Home Existence of positive solutions of discrete third-order three-point BVP with sign-changing Green's function
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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
Published/Copyright: October 13, 2022
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Open Mathematics
From the journal Open Mathematics Volume 20 Issue 1

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.

Keywords: third-order difference equation; three-point boundary conditions; sign-changing Green’s function; Guo-Krasnoselskii’s fixed-point theorem
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 ) , Δ ( T λ u ) ( 0 ) = − β ( T λ u ) ( 0 ) , ( T λ u ) ( N ) = 0 .

Therefore, T λ satisfies problem (2.1). Now, using similar arguments to the ones given in the proof of Lemma 2.5 and the fact that y ∈ K 0 , we know that T λ u ∈ K 0 and T λ u is concave on [ 0 , η + 1 ] Z .

Furthermore, by Lemma 2.6 and T λ u ∈ K 0 , we deduce that

min k ∈ [ N − θ , θ ] Z ( T λ u ) ( k ) ≥ θ ∗ ∥ T λ u ∥ .

Therefore, T λ u ∈ K and T λ : K → K is a completely continuous operator.□

Set

τ = max η ( 2 N − η − 1 ) − α η ( N − η − 1 ) ( N − 1 ) ρ , ( N − η − 1 ) ( α ( N − 1 ) η + ( N − η ) ) ρ , A = ∑ s = 1 N − 2 τ a ( s ) , B = ∑ s = N − θ θ G ( N − θ , s ) a ( s ) .

Theorem 3.2

Suppose that ( H 0 ) , ( H 1 ) , and ( H 2 ) hold. If there exist two positive constants r and R with r ≠ R such that

( A 1 ) f ( k , u ) ≤ r λ A , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ 0 , r ] ,

( A 2 ) f ( k , u ) ≥ R λ B , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ θ ∗ R , R ] ,

then problem (1.1) has at least one positive solution u ∈ K with min { r , R } ≤ ‖ u ‖ ≤ max { r , R } .

Proof

We only deal with the case r < R , since the case that r > R could be treated similarly.

Let Ω 1 = { u ∈ E : ‖ u ‖ < r } . From Lemma 2.3 (ii), we know that G ( k , s ) ≤ 0 for s ∈ [ η + 1 , N − 2 ] Z and G ( k , s ) ≥ 0 for s ∈ [ 1 , η ] Z . Then by ( A 1 ) , for u ∈ K ∩ ∂ Ω 1 , it is obvious that

‖ T λ u ‖ = λ max k ∈ [ 0 , N ] Z ∑ s = 1 N − 2 G ( k , s ) a ( s ) f ( s , u ( s ) ) ≤ λ max k ∈ [ 0 , N ] Z ∑ s = 1 η G ( k , s ) a ( s ) f ( s , u ( s ) ) + λ max k ∈ [ 0 , N ] Z ∑ s = η + 1 N − 2 G ( k , s ) a ( s ) f ( s , u ( s ) ) ≤ λ ∑ s = 1 η η ( 2 N − η − 1 ) − α η ( N − η − 1 ) ( N − 1 ) ρ a ( s ) f ( s , u ( s ) ) + λ ∑ s = η + 1 N − 2 ( N − η − 1 ) ( α ( N − 1 ) η + ( N − η ) ) ρ a ( s ) f ( s , u ( s ) ) ≤ λ ∑ s = 1 N − 2 τ a ( s ) f ( s , u ( s ) ) ≤ r .

Therefore,

(3.1) ‖ T λ u ‖ ≤ ‖ u ‖ , u ∈ K ∩ ∂ Ω 1 .

Let Ω 2 = { u ∈ E : ‖ u ‖ < R } . For u ∈ K ∩ ∂ Ω 2 and k ∈ [ N − θ , θ ] Z , we have

(3.2) T λ u ( N − θ ) = λ ∑ s = 1 N − 2 G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) ≥ λ ∑ s = N − θ θ G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) .

In fact,

∑ s = 1 N − θ − 1 G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) + ∑ s = θ + 1 N − 2 G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) ≥ ∑ s = 1 N − θ − 1 G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) + ∑ s = η + 1 N − 2 G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) ≥ a ( η ) f ( η , u ( η ) ) ∑ s = 1 N − θ − 1 G ( N − θ , s ) + ∑ s = η + 1 N − 2 G ( N − θ , s ) .

