Abstract

By applying a variant version of Mountain Pass Theorem in critical point theory, we prove the existence of homoclinic solutions for the following asymptotically 𝑝-linear difference system with 𝑝-Laplacian Δ(|Δ𝑢(𝑛1)|𝑝2Δ𝑢(𝑛1))+[𝐾(𝑛,𝑢(𝑛))+𝑊(𝑛,𝑢(𝑛))]=0, where 𝑝(1,+), 𝑛, 𝑢𝑁, 𝐾,𝑊×𝑁 are not periodic in 𝑛, and W is asymptotically 𝑝-linear at infinity.

1. Introduction

Consider the following 𝑝-Laplacian difference system:Δ||||Δ𝑢(𝑛1)𝑝2[]Δ𝑢(𝑛1)+𝐾(𝑛,𝑢(𝑛))+𝑊(𝑛,𝑢(𝑛))=0,𝑛,(1.1) where Δ is the forward difference operator defined by Δ𝑢(𝑛)=𝑢(𝑛+1)𝑢(𝑛), Δ2𝑢(𝑛)=Δ(Δ𝑢(𝑛)), 𝑝(1,+), 𝑛, 𝑢𝑁, 𝐾, 𝑊: ×𝑁 are not periodic in 𝑛, 𝑊 is asymptotically 𝑝-linear at infinity, and 𝐾 and 𝑊 are continuously differentiable in 𝑥. As usual, we say that a solution 𝑢(𝑛) of (1.1) is homoclinic (to 0) if 𝑢(𝑛)0 as 𝑛±. In addition, if 𝑢(𝑛)0, then 𝑢(𝑛) is called a nontrivial homoclinic solution.

When 𝑝=2, (1.1) can be regarded as a discrete analogue of the following second-order Hamiltonian system:[]̈𝑢(𝑡)+𝐾(𝑡,𝑢(𝑡))+𝑊(𝑡,𝑢(𝑡))=0,𝑡.(1.2)

The existence of homoclinic orbits for Hamiltonian systems is a classical problem and its importance in the study of the behavior of dynamical systems has been recognized by Poincaré [1]. If a system has the transversely intersected homoclinic orbits, then it must be chaotic. If it has the smoothly connected homoclinic orbits, then it cannot stand the perturbation and its perturbed system probably produces chaotic phenomenon. For the existence of homoclinic solutions of problem (1.2), one can refer to the papers [25].

Difference equations usually describe evolution of certain phenomena over the course of time. For example, if a certain population has discrete generations, the size of the (𝑛+1)th generation 𝑥(𝑛+1) is a function of the 𝑛th generation 𝑥(𝑛). In fact, difference equations provide a natural description of many discrete models in real world. Since discrete models exist in various fields of science and technology such as statistics, computer science, electrical circuit analysis, biology, neural network, and optimal control, it is of practical importance to investigate the solutions of difference equations. For more details about difference equations, we refer the readers to the books [68].

In some recent papers [920], the authors studied the existence of periodic solutions and subharmonic solutions of difference equations by applying critical point theory. These papers show that the critical point theory is an effective method to the study of periodic solutions for difference equations. Along this direction, several authors [2128] used critical point theory to study the existence of homoclinic orbits for difference equations. Motivated by the above papers, we consider the existence of homoclinic orbits for problem (1.1) by using the variant version of Mountain Pass Theorem. Our result is new, which seems not to have been considered in the literature. Here is our main result.

Theorem 1.1. Suppose that 𝐾 and 𝑊 satisfy the following conditions. (K1)There are two positive constants 𝑏1 and 𝑏2 such that 𝑏1||𝑥|𝑝𝐾(𝑛,𝑥)𝑏2||𝑥|𝑝,(𝑛,𝑥)×𝑁.(1.3)(K2)There is a positive constant 𝑏3 such that 𝑏3|𝑥|𝑝||||(𝐾(𝑛,𝑥),𝑥)𝐾(𝑛,𝑥)|𝑥|𝑝𝐾(𝑛,𝑥),(𝑛,𝑥)×𝑁.(1.4)(W1)𝑊(𝑛,0)=0, 𝑊(𝑛,𝑥)=𝑜(|𝑥|𝑝1) as |𝑥|0 uniformly for 𝑛.(W2)There exists a constant 𝑅>0 such that ||||𝑊(𝑛,𝑥)|𝑥|𝑝1𝑅,𝑛,𝑥𝑁.(1.5)(W3)There exists a function 𝑉𝑙(,+) such that lim|𝑥|||𝑊(𝑛,𝑥)𝑉(𝑛)|𝑥|𝑝2𝑥|||𝑥|𝑝1=0uniformlyfor𝑛,inf𝑉(𝑛)>max1,𝑝𝑏2.(1.6)(W4)𝑊(𝑛,𝑥)=(𝑊(𝑛,𝑥),𝑥)𝑝𝑊(𝑛,𝑥), lim|𝑥|𝑊(𝑛,𝑥)=+uniformlyfor𝑛,(1.7) and for any fixed 0<𝑐1<𝑐2<+, infn,𝑐1|𝑥|𝑐2𝑊(𝑛,𝑥)|𝑥|𝑝>0.(1.8)Then problem (1.1) has at least one nontrivial homoclinic solution.

