Infinite Homoclinic Solutions of the Discrete Partial Mean Curvature Problem with Unbounded Potential

Abstract

The mean curvature problem is an important class of problems in mathematics and physics. We consider the existence of homoclinic solutions to a discrete partial mean curvature problem, which is tied to the existence of discrete solitons. Under the assumptions that the potential function is unbounded and that the nonlinear term is superlinear at infinity, we obtain the existence of infinitely many homoclinic solutions to this problem by means of the fountain theorem in the critical point theory. In the end, an example is given to illustrate the applicability of our results.

1. Introduction

Many phenomena in the real world are modeled by difference equations, and the property of solutions of difference equations reflects complex phenomena in various fields. For example, in [1,2,3,4], the authors studied the discrete mosquito population inhibition model. In [5,6], difference equations were used to study economic problems. In recent years, the theory of difference equations has been greatly expanded in various aspects, such as stability [7], oscillation [8,9], periodic solutions [10,11], homoclinic solutions [12,13,14,15,16], and so on.

In the past few decades, the mean curvature equation and its various modified forms have received extensive attention, such as the dynamic problem of combustible gas [17,18], the capillarity problem in hydrodynamics [19,20], and flux-limited diffusion phenomenon [21]. For other applications of the mean curvature operator, also see [22,23].

Recently, the problem of discrete mean curvature has been discussed. In [24], Ling and Zhou considered the following Dirichlet boundary value problem involving the mean curvature operator:

$$\left\{\begin{array}{c}-\Delta {\varphi }_c(\Delta u(k-1))=\lambda f(k,u\left(k\right)),k\in \mathbb{Z}(1,N),\hfill \\ u\left(0\right)=u(N+1)=0,\hfill \end{array}\right.$$

where $N\in {\mathbb{Z}}^{+},\mathbb{Z}(1,N)=\{1,2,\cdots ,N\}$,$\Delta$ is the forward difference operator defined by $u\left(k\right)=u(k+1)-u\left(k\right)$, for each $k\in \mathbb{Z}(1,N),f(k,·)\in C(\mathbb{R},\mathbb{R}),\lambda >0$, and ${\varphi }_c$ is the mean curvature operator, which is defined as ${\varphi }_c\left(s\right)=s/\sqrt{1+s^2}$ for $s\in \mathbb{R}$ [25]. By using the critical point theory, they obtained some sufficient conditions on the existence of infinitely many positive solutions.

Moreover, owing to the important role of partial difference equation models in many studies, the study of difference equations involving function with two or more discrete variables has attracted great attention, and some excellent results have been published [26,27,28,29]. In [27], Du and Zhou proposed the following partial discrete Dirichlet boundary value problem with the ${\varphi }_c$-Laplacian:

$$\left\{\begin{array}{c}{\Delta }_1\left({\varphi }_c\left({\Delta }_{1}u(i-1,j)\right)\right)+{\Delta }_2\left({\varphi }_c\left({\Delta }_{2}u(i,j-1)\right)\right)+\lambda f((i,j),u(i,j))=0,(i,j)\in \mathbb{Z}(1,M)\times \mathbb{Z}(1,N),\hfill \\ u(i,0)=u(i,N+1)=0,i\in \mathbb{Z}(0,M+1),\hfill \\ u(0,j)=u(M+1,j)=0,j\in \mathbb{Z}(0,N+1),\hfill \end{array}\right.$$

where $M,N\in {\mathbb{Z}}^{+}$, $f\left(\right(i,j),u)\in C(\mathbb{R},\mathbb{R})$ for each $(i,j)\in \mathbb{Z}(1,M)\times \mathbb{Z}(1,N),{\Delta }_{1}u(i,j)=u(i+1,j)-u(i,j)$ and ${\Delta }_{2}u(i,j)=u(i,j+1)-u(i,j)$. By the critical point theory, they obtained the existence of at least two positive solutions and an unbounded sequence of positive solutions. There are also some studies involving the boundary value problem of this operator [30,31].

However, there are few studies on the homoclinic solutions of difference equations with the mean curvature operator. As we know, the discrete boundary value problem with the mean curvature operator is carried out in a finite dimensional space, while the homoclinic solution problem is carried out in an infinite dimensional space, which brings the problem that it is necessary to overcome the lack of compactness in infinite dimensional space.

Inspired by the above work, in this paper, we discuss the existence of homoclinic solutions of the following partial difference equation with the mean curvature operator:

(P)

$-{\Delta }_1\left({\varphi }_c\left({\Delta }_{1}u(i-1,j)\right)\right)-{\Delta }_2\left({\varphi }_c\left({\Delta }_{2}u(i,j-1)\right)\right)+q(i,j)u(i,j)=f((i,j),u(i,j)),(i,j)\in {\mathbb{Z}}^2,$

where the potential function q and the nonlinear term f satisfy the following conditions:

(Q)

$q(i,j)\ge q_{*}>0,q(i,j)\to +\infty \mathrm{as}\left|i\right|+\left|j\right|\to +\infty$;

($A_1$)

$f((i,j),-t)=-f((i,j),t)\mathrm{for}\mathrm{all}(i,j)\in {\mathbb{Z}}^2\mathrm{and}t\in \mathbb{R}$;

