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Advances in Difference Equations

, 2013:275 | Cite as

The existence of solutions to a class of boundary value problems with fractional difference equations

  • Yuanyuan Pan
  • Zhenlai HanEmail author
  • Shurong Sun
  • Chuanxia Hou
Open Access
Research
Part of the following topical collections:
  1. Recent Advances in Operator Equations, Boundary Value Problems, Fixed Point Theory and Applications, and General Inequalities
  2. Progress in Functional Differential and Difference Equations

Abstract

In this paper, we study the existence and uniqueness of solutions for the boundary value problem of fractional difference equations

{ Δ ν y ( t ) = f ( t + ν 1 , y ( t + ν 1 ) ) , y ( ν 3 ) = 0 , Δ y ( ν 3 ) = 0 , y ( ν + b ) = g ( y ) Open image in new window

and

{ Δ ν y ( t ) = λ f ( t + ν 1 , y ( t + ν 1 ) ) , y ( ν 3 ) = 0 , Δ y ( ν 3 ) = 0 , y ( ν + b ) = g ( y ) , Open image in new window

respectively, where t = 1 , 2 , , b Open image in new window, 2 < ν 3 Open image in new window, f : { ν 1 , , ν + b } × R R Open image in new window is a continuous function and g C ( [ ν 3 , ν + b ] Z ν 3 , R ) Open image in new window is a continuous functional. We prove the existence and uniqueness of a solution to the first problem by the contraction mapping theorem and the Brouwer theorem. Moreover, we present the existence and nonexistence of a solution to the second problem in terms of the parameter λ by the properties of the Green function and the Guo-Krasnosel’skii theorem. Finally, we present some examples to illustrate the main results.

MSC:34A08, 34B18, 39A12.

Keywords

discrete fractional equation boundary value problem existence and uniqueness of solution fixed point theorem eigenvalue 

1 Introduction

In recent years, fractional differential equations have been of great interest. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various sciences such as physics, mechanics, chemistry and engineering. Mathematicians have employed this fractional calculus in recent years to model and solve a variety of applied problems. Indeed, as Podlubny outlines in [1], fractional calculus aids significantly in the fields of viscoelasticity, capacitor theory, electrical circuits, electro-analytical chemistry, neurology, diffusion, control theory and statistics.

The continuous fractional calculus has developed greatly in the last decades. Some of the recent progress in the continuous fractional calculus includes the paper [2], in which the authors explored a continuous fractional boundary value problem of conjugate type using cone theory, they then deduced the existence of one or more positive solutions. Of particular interest with regard to the present paper is the recent work by Benchohra et al. [3]. In that paper, the authors considered a continuous fractional differential equation with nonlocal conditions. Other recent work in the direction of those articles may be found, for example, in [4, 5, 6, 7, 8, 9, 10, 11, 12].

In recent years, a number of papers on the discrete fractional calculus have appeared, such as [13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30], which has helped to build up some of the basic theory of this area. For example, Atici and Eloe discussed the properties of the generalized falling function, a corresponding power rule for fractional delta-operators and the commutativity of fractional sums in [13]. They presented in [13] more rules for composing fractional sums and differences. Goodrich studied a two-point fractional boundary value problem in [16], which gave the existence results for a certain two-point boundary value problem of right-focal type for a fractional difference equation. At the same time, a number of papers appeared, and these began to build up the theoretical foundations of the discrete fractional calculus. For example, a recent paper by Atici and Eloe [14] explored some of the theories of a conjugate discrete fractional boundary value problem. Discrete fractional initial value problems were considered in a paper by Atici and Eloe [15].

Atici and Eloe in [14] considered a two-point boundary value problem for the finite fractional difference equation
Δ ν y ( t ) = f ( t + ν 1 , y ( t + ν 1 ) ) , t = 1 , 2 , , b + 1 , y ( ν 2 ) = 0 , y ( ν + b + 1 ) = 0 , Open image in new window

where 1 < ν 2 Open image in new window is a real number, b 2 Open image in new window is an integer and f : [ ν , ν + b ] N ν 1 × R R Open image in new window is continuous. They analyzed the corresponding Green function, provided an application and obtained sufficient conditions for the existence of positive solutions for a two-point boundary value problem for a nonlinear finite fractional difference equation.

Goodrich in [18] considered a discrete fractional boundary value problem of the form
Δ ν y ( t ) = f ( t + ν 1 , y ( t + ν 1 ) ) , y ( ν 2 ) = 0 , y ( ν + b ) = g ( y ) , Open image in new window

where t [ 0 , b ] N 0 : = { 0 , 1 , , b } Open image in new window, f : [ ν 2 , ν 1 , , ν + b 1 ] N ν 2 × R R Open image in new window is a continuous function, g C ( [ ν 2 , ν + b ] Z ν 3 , R ) Open image in new window is a given functional, and 1 < ν 2 Open image in new window. He established the existence and uniqueness of a solution to this problem by the contraction mapping theorem, the Brouwer fixed point theorem and the Guo-Krasnosel’skii fixed point theorem.