Furthermore,

∑ s = 1 N − θ − 1 G ( N − θ , s ) + ∑ s = η + 1 N − 2 G ( N − θ , s ) ≥ ∑ s = 1 N − θ − 1 α ( N − s − 1 ) [ ( N − θ ) ( N − θ − 1 ) − ρ N − θ ρ N ( N − 1 ) ] 2 − 2 α ( N − 1 ) − ∑ s = 1 N − θ − 1 ρ N − θ ( N − s ) ( N − s − 1 ) 2 ρ + ∑ s = 1 N − θ − 1 ρ N − θ ρ N ( N − 1 ) − ( N − θ ) ( N − θ − 1 ) 2 − 2 α ( N − 1 ) + ∑ s = 1 N − θ − 1 ( N − θ − s ) ( N − θ − s − 1 ) 2 + ∑ s = θ N − 2 α ( N − s − 1 ) [ ( N − θ ) ( N − θ − 1 ) − ρ N − θ ρ N ( N − 1 ) ] 2 − 2 α ( N − 1 ) − ∑ s = θ N − 2 ρ N − θ ( N − s ) ( N − s − 1 ) 2 ρ = ρ N − θ N ( N − 1 ) − ( N − θ ) ( N − θ − 1 ) ρ ( 2 − 2 α ( N − 1 ) ) ρ [ 2 − α ( N − 2 + θ ) ] ( N − θ − 1 ) 2 − ρ N − θ ( N − θ − 1 ) 2 ρ N ( N − 1 ) − ( N − θ ) ( 2 N + θ − 1 ) 3 + ( N − θ ) ( N − θ − 1 ) ( N − θ − 2 ) 6 − ρ N − θ ( N − θ ) ( N − θ + 1 ) ( N − θ − 1 ) 6 ρ + ( N − θ ) ( N − θ − 1 ) ρ − N ( N − 1 ) ρ N − θ ( 2 − 2 α ( N − 1 ) ) ρ α ( N − θ ) ( N − θ − 1 ) 2

= ρ N − θ N ( N − 1 ) − ( N − θ ) ( N − θ − 1 ) ρ 2 ρ ( N − θ − 1 ) − ρ N − θ ( N − θ − 1 ) 2 ρ N ( N − 1 ) − ( N − θ ) ( 2 N + θ − 1 ) 3 + ( N − θ ) ( N − θ − 1 ) ( N − θ − 2 ) 6 − ρ N − θ ( N − θ ) ( N − θ + 1 ) ( N − θ − 1 ) 6 ρ = ρ N − θ 6 ρ ( N − θ − 1 ) ( N − θ ) ( N + 2 θ − 2 ) + ( N − θ ) ( N − θ − 1 ) ( − 2 N + 2 θ + 1 ) 6 ≥ ( N − θ − 1 ) ( N − θ ) 6 ( − N + 4 θ − 1 ) ≥ 0 .

Therefore, (3.2) holds, which implies that

T λ u ( N − θ ) ≥ λ ∑ s = N − θ θ G ( N − θ , s ) a ( s ) R λ B = R .

So, for u ∈ K ∩ ∂ Ω 2 , we have

(3.3) ‖ T λ u ‖ ≥ ‖ u ‖ .

From Theorem 1.1(i), (3.1), and (3.3), we found that T λ has a fixed point u ∈ K ∩ ( Ω ¯ 2 \ Ω 1 ) , and then u is a positive solution of (1.1) with r ≤ ‖ u ‖ ≤ R .□

Theorem 3.3

Suppose that ( H 0 ) , ( H 1 ) , and ( H 2 ) hold. If one of the following conditions holds:

( A 3 ) f 0 ≔ lim u → 0 + max k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = 0 , f ∞ ≔ lim u → ∞ min k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = ∞ , ( superlinear case ) , or ( A 4 ) f 0 ≔ lim u → 0 + min k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = ∞ , f ∞ ≔ lim u → ∞ max k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = 0 , ( sublinear case ) ,

then for any λ ∈ ( 0 , ∞ ) , problem (1.1) has at least one positive solution.

Proof

(Superlinear case). From ( A 3 ) , f 0 = 0 , then there exists a constant R 1 > 0 such that

f ( k , u ) ≤ R 1 λ A , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ 0 , R 1 ] .