($A_2$)

there are $r>2$ and $C>0$, such that $\left|f((i,j),t)\right|\le C\left(\right|t|+|t{|}^{r-1})$ for all $(i,j)\in {\mathbb{Z}}^2\mathrm{and}t\in \mathbb{R}$;

($A_3$)

${\displaystyle \frac{f\left(\right(i,j),t)}{t}\to +\infty }$ uniformly as $\left|t\right|\to +\infty$ for all $(i,j)\in {\mathbb{Z}}^2$;

($A_4$)

$f\left(\right(i,j),t)t\ge 2F\left(\right(i,j),t)\ge 0,\mathrm{and}f\left(\right(i,j),t)t-2F\left(\right(i,j),t)\to \infty$($\left|t\right|\to +\infty$) for all $(i,j)\in {\mathbb{Z}}^2$, where $F((i,j),u)={\int }_0^{u}f((i,j),s)ds.$

Among them, the coercive condition (Q) as well as $\left(A_4\right)$ will be used to obtain the compactness of Cerami sequences of the variational functional associated with problem $\left(P\right)$. Condition $\left(A_1\right)$ is a kind of symmetry conditions, which is essential when using the fountain theorem. Condition $\left(A_3\right)$ assumes that the nonlinear term is superlinear at infinity and it is used together with condition $\left(A_2\right)$ as more general conditions for improving the global Ambrosetti–Rabinowitz condition [32].

In general, a solution $u={\left\{u(i,j)\right\}}_{(i,j)\in {\mathbb{Z}}^2}$ is called a homoclinic solution of problem $\left(P\right)$ when it satisfies the following.

$$u(i,j)\to 0\mathrm{as}\left|i\right|+\left|j\right|\to \infty .$$

(1)

Moreover, it is called a nontrivial homoclinic solution when $u(i,j)\not\equiv 0$; otherwise u is called a trivial homoclinic solution.

Homoclinic solutions, a kind of solutions often related to solitons, have been one of the most important research fields in nonlinear difference equations. It can lead to system instability and chaos. In recent decades, by the critical point theory, many in-depth studies have been carried out to discuss the homoclinic solutions of discrete systems. For example, in [33,34,35,36], the authors discussed the existence of solitons to discrete nonlinear Schrödinger equations. In addition, the existence and multiplicity of homoclinic solutions of the discrete Hamiltonian systems [37,38] and difference equations with ${\varphi }_p$-Laplacian [14,15,16] have also attracted much attention.

The rest of this paper is organized as follows. In Section 2, we establish a space X and construct the variational framework corresponding to problem (P) in space X. In addition, two lemmas that are essential in this paper are proved, including the compactness of Cerami sequences and the compact embedding of spaces X to space $l^2$. Our main results will be shown in Section 3. In the last section, we provide a specific example to illustrate our main results.

2. Preliminaries

In order to use the critical point theory for problem (P), we introduce the definition of Cerami condition and the fountain theorem firstly.

Definition 1

(Cerami condition [32]). Let I be the $C^1$-functional defined on a real Banach space X, we say that I satisfies the Cerami condition if any Cerami sequence $\left\{u_j\right\}\subset X$, i.e., $I\left(u_j\right)\to c$ for some $c\in \mathbb{R}$ and $(1+\parallel u_j\parallel )\parallel I^{\prime }\left(u_j\right)\parallel \to 0$ as $j\to \infty$, has a convergent subsequences.

Lemma 1.

(Fountain theorem [32,39]). Suppose that X is a reflexive and separable Banach space, then $X=\overline{{\oplus }_{k\in \mathbb{N}}{Y}_k}$, where $Y_k$ is a finite-dimensional subspace of X for any $k\in \mathbb{N}$. We have the following.

$$H_n={\oplus }_{k=0}^n{Y}_k\mathrm{and}{I}_n=\overline{{\oplus }_{k=n}^{\infty }{Y}_k}.$$

Assume I is a continuously Gâteaux differentiable function on X and $I(-u)=I\left(u\right)$. If for each large enough $n\in \mathbb{N}$, there exists ${\omega }_n>{\tau }_n>0$ such that the following is the case:

($H_1$)  

$foranyc>0,IsatisfiestheCeramicondition$;

($H_2$)  

$a_n=\underset{u\in I_n,{\parallel u\parallel }_X={\tau }_n}{inf}I\left(u\right)\to +\infty asn\to +\infty$;

($H_3$) 

$b_n=\underset{u\in H_n,{\parallel u\parallel }_X={\omega }_n}{max}I\left(u\right)\le 0$;

then I has a critical point sequence $\left\{u_n\right\}$, and $I\left(u_n\right)\to +\infty$ as $n\to +\infty$.

The fountain theorem, a powerful variational tool for finding critical points [32,39], will be used in the next section to find the sequence of homoclinic solutions of problem $\left(P\right)$.

Now, we establish a space that will be used to discuss the homoclinic solutions of problem $\left(P\right)$ below.

