New potential condition on homoclinic orbits for a class of discrete Hamiltonian systems

Abstract

In the present paper, we establish an existence criterion to guarantee that the second-order self-adjoint discrete Hamiltonian system $\mathrm{△}[p(n)\mathrm{△}u(n-1)]-L(n)u(n)+\mathrm{\nabla }W(n,u(n))=0$ has a nontrivial homoclinic solution, which does not need periodicity and coercivity conditions on $L(n)$.

MSC:39A11, 58E05, 70H05.

1 Introduction

Consider the second-order self-adjoint discrete Hamiltonian system

$$\mathrm{△}[p(n)\mathrm{△}u(n-1)]-L(n)u(n)+\mathrm{\nabla }W(n,u(n))=0,$$

(1.1)

where $n\in \mathbb{Z}$, $u\in {\mathbb{R}}^{\mathcal{N}}$, $\mathrm{△}u(n)=u(n+1)-u(n)$ is the forward difference, $p,L:\mathbb{Z}\to {\mathbb{R}}^{\mathcal{N}\times \mathcal{N}}$ and $W:\mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}}\to \mathbb{R}$.

Discrete Hamiltonian systems can be applied in many areas, such as physics, chemistry, and so on. For more discussions on discrete Hamiltonian systems, we refer the reader to [1, 2].

As usual, we say that a solution $u(n)$ of system (1.1) is homoclinic (to 0) if $u(n)\to 0$ as $n\to \pm \mathrm{\infty }$. In addition, if $u(n)\not\equiv 0$ then $u(n)$ is called a nontrivial homoclinic solution.

The existence and multiplicity of homoclinic solutions of system (1.1) or its special forms have been investigated by many authors. Papers [3–8] deal with the periodic case where p, L and W are periodic in n or independent of n. In contrast, if the periodicity is lost, because of lack of compactness of the Sobolev embedding, up to our knowledge, all existence results require a coercivity condition on L:

$$\underset{|n|\to \mathrm{\infty }}{lim}\left[\underset{x\in {\mathbb{R}}^{\mathcal{N}},|x|=1}{inf}(L(n)x,x)\right]=\mathrm{\infty }.$$

(1.2)

For example, see [9–14]. In the above mentioned papers, except [14], L was always required to be positive definite.

In this paper, we derive an existence result which does not need periodicity and coercivity conditions on $L(n)$. To state our results precisely, we make the following assumptions.

• (P) $p(n)$ is $\mathcal{N}\times \mathcal{N}$ real symmetric positive definite matrix for all $n\in \mathbb{Z}$.

• (L) $L(n)$ is $\mathcal{N}\times \mathcal{N}$ real symmetric nonnegative definite matrix for all $n\in \mathbb{Z}$, and there exist a positive integer $N_0\in \mathbb{Z}$ and $\beta >0$ such that

  $$\underset{x\in {\mathbb{R}}^{\mathcal{N}},|x|=1}{min}(L(n)x,x)\ge \beta ,|n|\ge N_0,$$

where here and in the sequel, $(\cdot ,\cdot )$ denotes the standard inner product in ${\mathbb{R}}^{\mathcal{N}}$ and $|\cdot |$ is the induced norm.

• (W1) $W(n,x)$ is continuously differentiable in x for every $n\in \mathbb{Z}$, $W(n,0)=0$, $W(n,x)\ge 0$ for all $(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}}$.

• (W2) ${lim}_{|x|\to 0}\frac{\mathrm{\nabla }W(n,x)}{|x|}=0$ uniformly for all $n\in \mathbb{Z}$.

• (W3) ${lim}_{|x|\to \mathrm{\infty }}\frac{|W(n,x)|}{{|x|}^2}=\mathrm{\infty }$ uniformly for all $n\in \mathbb{Z}$.

