (1) can be regarded as a discrete analog of a partial differential equation. It is well known that, with the rapid development of critical point theory, it becomes a more and more powerful to deal with the existence and multiplicity solutions of both partial differential equations and partial difference equations [12–14]. As mentioned, the Morse theory is a very useful tool to study the existence of multiple solutions of differential equations having variational structure, and it has been applied successfully to study differential equations [15–18]. At the same time, difference equations, regarded as discretizations of differential equations, are considered and multiple solutions are achieved via the Morse theory in some literatures [19, 20]. However, there are few literatures using the Morse theory to study partial difference equations. Due to abovementioned reasons, we devote to studying the Dirichlet boundary value problem of second-order partial difference (equations (1) and (2)) by the Morse theory.The organization of the rest of this paper reads as follows. In Section2, we construct a suitable variational framework corresponding to (1) and (2) and reduce the existence of solutions of (1) and (2) to the existence of critical points of the associated functional. With preparation of Section 2, Section 3 not only displays the main result of this paper but also provides detailed proof of the main result. Finally, an example is exhibited to demonstrate our main result in Section 4.


## 2. Variational Structure and Some Auxiliary Results

In this section, we construct a variational functional corresponding to (1) and (2) on a suitable function space and state some basic facts.Let(4)S=u:ℤ0,T1+1×ℤ0,T2+1⟶ℝsuchthatui,0=ui,T2+1=0,i∈ℤ0,T1+1andu0,j=uT1+1,j=0,j∈ℤ0,T2+1.For anyu, v∈S, define an inner product ·,· by
(5)u,v=∑i=1T1∑j=1T2ui,jvi,j,then the induced norm is
(6)u=u,u=∑i=1T1∑j=1T2ui,j21/2,∀u∈S.Hence,S,·,· is a T1T2-dimensional Hilbert space.Consider the functionalI:S⟶ℝ as the following form:
(7)Iu=12∑i=1T1+1∑j=1T2pi−1,jΔ1ui−1,j2+12∑i=1T1∑j=1T2+1ri,j−1Δ2ui,j−12−12∑i=1T1∑j=1T2qi,jui,j2−∑i=1T1∑j=1T2Fi,j,ui,j,∀u∈S.Note thatfi,j,u is differentiable in u, which ensures Iu is twice differentiable. What is more, for any u, v∈S, make use of the boundary conditions (2), we have
(8)I′u,v=−∑i=1T1∑j=1T2Δ1pi−1,jΔ1ui−1,j+Δ2ri,j−1Δ2ui,j−1+qi,jui,j+fi,j,ui,jvi,j,and we transfer the existence of nontrivial solutions of (1) and (2) into the existence of critical points of I on S.In the following, we introduce some basic facts.Definition 1. [21].
The functionalI satisfies the weaker Cerami condition (Cc condition for short) at the level c∈ℝ if any sequence un⊆S satisfying Iun⟶c, 1+unI′un⟶0 as n⟶∞ has a convergent subsequence. I satisfies C condition if I satisfies Cc condition at any c∈ℝ.Definition 2 [16, 22].
Letu0 be an isolated critical group of I with Iu0=c∈ℝ and U be a neighborhood of u0, the group
(9)CqI,u0≔HqIc∩U,Ic∩U\u0,q∈ℤis called the q-th critical group of I at u0. Let κ=u∈S∣I′u=0. If Iκ is bounded from below by a∈ℝ and I satisfies Dc condition for all c≤a. Then, the group
(10)CqI,∞≔HqS,Ia,q∈ℤis called the q-th critical group of I at infinity.In applications of Morse theory, it is necessary to make the functional satisfy the deformation conditionD, which is introduced by [23]. And [24] proves that once the functional I satisfies the C condition, it must satisfy the deformation condition D. Let S be a real Hilbert space and I∈C2S,ℝ. Denote Morse index and zero dimension of u0 by μu0 and νu0, respectively. The following propositions are essential tools to verify our main result.Proposition 3 [12].
LetI satisfy the D condition. We have (J1) if CqI,∞≇0 holds for some q; then, I must have a critical point x such that CqI,x≇0;
(J2) if 0 is the isolated critical point of I and CqI,∞≇CqI,0 holds for some q; then, I has a nonzero critical point.Proposition 4 [18].
