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# Existence of positive solutions for boundary value problems of p-Laplacian difference equations

## Abstract

In this paper, by using the Avery-Peterson fixed point theorem, we investigate the existence of at least three positive solutions for a third order p-Laplacian difference equation. An example is given to illustrate our main results.

MSC:34B10, 34B15, 34J10.

## 1 Introduction

The aim of this paper is to study the existence of positive solutions for a third order p-Laplacian difference equation

${ Δ [ ϕ p ( Δ 2 y ( n ) ) ] + q ( n ) f ( n , y ( n ) , Δ y ( n ) ) = 0 , n ∈ [ 0 , N ] , a y ( 0 ) − b Δ y ( 0 ) = 0 , c y ( N + 3 ) + d Δ y ( N + 2 ) = 0 , Δ 2 y ( 0 ) = 0 ,$
(1.1)

where

• $N>1$ an integer;

• $a,c>0$, and $b,d≥0$ with $ad+ac(N+3)+bc>0$;

• f and q are continuous and positive;

• $ϕ p$ is called p-Laplacian, $ϕ p (x)= | x | p − 2 x$ with $p>1$, its inverse function is denoted by $ϕ q (x)$ with $ϕ q (x)= | x | q − 2 x$ with $1/p+1/q=1$;

• $∑ i = r s x(i)=0$ if $r,s∈Z$ and $s, where is the integer set, denote $[r,s]={r,r+1,…,s}$ for $r,s∈Z$ with $r≤s$.

Difference equations, the discrete analog of differential equations, have been widely used in many fields such as computer science, economics, neural network, ecology, cybernetics, etc. . In the past decade, the existence of positive solutions for the boundary value problems (BVPs) of the difference equations has been extensively studied; to mention a few references, see  and the references therein. Also there has been much interest shown in obtaining the existence of positive solutions for the third order p-Laplacian dynamic equations on time scales. To mention a few papers along these lines, see .

We now discuss briefly several of the appropriate papers on the topic.

Liu  studied the following second order p-Laplacian difference equation with multi-point boundary conditions:

${ Δ [ ϕ ( Δ x ( n ) ) ] + f ( n , x ( n + 1 ) , Δ x ( n ) , Δ x ( n + 1 ) ) = 0 , n ∈ [ 0 , N ] , x ( 0 ) − ∑ i = 1 m α i x ( η i ) = A , x ( N + 2 ) − β i x ( η i ) = B .$

The sufficient conditions to guarantee the existence of at least three positive solutions of the above multi-point boundary value problem were established by using a new fixed point theorem obtained in .

Liu  studied the following boundary value problem:

${ Δ [ ϕ ( Δ x ( n ) ) ] + f ( n , x ( n + 1 ) , Δ x ( n ) , Δ x ( n + 1 ) ) = 0 , n ∈ [ 0 , N ] , x ( 0 ) − ∑ i = 1 m α i x ( η i ) = 0 , x ( N + 2 ) − β i x ( η i ) = B .$

By using the five functionals fixed point theorem , Liu obtained the existence criteria of at least three positive solutions.

Therefore, in this paper, we will consider the existence of at least three positive solutions for the third order p-Laplacian difference equation (1.1) by using the Avery-Peterson fixed point theorem .

Throughout this paper we assume that the following condition holds:

(C1) $f:[0,N]×[0,+∞)×R→(0,+∞)$ and $q:[0,N]→(0,+∞)$ are continuous.

This paper is organized as follows. In Section 2, we give some preliminary lemmas which are key tools for our proof. The main result is given in Section 3. Finally, in Section 4, we give an example to demonstrate our result.

## 2 Preliminaries

In this section we present some lemmas, which will be needed in the proof of the main result.

Let γ and θ be nonnegative continuous convex functionals on $P$, α be a nonnegative continuous concave functional on $P$ and ψ be a nonnegative continuous functional on $P$. Then for positive real numbers t, v, w, and z, we define the following convex sets of $P$:

$P ( γ , z ) = { y ∈ P : γ ( y ) < z } , P ( α , v ; γ , z ) = { y ∈ P : v ≤ α ( y ) , γ ( y ) ≤ z } , P ( α , v ; θ , w ; γ , z ) = { y ∈ P : v ≤ α ( y ) , θ ( y ) ≤ w , γ ( y ) ≤ z } ,$

and a closed set

$R(ψ,t;γ,z)= { y ∈ P : t ≤ ψ ( y ) , γ ( y ) ≤ z } .$

The following fixed point theorem is fundamental and important to the proof of our main result.