The following is the case:

$$l^2\equiv l^2\left({\mathbb{Z}}^2\right)=\left\{u=\left\{u(i,j)\right\}:(i,j)\in {\mathbb{Z}}^2,u(i,j)\in \mathbb{R},\sum _{(i,j)\in {\mathbb{Z}}^2}{\left|u(i,j)\right|}^2<\infty \right\},$$

and endow $l^2$ with the following norm.

$${\parallel u\parallel }_{l^2}^2=\sum _{(i,j)\in {\mathbb{Z}}^2}{\left|u(i,j)\right|}^2\mathrm{and}{\parallel u\parallel }_{\infty }=\underset{(i,j)\in {\mathbb{Z}}^2}{max}\left\{\left|u(i,j)\right|:(i,j)\in {\mathbb{Z}}^2\right\}.$$

Consider the space as follows:

$$X=\left\{u\in l^2:\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){\left|u(i,j)\right|}^2<\infty \right\},$$

endowed with the following norm.

$${\parallel u\parallel }_X^2=\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){\left|u(i,j)\right|}^2\mathrm{for}\mathrm{all}u\in X.$$

Then, X is a Hilbert space with respect to the norm ${\parallel ·\parallel }_X$. In addition, for all $u\in X$, the following inequality holds clearly.

$${\parallel u\parallel }_{\infty }\le {\parallel u\parallel }_{l^2}\le \frac{1}{\sqrt{q_{*}}}{\parallel u\parallel }_X.$$

(2)

We define a functional $I:X\to \mathbb{R}$ as follows:

$$\begin{array}{cc}\hfill I\left(u\right)=& \sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\Phi }_c\left({\Delta }_{1}u(i-1,j)\right)+{\Phi }_c\left({\Delta }_{2}u(i,j-1)\right)\right]\hfill \\ & +\frac{1}{2}\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){\left|u(i,j)\right|}^2-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),u(i,j)),\hfill \end{array}$$

(3)

where $u\in X$ and ${\Phi }_c\left(s\right):={\int }_0^s{\varphi }_c\left(t\right)dt=\sqrt{1+s^2}-1$ for all $s\in \mathbb{R}$.

By using a regular argument, we see that $I\in C^1(X,\mathbb{R})$ and the following is the case:

$$\begin{array}{cc}\hfill I^{\prime }\left(u\right)\left(v\right)& =-\sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\Delta }_1{\varphi }_c\left({\Delta }_{1}u(i-1,j)\right)+{\Delta }_2{\varphi }_c\left({\Delta }_{2}u(i,j-1)\right)\right]v(i,j)\hfill \\ & +\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j)u(i,j)v(i,j)-\sum _{(i,j)\in {\mathbb{Z}}^2}f((i,j),u(i,j))v(i,j)\hfill \end{array}$$

for all $u,v\in X$. This implies that I is the variational functional associated with problem $\left(P\right)$, so the critical points of (3) are the solutions to problem $\left(P\right)$.

Lemma 2.

Assume that condition $\left(Q\right)$ holds, then for every sequence $\left\{u_n\right\}\subset X$ with $u_n\rightharpoonup u$ in X, we have $u_n\to u$ in $l^2$.

Proof. 

Without a loss of generality, we may assume that $u_n\rightharpoonup 0$ in X. Next, we prove that $u_n\to 0$ in $l^2$. It is easy to obtain the following:

$$\underset{n\in \mathbb{N}}{sup}{\parallel u_n\parallel }_X<\infty$$

by the Banach–Steinhauss theorem. Therefore, there is $\overline{s}>0$ such that the following is the case.

$$\parallel u_n{\parallel }_X^2=\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){\left|u_n(i,j)\right|}^2\le \overline{s}\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

By condition $\left(Q\right)$, for any $\epsilon >0$, there is $N>0$ such that when $\left|i\right|+\left|j\right|>N$, the following inequality holds.

$$q(i,j)\ge \frac{2\overline{s}}{\epsilon }.$$

Therefore, we have the following.

$$\sum _{\left|i\right|+\left|j\right|>N}|u_n{(i,j)|}^2\le \frac{\epsilon }{2\overline{s}}\sum _{\left|i\right|+\left|j\right|>N}q(i,j){|u_n(i,j)|}^2\le \frac{\epsilon }{2}.$$

On the other hand, since $u_n\rightharpoonup 0\mathrm{as}n\to +\infty$, then, for all $(i,j)\in {\mathbb{Z}}^2$, the following is the case.

$$u_n(i,j)\to 0\mathrm{as}n\to +\infty .$$

Therefore, there is $K>0$ such that when $n>K$, the sum of finite terms of sequence $\left\{u_n\right\}$ satisfies the following inequality.

$$\sum _{\left|i\right|+\left|j\right|\le N}{\left|u_n(i,j)\right|}^2<\frac{\epsilon }{2}.$$

In summary, when $n>K$, one has:

$$\parallel u_n{\parallel }_{l^2}^2=\sum _{(i,j)\in {\mathbb{Z}}^2}|u_n{(i,j)|}^2=\sum _{\left|i\right|+\left|j\right|\le N}{\left|u(i,j)\right|}^2+\sum _{\left|i\right|+\left|j\right|>N}{\left|u(i,j)\right|}^2<\epsilon ,$$

so sequence $\left\{u_n\right\}$ strongly converges to 0 in $l^2$. □

Next, we use the conclusion of Lemma 1 to prove that functional I satisfies the Cerami condition, which is a crucial step for our main results.

