1 Introduction

In recent years, we can see the nonlinear difference equations have been widely studied and applied by many research fields [1,2,3,4,5,6,7,8,9]. In particular, the discrete nonlinear Schrödinger equations have been studied very intensively in some related problems of physical and biological phenomena. For example, they can be used to describe how hight intensity light interacts with and propagates through matter [10], biomolecular chains [11], the self-focusing of beams and Bose-Einstein condensates [12], etc. Great effort has been made to study the excellent results for the discrete Schrödinger equations in the literature, such as the existence of gap solitons [13,14,15,16], the chaotic dynamics [17], and ground state solutions [18], etc.

In this paper, we shall study the standing wave problems for the following discrete nonlinear Schrödinger equation:

$$\begin{aligned} i{\dot{\psi }}_{n}=-\Delta \psi _{n}+v_{n}\psi _{n}-f_{n}(\psi _{n}), ~n\in {\mathbb {Z}}, \end{aligned}$$
(1)

where \(\Delta \psi _{n}=\psi _{n+1}+\psi _{n-1}-2\psi _{n}\) is the discrete Laplacian in one spatial dimension, the discrete potential \(V=\{v_{n}\}\) is a sequence of real numbers. The nonlinearity \(f_{n}(u)\) is continuous in u and \(f_{n}(0)=0\) for every \(n\in {\mathbb {Z}}\). Since \(f_{n}(0)=0\), then \(u=0\) is a trivial solution of problem (1).

In problem (1), the nonlinearity \(f_{n}(\cdot )\) is supposed to be gauge invariant, i.e.,

$$\begin{aligned} f_{n}\left( e^{i\theta }u\right) =e^{i\theta }f_{n}(u),~\theta \in {\mathbb {R}}. \end{aligned}$$
(2)

We consider the special solutions of problem (1) of the form

$$\begin{aligned} \psi _{n}=u_{n}e^{-i\omega t}, \end{aligned}$$
(3)

where \(\omega \in {\mathbb {R}}\) is the temporal frequency and \(\{u_{n}\}\) is a real valued sequence satisfying

$$\begin{aligned} \lim _{n\rightarrow \pm \infty }u_{n}=0. \end{aligned}$$
(4)

Such solutions are called standing wave solutions, inserting the standing wave Ansatz into problem (1), and thus, we get the following nonlinear difference equation:

$$\begin{aligned} -\Delta u_{n}+v_{n}u_{n}-\omega u_{n}=f_{n}(u_{n}), ~n\in {\mathbb {Z}} \end{aligned}$$
(5)

with the boundary condition (4). As usual, we also say that a solution of the equation (5) is homoclinic (to 0), that is, the standing wave solutions of problem (1) correspond to the homoclinic solutions of the equation (5) [19,20,21,22]. In particular, when the temporal frequency \(\omega \) belong to a spectral gap, a solution of the equation (5) is called a gap soliton, we refer readers to [13] for additional details.

In the past decade, most of the literature is devoted to the existence of the standing waves solutions for the discrete nonlinear Schrödinger equations with the superlinear or saturable nonlinearities. Pankov and Chen in [13, 22] considered spatially localized standing waves for the problem (1) with the superlinear nonlinearity function \(f_{n}(u)=\vert u\vert ^{2}u\). the superlinear nonlinearity has been considered very intensively in the physics fields [14]. In 2010, Zhou [23, 24] considered the standing waves solutions for the problem (1) with typical representatives of saturable nonlinearities \(f_{n}(u)=\left( 1-e^{-c_{n}u^{2}}\right) u\) and \(f_{n}(u)=u^{3}\left( 1+c_{n}u^{2}\right) ^{-1}\), respectively. The saturable nonlinearities can describe optical pulse propagation in various doped fibres [25, 26]. The interesting results of other standing wave solutions can be found in various models of nonlinear systems [27,28,29].

In addition, since some difficulties appeared in the process of proof, according to the lack of compactness on the set \({\mathbb {Z}}\). Most of authors considered to assume the potential \(V=\{v_{n}\}\) is real valued periodic sequence based on the above difficulties [13, 24, 30, 31], and such periodic case overcame these difficulties by means of the critical point theory in combination with periodic approximations. In recent years, there seems to be only a few papers discussed the existence of standing wave solutions for the discrete nonlinear Schrödinger equations with non-periodic coefficients and unbounded potentials, they can be found in the literature [27,28,29, 32]. The interesting results were dependent on a discrete version of compact embedding theorem [27]. Resonance is not only used in physics with a very high frequency, but also resonance phenomenon is also one of the most common and frequent natural phenomena in our life [33,34,35]. Everything we see with our eyes are receive signals according to the principle of resonance. Such as we watch TV and surf the internet, etc. To the best of our knowledge, most of these papers dealt with the superlinear or saturable nonlinearities without resonance. Motivated by the above work, we study the nonlinearity which satisfies the resonant case \(\lim \nolimits _{\vert t\vert \rightarrow \infty }f_{n}(t)/t=\lambda _{k}-\omega \) and the potential \(V=\{v_{n}\}\) is non-periodic and unbounded in this paper. We shall give sufficient conditions to prove the existence of standing wave solutions for the equation (1) with resonance, where it is vital to find a bounded critical sequence by means of linking methods rather than periodic approximations.

2 Preliminaries

In order to establish the variational framework associated with the problem (1) and apply the critical point theory, we give some basic notations and lemmas which will be used to prove our main results. We consider the real sequence spaces

$$\begin{aligned} l^{q}=\left\{ u=\{u_{n}\}\bigg \vert \forall n\in {\mathbb {Z}},u_{n}\in {\mathbb {R}}, \Vert u\Vert _{l^{q}}=\left( \sum \limits _{n\in {\mathbb {Z}}}\vert u_{n}\vert ^{q}\right) ^{\frac{1}{q}} <+\infty \right\} . \end{aligned}$$

The following elementary embedding relation holds

$$\begin{aligned} l^{q_{1}}\subset l^{q_{2}}, \Vert u\Vert _{l^{q_{2}}}\le \Vert u\Vert _{l^{q_{1}}}, 1\le q_{1}\le q_{2}\le \infty . \end{aligned}$$

We give the following assumption

  1. (i)

    The discrete potential \(V=\{v_{n}\}\) satisfies

    $$\begin{aligned} \lim _{n\rightarrow \pm \infty }v_{n}=+\infty . \end{aligned}$$

Without loss of generality, we assume \(v_{n}\ge 1\) for every \(n\in {\mathbb {Z}}\), and we also need the stronger assumption

  • (i’) \(V^{-1}=\{v_{n}^{-1}\}\in l^{1}\).

