Positive Solutions to the Discrete Boundary Value Problem of the Kirchhoff Type

Abstract

The paper aims to study a discrete boundary value problem of the Kirchhoff type based on the critical point theory and the strong maximum principle. Compared to the existing literature, the existence and multiplicity of positive solutions to the problem are considered according to the behavior of the nonlinear term f in some points between the zero and positive infinity, which is a new attempt. Under different assumptions of the nonlinear term f, we obtain the determined open intervals of the parameter $\lambda$, such that the problem has at least three positive solutions or at least two positive solutions in different intervals. In the end, two concrete examples are used to illustrate our main conclusions.

1. Introduction

Let $\mathbb{R}$ and $\mathbb{Z}$ be real numbers and integers, respectively. For any $v_1,v_2\in \mathbb{Z}$, with $v_1<v_2$, define $[v_1,v_2]=\left\{v_1,v_1+1,...,v_2\right\}$. In this paper, we consider the following discrete Dirichlet boundary value problem of the Kirchhoff type

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\left(a+b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right){\Delta }^2{u}_{k-1}=\lambda f\left(k,u_k\right),k\in [1,N],\hfill \\ u_0=u_{N+1}=0,\hfill \end{array}\right.\end{array}$$

(1)

where N is a positive integer, $f(k,·)\in C(\mathbb{R},\mathbb{R})$ for each $k\in [1,N]$, $\Delta$ is the forward difference operator defined by $\Delta u_k=u_{k+1}-u_k$, ${\Delta }^2{u}_k=\Delta (\Delta u_k)$, $a,b$ are two positive constants, and $\lambda$ is a positive parameter.

The Kirchhoff-type equation arises in various branches of mathematical physics, involved in modeling suspension bridges [1]. In particular, problem $\left(1\right)$ is regarded as the discrete analogue of the following Kirchhoff-type problem

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}-\left(a+b{\int }_{\Omega }{|\nabla u|}^{2}dx\right)\Delta u=\lambda f(x,u),\hfill & \mathrm{in}\Omega ,\hfill \\ u=0,\hfill & \mathrm{on}\partial \Omega ,\hfill \end{array}\right.\end{array}$$

(2)

which is associated with the stationary version of the Kirchhoff equation

$$\begin{array}{c}\hfill u_{tt}-\left(a+b{\int }_{\Omega }{|\nabla u|}^{2}dx\right)\Delta u=f(x,u)\end{array}$$

(3)

presented by Kirchhoff [2]. Equation (3), an extension of the classical D’Alembert’s wave equation, describes the changes in length of a stretched string by transversal oscillations. After Lions [3] introduced an abstract framework to problem (2), plenty of scholars were interested in it and obtained numerous interesting results in different ways. For instance, He and Zou [4] gained infinitely many positive solutions to problem (2) with the help of variational methods. In 2010, Yang and Zhang [5] employed the local linking theory to discuss a class of nonlocal quasilinear elliptic boundary value problems and successfully acquired nontrivial solutions. In addition, in [6,7,8], the authors obtained the existence of sign-changing solutions and nontrivial solutions for problem (2) with $\lambda =1$.

During the past decades, difference equations have received wide attention in a lot of research fields, such as computer science, neural network, biology, and so on [9,10,11,12,13,14,15], which is mainly because difference equations have strong realistic significance. There are a multitude of important tools in the study of the boundary value problems of difference equations, including fixed-point methods [16,17,18], the method of upper and lower solutions [19,20,21], and invariant sets of descent flow [22]. Similarly, critical point theory also plays an imperative role in the research of difference equations. Based on critical point theory, Guo and Yu [23] focused on a second-order difference equation and obtained the existence of periodic and subharmonic solutions for the first time. After that, more and more researchers have used the critical point theory to study difference equations, and many outstanding conclusions were obtained, including periodic solutions [24,25], homoclinic solutions [26,27,28], heteroclinic solutions [29], and boundary value problems [30,31,32,33].

Recently, the discrete boundary value problems of the Kirchhoff type have received increasing attention. Numerous scholars applied various ways to study problem (1). For example, when $\lambda =1$ in problem (1), the existence of nontrivial solutions was captured via critical groups in [34]. In addition, Long [35,36] investigated a class of partial discrete Kirchhoff-type problems and ensured the existence of nontrivial solutions by applying different methods (composing of variational technique, local linking theory, fountain theorem, and Morse theory). It is worth noticing that when the primary function of the nonlinear term f is oscillatory at the zero or at infinity, Zhang and Zhou [37] acquired the existence of either an unbounded sequence of solutions or a sequence of non-zero solutions that converge to zero for problem (1) by the utilizing critical point theory. However, there are rare works in the literature that investigate problem (1) based on the behavior of the nonlinear term f in some positive points.

Inspired by the above conclusion, the paper aims to establish the existence and multiplicity of positive solutions for problem (1) according to the behavior of the nonlinear term f in some points between the zero and positive infinity. The main tools are the critical point theory (Lemmas 1 and 2) and the strong maximum principle (Lemma 6).

For convenience, we restate the two crucial lemmas.

Lemma 1.

