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Positive solutions of a discrete nonlinear third-order three-point eigenvalue problem with sign-changing Green's function

  • Xueqin Cao , Chenghua Gao EMAIL logo and Duihua Duan
Published/Copyright: September 1, 2021
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Open Mathematics
From the journal Open Mathematics Volume 19 Issue 1

Abstract

In this paper, we discuss the existence of positive solutions to a discrete third-order three-point boundary value problem. Here, the weight function a ( t ) and the Green function G ( t , s ) both change their sign. Despite this, we also obtain several existence results of positive solutions by using the Guo-Krasnoselskii’s fixed-point theorem in a cone.

Keywords: third-order difference equation; three-point boundary conditions; sign-changing Green’s function; Guo-Krasnoselskii’s fixed-point theorem; positive solutions
MSC 2010: 39A10; 39A12

1 Introduction

Let [ a , b ] Z denote the integer set { a , a + 1 , … , b } with b > a . In this paper, we consider the existence of positive solutions for the discrete nonlinear third-order three-point boundary value problem (BVP)

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

where T > 5 is an integer, λ > 0 is a parameter, f : [ 1 , T − 1 ] Z × [ 0 , + ∞ ) → [ 0 , + ∞ ) is continuous, a : [ 1 , T − 1 ] Z → ( − ∞ , + ∞ ) , η ∈ 1 , T − 1 3 Z , α ∈ 0 , 1 T + 1 and β ∈ 0 , 2 2 ( T + 1 ) − α T ( T + 1 ) .

Difference equations arise in various fields, such as physics, applied mathematics and economy [1]. Positive solutions of a discrete problem have been discussed for a long time and several excellent results on this problem have been obtained, see [1,2,3, 4,5,6, 7,8,9, 10,11,12, 13,14] and references therein. For instance, Ma et al. [3] considered the existence of one-signed solutions of discrete second-order periodic BVP. Lu [4] considered positive periodic solutions of nonlinear first-order functional difference equations. Later, Jin and Luo [5] considered the existence of positive solutions of discrete second-order BVPs with fully nonlinear term. In recent years, by using fixed-point theory, upper and lower solution methods and critical point theory, the existence of positive solutions of the BVPs for third-order three-point equations has been discussed by several authors, see [15,16,17, 18,19,20]. Particularly, Agarwal and Henderson [21] considered the existence of positive solutions for discrete equations by using fixed-point theory in a cone. Later, Kong et al. [22] considered the positive solutions of BVPs for third-order functional difference equations. Gao [23] by using critical point theory considered the existence of positive solutions for difference equations. It is worth noting that the above literature are all positive solutions of the third-order three-point BVP with Green’s function G ( t , s ) ≥ 0 . For Green’s function G ( t , s ) changing sign case, can we get the existence of a positive solution? In fact, by using the Guo-Krasnoselskii’s fixed-point theorem in a cone, there are also several excellent results on the existence of positive solutions of third-order equations with three-point BVPs, see [24,25, 26,27,28, 29,30] and references therein. Particularly, Wang and Gao [27] discussed the existence of positive solutions of the following third-order difference equation BVP

(1.2) Δ 3 u ( t − 1 ) + a ( t ) f ( t , u ( t ) ) = 0 , t ∈ [ 1 , T − 1 ] Z , u ( 0 ) = Δ u ( T ) = Δ 2 u ( η ) = 0 ,

where a : [ 1 , T − 1 ] Z → ( 0 , + ∞ ) and Green’s function changes its sign. Furthermore, in 2016, Gao and Geng [29] considered positive solutions of the following a discrete nonlinear third-order three-point eigenvalue problem

(1.3) Δ 3 u ( t − 1 ) = λ a ( t ) f ( t , u ( t ) ) , t ∈ [ 1 , T − 2 ] Z , Δ u ( 0 ) = u ( T ) = Δ 2 u ( η ) = 0 ,

which sign-changing Green’s function. Later, Xu et al. [30] studied positive solutions of the following BVP:

(1.4) Δ 3 u ( t − 1 ) = λ a ( t ) f ( t , u ( t ) ) , t ∈ [ 1 , T − 2 ] Z , Δ u ( 0 ) = u ( T ) = 0 , Δ 2 u ( η ) − α Δ u ( T ) = 0 .

However, the above literature are considered the case that the weight function a ( t ) > 0 , when the weight function a ( t ) changes its sign, can we get the existence of a positive solution? Inspired by the work of above papers, we try to study a more general three-point BVP.

The rest of the present article is organized as follows. In Section 2, we study the linear problem, we deduce Green’s function G ( t , s ) changes its sign, this together with the sign-changing function a ( t ) will take lots of difficulties for us to obtain the existence of positive solutions. Meanwhile, we will point out that the condition for η to obtain the existence of positive solution of (1.1). In Section 3, we impose some conditions to obtain the existence of at least positive solution to the problem (1.1). The main tool we will use is the following fixed-point theorem in a cone.

Theorem 1.1

[31] Let E be a Banach space and K a cone in E . Assume that Ω 1 and Ω 2 are bounded open subsets of E such that 0 ∈ Ω 1 , Ω ¯ 1 ⊂ Ω 2 , and A : K ∩ ( Ω ¯ 2 \ Ω 1 ) → K is a completely continuous operator such that either

  1. ‖ A u ‖ ≤ ‖ u ‖ for u ∈ K ∩ ∂ Ω 1 and ‖ A u ‖ ≥ ‖ u ‖ for u ∈ K ∩ ∂ Ω 2 , or

  2. ‖ A u ‖ ≥ ‖ u ‖ for u ∈ K ∩ ∂ Ω 1 and ‖ A u ‖ ≤ ‖ u ‖ for u ∈ K ∩ ∂ Ω 2 .

Then A has a fixed-point in K ∩ ( Ω ¯ 2 \ Ω 1 ) .

2 Linear problem

At first, let us consider the following linear problem:

(2.1) Δ 3 u ( t − 1 ) = − y ( t ) , t ∈ [ 1 , T − 1 ] Z , u ( 0 ) = 0 , Δ 2 u ( η ) = α Δ u ( T ) , Δ u ( T ) = β u ( T + 1 ) .

Define Green’s function G ( t , s ) as follows.

If s > η , then

(2.2) G ( t , s ) = α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) t , t − 2 < s , 2 β ( T + 1 ) − 2 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 − s 2 2 − s 2 + 4 T + 2 − 2 β ( T + 1 ) 2 + ( α β − 2 β ( 1 − α T ) ) ( s 2 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t , s ≤ t − 2 .

If s ≤ η , then

(2.3) G ( t , s ) = 2 − 2 β ( T + 1 ) + α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + − 2 ( 2 T + 1 ) + 2 β ( T + 1 ) 2 + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) t , t − 2 < s , − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t − s 2 2 − s 2 , s ≤ t − 2 .

Remark 2.1

We point out that under conditions for α and β , we have 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) > 0 , Moreover, if β = 0 , α = 0 , we obtain the expression given in [27].