We denote an unbounded selfadjoint operator \(H=-\Delta +V\) in \(l^{2}\). Let \(E={\mathcal {D}}\left( H^{1/2}\right) \) be equipped with the inner product

$$\begin{aligned} (u,v)=\left( H^{1/2}u,H^{1/2}v\right) _{l^{2}}. \end{aligned}$$

This is a Hilbert space. We also define a norm

$$\begin{aligned} \Vert u\Vert _{E}=\Vert H^{1/2}u\Vert _{l^{2}}. \end{aligned}$$

Since the operator \(-\Delta \) is bounded, the following two norms are equivalent in E,

$$\begin{aligned} \Vert u\Vert _{E}\sim \Vert V^{1/2}u\Vert _{l^{2}}, \end{aligned}$$

where \(V^{\frac{1}{2}}u=\{v_{n}^{1/2}u_{n}\}\).

Firstly, we give the following two lemmas which will play an important role in the proof of our main results.

Lemma 1

[27] If V satisfies the condition (i), then

  1. (1)

    for any \(2\le p\le \infty \), the embedding map from E into \(l^{p}\) is compact, denote the best embedding constant \(c_{p}=\max _{\Vert u\Vert _{l^{p}}=1}\frac{1}{\Vert u\Vert _{E}}\);

  2. (2)

    the spectrum \(\sigma (H)\) is discrete, that is, it consists of eigenvalues with finite multiplicities.

From Lemma 1, we can assume that

$$\begin{aligned} 1\le \lambda _{1}<\lambda _{2}<\cdots<\lambda _{k}< \cdots \rightarrow \infty . \end{aligned}$$

On the Hilbert space E, we define the \(C^{1}\)-functional J as follows:

$$\begin{aligned} J(u)=\frac{1}{2}(Hu-\omega u,u)- \sum \limits _{n=-\infty }^{+\infty }F_{n}(u_{n}), \end{aligned}$$

where \((\cdot ,\cdot )\) is the inner product in \(l^{2}\), and \(F_{n}(t)=\int _{0}^{t}f_{n}(s)ds\) for all \(n\in {\mathbb {Z}}\). We can compute the Frećhet derivative as follows:

$$\begin{aligned} \langle J'(u),v\rangle =(Hu-\omega u,v)- \sum \limits _{n=-\infty }^{+\infty }f_{n}(u_{n})v_{n} \end{aligned}$$

for all \(u,~v\in E\).

It is clear that the critical points of J in E are exactly solutions of the equation (5).

Let X be a real Banach space. \(B_{\rho }\) is the open ball of radius \(\rho \) and center 0 in X and let \(\partial B_{\rho }\) denote its boundary. We give the Chapter II, Theorem 2.7.3 [36] as follows.

Lemma 2

[36] Let M, N are closed subspaces of X such that

$$\begin{aligned} X=M\oplus N,~M\ne X, ~N\ne X, \end{aligned}$$

with \(dimN<\infty \). Let \(G\in C^{1}(X,{\mathbb {R}})\) be such that

$$\begin{aligned}{} & {} G(v)\le \alpha , ~v\in N, \\{} & {} G(w)\ge \alpha , ~w\in \partial B_{\rho }\cap M \end{aligned}$$

and

$$\begin{aligned} G(sw_{0}+v)\le m_{R}, ~s\ge 0, ~v\in N, \Vert sw_{0}+v\Vert =R>R_{0}, \end{aligned}$$

for some \(w_{0}\in \partial B_{1}\cap M\), where \(0<\rho <R_{0}\), \(\alpha \in {\mathbb {R}}\).

Assume that \(m_{R}/R^{\theta +1}\rightarrow 0\) as \(R\rightarrow \infty \) holds for some \(\theta \ge 0\), then there is a sequence \(\{u^{(j)}\}\subset X\) such that

$$\begin{aligned} G\left( u^{(j)}\right) \rightarrow c,~ \alpha \le c\le \infty ,~ \frac{G'\left( u^{(j)}\right) }{\left( \Vert u^{(j)}\Vert +1\right) ^{\theta }}\rightarrow 0. \end{aligned}$$
(6)

Remark 1

( [36]) We note that if \(m_{R}\) is unbounded, then there is no upper bound on c . If \(m_{R}\) is bounded, this implies a bound on c.

3 Main results

In this section, we will look for one standing wave solution of the equation (5) in E.

Now, we give some basic assumptions on the nonlinearity \(f_{n}(\cdot )\):

  1. (ii)

    \(f_{n}(t)=o(t)\) as \(t\rightarrow 0\), uniformly for all \(n\in {\mathbb {Z}}\);

  2. (iii)

    there exists \(p\in (2,+\infty )\) such that \(f_{n}(t)=o\left( \vert t\vert ^{p-1}\right) \) as \(\vert t\vert \rightarrow \infty \), uniformly for all \(n\in {\mathbb {Z}}.\)

Lemma 3

Assume that the operator H and \(f_{n}(t)\) satisfy the conditions \((i)-(iii)\). Further, assume there exists a sequence \(\{u^{(j)}\}\subset E\) satisfying

$$\begin{aligned} J\left( u^{(j)}\right) \rightarrow c,~ \frac{J'\left( u^{(j)}\right) }{\left( \Vert u^{(j)}\Vert _{E}+1\right) ^{\theta }}\rightarrow 0 \end{aligned}$$
(7)

with

$$\begin{aligned} -\infty \le c\le \infty ,~-\infty< \theta <\infty . \end{aligned}$$

If \(\Vert u^{(j)}\Vert _{E}\) is bounded, then there is a \(u\in E\) such that

$$\begin{aligned} J(u)=c, ~J'(u)=0. \end{aligned}$$
(8)