Lemma 3.

If condition $\left(Q\right)\mathrm{and}\left(A_2\right)-\left(A_4\right)$ holds, then for any $c\in \mathbb{R}$, the functional $I\in C^1(X,\mathbb{R})$ satisfies the Cerami condition.

Proof. 

Set $\left\{u_n\right\}$ be a Cerami sequence.

$$I\left(u_n\right)\to c,(1+\parallel u_n{\parallel }_X)\parallel I^{\prime }\left(u_n\right){\parallel }_{X^{*}}\to 0\mathrm{as}n\to +\infty .$$

(4)

In fact, $\left\{u_n\right\}$ is a bounded sequence in X. Arguing by contradiction, assume that $\parallel u_n{\parallel }_X\to +\infty$ as $n\to +\infty$. Let ${\varrho }_n=u_n/{\parallel u_n\parallel }_X$, then $\left\{{\varrho }_n\right\}$ is a bounded sequence in X and has a weakly convergence subsequence. Assume that ${\varrho }_n\rightharpoonup \varrho \mathrm{in}X$. By Lemma 2, we have the following.

$${\varrho }_n\to \varrho \mathrm{in}{l}^2.$$

(5)

From condition $\left(A_2\right)$, the following inequality holds:

$$F((i,j),u)={\int }_0^{u}f((i,j),s)ds\le C{\int }_0^{1}u\left(\right|tu|+{\left|tu\right|}^{r-1})dt\le \frac{C}{2}{\left|u\right|}^2+\frac{C}{r}{\left|u\right|}^r$$

(6)

for all $(i,j)\in {\mathbb{Z}}^2$ and $u\in \mathbb{R}$.

In the case of $\varrho =0$, define the following.

$$I\left({\theta }_n{u}_n\right):=\underset{\theta \in [0,1]}{max}I\left(\theta u_n\right).$$

For any $M>1$, there is $n\in \mathbb{N}$ with $\parallel u_n{\parallel }_X\ge 2M^{\frac{1}{2}}$. Let the following be the case.

$$\widetilde{{\varrho }_n}=2M^{\frac{1}{2}}{\varrho }_n=2M^{\frac{1}{2}}\frac{u_n}{\parallel u_n{\parallel }_X}.$$

It is easy to obtain the following.

$$\begin{array}{cc}\hfill I\left({\theta }_n{u}_n\right)\ge & I\left(\frac{2M^{\frac{1}{2}}}{\parallel u_n{\parallel }_X}{u}_n\right)\hfill \\ \hfill =& \sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\Phi }_c\left({\Delta }_1\widetilde{{\varrho }_n}(i-1,j)\right)+{\Phi }_c\left({\Delta }_2\widetilde{{\varrho }_n}(i,j-1)\right)\right]\hfill \\ & +\frac{1}{2}\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){\left|\widetilde{{\varrho }_n}(i,j)\right|}^2-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),\widetilde{{\varrho }_n}(i,j))\hfill \\ \hfill \ge & 2M\parallel {\varrho }_n{\parallel }_X^2-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),\widetilde{{\varrho }_n}(i,j)).\hfill \end{array}$$

(7)

From (6), we can obtain the following.

$$\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),\widetilde{{\varrho }_n}(i,j))\le \frac{C}{2}\parallel \widetilde{{\varrho }_n}{\parallel }_{l^2}^2+\frac{C}{r}{\parallel \widetilde{{\varrho }_n}\parallel }_{l^2}^r\to 0\mathrm{as}n\to \infty .$$

Combined with (7), we have $I\left({\theta }_n{u}_n\right)\ge M$ with larger n. This inequality means the following.

$$I\left({\theta }_n{u}_n\right)\to \infty \mathrm{as}n\to \infty .$$

(8)

When n is large enough, $I\left(\theta u_n\right)$ reaches its maximum at ${\theta }_n\in (0,1)$ for the reason that $I\left(0\right)=0$ and $I\left(u_n\right)\to c$ as $n\to \infty$. Then, $\langle I^{\prime }\left({\theta }_n{u}_n\right),u_n\rangle =0$.

After a straightforward analysis, we know that ${\displaystyle {\Phi }_c\left(u\right)-\frac{1}{2}{\varphi }_c\left(u\right)u}$ is an even function and increases in $[0,\infty )$. By condition $\left(A_4\right)$, we have the following inequality.

