# Existence and Nonexistence of Solutions for Fourth-Order Nonlinear Difference Boundary Value Problems via Variational Methods

Authors: Xia Liu; Tao Zhou; Haiping Shi
Journal: Discrete Dynamics in Nature and Society
 (2018)

---

## Abstract

This paper is concerned with boundary value problems for a fourth-order nonlinear difference equation. Via variational methods and critical point theory, sufficient conditions are obtained for the existence of at least two nontrivial solutions, the existence ofn distinct pairs of nontrivial solutions, and nonexistence of solutions. Some examples are provided to show the effectiveness of the main results.


---

## Body

## 1. Introduction

Throughout this paper, we denote byN, Z, R the sets of all natural numbers, integers, and real numbers, respectively. Let the symbol ∗ denote the transpose of a vector. For any integers c and d with c≤d,  [c,d]Z is defined by the discrete interval {c,c+1,…,d}.Now, we are concerned with the existence and nonexistence solutions to the fourth-order nonlinear difference equation(1)Δ2pt-2Δ2xt-2+qtxt=ft,xt,t∈1,TZ,satisfying the boundary value conditions(2)Δkx-1=ΔkxT-1,k=0,1,2,3,where 1≤T∈Z,  Δ is the forward difference operator defined by Δx(t)=x(t+1)-x(t),  Δkx(t)=Δ(Δk-1x(t))(2≤k≤4),  Δ0x(t)=x(t),  p(t)∈C[-1,T]Z,R with p(-1)=p(T-1),  p(0)=p(T),  q(t)∈C[1,T]Z,R,  f(t,x)∈C([1,T]Z×R,R).As usual, a solution of (1), (2), in other words, a function x:[-1,T+2]Z→R, satisfies both (1) and (2).We may think of boundary value problem (BVP) (1), (2) as being a discrete analogue of the following fourth-order nonlinear differential equation:(3)psx′′s′′-qsxs=ft,xs,t∈0,1,with boundary value conditions(4)xk0=xk1,k=0,1,2,3.(3) includes the following differential equation:(5)x4s=fs,xs,s∈R,which is used to describe the bending of an elastic beam [1]. Equations similar in structure to (3) have been studied by many researchers using a variety of methods; see, for example, [2–12].It is well-known that the study of nonlinear difference equations [13–29] has long been an important one as a result of the fact that they arise in numerical solutions of both ordinary and partial differential equations as well as in applications to different areas of applied mathematics and physics.Domshlak and Matakaev [17] in 2001 investigated the oscillation properties of the delay difference equation(6)xn+1-xn+bnxn-k=0,bn>0,n≥1,for k=2 and k=3 near the 2-periodic critical states with respect to its oscillation properties. By making use of “the Sturmian comparison method: discrete version”, they obtained some conditions for the existence and for the nonexistence of eventually positive solution.Using the critical point theory, Yang [27] studied the following higher order nonlinear difference equation:(7)∑i=0nrixk-i+xk+i=fk,xk+M,…,xk,…,xk-M,n∈N,k∈Z1,T,with boundary value conditions(8)x1-m=x2-m=⋯=x0=0,xT+1=xT+2=⋯=xT+m=0.Some sufficient conditions for the existence of the solution to the boundary value problem (7), (8) are obtained.In 2010, He, Yang and Yang [18] considered the following second-order three-point discrete boundary value problem:(9)Δ2xt-1+ft,xt=0,t∈1,nZ,x0=0,xn+1=αxm.By using the topological degree theory and the fixed point index theory, they provided sufficient conditions for the existence of sign-changing solutions, positive solutions, and negative solutions.Investigating the high order difference equation(10)Δnrt-nϕcΔnxt-n=-1nft,xt+1,xk,xt-1,t∈Z,Leng [22] established some new criteria for the existence and multiplicity of periodic and subharmonic solutions of (10) based on the linking theorem in combination with variational technique.Leszczyński [23] considered the difference equation with both advance and retardation,(11)Δ2γt-1ϕptΔ2xt-2=ft,xt+1,xt,xt-1,t∈Z1,k,with boundary value conditions(12)Δx-1=Δx0=0,xk+1=xk+2=0.They applied the direct method of the calculus of variations and the mountain pass technique to prove the existence of at least one and at least two solutions. Nonexistence of nontrivial solutions is also undertaken.In this paper, we shall study the boundary value problem for a fourth-order nonlinear difference equation (1), (2). Via variational methods and critical point theory, sufficient conditions are obtained for the existence of at least two nontrivial solutions, the existence of n distinct pairs of nontrivial solutions, and nonexistence of solutions. The motivation for the present work stems from the recent papers [3, 5].Throughout the whole paper, we suppose that there exists a functionG(t,x) such that(13)Gt,x=∫0xft,vdv,for any (t,x)∈[1,T]Z×R.Let(14)q_=min⁡qt∣t=1,2,…,T,and(15)q-=max⁡qt∣t=1,2,…,T.The remainder of this article is organized as follows. In Section2, we shall give some preliminary lemmas and establish the variational structure of BVP (1), (2). In Section 3, we shall give sufficient conditions to the existence and nonexistence solutions. In Section 4, we shall complete the proofs of the main results. Some examples illustrating our main results are given in Section 5.For the basic knowledge of variational methods, the reader is referred to [30–32].