Proof of Lemma 3

Since \(\Vert u^{(j)}\Vert _{E}\) is bounded, there is a renamed subsequence \(\{ u^{(j)}\}\) which converges weakly to some u in E. From Lemma 1, we have

$$\begin{aligned} u^{(j)}\rightarrow u ~\text {in}~ l^{2}~\text {as}~ j\rightarrow \infty \end{aligned}$$

and

$$\begin{aligned} u^{(j)}_{n}\rightarrow u_{n} ~\text {as}~ j\rightarrow \infty ~\text {for}~ n\in {\mathbb {Z}}. \end{aligned}$$

For such a subsequence, we verify that

$$\begin{aligned} \sum \limits _{n=-\infty }^{+\infty }f_{n}\left( u^{(j)}_{n}\right) v_{n}\rightarrow \sum \limits _{n=-\infty }^{+\infty }f_{n}(u_{n})v_{n},~ \forall v\in E \end{aligned}$$
(9)

and

$$\begin{aligned} \sum \limits _{n=-\infty }^{+\infty }f_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n}\rightarrow \sum \limits _{n=-\infty }^{+\infty }f_{n}(u_{n})u_{n}. \end{aligned}$$
(10)

By (iii), let \(\varepsilon >0\), we take r sufficiently large such that

$$\begin{aligned} \vert f_{n}(t)\vert \le \varepsilon \vert t\vert ^{p-1}~~\text {for}~\vert t\vert \ge r \end{aligned}$$
(11)

and

$$\begin{aligned} \vert f_{n}(t)-f_{n}(w_{r}(t))\vert \le \varepsilon \vert t\vert ^{p-1}~~\text {for}~\vert t\vert \ge r, \end{aligned}$$
(12)

where \(w_{r}(t)\) is a continuous function defined by

$$\begin{aligned} w_{r}(t)={\left\{ \begin{array}{ll} \begin{array}{lll} -r,~t<-r,\\ t,~\vert t\vert \le r,\\ r,~t>r. \end{array} \end{array}\right. } \end{aligned}$$

For any \(v\in E\), we have

$$\begin{aligned} \begin{array}{lll} &{}\sum \limits _{n=-\infty }^{+\infty }\left( f_{n} \left( u^{(j)}_{n}\right) v_{n}-f_{n}(u_{n})v_{n}\right) \\ &{}\quad =\sum \limits _{\vert u^{(j)}_{n}\vert> r}\left( f_{n} \left( u^{(j)}_{n} \right) - f_{n}\left( w_{r}\left( u^{(j)}_{n}\right) \right) \right) v_{n}\\ &{}\qquad +\sum \limits _{n=-\infty }^{+\infty }\left( f_{n}\left( w_{r}\left( u^{(j)}_{n}\right) \right) - f_{n}(w_{r}(u_{n}))\right) v_{n}\\ &{}\qquad +\sum \limits _{{\vert u_{n}\vert >r}}\left( f_{n}(w_{r}(u_{n}))- f_{n}(u_{n})\right) v_{n}. \end{array} \end{aligned}$$

By combining (11) with the boundedness of the sequence \(\Vert u^{(j)}\Vert _{E}\), the first term on the right hand side can be estimated by

$$\begin{aligned} \begin{array}{lll} &{}\sum \limits _{\vert u^{(j)}_{n}\vert >r}\mid f_{n}\left( u^{(j)}_{n}\right) - f_{n}\left( w_{r}\left( u^{(j)}_{n}\right) \right) \mid \cdot \vert v_{n}\vert \\ &{}\quad \le \varepsilon \sum \limits _{n=-\infty }^{+\infty } \vert u^{(j)}_{n}\vert ^{p-1}\cdot \vert v_{n}\vert \\ &{}\quad \le \varepsilon \Vert u^{(j)}\Vert _{l^{2p-2}}^{p-1}\Vert v\Vert _{l^{2}}\\ &{}\quad \le \varepsilon \Vert u^{(j)}\Vert _{l^{2}}^{p-1}\Vert v\Vert _{l^{2}}\\ &{}\quad \le \varepsilon (c_{2})^{p-1}\Vert u^{(j)}\Vert _{E}^{p-1}\Vert v\Vert _{l^{2}}\\ &{}\quad \le \varepsilon C\Vert v\Vert _{l^{2}}. \end{array} \end{aligned}$$

Similarly we may consider the third term. Moreover, for every \(n\in {\mathbb {Z}}\), we notice that \(f_{n}(t)\) and \(w_{r}(t)\) are continuous, we have

$$\begin{aligned} f_{n}\left( w_{r}\left( u^{(j)}_{n}\right) \right) v_{n}\rightarrow f_{n}(w_{r}(u_{n}))v_{n} ~\text {as}~j\rightarrow \infty . \end{aligned}$$

From the assumptions \((ii)-(iii)\), it follows easily that for any \(\varepsilon >0\) there exists \(C_{\varepsilon }>0\) such that

$$\begin{aligned} \vert f_{n}(t)\vert \le \varepsilon \vert t\vert +C_{\varepsilon }\vert t\vert ^{p-1},~t\in {\mathbb {R}},n\in {\mathbb {Z}}. \end{aligned}$$
(13)

Then

$$\begin{aligned} \mid f_{n}\left( w_{r}\left( u^{(j)}_{n}\right) \right) v_{n}- f_{n}\left( w_{r}\left( u_{n}\right) \right) v_{n}\mid \le 2(\varepsilon \vert r\vert +C_{\varepsilon }\vert r\vert ^{p-1})\vert v_{n}\vert . \end{aligned}$$

By dominated convergence we obtain

$$\begin{aligned} \sum \limits _{n=-\infty }^{+\infty }\left( f_{n}\left( w_{r}\left( u^{(j)}_{n}\right) \right) v_{n}- f_{n}(w_{r}(u_{n}))v_{n}\right) \rightarrow 0. \end{aligned}$$

Hence (9) holds.