Furthermore, since f ∞ = ∞ , there exists R 2 > R 1 such that

f ( k , u ) ≥ u θ ∗ λ B ≥ θ ∗ R 2 θ ∗ λ B = R 2 λ B , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ θ ∗ R 2 , R 2 ] .

Now, by Theorem 3.2, problem (1.1) has a positive solution u ∈ K .

(Sublinear case). On the one hand, since f 0 = ∞ , there exists r 1 > 0 such that

f ( k , u ) ≥ u θ ∗ λ B , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ 0 , r 1 ] .

Set Ω 1 = { u ∈ E : ‖ u ‖ < r 1 } . If u ∈ K ∩ ∂ Ω 1 , then

min s ∈ [ N − θ , θ ] Z u ( s ) ≥ θ ∗ ‖ u ‖ = θ ∗ r 1 .

We obtain

T λ u ( N − θ ) = λ ∑ s = 1 N − 2 G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) ≥ λ ∑ s = N − θ θ G ( N − θ , s ) a ( s ) u ( s ) θ ∗ λ B ≥ r 1 .

This implies that

‖ T λ u ‖ ≥ ‖ u ‖ , u ∈ K ∩ ∂ Ω 1 .

On the other hand, since f ∞ = 0 , there exists r 2 > 0 such that

f ( k , u ) ≤ u λ A , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ r 2 , ∞ ) .

We consider two cases: f is bounded and f is unbounded. If f is bounded, i.e., there exists a constant M > 0 such that f ≤ M , then we take r 3 = max { 2 r 2 , λ M A } . If f is unbounded, then we take r 3 > max { 2 r 1 , r 2 } such that f ( k , u ) ≤ f ( k , r 2 ) , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ 0 , r 2 ] . Set Ω 2 = { u ∈ E : ‖ u ‖ < r 3 } . Now, similar to the proof of (3.1), we obtain

‖ T λ u ‖ ≤ ‖ u ‖ , u ∈ K ∩ ∂ Ω 2 .

Therefore, by Theorem 1.1, we obtain a positive solution u ∈ K of problem (1.1).□

Theorem 3.4

Suppose that ( H 0 ) , ( H 1 ) , and ( H 2 ) hold. If 0 < A f 0 < θ ∗ B f ∞ < ∞ , then for each λ ∈ 1 θ ∗ B f ∞ , 1 A f 0 , problem (1.1) has at least one positive solution.

Proof

For any λ ∈ 1 θ ∗ B f ∞ , 1 A f 0 , there exists ε > 0 such that

(3.4) 1 θ ∗ B ( f ∞ − ε ) ≤ λ ≤ 1 A ( f 0 + ε ) .

By the definition of f 0 , there exists r 4 > 0 such that f ( k , u ) ≤ ( f 0 + ε ) u , for ( k , u ) ∈ [ 1 , N − 2 ] Z × [ 0 , r 4 ] . Let Ω 3 = { u ∈ E : ‖ u ‖ < r 4 } , using similar arguments to these given in the proof of (3.1), we can deduce that

‖ T λ u ‖ ≤ λ ∑ s = 1 N − 2 τ a ( s ) ( f 0 + ε ) u .

Furthermore, by (3.4), we obtain

‖ T λ u ‖ ≤ ‖ u ‖ , u ∈ K ∩ ∂ Ω 3 .

By the definition of f ∞ , there exists R 3 such that f ( k , u ) ≥ ( f ∞ − ε ) u , ( k , u ) ∈ [ 1 , N − 2 ] Z × [ R 3 , ∞ ) . Let R 4 = max { 2 r 4 , R 3 θ ∗ } and Ω 4 = { u ∈ E : ‖ u ‖ < R 4 } . If u ∈ K with ‖ u ‖ = R 4 , then min s ∈ [ N − θ , θ ] Z u ( s ) ≥ θ ∗ ‖ u ‖ . Therefore, similar to the discussion from (3.2) and (3.3), we obtain

T λ u ( N − θ ) ≥ λ ∑ s = N − θ θ G ( N − θ , s ) ( f ∞ − ε ) θ ∗ ‖ u ‖ a ( s ) = λ θ ∗ B ( f ∞ − ε ) ‖ u ‖ .

