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BY 4.0 license Open Access Published by De Gruyter Open Access October 13, 2022

Existence of positive solutions of discrete third-order three-point BVP with sign-changing Green's function

  • Huijuan Li , Chenghua Gao EMAIL logo and Nikolay D. Dimitrov
From the journal Open Mathematics

Abstract

In this article, we consider a discrete nonlinear third-order boundary value problem

Δ 3 u ( k 1 ) = λ a ( k ) f ( k , u ( k ) ) , k [ 1 , N 2 ] Z , Δ 2 u ( η ) = α Δ u ( N 1 ) , Δ u ( 0 ) = β u ( 0 ) , u ( N ) = 0 ,

where N > 4 is an integer, λ > 0 is a parameter. f : [ 1 , N 2 ] Z × [ 0 , + ) [ 0 , + ) is continuous, a : [ 1 , N 2 ] Z ( 0 , + ) , α 0 , 1 N 1 , β 0 , 2 ( 1 α ( N 1 ) ) N ( 2 α ( N 1 ) ) , and η N 2 2 + 1 , N 2 Z . With the sign-changing Green’s function, we obtain not only the existence of positive solutions but also the multiplicity of positive solutions to this problem.

MSC 2010: 39A10; 39A12

1 Introduction

For any integers c and d with c d , let [ c , d ] Z = { c , c + 1 , , d } . In this article, we study existence and multiplicity of positive solutions for the following discrete nonlinear third-order boundary value problem (BVP)

(1.1) Δ 3 u ( k 1 ) = λ a ( k ) f ( k , u ( k ) ) , k [ 1 , N 2 ] Z , Δ 2 u ( η ) = α Δ u ( N 1 ) , Δ u ( 0 ) = β u ( 0 ) , u ( N ) = 0 ,

where N > 4 is an integer and λ > 0 is a parameter. f : [ 1 , N 2 ] Z × [ 0 , + ) [ 0 , + ) is continuous, a : [ 1 , N 2 ] Z ( 0 , + ) , α 0 , 1 N 1 , β 0 , 2 ( 1 α ( N 1 ) ) N ( 2 α ( N 1 ) ) , and η satisfies the condition:

( H 0 ) : η N 2 2 + 1 , N 2 Z .

BVPs for difference equations have been widely studied for different disciplines, such as the computer sciences, applied mathematics, economics, mechanical engineering and control systems, and so on, see [1,2,3]. Great effort has been made to study the existence, multiplicity, and uniqueness of solutions of BVPs by using fixed-point theorem [4,5,6, 7,8,9, 10,11,12], monotone iterative [13], degree theory [14], the method of upper and lower solutions [15], critical point theorem [16,17], and so forth. In recent years, the existence of positive solutions for third-order three-point BVPs has also been discussed by several authors, and see [8,9,10, 11,18,19]. However, it is necessary to point out that Green’s functions in the aforementioned literature are positive. As we all know, the positivity of Green’s function guarantees the positivity of the corresponding difference operator. Now, the natural question is: When the Green’s function changes its sign, how could we obtain the existence of positive solutions? Fortunately, there have been some papers on positive solution for third-order BVPs when the corresponding Green’s functions are sign-changing, and see [5,6,7,12,13,20]. For example, Gao and Geng [6] considered positive solutions of the following discrete nonlinear third-order three-point eigenvalue problem:

Δ 3 u ( k 1 ) = λ a ( k ) f ( k , u ( k ) ) , k [ 1 , T 2 ] Z , Δ u ( 0 ) = u ( T ) = Δ 2 u ( η ) = 0 ,

where η T 1 2 , , T 2 for odd T and η T 2 2 , , T 2 for even T , and the Green’s function is sign changing. Furthermore, by using Guo-Krasnoselskii’s fixed-point theorem, Xu et al. [12] considered the existence of positive solutions for third-order three-point BVP:

(1.2) Δ 3 u ( k 1 ) = a ( k ) f ( k , u ( k ) ) , k [ 1 , T 2 ] Z , Δ 2 u ( η ) α Δ u ( T 1 ) = 0 , Δ u ( 0 ) = u ( T ) = 0 ,

where η T 2 2 + 1 , T 2 Z and the Green’s function is sign changing. Clearly, BVP (1.2) is a special case of BVP (1.1), so BVP (1.1) is parallel to BVP (1.2) but more general. Recently, Cao et al. [5] discussed the existence of positive solutions for the following BVP:

Δ 3 u ( k 1 ) = λ a ( k ) f ( k , u ( k ) ) , k [ 1 , T 1 ] Z , u ( 0 ) = 0 , Δ 2 u ( η ) = α Δ u ( T ) , Δ u ( T ) = β u ( T + 1 ) ,

where both the weight function a ( t ) and the Green’s function G ( t , s ) change their sign. Inspired by the aforementioned works, in this article, we try to study problem (1.1).

The rest of this article is organized as follows. In Section 2, we study the linear problem. In particular, we will point out that the Green’s function changes its sign. Meanwhile, we also give a typical example to explain why we choose η N 2 2 + 1 , N 2 Z . In Section 3, we impose some conditions to obtain existence and multiplicity of positive solutions to problem (1.1). In Section 4, we add several specific examples to verify our main results. The main tool we will use is the following Guo-Krasnoselskii’s fixed-point theorem.

Theorem 1.1

[21] Let E be a Banach space and K E be a cone. Assume that Ω 1 and Ω 2 are open bounded subsets of E with 0 Ω 1 , Ω ¯ 1 Ω 2 . If T : K ( Ω ¯ 2 \ Ω 1 ) K is a completely continuous operator such that

(i) T u u , u K Ω ¯ 1 and T u u , u K Ω ¯ 2 ,

or

(ii) T u u , u K Ω ¯ 1 , and T u u , u K Ω ¯ 2 ,

then T has a fixed point in K ( Ω ¯ 2 \ Ω 1 ) .

2 Linear problem

Let us study the linear problem

(2.1) Δ 3 u ( k 1 ) = h ( k ) , k [ 1 , N 2 ] Z , Δ 2 u ( η ) = α Δ u ( N 1 ) , Δ u ( 0 ) = β u ( 0 ) , u ( N ) = 0 ,

with 1 N 1 > α 0 , 2 ( 1 α ( N 1 ) ) N ( 2 α ( N 1 ) ) > β 0 . We denote

ρ k = 2 ( 1 β k ) ( 1 α ( N 1 ) ) α β k ( k 1 )

and

ρ = 2 ( 1 β N ) ( 1 α ( N 1 ) ) α β N ( N 1 ) .

Lemma 2.1

Problem (2.1) has a unique solution

u ( k ) = s = 1 N 2 G ( k , s ) h ( s ) ,

where G ( k , s ) = u 1 ( k , s ) + u 2 ( k , s ) + u 3 ( k , s ) with

u 1 ( k , s ) = α ( N s 1 ) [ k ( k 1 ) ρ k ρ N ( N 1 ) ] 2 2 α ( N 1 ) ρ k ( N s ) ( N s 1 ) 2 ρ , u 2 ( k , s ) = u 2 ( k , s ) ρ k ρ N ( N 1 ) k ( k 1 ) 2 2 α ( N 1 ) , s η , 0 , s > η ,

and

u 3 ( k , s ) = u 3 ( k , s ) ( k s ) ( k s 1 ) 2 , 0 < s k 2 N 2 , 0 , 0 k 2 < s N 2 .

Proof

Summing both sides of Δ 3 u ( s 1 ) = h ( s ) from s = 1 to s = k 1 , we obtain

(2.2) Δ 2 u ( k 1 ) = Δ 2 u ( 0 ) + s = 1 k 1 h ( s ) ,

and

(2.3) Δ u ( k 1 ) = Δ u ( 0 ) + ( k 1 ) Δ 2 u ( 0 ) + s = 1 k 2 ( k s 1 ) h ( s ) .

Summing both sides of the aforementioned equation from τ = 1 to τ = k , we deduce that

(2.4) u ( k ) = u ( 0 ) + k Δ u ( 0 ) + k ( k 1 ) 2 Δ 2 u ( 0 ) + s = 1 k 2 ( k s ) ( k s 1 ) 2 h ( s ) .