Next, we show (10). We write

$$\begin{aligned} \begin{array}{lll} &{}\sum \limits _{n=-\infty }^{+\infty }\left( f_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n}-f_{n}(u_{n})u_{n}\right) \\ =&{}\sum \limits _{n=-\infty }^{+\infty } \left( f_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n}-f_{n}(u^{(j)}_{n})u_{n}\right) +\sum \limits _{n=-\infty }^{+\infty } \left( f_{n}\left( u^{(j)}_{n}\right) u_{n}-f_{n}(u_{n})u_{n}\right) . \end{array} \end{aligned}$$

In view of (9), we only need to show

$$\begin{aligned} \sum \limits _{n=-\infty }^{+\infty } \left( f_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n}-f_{n}\left( u^{(j)}_{n}\right) u_{n}\right) \rightarrow 0. \end{aligned}$$

Note that

$$\begin{aligned} \begin{array}{lll} &{}\left| \sum \limits _{n=-\infty }^{+\infty } \left( f_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n} -f_{n}\left( u^{(j)}_{n}\right) u_{n}\right) \right| \\ \le &{}\left| \sum \limits _{n=-\infty }^{+\infty }\left( f_{n}\left( u^{(j)}_{n}\right) -f_{n}(u_{n})\right) \left( u^{(j)}_{n}-u_{n}\right) \right| +\left| \sum \limits _{n=-\infty }^{+\infty } f_{n}(u_{n})\left( u^{(j)}_{n}-u_{n}\right) \right| . \end{array} \end{aligned}$$

We first show

$$\begin{aligned} \left| \sum \limits _{n=-\infty }^{+\infty }\left( f_{n}\left( u^{(j)}_{n}\right) -f_{n}(u_{n})\right) \left( u^{(j)}_{n}-u_{n}\right) \right| \rightarrow 0. \end{aligned}$$

By the Minkowski inequality, we have

$$\begin{aligned} \begin{array}{lll} &{}\left| \sum \limits _{n=-\infty }^{+\infty }\left( f_{n}\left( u^{(j)}_{n}\right) -f_{n}(u_{n})\right) \left( u^{(j)}_{n}-u_{n}\right) \right| \\ &{}\quad \le \sum \limits _{n=-\infty }^{+\infty }\left| \left( f_{n}(u^{(j)}_{n}) -f_{n}(u_{n})\right) \left( u^{(j)}_{n}-u_{n}\right) \right| \\ &{}\quad \le \sum \limits _{n=-\infty }^{+\infty } [\varepsilon \left( \vert u^{(j)}_{n}\vert +\vert u_{n}\vert \right) + C_{\varepsilon }\left( \vert u^{(j)}_{n}\vert ^{p-1}+\vert u_{n}\vert ^{p-1}\right) ] \vert u^{(j)}_{n}-u_{n}\vert \\ &{}\quad \le \varepsilon \left( \Vert u^{(j)}\Vert _{l^{2}}+\Vert u\Vert _{l^{2}}\right) \Vert u^{(j)}-u\Vert _{l^{2}}+ C_{\varepsilon }\left( \Vert u^{(j)}\Vert _{l^{p}}^{p-1}+\Vert u\Vert _{l^{p}}^{p-1})\right\| u^{(j)}-u\Vert _{l^{p}}. \end{array} \end{aligned}$$

By Lemma 1, we obtain

$$\begin{aligned} \left| \sum \limits _{n=-\infty }^{+\infty }\left( f_{n}\left( u^{(j)}_{n}\right) -f_{n}(u_{n})\right) \left( u^{(j)}_{n}-u_{n}\right) \right| \rightarrow 0~\text {as}~j\rightarrow \infty . \end{aligned}$$

Similarly to show that

$$\begin{aligned} \begin{array}{lll} \left| \sum \limits _{n=-\infty }^{+\infty } f_{n}(u_{n})\left( u^{(j)}_{n}-u_{n}\right) \right| &{}\le \sum \limits _{n=-\infty }^{+\infty } [\varepsilon \vert u_{n}\vert + C_{\varepsilon }\vert u_{n}\vert ^{p-1}] \vert u^{(j)}_{n}-u_{n}\vert \\ &{}\le \varepsilon \Vert u\Vert _{l^{2}} \Vert u^{(j)}-u\Vert _{l^{2}}+ C_{\varepsilon }\Vert u\Vert _{l^{p}}^{p-1} \Vert u^{(j)}-u\Vert _{l^{p}} \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{lll} \left| \sum \limits _{n=-\infty }^{+\infty } f_{n}(u_{n})\left( u^{(j)}_{n}-u_{n} \right) \right| \rightarrow 0~\text {as}~j\rightarrow \infty , \end{array} \end{aligned}$$

which show (10) holds.

For any \(v\in E\), we see, from (7) and (9), that

$$\begin{aligned} \begin{array}{lll} \langle J'\left( u^{(j)}\right) ,v\rangle &{}=\left( Hu^{(j)}-\omega u^{(j)},v\right) - \sum \limits _{n=-\infty }^{+\infty }f_{n}\left( u^{(j)}_{n}\right) v_{n}\\ &{}\rightarrow (Hu-\omega u,v)- \sum \limits _{n=-\infty }^{+\infty }f_{n}(u_{n})v_{n}\\ &{}=\langle J'(u),v\rangle =0. \end{array} \end{aligned}$$

Hence u is a solution of the second equation in (8).

Further, since

$$\begin{aligned} \langle J'\left( u^{(j)}\right) ,u^{(j)}\rangle =\left( Hu^{(j)}-\omega u^{(j)},u^{(j)}\right) - \sum \limits _{n=-\infty }^{+\infty }f_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n}\rightarrow 0, \end{aligned}$$
(14)

we have, by (10),

$$\begin{aligned} \Vert u^{(j)}\Vert _{E}^{2}\rightarrow \sum \limits _{n=-\infty }^{+\infty }f_{n}(u_{n}) u_{n}+(\omega u,u)=\Vert u\Vert _{E}^{2}. \end{aligned}$$

This implies that \(u^{(j)}\rightarrow u\) strongly in E, thus \(J(u)=c\). The proof is complete.

When \(\lim \nolimits _{\vert t\vert \rightarrow \infty }f_{n}(t)/t=\lambda _{k}-\omega \), we say the problem (1) is resonant at infinity, where \(\lambda _{k}\) is an eigenvalue of the operator H.