Remark 1.2. The function 𝑊(𝑛,𝑥) in this paper is asymptotically 𝑝-linear at infinity. The behavior of the gradient of 𝑊(𝑛,𝑥) at infinity is like that of a function 𝑉(𝑛)|𝑥|𝑝2𝑥, where 𝑉(𝑛) is a real function but not a matrix function. To the best of our knowledge, similar results of this kind of 𝑝-Laplacian difference systems with asymptotically 𝑝-linear 𝑊(𝑛,𝑥) at infinity cannot be found in the literature. From this point, our result is new.

2. Preliminaries

Let 𝑆={𝑢(𝑛)}𝑛𝑢(𝑛)𝑁,,𝑛𝐸=𝑢𝑆𝑛||||Δ𝑢(𝑛1)𝑝+||||𝑢(𝑛)𝑝,<+(2.1) and for 𝑢𝐸, let 𝑢=𝑛||||Δ𝑢(𝑛1)𝑝+||||𝑢(𝑛)𝑝<+1/𝑝.(2.2) Then 𝐸 is a uniform convex Banach space with this norm. As usual, for 1𝑝<+, let 𝑙𝑝,𝑁=𝑢𝑆𝑛||||𝑢(𝑛)𝑝<+,𝑙,𝑁=𝑢𝑆sup𝑛||||,𝑢(𝑛)<+(2.3) and their norms are given by 𝑢𝑙𝑝=𝑛||||𝑢(𝑛)𝑝1/𝑝,𝑢𝑙𝑝,𝑁,𝑢||𝑢||=sup(𝑛)𝑛,𝑢𝑙,𝑁,(2.4) respectively.

For any 𝑢𝐸, let1𝜑(𝑢)=𝑝𝑛||||Δ𝑢(𝑛1)𝑝𝑛[].𝐾(𝑛,𝑢(𝑛))+𝑊(𝑛,𝑢(𝑛))(2.5)

To prove our results, we need the following generalization of Lebesgue's dominated convergence theorem.

Lemma 2.1 (see [29]). Let {𝑓𝑘(𝑡)} and {𝑔𝑘(𝑡)} be two sequences of measurable functions on a measurable set 𝐴, and let ||𝑓𝑘||(𝑡)𝑔𝑘(𝑡),𝑎.𝑒.𝑡𝐴.(2.6) If lim𝑘𝑓𝑘(𝑡)=𝑓(𝑡),lim𝑘𝑔𝑘(𝑡)=𝑔(𝑡),𝑎.𝑒.𝑡𝐴,lim𝑘𝐴𝑔𝑘(𝑡)𝑑𝑡=𝐴𝑔(𝑡)𝑑𝑡<+,(2.7) then lim𝑘𝐴𝑓𝑘(𝑡)𝑑𝑡=𝐴𝑓(𝑡)𝑑𝑡.(2.8)

Lemma 2.2. For 𝑢𝐸, 𝑢𝑢𝑙𝑝2𝑢.(2.9)

Proof. Since 𝑢𝐸, it follows that lim|𝑛||𝑢(𝑛)|=0. Hence, there exists 𝑛 such that 𝑢=||𝑢𝑛||=max𝑛||𝑢||.(𝑛)(2.10) Hence, we have 𝑢𝑛||||𝑢(𝑛)𝑝1/𝑝=𝑢𝑙𝑝=𝑛||||𝑢(𝑛)𝑢(𝑛1)+𝑢(𝑛1)𝑝1/𝑝𝑛||||+||||𝑢(𝑛)𝑢(𝑛1)𝑢(𝑛1)𝑝1/𝑝2𝑝𝑛||||𝑢(𝑛)𝑢(𝑛1)𝑝+||||𝑢(𝑛1)𝑝1/𝑝=2𝑛||||Δ𝑢(𝑛1)𝑝+||||𝑢(𝑛1)𝑝1/𝑝=2𝑛||||Δ𝑢(𝑛1)𝑝+||||𝑢(𝑛)𝑝1/𝑝=2𝑢.(2.11)

Lemma 2.3. Suppose that (K1), (K2), and (W2) hold. If 𝑢𝑘𝑢 in 𝐸, then 𝐾(𝑛,𝑢𝑘)𝐾(𝑛,𝑢) and 𝑊(𝑛,𝑢𝑘)𝑊(𝑛,𝑢) in 𝑙𝑝(,𝑁), where 𝑝>1 satisfies 1/𝑝+1/𝑝=1.