• (W4) $\tilde{W}(n,x):=\frac{1}{2}(\mathrm{\nabla }W(n,x),x)-W(n,x)\ge 0$, $\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}}$, and there exist $\epsilon \in (0,1)$, $c_0>0$, and $R_0>0$ such that

  $$(\mathrm{\nabla }W(n,x),x)\le \frac{\beta (1-\epsilon )}{2}{|x|}^2,\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\le R_0$$

  and

  $$(\mathrm{\nabla }W(n,x),x)\le c_0{|x|}^2\tilde{W}(n,x),\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\ge R_0.$$

Now, we are ready to state the main result of this paper.

Theorem 1.1 Assume that p, L and W satisfy (P), (L), (W1), (W2), (W3), and (W4). If there exist $n_0\in \mathbb{Z}$ and $x_0\in {\mathbb{R}}^{\mathcal{N}}$ such that

$$\beta \ge 2c_0\underset{s\ge 0}{sup}[\frac{s^2}{2}((p(n_0)+p(n_0+1)+L(n_0))x_0,x_0)-W(n_0,s{x}_0)],$$

(1.3)

then system (1.1) possesses a nontrivial homoclinic solution.

In Theorem 1.1, we replace (L) and (W4) by the following assumptions:

(L′) $L(n)$ is $\mathcal{N}\times \mathcal{N}$ real symmetric nonnegative definite matrix for all $n\in \mathbb{Z}$, and it satisfies (1.2).

(W4′) $\tilde{W}(n,x):=\frac{1}{2}(\mathrm{\nabla }W(n,x),x)-W(n,x)\ge 0$, $\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}}$, and there exist $c_0>0$ and $R_0>0$ such that

$$(\mathrm{\nabla }W(n,x),x)\le c_0{|x|}^2\tilde{W}(n,x),\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\ge R_0.$$

Then we have the following corollary immediately.

Corollary 1.2 Assume that p, L and W satisfy (P), (L′), (W1), (W2), (W3) and (W4′). Then system (1.1) possesses a nontrivial homoclinic solution.

Remark 1.3 If $W(n,x)$ satisfies the well-known global Ambrosetti-Rabinowitz superquadratic condition:

(AR) there exists $\mu >2$ such that

$$0<\mu W(n,x)\le (\mathrm{\nabla }W(n,x),x),\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}}\setminus \{0\},$$

then there exists a constant $C_0>0$ such that

$$W(n,x)\ge C_0{|x|}^{\mu },\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\ge 1;$$

moreover $\tilde{W}(n,x)>0$ for all $(n,x)\in \mathbb{Z}\times ({\mathbb{R}}^{\mathcal{N}}\setminus \{0\})$, and

$$(\mathrm{\nabla }W(n,x),x)\le \frac{2\mu }{\mu -2}{|x|}^2\tilde{W}(n,x),\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\ge 1.$$

In addition, by virtue of (W2), there exists ${\beta }_1>0$ such that

$$(\mathrm{\nabla }W(n,x),x)\le \frac{{\beta }_1}{2}{|x|}^2,\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\le 1.$$

These show that (W3) and (W4) hold with $R_0=1$, $c_0=2\mu /(\mu -2)$ and $\beta >{\beta }_1$. Let $p(n)=I_{\mathcal{N}}$ and $L(n)=\lambda n^2/(1+n^2)I_{\mathcal{N}}$ and choose $n_0=0$ and $x_0=(1,0,\dots ,0)\in {\mathbb{R}}^{\mathcal{N}}$. In view of Theorem 1.1, if

$$\lambda >max\{\frac{4\mu }{\mu -2}\underset{s\ge 0}{sup}[s^2-W(0,s{x}_0)],{\beta }_1\},$$

then system (1.1) possesses a nontrivial homoclinic solution.