(From [38], Theorem 4.1) Let $(X,\parallel ·\parallel )$ be a real finite dimensional Banach space and let $\Phi ,\Psi :X\to \mathbb{R}$ be two continuously G$\widehat{a}$teaux differentiable functions with $\Phi$ coercive. In addition, $\Phi$ and $\Psi$ satisfy

$$\begin{array}{c}\hfill {\displaystyle \underset{X}{inf}}\Phi =\Phi \left(0\right)=\Psi \left(0\right)=0.\end{array}$$

Assume that there exist $r\in \mathbb{R}$ and $\overline{u}\in X$, with $0<r<\Phi \left(\overline{u}\right)$, such that:

(i) 

$\frac{{\displaystyle \underset{\Phi \left(u\right)\le r}{sup}}\Psi \left(u\right)}{r}<\frac{\Psi \left(\overline{u}\right)}{\Phi \left(\overline{u}\right)}$;

(ii) 

for each $\lambda \in \Lambda :=\left(\frac{\Phi \left(\overline{u}\right)}{\Psi \left(\overline{u}\right)},\frac{r}{{\displaystyle \underset{\Phi \left(u\right)\le r}{sup}}\Psi \left(u\right)}\right)$, the functional $I_{\lambda }=\Phi -\lambda \Psi$ is coercive.

Then, for each $\lambda \in \Lambda$, the functional $I_{\lambda }$ has at least three different critical points in X.

It is necessary to use (PS) condition in Lemma 2, so we firstly recall the definition of the (PS) condition. Assume that X is a real Banach space, and let $I\in C^1(X,\mathbb{R})$. We say that I satisfies the Palais–Smale condition ((PS) condition) if any sequence $\left\{u_n\right\}$ for I, such that $\left\{I\left(u_n\right)\right\}$ is bounded and $\left\{I^{{}^{\prime }}\left(u_n\right)\right\}$ is convergent to 0 in $X^{*}$, possesses a convergent subsequence.

Lemma 2.

(From [39], Theorem 2.1) Let X be a real Banach space and let $\Phi ,\Psi :X\to \mathbb{R}$ be two continuously G$\widehat{a}$teaux differentiable functions such that ${\displaystyle \underset{X}{inf}}\Phi =\Phi \left(0\right)=\Psi \left(0\right)=0$. Assume that there are $r\in \mathbb{R}$ and $\tilde{u}\in X$, with $0<\Phi \left(\tilde{u}\right)<r$, such that

$$\begin{array}{c}\hfill \frac{{\displaystyle \underset{\Phi \left(u\right)\le r}{sup}}\Psi \left(u\right)}{r}<\frac{\Psi \left(\tilde{u}\right)}{\Phi \left(\tilde{u}\right)}.\end{array}$$

(4)

Moreover, for each $\lambda \in \Lambda :=\left(\frac{\Phi \left(\tilde{u}\right)}{\Psi \left(\tilde{u}\right)},\frac{r}{{\displaystyle \underset{\Phi \left(u\right)\le r}{sup}}\Psi \left(u\right)}\right)$, the functional $I_{\lambda }=\Phi -\lambda \Psi$ satisfies the (PS) condition, and it is unbounded from below.

Then, for each $\lambda \in \Lambda$, the functional $I_{\lambda }$ admits at least two non-zero critical points $u_{\lambda ,1},u_{\lambda ,2}$ such that $I_{\lambda }\left(u_{\lambda ,1}\right)<0<I_{\lambda }\left(u_{\lambda ,2}\right)$.

The rest of this paper is arranged as follows. In Section 2, we show some definitions, notations, and important lemmas. Furthermore, we establish the variational framework corresponding to problem (1). In Section 3, Theorems 1 and 2 are proven by using Lemmas 1 and 2, respectively. Moreover, some important corollaries and remarks are presented. In Section 4, two concrete examples are used to illustrate our results. In Section 5, we share the main conclusions of the paper and our future work direction.

2. Preliminaries

First of all, we consider the N-dimensional Banach space $T=\{u:[0,N+1]\to \mathbb{R}:u_0=u_{N+1}=0\}$ endowed with the norm:

$$\begin{array}{c}\hfill \parallel u\parallel ={\left(\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)}^{\frac{1}{2}}.\end{array}$$

Next, we establish the variational framework corresponding to problem (1). For each $u\in T$, let

$$\begin{array}{c}\hfill \Phi \left(u\right)=\frac{a}{2}\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2+\frac{b}{4}{\left(\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)}^2,\Psi \left(u\right)=\sum _{k=1}^{N}F\left(k,u_k\right),\end{array}$$

where $F(k,\xi )={\int }_0^{\xi }f(k,t)dt$ for each $(k,\xi )\in [1,N]\times \mathbb{R}$, and define

$$\begin{array}{c}\hfill I_{\lambda }\left(u\right)=\Phi \left(u\right)-\lambda \Psi \left(u\right).\end{array}$$

Clearly, $\Phi ,\Psi \in C^1(T,\mathbb{R})$, so $I_{\lambda }\in C^1(T,\mathbb{R})$. According to the boundary condition and the summation-by-parts method, we show that

$$\begin{array}{cc}\hfill I_{\lambda }^{\prime }\left(u\right)\left(v\right)& ={\displaystyle \underset{t\to 0}{lim}}\frac{I_{\lambda }(u+tv)-I_{\lambda }\left(u\right)}{t}\hfill \\ \hfill & =a\sum _{k=1}^{N+1}\Delta u_{k-1}\Delta v_{k-1}+\left(b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)\sum _{k=1}^{N+1}\Delta u_{k-1}\Delta v_{k-1}-\lambda \sum _{k=1}^{N}f\left(k,u_k\right)v_k\hfill \\ \hfill & =\left(a+b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)\sum _{k=1}^{N+1}\Delta u_{k-1}\Delta v_{k-1}-\lambda \sum _{k=1}^{N}f\left(k,u_k\right)v_k\hfill \\ \hfill & =-\left(a+b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)\sum _{k=1}^N{\Delta }^2{u}_{k-1}{v}_k-\lambda \sum _{k=1}^{N}f\left(k,u_k\right)v_k\hfill \\ \hfill & =-\sum _{k=1}^N\left[\left(a+b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right){\Delta }^2{u}_{k-1}+\lambda f\left(k,u_k\right)\right]v_k\hfill \end{array}$$

for any $u,v\in T$. Therefore, it is obvious that critical points of $I_{\lambda }$ are solutions to problem (1), which illustrates that looking for solutions to problem (1) is equivalent to finding critical points of $I_{\lambda }$ on T.