Let \(g_{n}(t)=f_{n}(t)-(\lambda _{k}-\omega )t\). We denote \(G_{n}(t)=\int _{0}^{t}g_{n}(s)ds\) and consider the following assumptions:

  1. (iv)

    \(\lim \nolimits _{\vert t\vert \rightarrow \infty }f_{n}(t)/t=\lambda _{k}-\omega >0\) uniformly for all \(n\in {\mathbb {Z}}\);

  2. (v)

    There exists some \(\delta >0\) such that \(\{\sup _{\vert t\vert <\delta }\vert G_{n}(t)\vert \}_{n=-\infty }^{+\infty }\in l^{1}\) and \(G_{n}(t)\ge 0\) for all \(n\in {\mathbb {Z}}\) and \(\vert t\vert \ge \delta \). Moreover, for all \(n\in {\mathbb {Z}}\) and \(t\in {\mathbb {R}}\), we assume that \(tg_{n}(t)-2G_{n}(t)\le 0\) and

    $$\begin{aligned} \limsup \limits _{\vert t\vert \rightarrow +\infty } \frac{tg_{n}(t)-2G_{n}(t)}{\vert t\vert }=d_{n}<0; \end{aligned}$$
  3. (vi)

    \((\lambda _{k-1}-\omega )t^{2}\le 2F_{n}(t)\) for all \(n\in {\mathbb {Z}}\) and \(t\in {\mathbb {R}}\).

Example 1

Before proceeding further, we first give a function satisfying the conditions \((ii)-(vi)\) below. Put \(\lambda _{k}-\omega =2\ln 2-\frac{\sqrt{2}}{2}>0\) and \(\delta \) is taken sufficiently small. The function is defined by

$$\begin{aligned} f_{n}(t)={\left\{ \begin{array}{ll} \begin{array}{lll} \left( 2\ln 2-\frac{\sqrt{2}}{2}\right) t+\frac{t}{\sqrt{1+t^{2}}},~t<-1,\\ \frac{1}{n^{2}}\cdot \left( -t\ln (1-t)-3\sqrt{2}\ln 2\frac{ t^{2}}{\sqrt{1-t^{3}}}\right) ,~ -1\le t\le 0,\\ \frac{1}{n^{2}}\cdot \left( -t\ln (1+t)+3\sqrt{2}\ln 2\frac{ t^{2}}{\sqrt{1+t^{3}}}\right) ,~0<t\le 1,\\ \left( 2\ln 2-\frac{\sqrt{2}}{2}\right) t+\frac{t}{\sqrt{1+t^{2}}},~t>1. \end{array} \end{array}\right. } \end{aligned}$$

Then,

$$\begin{aligned} F_{n}(t)={\left\{ \begin{array}{ll} \begin{array}{lll} \left( \ln 2-\frac{\sqrt{2}}{4}\right) t^{2}+\sqrt{1+t^{2}}+3\ln 2-\frac{3\sqrt{2}}{4},~ t< -1,\\ \frac{1}{n^{2}}\cdot \left( \frac{1-t^{2}}{2}\ln (1-t)+\frac{(t+1)^{2}}{4} +2\sqrt{2}\ln 2\sqrt{1-t^{3}}\right) ,~-1\le t\le 0,\\ \frac{1}{n^{2}}\cdot \left( \frac{1-t^{2}}{2}\ln (1+t)+\frac{(t-1)^{2}}{4} +2\sqrt{2}\ln 2\sqrt{1+t^{3}}\right) ,~0<t\le 1,\\ \left( \ln 2-\frac{\sqrt{2}}{4}\right) t^{2}+\sqrt{1+t^{2}}+3\ln 2-\frac{3\sqrt{2}}{4},~t>1. \end{array} \end{array}\right. } \end{aligned}$$

Let \(E(\lambda _{k})\) be the finite dimensional eigenspace corresponding to the eigenvalue \(\lambda _{k}\), and let M denote the subspace of E spanned by the eigenfunctions corresponding to the eigenvalues which are greater than \(\lambda _{k}\).

Lemma 4

Assume that the conditons \((i)-(iv)\) hold, the following alternative hold:

  1. (a)

    there is a \({\bar{z}}\in E(\lambda _{k})\backslash \{0\}\) such that

    $$\begin{aligned} H{\bar{z}}-\omega {\bar{z}}=f({\bar{z}})=(\lambda _{k}-\omega ){\bar{z}}, \end{aligned}$$

    where f(u) be denoted by \((f(u))_{n}=f_{n}(u_{n})\).

  2. (b)

    for \(\rho >0\) sufficiently small, there exists an \(\varepsilon _{1}>0\) such that

    $$\begin{aligned} J(u')\ge \varepsilon _{1},~ \Vert u'\Vert _{E}=\rho ,~u'\in E(\lambda _{k})\bigoplus M. \end{aligned}$$

Proof of Lemma 4

Under the conditions (ii) and (iv), we get

$$\begin{aligned} \lim _{\vert t\vert \rightarrow 0}\frac{g_{n}(t)}{t}=-(\lambda _{k}-\omega )<0. \end{aligned}$$
(15)

Thus, there is a positive constant \(\delta >0\) such that

$$\begin{aligned} G_{n}(t)\le 0, \vert t\vert \le \delta , \end{aligned}$$

which implies that

$$\begin{aligned} 2F_{n}(t)\le (\lambda _{k}-\omega )\vert t\vert ^{2}, \vert t\vert \le \delta . \end{aligned}$$
(16)

Let \(u'=z+w\), \(z\in E(\lambda _{k})\), and \(w\in M\). Then

$$\begin{aligned} \lambda _{k+1}\Vert w\Vert _{l^{2}}^{2}\le \Vert w\Vert _{E}^{2},~ w\in M. \end{aligned}$$
(17)

For \(z\in E(\lambda _{k})\), we can find a \(\rho >0\) so small that

$$\begin{aligned} \Vert z\Vert _{E}\le \rho \Rightarrow \vert z_{n}\vert \le \frac{\delta }{2}, n\in {\mathbb {Z}}. \end{aligned}$$

We assume that \(u'\in E(\lambda _{k})\bigoplus M\) satisfies

$$\begin{aligned} \Vert u'\Vert _{E}\le \rho ~\text {and}~ \vert u'_{n}\vert \ge \delta \end{aligned}$$
(18)

for some \(n\in {\mathbb {Z}}\). Then for those \(n\in {\mathbb {Z}}\) satisfying (18) we have

$$\begin{aligned} \delta \le \vert u'_{n}\vert \le \vert z_{n}\vert +\vert w_{n}\vert \le \frac{\delta }{2}+\vert w_{n}\vert . \end{aligned}$$

Hence, \(\vert z_{n}\vert \le \frac{\delta }{2}\le \vert w_{n}\vert \), \(\vert u'_{n}\vert \le 2\vert w_{n}\vert \).