Proof. From (K1) and (K2), we have ||||𝐾(𝑛,𝑥)𝑝𝑏2|𝑥|𝑝1,(𝑛,𝑥)×𝑁.(2.12) Hence, from (2.12), we have ||𝐾𝑛,𝑢𝑘||(𝑛)𝐾(𝑛,𝑢(𝑛))𝑝𝑝𝑏2||𝑢𝑘||(𝑛)𝑝1+||||𝑢(𝑛)𝑝1𝑝𝑝𝑏22𝑝1||𝑢𝑘(||𝑛)𝑢(𝑛)𝑝1+𝑝𝑏21+2𝑝1||||𝑢(𝑛)𝑝1𝑝2𝑝𝑝𝑝𝑏2𝑝||𝑢𝑘||(𝑛)𝑢(𝑛)𝑝+2𝑝𝑝𝑏2𝑝1+2𝑝1𝑝||||𝑢(𝑛)𝑝=𝑔𝑘(𝑛).(2.13) Moreover, since 𝑢𝑘𝑢 in 𝑙𝑝(,𝑁) and 𝑢𝑘(𝑛)𝑢(𝑛) for almost every 𝑛, hence, lim𝑘𝑔𝑘(𝑛)=2𝑝𝑝𝑏2𝑝1+2𝑝1𝑝||||𝑢(𝑛)𝑝=𝑔(𝑛),a.e.𝑛,lim𝑘𝑛𝑔𝑘(𝑛)=lim𝑘𝑛2𝑝𝑝𝑝𝑏2𝑝||𝑢𝑘||(𝑛)𝑢(𝑛)𝑝+2𝑝𝑝𝑏2𝑝1+2𝑝1𝑝||||𝑢(𝑛)𝑝=2𝑝𝑝𝑝𝑏2𝑝lim𝑘𝑛||𝑢𝑘||(𝑛)𝑢(𝑛)𝑝+2𝑝𝑝𝑏2𝑝1+2𝑝1𝑝𝑛||𝑢||(𝑛)𝑝=2𝑝𝑝𝑏2𝑝1+2𝑝1𝑝𝑛||||𝑢(𝑛)𝑝=𝑛𝑔(𝑛)<+.(2.14) It follows from Lemma  2.1, (2.13), and the previous equations that lim𝑘𝑛||𝐾𝑛,𝑢𝑘||(𝑛)𝐾(𝑛,𝑢(𝑛))𝑝=0.(2.15) This shows that 𝐾(𝑛,𝑢𝑘)𝐾(𝑛,𝑢) in 𝑙𝑝(,𝑁). By a similar proof, we can prove that 𝑊(𝑛,𝑢𝑘)𝑊(𝑛,𝑢) in 𝑙𝑝(,𝑁). The proof is complete.

Lemma 2.4. Under the conditions of Theorem 1.1, one has 𝜑(𝑢),𝑣=𝑛||||Δ𝑢(𝑛1)𝑝2(Δ𝑢(𝑛1),Δ𝑣(𝑛1))+(𝐾(𝑛,𝑢(𝑛))𝑊(𝑛,𝑢(𝑛)),𝑣(𝑛))(2.16) for 𝑢,𝑣𝐸, which yields that 𝜑(𝑢),𝑢=𝑛||||Δ𝑢(𝑛1)𝑝.+(𝐾(𝑛,𝑢(𝑛)),𝑢(𝑛))(𝑊(𝑛,𝑢(𝑛)),𝑢(𝑛))(2.17) Moreover, 𝜑 is continuously Fréchet-differential defined on 𝐸; that is, 𝜑𝐶1(𝐸,) and any critical point 𝑢 of 𝜑 on 𝐸 is classical solution of (1.1) with 𝑢(±)=0.