Now, for any $u\in T$, define other norms:

$$\begin{array}{cc}\hfill & {\parallel u\parallel }_{\infty }={\displaystyle \underset{k\in [1,N]}{max}}\left|u_k\right|,\hfill \\ \hfill & {\parallel u\parallel }_p={\left(\sum _{k=1}^N{\left|u_k\right|}^p\right)}^{\frac{1}{p}}(1\le p<+\infty ).\hfill \end{array}$$

As we know, norms in the Banach space defined above are equivalent, and we acquire the relation of different norms from the following lemmas.

Lemma 3.

(From [33], Lemma 2.2) For any $u\in T$, the following relation holds:

$$\begin{array}{c}\hfill {\parallel u\parallel }_{\infty }\le \frac{\sqrt{N+1}}{2}\parallel u\parallel .\end{array}$$

Lemma 4.

(From [38], (2.2)) When p = 2, for any $u\in T$, the following relation holds:

$$\begin{array}{c}\hfill \sqrt{{\lambda }_1}{\parallel u\parallel }_2\le \parallel u\parallel \le \sqrt{{\lambda }_N}{\parallel u\parallel }_2,\end{array}$$

where ${\lambda }_1=4{sin}^2\frac{\pi }{2(N+1)}$ and ${\lambda }_N=4{sin}^2\frac{N\pi }{2(N+1)}$.

Lemma 5.

(From [34], (2.2)) When p = 4, for any $u\in T$, the following relation holds:

$$\begin{array}{c}\hfill {\parallel u\parallel }_4^2\le {\parallel u\parallel }_2^2\le \sqrt{N}{\parallel u\parallel }_4^2.\end{array}$$

Eventually, to obtain positive solutions for our problem, we present two important lemmas. The first is the following strong maximum principle.

Lemma 6.

(From [37], Lemma 2) Fix $u\in T$, such that either

$$\begin{array}{c}\hfill u_k>0or-\left(a+b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right){\Delta }^2{u}_{k-1}\ge 0\end{array}$$

for each $k\in [1,N]$. Then, either $u\equiv 0$ or $u_k>0$ for each $k\in [1,N]$.

Now, put

$$\begin{array}{c}\hfill F^{+}(k,\xi )={\int }_0^{\xi }f\left(k,t^{+}\right)dt,(k,\xi )\in [1,N]\times \mathbb{R},\end{array}$$

where $t^{+}$ = $max\{0,t\}$. Next, we define

$$\begin{array}{c}\hfill I_{\lambda }^{+}\left(u\right)=\Phi \left(u\right)-\lambda {\Psi }^{+}\left(u\right)and{\Psi }^{+}\left(u\right)=\sum _{k=1}^N{F}^{+}(k,u_k),\end{array}$$

(5)

where $\Phi$ is given as before. It is apparent that critical points of $I_{\lambda }^{+}$ are solutions to the following problem:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\left(a+b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right){\Delta }^2{u}_{k-1}=\lambda f\left(k,u_k^{+}\right),k\in [1,N],\hfill \\ u_0=u_{N+1}=0.\hfill \end{array}\right.\end{array}$$

(6)

Lemma 7.

(From [37], Lemma 3) If $f(k,0)\ge 0$ for each $k\in [1,N]$, then each non-zero critical point of $I_{\lambda }^{+}$ is a positive solution to problem (1).

3. Main Results

Denote

$$\begin{array}{c}\hfill F_x:=\sum _{k=1}^{N}F(k,x),\forall x\in \mathbb{R}.\end{array}$$

Now, we introduce the first theorem.

Theorem 1.

Assume that there exist two positive constants c and d with

$$\begin{array}{c}\hfill 2c^2[2b{c}^2+a(N+1)]<(a+b{d}^2)d^2{(N+1)}^2\end{array}$$

(7)

such that

(M1) 

$f(k,\xi )\ge 0$ for each $(k,\xi )\in [1,N]\times [-c,c]$;

(M2) 

$\frac{F_d}{(a+b{d}^2)d^2}>\frac{{(N+1)}^2{F}_c}{2c^2[2b{c}^2+a(N+1)]}$;

(M3) 

there is a positive constant ${\beta }_1$ with

$$\begin{array}{c}\hfill {\displaystyle \underset{k\in [1,N]}{max}}{\displaystyle \underset{\left|\xi \right|\to +\infty }{lim\; sup}}\frac{F(k,\xi )}{\left|\xi \right|}={\beta }_1,\end{array}$$

and it satisfies

$$\begin{array}{c}\hfill \frac{\left(a+b{d}^2\right)d^2}{F_d}<\frac{2a+b}{2{\beta }_{1}N\sqrt{N+1}}.\end{array}$$

Furthermore, put

$$\begin{array}{c}\hfill {\rho }_1=\frac{\left(a+b{d}^2\right)d^2}{F_d},\end{array}$$

$$\begin{array}{c}\hfill {\rho }_2=min\left\{\frac{2a+b}{2{\beta }_{1}N\sqrt{N+1}},\frac{2c^2\left[2b{c}^2+a(N+1)\right]}{{(N+1)}^2{F}_c}\right\}.\end{array}$$

Then, for each $\lambda \in \Lambda :=\left({\rho }_1,{\rho }_2\right)$, problem (1) has at least three solutions.