Next, refer equations (2.2) and (2.3) when Δ 2 u ( η ) = α Δ u ( N 1 ) is rewritten as follows:

Δ 2 u ( 0 ) + s = 1 η h ( s ) = α Δ u ( 0 ) + ( N 1 ) Δ 2 u ( 0 ) + s = 1 N 2 ( N s 1 ) h ( s ) ,

and thus,

(2.5) Δ 2 u ( 0 ) = α Δ u ( 0 ) + s = 1 N 2 ( N s 1 ) h ( s ) s = 1 η h ( s ) 1 α ( N 1 ) = α Δ u ( 0 ) 1 α ( N 1 ) + s = 1 N 2 α ( N s 1 ) 1 α ( N 1 ) h ( s ) s = 1 η 1 1 α ( N 1 ) h ( s ) .

Combining (2.4) and (2.5), we obtain

u ( k ) = u ( 0 ) + k Δ u ( 0 ) + k ( k 1 ) 2 Δ 2 u ( 0 ) + s = 1 k 2 ( k s ) ( k s 1 ) 2 h ( s ) = u ( 0 ) + k Δ u ( 0 ) + k ( k 1 ) 2 α Δ u ( 0 ) 1 α ( N 1 ) + s = 1 N 2 k ( k 1 ) 2 α ( N s 1 ) 1 α ( N 1 ) h ( s ) s = 1 η k ( k 1 ) 2 1 1 α ( N 1 ) h ( s ) + s = 1 k 2 ( k s ) ( k s 1 ) 2 h ( s ) .

By using boundary condition Δ u ( 0 ) = β u ( 0 ) , we deduce that

u ( 0 ) + k Δ u ( 0 ) + k ( k 1 ) 2 α Δ u ( 0 ) 1 α ( N 1 ) = u ( 0 ) ρ k 2 ( 1 α ( N 1 ) ) .

Then, by using u ( N ) = 0 , we obtain that

u ( 0 ) + N Δ u ( 0 ) + N ( N 1 ) 2 Δ 2 u ( 0 ) + s = 1 N 2 ( N s ) ( N s 1 ) 2 h ( s ) = 0 ,

which is the same as follows:

u ( 0 ) ( 1 β N ) = N ( N 1 ) 2 Δ 2 u ( 0 ) s = 1 N 2 ( N s ) ( N s 1 ) 2 h ( s ) = N ( N 1 ) 2 α β u ( 0 ) 1 α ( N 1 ) + s = 1 N 2 α ( N s 1 ) 1 α ( N 1 ) h ( s ) s = 1 η 1 1 α ( N 1 ) h ( s ) s = 1 N 2 ( N s ) ( N s 1 ) 2 h ( s ) .

Hence,

u ( 0 ) ρ 2 ( 1 α ( N 1 ) ) = s = 1 η N ( N 1 ) 2 ( 1 α ( N 1 ) ) h ( s ) s = 1 N 2 α N ( N 1 ) ( N s 1 ) 2 ( 1 α ( N 1 ) ) + ( N s ) ( N s 1 ) 2 h ( s )

and

u ( 0 ) = s = 1 η N ( N 1 ) ρ h ( s ) s = 1 N 2 ( N s 1 ) [ α N ( N 1 ) + ( N s ) ( 1 α ( N 1 ) ) ] ρ h ( s ) .

Finally, we obtain that

u ( k ) = ρ k u ( 0 ) 2 ( 1 α ( N 1 ) ) + s = 1 N 2 k ( k 1 ) 2 α ( N s 1 ) 1 α ( N 1 ) h ( s ) s = 1 η k ( k 1 ) 2 1 1 α ( N 1 ) h ( s ) + s = 1 k 2 ( k s ) ( k s 1 ) 2 h ( s ) = ρ k ρ s = 1 η N ( N 1 ) 2 2 α ( N 1 ) h ( s ) ρ k ρ s = 1 N 2 α N ( N 1 ) ( N s 1 ) 2 2 α ( N 1 ) h ( s ) ρ k ρ s = 1 N 2 ( N s ) ( N s 1 ) 2 h ( s ) + s = 1 N 2 α k ( k 1 ) ( N s 1 ) 2 2 α ( N 1 ) h ( s ) s = 1 η k ( k 1 ) 2 2 α ( N 1 ) h ( s ) + s = 1 k 2 ( k s ) ( k s 1 ) 2 h ( s )