Proof. Firstly, we show that 𝜑𝐸. Let 𝑢𝐸, by (2.9) and (K1), we have 𝑛𝐾(𝑛,𝑢(𝑛))𝑛𝑏2||||𝑢(𝑛)𝑝𝑏22𝑝𝑢𝑝.(2.18) By (W2), we get ||𝑊||=||||(𝑛,𝑥)10||||(𝑊(𝑛,𝑠𝑥),𝑥)𝑑𝑠𝑅|𝑥|𝑝,(𝑛,𝑥)×𝑁.(2.19) Hence, from (2.9) and (2.19), we have |||||𝑛|||||𝑊(𝑛,𝑢(𝑛))𝑛||||𝑊(𝑛,𝑢(𝑛))𝑛𝑅||||𝑢(𝑛)𝑝𝑅2𝑝𝑢𝑝.(2.20) It follows from (2.5), (2.18), and (2.20) that 𝜑𝐸. Next we prove that 𝜑𝐶1(𝐸,). Rewrite 𝜑 as follows: 𝜑(𝑢)=𝜑1(𝑢)+𝜑2(𝑢)𝜑3(𝑢),(2.21) where 𝜑11(𝑢)=𝑝𝑛||||Δ𝑢(𝑛1)𝑝,𝜑2(𝑢)=𝑛𝐾(𝑛,𝑢(𝑛)),𝜑3(𝑢)=𝑛𝑊(𝑛,𝑢(𝑛)).(2.22) It is easy to check that 𝜑1𝐶1(𝐸,) and 𝜑1=(𝑢),𝑣𝑛||||Δ𝑢(𝑛1)𝑝2(Δ𝑢(𝑛1),Δ𝑣(𝑛1)),𝑢,𝑣𝐸.(2.23) Next, we prove that 𝜑𝑖𝐶1(𝐸,),𝑖=2,3, and 𝜑2(=𝑢),𝑣𝑛(𝐾(𝑛,𝑢(𝑛)),𝑣(𝑛)),𝑢,𝑣𝐸,(2.24)𝜑3=(𝑢),𝑣𝑛(𝑊(𝑛,𝑢(𝑛)),𝑣(𝑛)),𝑢,𝑣𝐸.(2.25) For any 𝑢,𝑣𝐸 and for any function 𝜃(0,1), by (K2), we have 𝑛max[]0,1||||(𝐾(𝑛,𝑢(𝑛)+𝜃(𝑡)𝑣(𝑛)),𝑣(𝑛))𝑝𝑏2𝑛max[]0,1||||𝑢(𝑛)+𝜃(𝑡)𝑣(𝑛)𝑝1||||𝑣(𝑛)2𝑝1𝑝𝑏2𝑛||||𝑢(𝑛)𝑝1+||||𝑣(𝑛)𝑝1||||𝑣(𝑛)2𝑝1𝑝𝑏2𝑢𝑙𝑝1𝑝𝑣𝑙𝑝+𝑣𝑝𝑙𝑝<+.(2.26) Then by the previous equations and Lebesgue's dominated convergence theorem, we have 𝜑2(𝑢),𝑣=lim0+𝜑2(𝑢+𝑣)𝜑2(𝑢)=lim0+1𝑛[]𝐾(𝑛,𝑢(𝑛)+𝑣(𝑛))𝐾(𝑛,𝑢(𝑛))=lim0+𝑛=(𝐾(𝑡,𝑢(𝑛)+𝜃(𝑡)𝑣(𝑛)),𝑣(𝑛))𝑛(𝐾(𝑛,𝑢(𝑛)),𝑣(𝑛)),𝑢,𝑣𝐸.(2.27) Similarly, we can prove that (2.25) holds by using (W2) instead of (K2). Finally, we prove that 𝜑𝑖𝐶1(𝐸,),𝑖=2,3. Let 𝑢𝑘𝑢 in 𝐸; then by Lemma 2.3, we have ||𝜑2𝑢𝑘𝜑2||=|||||(𝑢),𝑣𝑛𝐾𝑛,𝑢𝑘|||||(𝑛)𝐾(𝑛,𝑢(𝑛)),𝑣(𝑛)𝑛||𝐾𝑛,𝑢𝑘||(𝑛)𝐾(𝑛,𝑢(𝑛))𝑣(𝑛)𝑣𝑛||𝐾𝑛,𝑢𝑘||(𝑛)𝐾(𝑛,𝑢(𝑛))𝑝1/𝑝0,𝑘,𝑣𝐸.(2.28) This shows that 𝜑2𝐶1(𝐸,). Similarly, we can prove that 𝜑3𝐶1(𝐸,). Furthermore, by a standard argument, it is easy to show that the critical points of 𝜑 in 𝐸 are classical solutions of (1.1) with 𝑢(±)=0. The proof is complete.

Lemma 2.5 (see [30]). Let 𝐸 be a real Banach space with its dual space 𝐸 and suppose that 𝜑𝐶1(𝐸,) satisfies max{𝜑(0),𝜑(𝑒)}𝜂0<𝜂inf𝑢=𝜌𝜑(𝑢),(2.29) for some 𝜂0<𝜂, 𝜌>0, and 𝑒𝐸 with 𝑒>𝜌. Let 𝑐𝜂 be characterized by 𝑐=infΥΓmax0𝜏1𝜑(Υ(𝜏)),(2.30) where Γ={Υ𝐶([0,1],𝐸)Υ(0)=0,Υ(1)=𝑒} is the set of continuous paths joining 0 to 𝑒; then there exists {𝑢𝑘}𝑘𝐸 such that 𝜑𝑢𝑘𝑢𝑐,1+𝑘𝜑(𝑢𝑘)𝐸0as𝑘.(2.31)