Lemma 2.2

The problem (2.1) has a unique solution

(2.4) u ( t ) = ∑ s = 1 T − 1 G ( t , s ) y ( s ) ,

where G ( t , s ) is defined in (2.2) and (2.3).

Proof

Summing from s = 1 to s = t − 1 at both sides of the equation (2.1), then we get

Δ 2 u ( t − 1 ) = Δ 2 u ( 0 ) − ∑ s = 1 t − 1 y ( s ) .

Repeating the above process, we obtain

Δ u ( t − 1 ) = Δ u ( 0 ) + ( t − 1 ) Δ 2 u ( 0 ) − ∑ s = 1 t − 2 ( t − s − 1 ) y ( s ) .

Summing from s = 1 to s = t at both sides of the above equation, we have

u ( t ) = t Δ u ( 0 ) + t ( t − 1 ) 2 Δ 2 u ( 0 ) − ∑ s = 1 t − 2 ( t − s ) ( t − s − 1 ) 2 y ( s ) .

By using the boundary condition Δ 2 u ( η ) = α Δ u ( T ) , we get that

Δ 2 u ( 0 ) − ∑ s = 1 η y ( s ) = α Δ u ( 0 ) + T Δ 2 u ( 0 ) − ∑ s = 1 T − 1 ( T − s ) y ( s ) ,

hence

Δ u ( 0 ) = 1 − α T α Δ 2 u ( 0 ) − 1 α ∑ s = 1 η y ( s ) + ∑ s = 1 T − 1 ( T − s ) y ( s ) .

Then, by using Δ u ( T ) = β u ( T + 1 ) , we obtain that

Δ u ( 0 ) + T Δ 2 u ( 0 ) − ∑ s = 1 T − 1 ( T − s ) y ( s ) = β ( T + 1 ) Δ u ( 0 ) + T ( T + 1 ) 2 Δ 2 u ( 0 ) − ∑ s = 1 T − 1 ( T − s ) ( T + 1 − s ) 2 y ( s ) .

Thus,

Δ 2 u ( 0 ) = 2 − 2 β ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 η y ( s ) + α β 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 T − 1 ( T − s ) ( T + 1 + s ) y ( s ) , Δ u ( 0 ) = 1 − α T α 2 − 2 β ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 η y ( s ) − 1 α ∑ s = 1 η y ( s ) + β ( 1 − α T ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 T − 1 ( T − s ) ( T + 1 + s ) y ( s ) + ∑ s = 1 T − 1 ( T − s ) y ( s ) .

Therefore,

(2.5) u ( t ) = 1 − β ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) t 2 ∑ s = 1 η y ( s ) + ( 1 − β ( T + 1 ) ) ( 2 − 2 α T − α ) α ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t ∑ s = 1 η y ( s ) − 1 α t ∑ s = 1 η y ( s ) + α β 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 ∑ s = 1 T − 1 ( T − s ) ( T + 1 + s ) y ( s ) + t ∑ s = 1 T − 1 ( T − s ) y ( s ) + 2 β ( 1 − α T ) − α β 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t ∑ s = 1 T − 1 ( T − s ) ( T + 1 + s ) y ( s ) − ∑ s = 1 t − 2 ( t − s ) ( t − s − 1 ) 2 y ( s ) .

This implies that (2.4) holds.□

Lemma 2.3

Assume η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z , β ∈ 0 , 2 2 ( T + 1 ) − α T ( T + 1 ) , and α ∈ 0 , 1 T + 1 . Then Green’s function G ( t , s ) has the following properties:

  1. If s ∈ [ 1 , η ] Z , then G ( t , s ) is decreasing with respect to t ∈ [ 0 , T + 1 ] Z ; if s ∈ [ η + 1 , T − 1 ] Z , then G ( t , s ) is increasing with respect to t ∈ [ 0 , T + 1 ] Z .

  2. G ( t , s ) changes its sign on [ 0 , T + 1 ] Z × [ 1 , T − 1 ] Z . In detail, if ( t , s ) ∈ [ 0 , T + 1 ] Z × [ 1 , η ] Z , then G ( t , s ) ≤ 0 ; if ( t , s ) ∈ [ 0 , T + 1 ] Z × [ η + 1 , T − 1 ] Z , then G ( t , s ) ≥ 0 .

  3. If s > η , then min t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( 0 , s ) = 0 and

    (2.6) max t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( T + 1 , s ) = T ( T + 1 ) − s − s 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ≤ T ( T + 1 ) − η − η 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) .

    If s ≤ η , then max t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( 0 , s ) = 0 and

    (2.7) min t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( T + 1 , s ) = − s + s 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ≥ − η + η 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) .

Proof

(i) As we know, if Δ t G ( t , s ) ≥ 0 ( ≤ 0 ) , then G ( t , s ) is increasing (decreasing) with respect to t . Now, we discuss the sign of Δ t G ( t , s ) and G ( t , s ) .

First, suppose that s > η and s > t − 2 . In this case, we have that

G ( t , s ) = α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) t

and

Δ t G ( t , s ) = α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) .

Thus, Δ t G ( t , s ) ≥ 0 is equivalent to

α β ( T − s ) ( T + 1 + s ) ( 2 t + 1 ) + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) + 2 ( T − s ) ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ≥ 0 ,

which is similar to

4 T − 4 s ≥ β [ 2 T ( T + 1 ) − 2 α T t ( T + 1 ) + s ( − 2 ( 2 T + 1 ) + α ( 2 t + 2 T 2 ) ) + s 2 ( 2 ( 1 − α T ) + 2 α t ) ] .

Using the fact that β < 2 2 ( T + 1 ) − α T ( T + 1 ) , we obtain

4 T − 4 s ≥ 2 2 ( T + 1 ) − α T ( T + 1 ) [ 2 T ( T + 1 ) − 2 α T t ( T + 1 ) + s ( − 2 ( 2 T + 1 ) + α ( 2 t + 2 T 2 ) ) + s 2 ( 2 ( 1 − α T ) + 2 α t ) ] ,

which is equivalent to

( s 2 + s − T ( T + 1 ) ) ( − 1 + α ( T − t ) ) ≥ 0 .

We know α ≤ 1 T + 1 < 1 T − t , then the last inequality holds.

As a result, in this case when s > η and s > t − 2 , we have that Δ t G ( t , s ) ≥ 0 , G ( 0 , s ) = 0 . Thus, G ( t , s ) ≥ 0 .

Second, suppose that s > η and s ≤ t − 2 . In this case, we have that

G ( t , s ) = 2 β ( T + 1 ) − 2 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + 4 T + 2 − 2 β ( T + 1 ) 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t − s 2 2 − s 2

and

Δ t G ( t , s ) = 2 β ( T + 1 ) − 2 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + 4 T + 2 − 2 β ( T + 1 ) 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) .