= s = 1 N 2 α ( N s 1 ) [ k ( k 1 ) ρ k ρ N ( N 1 ) ] 2 2 α ( N 1 ) ρ k ( N s ) ( N s 1 ) 2 ρ h ( s ) + s = 1 η ρ k ρ N ( N 1 ) k ( k 1 ) 2 2 α ( N 1 ) h ( s ) + s = 1 k 2 ( k s ) ( k s 1 ) 2 h ( s ) s = 1 N 2 u 1 ( k , s ) h ( s ) + s = 1 η u 2 ( k , s ) h ( s ) + s = 1 k 2 u 3 ( k , s ) h ( s ) .

Remark 2.2

We point out that under conditions for α and β , we have 2 ( 1 α ( N 1 ) ) ( 1 β N ) α β N ( N 1 ) > 0 . Moreover, if β = 0 and λ = 1 , we obtain the expression given in [12].

Lemma 2.3

The Green’s function G ( k , s ) satisfies the following properties:

  1. If s [ 1 , η ] Z , then G ( k , s ) is nonincreasing with respect to k [ 0 , N ] Z . If s [ η + 1 , N 2 ] Z , then G ( k , s ) is nondecreasing with respect to k [ 0 , N ] Z .

  2. G ( k , s ) changes its sign on [ 0 , N ] Z × [ 1 , N 2 ] Z . In details, G ( k , s ) 0 for ( k , s ) [ 0 , N ] Z × [ 1 , η ] Z , G ( k , s ) 0 for ( k , s ) [ 0 , N ] Z × [ η + 1 , N 2 ] Z .

  3. If s > η , then max k [ 0 , N ] Z G ( k , s ) = G ( N , s ) = 0 and

    min k [ 0 , N ] Z G ( k , s ) = G ( 0 , s ) = ( N s 1 ) ( α N ( N 1 ) + ( 1 α ( N 1 ) ) ( N s ) ) ρ ( N η 1 ) ( α N ( N 1 ) + ( 1 α ( N 1 ) ) ( N η ) ) ρ = ( N η 1 ) ( α ( N 1 ) η + ( N η ) ) ρ .

If s η , then min k [ 0 , N ] Z G ( k , s ) = G ( N , s ) = 0 and

max k [ 0 , N ] Z G ( k , s ) = G ( 0 , s ) = N ( N 1 ) ( N s 1 ) ( α N ( N 1 ) + ( 1 α ( N 1 ) ) ( N s ) ) ρ N ( N 1 ) ( N η 1 ) ( α N ( N 1 ) + ( 1 α ( N 1 ) ) ( N η ) ) ρ = η ( 2 N η 1 ) α η ( N η 1 ) ( N 1 ) ρ .

Proof

Now, we will study the sign properties of function G .

Firstly, suppose that s > η and s > k 2 . In this case, we obtain G ( k , s ) = u 1 ( k , s ) and

Δ k G ( k , s ) = Δ k u 1 ( k , s ) = α ( N s 1 ) 2 k N ( N 1 ) ρ Δ ρ k 2 2 α ( N 1 ) ( N s ) ( N s 1 ) Δ ρ k 2 ρ .

Notice that Δ ρ k = 2 β ( 1 α ( N 1 ) ) 2 α β k < 0 , we obtain that Δ k G ( k , s ) 0 for s > η and s > k 2 .

Secondly, suppose that s > η and s k 2 . In this case, it is fulfilled that G ( k , s ) = u 1 ( k , s ) + u 3 ( k , s ) and

Δ k G ( k , s ) = Δ k u 1 ( k , s ) + Δ k u 3 ( k , s ) = Δ k u 1 ( k , s ) + k s > 0 .

As a result, we deduce that Δ k G ( k , s ) 0 for s > η , which give us that

max k [ 0 , N ] Z G ( k , s ) = G ( N , s ) = 0 .