Example 1.4 Let $p(n)=I_{\mathcal{N}}$, $L(n)=[1+\lambda n^2/(1+n^2)]I_{\mathcal{N}}$ and

$$W(n,x)={|x|}^{2}ln(1+{|x|}^2).$$

(1.4)

Then

$$(\mathrm{\nabla }W(n,x),x)=2{|x|}^{2}ln(1+{|x|}^2)+\frac{2{|x|}^4}{1+{|x|}^2}.$$

It is easy to see that $\tilde{W}(n,x)\ge 0$ for all $(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}}$, and

$$\begin{array}{c}(\mathrm{\nabla }W(n,x),x)\le (2ln2+1){|x|}^2,\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\le 1,\hfill \\ (\mathrm{\nabla }W(n,x),x)\le 6{|x|}^2\tilde{W}(n,x),\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}},|x|\ge 1.\hfill \end{array}$$

These show that (W3) and (W4) hold with $R_0=1$, $c_0=6$ and $\beta >2(2ln2+1)$. We choose $n_0=0$ and $x_0=(1,0,\dots ,0)\in {\mathbb{R}}^{\mathcal{N}}$. Then

$$\begin{array}{r}\underset{s\ge 0}{sup}[\frac{s^2}{2}((p(n_0)+p(n_0+1)+L(n_0))x_0,x_0)-W(n_0,s{x}_0)]\\ =\underset{s\ge 0}{sup}[\frac{3s^2}{2}-s^{2}ln(1+s^2)]<6-ln2.\end{array}$$

In view of Theorem 1.1, if $\lambda \ge 12(6-ln2)$, then system (1.1) possesses a nontrivial homoclinic solution.

2 Preliminaries

Throughout this section, we always assume that p and L satisfy (P) and (L). Let

$$\begin{array}{c}S=\{{\{u(n)\}}_{n\in \mathbb{Z}}:u(n)\in {\mathbb{R}}^{\mathcal{N}},n\in \mathbb{Z}\},\hfill \\ E=\{u\in S:\sum _{n\in \mathbb{Z}}[(p(n+1)\mathrm{△}u(n),\mathrm{△}u(n))+(L(n)u(n),u(n))]<+\mathrm{\infty }\},\hfill \end{array}$$

and for $u,v\in E$, let

$$〈u,v〉=\sum _{n\in \mathbb{Z}}[(p(n+1)\mathrm{△}u(n),\mathrm{△}v(n))+(L(n)u(n),v(n))].$$

Then E is a Hilbert space with the above inner product, and the corresponding norm is

$$\parallel u\parallel ={\left\{\sum _{n\in \mathbb{Z}}[(p(n+1)\mathrm{△}u(n),\mathrm{△}u(n))+(L(n)u(n),u(n))]\right\}}^{1/2},u\in E.$$

As usual, for $1\le s<+\mathrm{\infty }$, set

$$l^s(\mathbb{Z},{\mathbb{R}}^{\mathcal{N}})=\{u\in S:\sum _{n\in \mathbb{Z}}{|u(n)|}^s<+\mathrm{\infty }\}$$

and

$$l^{\mathrm{\infty }}(\mathbb{Z},{\mathbb{R}}^{\mathcal{N}})=\{u\in S:\underset{n\in \mathbb{Z}}{sup}|u(n)|<+\mathrm{\infty }\},$$

and their norms are defined by

$$\begin{array}{c}{\parallel u\parallel }_s={\left(\sum _{n\in \mathbb{Z}}{|u(n)|}^s\right)}^{1/s},\mathrm{\forall }u\in l^s(\mathbb{Z},{\mathbb{R}}^{\mathcal{N}});\hfill \\ {\parallel u\parallel }_{\mathrm{\infty }}=\underset{n\in \mathbb{Z}}{sup}|u(n)|,\mathrm{\forall }u\in l^{\mathrm{\infty }}(\mathbb{Z},{\mathbb{R}}^{\mathcal{N}}),\hfill \end{array}$$

respectively.