Although the boundary value problem of fractional difference equations has been studied by several authors, the present works are almost all concerned with 1 < μ 2 Open image in new window, very little is known in the literature about a fractional difference equation with 2 < μ 3 Open image in new window.

Motivated by all the works above, in this paper, we first aim to study the following boundary value problem:
{ Δ ν y ( t ) = f ( t + ν 1 , y ( t + ν 1 ) ) , t = 0 , 1 , 2 , , b , y ( ν 3 ) = 0 , Δ y ( ν 3 ) = 0 , y ( ν + b ) = g ( y ) , Open image in new window
(1.1)

where t = 0 , 1 , 2 , , b Open image in new window, 2 < ν 3 Open image in new window, f : { ν 1 , , ν + b 1 } N ν 1 × R R Open image in new window is continuous and g C ( [ ν 3 , ν + b ] Z ν 3 , R ) Open image in new window is a continuous functional.

Our second aim is to investigate the boundary value problem of a fractional difference equation with parameter
{ Δ ν y ( t ) = λ f ( t + ν 1 , y ( t + ν 1 ) ) , t = 0 , 1 , 2 , , b , y ( ν 3 ) = 0 , Δ y ( ν 3 ) = 0 , y ( ν + b ) = g ( y ) , Open image in new window
(1.2)

where t = 0 , 1 , 2 , , b Open image in new window, 2 < ν 3 Open image in new window, f : { ν 1 , , ν + b 1 } N ν 1 × R R Open image in new window is continuous and g C ( [ ν 3 , ν + b ] Z ν 3 , R ) Open image in new window is a continuous functional, λ is a positive parameter. We establish some sufficient conditions for the nonexistence and existence of at least one or two positive solutions for the boundary value problem by considering the eigenvalue intervals of the nonlinear fractional differential equation with boundary conditions.

The plan of this paper is as follows. We first give the form of solutions of problem (1.1), second we prove the existence and uniqueness of a solution to problem (1.1) by the contraction mapping theorem and the Brouwer theorem, and then the eigenvalue intervals for the boundary value problem of nonlinear fractional difference equation (1.2) are considered by the properties of the Green function and the Guo-Krasnosel’skii fixed point theorem on cones. Finally we present some examples to illustrate the main results.

2 Preliminaries

For the convenience of the readers, we first present some useful definitions and fundamental facts of fractional calculus theory, which can be found in [13, 14].

Definition 2.1 [14]

We define t ν ̲ : = Γ ( t + 1 ) Γ ( t + 1 ν ) Open image in new window for any t and ν, for which the right-hand side is defined. We also appeal to the convention that if t + 1 ν Open image in new window is a pole of the gamma function and t + 1 Open image in new window is not a pole, then t ν ̲ = 0 Open image in new window.

Definition 2.2 [14]

The ν th fractional sum of a function f, for ν > 0 Open image in new window, is defined by
Δ ν f ( t ; a ) : = 1 Γ ( ν ) s = a t ν ( t s 1 ) ν 1 ̲ f ( s ) Open image in new window

for t { a + ν , a + ν + 1 , } : = N a + ν Open image in new window. We also define the ν th fractional difference for ν > 0 Open image in new window by Δ ν f ( t ) : = Δ N Δ ν N f ( t ) Open image in new window, where t N a + ν Open image in new window and ν N Open image in new window is chosen so that 0 N 1 < ν N Open image in new window.

Lemma 2.1 [13]

If t r Open image in new window, then t α ̲ r α ̲ Open image in new window for any α > 0 Open image in new window.

Lemma 2.2 [14]

Let 0 N 1 < ν N Open image in new window. Then Δ ν Δ ν y ( t ) = y ( t ) + C 1 t ν 1 ̲ + C 2 t ν 2 ̲ + + C N t ν N ̲ Open image in new window for some C i R Open image in new window with 1 i N Open image in new window.

Lemma 2.3 [17]

For t and s, for which both ( t s 1 ) ν ̲ Open image in new window and ( t s 2 ) ν ̲ Open image in new window are defined, we find that
Δ s [ ( t s 1 ) ν ̲ ] = ν ( t s 2 ) ν 1 ̲ . Open image in new window

Now let us consider a linear boundary value problem, which is important for us to facilitate the analysis of problems (1.1) and (1.2).