$$\begin{array}{cc}\hfill I\left({\theta }_n{u}_n\right)=& I\left({\theta }_n{u}_n\right)-\frac{1}{2}\langle I^{\prime }\left({\theta }_n{u}_n\right),{\theta }_n{u}_n\rangle \hfill \\ \hfill =& \sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\Phi }_c\left({\theta }_n{\Delta }_1{u}_n(i-1,j)\right)+{\Phi }_c\left({\theta }_n{\Delta }_2{u}_n(i,j-1)\right)\right]-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),{\theta }_n{u}_n(i,j))\hfill \\ & -\frac{1}{2}\sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\varphi }_c\left({\theta }_n{\Delta }_1{u}_n(i-1,j)\right){\theta }_n{\Delta }_1{u}_n(i-1,j)+{\varphi }_c\left({\theta }_n{\Delta }_2{u}_n(i,j-1)\right){\theta }_n{\Delta }_2{u}_n(i,j-1)\right]\hfill \\ & +\frac{1}{2}\sum _{(i,j)\in {\mathbb{Z}}^2}f((i,j),{\theta }_n{u}_n(i,j)){\theta }_n{u}_n(i,j)\hfill \\ \hfill \le & \sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\Phi }_c\left({\Delta }_{1}u(i-1,j)\right)-\frac{1}{2}{\varphi }_c\left({\Delta }_1{u}_n(i-1,j)\right){\Delta }_1{u}_n(i-1,j)\right]\hfill \\ & +\sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\Phi }_c\left({\Delta }_{2}u(i,j-1)\right)-\frac{1}{2}{\varphi }_c\left({\Delta }_2{u}_n(i,j-1)\right){\Delta }_2{u}_n(i,j-1)\right]\hfill \\ & +2\sum _{(i,j)\in {\mathbb{Z}}^2}\left[\frac{1}{2}f((i,j),u_n(i,j))u_n(i,j)-F((i,j),u_n(i,j))\right]\hfill \\ \hfill \le & 2\left(I\left(u_n\right)-\frac{1}{2}\langle I^{\prime }\left(u_n\right),u_n\rangle \right).\hfill \end{array}$$

Consider this inequality with (4), we can obtain that

$$I\left({\theta }_n{u}_n\right)\le 2I\left(u_n\right)-\langle I^{\prime }\left(u_n\right),u_n\rangle \to 2c,$$

as $n\to \infty$, which is in contradiction with (8).

In another case, that is $\varrho \ne 0$, the following set is nonempty.

$$\Gamma :=\{(i,j)\in {\mathbb{Z}}^2:\varrho (i,j)\ne 0\}.$$

Now, we prove that sequence $\left\{u_n\right\}$ is bounded. By the condition $\left(A_3\right)$, we know the following.

$$\begin{array}{cc}\hfill I\left(u_n\right)& \le \frac{1}{2}\left(\sum _{(i,j)\in {\mathbb{Z}}^2}|{\Delta }_1{u}_n{(i-1,j)|}^2+\sum _{(i,j)\in {\mathbb{Z}}^2}{\left|{\Delta }_2{u}_n(i,j-1)\right|}^2\right)\hfill \\ & +\frac{1}{2}\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){\left|u_n(i,j)\right|}^2-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),u_n(i,j))\hfill \\ & \le \left(\frac{4}{q_{*}}+\frac{1}{2}\right){\parallel u_n\parallel }_X^2-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),u_n(i,j)).\hfill \end{array}$$

(9)

In addition, there is a constant $\tilde{c}$ with $I\left(u_n\right)\ge \tilde{c}$ from (4), so we have the following.

$$\underset{n\to \infty }{lim}\frac{\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),u_n(i,j))}{\parallel u_n{\parallel }_X^2}\le \underset{n\to \infty }{lim}\left(\frac{4}{q_{*}}+\frac{1}{2}-\frac{\tilde{c}}{\parallel u_n{\parallel }_X^2}\right)<+\infty .$$

(10)

However, fixing $(i_0,j_0)\in \Gamma$, from (5), we have the following.

$$u_n(i_0,j_0)={\varrho }_n(i_0,j_0){\parallel u_n\parallel }_X\to \infty \mathrm{as}n\to \infty .$$

As the following is the case:

$$\underset{\left|t\right|\to +\infty }{lim}\frac{F\left(\right(i,j),t)}{{\left|t\right|}^2}=+\infty$$

for all $(i,j)\in {\mathbb{Z}}^2$.

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}\frac{\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),u_n(i,j))}{\parallel u_n{\parallel }_X^2}& \ge \underset{n\to \infty }{lim}\frac{F((i_0,j_0),u_n(i_0,j_0))}{\parallel u_n{\parallel }_X^2}\hfill \\ & =\underset{n\to \infty }{lim}\frac{F((i_0,j_0),u_n(i_0,j_0))}{|u_n(i_0,j_0){|}^2}{\left|{\varrho }_n(i_0,j_0)\right|}^2\hfill \\ & =+\infty ,\hfill \end{array}$$

which is contrary to (10).

Until now, we have proved that the Cerami sequence $\left\{u_n\right\}$ is bounded in X. We still have to show that $\left\{u_n\right\}$ has a convergent subsequence. Clearly, $\left\{u_n\right\}$ has a weakly convergent subsequence, which might as well be denoted as $\left\{u_n\right\}$, and suppose $u_n\rightharpoonup u_0$ in X. Applying Lemma 2, we have the following.

$$u_n\to u_0\mathrm{in}{l}^2.$$

(11)

Consider (13) together with the monotonicity of ${\varphi }_c\left(u\right)$, we have the following relation.

$$\left({\varphi }_c\left(x\right)-{\varphi }_c\left(y\right)\right)(x-y)\ge 0\mathrm{for}\mathrm{any}x,y\in \mathbb{R}.$$

(12)

Hence, it can be obtained by direct calculations that the following is the case.