Now, if s η and s k 2 , we have G ( k , s ) = u 1 ( k , s ) + u 2 ( k , s ) + u 3 ( k , s ) and

Δ k G ( k , s ) = Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) + Δ k u 3 ( k , s ) = α s k + s ( α ( N 1 ) 1 ) 1 α ( N 1 ) ( α ( N s 1 ) 1 ) N ( N 1 ) Δ ρ k ( 2 2 α ( N 1 ) ) ρ ( N s ) ( N s 1 ) Δ ρ k 2 ρ = 1 2 ρ ( 1 α ( N 1 ) ) { [ α s k + s ( α ( N 1 ) 1 ) ] 2 ρ ( α ( N s 1 ) 1 ) N ( N 1 ) Δ ρ k ( N s ) ( N s 1 ) ( 1 α ( N 1 ) ) Δ ρ k } = 1 2 ρ ( 1 α ( N 1 ) ) { [ 1 α ( N 1 ) ] Δ ρ k [ N ( N 1 ) ( N s ) ( N s 1 ) ] + α s N ( N 1 ) Δ ρ k + [ α s k + s ( α ( N 1 ) 1 ) ] 2 ρ } 0 .

Finally, suppose that s η and s > k 2 . We obtain that

Δ k G ( k , s ) = Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) = ( α ( N s 1 ) 1 ) ( 2 k N ( N 1 ) Δ ρ k ρ ) 2 2 α ( N 1 ) ( N s ) ( N s 1 ) Δ ρ k 2 ρ = 1 2 ρ ( 1 α ( N 1 ) ) { [ 1 α ( N 1 ) ] [ N ( N 1 ) ( N s ) ( N s 1 ) ] Δ ρ k + 2 k ρ [ α ( N 1 ) 1 ] α s [ 2 k ρ N ( N 1 ) Δ ρ k ] } 0 .

As a result, we have Δ k G ( k , s ) 0 for s η . Moreover,

min k [ 0 , N ] Z G ( k , s ) = G ( N , s ) = 0 .

Summarizing the aforementioned results, if s [ 1 , η ] Z , then G ( k , s ) 0 and G ( k , s ) is nonincreasing. Moreover, if s [ η + 1 , N 2 ] Z , then G ( k , s ) 0 and G ( k , s ) is nondecreasing.□

Remark 2.4

Let us give some reasons for why we choose

η N 2 2 + 1 , N 2 Z .

To obtain it, let us consider the following BVP:

Δ 3 u ( k 1 ) = 1 , k [ 1 , N 2 ] Z , Δ 2 u ( η ) = α Δ u ( N 1 ) , Δ u ( 0 ) = β u ( 0 ) , u ( N ) = 0 .

From Lemma 2.1, we obtain

u ( k ) = s = 1 N 2 u 1 ( k , s ) + s = 1 η u 2 ( k , s ) + s = 1 k 2 u 3 ( k , s ) = φ ( k ) 12 ρ ( 1 α ( N 1 ) ) ,

where

φ ( k ) = 3 α ( N 1 ) ( N 2 ) [ k ( k 1 ) ρ ρ k N ( N 1 ) ] 2 N ( N 1 ) ( N 2 ) ( 1 α ( N 1 ) ) ρ k + 6 η [ ρ k N ( N 1 ) k ( k 1 ) ρ ] + 2 k ( k 1 ) ( k 2 ) ρ ( 1 α ( N 1 ) ) .

Clearly, u ( k ) 0 is equivalent to φ ( k ) 0 , and

Δ φ ( k ) = 3 α ( N 1 ) ( N 2 ) [ 2 k ρ Δ ρ k N ( N 1 ) ] 2 N ( N 1 ) ( N 2 ) ( 1 α ( N 1 ) ) Δ ρ k + 6 η [ Δ ρ k N ( N 1 ) 2 k ρ ] + 6 k ( k 1 ) ρ ( 1 α ( N 1 ) ) = 6 ρ k [ ( k 1 ) ( 1 α ( N 1 ) ) + α ( N 1 ) ( N 2 ) 2 η ] + Δ ρ k N ( N 1 ) [ 6 η ( N 2 ) ( α ( N 1 ) + 2 ) ] .

Obviously, if Δ φ ( 0 ) 0 and Δ φ ( N 1 ) 0 , then Δ φ ( k ) 0 , k [ 0 , N 1 ] Z . By direct computation, if Δ φ ( 0 ) 0 , then

η [ α ( N 1 ) + 2 ] ( N 2 ) 6 C 1 .