Lemma 2.4 Let h : [ ν 1 , ν + b 1 ] Z ν 3 R Open image in new window and g : R b + 4 R Open image in new window. A function y is a solution of the problem
Δ ν y ( t ) = h ( t + ν 1 ) , Open image in new window
(2.1)
y ( ν 3 ) = 0 , Open image in new window
(2.2)
Δ y ( ν 3 ) = 0 , Open image in new window
(2.3)
y ( ν + b ) = g ( y ) , Open image in new window
(2.4)
where t [ 0 , b ] Z 0 Open image in new window if and only if y ( t ) Open image in new window, t [ ν 3 , ν + b ] Z ν 3 Open image in new window, has the form
y ( t ) = s = 0 b G ( t , s ) h ( s + ν 1 ) + t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) , Open image in new window
where
G ( t , s ) = 1 Γ ( ν ) { t ν 1 ̲ ( ν + b s 1 ) ν 1 ̲ ( ν + b ) ν 1 ̲ ( t s 1 ) ν 1 ̲ , s < t ν + 1 b , t ν 1 ̲ ( ν + b s 1 ) ν 1 ̲ ( ν + b ) ν 1 ̲ , t ν + 1 s b . Open image in new window
Proof By Lemma 2.2, we obtain
y ( t ) = Δ ν h ( t + ν 1 ) + C 1 t ν 1 ̲ + C 2 t ν 2 ̲ + C 3 t ν 3 ̲ , t [ ν 3 , ν + b ] Z ν 3 . Open image in new window
(2.5)
From (2.2) we get
y ( ν 3 ) = Δ ν h ( t + ν 1 ) | t = ν 3 + C 1 ( ν 3 ) ν 1 ̲ + C 2 ( ν 3 ) ν 2 ̲ + C 3 ( ν 3 ) ν 3 ̲ . Open image in new window
Noting that
Δ ν h ( t + ν 1 ) | t = ν 3 = 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ h ( s + ν 1 ) | t = ν 3 = 0 , ( ν 3 ) ν 1 ̲ = 0 , ( ν 3 ) ν 2 ̲ = 0 , Open image in new window

we deduce C 3 = 0 Open image in new window.

By (2.3) we have
Δ y ( ν 3 ) = y ( ν 2 ) y ( ν 3 ) = 0 , Open image in new window
then y ( ν 2 ) = 0 Open image in new window. We note that
y ( ν 2 ) = Δ ν h ( t + ν 1 ) | t = ν 2 + C 1 ( ν 2 ) ν 1 ̲ + C 2 ( ν 2 ) ν 2 ̲ Open image in new window
and
Δ ν h ( t + ν 1 ) | t = ν 2 = 0 , ( ν 2 ) ν 1 ̲ = 0 , Open image in new window

it follows that C 2 = 0 Open image in new window.

From (2.4) we know
y ( ν + b ) = Δ ν h ( t + ν 1 ) | t = ν + b + C 1 ( ν + b ) ν 1 ̲ = g ( y ) , Open image in new window
(2.6)
which implies that
C 1 = 1 ( ν + b ) ν 1 ̲ [ 1 Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ h ( s + ν 1 ) + g ( y ) ] . Open image in new window
Consequently, we deduce that y ( t ) Open image in new window has the form
y ( t ) = 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ h ( s + ν 1 ) + 1 ( ν + b ) ν 1 ̲ [ 1 Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ h ( s + ν 1 ) + g ( y ) ] t ν 1 ̲ = 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ h ( s + ν 1 ) + t ν 1 ̲ ( ν + b ) ν 1 ̲ Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ h ( s + ν 1 ) + t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) = s = 0 b G ( t , s ) h ( s + ν 1 ) + t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) Open image in new window
(2.7)
for t [ ν 3 , ν + b ] Z ν 3 Open image in new window, where
G ( t , s ) = 1 Γ ( ν ) { t ν 1 ̲ ( ν + b s 1 ) ν 1 ̲ ( ν + b ) ν 1 ̲ ( t s 1 ) ν 1 ̲ , s < t ν + 1 b , t ν 1 ̲ ( ν + b s 1 ) ν 1 ̲ ( ν + b ) ν 1 ̲ , t ν + 1 s b . Open image in new window

This shows that if (2.1)-(2.4) has a solution, then it can be represented by (2.7) and that every function of the form (2.7) is a solution of (2.1)-(2.4), which completes the proof. □

Theorem 2.1 The Green function G ( t , s ) Open image in new window satisfies the following conditions:
  1. (i)

    G ( t , s ) > 0 Open image in new window for ( t , s ) [ ν 3 , ν + b ] Z ν 3 × [ 0 , b ] Open image in new window.

     
  2. (ii)

    max t [ ν 3 , ν + b ] Z ν 3 G ( t , s ) = G ( s + ν 1 , s ) Open image in new window for s [ 0 , b ] Open image in new window.