Thus, Δ t G ( t , s ) ≥ 0 is equivalent to

( 2 β ( T + 1 ) − 2 − α β s − α β s 2 ) ( 2 t + 1 ) + 4 T + 2 − 2 β ( T + 1 ) 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) ≥ 0 ,

which is the same as

4 T − 4 t ≥ β [ 2 ( T + 1 ) 2 − 2 ( 2 t + 1 ) ( T + 1 ) + s ( 2 + 2 α ( t − T ) ) + s 2 ( 2 + 2 α ( t − T ) ) ] .

Using the fact that β < 2 2 ( T + 1 ) − α T ( T + 1 ) , we obtain

4 T − 4 t ≥ 2 2 ( T + 1 ) − α T ( T + 1 ) [ 2 ( T + 1 ) 2 − 2 ( 2 t + 1 ) ( T + 1 ) + s ( 2 + 2 α ( t − T ) ) + s 2 ( 2 + 2 α ( t − T ) ) ] ,

which is equivalent to

( s 2 + s − T ( T + 1 ) ) ( − 1 + α ( T − t ) ) ≥ 0 .

We know α ≤ 1 T + 1 < 1 T − t , then the last inequality holds.

Moreover,

G ( s , s ) = 2 β ( T + 1 ) − 2 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) s 2 + 4 T + 2 − 2 β ( T + 1 ) 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) s − s 2 2 − s 2 ,

thus, G ( s , s ) ≥ 0 is equivalent to

s 2 ( 2 β ( T + 1 ) − 2 − α β s − α β s 2 ) + s ( 4 T + 2 − 2 β ( T + 1 ) 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) ) ≥ ( s 2 + s ) ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ,

which is the same as

4 s ( T − s ) ≥ β s [ ( T − s ) ( α + α T + 2 ) + ( α s + 2 ) ( T − s ) 2 ] .

Using the fact that β < 2 2 ( T + 1 ) − α T ( T + 1 ) , we obtain

4 s ( T − s ) ≥ 2 2 ( T + 1 ) − α T ( T + 1 ) s [ ( T − s ) ( α + α T + 2 ) + ( α s + 2 ) ( T − s ) 2 ] ,

which is equivalent to

4 ( T + 1 ) + 4 s + 2 α s 2 ≥ 2 α s T + 2 α T + 2 α + 4 α T ( T + 1 ) .

The last one holds due to 4 ( T + 1 ) > 4 α T ( T + 1 ) and 4 s > 2 α s T + 2 α T .

As a result, in this case when s > η and s ≤ t − 2 , we have that Δ t G ( t , s ) ≥ 0 , G ( s , s ) ≥ 0 . Thus, G ( t , s ) ≥ 0 .

Now, suppose that s ≤ η and s > t − 2 . In this case, we have that

G ( t , s ) = 2 − 2 β ( T + 1 ) + α β ( T + 1 + s ) ( T − s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + − 4 T − 2 + 2 β ( T + 1 ) 2 + ( 2 β ( 1 − α T ) − α β ) ( T + 1 + s ) ( T − s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) t

and

Δ t G ( t , s ) = 2 − 2 β ( T + 1 ) + α β ( T + 1 + s ) ( T − s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + − 4 T − 2 + 2 β ( T + 1 ) 2 + ( 2 β ( 1 − α T ) − α β ) ( T + 1 + s ) ( T − s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) .

Thus, Δ t G ( t , s ) ≤ 0 is equivalent to

[ 2 − 2 β ( T + 1 ) + α β ( T + 1 + s ) ( T − s ) ] ( 2 t + 1 ) + 2 ( T − s ) ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) − 4 T − 2 + 2 β ( T + 1 ) 2 + ( 2 β ( 1 − α T ) − α β ) ( T + 1 + s ) ( T − s ) ≤ 0 ,

which is the same as

( t − s ) ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( s + s 2 ) ( α β ( T − t ) − β ) ≤ 0 .

We know α ≤ 1 T + 1 < 1 T − t , then the last inequality holds.

As a result, in this case when s ≤ η and s > t − 2 , we have that Δ t G ( t , s ) ≤ 0 , G ( 0 , s ) = 0 . Thus, G ( t , s ) ≤ 0 .

Finally, suppose that s ≤ η and s ≤ t − 2 . In this case, we have that

G ( t , s ) = − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t − s 2 2 − s 2 ≤ 0 .

Obviously, since α ≤ 1 T + 1 < 2 1 + 2 T , we have that G ( t , s ) ≤ 0 . Moreover,

Δ t G ( t , s ) = − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ≤ 0 .

As a result, in this case when s ≤ η and s ≤ t − 2 , we have that Δ t G ( t , s ) ≤ 0 , G ( t , s ) ≤ 0 .

(iii) If s > η , from (i), we know G ( t , s ) is increasing with respect to t , then we have min t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( 0 , s ) = 0 and

max t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( T + 1 , s ) = T ( T + 1 ) − s − s 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ≤ T ( T + 1 ) − η − η 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) .

Furthermore, we get the inequality (2.6).

Meanwhile, if s ≤ η , G ( t , s ) is decreasing with respect to t . Then max t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( 0 , s ) = 0 and

min t ∈ [ 0 , T + 1 ] Z G ( t , s ) = G ( T + 1 , s ) = − s + s 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ≥ − η + η 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) .

Furthermore, we get the inequality (2.7).□

Remark 2.4

Let us give some reasons for why η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z . To get it, let us consider the following BVPs:

(2.8) Δ 3 u ( t − 1 ) = − 1 , t ∈ [ 1 , T − 1 ] Z , u ( 0 ) = 0 , Δ 2 u ( η ) = α Δ u ( T ) , Δ u ( T ) = β u ( T + 1 ) .

From Lemma 2.2, we know that (2.8) has a solution u ( t ) as follows:

u ( t ) = − 1 6 t 3 + 1 2 t 2 − 1 3 t + − 4 T η − 2 η + 2 β η ( T + 1 ) 2 + ( 2 β ( 1 − α T ) − α β ) T ( T + 1 ) ( T − 1 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t + T ( T − 1 ) 2 t + α β T ( T + 1 ) ( T − 1 ) + 2 η ( 1 − β ( T + 1 ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 − α β t 2 + ( 2 β ( 1 − α T ) − α β ) t 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) T ( T + 1 ) ( T − 1 ) 3 .

For the sake of convenience, let

ϕ ( t ) = t 2 − 3 t + 2 + 12 T η + 6 η − 6 β η ( T + 1 ) 2 + 2 ( 2 β ( 1 − α T ) − α β ) T ( T + 1 ) ( 1 − T ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) − 3 T ( T − 1 ) + 2 α β T ( T + 1 ) ( 1 − T ) + 6 η ( β ( T + 1 ) − 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) t .

Obviously, u ( t ) ≥ 0 ⇔ ϕ ( t ) ≤ 0 and

Δ ϕ ( t ) = 2 t − 2 + 2 α β T ( T + 1 ) ( 1 − T ) + 6 η ( β ( T + 1 ) − 1 ) ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) .

Consequently, if ϕ ( 0 ) ≤ 0 and ϕ ( T + 1 ) ≤ 0 imply that ϕ ( t ) ≤ 0 . By the direct computation, ϕ ( 0 ) ≤ 0 for η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z and ϕ ( T + 1 ) ≤ 0 for η ∈ 1 , 2 T − 2 3 Z . Combining with the fact 2 T − 2 3 > 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 , we get η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z .