Supposeu0 is the isolated critical point of I and I′′u0 is a Fredholm operator. Further, if μu0 and νu0 are finite, there holds
(J3) if u0 is the local minimum point of I, then
(11)CqI,u0≅δq,0ℤ,q=0,1,2,⋯.Proposition 5 [18].
LetA:S⟶S be a self-adjoint linear operator with the isolated spectral point 0 and write I in the form of
(12)Iu=12Au,u+Qu,where Q∈C1S,ℝ such that limu⟶∞Q′u/u=0. Define V≔kerA,W≔V⊥=W+⊕W−, and μ=dimW− and ν=dimV≠0 are finite numbers. Suppose f satisfies the D condition and I satisfies the angle condition at infinity:
AC∞± there exist constants M>0 and α∈0,1 such that
(13)±I′u,φ≥0,∀u=v+w,u≥M,w≤αu.
Then,(14)CqI,∞≅δq,k±ℤ,where k+=μ, k−=μ+ν, v∈V, w∈W.In our proofs, we also need the following Mountain Pass Lemma.Proposition 6 [22].
LetS be a real Banach space and I∈C1S,ℝ satisfy the Palais-Smale (PS in short) condition. Further, if I0=0 and
(J4) there exist constants ρ, a>0 such that I∂Bρ≥a,
(J5) there is e∈S\Bρ such that Ie≤0.
Then,I possesses a critical value c≥a given by
(15)c=infh∈Γsupx∈0,1fhx,where
(16)Γ=h∈C0,1,S∣h0=0,h1=e.Denoted byλk be the k-th eigenvalue corresponding to linear eigenvalue problem of the equation (1), namely,
(17)Δ1pi−1,jΔ1ui−1,j+Δ2ri,j−1Δ2ui,j−1+qi,jui,j+λui,j=0.We claimLemma 7.
For1≤k≤T1T2, the eigenvalue λk of (17)-(2) is positive.Proof.
Rewriteu as
(18)u=u0,1,⋯,uT1,1,⋯,uT1+1,T2tr,where ·tr denotes the transpose of vector·. Then, the functional I, defined by (7), can be expressed by
(19)Iu=12Au,u−Fu,∀u∈S,where Fu=∑i=1T1∑j=1T2Fi,j,ui,j, A is the T1T2×T1T2 matrix corresponding to the quadratic form
(20)∑i=1T1+1∑j=1T2pi−1,jΔ1ui−1,j2+∑i=1T1∑j=1T2+1ri,j−1Δ2ui,j−12−∑i=1T1∑j=1T2qi,jui,j2.At first, it is easy to get thatA possesses at most T1T2 eigenvalues. Subsequently, we need to prove that A is a positive definite matrix.In fact, for allu∈ℝ and u≠0, we have
(21)uTAu=∑i=1T1+1∑j=1T2pi−1,jΔ1ui−1,j2+∑i=1T1∑j=1T2+1ri,j−1Δ2ui,j−12−∑i=1T1∑j=1T2qi,jui,j2≥0.IfuTAu=0, according to p, there hold
(22)pi−1,jΔ1ui−1,j2=0,∀i,j∈ℤ1,T1+1×ℤ1,T2,ri,j−1Δ2ui,j−12=0,∀i,j∈ℤ1,T1×ℤ1,T2+1,qi,jui,j2=0,∀i,j∈Ω,which lead to thatui,j=ui−1,j=ui,j−1 for all i,j∈Ω and there exists some i0,j0∈Ω such that ui0,j0=0. Together with Dirichlet boundary conditions (2), we can deduce that u1,1=⋯=uT1,T2=0, which show a contradiction with respect to u≠0. Therefore, A is positive define, which ensures that λk>0 for all k∈1,T1T2. Without loss of generality, we can rearrange all eigenvalues of A as 0<λ1≤λ2≤⋯≤λT1T2.