Proof. 

We use Lemma 1 to prove our result. Let $T,\Phi ,\Psi$, and $I_{\lambda }$ be defined as in Section 2. Therefore, we know that $\Phi$ and $\Psi$ are two continuously G$\widehat{a}$teaux differentiable functions, and one has ${\displaystyle \underset{T}{inf}}\Phi =\Phi \left(0\right)=\Psi \left(0\right)=0$.

Now, we have

$$\begin{array}{c}\hfill \Phi \left(u\right)=\frac{a}{2}\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2+\frac{b}{4}{\left(\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)}^2=\frac{a}{2}{\parallel u\parallel }^2+\frac{b}{4}{\parallel u\parallel }^4.\end{array}$$

Owing to constants $a,b>0$, it is easy to understand that when $\parallel u\parallel \to +\infty$, we can acquire $\Phi \left(u\right)\to +\infty$. Hence, we verify the coercivity of $\Phi$.

Put

$$\begin{array}{c}\hfill r=\frac{2c^2[2b{c}^2+a(N+1)]}{{(N+1)}^2}.\end{array}$$

If $\Phi \left(u\right)\le r$, one has $\Phi \left(u\right)=\frac{a}{2}{\parallel u\parallel }^2+\frac{b}{4}{\parallel u\parallel }^4\le r$, which means $\parallel u\parallel \le {\left(\frac{\sqrt{a^2+4br}-a}{b}\right)}^{\frac{1}{2}}$. According to Lemma 3, we can obtain ${\parallel u\parallel }_{\infty }\le \frac{\sqrt{N+1}}{2}\parallel u\parallel \le c$. From condition ($M_1$) of Theorem 1, we have

$$\begin{array}{cc}\hfill \frac{{sup}_{\Phi \left(u\right)\le r}\Psi \left(u\right)}{r}& \le \frac{{sup}_{{\parallel u\parallel }_{\infty }\le c}{\sum }_{k=1}^{N}F(k,u_k)}{r}\hfill \\ \hfill & \le \frac{{(N+1)}^2{\sum }_{k=1}^N{max}_{\left|\xi \right|\le c}F(k,\xi )}{2c^2[2b{c}^2+a(N+1)]}\hfill \\ \hfill & =\frac{{(N+1)}^2{F}_c}{2c^2[2b{c}^2+a(N+1)]}.\hfill \end{array}$$

Owing to $\lambda \in \Lambda$, it is easy to gain

$$\begin{array}{c}\hfill \frac{{sup}_{\Phi \left(u\right)\le r}\Psi \left(u\right)}{r}<\frac{1}{\lambda }.\end{array}$$

(8)

Now, let $\overline{u}\in T$ be given by

$$\begin{array}{c}\hfill {\overline{u}}_k=\left\{\begin{array}{cc}d,\hfill & \mathrm{if}k\in [1,N],\hfill \\ 0,\hfill & \mathrm{if}k=0\mathrm{or}k=N+1.\hfill \end{array}\right.\end{array}$$

From the definition of $\overline{u}$ and (7), we could see that

$$\begin{array}{c}\hfill \Phi \left(\overline{u}\right)=(a+b{d}^2)d^2>\frac{2c^2[2b{c}^2+a(N+1)]}{{(N+1)}^2}=r>0.\end{array}$$

Moreover, we obtain that

$$\begin{array}{c}\hfill \frac{\Psi \left(\overline{u}\right)}{\Phi \left(\overline{u}\right)}=\frac{{\sum }_{k=1}^{N}F(k,{\overline{u}}_k)}{(a+b{d}^2)d^2}=\frac{F_d}{(a+b{d}^2)d^2}.\end{array}$$

Thus, the following holds

$$\begin{array}{c}\hfill \frac{\Psi \left(\overline{u}\right)}{\Phi \left(\overline{u}\right)}>\frac{1}{\lambda }.\end{array}$$

(9)

As a result, due to (8) and (9), condition $\left(i\right)$ of Lemma 1 follows.

Next, we illustrate the coercivity of the functional $I_{\lambda }$. Because of $\lambda <\frac{2a+b}{2{\beta }_{1}N\sqrt{N+1}}$, when we fix $\lambda \in \Lambda$, it is clear that there is a positive constant ${\epsilon }_1$ such that

$$\begin{array}{c}\hfill \lambda <\frac{2a+b}{2({\beta }_1+{\epsilon }_1)N\sqrt{N+1}}.\end{array}$$

(10)

From condition $\left(M_3\right)$ of Theorem 1, we could observe that

$$\begin{array}{c}\hfill {\displaystyle \underset{\left|\xi \right|\to +\infty }{lim\; sup}}\frac{F(k,\xi )}{\left|\xi \right|}\le {\beta }_1<{\beta }_1+{\epsilon }_1.\end{array}$$

