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# Large Constant-Sign Solutions of Discrete Dirichlet Boundary Value Problems with p-Mean Curvature Operator

by Jianxia Wang 1,2 and Zhan Zhou 1,2,* 1
School of Mathematics and Information Science, Guangzhou University, Guangdong 510006, China
2
Center for Applied Mathematics, Guangzhou University, Guangdong 510006, China
*
Author to whom correspondence should be addressed.
Mathematics 2020, 8(3), 381; https://doi.org/10.3390/math8030381
Received: 3 February 2020 / Revised: 4 March 2020 / Accepted: 5 March 2020 / Published: 9 March 2020

## Abstract

In this paper, we consider the existence of infinitely many large constant-sign solutions for a discrete Dirichlet boundary value problem involving $p$ -mean curvature operator. The methods are based on the critical point theory and truncation techniques. Our results are obtained by requiring appropriate oscillating behaviors of the non-linear term at infinity, without any symmetry assumptions.

## 1. Introduction

Let $Z , N$ and $R$ denote the sets of integer numbers, natural numbers and real numbers, respectively. For $a , b ∈ Z$, define $Z ( a ) = { a , a + 1 , ⋯ }$, and $Z ( a , b ) = { a , a + 1 , ⋯ , b }$ when $a ≤ b$.
Consider the following Dirichlet boundary value problem of the nonlinear difference equation
$( D p λ , f ) − ▵ ϕ p , c ▵ u ( k − 1 ) = λ f ( k , u ( k ) ) , k ∈ Z ( 1 , T ) , u ( 0 ) = u ( T + 1 ) = 0 ,$
where T is a given positive integer, $λ$ is a positive real parameter, is the forward difference operator defined by $▵ u ( k ) = u ( k + 1 ) − u ( k )$, $f ( k , · ) : R → R$ is a continuous function for each $k ∈ Z ( 1 , T )$ and $ϕ p , c ( s ) : = ( 1 + | s | 2 ) p − 2 2 s , p ∈ [ 1 , + ∞ )$. Here, $▵ ϕ p , c ▵ u ( k − 1 )$ may be seen as a discretization of the p-mean curvature operator.
We may think problem $( D p λ , f )$ as being a discrete analog of one-dimensional case of the following problem
$− div ϕ p , c ▿ u = λ f ( x , u ) , x ∈ Ω ⊂ R n , u = 0 , x ∈ ∂ Ω ,$
where $d i v ϕ p , c ▿ u$ is named $p$-mean curvature operator, which is a generalization of mean curvature operator; see [1,2]. If $p = 1$, it reduces to the mean curvature operator. If $p = 2$, it reduces to the Laplacian operator. The above problem arises from differential geometry and physics such as capillarity; see [3,4,5] and references therein. When $p = 1$ and $f ( x , u ) = u$, the above problem describes the free surface of a pendent drop filled with liquid under gravitational field . In the past decades, several authors have discussed the existence and multiplicity of solutions of Problem (1); see [1,6,7,8,9,10,11,12]. For example, Chen and Shen in  have obtained the existence of infinitely many solutions of Problem (1) with $λ = 1$ via a symmetric version of Mountain Pass Theorem. When $p = 1$ and $Ω = ( 0 , 1 )$, Obersnel and Omari in  have established the existence and multiplicity of positive solutions of Problem (1), which depend on the behavior of f at zero or at infinity. G. A. Afrouzi et al. in  have acquired a sequence of nonnegative and nontrivial solutions strongly converging to zero in $C 1 ( [ 0 , 1 ] )$, under suitable oscillating behavior of the nonlinear term f at zero. However, the results on the existence of solutions for problem $( D p λ , f )$ are scarce in the literature besides the case of $p = 1$.