## 3. Main Result and Its Proof

Thanks to above preparations, we are ready to establish our main result and state the detailed proof of it.First, we give some notations.Letϕk=ϕk1,ϕk2,⋯,ϕkT1T2tr and 1≤k≤T1T2, represent an eigenvector corresponding to the eigenvalue λk. Write W−=spanϕ1,⋯,ϕk−1, W0=spanϕk, and W+=W−⊕W0⊥; then, S can be split into
(23)S=W−⊕W+⊕W0.Denote(24)limu⟶∞fi,j,uu=λk,∀i,j∈Ω,andgi,j,u=fi,j,u−λku. Then, it yields that
(25)lim∣u∣⟶∞gi,j,uu=0,∀i,j∈Ω.We also need the following denotation.g± If un⟶∞ such that vn/un⟶1 as n⟶∞, then there exist δ>0 and N∈ℕ such that
(26)±∑i=1T1∑j=1T2gi,j,uni,j,vni,j≥δ,n≥N.where un=vn+wn, vn∈W0, wn∈W+⊕W−.Now we state our main result as the following.Theorem 8.
Letf0, (p), and (24) hold. If, for all i,j∈Ω,
(V1) F∈C2Ω,ℝ and ∂2Fi,j,u/∂u2>0, u∈ℝ,
(V2) ∂2Fi,j,u/∂u2u=0<λ1
are satisfied. Then, (1) and (2) possess at least three nontrivial solutions, including a positive solution and a negative solution under either 1̇g+ with k≥2 or 1̇1̇g− with k≥3 is fulfilled.According to Proposition3, we are to verify the compactness conditions (the Cc condition) of I under the assumptions given in our theorem.Lemma 9.
Letf0 and (24) hold. If g± holds, then I satisfies the (C) condition.Proof.
Suppose that there exists a sequenceum⊆S such that
(27)Ium⟶c,1+umI′um⟶0,asm⟶∞.
Due toS,·,· is a T1T2-dimensional real Hilbert space, it suffices to verify that um is bounded. Arguing indirectly, suppose um is unbounded, namely,
(28)um⟶∞,asm⟶∞.
Denoteu¯m=um/um, then u¯m=1. As a result, there is a convergent subsequence for u¯m. It might as well be set as itself, and there exists u¯∈S, u¯=1 such that u¯m⟶u¯.
Recallgi,j,u=fi,j,u−λku, then for all φ∈S, we have
(29)I′um,φum=−∑i=1T1∑j=1T2Δ1pi−1,jΔ1u¯mi−1,j+Δ2ri,j−1Δ2u¯mi,j−1+qi,ju¯mi,j+gi,j,umi,jum+λku¯mi,jφi,j.
Meanwhile, (25) implies that gi,j,um/um⟶0 as m⟶∞. Thus, (29) gives
(30)Δ1pi−1,jΔ1u¯i−1,j+Δ2ri,j−1Δ2u¯i,j−1+qi,ju¯i,j+λku¯i,j=0.
Denoteum=vm+wm++wm−, where vm∈W0, wm+∈W+, and wm−∈W−, then according to I′u=Au−fu=Au−gu−λku we get
(31)λk+1−λkwm+2=λk+1wm+,wm+−λkwm+,wm+≤Awm+,wm+−λkwm+,wm+=Aum,wm+−λkum,wm+=gum,wm++I′um,wm+.λk−1−λkwm−2=λk−1wm−,wm−−λkwm−,wm−≥Awm−,wm−−λkwm−,wm−=Aum,wm−−λkum,wm−=gum,wm−+I′um,wm−.
Hence,(32)wm+um⟶0,wm−um⟶0,m⟶∞.
Therefore, we can deduce(33)vmum⟶1,m⟶∞.
According tog+, there exist δ>0 and N∈ℕ such that
(34)∑i=1T1∑j=1T2gi,j,umi,j,vmi,j≥δ,∀m≥N.
Making use of (30), we get
(35)I′um,vm=−∑i=1T1∑j=1T2Δ1pi−1,jΔ1umi−1,j+Δ2ri,j−1Δ2umi,j−1+qi,jumi,j+fi,j,umi,jvmi,j=−∑i=1T1∑j=1T2Δ1pi−1,jΔ1umi−1,j+Δ2ri,j−1Δ2umi,j−1+qi,jumi,j+gi,j,umi,j+λkumi,jvmi,j=−∑i=1T1∑j=1T2gi,j,umi,j,vmi,j≤−δ,∀m≥N.
Furthermore, sinceum=vm+wm, it follows that
(36)I′umum≥I′umvm≥I′um,vm≥δ,∀m≥N,which contradicts the hypothesis. Therefore, um is bounded.Now, we will calculate its critical groups at infinity,CqI,∞, via Proposition 5.Lemma 10.