Hence, there is a positive constant $h_1$ such that

$$\begin{array}{c}\hfill F(k,\xi )\le ({\beta }_1+{\epsilon }_1)\left|\xi \right|+h_1\end{array}$$

for each $(k,\xi )\in [1,N]\times \mathbb{R}$. Then, by using Lemma 3, we have

$$\begin{array}{cc}\hfill \lambda \sum _{k=1}^{N}F(k,u_k)& \le \lambda \sum _{k=1}^N[({\beta }_1+{\epsilon }_1)|u_k|+h_1]\hfill \\ \hfill & \le \lambda \sum _{k=1}^N[({\beta }_1+{\epsilon }_1){\parallel u\parallel }_{\infty }+h_1]\hfill \\ \hfill & \le \frac{\lambda ({\beta }_1+{\epsilon }_1)N\sqrt{N+1}}{2}\parallel u\parallel +\lambda N{h}_1\hfill \end{array}$$

for each $u\in T$. Thus, we can determine that

$$\begin{array}{cc}\hfill I_{\lambda }\left(u\right)& =\Phi \left(u\right)-\lambda \Psi \left(u\right)\hfill \\ \hfill & =\frac{a}{2}\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2+\frac{b}{4}{\left(\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)}^2-\lambda \sum _{k=1}^{N}F(k,u_k)\hfill \\ \hfill & \ge \frac{a}{2}{\parallel u\parallel }^2+\frac{b}{4}{\parallel u\parallel }^4-\frac{\lambda ({\beta }_1+{\epsilon }_1)N\sqrt{N+1}}{2}\parallel u\parallel -\lambda N{h}_1\hfill \\ \hfill & \ge \frac{a}{2}\parallel u\parallel +\frac{b}{4}\parallel u\parallel -\frac{\lambda ({\beta }_1+{\epsilon }_1)N\sqrt{N+1}}{2}\parallel u\parallel -\lambda N{h}_1\hfill \\ \hfill & =\left(\frac{2a+b}{4}-\frac{\lambda ({\beta }_1+{\epsilon }_1)N\sqrt{N+1}}{2}\right)\parallel u\parallel -\lambda N{h}_1\hfill \end{array}$$

for all $\parallel u\parallel \ge 1$. From (10), it is evident that $\left(\frac{2a+b}{4}-\frac{\lambda ({\beta }_1+{\epsilon }_1)N\sqrt{N+1}}{2}\right)>0$, and when $\parallel u\parallel \to +\infty$, one has $I_{\lambda }\left(u\right)\to +\infty$. All assumptions of Lemma 1 are indeed proven. Therefore, problem (1) has at least three solutions. □

Next, we would like to obtain positive solutions to problem (1), which is the following corollary.

Corollary 1.

Assume that condition $\left(M_1\right)$ of Theorem 1 transforms into

$$\begin{array}{c}\hfill f(k,\xi )>0,(k,\xi )\in [1,N]\times [-c,c],\end{array}$$

(11)

and other conditions of Theorem 1 remain. Then, for each $\lambda \in \Lambda :=\left({\rho }_1,{\rho }_2\right)$, problem (1) has at least three positive solutions.

Proof. 

Let

$$\begin{array}{c}\hfill f^{+}(k,\xi )=\left\{\begin{array}{cc}f(k,\xi ),\hfill & \mathrm{if}\xi >0,\hfill \\ f(k,0),\hfill & \mathrm{if}\xi \le 0,\hfill \end{array}\right.\end{array}$$

(12)

and we firstly discuss problem (6). According to (11), we distinctly know $f(k,0)>0$ for each $k\in [1,N]$, so it is easy to observe that $u=0$ is not a solution to problem (6). Assume that $u=\left\{u_k\right\}$, for $k\in [1,N]$, is a nontrivial solution to problem (6), and one has either $u_k>0$ or

$$\begin{array}{c}\hfill -\left(a+b\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right){\Delta }^2{u}_{k-1}=f(k,0)>0.\end{array}$$

Then, we obtain that u is a positive solution to problem (6) by employing Lemma 6. Moreover, if u is a positive solution to problem (6), it is apparent that u is a positive solution to problem (1), and the result of Corollary 1 follows. □

Next, we introduce another theorem.

Theorem 2.

Assume that there exist two positive constants c and d with

$$\begin{array}{c}\hfill 2c^2[2b{c}^2+a(N+1)]>(a+b{d}^2)d^2{(N+1)}^2\end{array}$$

(13)

such that

(N1) 

$f(k,0)\ge 0$ for each $k\in [1,N]$;

(N2) 

$\frac{F_d}{(a+b{d}^2)d^2}>\frac{{(N+1)}^2{\sum }_{k=1}^N{max}_{\left|\xi \right|\le c}F(k,\xi )}{2c^2[2b{c}^2+a(N+1)]}$;

(N3) 

there is a positive constant ${\beta }_2$ such that

$$\begin{array}{c}\hfill {\displaystyle \underset{k\in [1,N]}{min}}{\displaystyle \underset{\left|\xi \right|\to +\infty }{lim\; inf}}\frac{F(k,\xi )}{{\xi }^4}={\beta }_2,\end{array}$$

and it satisfies

$$\begin{array}{c}\hfill \frac{N{\lambda }_N^2(2a+b)}{4{\beta }_2}<\frac{2c^2\left[2b{c}^2+a(N+1)\right]}{{(N+1)}^2{\sum }_{k=1}^N{max}_{\left|\xi \right|\le c}F(k,\xi )}.\end{array}$$

Furthermore, put

$$\begin{array}{c}\hfill {\tau }_1=max\left\{\frac{\left(a+b{d}^2\right)d^2}{F_d},\frac{N{\lambda }_N^2(2a+b)}{4{\beta }_2}\right\},\end{array}$$

$$\begin{array}{c}\hfill {\tau }_2=\frac{2c^2\left[2b{c}^2+a(N+1)\right]}{{(N+1)}^2{\sum }_{k=1}^N{max}_{\left|\xi \right|\le c}F(k,\xi )}.\end{array}$$

Then, for each $\lambda \in {\Lambda }_1:=\left({\tau }_1,{\tau }_2\right)$, problem (1) admits at least two positive solutions.