And if Δ φ ( N 1 ) 0 , then

η [ 3 ρ + β N ( α ( N 1 ) + 2 ) ] ( N 2 ) 6 ( ρ + β N ) C 2 .

Combining with the fact C i N 2 2 , i = 1 , 2 , we obtain η N 2 2 + 1 , N 2 Z , Δ φ ( k ) 0 , k [ 0 , N 1 ] Z and φ ( k ) 0 , k [ 0 , N ] Z . Hence, we have u ( k ) 0 , k [ 0 , N ] Z and Δ u ( k ) 0 , k [ 0 , N 1 ] Z .

Let

E = { u : [ 0 , N ] Z R Δ 2 u ( η ) = α Δ u ( N 1 ) , Δ u ( 0 ) = β u ( 0 ) , u ( N ) = 0 } .

Then E is a Banach space with the norm u = max k [ 0 , N ] Z u ( k ) . Let

K 0 = { h E : h ( k ) 0 , k [ 0 , N ] Z ; Δ h ( k ) 0 , k [ 0 , N 1 ] Z } .

Then K 0 is a cone in E .

Lemma 2.5

Assume ( H 0 ) hold. If h K 0 , then the solution u ( k ) of (2.1) belongs to K 0 , i.e., u K 0 . Furthermore, u is concave on [ 0 , η + 1 ] Z .

Proof

Suppose that 0 k 2 η . Then

u ( k ) = s = 1 N 2 u 1 ( k , s ) h ( s ) + s = 1 η u 2 ( k , s ) h ( s ) + s = 1 k 2 u 3 ( k , s ) h ( s ) = s = 1 k 2 ( u 1 ( k , s ) + u 2 ( k , s ) + u 3 ( k , s ) ) h ( s ) + s = k 1 η ( u 1 ( k , s ) + u 2 ( k , s ) ) h ( s ) + s = η + 1 N 2 u 1 ( k , s ) h ( s ) .

By using the fact that u 3 ( k + 1 , k 1 ) = 1 , we obtain that

Δ u ( k ) = s = 1 k 2 ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) + Δ k u 3 ( k , s ) ) h ( s ) + ( u 1 ( k + 1 , k 1 ) + u 2 ( k + 1 , k 1 ) + u 3 ( k + 1 , k 1 ) ) h ( k 1 ) + s = k 1 η ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) ) h ( s ) ( u 1 ( k + 1 , k 1 ) + u 2 ( k + 1 , k 1 ) ) h ( k 1 ) + s = η + 1 N 2 Δ k u 1 ( k , s ) h ( s ) = s = 1 k 2 ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) + Δ k u 3 ( k , s ) ) h ( s ) + h ( k 1 ) + s = k 1 η ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) ) h ( s ) + s = η + 1 N 2 Δ k u 1 ( k , s ) h ( s ) .

Now, we will show that Δ u ( k ) < 0 for 0 k 2 η . If k 1 η , then

Δ u ( k ) = s = 1 k 1 ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) + Δ k u 3 ( k , s ) ) h ( s ) + ( u 1 ( k , k 1 ) + u 2 ( k , k 1 ) + u 3 ( k , k 1 ) ) h ( k 1 ) + s = k η ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) ) h ( s ) ( u 1 ( k , k 1 ) + u 2 ( k , k 1 ) ) h ( k 1 ) + s = η + 1 N 2 Δ k u 1 ( k , s ) h ( s )

= s = 1 k 1 ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) + Δ k u 3 ( k , s ) ) h ( s ) + s = k η ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) ) h ( s ) + s = η + 1 N 2 Δ k u 1 ( k , s ) h ( s ) h ( η ) s = 1 k 1 ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) + Δ k u 3 ( k , s ) ) + s = k η ( Δ k u 1 ( k , s ) + Δ k u 2 ( k , s ) ) + s = η + 1 N 2 Δ k u 1 ( k , s ) = h ( η ) s = 1 N 2 Δ k u 1 ( k , s ) + s = 1 η Δ k u 2 ( k , s ) + s = 1 k 1 Δ k u 3 ( k , s ) = h ( η ) s = 1 N 2 α ( N s 1 ) 2 k N ( N 1 ) ρ Δ ρ k 2 2 α ( N 1 ) s = 1 N 2 ( N s ) ( N s 1 ) Δ ρ k 2 ρ + s = 1 η N ( N 1 ) ρ Δ ρ k 2 k 2 2 α ( N 1 ) + s = 1 k 1 ( k s ) .