Nonlinear discrete problems appear in many mathematical models, such as computer science, mechanical engineering, control systems, artificial or biological neural networks, economics, fluid mechanics and many others; see [13,14,15,16,17]. Many authors have discussed the existence and multiplicity of solutions for difference equations through classical tools of nonlinear analysis: Fixed point theorems, upper and lower solutions techniques; see [7,9] and the references given therein. Since 2003, by starting from the seminal paper , variational methods have been used to investigate nonlinear difference equations, which have obtained various results; see [19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34].
In paper , the authors have considered problem $( D 1 λ , f )$, obtaining infinitely many positive solutions when $λ$ belongs to a precise real interval. It is worth noticing that the suitable oscillating behaviors of the nonlinear term f at infinity play a key role. Inspired by [19,32,35,36,37,38,39,40], the main purpose of this paper is to investigate the existence conditions of infinitely many constant-sign solutions for problem $( D p λ , f )$, without any symmetry hypothesis. Here, a solution ${ u ( k ) }$ of $( D p λ , f )$ is called a constant-sign solution, if $u ( k ) > 0$ for all $k ∈ Z ( 1 , T )$ or $u ( k ) < 0$ for all $k ∈ Z ( 1 , T )$. Compared to problem $( D 1 λ , f )$, problem $( D p λ , f )$ is more difficult to handle. To facilitate the analysis, we have to divide the problem into two categories: $1 ≤ p < 2$ and $2 ≤ p < + ∞$. We believe that this is the first time to discuss the existence of infinitely many solutions for a non-linear second order difference equation with $p$-mean curvature operator.
A special case of our results is the following.
Theorem 1.
Let $g : R → R$ be a continuous function such that $g ( t ) t ≥ 0$ for $t ≠ 0$. Assume that
$lim inf t → ∞ ∫ 0 t g ( τ ) d τ | t | p = 0 , and lim sup t → ∞ ∫ 0 t g ( τ ) d τ | t | p = + ∞ .$
Then, for every $λ > 0$, the problem
$− ▵ ϕ p , c ▵ u ( k − 1 ) = λ g ( u ( k ) ) , k ∈ Z ( 1 , T ) , u ( 0 ) = u ( T + 1 ) = 0 ,$
admits two unbounded sequences of constant-sign solutions (one positive and one negative).
This paper is organized as follows. In Section 2, we introduce the the suitable Banach space and appropriate functional corresponding to problem $( D p λ , f )$. To obtain sequences of constant-sign solutions of problem $( D p λ , f )$, three basic lemmas are introduced. In Section 3, under suitable hypotheses on f, we obtain the existence of infinitely many constant-sign solutions for problem $( D p λ , f )$. In Section 4, we give two examples to demonstrate our results. Finally, conclusions are given for this paper.