Let E = { u : [ 0 , T + 1 ] Z → R ∣ u ( 0 ) = 0 , Δ 2 u ( η ) = α Δ u ( T ) , Δ u ( T ) = β u ( T + 1 ) } . Then E is a Banach space under the norm ‖ u ‖ = max t ∈ [ 0 , T + 1 ] Z ∣ u ( t ) ∣ . Define a subset P ^ ⊂ E as follows:

P ^ = { y ∈ E : y ( t ) ≥ 0 , Δ y ( t ) ≥ 0 , t ∈ [ 0 , T ] Z } ,

then P ^ is a cone in E .

Lemma 2.5

If y ∈ P ^ , then the unique solution u ( t ) of (2.1) belongs to P ^ , where u ( t ) is defined as (2.4).

Proof

First, if 0 ≤ t − 2 ≤ η , then

u ( t ) = ∑ s = 1 t − 2 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t − s 2 2 − s 2 y ( s ) + ∑ s = t − 1 η 2 − 2 β ( T + 1 ) + α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( T − s ) t + − 2 ( 2 T + 1 ) + 2 β ( T + 1 ) 2 + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t y ( s ) + ∑ s = η + 1 T − 1 α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t + ( T − s ) t y ( s ) .

Since η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z , we get

Δ u ( t ) = u ( t + 1 ) − u ( t ) = ∑ s = 1 t − 1 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) y ( s ) + ∑ s = t η 2 − 2 β ( T + 1 ) + α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + ( T − s ) + − 2 ( 2 T + 1 ) + 2 β ( T + 1 ) 2 + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) y ( s ) + ∑ s = η + 1 T − 1 α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) y ( s ) ≥ y ( η ) ( η − t + 1 ) ( t − T ) + T ( η − t + 1 ) − ( t + η ) ( η − t + 1 ) 2 + ( T − η ) ( T − η − 1 ) 2 + ( 2 α β ( t − T ) + 2 β ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) T ( T + 1 ) ( T − η − 1 ) − T ( T + 1 ) ( T − 1 ) 3 = y ( η ) ( t − η ) ( η − t + 1 ) 2 + ( T − η ) ( T − η − 1 ) 2 + ( 2 α β ( t − T ) + 2 β ) T ( T + 1 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( T − η − 1 ) − T − 1 3 = y ( η ) ( T − t ) ( T + t − 1 − 2 η ) 2 + ( 2 α β ( t − T ) + 2 β ) T ( T + 1 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( T − η − 1 ) − T − 1 3 ≥ 0 .

Furthermore, we get

Δ 2 u ( t − 1 ) = Δ u ( t ) − Δ u ( t − 1 ) = ∑ s = 1 t − 1 − α β s − α β s 2 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) y ( s ) + ∑ s = t η 2 − 2 β ( T + 1 ) + α β ( T − s ) ( T + 1 + s ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) y ( s ) + ∑ s = η + 1 T − 1 α β ( T − s ) ( T + 1 + s ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) y ( s ) ≥ y ( t ) 2 − 2 β ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ( η − t + 1 ) − α β 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) T ( T − 1 ) ( T + 1 ) 3 + α β T ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ( ( T − η − 1 ) + ( η − t + 1 ) ) = y ( t ) ( η − t + 1 ) + α β T ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) T − η − 1 − T − 1 3 ≥ 0 .

We know η < 2 T − 2 3 , then it clearly holds.

Second, if η < t − 2 ≤ T − 2 , then

u ( t ) = ∑ s = 1 η − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t − s 2 2 − s 2 y ( s ) + ∑ s = η + 1 t − 2 2 β ( T + 1 ) − 2 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + 4 T + 2 − 2 β ( T + 1 ) 2 + ( α β − 2 β ( 1 − α T ) ) ( s 2 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t − s 2 2 − s 2 y ( s )

+ ∑ s = t − 1 T − 1 α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t 2 + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t + ( T − s ) t y ( s ) .

Since η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z , we get

Δ u ( t ) = u ( t + 1 ) − u ( t ) = ∑ s = 1 η − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + ( s + s 2 ) ( α β − 2 β ( 1 − α T ) ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) y ( s ) + ∑ s = η + 1 t − 1 2 β ( T + 1 ) − 2 − α β s − α β s 2 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + 4 T + 2 − 2 β ( T + 1 ) 2 + ( α β − 2 β ( 1 − α T ) ) ( s 2 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) y ( s ) + ∑ s = t T − 1 α β ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( 2 t + 1 ) + ( 2 β ( 1 − α T ) − α β ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) + ( T − s ) y ( s ) ≥ y ( η ) 2 α β ( T − t ) − 2 β 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) T ( T − 1 ) ( T + 1 ) 3 + ( T − t ) ( T − t + 1 ) 2 + ( 2 α β ( t − T ) + 2 β ) T ( T + 1 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( T − t ) + 4 β t ( T + 1 ) − 2 β T ( T + 1 ) + 4 T − 4 t 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( t − η − 1 ) ≥ 0 .

The last inequality holds since

( 4 T − 4 t ) ( t − η − 1 ) + 2 ( T − t ) ( T − t + 1 ) ≥ β ( 2 T ( T + 1 ) − 4 t ( T + 1 ) ) ( t − η − 1 ) + ( 2 ( T + 1 ) − α T ( T + 1 ) ) ( T − t ) ( T − t + 1 ) + ( 2 α ( T − t ) − 2 ) T ( T + 1 ) 2 T − 3 t + 1 3 .

Using the fact that β < 2 2 ( T + 1 ) − α T ( T + 1 ) , then

( 4 T − 4 t ) ( t − η − 1 ) + 2 ( T − t ) ( T − t + 1 ) ≥ 2 2 ( T + 1 ) − α T ( T + 1 ) ( 2 T ( T + 1 ) − 4 t ( T + 1 ) ) ( t − η − 1 ) + ( 2 ( T + 1 ) − α T ( T + 1 ) ) ( T − t ) ( T − t + 1 ) + ( 2 α ( T − t ) − 2 ) T ( T + 1 ) 2 T − 3 t + 1 3 ,

which is equivalent to

3 ( t − η − 1 ) ≥ − ( 2 T − 3 t + 1 ) .

Then, it clearly holds.□

In order to deduce existence results for the nonlinear problem, we will work in more restrictive cones. To this end, from the fact that

(2.9) Δ s G ( t , s ) G ( T + 1 , s ) = − t ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( T − s ) ( T − 1 − s ) ( T ( T + 1 ) − ( s + 1 ) 2 − ( s + 1 ) ) ( T ( T + 1 ) − s 2 − s ) , s > η and s > t − 2 , − ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( T − t ) ( T − t + 1 ) ( T ( T + 1 ) − ( s + 1 ) 2 − ( s + 1 ) ) ( T ( T + 1 ) − s 2 − s ) , s > η and s ≤ t − 2 , − t ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( s + 1 − t ) s ( s + 1 ) ( s + 2 ) , s ≤ η and s > t − 2 , 0 , s ≤ η and s ≤ t − 2 ,

we deduce that Δ s G ( t , s ) G ( T + 1 , s ) ≤ 0 for all t , s ∈ [ 1 , T − 1 ] Z .