3. Proof of Theorem 1.1

Proof of Theorem 1.1. We divide the proof of Theorem 1.1 into three steps.
Step 1. From (W1), there exists 𝜌0>0 such that 𝐶𝑊(𝑛,𝑥)12𝑝|𝑥|𝑝1,𝑛,|𝑥|𝜌0,(3.1) where 𝐶1=min{1/𝑝,𝑏1}. From (3.1), we have 𝑊(𝑛,𝑥)=10(𝑊(𝑛,𝑠𝑥),𝑥)𝑑𝑠10𝐶12𝑝|𝑥|𝑝𝑠𝑝1𝐶𝑑𝑠=1𝑝2𝑝|𝑥|𝑝,𝑛,|𝑥|𝜌0.(3.2) Let 𝜌=𝜌0/2 and 𝑆={𝑢𝐸𝑢=𝜌}; then from (2.9), we obtain 𝑢𝜌0,𝑢𝑙𝑝2𝜌,𝑢𝑆,(3.3) which together with (2.9), (3.2), and (K1) implies that 1𝜑(𝑢)=𝑝𝑛||||Δ𝑢(𝑛1)𝑝𝑛[]1𝐾(𝑛,𝑢(𝑛))+𝑊(𝑛,𝑢(𝑛))𝑝𝑛||||Δ𝑢(𝑛1)𝑝+𝑏1𝑛||||𝑢(𝑛)𝑝𝑛𝐶1𝑝2𝑝||||𝑢(𝑛)𝑝1min𝑝,𝑏1𝑢𝑝𝐶1𝑝2𝑝𝑢𝑝𝑙𝑝𝐶1𝑢𝑝𝐶1𝑝𝑢𝑝=(𝑝1)𝐶1𝑝𝑢𝑝=𝛼1>0,𝑢𝑆.(3.4)Step 2. From (K1), we have 1𝜑(𝑢)=𝑝𝑛||||Δ𝑢(𝑛1)𝑝𝑛[]1𝐾(𝑛,𝑢(𝑛))+𝑊(𝑛,𝑢(𝑛))𝑝𝑛||||Δ𝑢(𝑛1)𝑝+𝑏2𝑛||||𝑢(𝑛)𝑝𝑛1𝑊(𝑛,𝑢(𝑛))max𝑝,𝑏2𝑢𝑝𝑛𝑊(𝑛,𝑢(𝑛))𝐶2𝑢𝑝𝑛𝑊(𝑛,𝑢(𝑛)).(3.5) By (W2) and (W3), we get lim|𝑥|𝑝𝑊(𝑛,𝑥)|𝑥|𝑝=𝑉(𝑛)uniformlyfor𝑛.(3.6) Let 𝑊(𝑛,𝑥)=𝑝𝑊(𝑛,𝑥)𝑉(𝑛)|𝑥|𝑝; it follows from (W2), (W3), (2.19), and (3.6) that 𝑊(𝑛,𝑥)𝑝𝑅+sup𝑉(|𝑛)𝑥|𝑝,𝑥𝑁,lim|𝑥|𝑊(𝑛,𝑥)|𝑥|𝑝=0.(3.7) Define 𝐸1={𝑢(𝑛)=𝑥𝑒|𝑛|𝑥𝑁,𝑛}𝐸 with inf𝑉(𝑛)>max1,𝑝𝑏2||1+1𝑒|𝑛||𝑛1|||𝑝.(3.8) By an easy calculation, we have 𝑢𝑝=||1+1𝑒|𝑛||𝑛1|||𝑝𝑢𝑝𝑙𝑝.(3.9) In what follows, we prove that for some 𝑢𝐸1 with 𝑢=1, 𝜑(𝑠𝑢) as 𝑠. Otherwise, there exist a sequence {𝑠𝑘} with 𝑠𝑘 as 𝑘 and a positive constant 𝐶3 such that 𝜑(𝑠𝑘𝑢)𝐶3 for all 𝑘. From (3.5), we obtain 𝐶3𝑠𝑝𝑘𝜑𝑠𝑘𝑢𝑠𝑝𝑘𝐶21𝑝𝑛𝑊𝑛,𝑠𝑘𝑢(𝑛)𝑠𝑝𝑘1𝑝𝑛𝑉||||(𝑛)𝑢(𝑛)𝑝𝐶21𝑝𝑛𝑊𝑛,𝑠𝑘𝑢(𝑛)𝑠𝑝𝑘1𝑝inf𝑉(𝑛)𝑢𝑝𝑙𝑝.(3.10) It follows from (3.7) that 𝑊𝑛,𝑠𝑘𝑢(𝑛)𝑠𝑝𝑘𝑝𝑅+sup𝑉||||(𝑛)𝑢(𝑛)𝑝,𝑊𝑛,𝑠𝑘𝑢(𝑛)||𝑠𝑘||𝑝0as𝑘.(3.11) Hence, from Lebesgue's dominated theorem and (3.11), we have 𝑛𝑊𝑛,𝑠𝑛𝑢(𝑛)||𝑠𝑘||𝑝0as𝑘.(3.12) It follows from (3.8), (3.9), (3.10), and (3.12) that 𝐶03𝑠𝑝𝑘𝐶21𝑝||1+1𝑒|𝑛||𝑛1|||𝑝inf𝑉(𝑛)<0as𝑘,(3.13) which is a contradiction. Hence, there exists 𝑒𝐸 with 𝑒>𝜌 such that 𝜑(𝑒)0.Step 3. From Step 1, Step 2, and Lemma 2.5, we know that there is a sequence {𝑢𝑘}𝑘𝐸 such that 𝜑𝑢𝑘𝑢𝑐,1+𝑘𝜑𝑢𝑘𝐸0as𝑘,(3.14) where 𝐸 is the dual space of 𝐸. In the following, we will prove that {𝑢𝑘}𝑘 is bounded in 𝐸. Otherwise, assume that 𝑢𝑘 as 𝑘. Let 𝑧𝑘=𝑢𝑘/𝑢𝑘; we have 𝑧𝑘=1. It follows from (2.5), (2.16), (3.14), and (K2) that 𝐶4𝑢𝑝𝜑𝑘𝑢𝜑𝑘,𝑢𝑘=𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)𝑝𝑊𝑛,𝑢𝑘+(𝑛)𝑛𝑝𝐾𝑛,𝑢𝑘(𝑛)𝐾𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)𝑝𝑊𝑡,𝑢𝑘(𝑛)=𝑛𝑊𝑛,𝑢𝑘.