Lemma 2.1 Suppose that (L) is satisfied. Then

$${\parallel u\parallel }_{\mathrm{\infty }}\le \frac{1}{\sqrt{\beta }}\parallel u\parallel +\sum _{|s|\le N_0-1}|\mathrm{△}u(s)|,u\in E,$$

(2.1)

and

$${\parallel u\parallel }_{\mathrm{\infty }}\le max\{\sqrt{\frac{2}{\beta }},\sqrt{\frac{2N_0}{\alpha }}\}\parallel u\parallel ,u\in E,$$

(2.2)

where $\alpha ={min}_{|n|\le N_0,|x|=1}(p(n)x,x)$.

Proof Since $u\in E$, it follows that ${lim}_{|n|\to \mathrm{\infty }}|u(n)|=0$. Hence, there exists $n_{\ast }\in \mathbb{Z}$ such that ${\parallel u\parallel }_{\mathrm{\infty }}=|u(n_{\ast })|$. There are two possible cases.

Case (i). $|n_{\ast }|\ge N_0$. According to (L), one has

$${\parallel u\parallel }_{\mathrm{\infty }}^2={|u(n_{\ast })|}^2\le \frac{1}{\beta }\sum _{|s|\ge N_0}(L(s)u(s),u(s))\le \frac{1}{\beta }\parallel u\parallel .$$

Case (ii). $|n_{\ast }|<N_0$. Without loss of generality, we can assume that $n_{\ast }\ge 0$, then

$$\begin{array}{rcl}{\parallel u\parallel }_{\mathrm{\infty }}& \le & |u(N_0)|+\sum _{s=n_{\ast }}^{N_0-1}|\mathrm{△}u(s)|\\ \le & {\left[\frac{1}{\beta }\sum _{|s|\ge N_0}(L(s)u(s),u(s))\right]}^{1/2}+\sqrt{\frac{N_0}{\alpha }}{\left(\sum _{s=n_{\ast }}^{N_0-1}\alpha {|\mathrm{△}u(s)|}^2\right)}^{1/2}\\ \le & \sqrt{2}{[\frac{1}{\beta }\sum _{|s|\ge N_0}(L(s)u(s),u(s))+\frac{N_0}{\alpha }\sum _{s=n_{\ast }}^{N_0-1}(p(s+1)\mathrm{△}u(s),\mathrm{△}u(s))]}^{1/2}\\ \le & max\{\sqrt{\frac{2}{\beta }},\sqrt{\frac{2N_0}{\alpha }}\}\parallel u\parallel .\end{array}$$

(2.3)

Cases (i) and (ii) imply that (2.1) and (2.2) hold. □

Now we define a functional Φ on E by

$$\mathrm{\Phi }(u)=\frac{1}{2}\sum _{n\in \mathbb{Z}}[(p(n+1)\mathrm{△}u(n),\mathrm{△}u(n))+(L(n)u(n),u(n))]-\sum _{n\in \mathbb{Z}}W(n,u(n)).$$

(2.4)

For any $u\in E$, there exists an $n_1\in \mathbb{N}$ such that $|u(n)|\le 1$ for $|n|\ge n_1$. Hence, under assumptions (P), (L), (W1), and (W2), the functional Φ is of class $C^1(E,\mathbb{R})$. Moreover,

$$\mathrm{\Phi }(u)=\frac{1}{2}{\parallel u\parallel }^2-\sum _{n\in \mathbb{Z}}W(n,u),\mathrm{\forall }u\in E$$

(2.5)

and

$$〈{\mathrm{\Phi }}^{\prime }(u),v〉=〈u,v〉-\sum _{n\in \mathbb{Z}}(\mathrm{\nabla }W(n,u),v),\mathrm{\forall }u,v\in E.$$

(2.6)

Furthermore, the critical points of Φ in E are solutions of system (1.1) with $u(\pm \mathrm{\infty })=0$, see [5, 6].

Let $e={\{e(n)\}}_{n\in \mathbb{Z}}\in E$ with $e(n_0)=x_0$ and $e(n)=0\in {\mathbb{R}}^{\mathcal{N}}$ for $n\ne n_0$.