## 2. Mathematical Background

To solve problem $( D p λ , f )$, we naturally select the T-dimensional Banach space
$X = { u : Z ( 0 , T + 1 ) → R : u ( 0 ) = u ( T + 1 ) = 0 } ,$
endowed with the norm
$∥ u ∥ : = ∑ k = 1 T ( ▵ u ( k ) ) 2 1 2 f o r a l l u ∈ X .$
Another useful norm on X is
$| | u | | ∞ : = max k ∈ Z ( 1 , T ) | u ( k ) | for all u ∈ X .$
In the sequel, we will use the following inequalities.
For $0 < r < s , x k ≥ 0 , k ∈ Z ( 1 , n )$, one has
$∑ k = 1 n x k s 1 / s ≤ ∑ k = 1 n x k r 1 / r ,$
see .
$| | u | | ∞ ≤ T + 1 2 ∥ u ∥ ,$
for every $u ∈ X$, it can follow from Lemma 2.2 of .
For all $u ∈ X$, let
$Φ ( u ) : = 1 p ∑ k = 0 T 1 + ( ▵ u ( k ) ) 2 p 2 − 1 , a n d Ψ ( u ) : = ∑ k = 1 T F ( k , u ( k ) ) ,$
where $F ( k , t ) : = ∫ 0 t f ( k , τ ) d τ$ for every $t ∈ R$ and $k ∈ Z ( 1 , T )$. Further, let us denote $I λ ( u ) : = Φ ( u ) − λ Ψ ( u )$ for $u ∈ X .$ Through standard arguments, we follow that $I λ ∈ C 1 ( S , R )$, and the critical points of $I λ$ are exactly the solutions of problem $( D p λ , f )$. In fact, one has
$I λ ′ ( u ) ( v ) = ∑ k = 0 T ϕ p , c ( ▵ u ( k ) ▵ v ( k ) − λ ∑ k = 1 T f ( k , u ( k ) ) v ( k ) = ∑ k = 0 T ϕ p , c ( ▵ u ( k ) v ( k + 1 ) − ∑ k = 0 T ϕ p , c ( ▵ u ( k ) v ( k ) − λ ∑ k = 1 T f ( k , u ( k ) ) v ( k ) = ∑ k = 1 T ϕ p , c ( ▵ u ( k − 1 ) v ( k ) − ∑ k = 1 T ϕ p , c ( ▵ u ( k ) v ( k ) − λ ∑ k = 1 T f ( k , u ( k ) ) v ( k ) = − ∑ k = 1 T [ ▵ ( ϕ p , c ( ▵ u ( k ) − λ f ( k , u ( k ) ) ] v ( k ) ,$
for all $u , v ∈ X .$
Next, we need to establish the following strong maximum principle to obtain the positive solutions of problem $( D p λ , f )$, i.e., $u ( k ) > 0$ for each $k ∈ Z ( 1 , T )$.
Lemma 1.
Assume $u ∈ X$ such that either
$u ( k ) > 0 o r − ▵ φ p , c ( ▵ u ( k − 1 ) ≥ 0 ,$
for any $k ∈ Z ( 1 , T )$. Then, either $u > 0$ in $Z ( 1 , T )$ or $u ≡ 0$.
Proof. For $u ∈ X$, put $m = min { u ( k ) , k ∈ Z ( 0 , T + 1 ) }$, then $m ≤ 0$.
If there exists $j ∈ Z ( 1 , T )$ such that $u ( j ) = m$, we claim that $u ≡ 0$. Indeed, since $Δ u ( j − 1 ) = u ( j ) − u ( j − 1 ) ≤ 0$ and $Δ u ( j ) = u ( j + 1 ) − u ( j ) ≥ 0$, $φ p , c ( s )$ is strictly monotone increasing in s, and $φ p , c ( 0 ) = 0$, we have
$φ p , c ( ▵ u ( j ) ) ≥ 0 ≥ φ p , c ( ▵ u ( j − 1 ) ) .$
On the other hand, by (5), let $k = j$, we obtain
$φ p , c ( ▵ u ( j ) ) ≤ φ p , c ( ▵ u ( j − 1 ) ) .$