Proof. 

Ultimately, our aim is to demonstrate that there are at least two non-zero solutions to problem (1) based on Lemma 2. Now, let $T,\Phi ,\Psi$, and $I_{\lambda }$ be defined as in Section 2, so it is obvious that $\Phi$ and $\Psi$ are two continuously G$\widehat{a}$teaux differentiable functions, and we have ${\displaystyle \underset{T}{inf}}\Phi =\Phi \left(0\right)=\Psi \left(0\right)=0$.

Provided that the definitions of $r_1$ and $\tilde{u}$ are the same as r and $\overline{u}$ in Theorem 1, respectively, then, from condition $\left(N_2\right)$ of Theorem 2 and $\lambda \in {\Lambda }_1$, one has

$$\begin{array}{cc}\hfill \frac{{sup}_{\Phi \left(u\right)\le r_1}\Psi \left(u\right)}{r_1}& \le \frac{{sup}_{{\parallel u\parallel }_{\infty }\le c}{\sum }_{k=1}^{N}F(k,u_k)}{r_1}\hfill \\ \hfill & \le \frac{{(N+1)}^2{\sum }_{k=1}^N{max}_{\left|\xi \right|\le c}F(k,\xi )}{2c^2[2b{c}^2+a(N+1)]}\hfill \\ \hfill & <\frac{1}{\lambda }\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \frac{\Psi \left(\tilde{u}\right)}{\Phi \left(\tilde{u}\right)}=\frac{{\sum }_{k=1}^{N}F(k,{\tilde{u}}_k)}{(a+b{d}^2)d^2}=\frac{F_d}{(a+b{d}^2)d^2}>\frac{1}{\lambda }.\end{array}$$

In addition, from (13), we have

$$\begin{array}{c}\hfill 0<\Phi \left(\tilde{u}\right)=(a+b{d}^2)d^2<\frac{2c^2[2b{c}^2+a(N+1)]}{{(N+1)}^2}=r_1.\end{array}$$

Therefore, (4) of Lemma 2 is true.

Now, we prove that the functional $I_{\lambda }$ is unbounded from below. Owing to $\lambda >\frac{N{\lambda }_N^2(2a+b)}{4{\beta }_2}$, when we fix $\lambda \in {\Lambda }_1$, it is apparent that there is a positive constant ${\epsilon }_2$$({\epsilon }_2<{\beta }_2)$ such that

$$\begin{array}{c}\hfill \lambda >\frac{N{\lambda }_N^2(2a+b)}{4({\beta }_2-{\epsilon }_2)}.\end{array}$$

(14)

According to condition $\left(N_3\right)$ of Theorem 2, we could observe that

$$\begin{array}{c}\hfill {\displaystyle \underset{\left|\xi \right|\to +\infty }{lim\; inf}}\frac{F(k,\xi )}{{\xi }^4}\ge {\beta }_2>{\beta }_2-{\epsilon }_2.\end{array}$$

Consequently, there is a positive constant $h_2$ such that

$$\begin{array}{c}\hfill F(k,\xi )\ge ({\beta }_2-{\epsilon }_2){\xi }^4-h_2\end{array}$$

for each $(k,\xi )\in [1,N]\times \mathbb{R}$. Then, from Lemmas 4 and 5, we have

$$\begin{array}{cc}\hfill \lambda \sum _{k=1}^{N}F(k,u_k)& \ge \lambda \sum _{k=1}^N(({\beta }_2-{\epsilon }_2)u_k^4-h_2)\hfill \\ \hfill & =\lambda ({\beta }_2-{\epsilon }_2){\parallel u\parallel }_4^4-\lambda N{h}_2\hfill \\ \hfill & \ge \frac{\lambda ({\beta }_2-{\epsilon }_2)}{N}{\parallel u\parallel }_2^4-\lambda N{h}_2\hfill \\ \hfill & \ge \frac{\lambda ({\beta }_2-{\epsilon }_2)}{N{\lambda }_N^2}{\parallel u\parallel }^4-\lambda N{h}_2\hfill \end{array}$$

for each $u\in T$. Thus, we can determine that

$$\begin{array}{cc}\hfill I_{\lambda }\left(u\right)& =\Phi \left(u\right)-\lambda \Psi \left(u\right)\hfill \\ \hfill & =\frac{a}{2}\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2+\frac{b}{4}{\left(\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right)}^2-\lambda \sum _{k=1}^{N}F(k,u_k)\hfill \\ \hfill & \le \frac{a}{2}{\parallel u\parallel }^2+\frac{b}{4}{\parallel u\parallel }^4-\frac{\lambda ({\beta }_2-{\epsilon }_2)}{N{\lambda }_N^2}{\parallel u\parallel }^4+\lambda N{h}_2\hfill \\ \hfill & \le \frac{a}{2}{\parallel u\parallel }^4+\frac{b}{4}{\parallel u\parallel }^4-\frac{\lambda ({\beta }_2-{\epsilon }_2)}{N{\lambda }_N^2}{\parallel u\parallel }^4+\lambda N{h}_2\hfill \\ \hfill & =\left(\frac{2a+b}{4}-\frac{\lambda ({\beta }_2-{\epsilon }_2)}{N{\lambda }_N^2}\right){\parallel u\parallel }^4+\lambda N{h}_2\hfill \end{array}$$

for all $\parallel u\parallel \ge 1$. From (14), it is true that $\left(\frac{2a+b}{4}-\frac{\lambda ({\beta }_2-{\epsilon }_2)}{N{\lambda }_N^2}\right)<0$. When $\parallel u\parallel \to +\infty$, we see that $I_{\lambda }\to -\infty$, meaning that $I_{\lambda }$ is unbounded from below. Further, we know that $-I_{\lambda }$ is coercive, which illustrates that the functional $I_{\lambda }$ satisfies the $\left(PS\right)$ condition. All assumptions of Lemma 2 are thus proven, so problem (1) admits at least two non-zero solutions.