$$\begin{array}{cc}\hfill & 〈I^{\prime }\left(u_n\right)-I^{\prime }\left(u_0\right),u_n-u_0〉\hfill \\ \hfill =& \sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\varphi }_c\left({\Delta }_1{u}_n(i-1,j)\right)-{\varphi }_c\left({\Delta }_1{u}_0(i-1,j)\right)\right][{\Delta }_1{u}_n(i-1,j)-{\Delta }_1{u}_0(i-1,j)]\hfill \\ & +\sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\varphi }_c\left({\Delta }_2{u}_n(i,j-1)\right)-{\varphi }_c\left({\Delta }_2{u}_0(i,j-1)\right)\right][{\Delta }_2{u}_n(i,j-1)-{\Delta }_2{u}_0(i,j-1)]\hfill \\ & +\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){|u_n(i,j)-u_0(i,j)|}^2-\sum _{(i,j)\in {\mathbb{Z}}^2}[f\left(u_n(i,j)\right)-f\left(u_0(i,j)\right)][u_n(i,j)-u_0(i,j)]\hfill \\ \hfill \ge & \sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){|u_n(i,j)-u_0(i,j)|}^2-\sum _{(i,j)\in {\mathbb{Z}}^2}[f\left(u_n(i,j)\right)-f\left(u_0(i,j)\right)][u_n(i,j)-u_0(i,j)].\hfill \end{array}$$

(13)

Proceeding a step further, we have the following.

$$\begin{array}{cc}\hfill & \sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){|u_n(i,j)-u_0(i,j)|}^2\hfill \\ & \le 〈I^{\prime }\left(u_n\right)-I^{\prime }\left(u_0\right),u_n-u_0〉+\sum _{(i,j)\in {\mathbb{Z}}^2}[f\left(u_n(i,j)\right)-f\left(u_0(i,j)\right)][u_n(i,j)-u_0(i,j)].\hfill \end{array}$$

(14)

Let $\overline{r}\in {\mathbb{R}}^{+}$ satisfy $\frac{1}{\overline{r}}+\frac{1}{r}=1$, from the Hölder inequality, we have the following.

$$\begin{array}{cc}\hfill & \sum _{(i,j)\in {\mathbb{Z}}^2}|u_n{(i,j)|}^{r-1}|u_n(i,j)-u_0(i,j)|\hfill \\ \hfill \le & {\left(\sum _{(i,j)\in {\mathbb{Z}}^2}{\left(|u_n{(i,j)|}^{r-1}\right)}^{\overline{r}}\right)}^{1/\overline{r}}{\left(\sum _{(i,j)\in {\mathbb{Z}}^2}{|u_n(i,j)-u_0(i,j)|}^r\right)}^{1/r}\hfill \\ \hfill =& \parallel u_n{\parallel }_{l^r}^{r-1}{\parallel u_n-u_0\parallel }_{l^r}.\hfill \end{array}$$

Therefore, combined with (6), we have the following.

$$\begin{array}{cc}\hfill & \left|\sum _{(i,j)\in {\mathbb{Z}}^2}[f((i,j),u_n(i,j))-f((i,j),u_0(i,j))][u_n(i,j)-u_0(i,j)]\right|\hfill \\ \hfill \le & C\sum _{(i,j)\in {\mathbb{Z}}^2}\left(|u_n(i,j)|+|u_n{(i,j)|}^{r-1}\right)|u_n(i,j)-u_0(i,j)|\hfill \\ & +C\sum _{(i,j)\in {\mathbb{Z}}^2}\left(|u_0(i,j)|+|u_0{(i,j)|}^{r-1}\right)|u_n(i,j)-u_0(i,j)|.\hfill \\ \hfill \le & \left(\parallel u_n{\parallel }_{l^2}\parallel u_n-u_0{\parallel }_{l^2}+\parallel u_n{\parallel }_{l^r}^{r-1}{\parallel u_n-u_0\parallel }_{l^r}\right)+\left(\parallel u_0{\parallel }_{l^2}\parallel u_n-u_0{\parallel }_{l^2}+\parallel u_0{\parallel }_{l^r}^{r-1}{\parallel u_n-u_0\parallel }_{l^r}\right)\hfill \\ \hfill \le & \left(\parallel u_n{\parallel }_{l^2}+\parallel u_0{\parallel }_{l^2}+\parallel u_n{\parallel }_{l^r}^{r-1}+{\parallel u_0\parallel }_{l^r}^{r-1}\right){\parallel u_n-u_0\parallel }_{l^2}.\hfill \end{array}$$

(15)

Associating (2), (11) with the boundedness of the sequence $\left\{u_n\right\}$ in X, inequality (15) means the following.

$$\begin{array}{c}\hfill \sum _{(i,j)\in {\mathbb{Z}}^2}[f((i,j),u_n(i,j))-f((i,j),u(i,j))][u_n(i,j)-u_0(i,j)]\to 0\mathrm{as}n\to +\infty .\end{array}$$

(16)

In addition, the following is the case.