Combining inequalities (6) and (7), we get that $φ p , c ( ▵ u ( j ) ) = 0 = φ p , c ( ▵ u ( j − 1 ) )$. That is $u ( j + 1 ) = u ( j − 1 ) = u ( j ) = m$. By iterating this argument, we obtain easily $u ( 0 ) = u ( 1 ) = u ( 2 ) = … = u ( T ) = u ( T + 1 )$. Thus $u ≡ 0$.
If $u ( j ) > m$ for every $j ∈ Z ( 1 , T )$, then $u ( 0 ) = u ( T + 1 ) = m = 0$. It follows that $u ( j ) > 0$, for all $j ∈ Z ( 1 , T )$. The proof is complete.
In the same way, we have the following result to get negative solutions problem $( D p λ , f )$, i.e., $u ( k ) < 0$ for each $k ∈ Z ( 1 , T )$.
Lemma 2.
Assume $u ∈ X$ such that either
$u ( k ) < 0 o r − ▵ φ p , c ( ▵ u ( k − 1 ) ≤ 0 ,$
for any $k ∈ Z ( 1 , T )$. Then, either $u < 0$ in $Z ( 1 , T )$ or $u ≡ 0$.
Truncation techniques are usually used to discuss the existence of constant-sign solutions. To the end, we introduce the following truncations of the functions $f ( k , t )$ for every $k ∈ Z ( 1 , T ) .$
If $f ( k , 0 ) ≥ 0$ for each $k ∈ Z ( 1 , T )$. Set
$f + ( k , t ) : = f ( k , t ) , if t ≥ 0 , f ( k , 0 ) , if t < 0 .$
Clearly, $f + ( k , · )$ is also continuous, for every $k ∈ Z ( 1 , T )$. By Lemma 1, all solutions of problem $( D p λ , f + )$ are also solutions of problem $( D p λ , f )$. Therefore, when problem $( D p λ , f + )$ has non-zero solutions, then problem $( D p λ , f )$ possesses positive solutions.
If $f ( k , 0 ) ≤ 0$ for each $k ∈ Z ( 1 , T )$. Set
$f − ( k , t ) : = f ( k , 0 ) , if t > 0 , f ( k , t ) , if t ≤ 0 .$
When problem $( D p λ , f − )$ has non-zero solutions, then problem $( D p λ , f )$ possesses negative solutions.
Here, we introduce a lemma (Theorem 4.3 of ) which is the main tool used to research problem $( D p λ , f ) .$
Lemma 3.
Let X be a finite dimensional Banach space and let $I λ : X → R$ be a function satisfying the following structure hypothesis:
(H)$I λ ( u ) : = Φ ( u ) − λ Ψ ( u )$ for all $u ∈ X$, where $Φ , Ψ : X → R$ be two continuously G$a ^$teux differentiable functions with Φ coercive, i.e., $lim ∥ u ∥ → + ∞ Φ ( u ) = + ∞$, and such that $inf X Φ = Φ ( 0 ) = Ψ ( 0 ) = 0 .$
For all $r > 0$, put
$φ ( r ) : = sup Φ − 1 [ 0 , r ] Ψ r , a n d φ ∞ : = lim inf r → + ∞ φ ( r ) .$
Assume that $φ ∞ < + ∞$ and for each $λ ∈ ( 0 , 1 φ ∞ )$$I λ$ is unbounded from below. Then, there is a sequence ${ u n }$ of critical points $( l o c a l m i n i m a )$ of $I λ$ such that $lim n → + ∞ Φ ( u n ) = + ∞$.