Next, we want to acquire positive solutions to problem (1). When the assumption for f is the same as (12), according to the $I_{\lambda }^{+}$ as in (5) and condition $\left(N_1\right)$ of Theorem 2, it follows that each non-zero critical point on T of the functional $I_{\lambda }^{+}$ is a positive solution to problem (1) by using Lemma 7. Hence, problem (1) admits at least two positive solutions. □

Remark 1.

If $f(k,\xi )$ is non-negative for each $(k,\xi )\in [1,N]\times [0,+\infty ]$, we could reduce assumptions of Theorem 2. Clearly, condition $\left(N_1\right)$ follows, and condition $\left(N_2\right)$ becomes

$$\begin{array}{c}\hfill \frac{F_d}{(a+b{d}^2)d^2}>\frac{{(N+1)}^2{F}_c}{2c^2[2b{c}^2+a(N+1)]}.\end{array}$$

A consequence of Theorem 2 is the following corollary.

Corollary 2.

Assume that f is a continuous function such that condition ($N_1$) of Theorem 2 holds, and

$$\begin{array}{c}\hfill {\displaystyle \underset{\xi \to 0^{+}}{lim\; sup}}\frac{F(k,\xi )}{{\xi }^2}=+\infty ,\end{array}$$

(15)

and

$$\begin{array}{c}\hfill {\displaystyle \underset{\xi \to +\infty }{lim\; inf}}\frac{F(k,\xi )}{{\xi }^4}=+\infty ,\end{array}$$

(16)

for all $k\in [1,N]$. Moreover, put

$$\begin{array}{c}\hfill {\lambda }^{*}=\frac{1}{{(N+1)}^2}{\displaystyle \underset{c>0}{sup}}\frac{2c^2[2b{c}^2+a(N+1)]}{{\sum }_{k=1}^N{max}_{\left|\xi \right|\le c}F(k,\xi )},\end{array}$$

then, for each $\lambda \in (0,{\lambda }^{*})$, problem (1) admits at least two positive solutions.

Proof. 

Fix $\lambda \in (0,{\lambda }^{*})$, and there is a positive constant c, such that

$$\begin{array}{c}\hfill \lambda <\frac{1}{{(N+1)}^2}\frac{2c^2[2b{c}^2+a(N+1)]}{{\sum }_{k=1}^N{max}_{\left|\xi \right|\le c}F(k,\xi )}.\end{array}$$

From (15), we could obtain

$$\begin{array}{c}\hfill {\displaystyle \underset{\xi \to 0^{+}}{lim\; sup}}\frac{{\sum }_{k=1}^{N}F(k,\xi )}{{\xi }^2}=+\infty .\end{array}$$

Then, there is a positive constant d, satisfying (13), such that $\frac{F_d}{(a+b{d}^2)d^2}>\frac{1}{\lambda }$. Consequently, by applying Theorem 2, the conclusion follows. □

Remark 2.

If $f(k,\xi )$ is non-negative for each $(k,\xi )\in [1,N]\times [0,+\infty ]$, it is sufficient to conclude that condition (15) of Corollary 2 holds only for at least one $k\in [1,N]$, as the same proof displays. The conclusion of Corollary 2 follows when each

$$\begin{array}{c}\hfill \lambda \in \left(0,\frac{1}{{(N+1)}^2}{\displaystyle \underset{c>0}{sup}}\frac{2c^2[2b{c}^2+a(N+1)]}{F_c}\right).\end{array}$$

Remark 3.

Assume that $f(k,\xi )=f\left(\xi \right)$, for each $k\in [1,N]$, is a continuous function. Then, we could obtain the result of Corollary 2, if f satisfies

$$\begin{array}{c}\hfill {\displaystyle \underset{\xi \to 0^{+}}{lim}}\frac{f\left(\xi \right)}{\xi }=+\infty \end{array}$$

and

$$\begin{array}{c}\hfill {\displaystyle \underset{\xi \to +\infty }{lim}}\frac{f\left(\xi \right)}{{\xi }^3}=+\infty .\end{array}$$

Moreover, if f is a non-negative function, it is enough to acquire at least two positive solutions to problem (1) for each

$$\begin{array}{c}\hfill \lambda \in \left(0,\frac{1}{N{(N+1)}^2}{\displaystyle \underset{c>0}{sup}}\frac{2c^2[2b{c}^2+a(N+1)]}{{\int }_0^{c}f\left(t\right)dt}\right).\end{array}$$

4. Examples

In this section, we show two simple examples to illustrate our conclusions.

Example 1.