Lemma 2.6

Assume α ∈ 0 , 1 T + 1 , β ∈ 0 , 2 2 ( T + 1 ) − α T ( T + 1 ) , and η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z . Then, the following inequality holds.

G ( t , s ) G ( T + 1 , s ) ≥ G ( t , T + 1 ) G ( T + 1 , T + 1 ) ≥ 1 2 α β ( t 2 − t ) ,

where G ( t , s ) is defined in (2.2) and (2.3). Now, by defining g ( t ) = 1 2 α β ( t 2 − t ) and H ( t , s ) ≔ G ( t , s ) − g ( t ) G ( T + 1 , s ) , we deduce the following direct consequence.

Corollary 2.7

Assume η ∈ 1 , 6 T ( T − 1 ) − 4 − α β T 2 ( T + 1 ) 2 − 2 β ( T + 1 ) ( T 2 − T − 2 ) 12 T + 6 − 6 β ( T + 1 ) 2 Z , β ∈ 0 , 2 2 ( T + 1 ) − α T ( T + 1 ) , and α ∈ 0 , 1 T + 1 . Then

G ( t , s ) ≤ g ( t ) G ( T + 1 , s ) , 1 ≤ s ≤ η , G ( t , s ) ≥ g ( t ) G ( T + 1 , s ) , η < s ≤ T − 1 .

We are in a position to introduce a more restrictive cone than P ^ , as follows:

P 0 ^ = { y ∈ E : y ∈ P ^ , y ( t ) ≥ g ( t ) ‖ y ‖ } .

So, we assume that the solution of (2.1) belongs to the previous cone, when η is in the more restrictive interval 1 , T − 1 3 Z . The result is the following.

Lemma 2.8

If y ∈ P 0 ^ , then the unique solution u ( t ) of (2.1) belongs to P 0 ^ , where u ( t ) is defined as (2.4).

Proof

We assume that

(2.10) H ( t , s ) ≔ ( T − s ) t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t , s > η and s > t − 2 , − t 2 2 − s 2 + s 2 + ( 2 T + 1 ) 2 t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t , s > η and s ≤ t − 2 , t 2 2 − s t − 1 2 t − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t , s ≤ η and s > t − 2 , − s 2 + s 2 − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t , s ≤ η and s ≤ t − 2 .

First, if 0 ≤ t − 2 ≤ η , then

u ( t ) − g ( t ) ‖ u ‖ = u ( t ) − g ( t ) u ( T + 1 ) = ∑ s = 1 t − 2 − s 2 + s 2 − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t y ( s ) + ∑ s = t − 1 η t 2 2 − s t − 1 2 t − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t y ( s ) + ∑ s = η + 1 T − 1 ( T − s ) t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t y ( s ) ≥ y ( η ) ∑ s = 1 t − 2 − s 2 + s 2 − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t + ∑ s = t − 1 η t 2 2 − s t − 1 2 t − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t + ∑ s = η + 1 T − 1 ( T − s ) t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t

= y ( η ) − t ( t − 1 ) ( t − 2 ) 6 + t 2 2 ( η − t + 2 ) − t 2 ( η − t + 2 ) + T t ( T − η − 1 ) − ( T + t − 2 ) ( T − t + 1 ) 2 t + 2 β ( 1 − α T ) T ( T + 1 ) t 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( T − η − 1 ) − T − 1 3 ≥ 0 .

The last inequality holds since

− t ( t − 1 ) ( t − 2 ) + 3 t 2 ( η − t + 2 ) − 3 t ( η − t + 2 ) + 6 T t ( T − η − 1 ) − 3 ( T + t − 2 ) ( T − t + 1 ) t + 2 β ( 1 − α T ) T ( T + 1 ) t 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ( 2 T − 3 η − 2 ) ≥ 0 ,

which is equivalent to

3 T ( T − 1 ) − t 2 + 3 t − 2 − 6 T η − 3 η + 3 t η + 2 β ( 1 − α T ) T ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ( 2 T − 3 η − 2 ) ≥ 0 .

Combining this with η ∈ 1 , T − 1 3 Z , we know the last inequality holds.

Second, if η < t − 2 ≤ T − 2 , then

u ( t ) − g ( t ) ‖ u ‖ = u ( t ) − g ( t ) u ( T + 1 ) = ∑ s = 1 η − s 2 + s 2 − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t y ( s ) + ∑ s = η + 1 t − 2 − t 2 2 − s 2 + s 2 + ( 2 T + 1 ) 2 t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t y ( s ) + ∑ s = t − 1 T − 1 ( T − s ) t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t y ( s ) , ≥ y ( η ) ∑ s = 1 η − s 2 + s 2 − 2 β ( 1 − α T ) ( s + s 2 ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t + ∑ s = η + 1 t − 2 − t 2 2 − s 2 + s 2 + ( 2 T + 1 ) 2 t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t + ∑ s = t − 1 T − 1 ( T − s ) t + 2 β ( 1 − α T ) ( T − s ) ( T + 1 + s ) 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) t = y ( η ) − t 2 2 ( t − η − 2 ) − t ( t − 1 ) ( t − 2 ) 6 + T t ( t − η − 2 ) + t 2 ( t − η − 2 ) + T t ( T − t + 1 ) − t ( T + t − 2 ) ( T − t + 1 ) 2 + 2 β ( 1 − α T ) T ( T + 1 ) t 2 ( 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ) ( T − η − 1 ) − T − 1 3 ≥ 0 .

The last inequality holds since

( − 3 t 2 + 6 T t + 3 t ) ( t − η − 2 ) − t ( t − 1 ) ( t − 2 ) + 6 T t ( T − t + 1 ) − 3 ( T + t − 2 ) ( T − t + 1 ) t + 2 β ( 1 − α T ) T ( T + 1 ) t 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ( 2 T − 3 η − 2 ) ≥ 0 ,

which is equivalent to

3 T ( T − 1 ) − t 2 + 3 t − 2 + 3 t η − 3 η − 6 T η + 2 β ( 1 − α T ) T ( T + 1 ) 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ( 2 T − 3 η − 2 ) ≥ 0 .

Combining this with η ∈ 1 , T − 1 3 Z , we know the last inequality holds. Overall, u ∈ P 0 ^ .□

3 Existence results

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

  1. f : [ 1 , T − 1 ] Z × [ 0 , ∞ ) → [ 0 , ∞ ) is continuous, the mapping t → f ( t , u ) is increasing for each u ∈ [ 0 , ∞ ) and the mapping u → f ( t , u ) is increasing for each t ∈ [ 1 , T − 1 ] Z ;

  2. a : [ 1 , T − 1 ] Z → ( − ∞ , + ∞ ) is increasing and satisfies a < 0 on [ 0 , η ] Z , a > 0 on [ η + 1 , T − 1 ] Z .