## 2. Preliminary Lemmas

Assume thatX is a real Banach space and I∈C1(X,R) is a continuously Fréchet differentiable functional defined on X. As usual, I is said to satisfy the Palais-Smale condition if any sequence xkk=1∞⊂X for which Ixkk=1∞ is bounded and I′xk→0 as k→∞ possesses a convergent subsequence. Here, the sequence xkk=1∞ is called a Palais-Smale sequence.LetX be a real Banach space. We denote by the symbol Br the open ball in X about 0 of radius r, ∂Br its boundary, and B-r its closure.In the present article, we define a vector spaceX by(16)X≔x:-1,T+2Z→R∣Δkx-1=ΔkxT-1,k=0,1,2,3,and for any x∈X, define(17)x,y≔∑t=1Txtyt,∀x,y∈X,and(18)x≔∑t=1Tx2t1/2,∀x∈X.Remark 1.
For anyx∈X, it is easy to see that(19)x-1=xT-1,x0=xT,x1=xT+1,x2=xT+2.As the case stands, X is isomorphic to RT. In the following and in the sequel, when we write x=x(1),x(2),…,x(T)∈RT, we always imply that x can be extended to a vector in X so that (19) is satisfied.For anyx∈X, let the functional I be denoted by(20)Ix≔12∑t=1Tpt-2Δ2xt-22+12∑t=1Tqtxt2-∑t=1TGt,xt.Then I∈C1(X,R) and(21)∂I∂xt=Δ2pt-2Δ2xt-2+qtxt-ft,xt,t∈1,TZ.Thus,I′(x)=0 if and only if(22)Δ2pt-2Δ2xt-2+qtxt=ft,xt,t∈1,TZ.Thereupon a functionx∈X is a critical point of the functional I on X if and only if x is a solution of BVP (1), (2).LetP be the T×T matrix. If T≥5, let(23)P=b1a1p10⋯0pT-1aTa1b2a2p2⋯00pTp1a2b3a3⋯0000p2a3b4⋯000⋯⋯⋯⋯⋯⋯⋯⋯0000⋯aT-3pT-300000⋯bT-2aT-2pT-2pT-1000⋯aT-2bT-1aT-1aTpT00⋯pT-2aT-1bT,where b(k)=p(k)+4p(k-1)+p(k-2),a(k)=-2(p(k-1)+p(k)),k=1,2,…,T.IfT=4, let(24)P=p-1+4p0+p1-2p0+p1p1+p3-2p3+p4-2p0+p1p0+4p1+p2-2p1+p2p2+p4p1+p3-2p1+p2p1+4p2+p3-2p2+p3-2p3+p4p2+p4-2p2+p3p2+4p3+p4.IfT=3, let(25)P=p-1+4p0+p1p2-2p0+p1p1-2p2+p3p2-2p0+p1p0+4p1+p2p3-2p1+p2p1-2p2+p3p3-2p1+p2p1+4p2+p3.IfT=2, let(26)P=p-1+4p0+3p1-2p0-4p1-2p2-2p0-4p1-2p2p0+4p1+3p2.IfT=1, let P=(0).Then the functionalI(x) can be rewritten as(27)Ix=12x∗Px+12∑t=1Tqtxt2-∑t=1TGt,xt.Lemma 2 (linking theorem [30]).
LetX be a real Banach space, X=X1⊕X2, where X1 is finite dimensional. Assume that I∈C1(X,R) satisfies the Palais-Smale condition and the following:(I1)
There are positive constantsr and ρ such that I|∂Bρ∩X2≥r.(I2)
There is aν∈∂B1∩X2 and a positive constant c~≥ρ such that I|∂Φ≤0, where Φ=(B-c~∩X1)⊕{sν∣0<s<c~}.ThenI possesses a critical value c≥r, where(28)c=infφ∈Π⁡supx∈Π⁡Iφx,andΠ={φ∈C(Φ-,X)∣φ|∂Ω=id}, where id denotes the identity operator.Lemma 3 (Clark theorem [30]).
LetX be a real Banach space, I∈C1(X,R), with I being even, bounded from below, and satisfying Palais-Smale condition. Suppose I(0)=0, there is a set Ω⊂X such that Ω is homeomorphic to ST-1 (T-1 dimension unit sphere) by an odd map, and supΩ⁡I<0. Then I has at least T distinct pairs of nonzero critical points.