## 3. Main Results

In the following, we will discuss the existence of constant-sign solutions of problem $( D p λ , f )$. Our purpose is to apply Lemma 3 to the function $I λ ± : X → R ,$$I λ ± ( u ) : = Φ ( u ) − λ Ψ ± ( u ) ,$ where $Ψ ± ( u ) = ∑ k = 1 T F ± ( k , u ( k ) )$ and $F ± ( k , t ) : = ∫ 0 t f ± ( k , τ ) d τ$ for every $k ∈ Z ( 1 , T )$ and then exploit Lemma 1 or Lemma 2 to get our results.
Let
$A ± ∞ : = lim inf t → + ∞ ∑ k = 1 T max 0 ≤ s ≤ t F ( k , ± s ) t p , and B ± ∞ : = lim sup t → ± ∞ ∑ k = 1 T F ( k , t ) | t | p .$
Considering the functional $I λ +$, we have the following conclusions.
Theorem 2.
Let $1 ≤ p < 2$ and $f ( k , · ) : R → R$ to be a continuous function with $f ( k , 0 ) ≥ 0$ for each $k ∈ Z ( 1 , T )$. Assume that
$( i 1 ) A + ∞ < 2 p − 1 ( T + 1 ) p 2 B + ∞ .$
Then, for each $λ ∈ ( 2 p B + ∞ , 2 p p ( T + 1 ) p 2 A + ∞ )$, problem $( D p λ , f )$ has an unbounded sequence of positive solutions.
Proof. Consider the auxiliary problem
$( D p λ , f + ) − ▵ ϕ p , c ▵ u ( k − 1 ) = λ f + ( k , u ( k ) ) , k ∈ Z ( 1 , T ) , u ( 0 ) = u ( T + 1 ) = 0 .$
Obviously $Φ$ and $Ψ +$ satisfy hypothesis required in Lemma 3. For $t > 0$, set
$r = 1 p 4 t 2 T + 1 + ( T + 1 ) 2 p − 2 p − ( T + 1 ) p − 1 p p .$
Assume $u ∈ X$ and
$Φ ( u ) = 1 p ∑ k = 0 T 1 + ( ▵ u ( k ) ) 2 p 2 − 1 ≤ r .$
Put $v ( k ) = 1 + ( ▵ u ( k ) ) 2 p 2 − 1$, for every $k ∈ Z ( 0 , T )$, then $∑ k = 0 T v ( k ) ≤ p r$.
By (2) and Hölder inequality as well, we have
$∑ k = 0 T ( ▵ u ( k ) ) 2 = ∑ k = 0 T ( 1 + v ( k ) ) 1 p 2 − 1 ≤ ∑ k = 0 T v ( k ) 2 p + 2 ( T + 1 ) p − 1 p ∑ k = 0 T v ( k ) 1 p ≤ ( p r ) 2 p + 2 ( T + 1 ) p − 1 p ( p r ) 1 p = 4 t 2 T + 1 .$
Owing to (3), it follows
$| | u | | ∞ ≤ T + 1 2 ∑ k = 0 T ( ▵ u ( k ) ) 2 1 2 ≤ t .$
Thus, one has $Φ − 1 [ 0 , r ] ⊆ { u ∈ X : | | u | | ∞ ≤ t }$
By the definition of $φ$, we obtain
$φ ( r ) = sup Φ − 1 [ 0 , r ] Ψ + r ≤ sup | | u | | ∞ ≤ t ∑ k = 0 T F + ( k , u ( k ) ) r ≤ p ∑ k = 1 T max 0 ≤ s ≤ t F ( k , s ) 4 t 2 T + 1 + ( T + 1 ) 2 p − 2 p − ( T + 1 ) p − 1 p p .$
Bearing in mind condition$( i 1 )$, we follow that $φ ∞ ≤ p ( T + 1 ) p 2 2 p A + ∞ < + ∞ .$
In the next step, we need to prove that $I λ +$ is unbounded from below. To this end, we consider two cases: $B + ∞ = + ∞$ and $B + ∞ < + ∞$. If $B + ∞ = + ∞$, let ${ c n }$ be a sequence of positive numbers, with $lim n → + ∞ c n = + ∞$, such that
$∑ k = 1 T F + ( k , c n ) = ∑ k = 1 T F ( k , c n ) ≥ ( 2 + p ) λ p c n p , f o r e v e r y n ∈ N .$
In the following, we take in X the sequence ${ ω n }$ defined by putting $ω n ( k ) = c n$, for $k ∈ Z ( 1 , T )$.
Using again (2), one has
$I λ + ( ω n ) = 2 p 1 + c n 2 p 2 − 1 − λ ∑ k = 1 T F + ( k , c n ) ≤ 2 p c n p − 2 + p p c n p = − c n p ,$
which implies that $lim n → + ∞ I λ + ( ω n ) = − ∞$. If $B + ∞ < + ∞$, since $λ > 2 p B + ∞$, we may take $ϵ 0 > 0$ such that $2 p − λ B + ∞ + λ ϵ 0 < 0$. Then there exists a sequence of positive numbers ${ c n }$ such that $lim n → + ∞ c n = + ∞$ and