Letf0 and (24) hold, then
(V3) CqI,∞≅δq,kℤ if g+ holds;
(V4) CqI,∞≅δq,k−1ℤ if g− holds.Proof.
SinceS is a T1T2-dimensional Hilbert space and A is a positive definite matrix, there exists a self-adjoint linear operator, which can be still represented by A. Write
(37)Ju,v=λk∑i=1T1∑j=1T2ui,j,vi,j.Then,(38)Iu=12A−Ju,u−Gu,which has the form (12) with Qu=−Gu. According to (25), Qu satisfies
(39)limu⟶∞Q′uu=−limu⟶∞guu=0.Moreover,I satisfies the C condition guarantees that the D condition is fulfilled and kerA−J=spanϕk. Subsequently, what we need to do is to show AC∞− is met with the condition g+. Otherwise, for every natural number m and every αm=1/m, there exists um=vm+wm∈W0⊕W+⊕W−, where vm∈W0, wm∈W+⊕W− such that um≥m, wm≤1/mum. Therefore, it yields that
(40)um⟶∞,vmum⟶1,m⟶∞.Byg+, there exist δ>0 and N∈ℕ such that
(41)∑i=1T1∑j=1T2gi,j,umi,j,vmi,j≥δ,∀m≥N.Therefore,(42)I′um,vm=−∑i=1T1∑j=1T2gi,j,umi,j,vmi,j≤−δ,∀m≥N.Meanwhile, the assumptionAC∞− indicates that
(43)I′um,vm>0,which is inconsistent with (42). As a result, I satisfy AC∞− condition.In order to gain mountain pass type critical points, we need the following Lemmas.Lemma 11.
Let(44)f+i,j,u=fi,j,u,u≥0,0,u<0,
whereF+=i,j,u=∫0uf+i,j,τdτ. If
(45)limu⟶+∞f+i,j,uu=σ>λ1,then the functional
(46)I+u=∑i=1T1+1∑j=1T212pi−1,jΔ1ui−1,j2+∑i=1T1∑j=1T2+112ri,j−1Δ2ui,j−12−∑i=1T1∑j=1T212qi,jui,j2−∑i=1T1∑j=1T2F+i,j,ui,jsatisfies the PS condition.Proof.
Suppose that there exists a sequenceum⊆S satisfying
(47)I+um⟶c,I′+um⟶0,m⟶∞.SinceS,·,· is a T1T2-dimensional real Hilbert space, we only need to verify that um is bounded. On the contrary, suppose um is unbounded, that is, umi,j⟶∞. Denote that vm=um/um, then vm=1 and exists v∈S, v=1 such that vm⟶v. For any φ∈S, we get
(48)I′+um,φum=−∑i=1T1∑j=1T2Δ1pi−1,jΔ1vmi−1,j+Δ2ri,j−1Δ2vmi,j−1+qi,jvmi,j+f+i,j,umi,jumφi,j.Sinceumi,j⟶∞ for every i,j∈Ω, we get
(49)limm⟶∞f+i,j,umi,jum=limm⟶∞f+i,j,umi,jumi,jvmi,j=σv+i,j,where v+i,j=maxvi,j,0, i,j∈Ω. Therefore,
(50)−∑i=1T1∑j=1T2Δ1pi−1,jΔ1vi−1,j+Δ2ri,j−1Δ2vi,j−1+qi,jvi,j+σv+i,jφi,j=0,which implies that v≠0 satisfies
(51)Δ1pi−1,jΔ1vi−1,j+Δ2ri,j−1Δ2vi,j−1+qi,jvi,j+σv+i,j=0,i,j∈Ω,(52)vi,0=vi,T2+1=0i∈ℤ1,T1,v0,j=vT1+1,j=0j∈ℤ1,T2.Next, we provevi,j>0. Write vi0,j0=minvi,j∣i,j∈Ω, it is necessary to show vi0,j0>0. If vi0,j0≤0, (51) and (52) imply that
(53)Δ1pi0−1,j0Δ1vi0−1,j0+Δ2ri0,j0−1Δ2vi0,j0−1+qi0,j0vi0,j0=0.Together with(p), it follows that
(54)Δ1vi0−1,j0=Δ2vi0,j0−1=0.Thus,(55)vi0−1,j0=vi0,j0=vi0,j0−1,which leads to v=0 for all i,j∈Ω. Obviously, it contradicts with v≠0, then vi,j>0. Therefore, um is bounded.Similar to Lemma11, we haveLemma 12.