Define the cone K by

K = { u ∈ P 0 ^ ∣ min t ∈ [ 0 , T + 1 ] Z u ( t ) ≥ g ( t ) ‖ u ‖ }

and the operator T λ : K → E by

T λ u ( t ) = λ ∑ s = 1 T − 1 G ( t , s ) a ( s ) f ( s , u ( s ) ) .

From Lemmas 2.5 to 2.8, we know that T λ : K → K . Meanwhile, since E is finite dimensional, T λ : K → K is a completely continuous operator. Therefore, if u is a fixed-point of T λ in K , then u is positive solution of (1.1). Set

A = ∑ s = 1 T − 1 T ( T + 1 ) − η 2 − η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∣ a ( s ) ∣ , B = ∑ s = 1 T − 1 G ( T + 1 , s ) a ( s ) g ( s ) .

Theorem 3.1

Suppose that ( H 1 ) and ( H 2 ) hold. Assume that

( A 1 ) f 0 ≔ lim u → 0 + min t ∈ [ 1 , T − 1 ] Z f ( t , u ) u = ∞ , f ∞ ≔ lim u → ∞ max t ∈ [ 1 , T − 1 ] Z f ( t , u ) u = 0 ( sublinear case ) .

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

Proof

(sublinear case). Since f 0 = ∞ , there exists r 1 > 0 such that

(3.1) f ( t , u ) ≥ u λ B , ( t , u ) ∈ [ 1 , T − 1 ] Z × [ 0 , r 1 ] .

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

(3.2) min s ∈ [ 0 , T + 1 ] Z u ( s ) ≥ g ( s ) ‖ u ‖ .

Then, we get

(3.3) T λ u ( T + 1 ) = λ ∑ s = 1 T − 1 G ( T + 1 , s ) a ( s ) f ( s , u ( s ) ) ≥ λ ∑ s = 1 T − 1 G ( T + 1 , s ) a ( s ) u ( s ) λ B ≥ λ ∑ s = 1 T − 1 g ( s ) G ( T + 1 , s ) a ( s ) ‖ u ‖ λ B = ‖ u ‖ .

This combines with the fact that T λ u ( t ) ≥ 0 for t ∈ [ 0 , T + 1 ] Z , we have

(3.4) ‖ T λ u ‖ ≥ ‖ u ‖ , for u ∈ K ∩ ∂ Ω 1 .

Since f ∞ = 0 , there exists r 2 > 0 such that

(3.5) f ( t , u ) ≤ u λ A , ( t , u ) ∈ [ 1 , T − 1 ] Z × [ r 2 , ∞ ) .

We consider two cases: f is bounded and f is unbounded.

Case I. f is bounded, there exists a constant M > 0 , such that f ≤ M , then we take R = max { 2 r 1 , λ M A } and Ω 2 = { u ∈ K : ‖ u ‖ < R } . Now, from Lemma 2.3(ii), we know that G ( t , s ) ≥ 0 for s ∈ [ η + 1 , T − 1 ] Z and G ( t , s ) ≤ 0 for s ∈ [ 1 , η ] Z . Then, for u ∈ K ∩ ∂ Ω 2 ,

‖ T λ u ‖ = λ max t ∈ [ 0 , T + 1 ] Z ∑ s = 1 T − 1 G ( t , s ) a ( s ) f ( s , u ( s ) ) ≤ λ max t ∈ [ 0 , T + 1 ] Z ∑ s = 1 η G ( t , s ) a ( s ) f ( s , u ( s ) ) + λ max t ∈ [ 0 , T + 1 ] Z ∑ s = η + 1 T − 1 G ( t , s ) a ( s ) f ( s , u ( s ) ) ≤ λ max t ∈ [ 0 , T + 1 ] Z ∑ s = 1 η ∣ G ( t , s ) ∣ ∣ a ( s ) ∣ f ( s , u ( s ) ) + λ max t ∈ [ 0 , T + 1 ] Z ∑ s = η + 1 T − 1 ∣ G ( t , s ) ∣ ∣ a ( s ) ∣ f ( s , u ( s ) ) ≤ λ η 2 + η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 η ∣ a ( s ) ∣ f ( s , u ( s ) ) + λ T ( T + 1 ) − η 2 − η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = η + 1 T − 1 ∣ a ( s ) ∣ f ( s , u ( s ) ) .

Since η ∈ 1 , T − 1 3 Z , we get that T ( T + 1 ) − η 2 − η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ≥ η 2 + η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) . Then

‖ T λ u ‖ ≤ λ T ( T + 1 ) − η 2 − η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 T − 1 ∣ a ( s ) ∣ f ( s , u ( s ) ) ≤ λ T ( T + 1 ) − η 2 − η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 T − 1 ∣ a ( s ) ∣ R λ A = R .

Therefore,

(3.6) ‖ T λ u ‖ ≤ ‖ u ‖ , for u ∈ K ∩ ∂ Ω 2 .

Case II. f is unbounded, then we take R > max { 2 r 1 , r 2 } , for ( t , u ) ∈ [ 1 , T − 1 ] Z × [ 0 , R ] and Ω 2 = { u ∈ K : ‖ u ‖ < R } . Now, similar to the proof of (3.6), we will get

(3.7) ‖ T λ u ‖ ≤ ‖ u ‖ , for u ∈ K ∩ ∂ Ω 2 .

Therefore, applying Theorem 1.1(ii), we get T λ has a fixed-point u ∈ K ∩ ( Ω ¯ 2 \ Ω 1 ) . Now, (1.1) has a positive solution u ∈ K .□

Theorem 3.2

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

Proof

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

(3.8) 1 B ( f 0 − ε ) ≤ λ ≤ 1 A ( f ∞ + ε ) .

On the one hand, by the definition of f 0 , there exists r 3 > 0 such that f ( t , u ) ≥ ( f 0 − ε ) u , for ( t , u ) ∈ [ 1 , T − 1 ] Z × [ 0 , r 3 ] . Let Ω 1 = { u ∈ K : ‖ u ‖ < r 3 } . Therefore, similar to the discussion from (3.3)–(3.4), we get that

T λ u ( T + 1 ) ≥ λ ∑ s = 1 T − 1 G ( T + 1 , s ) ( f 0 − ε ) a ( s ) u ( s ) ≥ λ B ( f 0 − ε ) ‖ u ‖ .

Combining with (3.8), we get that

(3.9) ‖ T λ u ‖ ≥ ‖ u ‖ , for u ∈ K ∩ ∂ Ω 1 .

On the other hand, by the definition of f ∞ , there exists r 4 such that f ( t , u ) ≤ ( f ∞ + ε ) u , for ( t , u ) ∈ [ 1 , T − 1 ] Z × [ r 4 , ∞ ) . Let R = m a x { 2 r 3 , r 4 } , Ω 2 = { u ∈ K : ‖ u ‖ < R } . Then similar to the proof of (3.6), if u ∈ K ∩ ∂ Ω 2 , then

(3.10) ‖ T λ u ‖ ≤ λ T ( T + 1 ) − η 2 − η 2 + α β T ( T + 1 ) − 2 β ( T + 1 ) ∑ s = 1 T − 1 ∣ a ( s ) ∣ ( f ∞ + ε ) ‖ u ‖ .