$( B + ∞ − ϵ 0 ) c n p ≤ ∑ k = 1 T F + ( k , c n ) = ∑ k = 1 T F ( k , c n ) ≤ ( B + ∞ + ϵ 0 ) c n p .$
Arguing as before and by choosing ${ ω n }$ in X as above, we have
$I λ + ( ω n ) = 2 p 1 + c n 2 p 2 − 1 − λ ∑ k = 1 T F + ( k , c n ) ≤ 2 p c n p − λ ( B + ∞ − ϵ 0 ) c n p = ( 2 p − λ B + ∞ + λ ϵ 0 ) c n p .$
Since $2 p − λ B + ∞ + λ ϵ 0 < 0$, it is clear that $lim n → + ∞ I λ + ( ω n ) = − ∞$. Considering the above two cases, we follow that $I λ +$ is unbounded from below.
According to Lemma 3, there exist a sequence ${ u n }$ of critical points (local minima) of $I λ +$ such that $lim n → + ∞ Φ ( u n ) = + ∞$. Hence, for every $n ∈ N$, $u n$ is a non-zero solution of problem $( D p λ , f + )$, by Lemma 1, $u n$ is a positive solution of problem $( D p λ , f )$. Since $Φ$ is bounded on bounded sets and $lim n → + ∞ Φ ( u n ) = + ∞$, ${ u n }$ must be unbounded. So Theorem 2 holds and the proof is complete.
Theorem 3.
Let $2 ≤ p < + ∞$ and $f ( k , · ) : R → R$ to be a continuous function with $f ( k , 0 ) ≥ 0$ for each $k ∈ Z ( 1 , T )$. Assume that
$( i 2 ) A + ∞ < ( 2 ) p ( T + 1 ) p − 1 B + ∞ .$
Then, for each $λ ∈ ( ( 2 ) p p B + ∞ , 2 p p ( T + 1 ) p − 1 A + ∞ )$, problem $( D p λ , f )$ has an unbounded sequence of positive solutions.
Proof. We sketch only the differences with the proof of Theorem 2. For $t > 0$, make
$r = ( 2 t ) p p ( T + 1 ) p − 1 .$
Assume $u ∈ X$ and
$Φ ( u ) = 1 p ∑ k = 0 T 1 + ( ▵ u ( k ) ) 2 p 2 − 1 ≤ r .$
Denote $v ( k ) = 1 + ( ▵ u ( k ) ) 2 p 2 − 1$, for every $k ∈ Z ( 0 , T )$, then $∑ k = 0 T v ( k ) ≤ p r$.
Noting the inequality $( x + y ) θ ≤ x θ + y θ$, for $0 < θ ≤ 1 , x ≥ 0 , y ≥ 0$ and Hölder inequality, one has
$∑ k = 0 T ( ▵ u ( k ) ) 2 = ∑ k = 0 T 1 + v ( k ) 2 p − 1 ) ≤ ∑ k = 0 T ( v ( k ) ) 2 p ≤ ( T + 1 ) p − 2 p ∑ k = 0 T v ( k ) 2 p ≤ ( T + 1 ) p − 2 p ( p r ) 2 p = 4 t 2 T + 1 .$
Applying (3), we have
$| | u | | ∞ ≤ T + 1 2 ∑ k = 0 T ( ▵ u k ) 2 1 2 ≤ t .$
By the definition of $φ$, we have
$φ ( r ) = sup Φ − 1 [ 0 , r ] Ψ + r ≤ sup | | u | | ∞ ≤ t ∑ k = 0 T F + ( k , u ( k ) ) r ≤ p ( T + 1 ) p − 1 ∑ k = 1 T max 0 ≤ s ≤ t F ( k , s ) 2 p t p .$
Using condition$( i 2 )$, $φ ∞ ≤ p ( T + 1 ) p − 1 2 p A + ∞ < + ∞$ holds.
Now, we verify that $I λ +$ is unbounded form blow. Fist, assume that $B + ∞ = + ∞$. Let ${ c n }$ be a sequence of positive numbers, with $lim n → + ∞ c n = + ∞$, such that
$∑ k = 1 T F + ( k , c n ) = ∑ k = 1 T F ( k , c n ) ≥ ( 2 ) p + p λ p c n p , f o r n ∈ N .$
Picking the sequence ${ ω n }$ in X by $ω n ( k ) = c n ,$ for $k ∈ Z ( 1 , T ) .$ Exploiting the inequality $( x + y ) θ ≤ 2 θ − 1 ( x θ + y θ )$ for $θ ≥ 1 , x ≥ 0 , y ≥ 0$ , we get
$I λ + ( ω n ) = 2 p 1 + c n 2 p 2 − 1 − λ ∑ k = 1 T F + ( k , c n ) ≤ ( 2 ) p p c n p + ( 2 ) p − 2 p − ( 2 ) p + p p c n p = − c n p + ( 2 ) p − 2 p ,$
which implies that $lim n → + ∞ I λ ( ω n ) = − ∞$.
Next, assume that $B + ∞ < + ∞$. Since $λ > ( 2 ) p p B + ∞$, we may take $ϵ 0 > 0$ such that $( 2 ) p p − λ B ∞ + λ ϵ 0 < 0$. Then there exists a sequence of positive numbers ${ c n }$ such that $lim n → + ∞ c n = + ∞$ and