Lemma 2.1 ()

Let $B$ be a real Banach space and $P⊂B$ be a cone in $B$. Let γ, θ be nonnegative continuous convex functionals on $P$, let α be a nonnegative continuous concave functional on $P$, and let ψ be a nonnegative continuous functional on $P$ satisfying $ψ(λy)≤λψ(y)$ for all $0≤λ≤1$, such that for some positive numbers z and M,

Suppose that $T: P ( γ , z ) ¯ → P ( γ , z ) ¯$ is completely continuous and there exist positive numbers t, v, and w with $0 such that

1. (i)

${y∈P(α,v;θ,w;γ,z)|α(y)>v}≠∅$ and $α(Ty)>v$ for $y∈P(α,v;θ,w;γ,z)$;

2. (ii)

$α(Ty)>v$ for $y∈P(α,v;γ,z)$ with $θ(Ty)>w$;

3. (iii)

$0∉R(ψ,t;γ,z)$ and $ψ(Ty) for $y∈R(ψ,t;γ,z)$ with $ψ(y)=t$.

Then T has at least three fixed points $y 1 , y 2 , y 3 ∈ P ( γ , z ) ¯$ such that

Let $h(n)$ ($n∈[0,N]$) be a positive sequence. Consider the following BVP:

${ Δ [ ϕ p ( Δ 2 y ( n ) ) ] + h ( n ) = 0 , n ∈ [ 0 , N ] , a y ( 0 ) − b Δ y ( 0 ) = 0 , c y ( N + 3 ) + d Δ y ( N + 2 ) = 0 , Δ 2 y ( 0 ) = 0 .$
(2.1)

Lemma 2.2 If y is a solution of BVP (2.1), then there exists unique $n 0 ∈[0,N+1]$ such that $Δy( n 0 )>0$ and $Δy( n 0 +1)≤0$.

Proof Suppose y satisfies (2.1). It follows that

$y(n)=y(0)+nΔy(0)− ∑ r = 0 n − 1 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ,n∈[0,N+3].$
(2.2)

The BCs in (2.1) imply that

$ay(0)=bΔy(0)$

and

$cy(0)+c(N+3)Δy(0)−c ∑ r = 0 N + 2 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) +dΔy(0)−d ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) =0.$

It follows that

$y(0)= b a Δy(0)$

and

$Δy(0)= a a d + a c ( N + 3 ) + b c ( c ∑ r = 0 N + 2 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) + d ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ) .$

Similarly, we get

$y ( n ) = y ( N + 3 ) − ( N + 3 − n ) Δ y ( N + 2 ) − ∑ r = n N + 2 ∑ i = r N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) , n ∈ [ 0 , N + 3 ] .$
(2.3)

The BCs in (2.1) imply that

$y(N+3)=− d c Δy(N+2)$

and

$Δ y ( N + 2 ) = − c a d + a c ( N + 3 ) + b c ( a ∑ r = 0 N + 2 ∑ i = r N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) + b ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ) .$

Since $a,b≥0$, and $c,d>0$ with $ad+ac(N+3)+bc>0$ and $h(n)$ a positive sequence, one can easily see that $Δy(0)>0$ and $Δy(N+2)≤0$. It follows from $Δy(0)>0$, $Δy(N+2)≤0$, and the fact that $Δy(n)$ is decreasing on $[0,N+2]$ that there exists unique $n 0 ∈[0,N+1]$ such that $Δy( n 0 )>0$ and $Δy( n 0 +1)≤0$. The proof is complete. □

Lemma 2.3 If y is a solution of BVP (2.1), then $y(0)≥0$, $y(N+3)≥0$, and $y(n)>0$ for all $n∈[1,N+2]$.