Lemma 2.2 Suppose that (L), (W1) and (W2) are satisfied. Then

$$sup\{\mathrm{\Phi }(se):s\ge 0\}\le \underset{s\ge 0}{sup}[\frac{s^2}{2}((p(n_0)+p(n_0+1)+L(n_0))x_0,x_0)-W(n_0,s{x}_0)].$$

(2.7)

Proof From (2.4) and the definition of e, we get

$$\begin{array}{rcl}\mathrm{\Phi }(se)& =& \frac{s^2}{2}\sum _{n\in \mathbb{Z}}[(p(n+1)\mathrm{△}e(n),\mathrm{△}e(n))+(L(n)e(n),e(n))]-\sum _{n\in \mathbb{Z}}W(n,se(n))\\ =& \frac{s^2}{2}[(p(n_0+1)\mathrm{△}e(n_0),\mathrm{△}e(n_0))+(p(n_0)\mathrm{△}e(n_0-1),\mathrm{△}e(n_0-1))\\ +(L(n_0)e(n_0),e(n_0))]-W(n_0,se(n_0))\\ =& \frac{s^2}{2}((p(n_0)+p(n_0+1)+L(n_0))x_0,x_0)-W(n_0,s{x}_0).\end{array}$$

(2.8)

Now the conclusion of Lemma 2.1 follows by (2.8). □

Applying the mountain-pass lemma without the (PS) condition, by standard arguments, we can prove the following lemma.

Lemma 2.3 Let $W(n,x)\ge 0$, $\mathrm{\forall }(n,x)\in \mathbb{Z}\times {\mathbb{R}}^{\mathcal{N}}$. Suppose that (P), (L), (W1), (W2) and (W3) are satisfied. Then there exist a constant $c\in (0,{sup}_{s\ge 0}\mathrm{\Phi }(se)]$ and a sequence $\{u_k\}\subset E$ satisfying

$$\mathrm{\Phi }(u_k)\to c,\parallel {\mathrm{\Phi }}^{\prime }(u_k)\parallel (1+\parallel u_k\parallel )\to 0.$$

(2.9)

Lemma 2.4 Suppose that (P), (L), (W1), (W2), (W3), and (W4) are satisfied. Then any sequence $\{u_k\}\subset E$ satisfying

$$\mathrm{\Phi }(u_k)\to c>0,〈{\mathrm{\Phi }}^{\prime }(u_k),u_k〉\to 0$$

(2.10)

is bounded in E.

Proof To prove the boundedness of $\{u_k\}$, arguing by contradiction, suppose that $\parallel u_k\parallel \to \mathrm{\infty }$. Let $v_k=u_k/\parallel u_k\parallel$. Then $\parallel v_k\parallel =1$. By virtue of (2.5), (2.6), and (2.10), we have

$$\mathrm{\Phi }(u_k)-\frac{1}{2}〈{\mathrm{\Phi }}^{\prime }(u_k),u_k〉=\sum _{n\in \mathbb{Z}}\tilde{W}(n,u_k)=c+o(1).$$

(2.11)

If $\delta :={lim\hspace{0.17em}sup}_{k\to \mathrm{\infty }}{\parallel v_k\parallel }_{\mathrm{\infty }}=0$, then it follows from (L), (W4) and (2.11) that

$$\begin{array}{rcl}\sum _{|u_k|<R_0}\left|(\mathrm{\nabla }W(n,u_k),u_k)\right|& \le & \frac{\beta }{2}\sum _{|u_k|<R_0}{|u_k|}^2\le \frac{\beta }{2}\sum _{|s|\ge N_0}{|u_k(s)|}^2+\frac{\beta }{2}\sum _{|s|<N_0}{|u_k(s)|}^2\\ \le & \frac{1}{2}{\parallel u_k\parallel }^2+N_0\beta {\parallel u_k\parallel }^2{\parallel v_k\parallel }_{\mathrm{\infty }}^2\le [\frac{1}{2}+o(1)]{\parallel u_k\parallel }^2\end{array}$$