$$\begin{array}{c}\hfill 〈I^{\prime }\left(u_n\right)-I^{\prime }\left(u_0\right),u_n-u_0〉\le \parallel I^{\prime }\left(u_n\right){\parallel }_{X^{*}}\parallel u_n{\parallel }_X+\parallel I^{\prime }\left(u_n\right){\parallel }_{X^{*}}\parallel u_0{\parallel }_X+\left|I^{\prime }\left(u_0\right)(u_n-u_0)\right|.\end{array}$$

Due to (4) and the weak convergence and boundedness of $\left\{u_n\right\}$, we have the following.

$$〈I^{\prime }\left(u_n\right)-I^{\prime }\left(u_0\right),u_n-u_0〉\to 0\mathrm{as}n\to +\infty .$$

(17)

Finally, we can conclude that $\parallel u_n-u_0{\parallel }_X\to 0$ as $n\to \infty$ from (14), (16), and (17). The proof is completed. □

3. Main Results

As a matter of convenience, we provide the signs and conclusions that need to be used in the main conclusions. For $(s,t)\in {\mathbb{Z}}^2$, define ${\xi }_{(s,t)}=\left\{{\xi }_{(s,t)}(i,j)\right\}$ as follows.

$${\xi }_{(s,t)}(i,j)=\left\{\begin{array}{cc}1,\hfill & (i,j)=(s,t),\hfill \\ 0,\hfill & (i,j)\ne (s,t).\hfill \end{array}\right.$$

Then, $X=\overline{\mathrm{span}\{{\xi }_{(s,t)}:(s,t)\in {\mathbb{Z}}^2\}}$. Let the following be the case.

$$E_{(s,t)}=\mathrm{span}\{{\xi }_{(s,t)}:(s,t)\in {\mathbb{Z}}^2\},$$

$$H_n={\oplus }_{\left|s\right|+\left|t\right|=0}^n{E}_{(s,t)}\mathrm{and}{I}_n=\overline{{\oplus }_{\left|s\right|+\left|t\right|=n}^{\infty }{E}_{(s,t)}}.$$

Since space $H_n$ is a finite dimensional space for each $n\in \mathbb{N}$, there exists a constant ${\lambda }_n>0$ such that the following is the case.

$${\parallel u\parallel }_X\le {\lambda }_n{\parallel u\parallel }_{\infty }\mathrm{for}\mathrm{all}u\in H_n.$$

(18)

We use the following:

$${\zeta }_n=\underset{u\in I_n,{\parallel u\parallel }_X=1}{sup}{\parallel u\parallel }_{l^2},$$

and the following is also the case.

$$\underset{n\to \infty }{lim}{\zeta }_n={0\mathrm{and}\parallel u\parallel }_{l^2}\le {\zeta }_n{\parallel u\parallel }_X.$$

(19)

Now, we state our main results.

Theorem 1.

Assume that conditions $\left(Q\right)$ and $\left(A_1\right)$–$\left(A_4\right)$ are true. Then, the mean curvature problem $\left(P\right)$ has a sequence of non-trivial homoclinic solutions $\left\{u_n\right\}\subset X$ such that $I\left(u_n\right)\to +\infty$ as $n\to +\infty$.

Proof. 

For $\left(A_1\right)$ and Lemma 3, $I\in C^1(X,\mathbb{R})$ is an even function that satisfies condition $\left(H_1\right)$. It remains to prove that conditions $\left(H_2\right)$ and $\left(H_3\right)$ of Lemma 1 hold.

Let ${\tau }_n={\left(\frac{r}{4C{\zeta }_n^r}\right)}^{\frac{1}{r-2}}$, then $\underset{n\to \infty }{lim}{\tau }_n=+\infty$. For any $u\in I_n$, we have the following.

$$\begin{array}{cc}\hfill I\left(u\right)& =\sum _{(i,j)\in {\mathbb{Z}}^2}\left[{\Phi }_c\left({\Delta }_{1}u(i-1,j)\right)+{\Phi }_c\left({\Delta }_{2}u(i,j-1)\right)\right]\hfill \\ & +\frac{1}{2}\sum _{(i,j)\in {\mathbb{Z}}^2}q(i,j){\left|u(i,j)\right|}^2-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),u(i,j))\hfill \\ & \ge \frac{1}{2}{\parallel u\parallel }_X^2-\frac{C}{2}{\parallel u\parallel }_{l^2}^2-\frac{C}{r}{\parallel u\parallel }_{l^r}^r.\hfill \end{array}$$

Thus, from (19), we have the following:

$$a_n=\underset{u\in I_n,{\parallel u\parallel }_X={\tau }_n}{inf}I\left(u\right)\ge \left(\frac{1}{2}-\frac{C{\zeta }_n^2}{2}-\frac{C{\zeta }_n^r}{r}{\tau }_n^{r-2}\right){\tau }_n^2=\left(\frac{1}{4}-\frac{C{\zeta }_n^2}{2}\right){\tau }_n^2,$$

which means $a_n\to +\infty$ as $n\to +\infty$. Thus, the condition $\left(H_2\right)$ of Lemma 1 holds.