Proof We get from Lemma 2.2 the result that there exists unique $n 0 ∈[0,N]$ such that $Δy( n 0 )>0$ and $Δy( n 0 +1)≤0$. It follows from (2.1) that

$Δy(n)= { Δ y ( n 0 + 1 ) − ∑ i = n 0 + 1 n − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) , n ∈ [ n 0 + 1 , N + 2 ] , Δ y ( n 0 ) + ∑ i = n n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) , n ∈ [ 0 , n 0 ] .$

Then

$y(n)= { y ( N + 3 ) − ( N + 3 − n ) Δ y ( n 0 + 1 ) + ∑ r = n N + 2 ∑ i = n 0 + 1 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) , n ∈ [ n 0 + 1 , N + 3 ] , y ( 0 ) + n Δ y ( n 0 ) + ∑ r = 0 n − 1 ∑ i = r n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) , n ∈ [ 0 , n 0 + 1 ] ,$

with

$y ( n 0 + 1 ) = y ( 0 ) + ( n 0 + 1 ) Δ y ( n 0 ) + ∑ r = 0 n 0 ∑ i = r n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) = y ( N + 3 ) − ( N + 2 − n 0 ) Δ y ( n 0 + 1 ) ∑ r = n 0 + 1 N + 2 ∑ i = n 0 + 1 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

It follows from $h(n)$ being positive, $Δy( n 0 )>0$, and $Δy( n 0 +1)≤0$ that

$y(n)> { y ( N + 3 ) , n ∈ [ n 0 + 1 , N + 2 ] , y ( 0 ) , n ∈ [ 1 , n 0 + 1 ] .$

So $y(n)≥min{y(0),y(N+3)}$ for all $n∈[0,N+3]$. From BCs in BVP (2.1), we get

$a y ( 0 ) = b Δ y ( 0 ) ≥ 0 , c y ( N + 3 ) = − d Δ y ( N + 2 ) ≥ 0 .$

Then

$min { y ( 0 ) , y ( N + 3 ) } ≥0.$

Hence $y(n)>0$ for all $n∈[1,N+2]$. The proof is complete. □

Lemma 2.4 If y is a solution of BVP (2.1), then

(2.4)

where $σ n =min{ n N + 3 , N + 3 − n N + 3 }$.

Proof It follows from Lemma 2.2 and Lemma 2.3 that $y(n)≥0$ for $n∈[0,N+3]$. Suppose that $y( n 0 )=max{y(n):n∈[0,N+3]}$. Since $Δy(0)>0$ and $Δy(N+1)≤0$, we get $n 0 ∈[1,N+1]$. For $n∈[1, n 0 ]$, it is easy to see that

$y ( n 0 ) − y ( 0 ) n 0 n + y ( 0 ) − y ( n ) = n ∑ s = 0 n 0 − 1 Δ y ( s ) − n 0 ∑ s = 0 n − 1 Δ y ( s ) n 0 = − ( n 0 − n ) ∑ s = 0 n − 1 Δ y ( s ) + n ∑ s = n n 0 − 1 Δ y ( s ) n 0 .$

Since $Δ 2 y(n)=− ϕ q ( ∑ s = 0 n − 1 h(s))<0$ for all $n∈[0,N+1]$, we get $Δy(s)≤Δy(j)$ for all $s≥j$. Then $−( n 0 −n) ∑ s = 0 n − 1 Δy(s)+n ∑ s = n n 0 − 1 Δy(s)≤0$. It follows that $y ( n 0 ) − y ( 0 ) n 0 n+y(0)−y(n)≤0$. Then

Similarly, if $n∈[ n 0 ,N+2]$, we get

Then

□

Lemma 2.5 If y is a solution of BVP (2.1), then

$y(n)= b + n a a A h − ∑ r = 0 n − 1 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ,$
(2.5)

where $A h$ satisfies the equation

$A h = a a d + a c ( N + 3 ) + b c ( c ∑ r = 0 N + 2 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) + d ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ) .$

Proof The proof follows from Lemma 2.2 and is omitted. □

Lemma 2.6 If y is a solution of BVP (2.1), then there exists an $n 0 ∈[0,N]$ such that

$max n ∈ [ 0 , N + 3 ] y ( n ) = y ( n 0 + 1 ) ≥ max { ∑ r = 0 n 0 ∑ i = r n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) , ∑ r = n 0 + 1 N + 2 ∑ i = n 0 + 1 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) } .$