In view of (16), (17) and (13), we let  \(\varepsilon =\frac{\lambda _{k+1}-\lambda _{k}}{16}>0\). For each \(u'\in E(\lambda _{k})\bigoplus M\), we get

$$\begin{aligned} \begin{array}{lll} J(u')&{}=\frac{1}{2}(Hu'-\omega u',u')- \sum \limits _{n=-\infty }^{+\infty }F_{n}(u'_{n})\\ &{}\ge \frac{1}{2}\Vert u'\Vert _{E}^{2}-\frac{\omega }{2}\Vert u'\Vert _{l^{2}}^{2} -\frac{\lambda _{k}-\omega }{2} \sum \limits _{\vert u'_{n}\vert \le \delta }\vert u'_{n}\vert ^{2}\\ &{}~~~~-\frac{\lambda _{k+1}-\lambda _{k}}{16}\sum \limits _{\vert u'_{n}\vert>\delta }\vert u'_{n}\vert ^{2} -C_{(\lambda _{k+1}-\lambda _{k})/16}\sum \limits _{\vert u'_{n}\vert>\delta }\vert u'_{n}\vert ^{p}\\ &{}\ge \frac{1}{2}\Vert u'\Vert _{E}^{2}-\frac{\omega }{2}\Vert u'\Vert _{l^{2}}^{2} -\frac{\lambda _{k}-\omega }{2}\Vert u'\Vert _{l^{2}}^{2}\\ &{}~~~~-\frac{\lambda _{k+1}-\lambda _{k}}{16}\sum \limits _{\vert u'_{n}\vert>\delta }\vert u'_{n}\vert ^{2} -C_{(\lambda _{k+1}-\lambda _{k})/16}\sum \limits _{\vert u'_{n}\vert>\delta }\vert u'_{n}\vert ^{p}\\ &{}\ge \frac{1}{2}\Vert w\Vert _{E}^{2}-\frac{\lambda _{k}}{2}\Vert w\Vert _{l^{2}}^{2}\\ &{}~~~~-\frac{\lambda _{k+1}-\lambda _{k}}{4}\sum \limits _{2\vert w_{n}\vert>\delta }\vert w_{n}\vert ^{2} -2^{p}C_{(\lambda _{k+1}-\lambda _{k})/16}\sum \limits _{2\vert w_{n}\vert>\delta }\vert w_{n}\vert ^{p}\\ &{}\ge \frac{1}{2}\Vert w\Vert _{E}^{2}-\frac{\lambda _{k}}{2}\Vert w\Vert _{l^{2}}^{2}\\ &{}~~~~-\frac{\lambda _{k+1}-\lambda _{k}}{4}\Vert w\Vert _{l^{2}}^{2} -2^{p}C_{(\lambda _{k+1}-\lambda _{k})/16}\sum \limits _{2\vert w_{n}\vert>\delta }\vert w_{n}\vert ^{p}\\ &{}\ge \frac{1}{4}\left( 1-\frac{\lambda _{k}}{\lambda _{k+1}} \right) \Vert w\Vert _{E}^{2} -2^{p}C_{(\lambda _{k+1}-\lambda _{k})/16} \sum \limits _{2\vert w_{n}\vert >\delta }\vert w_{n}\vert ^{p}. \end{array} \end{aligned}$$

It follows from the elementary embedding relation that

$$\begin{aligned} 2^{p}C_{(\lambda _{k+1}-\lambda _{k})/16} \sum \limits _{2\vert w_{n}\vert >\delta }\vert w_{n}\vert ^{p}=o(\Vert w\Vert _{E}^{2})~\text {as}~ \Vert w\Vert _{E}\rightarrow 0. \end{aligned}$$

Thus,

$$\begin{aligned} J(u')\ge \frac{1}{4}\left( 1-\frac{\lambda _{k}}{\lambda _{k+1}} -\frac{o(\Vert w\Vert _{E}^{2})}{\Vert w\Vert _{E}^{2}}\right) \Vert w\Vert _{E}^{2},~\Vert u'\Vert _{E}\le \rho . \end{aligned}$$
(19)

Now we assume that (b) is not true. Then there is a sequence \(\{u^{(j)}\}\subset E(\lambda _{k})\bigoplus M\) such that

$$\begin{aligned} J(u^{(j)})\rightarrow 0,~\Vert u^{(j)}\Vert _{E}=\rho ,~j\rightarrow \infty . \end{aligned}$$

If \(\rho \) is sufficiently small, then (19) implies \(\Vert w^{(j)}\Vert _{E}\rightarrow 0\). Consequently, \(\Vert z^{(j)}\Vert _{E}\rightarrow \rho \). We note that \(E(\lambda _{k})\) is finite dimensional. We can take a renamed subsequence \( \{z^{(j)}\}\) such that \(z^{(j)}\rightarrow {\bar{z}}\in E(\lambda _{k})\), and we obtain

$$\begin{aligned} J({\bar{z}})=0,~\Vert {\bar{z}}\Vert _{E}=\rho ,~\vert {\bar{z}}_{n}\vert \le \frac{\delta }{2}, n\in {\mathbb {Z}}. \end{aligned}$$
(20)

Further, from (16), we have

$$\begin{aligned} 2F_{n}({\bar{z}}_{n}) \le (\lambda _{k}-\omega )\vert {\bar{z}}_{n}\vert ^{2}, \vert {\bar{z}}_{n}\vert \le \delta ,~n\in {\mathbb {Z}}. \end{aligned}$$
(21)

By the definition J and (20),

$$\begin{aligned} 2J({\bar{z}})= \sum \limits _{n=-\infty }^{+\infty }\left( (\lambda _{k}-\omega ) \vert {\bar{z}}_{n}\vert ^{2} -2F_{n}({\bar{z}}_{n})\right) =0. \end{aligned}$$

Thus, by (21), we get

$$\begin{aligned} 2F_{n}({\bar{z}}_{n}) =(\lambda _{k}-\omega )\vert {\bar{z}}_{n}\vert ^{2},~n\in {\mathbb {Z}}. \end{aligned}$$

Let \(\nu \) be any element in E. Then for \(t>0\) sufficiently small, we have

$$\begin{aligned} (\lambda _{k}-\omega )(\vert {\bar{z}}_{n} +t\nu _{n}\vert ^{2}-\vert {\bar{z}}_{n}\vert ^{2}) -(2F_{n}({\bar{z}}_{n}+t\nu _{n})-2F_{n}({\bar{z}}_{n}))\ge 0. \end{aligned}$$