(2.12)

and

$$\begin{array}{rcl}\sum _{|u_k|\ge R_0}\frac{|(\mathrm{\nabla }W(n,u_k),u_k)|}{{\parallel u_k\parallel }^2}& \le & c_0\sum _{|u_k|\ge R_0}{|v_k|}^2\tilde{W}(n,u_k)\le c_0{\parallel v_k\parallel }_{\mathrm{\infty }}^2\sum _{|u_k|\ge R_0}\tilde{W}(n,u_k)\\ \le & c_0(c+1){\parallel v_k\parallel }_{\mathrm{\infty }}^2\to 0,k\to \mathrm{\infty }.\end{array}$$

(2.13)

Combining (2.12) with (2.13) and using (2.5) and (2.10), we have

$$\begin{array}{rcl}1+o(1)& \le & \frac{{\parallel u_k\parallel }^2-〈{\mathrm{\Phi }}^{\prime }(u_k),u_k〉}{{\parallel u_k\parallel }^2}\le \sum _{n\in \mathbb{Z}}\frac{|(\mathrm{\nabla }W(n,u_k),u_k)|}{{\parallel u_k\parallel }^2}\\ =& \sum _{|u_k|<R_0}\frac{|(\mathrm{\nabla }W(n,u_k),u_k)|}{{\parallel u_k\parallel }^2}+\sum _{|u_k|\ge R_0}\frac{|(\mathrm{\nabla }W(n,u_k),u_k)|}{{\parallel u_k\parallel }^2}\le \frac{1}{2}+o(1).\end{array}$$

(2.14)

This contradiction shows that $\delta >0$.

Going if necessary to a subsequence, we may assume the existence of $n_k\in \mathbb{Z}$ such that

$$|v_k(n_k)|={\parallel v_k\parallel }_{\mathrm{\infty }}>\frac{\delta }{2}.$$

Let $w_k(n)=v_k(n+n_k)$, then

$$|w_k(0)|>\frac{\delta }{2},\mathrm{\forall }k\in \mathbb{N}.$$

(2.15)

Now we define ${\tilde{u}}_k(n)=u_k(n+n_k)$. Then ${\tilde{u}}_k(n)/\parallel u_k\parallel =w_k(n)$ and ${\parallel w_k\parallel }_2={\parallel v_k\parallel }_2$. Passing to a subsequence, we have $w_k\rightharpoonup w$ in $l^2(\mathbb{Z},{\mathbb{R}}^{\mathcal{N}})$, then $w_k(n)\to w(n)$ for all $n\in \mathbb{Z}$. Clearly, (2.15) implies that $w(0)\ne 0$.

It is obvious that $w(n)\ne 0$ implies ${lim}_{k\to \mathrm{\infty }}|{\tilde{u}}_k(n)|=\mathrm{\infty }$. Hence, it follows from (2.5), (2.10), and (W3) that

$$\begin{array}{rcl}0& =& \underset{k\to \mathrm{\infty }}{lim}\frac{c+o(1)}{{\parallel u_k\parallel }^2}=\underset{k\to \mathrm{\infty }}{lim}\frac{\mathrm{\Phi }(u_k)}{{\parallel u_k\parallel }^2}\\ =& \underset{k\to \mathrm{\infty }}{lim}[\frac{1}{2}-\sum _{n\in \mathbb{Z}}\frac{W(n,u_k)}{{|u_k|}^2}{|v_k|}^2]\\ =& \underset{k\to \mathrm{\infty }}{lim}[\frac{1}{2}-\sum _{n\in \mathbb{Z}}\frac{W(n+k_n,{\tilde{u}}_k)}{{|{\tilde{u}}_k|}^2}{|w_k|}^2]\\ \le & \frac{1}{2}-\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\sum _{n\in \mathbb{Z}}\frac{W(n+k_n,{\tilde{u}}_k)}{{|{\tilde{u}}_k|}^2}{|w_k|}^2\\ =& -\mathrm{\infty },\end{array}$$

which is a contradiction. Thus $\{u_k\}$ is bounded in E. □

3 Proof of theorem

Proof of Theorem 1.1 Applying Lemmas 2.3 and 2.4, we deduce that there exists a bounded sequence $\{u_k\}\subset E$ satisfying (2.9). By Lemma 2.2 and (1.3), one has

$$c\le \frac{\beta }{2c_0}.$$

(3.1)