Proof It follows from Lemma 2.2 that there is $n 0 ∈[0,N+1]$ such that $Δy( n 0 )>0$ and $Δy( n 0 +1)≤0$, $Δy(n)>0$ for all $n∈[0, n 0 ]$ and $Δy(n)≤0$ for all $n∈[ n 0 +1,N+2]$. Then

$max n ∈ [ 0 , N + 3 ] y(n)=y( n 0 +1),$

there exists $ξ∈( n 0 , n 0 +1]$ such that

$Δ y ( n 0 + 1 ) − Δ y ( n 0 ) n 0 + 1 − n 0 = 0 − Δ y ( n 0 ) ξ − n 0 .$

Then

$Δy( n 0 +1)=− n 0 + 1 − ξ ξ − n 0 Δy( n 0 ).$
(2.6)

It is easy to see from (2.1) that

$0<Δy( n 0 )= A h − ∑ i = 0 n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ,$
(2.7)
$0≥Δy( n 0 +1)= A h − ∑ i = 0 n 0 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$
(2.8)

Here $A h =Δy(0)$. So (2.6)-(2.8) imply that

$A h − ∑ i = 0 n 0 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) =− n 0 + 1 − ξ ξ − n 0 A h + n 0 + 1 − ξ ξ − n 0 ∑ i = 0 n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

Then

$A h =( n 0 +1−ξ) ∑ i = 0 n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) +(ξ− n 0 ) ∑ i = 0 n 0 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

We get

$∑ i = 0 n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ≤ A h ≤ ∑ i = 0 n 0 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$
(2.9)

Lemma 2.3 implies that $B h =y(0)≥0$. Furthermore, one has from (2.6)

$y ( n 0 + 1 ) = B h + ( n 0 + 1 ) A h − ∑ r = 0 n 0 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ≥ ∑ r = 0 n 0 ∑ i = 0 n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) − ∑ r = 0 n 0 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) = ∑ r = 0 n 0 ∑ i = r n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

On the other hand, by a discussion similar to Lemma 2.2 and Lemma 2.3, we have $A ¯ h =Δy(N+2)$, $B ¯ h =y(N+3)$ with

$Δy(n)= A ¯ h + ∑ i = n N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) )$

and

$y(n)= B ¯ h −(N+3−n) A ¯ h − ∑ r = n N + 2 ∑ i = r N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

It follows that

$Δ y ( n 0 ) = A ¯ h + ∑ i = n 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) > 0 , Δ y ( n 0 + 1 ) = A ¯ h + ∑ i = n 0 + 1 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ≤ 0 .$

So

$A ¯ h + ∑ i = n 0 + 1 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) =− n 0 + 1 − ξ ξ − n 0 A ¯ h − n 0 + 1 − ξ ξ − n 0 ∑ i = n 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

Then

$A ¯ h =−(ξ− n 0 ) ∑ i = n 0 + 1 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) −( n 0 +1−ξ) ∑ i = n 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

We get

$− ∑ i = n 0 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ≥ A ¯ h ≥− ∑ i = n 0 + 1 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

One has from Lemma 2.3 $B ¯ h =y(N+3)≥0$. Therefore

$y ( n 0 + 1 ) = B ¯ h − ( N + 2 − n 0 ) A ¯ h − ∑ r = n 0 + 1 N + 2 ∑ i = n 0 + 1 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) ≥ ( N + 2 − n 0 ) ∑ i = n 0 + 1 N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) − ∑ r = n 0 + 1 N + 2 ∑ i = r N + 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) = ∑ r = n 0 + 1 N + 2 ∑ i = n 0 + 1 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) .$

Hence

$max n ∈ [ 0 , N + 3 ] y ( n ) = y ( n 0 + 1 ) ≥ max { ∑ r = 0 n 0 ∑ i = r n 0 − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) , ∑ r = n 0 + 1 N + 2 ∑ i = n 0 + 1 r − 1 ϕ q ( ∑ s = 0 i − 1 h ( s ) ) } .$

□

Let $h(n)=q(n)f(n,y(n),Δy(n))$. Then $A y$ satisfies the following equation:

$A y = a a d + a c ( N + 3 ) + b c ( c ∑ r = 0 N + 2 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) + d ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) ) .$

Let $B= R N + 4$. We call $x≤y$ for $x,y∈B$ if $x(n)≤y(n)$ for all $n∈[0,N+3]$.