Let(56)f−i,j,u=fi,j,u,u≤0,0,u>0,
whereF−=i,j,u=∫0uf−i,j,τdτ. If
(57)limu⟶−∞f−i,j,uu=σ>λ1,then the functional
(58)I−u=∑i=1T1+1∑j=1T212pi−1,jΔ1ui−1,j2+∑i=1T1∑j=1T2+112ri,j−1Δ2ui,j−12−∑i=1T1∑j=1T212qi,jui,j2−∑i=1T1∑j=1T2F−i,j,ui,jsatisfies the PS condition.Lemma 13.
Under the condition of Theorem8, the functional I+ possesses a positive critical point u+ and CqI+,u+≅δq,1ℤ, the functional I− possesses a negative critical point u− and CqI−,u−≅δq,1ℤ.Proof.
We only prove the case ofI+ and the proof of the case of I− can be obtained similarily. First of all, we need to prove that I+ satisfies the Proposition 6 so that it has a nonzero critical point u+. As a matter of fact, I+0=0 and according to the conclusion of Lemma 11, I+ satisfies the PS condition. Now, we must prove that I+ satisfies J4 and J5.On one hand, sinceV2 is satisfied, there exist ρ>0 and ρ1>0 satisfying F′′i,j,0<ρ1<λ1 such that
(59)Fi,j,u≤12ρ1u2,u≤ρ.Then,(60)I+u=12Au,u−Fu≥12λ1u2−∑i=1T1∑j=1T2Fi,j,ui,j≥12λ1u2−12ρ1∑i=1T1∑j=1T2ui,j2=12λ1u2−12ρ1u2=12λ1−ρ1u2>0.Therefore,J4 is satisfied. On the other hand, (24) means that there exists γ>λk−1>λ1, b∈ℝ such that
(61)Ft≥γ2t2+b,∀t∈ℝ.Choosee∈spanϕ1, where ϕ1 is the eigenvector corresponding to λ1, we have
(62)I+te=12Ate,te−Fte≤12λ1te2−γte22−bT1T2=t22λ1−γe2−bT1T2.Therefore, there exists a sufficiently larget>ρ such that I+u0≤0 for u0=te∈S, which makes J5 is fulfilled. By Proposition 6, I+ possesses a critical point u+≠0.In the following, we devote to verifyingu+>0. Due to I′+u+=0, there exists a sequence um+ such that I′+um+⟶0=I′+u+ as m⟶∞. Then, for all φ∈S, there holds
(63)I′+um+,φ=−∑i=1T1∑j=1T2Δ1pi−1,jΔ1um+i−1,j+Δ2ri,j−1Δ2um+i,j−1+qi,jum+i,j+f+i,j,um+i,jφi,j.Letm⟶∞ and u+i,j=maxi,j∈Ωui,j,0, we have
(64)limm⟶∞f+i,j,um+i,j=limm⟶∞f+i,j,um+i,jum+i,jum+i,j=σu+i,j,Δ1pi−1,jΔ1u+i−1,j+Δ2ri,j−1Δ2u+i,j−1+qi,ju+i,j+σu+i,j=0.Denoteui0,j0=mini,j∈Ωui,j. Now, we need to prove ui0,j0>0 to get ui,j>0. Arguing indirectly, let ui0,j0≤0, then
(65)Δ1pi0−1,j0Δ1u+i0−1,j0+Δ2ri0,j0−1Δ2u+i0,j0−1=0.Applying(p), we get
(66)Δ1u+i0−1,j0=Δ2u+i0,j0−1=0,which means u+i0−1,j0=u+i0,j0=u+i0,j0−1. Then,
(67)u+=0,∀i,j∈Ωcontradicts with u+≠0. Consequently, u+>0.Now, we calculateCqI+,u+. On account of
(68)I′′uv,w=wTAv−∑i=1T1∑j=1T2F′′i,j,ui,jvi,jwi,j,for any v∈S, we have I′′u+v,v≥0 and there exists v0≠0 such that I′′u+v0,v=0, which indicates that v0 is a solution of
(69)Δ1pi−1,jΔ1v0i−1,j+Δ2ri,j−1Δ2v0i,j−1+qi,jv0i,j−F′′i,j,u+v0i,j=0,i,j∈Ω,v0i,0=v0i,T2+1=v00,j=v0T1+1,j=0,i∈ℤ1,T1,j∈ℤ1,T2.Therefore, the corresponding Dirichlet eigenvalue problem(70)Δ1pi−1,jΔ1vi−1,j+Δ2ri,j−1Δ2vi,j−1+qi,jvi,j−λF′′i,j,u+vi,j=0,i,j∈Ωadmits an eigenvalue λ=1. According to V1, 1 is a single eigenvalue, which means that dimkerI′′u+=1. Combining with I′′u+:S⟶ℝ is a surjection, I′′u+ is a Fredholm operator, we conclude that
(71)CqI+,u+≅δq,1ℤCqI,u+≅CqI+,u+≅δq,1ℤ.With the aid of preceding preparations, now we are in the position to complete the verification of Theorem8 via Proposition 3.Proof of Theorem 14.