$( B + ∞ − ϵ 0 ) c n p ≤ ∑ k = 1 T F + ( k , c n ) = ∑ k = 1 T F ( k , c n ) ≤ ( B + ∞ + ϵ 0 ) c n p .$
Define the sequence ${ ω n }$ in S as above, we obtain
$I λ + ( ω n ) = 2 p 1 + c n 2 p 2 − 1 − λ ∑ k = 1 T F + ( k , c n ) ≤ ( 2 ) p p c n p + ( 2 ) p − 2 p − λ ( B + ∞ − ϵ 0 ) c n p = ( ( 2 ) p p − λ B + ∞ + λ ϵ 0 ) c n p + ( 2 ) p − 2 p .$
Since $( 2 ) p p − λ B + ∞ + λ ϵ 0 < 0$, it is obvious that $lim n → + ∞ I λ + ( ω n ) = − ∞$.
Thus, we follow that $I λ +$ is unbounded from below. According to Lemmas 1 and 3, we have finished the proof of the theorem.
Similarly, considering the functional $I λ −$, we can achieve the following results.
Theorem 4.
Let $1 ≤ p < 2$ and $f ( k , · ) : R → R$ to be a continuous function with $f ( k , 0 ) ≤ 0$ for each $k ∈ Z ( 1 , T )$. Assume that
$( i 3 ) A − ∞ < 2 p − 1 ( T + 1 ) p 2 B − ∞ .$
Then, for each $λ ∈ ( 2 p B − ∞ , 2 p p ( T + 1 ) p 2 A − ∞ )$, problem $( D p λ , f )$ has an unbounded sequence of negative solutions.
Theorem 5.
Let $2 ≤ p < + ∞$ and $f ( k , · ) : R → R$ to be a continuous function with $f ( k , 0 ) ≤ 0$ for each $k ∈ Z ( 1 , T )$. Assume that
$( i 4 ) A − ∞ < ( 2 ) p ( T + 1 ) p − 1 B − ∞ .$
Then, for each $λ ∈ ( ( 2 ) p p B − ∞ , 2 p p ( T + 1 ) p − 1 A − ∞ )$, problem $( D p λ , f )$ has an unbounded sequence of negative solutions.
Combining Theorems 2 and 4, we have the following corollary.
Corollary 1.
Let $1 ≤ p < 2$ and $f ( k , · ) : R → R$ to be a continuous function with $f ( k , 0 ) = 0$ for each $k ∈ Z ( 1 , T )$. Assume that
$( i 5 ) max { A + ∞ , A − ∞ } < 2 p − 1 ( T + 1 ) p 2 min { B + ∞ , B − ∞ } .$
Then, for each $λ ∈ ( 2 p min { B + ∞ , B − ∞ } , 2 p p ( T + 1 ) p 2 max { A + ∞ , A − ∞ } )$, problem $( D p λ , f )$ admits two unbounded sequences of constant-sign solutions ( one positive and one negative ).
Similarly, combining Theorems 3 and 5, we have the following corollary.
Corollary 2.
Let $2 ≤ p < + ∞$ and $f ( k , · ) : R → R$ to be a continuous function with $f ( k , 0 ) = 0$ for each $k ∈ Z ( 1 , T )$. Assume that
$( i 6 ) max { A + ∞ , A − ∞ } < ( 2 ) p ( T + 1 ) p − 1 min { B + ∞ , B − ∞ } .$
Then, for each $λ ∈ ( ( 2 ) p p min { B + ∞ , B − ∞ } , 2 p p ( T + 1 ) p − 1 max { A + ∞ , A − ∞ } )$, problem $( D p λ , f )$ admits admits two unbounded sequences of constant-sign solutions ( one positive and one negative ).
Remark 1.
If we let $p → 2 −$ in Theorem 2, we find that the conditions and consequence of Theorem 2 is the same as those of Theorem 3 for $p = 2$. Moreover the results are consistent with results in . For the special case, $p = 1$, Theorem 2 reduces to Corollary 2.1 of .
Remark 2.
We note that, if for each $k ∈ Z ( 1 , T ) , f ( k , · ) : R → R$ is a continuous function satisfying $f ( k , t ) t ≥ 0$ for all $t ∈ R ∖ { 0 }$, then
$A + ∞ = lim inf t → + ∞ ∑ k = 1 T F ( k , t ) t p , a n d A − ∞ = lim inf t → − ∞ ∑ k = 1 T F ( k , t ) | t | p .$
Consequently, Theorem 1 immediately follows by Corollaries 1 and 2.