Define the norm

$∥y∥=max { max n ∈ [ 0 , N + 3 ] y ( n ) , max n ∈ [ 0 , N + 2 ] | Δ y ( n ) | } .$

It is easy to see that $B$ is a semi-ordered real Banach space.

Choose

(2.10)

where $σ n =min{ n N + 3 , N + 3 − n N + 3 }$. Then $P$ is a cone in $B$.

Define the operator $T:P→B$ by

$(Ty)(n)= b + n a a A y − ∑ r = 0 n − 1 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) ,$

for $y∈P$, $n∈[0,N+3]$. Then

$( T y ) ( n ) = b + n a a d + a c ( N + 3 ) + b c ( c ∑ r = 0 N + 2 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) + d ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) ) − ∑ r = 0 n − 1 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) .$

Lemma 2.7 Suppose that (C1) holds. Then

1. (i)

Ty satisfies the following:

${ Δ [ ϕ p ( Δ 2 ( T y ) ( n ) ) ] + q ( n ) f ( n , y ( n ) , Δ y ( n ) ) = 0 , 0 < n < N , a ( T y ) ( 0 ) − b Δ ( T y ) ( 0 ) = 0 , c ( T y ) ( N + 3 ) + d Δ ( T y ) ( N + 2 ) = 0 , Δ 2 ( T y ) ( 0 ) = 0 .$
(2.11)
2. (ii)

$Ty∈P$ for each $y∈P$.

3. (iii)

y is a solution of BVP (1.1) if and only if y is a solution of the operator equation $y=Ty$.

4. (iv)

$T:P→P$ is completely continuous.

Proof

1. (i)

By the definition of Ty, we get (2.11).

2. (ii)

Note the definition of $P$. Since (C1) holds, for $y∈P$, (2.11), Lemma 2.2, Lemma 2.3 and Lemma 2.4 imply that $Δ(Ty)(n)$ is decreasing on $[0,N+2]$ and $(Ty)(n)≥ σ n max n ∈ [ 0 , N + 3 ] (Ty)(n)$ for all $n∈[0,N+3]$. Together with (2.11), it follows that $Ty∈P$.

3. (iii)

It is easy to see from (2.11) that y is a solution of BVP (1.1) if and only if y is a solution of the operator equation $y=Ty$.

4. (iv)

It suffices to prove that T is continuous on $P$ and T is relative compact.

We divide the proof into three steps:

Step 1. For each bounded subset $D⊂P$, prove that $A y$ is bounded in for $y∈ D ¯$

Denote

$L 1 =max { max n ∈ [ 0 , N + 3 ] y ( n ) , max n ∈ [ 0 , N + 1 ] | Δ y ( n ) | : x ∈ D ¯ }$

and

$L 2 = max j ∈ [ 0 , N ] f L 1 (j)= max j ∈ [ 0 , N ] max u , | x | ≤ L 1 q(j)f(j,u,x).$

It follows from (2.9) in the proof of Lemma 2.6 that

$0≤ A y ≤ ∑ i = 0 N ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) ≤(N+1) ϕ q (N L 2 ).$

Hence $A y$ is bounded in .

Step 2. For each bounded subset $D⊂P$, and each $y 0 ∈D$, it is easy to prove that T is continuous at $y 0$.

Step 3. For each bounded subset $D⊂P$, prove that T is relative compact on D.

In fact, for each bounded subset $Ω⊆D$ and $y∈Ω$. Suppose

$∥y∥=max { max n ∈ [ 0 , N + 3 ] y ( n ) , max n ∈ [ 0 , N + 2 ] | Δ y ( n ) | } < M 1 ,$

and Step 1 implies that there exists a constant $M 2 >0$ such that $A y < M 2$. Then