(𝑛)(3.15) Set Ω𝑘(𝛼,𝛽)={𝑛𝛼|𝑢𝑘(𝑛)|𝛽} for 0<𝛼<𝛽. Then from (3.15), we have 𝐶4𝑛Ω𝑘(0,𝛼)𝑊𝑛,𝑢𝑘(+𝑛)𝑛Ω𝑘(𝛼,𝛽)𝑊𝑛,𝑢𝑘(+𝑛)𝑛Ω𝑘(𝛽,+)𝑊𝑛,𝑢𝑘(.𝑛)(3.16) From (K1), (K2), and (3.14), we get 𝑜𝑢(1)=𝜑𝑘,𝑢𝑘=𝑛||Δ𝑢𝑘||(𝑛1)𝑝+𝐾𝑛,𝑢𝑘(𝑛)𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)𝑛||Δ𝑢𝑘(||𝑛1)𝑝+𝑏3||𝑢𝑘(||𝑛)𝑝𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)min1,𝑏3𝑢𝑘𝑝𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)=𝐶5𝑢𝑘𝑝𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘=𝑢(𝑛)𝑘𝑝𝐶5𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)𝑢𝑘𝑝,(3.17) which implies that limsup𝑘𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)||𝑢𝑘||(𝑛)𝑝||𝑧𝑘||(𝑛)𝑝=limsup𝑘𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)𝑢𝑘𝑝𝐶5.(3.18) Let 0<𝜀<𝐶5/3. From (W1), there exists 𝛼𝜀>0 such that ||||𝜀𝑊(𝑛,𝑥)2𝑝|𝑥|𝑝1for|𝑥|𝛼𝜀uniformlyfor𝑛.(3.19) Since 𝑧𝑘=1, it follows from (2.9) and (3.19) that 𝑛Ω𝑘(0,𝛼𝜀)||𝑊𝑛,𝑢𝑘||(𝑛)||𝑢𝑘||(𝑛)𝑝1||𝑧𝑘||(𝑛)𝑝𝑛Ω𝑘(0,𝛼𝜀)𝜀2𝑝||𝑧𝑘||(𝑛)𝑝𝜀,𝑘.(3.20) For 𝑠>0, let (𝑠)=inf𝑊(𝑛,𝑥)𝑛,𝑥𝑁with.|𝑥|𝑠(3.21) Thus, from (W4), we have (𝑠)+ as 𝑠+, which together with (3.16) implies that Ωmeas𝑘𝐶(𝛽,+)6(𝛽)0,as𝛽+.(3.22) Hence, we can take 𝛽𝜀 sufficiently large such that 𝑛Ω𝑘𝛽𝜀,+||𝑧𝑘||(𝑛)𝑝<𝜀𝑅.(3.23) The previous inequality and (W2) imply that 𝑛Ω𝑘𝛽𝜀,+||𝑊𝑛,𝑢𝑘||(𝑛)||𝑢𝑘||(𝑛)𝑝1||𝑧𝑘||(𝑛)𝑝𝑅𝑛Ω𝑘𝛽𝜀,+||𝑧𝑘||(𝑛)𝑝<𝜀,𝑘.(3.24) Next, for the previous 0<𝛼𝜀<𝛽𝜀, let 𝑐𝜀=inf𝑊(𝑛,𝑥)|𝑥|𝑝𝑛,𝑥𝑁with𝛼𝜀|𝑥|𝛽𝜀,𝑑𝜀||||=max𝑊(𝑛,𝑥)|𝑥|𝑝1𝑛,𝑥𝑁with𝛼𝜀|𝑥|𝛽𝜀.(3.25) From (W4), we have 𝑐𝜀>0 and 𝑊𝑛,𝑢𝑘(𝑛)𝑐𝜀||𝑢𝑘(||𝑛)𝑝,𝑛Ω𝑘𝛼𝜀,𝛽𝜀.(3.26) From (3.15) and (3.26), we get 𝑛Ω𝑘(𝛼𝜀,𝛽𝜀)||𝑧𝑘||(𝑛)𝑝=1𝑢𝑘𝑝𝑛Ω𝑘(𝛼𝜀,𝛽𝜀)||𝑢𝑘||(𝑛)𝑝1𝑢𝑘𝑝𝑛Ω𝑘𝛼𝜀,𝛽𝜀1𝑐𝜀𝑊𝑛,𝑢𝑘𝐶(𝑛)4𝑐𝜀𝑢𝑘𝑝0as𝑘,(3.27) which implies that 𝑛Ω𝑘𝛼𝜀,𝛽𝜀||𝑊𝑛,𝑢𝑘||(𝑛)||𝑢𝑘||(𝑛)𝑝1||𝑧𝑘||(𝑛)𝑝𝑑𝜀𝑛Ω𝑘𝛼𝜀,𝛽𝜀||𝑧𝑘||(𝑛)𝑝0as𝑘.(3.28) Therefore, there exists 𝑘0>0 such that 𝑛Ω𝑘𝛼𝜀,𝛽𝜀||𝑊𝑛,𝑢𝑘||(𝑛)||𝑢𝑘||(𝑛)𝑝1||𝑧𝑘||(𝑛)𝑝𝜀,𝑘𝑘0.(3.29) It follows from (3.20), (3.24), and (3.29) that 𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)||𝑢𝑘||(𝑛)𝑝||𝑧𝑘||(𝑛)𝑝𝑛||𝑊𝑡,𝑢𝑘||(𝑛)||𝑢𝑘||(𝑛)𝑝1||𝑧𝑘||(𝑛)𝑝<3𝜀<𝐶5,𝑘𝑘0,(3.30) which implies that limsup𝑛𝑛𝑊𝑛,𝑢𝑘(𝑛),𝑢𝑘(𝑛)||𝑢𝑘||(𝑛)𝑝||𝑧𝑘||(𝑛)𝑝<𝐶5,(3.31) but this contradicts to (3.18). Hence, 𝑢𝑘 is bounded in 𝐸.