Let1̇ [1̇1̇] holds. On one hand, through Lemma 10 we get
(72)CqI,∞≅δq,kℤ,CqI,∞≅δq,k−1ℤ.
On the other hand, because ofF′′0<λ1, I′0=0 and
(73)I′′0u,u≥λ1−F′′0u2,we can refer that 0 is the local minimum of I. Moreover, 0 is the isolated critical point of I which implies that I′′0 is a Fredholm operator with finite Morse index and zero dimension. From Proposition 4, it gives that
(74)CqI,0≅δq,0ℤ.
For the reason thatI satisfies the D condition from Lemma 9, we can deduce I possesses a critical point u1≠0 such that
(75)CkI,u1≇0Ck−1I,u1≇0.
As a result, we conclude that whenk≥2k≥3, u+, u−, and u1 are the nontrivial critical points of I. Thus, the proof of Theorem 14 is finished.


## 4. An Example

As an application of our result, an example is elaborated here.Example 15.
TakeT1=3, T2=2, then for i,j∈ℤ1,3×ℤ1,2, pi,j=1, qi,j=−1, ri,j=1. Consider
(76)Δ12ui−1,j+Δ22ui,j−1−ui,j+λ1/2−λkui,j1+ui,j2+λkui,j=0,with the boundary value conditions (2).According to the expression offi,j,u, it follows that
(77)Fi,j,u=λ1/2−λk2ln1+u2+12λku2.F′′i,j,u=3λk−λ1/2u2+λku4+λ1/21+u22>0.It is not difficult to verify thatfi,j,0=0, Fi,j,0=0, F′′i,j,0=λ1/2<λ1, and
(78)limu⟶∞fi,j,uu=limu⟶∞λ1/2−λk1+ui,j2+λk=λk.Sincepi,j=1, qi,j=−1, and ri,j=1 for each i,j∈ℤ1,3×ℤ1,2, then the matrix
(79)A=5−10−100−15−10−100−1500−1−1005−100−10−15−100−10−15is positive define, and the eigenvalues are given by
(80)λ1=4−2,λ2=4,λ3=4+2,λ4=6−2,λ5=6,λ6=6+2.Next, we check1̇1̇. If um⟶∞ such that vm/um⟶1 with k≥3, then there exist δ>0 and M∈ℕ such that
(81)−∑i=1T1∑j=1T2gumi,j,vmi,j≥δ,m≥M.In fact, we can choosek=4, δ=λ4−λ1=2>0. Since gi,j,u=λ1/2−λku/1+u2, then
(82)−∑i=13∑j=12gumi,j,vmi,j=−∑i=13∑j=12gvmi,j,vmi,j=−∑i=13∑j=12λ1/2−λ4vmi,j1+vmi,j2·vmi,j=−∑i=13∑j=12λ1/2−λ41/vmi,j2+1=6λ4−λ12>δ.Therefore, all conditions of Theorem8 are satisfied and (76)-(2) admits at least three nontrivial solutions, which include a positive solution and a negative solution.



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*Source: 1564961-2022-03-08.xml*