$( T y ) ( n ) = b + n a a A y − ∑ r = 0 n − 1 ∑ i = 0 r − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) ≤ b + n a a M 2 + ∑ r = 0 N + 2 ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 N q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) ≤ b + ( N + 3 ) a a M 2 + ( N + 3 ) ( N + 2 ) ϕ q ( ∑ s = 0 N f M 1 ( s ) ) : = M 3 ,$

where $f M 1 (s)= max u ≤ M 1 , | x | ≤ M 1 q(j)f(s,u,x)$. Similarly, one has

$| Δ ( T y ) ( n ) | = | A y − ∑ i = 0 n − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) | ≤ M 2 + ∑ i = 0 N + 1 ϕ q ( ∑ s = 0 N f M 1 ( s ) ) = M 2 + ( N + 2 ) ϕ q ( ∑ s = 0 N f M 1 ( s ) ) : = M 4 .$

It follows that T Ω is bounded. Since $B= R N + 4$, one knows that T Ω is relative compact. Steps 1, 2, and 3 imply that T is completely continuous. □

## 3 Main result

In this section, our objective is to establish the existence of at least three positive solutions for BVP (1.1) by using the Avery-Peterson fixed point theorem .

Choose $[ N + 3 2 ]>k>0$, where $[x]$ denotes the largest integer not greater than x, and denote $σ k =min{ k N + 3 , N + 3 − k N + 3 }$.

Define the functionals on $P:P→[0,+∞)$ by

$γ ( y ) = max n ∈ [ 0 , N + 2 ] | Δ y ( n ) | , θ ( y ) = ψ ( y ) = max n ∈ [ 0 , N + 3 ] y ( n ) , α ( y ) = min n ∈ [ k , N + 3 − k ] y ( n ) .$

For $y∈P$ and $n∈[0,N+3]$ we have

$y ( n ) = y ( n ) − y ( 0 ) + y ( 0 ) ≤ | ∑ i = 0 n − 1 Δ y ( i ) | + y ( 0 ) = | ∑ i = 0 n − 1 Δ y ( i ) | + b a Δ y ( 0 ) ≤ | ∑ i = 0 N + 2 Δ y ( i ) | + b a Δ y ( 0 ) ≤ ( N + 3 + b a ) max n ∈ [ 0 , N + 2 ] | Δ y ( n ) | ,$

i.e.,

$max n ∈ [ 0 , N + 3 ] y(n)≤ ( N + 3 + b a ) max n ∈ [ 0 , N + 2 ] |Δy(n)|.$
(3.1)

So,

$∥ y ∥ = max { max n ∈ [ 0 , N + 3 ] y ( n ) , max n ∈ [ 0 , N + 2 ] | Δ y ( n ) | } ≤ ( N + 3 + b a ) max n ∈ [ 0 , N + 2 ] | Δ y ( n ) | .$

Hence, we obtain

$∥y∥≤ ( N + 3 + b a ) γ(y),y∈P.$
(3.2)

Let

$Ω = 2 ∑ i = 0 N + 2 ϕ q ( ∑ s = 0 i − 1 q ( s ) ) , Λ = σ k min { ∑ r = k [ N + 3 2 ] ∑ i = r [ N + 3 2 ] − 1 ϕ q ( ∑ s = k i − 1 q ( s ) ) , ∑ r = [ N + 3 2 ] N + 3 − k ∑ i = [ N + 3 2 ] r − 1 ϕ q ( ∑ s = k i − 1 q ( s ) ) } .$

Theorem 3.1 Suppose that (C1) holds. If there are positive numbers

such that the following conditions are satisfied:

(C2) $f(n,y(n),Δy(n))≤ ϕ p ( z Ω )$ for all $(n,y(n),Δy(n))∈[0,N+3]×[0,(N+3+ b a )z]×[−z,z]$;

(C3) $f(n,y(n),Δy(n))> ϕ p ( v Λ )$ for all $(n,y(n),Δy(n))∈[k,N+3−k]×[v, v σ k ]×[−z,z]$;

(C4) $f(n,y(n),Δy(n))< a ( N + 3 ) a + b ϕ p ( t Ω )$ for all $(n,y(n),Δy(n))∈[0,N+3]×[0,t]×[−z,z]$,

then BVP (1.1) has at least three positive solutions.

Proof We choose positive numbers t, v, $w= v σ k$, z with $t, $ϕ p ( v Λ )≤min{ ϕ p ( z Ω ), a ( N + 3 ) a + b ϕ p ( t Ω )}$. Next we show that all the conditions of Lemma 2.1 are satisfied.