Going to a subsequence if necessary, we may assume that there exists 𝑢𝐸 such that 𝑢𝑘𝑢 as 𝑘. In order to prove our theorem, it is sufficient to show that 𝜑(𝑢)=0. For any 𝑎 with 𝑎>0, let 𝜒𝑎(𝑡)=1 for 𝑡[𝑎,𝑎] and let 𝜒𝑎(𝑡)=0 for 𝑡(,𝑎)(𝑎,). Then from (2.16), we have 𝑢𝜑𝑘𝜑(𝑢),𝜒𝑎𝑢𝑘=𝑢[]𝑛𝑎,𝑎||Δ𝑢𝑘||(𝑛1)𝑝2Δ𝑢𝑘(𝑛1),Δ𝑢𝑘(𝑛1)Δ𝑢(𝑛1)[]𝑛𝑎,𝑎||||Δ𝑢(𝑛1)𝑝2Δ𝑢(𝑛1),Δ𝑢𝑘+(𝑛1)Δ𝑢(𝑛1)[]𝑛𝑎,𝑎𝐾𝑛,𝑢𝑘(𝑛)𝐾(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)[]𝑛𝑎,𝑎𝑊𝑛,𝑢𝑘(𝑛)𝑊(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)Δ𝑢𝑘𝑝𝑙𝑝[]𝑎,𝑎+Δ𝑢𝑝𝑙𝑝[]𝑎,𝑎[]𝑛𝑎,𝑎||Δ𝑢𝑘||(𝑛1)𝑝1||||Δ𝑢(𝑛1)[]𝑛𝑎,𝑎||||Δ𝑢(𝑛1)𝑝1||Δ𝑢𝑘||+(𝑛1)[]𝑛𝑎,𝑎𝐾𝑛,𝑢𝑘(𝑛)𝐾(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)[]𝑛𝑎,𝑎𝑊𝑛,𝑢𝑘(𝑛)𝑊(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)Δ𝑢𝑘𝑝𝑙𝑝[]𝑎,𝑎+Δ𝑢𝑝𝑙𝑝[𝑎,𝑎]Δ𝑢𝑙𝑝[𝑎,𝑎]Δ𝑢𝑘𝑙𝑝1𝑝[𝑎,𝑎]Δ𝑢𝑘𝑙𝑝[𝑎,𝑎]Δ𝑢𝑙𝑝1𝑝[𝑎,𝑎]+𝑛[𝑎,𝑎]𝐾𝑛,𝑢𝑘(𝑛)𝐾(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)[]𝑛𝑎,𝑎𝑊𝑛,𝑢𝑘(𝑛)𝑊(𝑛,𝑢(𝑛)),𝑢𝑘=(𝑛)𝑢(𝑛)Δ𝑢𝑘𝑙𝑝1𝑝[]𝑎,𝑎Δ𝑢𝑙𝑝1𝑝[]𝑎,𝑎Δ𝑢𝑘𝑙𝑝[𝑎,𝑎]Δ𝑢𝑙𝑝[𝑎,𝑎]+[]𝑛𝑎,𝑎𝐾𝑛,𝑢𝑘(𝑛)𝐾(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)[]𝑛𝑎,𝑎𝑊𝑛,𝑢𝑘(𝑛)𝑊(𝑛,𝑢(𝑛)),𝑢𝑘.(𝑛)𝑢(𝑛)(3.32) Since 𝜑(𝑢𝑘)0 as 𝑘+ and 𝑢𝑘𝑢 in 𝐸, it follows from (3.14) that 𝑢𝜑𝑘𝜑(𝑢),𝜒𝑎𝑢𝑘𝑢0as𝑘,𝑛[𝑎,𝑎]𝐾𝑛,𝑢𝑘(𝑛)𝐾(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)0as𝑘,𝑛[𝑎,𝑎]𝑊𝑛,𝑢𝑘(𝑛)𝑊(𝑛,𝑢(𝑛)),𝑢𝑘(𝑛)𝑢(𝑛)0as𝑘.(3.33) It follows from (3.32) and (3.33) that Δ𝑢𝑘𝑙𝑝[𝑎,𝑎]Δ𝑢𝑙𝑝[𝑎,𝑎] as 𝑘+.
For any 𝑤𝐶0(,𝑁), and assume that for some 𝐴 with 𝐴>0, supp(𝑤)[𝐴,𝐴]. Since lim𝑘Δ𝑢𝑘(𝑛1)=Δ𝑢(𝑛1),a.e|||||.𝑛,Δ𝑢𝑘||(𝑛1)𝑝2Δ𝑢𝑘|||(𝑛1),Δ𝑤(𝑛1)𝑝1𝑝||Δ𝑢𝑘||(𝑛1)𝑝+1𝑝||||Δ𝑤(𝑛1)𝑝,𝑛,𝑘=1,2,,lim𝑘[]𝑛𝐴,𝐴𝑝1𝑝||Δ𝑢𝑘||(𝑛1)𝑝+1𝑝||||Δ𝑤(𝑛1)𝑝=𝑝1𝑝lim𝑘Δ𝑢𝑘𝑝𝑙𝑝[]𝐴,𝐴+1𝑝Δ𝑤𝑝𝑙𝑝[]𝐴,𝐴=𝑝1𝑝Δ𝑢𝑝𝑙𝑝[]𝐴,𝐴+1𝑝Δ𝑤𝑝𝑙𝑝[]𝐴,𝐴=[]𝑛𝐴,𝐴𝑝1𝑝||||Δ𝑢(𝑛1)𝑝+1𝑝||||Δ𝑤(𝑛1)𝑝<+,(3.34) then, we have 𝑛[𝐴,𝐴]||Δ𝑢𝑘||(𝑛1)𝑝2Δ𝑢𝑘(𝑛1),Δ𝑤(𝑛1)𝑛[𝐴,𝐴]||||Δ𝑢(𝑛1)𝑝2Δ𝑢(𝑛1),Δ𝑤(𝑛1)(3.35) as 𝑘. Noting that []𝑛𝐴,𝐴𝐾𝑛,𝑢𝑘(𝑛),𝑤(𝑛)𝑛[𝐴,𝐴](𝐾(𝑛,𝑢(𝑛)),𝑤(𝑛))as𝑘,𝑛[𝐴,𝐴]𝑊𝑛,𝑢𝑘(𝑛),𝑤(𝑛)𝑛[𝐴,𝐴](𝑊(𝑛,𝑢(𝑛)),𝑤(𝑛))as𝑘.(3.36) Hence, we have 𝜑(𝑢),𝑤=lim𝑘𝑢𝜑𝑘,𝑤=0,(3.37) which implies that 𝜑(𝑢)=0; that is, 𝑢 is a critical point of 𝜑. From (K1) and (W1), we know that 𝑢0. In fact, if 𝑢=0, we have from (2.5), (K1), and (W1) that 𝜑(𝑢)=0. On the other hand, from Step 1, Step 2, and Lemma 2.5, we know that 𝜑(𝑢)=𝑐>0. This is a contradiction. The proof of Theorem 1.1 is complete.