By (3.9), we obtain

(3.5) ‖ T λ u ‖ ≥ ‖ u ‖ , u ∈ K ∩ ∂ Ω 4 .

By Theorem 1.1 (i), problem (1.1) has at least one positive solution u ∈ K .□

Similar to the discussion of Theorems 3.2–3.4, we could obtain the following two theorems. So, we just state them here without any proof.

Theorem 3.5

Suppose that ( H 0 ) , ( H 1 ) , and ( H 2 ) hold. If 0 < A f ∞ < θ ∗ B f 0 < ∞ , then for each λ ∈ 1 θ ∗ B f 0 , 1 A f ∞ , problem (1.1) has at least one positive solution.

Theorem 3.6

Suppose ( H 0 ) , ( H 1 ) , and ( H 2 ) hold. Then the following results hold.

  1. If f ∞ = ∞ , 0 < f 0 < ∞ , then for each λ ∈ 0 , 1 A f 0 , problem (1.1) has at least one positive solution.

  2. If f 0 = ∞ , 0 < f ∞ < ∞ , then for each λ ∈ 0 , 1 A f ∞ , problem (1.1) has at least one positive solution.

  3. If f 0 = 0 , 0 < f ∞ < ∞ , then for each λ ∈ 1 θ ∗ B f ∞ , ∞ , problem (1.1) has at least one positive solution.

  4. If f ∞ = 0 , 0 < f 0 < ∞ , then for each λ ∈ 1 θ ∗ B f 0 , ∞ , problem (1.1) has at least one positive solution.

Theorem 3.7

Assume that ( H 0 ) , ( H 1 ) , and ( H 2 ) hold. If

( F 1 ) f 0 ≔ lim u → 0 + min k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = + ∞ , f ∞ ≔ lim u → ∞ min k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = + ∞ , and

( F 2 ) There exists a constant p > 0 such that f ( s , u ) ≤ δ p for 0 ≤ u ≤ p and s ∈ [ 1 , N − 2 ] Z , where

δ = λ ∑ s = 1 N − 2 τ a ( s ) − 1 ,

then problem ( 1.1 ) has at least two positive solutions u 1 and u 2 with 0 ≤ ‖ u 1 ‖ ≤ p ≤ ‖ u 2 ‖ .

Proof

Since f 0 = + ∞ , there exists a constant r with 0 < r < p such that f ( t , u ) ≥ M u for 0 ≤ u ≤ r , where M > 0 satisfies

λ M θ ∗ ∑ s = N − θ θ G ( N − θ , s ) a ( s ) ≥ 1 .

Let K r = { u ∈ K : ‖ u ‖ < r } . Then, for ∀ u ∈ ∂ K r , we have

T λ u ( N − θ ) = λ ∑ s = 1 N − 2 G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) ≥ λ ∑ s = N − θ θ G ( N − θ , s ) a ( s ) f ( s , u ( s ) ) ≥ λ M θ ∗ ∑ s = N − θ θ G ( N − θ , s ) a ( s ) ‖ u ‖ ≥ ‖ u ‖ .

Since f ∞ = + ∞ , there exists a constant R 1 ≥ 0 such that f ( t , u ) ≥ M u for u ≥ R 1 . Let K R = { u ∈ K : ‖ u ‖ < R } . Choose R ≥ max p , R 1 θ ∗ , then for u ∈ ∂ K R , min t ∈ [ N − θ , θ ] Z u ( t ) ≥ θ ∗ ‖ u ‖ > R 1 . Similar to the aforementioned proof, we have ‖ T u ‖ ≥ ‖ u ‖ for u ∈ ∂ K R . Let K p = { u ∈ K : ‖ u ‖ < p } . From ( F 2 ) , for u ∈ ∂ K p ,

‖ T λ u ‖ = λ max k ∈ [ 0 , N ] Z ∣ ∑ s = 1 N − 2 G ( k , s ) a ( s ) f ( s , u ( s ) ) ∣ ≤ λ ∑ s = 1 N − 2 τ a ( s ) f ( s , u ( s ) ) ≤ ‖ u ‖ .