It is clear that for $y∈P$ and $λ∈[0,1]$, there are $α(y)≤ψ(y)$, $ψ(λy)=λψ(y)$. From (3.2), we have $∥y∥≤(N+3+ b a )γ(y)$. Furthermore, $ψ(0)=0 and therefore $0∉R(ψ,t;γ,z)$.

Now the proof is divided into four steps.

Step 1. We will show that (C2) implies that

$T: P ( γ , z ) ¯ → P ( γ , z ) ¯ .$

For $y∈ P ( γ , z ) ¯$, we have $γ(y)= max n ∈ [ 0 , N + 2 ] |Δy(n)|≤z$. From (3.2) we get

$( n , y ( n ) , Δ y ( n ) ) ∈[0,N+3]× [ 0 , ( N + 3 + b a ) z ] ×[−z,z].$

This implies that (C2) holds. Then one has from (C2) and (2.9) in the proof of Lemma 2.6

$γ ( T y ) = max n ∈ [ 0 , N + 2 ] | Δ ( T y ) ( n ) | = max n ∈ [ 0 , N + 2 ] | A y − ∑ i = 0 n − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) | ≤ max n ∈ [ 0 , N + 2 ] | ∑ i = 0 N + 2 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) + ∑ i = 0 n − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) | ≤ 2 ∑ i = 0 N + 2 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) ≤ 2 z Ω ∑ i = 0 N + 2 ϕ q ( ∑ s = 0 i − 1 q ( s ) ) = z .$

Therefore, $T: P ( γ , z ) ¯ → P ( γ , z ) ¯$. Hence, by Lemma 2.7, we know that $T: P ( γ , z ) ¯ → P ( γ , z ) ¯$ is completely continuous.

Step 2. We show that condition (i) in Lemma 2.1 holds.

Choose $y(n)= v σ k =w$ for all $n∈[0,N+3]$. It is easy to see that

$α ( y ) = min n ∈ [ k , N + 3 − k ] y ( n ) = v σ k > v , θ ( y ) = max n ∈ [ 0 , N + 3 ] y ( n ) = v σ k = w , γ ( y ) = max n ∈ [ 0 , N + 2 ] | Δ y ( n ) | = 0 < z ,$

since $σ k =min{ k N + 3 , N + 3 − k N + 3 }< 1 2$. Hence ${y∈P(α,v;θ,w;γ,z)|α(y)>v}≠∅$.

For $y∈P(α,v;θ,w;γ,z)$, we have $v≤y(n)≤ v σ k$ and $−z≤Δy(n)≤z$ for $n∈[k,N+3−k]$.

It follows from (C3) that

$f ( n , y ( n ) , Δ y ( n ) ) > ϕ p ( v Λ ) , ( n , y ( n ) , Δ y ( n ) ) ∈[k,N+3−k]× [ v , v σ k ] ×[−z,z].$

Similarly to Lemma 2.6 there exists $n 0 ∈[k,N+3−k]$ such that $Δy( n 0 )>0$ and $Δy( n 0 +1)≤0$ and

$max n ∈ [ 0 , N + 3 ] (Ty)(n)=(Ty)( n 0 +1)≥max { ( T y ) ( n 0 ) , ( T y ) ( n 0 + 2 ) } ,$

we get from (2.4), (C1), (C3), and Lemma 2.6

$α ( T y ) = min n ∈ [ k , N + 3 − k ] ( T y ) ( n ) ≥ σ k max n ∈ [ 0 , N + 3 ] ( T y ) ( n ) = σ k ( T y ) ( n 0 ) ≥ σ k max { ∑ r = 0 n 0 ∑ i = r n 0 − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) , ∑ r = n 0 + 1 N + 2 ∑ i = n 0 + 1 r − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) } ≥ σ k max { ∑ r = 0 [ N + 3 2 ] ∑ i = r [ N + 3 2 ] − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) , ∑ r = [ N + 3 2 ] N + 2 ∑ i = [ N + 3 2 ] r − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) } ≥ σ k min { ∑ r = k [ N + 3 2 ] ∑ i = r [ N + 3 2 ] − 1 ϕ q ( ∑ s = k i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) , ∑ r = [ N + 3 2 ] N + 2 ∑ i = [ N + 3 2 ] r − 1 ϕ q ( ∑ s = k i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) } ≥ v σ k Λ min { ∑ r = k [ N + 3 2 ] ∑ i = r [ N + 3 2 ] − 1 ϕ q ( ∑ s = k i − 1 q ( s ) ) , ∑ r = [ N + 3 2 ] N + 3 − k ∑ i = [ N + 3 2 ] r − 1 ϕ q ( ∑ s = k i − 1 q ( s ) ) } = v .$