Going if necessary to a subsequence, we can assume that $u_k\rightharpoonup \overline{u}$ in E and ${\mathrm{\Phi }}^{\prime }(u_k)\to 0$. Next, we prove that $\overline{u}\ne 0$.

Arguing by contradiction, suppose that $\overline{u}=0$, i.e. $u_k\rightharpoonup 0$ in E, and so $u_k(n)\to 0$ for every $n\in \mathbb{Z}$. Hence,

$${\parallel u_k\parallel }_2^2=\sum _{|n|\ge N_0}{|u_k(n)|}^2+\sum _{|n|<N_0}{|u_k(n)|}^2\le \frac{1}{\beta }{\parallel u_k\parallel }^2+o(1).$$

(3.2)

According to (W4) and (3.2), one gets

$$\sum _{|u_k|<R_0}(\mathrm{\nabla }W(n,u_k),u_k)\le \frac{\beta (1-\epsilon )}{2}\sum _{|u_k|<R_0}{|u_k|}^2\le \frac{1-\epsilon }{2}{\parallel u_k\parallel }^2+o(1).$$

(3.3)

By virtue of (2.5), (2.6), and (2.9), we have

$$\mathrm{\Phi }(u_k)-\frac{1}{2}〈{\mathrm{\Phi }}^{\prime }(u_k),u_k〉=\sum _{n\in \mathbb{Z}}\tilde{W}(n,u_k)=c+o(1).$$

(3.4)

Using (W4), (2.1), (3.1), (3.2), and (3.4), we obtain

$$\begin{array}{rcl}\sum _{|u_k|\ge R_0}(\mathrm{\nabla }W(n,u_k),u_k)& \le & c_0\sum _{|u_k|\ge R_0}{|u_k|}^2\tilde{W}(n,u_k)\\ \le & c_0{\parallel u_k\parallel }_{\mathrm{\infty }}^2\sum _{|u_k|\ge R_0}\tilde{W}(n,u_k)\\ \le & c_{0}c{\parallel u_k\parallel }_{\mathrm{\infty }}^2+o(1)\\ \le & c_{0}c{(\frac{1}{\sqrt{\beta }}\parallel u_k\parallel +\sum _{|s|\le N_0-1}|\mathrm{△}{u}_k(s)|)}^2+o(1)\\ =& \frac{c_{0}c}{\beta }{\parallel u_k\parallel }^2+o(1)\\ \le & \frac{1}{2}{\parallel u_k\parallel }^2+o(1),\end{array}$$

(3.5)

which, together with (2.6), (2.9), and (3.3), yields

$$\begin{array}{rcl}o(1)& =& 〈{\mathrm{\Phi }}^{\prime }(u_k),u_k〉={\parallel u_k\parallel }^2-\sum _{n\in \mathbb{Z}}(\mathrm{\nabla }W(n,u_k),u_k)\\ \ge & \frac{\epsilon }{2}{\parallel u_k\parallel }^2+o(1),\end{array}$$

(3.6)

resulting in the fact that $\parallel u_k\parallel \to 0$. Consequently, it follows from (W1), (2.5), and (2.9) that

$$0<c=\underset{k\to \mathrm{\infty }}{lim}\mathrm{\Phi }(u_k)=\mathrm{\Phi }(0)=0.$$

This contradiction shows $\overline{u}\ne 0$. By standard arguments, we easily prove that $\overline{u}$ is a nontrivial solution of (1.1). □