## 3. Main Results

We now state our main theorems in this paper.Theorem 4.
Suppose that the functionG(t,x)≥0 and the following conditions are satisfied:(p)
For anyt∈[-1,T]Z,  p(t)>0.(q)
For anyt∈[1,T]Z,  q(t)≤0,  q(0)=0 and λ_+q_>0.(W)
b(1)+a(1)+p(1)+p(T-1)+a(T)=0,  b(2)+a(2)+p(2)+p(T)+a(1)=0,  b(k)+a(k)+p(k)+p(k-2)+a(k-1)=0,k=3,4,…,T.(G1)
There are constantsρ1>0 and c1∈0,1/2(λ_+q_) such that(29)Gt,x≤c1x2,∀t∈1,TZ,x≤ρ1.(G2)
There are constantsc2∈1/2λ-+q-,+∞ and c3>0 such that(30)Gt,x≥c2x2-c3,∀t∈1,TZ,x∈R.Hereλ_ and λ- are constants which can be referred to (32) and (33).Then BVP (1), (2) has at least three solutions.Corollary 5.
Suppose that the functionG(t,x)≥0 and the conditions (p), (q), (W), (G1), and (G2) are satisfied. Then BVP (1), (2) has at least two nontrivial solutions.Theorem 6.
Suppose that the functionG(t,x)≥0,  (p),  (q),  (W) and the following conditions are satisfied:(G3)
For any(t,x)∈[1,T]Z×R, lim|x|→0⁡G(t,x)/x2=0.(G4)
For any(t,x)∈[1,T]Z×R, there are constants c4>0, α>2 and c5>0 such that(31)Gt,x≥c4xα-c5.Then BVP (1), (2) has at least three solutions.Corollary 7.
Suppose that the functionG(t,x)≥0, (p), (q), (W) and the conditions (G3) and (G4) are satisfied. Then BVP (1), (2) has at least two nontrivial solutions.Theorem 8.
Suppose that the functionG(t,x)≥0, (p), (q), (W), (G1), (G2) and the following condition are satisfied:(ψ)
f(t,-x)=-f(t,x),∀(t,x)∈[1,T]Z×R.Then BVP (1), (2) has at least n distinct pairs of nontrivial solutions, where n is the dimension of X2 which can be referred to (34).Theorem 9.
Suppose that the following conditions are satisfied:(p′)
For anyt∈[-1,T]Z,  p(t)≤0.(q′)
For anyt∈[1,T]Z,  q(t)≤0.(ϕ)
For anyx∉0,t∈[1,T]Z,  xf(t,x)>0.Then BVP (1), (2) has no nontrivial solutions.