We conclude that condition (i) of Lemma 2.1 holds.

Step 3. We prove that condition (ii) of Lemma 2.1 holds. If $y∈P(α,v;γ,z)$ and $θ(Ty)>w= v σ k$, then we have

$α(Ty)= min n ∈ [ k , N + 3 − k ] (Ty)(n)≥ σ k max n ∈ [ 0 , N + 3 ] (Ty)(n)= σ k θ(Ty)>v.$

Then condition (ii) of Lemma 2.1 is satisfied.

Step 4. Finally, we verify that (iii) of Lemma 2.1 also holds. Clearly, $0∉R(ψ,t;γ,z)$. Suppose that $y∈R(ψ,t;γ,z)$ with $ψ(y)=t$. Then by condition (C4) and (3.1), we obtain

$ψ ( T y ) = max n ∈ [ 0 , N + 3 ] ( T y ) ( n ) ≤ ( N + 3 + b a ) max n ∈ [ 0 , N + 2 ] | Δ ( T y ) ( n ) | = ( N + 3 + b a ) max n ∈ [ 0 , N + 2 ] | A y − ∑ i = 0 n − 1 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) | ≤ ( N + 3 + b a ) 2 ∑ i = 0 N + 2 ϕ q ( ∑ s = 0 i − 1 q ( s ) f ( s , y ( s ) , Δ y ( s ) ) ) < ( N + 3 + b a ) ( a ( N + 3 ) a + b ) 2 t Ω ∑ i = 0 N + 2 ϕ q ( ∑ s = 0 i − 1 q ( s ) ) = t .$

Thus, condition (iii) of Lemma 2.1 is satisfied.

From Steps 1-4 together with Lemma 2.1 we find that the operator T has at least three fixed points which are positive solutions $y 1$, $y 2$, and $y 3$ belonging to $P ( γ , z ) ¯$ of (1.1) such that

□

## 4 An example

Example 4.1 Consider the following BVP:

${ Δ 3 y ( n ) + f ( n , y ( n ) , Δ y ( n ) ) = 0 , n ∈ [ 0 , 99 ] , 2 y ( 0 ) − Δ y ( 0 ) = 0 , 2 y ( 102 ) + Δ y ( 101 ) = 0 , Δ 2 y ( 0 ) = 0 ,$
(4.1)

where $f(n,y(n),Δy(n))$ is continuous and positive for all $(n,y(n),Δy(n))∈[0,N]×[0,+∞)×R$. Corresponding to BVP (1.1), we have $N=99$, $p=2$, $q(n)=1$, $n∈[0,N]$, $a=c=2$, $b=d=1$, $ϕ 2 (y)=y$.

It is easy to see that (C1) holds.

Choose the constant $k=49$, then $σ 49 =min{ 49 102 , 53 102 }= 49 102$, $Ω=10,302$, $Λ= 49 51$. Taking $t=10$, $v=50$, and $z=600,000$, it is easy to check that

$10=t

If

$f(n,y(n),Δy(n))≤ 300 , 000 5 , 151$, for all $(n,y(n),Δy(n))∈[0,102]×[0,62,400,000]×[−600,000,600,000]$;

$f(n,y(n),Δy(n))> 2 , 550 49$ for all $(n,y(n),Δy(n))∈[49,53]×[50, 5 , 100 49 ]×[−600,000,600,000]$;

$f(n,y(n),Δy(n))< 5 535 , 704$, for all $(n,y(n),Δy(n))∈[0,102]×[0,10]×[−600,000,600,000]$,

then Theorem 3.1 implies that BVP (4.1) has at least three positive solutions such that

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