If k 1 > η , then h ( k 1 ) < h ( η ) , so no matter either k 1 < η or k 1 > η , we always have

Δ u ( k ) h ( η ) s = 1 N 2 α ( N s 1 ) ( 2 k N ( N 1 ) ρ Δ ρ k ) 2 2 α ( N 1 ) s = 1 N 2 ( N s ) ( N s 1 ) Δ ρ k 2 ρ + s = 1 η N ( N 1 ) ρ Δ ρ k 2 k 2 2 α ( N 1 ) + s = 1 k 1 ( k s ) = h ( η ) 12 ρ ( 1 α ( N 1 ) ) [ 6 ρ k ( ( k 1 ) ( 1 α ( N 1 ) ) + α ( N 1 ) ( N 2 ) 2 η ) + Δ ρ k N ( N 1 ) ( 6 η ( N 2 ) ( α ( N 1 ) + 2 ) ) ] .

By using Remark 2.4, we can deduce that Δ u ( k ) 0 .

Moreover, if k 1 η , then

Δ 2 u ( k 1 ) = s = 1 k 1 ( Δ k 2 u 1 ( k 1 , s ) + Δ k 2 u 2 ( k 1 , s ) + Δ k 2 u 3 ( k 1 , s ) ) h ( s ) + s = k η ( Δ k 2 u 1 ( k 1 , s ) + Δ k 2 u 2 ( k 1 , s ) ) h ( s ) + s = η + 1 N 2 Δ k 2 u 1 ( k 1 , s ) h ( s ) = s = 1 k 1 α s 1 α ( N 1 ) ( α ( N s 1 ) 1 ) N ( N 1 ) Δ 2 ρ k 1 ( 2 2 α ( N 1 ) ) ρ ( N s ) ( N s 1 ) Δ 2 ρ k 1 2 ρ h ( s ) s = k η ( 1 α ( N s 1 ) ) ( 2 N ( N 1 ) Δ 2 ρ k 1 ρ ) 2 2 α ( N 1 ) h ( s ) s = k η ( N s ) ( N s 1 ) Δ 2 ρ k 1 2 ρ h ( s ) + s = η + 1 N 2 α ( N s 1 ) [ 2 N ( N 1 ) ρ Δ 2 ρ k 1 ] 2 2 α ( N 1 ) ( N s ) ( N s 1 ) Δ 2 ρ k 1 2 ρ h ( s ) h ( η ) s = 1 N 2 α ( N s 1 ) [ 2 N ( N 1 ) ρ Δ 2 ρ k 1 ] 2 2 α ( N 1 ) ( N s ) ( N s 1 ) Δ 2 ρ k 1 2 ρ + s = 1 η N ( N 1 ) ρ Δ 2 ρ k 1 2 2 2 α ( N 1 ) + s = 1 k 1 1 = h ( η ) 2 ρ ( 1 α ( N 1 ) ) ρ { 2 η + α ( N 1 ) ( N 2 ) + 2 ( k 1 ) ( 1 α ( N 1 ) ) } + Δ 2 ρ k 1 N ( N 1 ) { ( N 2 ) ( 1 α ( N 1 ) ) 3 α ( N 1 ) ( N 2 ) 2 + η } .

Since η > N 2 2 , it is enough to show that

Δ 2 u ( k 1 ) h ( η ) 2 ρ ( 1 α ( N 1 ) ) ρ { 2 η + α ( N 1 ) ( N 2 ) + 2 η ( 1 α ( N 1 ) ) } + Δ 2 ρ k 1 N ( N 1 ) { ( N 2 ) ( 1 α ( N 1 ) ) 3 α ( N 1 ) ( N 2 ) 2 + η } 0 .