Therefore, for u ∈ ∂ K p , ‖ T u ‖ ≤ ‖ u ‖ . So, T has a fixed point u 1 in K p \ K ˚ r and anther fixed point u 2 in K R \ K ˚ p . Therefore, problem ( 1.1 ) has at least two positive solutions u 1 and u 2 with 0 ≤ ‖ u 1 ‖ ≤ p ≤ ‖ u 2 ‖ .□

Similar to the discussion of Theorem 3.7, we could obtain the following theorem. So, we just state it here without any proof.

Theorem 3.8

Assume that ( H 0 ) , ( H 1 ) , and ( H 2 ) hold. If

( L 1 ) f 0 ≔ lim u → 0 + max k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = 0 , f ∞ ≔ lim u → ∞ max k ∈ [ 1 , N − 2 ] Z f ( k , u ) u = 0 , and

( L 2 ) there exists a constant q > 0 such that f ( s , u ) > β q for θ ∗ q ≤ u ≤ q and s ∈ [ 1 , N − 2 ] Z , where

β = λ θ ∗ ∑ s = N − θ θ G ( N − θ , s ) a ( s ) − 1 ,

then problem (1.1) has at least two solutions u 1 and u 2 with 0 ≤ ‖ u 1 ‖ ≤ q ≤ ‖ u 2 ‖ .

4 Example

Example 4.1

Consider the following discrete third-order three-point BVP:

(4.1) Δ 3 u ( k − 1 ) = λ a ( k ) f ( k , u ( k ) ) , k ∈ [ 1 , 9 ] Z , Δ 2 u ( η ) = 1 12 Δ u ( 10 ) , Δ u ( 0 ) = − 1 44 u ( 0 ) , u ( 11 ) = 0 ,

where λ = 4 5 , a ( k ) = 10 − k 10 , f ( k , u ) = u 3 + 20 − 2 k . Now, according to Remark 2.4, let η = 6 . Furthermore, we obtain the expression of G ( t , s ) as follows.

If s > 6 , then

G ( k , s ) = ( 10 − s ) ( 77 − s ) 44 k 2 + ( 10 − s ) ( 187 − 3 s ) 44 k − ( 10 − s ) ( 264 − 4 s ) , s > k − 2 , s 2 − 87 s + 792 44 k 2 + 3 s 2 − 261 s + 1848 44 k − 7 2 s 2 + 609 2 s − 2640 , s ≤ k − 2 .

If s ≤ 6 , then

G ( k , s ) = s 2 − 87 s − 22 44 k 2 + 3 s 2 − 217 s + 22 44 k − 4 s 2 + 304 s , s > k − 2 , s 2 − 87 s 44 k 2 + 3 s 2 − 261 s 44 k − 7 2 s 2 + 609 2 s , s ≤ k − 2 .

Obviously, if ( k , s ) ∈ [ 0 , 11 ] Z × [ 1 , 6 ] Z , then Δ k G ( k , s ) ≤ 0 . This implies that G ( k , s ) is nonincreasing with respect to k in this case. Hence, max k ∈ [ 0 , N ] Z G ( k , s ) = G ( 0 , s ) = − 4 s + 304 s ≤ 1,680 , min k ∈ [ 0 , N ] Z G ( k , s ) = G ( N , s ) = 0 and G ( k , s ) ≥ 0 . If ( k , s ) ∈ [ 0 , 11 ] Z × [ 7 , 9 ] Z , then Δ k G ( k , s ) ≥ 0 . This implies that G ( k , s ) is nondecreasing with respect to k in this case. Therefore, min k ∈ [ 0 , N ] Z G ( k , s ) = G ( 0 , s ) = − ( 10 − s ) ( 264 − 4 s ) ≥ − 960 , max k ∈ [ 0 , N ] Z G ( k , s ) = G ( N , s ) = 0 , and G ( k , s ) ≤ 0 .

Now, by direct calculation, we choose θ = 6 , then θ ∗ = η + 1 − θ η + 1 = 1 7 . So

τ = max { 1,680 , 960 } = 1,680 , A = ∑ s = 1 N − 2 τ a ( s ) = 1,680 ∑ s = 1 9 10 − s 10 = 7,560 , B = ∑ s = N − θ θ G ( N − θ , s ) a ( s ) = ∑ s = 5 6 25 ( s 2 − 87 s − 22 ) 44 + 5 ( 3 s 2 − 217 s + 22 ) 44 − 4 s 2 + 304 s 10 − s 10 = 22,757 22 .