We consider problem (1) with $N=9$, and let

$$\begin{array}{c}\hfill f(k,\xi )=f\left(\xi \right)=\left\{\begin{array}{cc}{e}^{\xi },\hfill & \xi \le 4\pi ,\hfill \\ e^{4\pi }cos\xi ,\hfill & \xi >4\pi ,\hfill \end{array}\right.\end{array}$$

for each $k\in [1,9]$. Then, we have

$$\begin{array}{c}\hfill F(k,\xi )=F\left(\xi \right)=\left\{\begin{array}{cc}{e}^{\xi }-1,\hfill & \xi \le 4\pi ,\hfill \\ e^{4\pi }(1+sin\xi )-1,\hfill & \xi >4\pi .\hfill \end{array}\right.\end{array}$$

Let a = b = 1, c = 2, d = 13. Then, condition (7) of Theorem 1 holds, since

$$\begin{array}{c}\hfill 8\times (8+9+1)=144<(1+169)\times 169\times 100=2873000.\end{array}$$

Clearly, $f\left(\xi \right)>0$ holds for each $\xi \in [-2,2]$. In addition, we find that

$$\begin{array}{c}\hfill \frac{F_d}{(a+b{d}^2)d^2}=\frac{N[e^{4\pi }(1+sind)-1]}{(a+b{d}^2)d^2}\approx 127.571,\end{array}$$

and

$$\begin{array}{c}\hfill \frac{{(N+1)}^2{F}_c}{2c^2[2b{c}^2+a(N+1)]}=\frac{N{(N+1)}^2(e^c-1)}{2c^2[2b{c}^2+a(N+1)]}\approx 39.932.\end{array}$$

Therefore, condition $\left(M_2\right)$ of Theorem 1 follows. Moreover, one has

$$\begin{array}{c}\hfill {\beta }_1={\displaystyle \underset{\left|\xi \right|\to +\infty }{lim\; sup}}\frac{F(k,\xi )}{\left|\xi \right|}={\displaystyle \underset{\left|\xi \right|\to +\infty }{lim\; sup}}\frac{e^{4\pi }(1+sin\xi )-1}{\left|\xi \right|}=0,\end{array}$$

and we can consider that

$$\begin{array}{c}\hfill {\displaystyle \underset{{\beta }_1\to 0^{+}}{lim}}\frac{2a+b}{2{\beta }_{1}N\sqrt{N+1}}=+\infty .\end{array}$$

As a result, one has

$$\begin{array}{c}\hfill \frac{(a+b{d}^2)d^2}{F_d}\approx 0.008<\frac{2a+b}{2{\beta }_{1}N\sqrt{N+1}}=+\infty \end{array}$$

and

$$\begin{array}{c}\hfill \frac{(a+b{d}^2)d^2}{F_d}\approx 0.008<\frac{2c^2[2b{c}^2+a(N+1)]}{{(N+1)}^2{F}_c}\approx 0.025.\end{array}$$

Consequently, applying the result of Corollary 1, for each $\lambda \in (0.008,0.025)$, the problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\left(1+\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right){\Delta }^2{u}_{k-1}=\lambda f\left(u_k\right),k\in [1,9],\hfill \\ u_0=u_{10}=0,\hfill \end{array}\right.\end{array}$$

admits at least three positive solutions.

Example 2.

We consider problem (1) with $N=3$, and let

$$\begin{array}{c}\hfill g\left(t\right)=\left\{\begin{array}{cc}0,\hfill & t\le 0,\hfill \\ t^{\frac{1}{2}},\hfill & 0<t\le 1,\hfill \\ t^4,\hfill & t>1.\hfill \end{array}\right.\end{array}$$

Then, it is apparent that g is a continuous and non-negative function, and we have

$$\begin{array}{c}\hfill {\displaystyle \underset{t\to 0^{+}}{lim}}\frac{g\left(t\right)}{t}={\displaystyle \underset{t\to 0^{+}}{lim}}\frac{t^{\frac{1}{2}}}{t}=+\infty \end{array}$$

and

$$\begin{array}{c}\hfill {\displaystyle \underset{t\to +\infty }{lim}}\frac{g\left(t\right)}{t^3}={\displaystyle \underset{t\to +\infty }{lim}}\frac{t^4}{t^3}=+\infty .\end{array}$$

Let a = b = 1, c = 2. Then, for each $\lambda \in \left(0,\frac{1}{N{(N+1)}^2}\frac{2c^2[2b{c}^2+a(N+1)]}{{\int }_0^{c}g\left(t\right)dt}\right)\approx (0,0.291)$, the problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}-\left(1+\sum _{k=1}^{N+1}{\left|\Delta u_{k-1}\right|}^2\right){\Delta }^2{u}_{k-1}=\lambda g\left(u_k\right),k\in [1,3],\hfill \\ u_0=u_4=0\hfill \end{array}\right.\end{array}$$

admits at least two positive solutions by using Remark 3.

5. Conclusions

In this paper, the discrete Dirichlet boundary value problem of the Kirchhoff type is studied by using critical point theory. Based on the behavior of the nonlinear term f in some points between the zero and positive infinity, we search for positive solutions to problem (1). Firstly, we could acquire at least three solutions to problem (1) in Theorem 1. Moreover, by strengthening the condition ($M_1$) of Theorem 1, we easily employ the strong maximum principle to obtain at least three positive solutions to problem (1). In addition, we know that problem (1) has at least two positive solutions according to Theorem 2. It is worth noting that Corollary 2 is a simpler form of Theorem 2. Finally, if we assume f is a non-negative function, the conditions of Theorem 2 and Corollary 2 are simplified.

Although we acquire the existence and multiplicity of positive solutions to problem (1), when the parameter $\lambda$ lies in the determined open intervals, it is not clear for the parameter $\lambda$ lying in other intervals. On the other hand, while the aim of the paper is to study the existence and multiplicity of positive solutions to problem (1), we do not further explore whether positive solutions are stable or optimal. That is a regret and a limitation of this study and is therefore our future work direction.