Taking the limit as \(t\rightarrow 0\), we see that

$$\begin{aligned} \begin{array}{lll} &{}~~\sum \limits _{n=-\infty }^{+\infty }\lim \limits _{t\rightarrow 0}\frac{ \frac{(\lambda _{k}-\omega )}{2}(\vert {\bar{z}}_{n} +t\nu _{n}\vert ^{2}-\vert {\bar{z}}_{n}\vert ^{2}) -(F_{n}({\bar{z}}_{n}+t\nu _{n})-F_{n}({\bar{z}}_{n}))}{t}\\ &{}=\left( (\lambda _{k}-\omega ){\bar{z}}-f({\bar{z}}),\nu \right) \\ &{}\ge 0, \end{array} \end{aligned}$$

that is,

$$\begin{aligned} (\lambda _{k}-\omega ){\bar{z}}=f({\bar{z}}). \end{aligned}$$

This completes the proof.

Now, we state our main results.

Theorem 1

Assume that \((i)-(vi)\) and \((i')\) hold, then problem (5) has at least one nontrivial solution \(u_{0}\in E\).

Proof of Theorem 1

We will use Lemma 2, Lemma 3 and Lemma 4 to prove our results. Let N denote the finite dimensional subspace of E spanned by the eigenfunctions corresponding to the eigenvalues which is less than \(\lambda _{k}\). Thus we have \(E=N\bigoplus E(\lambda _{k})\bigoplus M\).

By (vi), we have

$$\begin{aligned} \begin{array}{lll} J(v)&{}=\frac{1}{2}(Hv-\omega v,v)- \sum \limits _{n=-\infty }^{+\infty }F_{n}(v_{n})\\ &{}\le \frac{1}{2}\Vert v\Vert _{E}^{2} -\frac{\lambda _{k-1}}{2}\Vert v\Vert _{l^{2}}^{2}\\ &{}\le 0,~v\in N. \end{array} \end{aligned}$$

On the other hand, for \(\forall u'\in E(\lambda _{k})\bigoplus M\), if the result (a) of Lemma 4 holds, then problem (5) has at least one nontrivial \(u_{0}={\bar{z}}\in E(\lambda _{k})\). If not, we have, for \(\rho \) sufficiently small,

$$\begin{aligned} J(u')\ge \varepsilon _{1},~ \Vert u'\Vert _{E}=\rho ,~u'\in E(\lambda _{k})\bigoplus M, \end{aligned}$$

for some positive \(\varepsilon _{1}\).

Let \(e^{k}\) be an eigenfunction corresponding to \(\lambda _{k}\) such that \(e^{k}\in \partial B_{1}\bigcap (E(\lambda _{k})\bigoplus M)\). For every \(v\in N\) and \(s>0\), let \(v'=se^{k}+v\) satisfy \(\Vert v'\Vert _{E}=\Vert se^{k}+v\Vert _{E}=R>R_{0}>\rho \). We have

$$\begin{aligned} \begin{array}{lll} J(v')&{}=\frac{1}{2}\Vert v'\Vert _{E}^{2} -\frac{\lambda _{k}}{2}\Vert v'\Vert _{l^{2}}^{2} -\sum \limits _{n=-\infty }^{+\infty }G_{n}\left( v'_{n}\right) \\ &{}=\frac{1}{2}\Vert v\Vert _{E}^{2} -\frac{\lambda _{k}}{2}\Vert v\Vert _{l^{2}}^{2} -\sum \limits _{n=-\infty }^{+\infty }G_{n}\left( v'_{n}\right) \\ &{}\le -\sum \limits _{n=-\infty }^{+\infty }G_{n}\left( v'_{n}\right) . \end{array} \end{aligned}$$

From (v), there exists \(\phi \in l^{1}\) such that \(\vert G_{n}(t)\vert \le \phi _{n}\) for all \(n\in {\mathbb {Z}}\) and \(\vert t\vert <\delta \), then

$$\begin{aligned} \begin{array}{lll} J(v')&{}\le -\sum \limits _{n=-\infty }^{+\infty }G_{n}\left( v'_{n}\right) \\ &{}\le \sum \limits _{\vert v'_{n}\vert <\delta }\vert G_{n}\left( v'_{n}\right) \vert -\sum \limits _{\vert v'_{n}\vert \ge \delta } G_{n}\left( v'_{n}\right) \\ &{}=\Vert \phi \Vert _{l^{1}}. \end{array} \end{aligned}$$

Let \(m_{R}=\Vert \phi \Vert _{l^{1}}\). Obviously, \(m_{R}/R\rightarrow 0(R\rightarrow \infty )\). It follows from Lemma 2 and Remark 1 that there exists a sequence \(\{u^{(j)}\}\subset E\) satisfying

$$\begin{aligned} J\left( u^{(j)}\right) \rightarrow c,~ \varepsilon _{1}\le c<\infty ,~ J'\left( u^{(j)}\right) \rightarrow 0. \end{aligned}$$
(22)

Next, we will show that {\(u^{(j)}\}\) is bounded. If not, that is \(\rho _{j}=\Vert u^{(j)}\Vert _{E}\rightarrow \infty \) as \(j\rightarrow \infty \). Let \({\tilde{u}}^{(j)}=\frac{u^{(j)}}{\rho _{j}}\). Then \(\Vert {\tilde{u}}^{(j)}\Vert _{E}=1\). Consequently, there is renamed subsequence such that \({\tilde{u}}^{(j)}\rightarrow {\tilde{u}}\) weakly in E, strongly in \(l^{2}\) and \({\tilde{u}}^{(j)}_{n}\rightarrow {\tilde{u}}_{n}\) as \(j\rightarrow \infty \) for all \(n\in {\mathbb {Z}}\).