## 4. Proofs of the Main Results

In this section, we shall finish proofs of the main results via variational methods.Proof of Theorem4.
The matrixP satisfies that P is positive semidefinite. In fact, from (W), it is obvious that 0 is an eigenvalue of P with an eigenvector (1,1,…,1)∗. Define the eigenvalues of P by λ1,λ2,…,λT.
Denote(32)λ_=min⁡λk∣λk≠0,k=1,2,…,T,and(33)λ-=max⁡λk∣λk≠0,k=1,2,…,T.
LetX1={(d,d,…,d)∗∈X∣d∈R}. Clearly, X1 is an invariant subspace of X. The space X2 is defined by(34)X=X1⊕X2.
Letxkk∈N⊂X be such that Ixkk∈N is bounded and I′xk→0 as k→∞. Thus, for any k∈N, there exists a positive constant C such that(35)-C≤Ixk≤C.For any xkk∈N⊂X, it comes from (27) and (G2) that(36)-C≤Ixk=12xk∗Pxk+12∑t=1Tqtxkt2-∑t=1TGt,xkt≤12λ-xk2+12q-xk2-∑t=1Tc2xkt2-c3=12λ-+q--c2xk2+c3T.Therefore,(37)c2-12λ-+q-xk2≤C+c3T.From (G2), we have c2>1/2λ-+q-. (37) implies that xkk∈N is a bounded in X. Since the dimension of X is finite, xkk∈N possesses a convergent subsequence. Consequently, the functional I(x) satisfies the Palais-Smale condition. Therefore, it suffices to prove that I(x) satisfies the conditions (I1) and (I2) of Lemma 2.
For anyt∈Z[1,T], it follows from (q), (G1), and (G2) that G(t,0)=0 and f(t,0)=0. Hence, x=0 is a trivial solution of BVP (1), (2).
From the proof of (36), we have that I(x) is bounded from above in X. Denote(38)c0=supx∈X⁡Ix.As a consequence, on the one hand, there is a sequence xi in X such that(39)c0=limi→∞⁡Ixi.By (27), on the other hand, the functional I(x) satisfies(40)Ix≤12λ-+q--c2x2+c3T,∀x∈X.(40) means that lim‖x‖→+∞⁡I(x)=-∞ which implies that xi is bounded. As a result, xi has a convergent subsequence in X denoted by xik. Let(41)x0=limk→+∞⁡xik.By the reason of the continuity of I(x) in x, it is easy to see that I(x0)=c0. That is, x0∈X is a critical point of I(x).
For anyx∈X2,‖x‖≤ρ1, via (27) and (G1), the functional I(x) satisfies(42)Ix=12x∗Px+12∑t=1Tqtxt2-∑t=1TGt,xt≥12λ_x2+12q_x2-c1∑t=1kxt2≥12λ_+q_-c1x2.Take(43)r=12λ_+q_-c1ρ12.Then(44)Ix≥r,∀x∈X2∩∂Bρ1.In other words, there exist two positive constants r and ρ1 such that I|∂Bρ1∩X2≥r. Hence, I(x) satisfies (I1) of Lemma 2.
For anyx∈X1,  Px=0. Combining with (q), we have(45)Ix=12x∗Px+12∑t=1Tqtxt2-∑t=1TGt,xt≤-∑t=1TGt,xt≤0.Therefore, x0∉X1 and the critical point x0 of I(x) corresponding to the critical value c0 is a nontrivial solution of BVP (1), (2).
Next, we shall prove(I2) of Lemma 2.
Takey∈∂B1∩X2; for any z∈X1 and ξ∈R, let x=ξy+z. By (F2), we have(46)Ix=12ξy+z∗Pξy+z+12∑t=1Tqtξyt+zt2-∑t=1TGt,ξyt+zt≤12ξy∗Pξy+12∑t=1Tqtξyt+zt2-∑n=1kc2ξyt+zt2-c3≤12λ-ξ2+12q-∑t=1Tξyt+zt2-c2∑t=1Tξyt+zt2+c3T=12λ-+q--c2ξ2+12q--c2z2+c3T≤12q--c2z2+c3T.Thereby, there exists a constant c~>ρ1 such that(47)Ix≤0,∀x∈∂Φ,where Φ=(B-c~∩X1)⊕{sν∣0<s<c~}. According to Lemma 2, I(x) has a critical value c≥r>0, where(48)c=infφ∈Π⁡supx∈Π⁡Iφx,and Π={φ∈C(Φ-,X)∣φ|∂Ω=id}.
Similar to the proof of Theorem 1.1 in [22], we can prove that BVP (1), (2) has at least three solutions. For simplicity, its proof is omitted.Remark 10.
From the course of the proof of Theorem4, the conclusion of Corollary 5 is evidently correct.Remark 11.
The techniques of the proof of Theorem6 are just the same as those carried out in the proof of Theorem 4. We do not repeat them here.Remark 12.
According to Theorem6, it is easy to see that the conclusion of Corollary 7 is true.Proof of Theorem8.
For anyt∈[1,T]Z, by the continuity of f(t,x) in x, I(x) can be viewed as a continuously differentiable functional defined on X. It comes from (q) and (G1) that I(0)=0. Owing to the condition (ψ), I(x) is even. From the process of proof of Theorem 4, I(x) is bounded from below and satisfies the Palais-Smale condition. Next, in the light of Clark Theorem, we shall find a set Ω and an odd map such that Ω is homeomorphic to ST-1 by an odd map.
Set(49)Ω=∂Bρ1∩X2.It is obvious that Ω is homeomorphic to ST-1 by an odd map. (44) implies that supΩ⁡(-I(x))<0. As a result of Clark Theorem, I(x) has at least T distinct pairs of nonzero critical points. As a consequence, BVP (1), (2) has at least T distinct pairs of nontrivial solutions. The desired result is obtained.Proof of Theorem9.
On the contrary, we suppose that BVP (1), (2) has a nontrivial solution. Therefore, I(x) has a nonzero critical point x~. Since(50)∂I∂x~t=Δ2ptΔ2x~t-2+qtx~t-ft,x~t,t∈1,TZ,we have(51)∑t=1Tft,x~tx~t=∑t=1TΔ2ptΔ2x~t-2+qtx~tx~t=∑t=1TptΔ2x~t2+∑t=1Tqtx~t2.On the one hand, it follows from (p′) and (q′) that(52)∑t=1Tft,x~tx~t≤0.On the other hand, by (ϕ), we have(53)∑t=1Tft,x~tx~t>0,which is a contradiction with (52).