For any $n\in \mathbb{N}$, by $\left(A_3\right)$, there exists a positive constant $K_n$, such that the following is the case.

$$F((i,j),u)>{\lambda }_n\left(\frac{4}{q_{*}}+\frac{1}{2}\right){\left|u\right|}^2\mathrm{for}(i,j)\in {\mathbb{Z}}^2\mathrm{and}\left|u\right|>K_n.$$

Let ${\omega }_n={\lambda }_n{K}_n+{\tau }_n$. Then, we have ${\omega }_n>{\tau }_n>0$. For any $u\in H_n$ with ${\parallel u\parallel }_X={\omega }_n$, the following is the case.

$${\parallel u\parallel }_{\infty }\ge \frac{1}{{\lambda }_n}{\parallel u\parallel }_X>K_n$$

(20)

by (18). There is $(i_0,j_0)\in {\mathbb{Z}}^2$ such that ${\parallel u\parallel }_{\infty }=\left|u(i_0,j_0)\right|$ and the following is the case:

$$\begin{array}{cc}\hfill I\left(u\right)& \le \left(\frac{4}{q_{*}}+\frac{1}{2}\right){\parallel u\parallel }_X^2-\sum _{(i,j)\in {\mathbb{Z}}^2}F((i,j),u(i,j))\hfill \\ & \le \left(\frac{4}{q_{*}}+\frac{1}{2}\right){\lambda }_n{\parallel u\parallel }_{\infty }^2-F((i_0,j_0),u(i_0,j_0))\hfill \\ & <{\lambda }_n\left(\frac{4}{q_{*}}+\frac{1}{2}\right)|u(i_0,j_0){|}^2-{\lambda }_n\left(\frac{4}{q_{*}}+\frac{1}{2}\right){\left|u(i_0,j_0)\right|}^2\hfill \\ & =0\hfill \end{array}$$

from (9), (18) and (20). Therefore, we have the following inequality:

$$b_n=\underset{u\in H_n,{\parallel u\parallel }_X={\omega }_n}{max}I\left(u\right)\le 0.$$

Thus, the condition $\left(H_3\right)$ of Lemma 1 holds as well.

In summary, the proof of all conditions of Lemma 1 is completed. From Lemma 1, we know I has a critical point sequence $\left\{u_n\right\}\subset X$ and $I\left(u_n\right)\to +\infty$ as $n\to +\infty$. Therefore, problem $\left(P\right)$ has a non-trivial sequence of homoclinic solutions. □

4. An Example

In this section, we provide an example to show the applicability of the main results.

Example 1.

For any $(i,j)\in {\mathbb{Z}}^2$, let the following be the case.

$$f((i,j),u)=f\left(u\right)=u^{2}arctanu\mathrm{for}\mathrm{all}u\in \mathbb{R}.$$

Then, condition $\left(A_1\right)$ holds and the following is the case:

$$F((i,j),u)=\frac{1}{3}{u}^{3}arctanu-\frac{1}{6}{u}^2+\frac{1}{6}ln(1+u^2)\ge 0$$

for any $(i,j)\in {\mathbb{Z}}^2$ and $u\in \mathbb{R}$. Choose $r=3$ and $C=1$; clearly, we have the following.

$$\left|f((i,j),u)\right|\le \left|u\right|+{\left|u\right|}^3\mathrm{for}\mathrm{all}(i,j)\in {\mathbb{Z}}^2,u\in \mathbb{R}.$$

This means the condition $\left(A_2\right)$ holds. Furthermore, for all $(i,j)\in {\mathbb{Z}}^2$, we have the following.

$$\frac{f\left(\right(i,j),u)}{u}=uarctanu\to +\infty \mathrm{uniformly}\mathrm{as}\left|u\right|\to +\infty .$$

Therefore, condition $\left(A_3\right)$ holds. For any $(i,j)\in {\mathbb{Z}}^2$, we have the following.

$$G((i,j),u):=f((i,j),u)u-2F((i,j),u)=\frac{1}{3}\left(u^{3}arctanu+u^2-ln(1+u^2)\right)\ge \frac{1}{3}{u}^{3}arctanu.$$

Obviously, for all $(i,j)\in {\mathbb{Z}}^2$, the following is the case.

$$G\left(\right(i,j),u)\to +\infty \mathrm{as}\left|u\right|\to +\infty .$$

Thus, condition $\left(A_4\right)$ holds. In conclusion, take a function satisfying condition $\left(Q\right)$; problem $\left(P\right)$ has a sequence of homoclinic solutions by Theorem 1.

5. Conclusions

For a long period of time, the problem of homoclinic solutions is one of the most important research fields for differential equations [32,38,39]. Due to real applications, many scholars have devoted themselves to the study of the existence of homoclinic solutions of difference equations, especially the discrete solitons of discrete nonlinear Schrödinger equations in recent years [11,13,34,35]. The mean curvature problem arises from differential geometry and physics, such as dynamic problem of combustible gas [17,18] and flux-limited diffusion phenomenon [21]. In this paper, by using the fountain theorem, we obtain the existence of a sequence of homoclinic solutions for the discrete partial mean curvature problem with unbounded potential function. As far as we know, this is the first time for a discussion of the homoclinic solutions of the discrete partial mean curvature problem. At the end of the article, an example is given to illustrate the applicability of the conclusion of this article. The method used in this paper can be used for discussing the existence of homoclinic solutions to the higher-dimensional mean curvature problem or the existence of periodic solutions to such problems when the potential term is periodic, which will be our future work.