Furthermore, by (3.8), we get

(3.11) ‖ T λ u ‖ ≤ ‖ u ‖ , for u ∈ K ∩ ∂ Ω 2 .

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

Similar to the discussion of Theorems 3.1 and 3.2, we could obtain the following theorem. So, we just state them here without any proof.

Theorem 3.3

Suppose ( H 1 ) and ( H 2 ) hold. Then

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

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

Remark 3.4

Suppose that ( H 1 ) and ( H 2 ) hold. Note that similar to Theorems 3.1, 3.2 and 3.3, we can also get

  1. Assume that

    ( A 2 ) f 0 ≔ lim u → 0 + max t ∈ [ 1 , T − 1 ] Z f ( t , u ) u = 0 , f ∞ ≔ lim u → ∞ min t ∈ [ 1 , T − 1 ] Z f ( t , u ) u = ∞ (super linear case) .

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

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

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

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

Example 3.5

Consider the following discrete nonlinear third-order three-point eigenvalue problem:

(3.12) Δ 3 u ( t − 1 ) = − λ a ( t ) f ( t , u ( t ) ) , t ∈ [ 1 , 7 ] Z , u ( 0 ) = 0 , Δ 2 u ( η ) = α Δ u ( 8 ) , Δ u ( 8 ) = β u ( 9 ) ,

where a ( t ) = t − 2 , f ( t , u ) satisfies the condition ( H 1 ) , we will give f ( t , u ) the specific forms. Now, according to Remark 2.4, we let η = 2 , α = 1 9 , and β = 1 8 . Then finding the positive solutions of the problem (3.12) is equivalent to finding the fixed-point operator equation

T λ u ( t ) = λ ∑ s = 1 7 G ( t , s ) a ( s ) f ( s , u ( s ) ) .

Furthermore, we get the expression of G ( t , s ) as follows.

If s > 2 , then

G ( t , s ) = 72 − s − s 2 108 t 2 + 72 − s − s 2 108 + ( 8 − s ) t , t − 2 < s , 1 6 − s + s 2 108 t 2 − s 2 2 − s 2 + 55 6 − s + s 2 108 t , s ≤ t − 2 .

If s ≤ 2 , then

G ( t , s ) = − 1 6 + 72 − s − s 2 108 t 2 + − 55 6 + 72 − s − s 2 108 + ( 8 − s ) t , t − 2 < s , − s − s 2 108 t 2 + − s − s 2 108 t − s 2 2 − s 2 , s ≤ t − 2 .

Obviously, if ( t , s ) ∈ [ 0 , 9 ] Z × [ 1 , 2 ] Z , then G ( t , s ) ≤ 0 . Meanwhile, Δ t G ( t , s ) = G ( t + 1 , s ) − G ( t , s ) < 0 , we see that G ( t , s ) is decreasing with respect to t ; if ( t , s ) ∈ [ 0 , 9 ] Z × [ 3 , 7 ] Z , then G ( t , s ) ≥ 0 . Meanwhile, Δ t G ( t , s ) = G ( t + 1 , s ) − G ( t , s ) > 0 , we see that G ( t , s ) is increasing with respect to t . Therefore, if s > 2 , then max t ∈ [ 0 , 9 ] Z G ( t , s ) = G ( 9 , s ) = 72 − s − s 2 0.75 , min t ∈ [ 0 , 9 ] Z G ( t , s ) = G ( 0 , s ) = 0 . If s ≤ 2 , then max t ∈ [ 0 , 9 ] Z G ( t , s ) = G ( 0 , s ) = 0 , min t ∈ [ 0 , 9 ] Z G ( 9 , s ) = − s − s 2 0.75 .

Now, according to Lemma 2.6, we let g ( s ) = 1 144 ( s 2 − s ) , then

A = ∑ s = 1 7 8 × 9 − 4 − 2 2 + 1 9 × 1 8 × 8 × 9 − 2 × 1 8 × 9 ∣ s − 2 ∣ = 1,408 , B = ∑ s = 1 7 G ( 9 , s ) ( s − 2 ) s ( s − 1 ) = 308 3 .

If f ( t , u ) = ( u 1 3 + 2 ) t , we have f 0 = ∞ , f ∞ = 0 . Furthermore, by Theorem 3.1, for λ ∈ ( 0 , ∞ ) , (3.12) has at least one positive solution.

If f ( t , u ) = w 1 ( u ) t , where

w 1 ( u ) = sin u , ( t , u ) ∈ [ 1 , 7 ] Z × 0 , π 2 , 2 4 − π u − π 2 + 1 , ( t , u ) ∈ [ 1 , 7 ] Z × π 2 , 2 , u 2 + 2000 500 u + 2 , ( t , u ) ∈ [ 1 , 7 ] Z × [ 2 , ∞ ) .

Then f ( t , u ) satisfies the condition ( H 1 ), f 0 = 1 , f ∞ = 7 500 . This implies that 0 < A f ∞ = 19.712 < B f 0 = 308 3 < ∞ . Furthermore, by Theorem 3.2, for λ ∈ 1 B f 0 , ∞ , (3.12) has at least one positive solution.

If f ( t , u ) = w 2 ( u ) t , where

w 2 ( u ) = ( u + 3 ) 1 2 , ( t , u ) ∈ [ 1 , 7 ] Z × [ 0 , 1 ] , u + 1 , ( t , u ) ∈ [ 1 , 7 ] Z × [ 1 , 2 ] , u 3 + 10 u 2 + 2 , ( t , u ) ∈ [ 1 , 7 ] Z × [ 2 , ∞ ) .

Then f ( t , u ) satisfies the condition ( H 1 ), f 0 = ∞ , f ∞ = 7 . This implies that 0 < f ∞ < ∞ . Furthermore, by Theorem 3.3 for λ ∈ 0 , 1 A f ∞ , (3.12) has at least one positive solution.

If f ( t , u ) = w 3 ( u ) t , where

w 3 ( u ) = tan u , ( t , u ) ∈ [ 1 , 7 ] Z × 0 , π 4 , 1 4 − π u − π 4 + 1 , ( t , u ) ∈ [ 1 , 7 ] Z × [ π 4 , 1 ] , u − 1 2 1 2 + 1 , ( t , u ) ∈ [ 1 , 7 ] Z × [ 1 , ∞ ) .

Then f ( t , u ) satisfies the condition ( H 1 ), f 0 = 1 , f ∞ = 0 . This implies that 0 < f 0 < ∞ . Furthermore, by Theorem 3.3, for λ ∈ 1 B f 0 , ∞ , (3.12) has at least one positive solution.

Acknowledgements

The authors are very grateful to the anonymous referees for their valuable suggestions.

  1. Funding information: This work was supported by the National Science Foundation China (11961060) and the key project of Natural Science Foundation of Gansu Province (18JR3RA084).