## 5. Some Examples

In this section, we shall provide three examples to illustrate our main results.Example 1.
Fort∈[1,3]Z, assume that(54)Δ2t-22Δ2xt-2-t3xt=15xt4,satisfying the boundary value conditions(55)x-1=x2,Δx-1=Δx2,Δ2x-1=Δ2x2,Δ3x-1=Δ3x2.We have(56)pt=t2,qt=-t3,ft,ut=15xt4,t∈1,3Z,with(57)p-1=4,p0=9,and(58)Gt,ut=xt5,t∈1,3Z.Besides,(59)P=41-16-25-1617-1-25-126,and the eigenvalues of P are λ1=0,λ2=21, and λ3=63. It is easy to verify that all the suppositions of Theorem 4 are satisfied and then BVP (54), (55) has at least three solutions.Example 2.
Fort∈[1,4]Z, assume that(60)Δ2t-2Δ2xt-2-t6xt=xt3,satisfying the boundary value conditions(61)x-1=x3,Δx-1=Δx3,Δ2x-1=Δ2x3,Δ3x-1=Δ3x3.We have(62)pt=t,qt=-t6,ft,ut=xt3,t∈1,4Z,with(63)p-1=3,p0=4,and(64)Gt,ut=14xt4,t∈1,4Z.Besides,(65)P=20-104-14-1010-664-612-10-146-1018,and the eigenvalues of P are λ1=0,λ2≈6.8154,λ3≈11.1773, and λ4≈42.0074. It is easy to verify that all the suppositions of Theorem 8 are satisfied and then BVP (60), (61) has at least 3 distinct pairs of nontrivial solutions.Example 3.
Fort∈[1,5]Z, assume that(66)-Δ2t-2Δ2xt-2-t6xt=xt5,satisfying the boundary value conditions(67)x-1=x4,Δx-1=Δx4,Δ2x-1=Δ2x4,Δ3x-1=Δ3x4.We have(68)pt=-t,qt=-t6,ft,ut=xt5,t∈1,5Z,with(69)p-1=-4,p0=-5,and(70)Gt,ut=16xt6,t∈1,5Z.It is easy to verify that all the suppositions of Theorem 9 are satisfied and then BVP (66), (67) has no nontrivial solutions.



---

*Source: 6703503-2018-09-16.xml*