Suppose that η k 2 N 2 . We have

u ( k ) = s = 1 η ( u 1 ( k , s ) + u 2 ( k , s ) + u 3 ( k , s ) ) h ( s ) + s = η + 1 N 2 u 1 ( k , s ) h ( s ) + s = η + 1 k 2 u 3 ( k , s ) h ( s ) .

We will show that Δ u ( k ) < 0 for η k 2 N 2 . Indeed, using the same arguments as mentioned earlier, we obtain that

Δ u ( k ) = s = 1 η α s k + s ( α ( N 1 ) 1 ) 1 α ( N 1 ) ( α ( N s 1 ) 1 ) N ( N 1 ) Δ ρ k ( 2 2 α ( N 1 ) ) ρ ( N s ) ( N s 1 ) Δ ρ k 2 ρ h ( s ) + s = η + 1 k 1 ( k s ) h ( s ) + s = η + 1 N 2 α ( N s 1 ) [ 2 k N ( N 1 ) ρ Δ ρ k ] 2 2 α ( N 1 ) ( N s ) ( N s 1 ) Δ ρ k 2 ρ h ( s ) h ( η ) 12 ρ ( 1 α ( N 1 ) ) { 6 ρ k ( ( k 1 ) ( 1 α ( N 1 ) ) + α ( N 1 ) ( N 2 ) 2 η ) + Δ ρ k N ( N 1 ) [ 6 η ( N 2 ) ( α ( N 1 ) + 2 ) ] } 0 .

So, Δ u ( k ) 0 for k [ 0 , N 1 ] Z , which implies that u ( k ) is nonincreasing. Since u ( N ) = 0 , we have u ( k ) 0 , k [ 0 , N ] Z . For k [ 1 , η ] Z , Δ 2 u ( k 1 ) 0 , we found that u is concave on [ 0 , η + 1 ] Z .□

Lemma 2.6

Let ( H 0 ) hold. Assume that h K 0 and u is the solution of (2.1). Then

min k [ N θ , θ ] Z u ( k ) θ u ,

where θ = η + 1 θ η + 1 , θ N 2 + 1 , η + 1 Z .

Proof

It follows from Lemma 2.5 that u ( k ) is concave on [ 0 , η + 1 ] Z . Therefore,

u ( k ) u ( 0 ) k u ( η + 1 ) u ( 0 ) η + 1 , k [ 0 , η + 1 ] Z .

Meanwhile, u ( k ) is nonincreasing on [ 0 , N ] Z , which implies that u ( 0 ) = u . Therefore,

u ( k ) η + 1 k η + 1 u ( 0 ) = η + 1 k η + 1 u .

Thus,

min k [ N θ , θ ] Z u ( k ) = u ( θ ) η + 1 θ η + 1 u .

3 Main results

In this section, we consider the existence of at least one positive solution of the problem (1.1). Assume that

( H 1 ) f : [ 1 , N 2 ] Z × [ 0 , ) [ 0 , ) is continuous, the mapping k f ( k , u ) is decreasing for each u [ 0 , ) and the mapping u f ( k , u ) is increasing for each k [ 1 , N 2 ] Z ;

( H 2 ) a : [ 1 , N 2 ] Z [ 0 , + ) is decreasing.

Define the cone K by

K = { u K 0 min k [ N θ , θ ] Z u ( k ) θ u }

and the operator T λ : K E by

T λ u ( k ) = λ s = 1 N 2 G ( k , s ) a ( s ) f ( s , u ( s ) ) .

Lemma 3.1

T λ : K K is a completely continuous operator.

Proof

It is obvious that T λ : K E is a completely continuous operator. Now, let us prove that T λ : K K . We will show that for any u K , we have T λ u K .

Let u K . Then u K 0 , which gives us that Δ u ( k ) 0 . In other words u is decreasing. Then, by (H1), we have that f ( k , u ) is also decreasing on k . Denote y ( k ) a ( k ) f ( k , u ( k ) ) . By using (H1) and (H2), we obtain that y ( k ) 0 and y ( k ) is decreasing. Thus, y K 0 .

Moreover, by using the definition of T λ , one can check that

Δ 3 ( T λ u ) ( k 1 ) = λ y ( k ) , k [ 1 , N 2 ] Z

and

Δ 2 ( T λ u ) ( η ) = α Δ ( T λ u ) ( N 1 )