We choose R = 448 , r = 1,000,000 , and from Theorem 3.2, problem (4.1) has at least one positive solution.

Example 4.2

Consider the following discrete third-order three-point BVP:

(4.2) Δ 3 u ( k − 1 ) = λ a ( k ) f ( k , u ( k ) ) , k ∈ [ 1 , 5 ] Z , Δ 2 u ( η ) = 1 7 Δ u ( 6 ) , Δ u ( 0 ) = − 1 56 u ( 0 ) , u ( 7 ) = 0 ,

where η = 4 , a ( k ) = 9 − k 7 , f ( k , u ) = ( 8 − k ) u 2 log 3 ( u + 1 ) 11 . By direct computation, we have f 0 = 0 , f ∞ = ∞ . Furthermore, by Theorem 3.3, for λ ∈ ( 0 , ∞ ) , (4.2) has at least one positive solution.

Example 4.3

In this example, we continue to discuss problem ( 4.1 ) with

f ( k , u ) = u 2 + ( 10 − k ) 6,323 , ( k , u ) ∈ [ 1 , 9 ] Z × [ 0 , 1 ] , u 3 + ( 10 − k ) 6,323 , ( k , u ) ∈ [ 1 , 9 ] Z × [ 1 , ∞ ) .

We take a ( k ) = 10 − k 10 , λ = 1 120 , and η = 6 . Then

δ = λ ∑ s = 1 N − 2 τ a ( s ) − 1 = 1 63 .

Furthermore, if we choose p = 1 , then f ( k , u ) ≤ δ p for 0 ≤ u ≤ p . From Theorem 3.7, problem ( 4.1 ) has at least two positive solutions u 1 and u 2 with 0 ≤ ‖ u 1 ‖ ≤ p ≤ ‖ u 2 ‖ .

Acknowledgements

The authors are very grateful to the editor and the reviewers for their valuable suggestions.

  1. Funding information: Huijuan Li and Chenghua Gao were supported by National Natural Science Foundation of China (No. 11961060) and Natural Science Foundation of Gansu Province (No. 18JR3RA084). Nikolay D. Dimitrov was supported by the Bulgarian National Science Foundation under Project DN 12/4 “Advanced analytical and numerical methods for nonlinear differential equations with applications in finance and environmental pollution,” 2017.

  2. Author contributions: All authors contributed equally to the writing of this article. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors state no conflict of interest.

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Received: 2021-11-29
Revised: 2022-04-28
Accepted: 2022-09-13
Published Online: 2022-10-13

© 2022 Huijuan Li et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2022-0510
Keywords for this article
third-order difference equation; three-point boundary conditions; sign-changing Green’s function; Guo-Krasnoselskii’s fixed-point theorem
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Articles in the same Issue

  1. Regular Articles
  2. A random von Neumann theorem for uniformly distributed sequences of partitions
  3. Note on structural properties of graphs
  4. Mean-field formulation for mean-variance asset-liability management with cash flow under an uncertain exit time
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  135. Multiple periodic solutions for discrete boundary value problem involving the mean curvature operator
  136. Special Issue on Evolution Equations, Theory and Applications (Part II)
  137. Coupled measure of noncompactness and functional integral equations
  138. Existence results for neutral evolution equations with nonlocal conditions and delay via fractional operator
  139. Global weak solution of 3D-NSE with exponential damping
  140. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  141. Ground state solutions of nonlinear Schrödinger equations involving the fractional p-Laplacian and potential wells
  142. A class of p1(x, ⋅) & p2(x, ⋅)-fractional Kirchhoff-type problem with variable s(x, ⋅)-order and without the Ambrosetti-Rabinowitz condition in ℝN
  143. Jensen-type inequalities for m-convex functions
  144. Special Issue on Problems, Methods and Applications of Nonlinear Analysis (Part III)
  145. The influence of the noise on the exact solutions of a Kuramoto-Sivashinsky equation
  146. Basic inequalities for statistical submanifolds in Golden-like statistical manifolds
  147. Global existence and blow up of the solution for nonlinear Klein-Gordon equation with variable coefficient nonlinear source term
  148. Hopf bifurcation and Turing instability in a diffusive predator-prey model with hunting cooperation
  149. Efficient fixed-point iteration for generalized nonexpansive mappings and its stability in Banach spaces
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