4. An Example

Example 4.1. In problem (1.1), let 𝑝=3/2, and 1𝐾(𝑛,𝑥)=1+|𝑥|3/2+1|𝑥|3/2,𝑊(𝑛,𝑥)=𝑎(𝑛)|𝑥|3/211(ln(𝑒+|𝑥|))1/2,(4.1) where 𝑎𝑙(,+) with inf𝑎(𝑛)>3. One can easily check that 𝐾 satisfies conditions (K1) and (K2) with 𝑏1=1, 𝑏2=2, and 𝑏3=3/2. An easy computation shows that 3𝑊(𝑛,𝑥)=2𝑎(𝑛)|𝑥|1/2𝑥11(ln(𝑒+|𝑥|))1/2+𝑎(𝑛)|𝑥|1/2𝑥2(𝑒+|𝑥|)(ln(𝑒+|𝑥|))3/2,3(𝑊(𝑛,𝑥),𝑥)2𝑊(𝑛,𝑥)=𝑎(𝑛)|𝑥|5/22(𝑒+|𝑥|)(ln(𝑒+|𝑥|))3/2.(4.2) Then it is easy to check that 𝑊 satisfies (W1)–(W4). Hence, 𝐾(𝑛,𝑥) and 𝑊(𝑛,𝑥) satisfy all the conditions of Theorem 1.1 and then problem (1.1) has at least one nontrivial homoclinic solution.

Acknowledgment

This work is partially supported by the NNSF (no. 10771215) of China.