In view of (22), we obtain

$$\begin{aligned} \langle J'\left( u^{(j)}\right) , u^{(j)}\rangle =\Vert u^{(j)}\Vert _{E}^{2}-\lambda _{k}\Vert u^{(j)}\Vert _{l^{2}}^{2} -\sum \limits _{n=-\infty }^{+\infty } g_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n} =o(\rho _{j}). \end{aligned}$$
(23)

Hence, if j is taken sufficiently large,

$$\begin{aligned} \begin{array}{lll} 2\varepsilon _{1} &{}\le 2J\left( u^{(j)}\right) \\ &{}=\Vert u^{(j)}\Vert _{E}^{2}-\lambda _{k}\Vert u^{(j)}\Vert _{l^{2}}^{2} -\sum \limits _{n=-\infty }^{+\infty } 2G_{n}\left( u^{(j)}_{n}\right) \\ &{}=\sum \limits _{n=-\infty }^{+\infty } \left( g_{n}\left( u^{(j)}_{n}\right) u^{(j)}_{n}-2G_{n}\left( u^{(j)}_{n}\right) \right) +o(\rho _{j}). \end{array} \end{aligned}$$
(24)

In addition, if \(\rho _{j}=\Vert u^{(j)}\Vert _{E}\rightarrow \infty \), then there exists some  \(n_{0}\in {\mathbb {Z}}\) such that

$$\begin{aligned} \vert u^{(j)}_{n_{0}}\vert \rightarrow \infty ~\text {as}~j\rightarrow \infty , ~\text {and}~\vert {\tilde{u}}_{n_{0}}\vert \ne 0. \end{aligned}$$
(25)

In fact, if \(\rho _{j}=\Vert u^{(j)}\Vert _{E}\rightarrow \infty \), then there would be two positive constants \({\overline{M}}\) and T such that \(\Vert u^{(j)}\Vert _{E}>{\overline{M}}\) for \(j>T\).

Let \(A_{j}=\{n\in {\mathbb {Z}}\vert \vert u^{(j)}_{n}\vert >{\overline{M}}\}\) for \(j>T\). We introduce the set \(A=\{m\vert m\in \bigcap \limits _{i=1}^{+\infty }A_{j_{i}} \}\), where \(\{j_{i}\}\) is any subsequence of the sequence \(\{j\}\) with \(j>T\) and \(i\in {\mathbb {N}}\). If A is an empty set, then we can choose a subsequence of \(\{u^{(j)}\}\), still denoted by\(\{u^{(j)}\}\), such that \(\vert u^{(j)}_{n}\vert \le {\overline{M}}\) for all \(n\in {\mathbb {Z}}\) when j is sufficiently large. It follows that \({\tilde{u}}^{(j)}_{n}\rightarrow {\tilde{u}}_{n}=0\) as \(j\rightarrow \infty \), and \({\tilde{u}}^{(j)}\rightarrow {\tilde{u}}=0\) strongly in \(l^{2}\). By (iv) and (v), we have

$$\begin{aligned} \frac{\vert G_{n}(t)\vert }{t^{2}}\le M_{1}, \forall n\in {\mathbb {Z}}, t\in {\mathbb {R}}. \end{aligned}$$

Combining this with (22),

$$\begin{aligned} \begin{array}{lll} \frac{2J(u^{(j)})}{\rho _{j}^{2}} &{}=1-\lambda _{k}\Vert {\tilde{u}}^{(j)}\Vert _{l^{2}}^{2} -2\sum \limits _{n=-\infty }^{+\infty } \frac{G_{n}\left( {\tilde{u}}^{(j)}_{n}\rho _{j}\right) }{\left( {\tilde{u}}^{(j)}_{n}\rho _{j}^{2}\right) ^{2}}\left( {\tilde{u}}^{(j)}_{n}\right) ^{2}\\ &{}\ge 1-\lambda _{k}\Vert {\tilde{u}}^{(j)}\Vert _{l^{2}}^{2} -2M_{1}\Vert {\tilde{u}}^{(j)}\Vert _{l^{2}}^{2}. \end{array} \end{aligned}$$
(26)

Taking the limit as \(j\rightarrow \infty \), we obtain \(0\ge 1\). This contradiction tells us that A is a nonempty set. Thus there exist some \(m\in A\) suth that \(\vert u^{(j)}_{m}\vert \rightarrow \infty \) as \(j\rightarrow \infty \). If  \(\vert {\tilde{u}}^{(j)}_{m}\vert \rightarrow \infty \) as \(j\rightarrow \infty \), in this case  \(\Vert {\tilde{u}}^{(j)}\Vert _{E}\rightarrow \infty \). This is a contradiction. Hence, \(\vert {\tilde{u}}^{(j)}_{m}\vert \) is a finite number.

We notice that \({\tilde{u}}\ne 0\) form (26). If we suppose \({\tilde{u}}_{m}^{(j)}=u_{m}^{(j)}/\rho _{j}\rightarrow {\tilde{u}}_{m}=0\) for all \(m\in A\), then it is easy to verify that \({\tilde{u}}_{n}^{(j)}\rightarrow {\tilde{u}}_{n}=0\) as \(j\rightarrow \infty \) for all \(n\in {\mathbb {Z}}\). This contradicts with \({\tilde{u}}\ne 0\). Thus, there is a  \(n_{0}\in A\) such that (25) holds.

In view of (24), (25) and (v), we see that

$$\begin{aligned} \begin{array}{lll} 0=\lim \limits _{j\rightarrow \infty }\frac{2\varepsilon _{1}}{\rho _{j}} &{}\le \limsup \limits _{j\rightarrow \infty }\frac{g_{n_{0}}\left( u^{(j)}_{n_{0}}\right) u^{(j)}_{n_{0}}-2G_{n_{0}}\left( u^{(j)}_{n_{0}}\right) }{\rho _{j}}\\ &{}=\limsup \limits _{j\rightarrow \infty } \frac{g_{n_{0}}\left( {\tilde{u}}^{(j)}_{n_{0}}\rho _{j}\right) {\tilde{u}}^{(j)}_{n_{0}}\rho _{j} -2G_{n_{0}}\left( {\tilde{u}}^{(j)}_{n_{0}}\rho _{j}\right) }{\vert {\tilde{u}}^{(j)}_{n_{0}}\rho _{j}\vert } \cdot \vert {\tilde{u}}^{(j)}_{n_{0}}\vert \\ &{}=d_{n_{0}}\vert {\tilde{u}}_{n_{0}}\vert \\ &{}<0. \end{array} \end{aligned}$$

This is a contradiction. Thus the critical sequence \(\{u^{(j)}\}\) is bounded. By (22) and Lemma 3, problem (5) possesses at least one nontrivial solution u in E. This completes the proof.