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

  3. Conflict of interest: Authors state no conflict of interest.

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Received: 2021-04-19
Revised: 2021-07-18
Accepted: 2021-07-26
Published Online: 2021-09-01

© 2021 Xueqin Cao 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-2021-0085
Keywords for this article
third-order difference equation; three-point boundary conditions; sign-changing Green’s function; Guo-Krasnoselskii’s fixed-point theorem; positive solutions
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  59. A note on polyexponential and unipoly Bernoulli polynomials of the second kind
  60. On the type 2 poly-Bernoulli polynomials associated with umbral calculus
  61. Some estimates for commutators of Littlewood-Paley g-functions
  62. Construction of a family of non-stationary combined ternary subdivision schemes reproducing exponential polynomials
  63. On the evolutionary bifurcation curves for the one-dimensional prescribed mean curvature equation with logistic type
  64. On intersections of two non-incident subgroups of finite p-groups
  65. Global existence and boundedness in a two-species chemotaxis system with nonlinear diffusion
  66. Finite groups with 4p2q elements of maximal order
  67. Positive solutions of a discrete nonlinear third-order three-point eigenvalue problem with sign-changing Green's function
  68. Power moments of automorphic L-functions related to Maass forms for SL3(ℤ)
  69. Entire solutions for several general quadratic trinomial differential difference equations
  70. Strong consistency of regression function estimator with martingale difference errors
  71. Fractional Hermite-Hadamard-type inequalities for interval-valued co-ordinated convex functions
  72. Montgomery identity and Ostrowski-type inequalities via quantum calculus
  73. Universal inequalities of the poly-drifting Laplacian on smooth metric measure spaces
  74. On reducible non-Weierstrass semigroups
  75. so-metrizable spaces and images of metric spaces
  76. Some new parameterized inequalities for co-ordinated convex functions involving generalized fractional integrals
  77. The concept of cone b-Banach space and fixed point theorems
  78. Complete consistency for the estimator of nonparametric regression model based on m-END errors
  79. A posteriori error estimates based on superconvergence of FEM for fractional evolution equations
  80. Solution of integral equations via coupled fixed point theorems in 𝔉-complete metric spaces
  81. Symmetric pairs and pseudosymmetry of Θ-Yetter-Drinfeld categories for Hom-Hopf algebras
  82. A new characterization of the automorphism groups of Mathieu groups
  83. The role of w-tilting modules in relative Gorenstein (co)homology
  84. Primitive and decomposable elements in homology of ΩΣℂP∞
  85. The G-sequence shadowing property and G-equicontinuity of the inverse limit spaces under group action
  86. Classification of f-biharmonic submanifolds in Lorentz space forms
  87. Some new results on the weaving of K-g-frames in Hilbert spaces
  88. Matrix representation of a cross product and related curl-based differential operators in all space dimensions
  89. Global optimization and applications to a variational inequality problem
  90. Functional equations related to higher derivations in semiprime rings
  91. A partial order on transformation semigroups with restricted range that preserve double direction equivalence
  92. On multi-step methods for singular fractional q-integro-differential equations
  93. Compact perturbations of operators with property (t)
  94. Entire solutions for several complex partial differential-difference equations of Fermat type in ℂ2
  95. Random attractors for stochastic plate equations with memory in unbounded domains
  96. On the convergence of two-step modulus-based matrix splitting iteration method
  97. On the separation method in stochastic reconstruction problem
  98. Robust estimation for partial functional linear regression models based on FPCA and weighted composite quantile regression
  99. Structure of coincidence isometry groups
  100. Sharp function estimates and boundedness for Toeplitz-type operators associated with general fractional integral operators
  101. Oscillatory hyper-Hilbert transform on Wiener amalgam spaces
  102. Euler-type sums involving multiple harmonic sums and binomial coefficients
  103. Poly-falling factorial sequences and poly-rising factorial sequences
  104. Geometric approximations to transition densities of Jump-type Markov processes
  105. Multiple solutions for a quasilinear Choquard equation with critical nonlinearity
  106. Bifurcations and exact traveling wave solutions for the regularized Schamel equation
  107. Almost factorizable weakly type B semigroups
  108. The finite spectrum of Sturm-Liouville problems with n transmission conditions and quadratic eigenparameter-dependent boundary conditions
  109. Ground state sign-changing solutions for a class of quasilinear Schrödinger equations
  110. Epi-quasi normality
  111. Derivative and higher-order Cauchy integral formula of matrix functions
  112. Commutators of multilinear strongly singular integrals on nonhomogeneous metric measure spaces
  113. Solutions to a multi-phase model of sea ice growth
  114. Existence and simulation of positive solutions for m-point fractional differential equations with derivative terms
  115. Bernstein-Walsh type inequalities for derivatives of algebraic polynomials in quasidisks
  116. Review Article
  117. Semiprimeness of semigroup algebras
  118. Special Issue on Problems, Methods and Applications of Nonlinear Analysis (Part II)
  119. Third-order differential equations with three-point boundary conditions
  120. Fractional calculus, zeta functions and Shannon entropy
  121. Uniqueness of positive solutions for boundary value problems associated with indefinite ϕ-Laplacian-type equations
  122. Synchronization of Caputo fractional neural networks with bounded time variable delays
  123. On quasilinear elliptic problems with finite or infinite potential wells
  124. Deterministic and random approximation by the combination of algebraic polynomials and trigonometric polynomials
  125. On a fractional Schrödinger-Poisson system with strong singularity
  126. Parabolic inequalities in Orlicz spaces with data in L1
  127. Special Issue on Evolution Equations, Theory and Applications (Part II)
  128. Impulsive Caputo-Fabrizio fractional differential equations in b-metric spaces
  129. Existence of a solution of Hilfer fractional hybrid problems via new Krasnoselskii-type fixed point theorems
  130. On a nonlinear system of Riemann-Liouville fractional differential equations with semi-coupled integro-multipoint boundary conditions
  131. Blow-up results of the positive solution for a class of degenerate parabolic equations
  132. Long time decay for 3D Navier-Stokes equations in Fourier-Lei-Lin spaces
  133. On the extinction problem for a p-Laplacian equation with a nonlinear gradient source
  134. General decay rate for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping
  135. On hyponormality on a weighted annulus
  136. Exponential stability of Timoshenko system in thermoelasticity of second sound with a memory and distributed delay term
  137. Convergence results on Picard-Krasnoselskii hybrid iterative process in CAT(0) spaces
  138. Special Issue on Boundary Value Problems and their Applications on Biosciences and Engineering (Part I)
  139. Marangoni convection in layers of water-based nanofluids under the effect of rotation
  140. A transient analysis to the M(τ)/M(τ)/k queue with time-dependent parameters
  141. Existence of random attractors and the upper semicontinuity for small random perturbations of 2D Navier-Stokes equations with linear damping
  142. Degenerate binomial and Poisson random variables associated with degenerate Lah-Bell polynomials
  143. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  144. On the mixed fractional quantum and Hadamard derivatives for impulsive boundary value problems
  145. The Lp dual Minkowski problem about 0 < p < 1 and q > 0
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