## 4. Two Examples

Example 1.
For $1 ≤ p < 2$, we consider the boundary value problem $( D p λ , f )$ with
$f ( k , t ) = p | t | p − 1 sign ( t ) T + 1 T + sin 1 2 T ln ( | t | p + 1 ) + 1 2 T cos 1 2 T ln ( | t | p + 1 ) ,$
for $k ∈ Z ( 1 , T )$, then
$F ( k , t ) = ∫ 0 t f ( k , τ ) d τ = T + 1 T | t | p + ( | t | p + 1 ) sin 1 2 T ln ( | t | p + 1 ) , f o r t ∈ R .$
Since $f ( k , t ) ≥ p t p − 1 T + 1 T − 1 − 1 2 T = p 2 T t p − 1 > 0$, for $t > 0$ and $f ( k , 0 ) = 0$, we follow that for each fixed $k ∈ Z ( 1 , T )$, $F ( k , t )$ is strictly monotone increasing on $[ 0 , + ∞ )$. One has $max 0 ≤ s ≤ t F ( k , s ) = F ( k , t )$, for each $t ≥ 0$. Clearly,
$A + ∞ = lim inf t → + ∞ T F ( k , t ) t p = lim inf t → + ∞ ( T + 1 ) t p + T ( t p + 1 ) sin ( 1 2 T ln ( t p + 1 ) ) t p = 1 ,$
and
$B + ∞ = lim sup t → + ∞ T F ( k , t ) t p = lim sup t → + ∞ ( T + 1 ) t p + T ( t p + 1 ) sin ( 1 2 T ln ( t p + 1 ) ) t p = 2 T + 1 .$
In view of $1 ≤ p < 2$ , we follow that $A + ∞ < 2 p − 1 ( T + 1 ) p 2 B + ∞$. Applying to Theorem 2, problem $( D p λ , f )$ admits an unbounded sequence of positive solutions.
Let us consider another example.
Example 2.
Let $T = 4 , p = 3$ andf be a function defined as follows
$f ( k , t ) = 3 | t | t 5 4 + sin ( 1 8 ln ( | t | 3 + 1 ) ) + 1 8 cos ( 1 8 ln ( | t | 3 + 1 ) ) , k ∈ Z ( 1 , 4 )$
Then, for every $λ ∈ ( 2 2 27 , 8 75 )$, the problem
$− ▵ ϕ 3 , c ▵ u ( k − 1 ) = λ f ( k , u ( k ) ) , k ∈ Z ( 1 , 4 ) , u ( 0 ) = u ( 5 ) = 0 ,$
Admits an unbounded sequence of positive solutions and an unbounded sequence of negative solutions. Indeed, $f ( k , t ) ≥ 3 t 2 5 4 − 1 − 1 8 = 3 8 t 2 > 0$, for $t > 0$ and $f ( k , 0 ) = 0 .$
$F ( k , t ) = ∫ 0 t f ( k , τ ) d τ = 5 4 | t | 3 + ( | t | 3 + 1 ) sin ( 1 8 ln ( | t | 3 + 1 ) ) , f o r t ∈ R .$
Since $f ( k , t ) ≥ 3 t 2 5 4 − 1 − 1 8 = 3 8 t 2 > 0$, for $t > 0$, we follow that for each fixed $k ∈ Z ( 1 , 4 )$, $F ( k , t )$ is strictly monotone increasing on $[ 0 , + ∞ )$. Thus, $max | s | ≤ t F ( k , s ) = F ( k , t )$, for each $t ≥ 0$. Obviously,
$A ± ∞ = lim inf t → + ∞ 4 F ( k , t ) t 3 = lim inf t → + ∞ 5 t 3 + 4 ( t 3 + 1 ) sin ( 1 8 ln ( t 3 + 1 ) ) t 3 = 1 ,$
and
$B ± ∞ = lim sup t → + ∞ 4 F ( k , t ) t 3 = lim sup t → + ∞ 5 t 3 + 4 ( t 3 + 1 ) sin ( 1 8 ln ( t 3 + 1 ) ) t 3 = 9 .$
Through simple computation, $max { A + ∞ , A − ∞ } < ( 2 ) p ( T + 1 ) p − 1 min { B + ∞ , B − ∞ }$ holds. Corollary 2 ensures our claim.

## 5. Conclusions

In this paper, we have discussed the Dirichlet boundary value problem of the difference equation with p-mean curvature operator. Some sufficient conditions are derived for the existence of sequences of constant-sign solutions to the problem. Two examples are given to show the effectiveness of our results.
To solve problem $( D p λ , f )$, we further develop the methods adopted in . The approaches can be used for the boundary value problems of differential equations involving p-mean curvature operator. Therefore, our work has both theoretical and practical significance.

## Author Contributions

Conceptualization, J.W.; Formal analysis, J.W. and Z.Z.; Funding acquisition, Z.Z.; Investigation, Z.Z.; Methodology, J.W.; Supervision, Z.Z.; Writing–original draft, J.W.; Writing–review and editing, Z.Z. All authors have read and agreed to the published version of the manuscript.

## Funding

This work is supported by the National Natural Science Foundation of China (Grant No. 11971126) and the Program for Changjiang Scholars and Innovative Research Team in University (Grant No. IRT$−$16R16).

## Acknowledgments

The authors wish to thank three anonymous reviewers for their valuable comments and suggestions.

## Conflicts of Interest

The authors declare that they have no conflicts of interest.

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