     
  3. (iii)
    There exists a positive number γ ( 0 , 1 ) Open image in new window such that
    min t [ b + ν 4 , 3 ( b + ν ) 4 ] G ( t , s ) γ max t [ ν 3 , ν + b ] Z ν 3 G ( t , s ) = γ G ( s + ν 1 , s ) for  s [ 0 , b ] . Open image in new window
     

The proof of this theorem is similar to that of Theorem 3.2 in [14]. Hence, we omit the proof here.

3 Existence and uniqueness of solution

In this section, we wish to show that under certain conditions, problem (1.1) has at least one solution. We know that problem (1.1) can be recast as an equivalent summation equation. It follows from Lemma 2.4 that y is a solution of (1.1) if and only if y is a fixed point of the operator T : R b + 4 R Open image in new window, where
( T y ) ( t ) : = 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ f ( t + ν 1 , y ( t + ν 1 ) ) + t ν 1 ̲ ( ν + b ) ν 1 ̲ Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ f ( t + ν 1 , y ( t + ν 1 ) ) + t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) Open image in new window
(3.1)

for t [ ν 3 , ν + b ] Z ν 3 Open image in new window. We use this fact to prove the first existence theorem.

Theorem 3.1 Define y = max t [ ν 3 , ν + b ] Z ν 3 | y ( t ) | Open image in new window. Suppose that f ( t , y ) Open image in new window and g ( y ) Open image in new window are Lipschitz in y. That is, there exist α , β > 0 Open image in new window such that | f ( t , y 1 ) f ( t , y 2 ) | α y 1 y 2 Open image in new window, | g ( y 1 ) g ( y 2 ) | β y 1 y 2 Open image in new window for any functions y 1 Open image in new window, y 2 Open image in new window defined on [ ν 3 , ν + b ] Z ν 3 Open image in new window. Then if the condition
2 α j = 1 b ( ν + j j ) + β < 1 Open image in new window
(3.2)

holds, then problem (1.1) has a unique solution.

Proof We show that T is a contraction mapping. To achieve this, we notice that for given y 1 Open image in new window and y 2 Open image in new window,
T y 1 T y 2 α y 1 y 2 max t [ ν 3 , ν + b ] Z ν 3 [ 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ ] + α y 1 y 2 max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ Γ ( ν ) ( ν + b ) ν 1 s = 0 b ( ν + b s 1 ) ν 1 ̲ + β y 1 y 2 max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ . Open image in new window
(3.3)
By an application of Lemma 2.3, we get
1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ = 1 Γ ( ν ) [ 1 ν ( t s ) ν ̲ ] s = 0 t ν + 1 = Γ ( t + 1 ) Γ ( ν + 1 ) Γ ( t + 1 ν ) Γ ( ν + b + 1 ) Γ ( b + 1 ) Γ ( ν + 1 ) = j = 1 b ( ν + j j ) . Open image in new window
(3.4)
Similar to the above inequality, we have
t ν 1 ̲ Γ ( ν ) ( ν + b ) ν 1 s = 0 b ( ν + b s 1 ) ν 1 ̲ 1 Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ = 1 Γ ( ν ) [ 1 ν ( ν + b s ) ν ̲ ] s = 0 b + 1 = Γ ( ν + b + 1 ) Γ ( b + 1 ) Γ ( ν + 1 ) = j = 1 b ( ν + j j ) . Open image in new window
(3.5)
From another application of Lemma 2.1, we obtain
t ν 1 ̲ ( ν + b ) ν 1 ( ν + b ) ν 1 ( ν + b ) ν 1 = 1 . Open image in new window
(3.6)
So, putting (3.4)-(3.6) in (3.3), we conclude that
T y 1 T y 2 { 2 α j = 1 b ( ν + j j ) + β } y 1 y 2 . Open image in new window

Then condition (3.2) holds. We find that (1.1) has a unique solution, which completes the proof of the theorem. □

By weakening the conditions imposed on f ( t , y ) Open image in new window and g ( y ) Open image in new window, we can still obtain the existence of a solution to (1.1). We apply the Brouwer theorem to accomplish this.

Theorem 3.2 Suppose that there exists a constant K > 0 Open image in new window such that f ( t , y ) Open image in new window satisfies the inequality
max ( t , y ) [ ν 3 , ν + b ] Z ν 3 × [ K , K ] | f ( t , y ) | K 2 Γ ( ν + b + 1 ) Γ ( ν + 1 ) Γ ( b + 1 ) + 1 Open image in new window
(3.7)
and g ( y ) Open image in new window satisfies the inequality
max y [ K , K ] | g ( y ) | K 2 Γ ( ν + b + 1 ) Γ ( ν + 1 ) Γ ( b + 1 ) + 1 . Open image in new window
(3.8)

Then (1.1) has at least one solution y 0 Open image in new window satisfying | y 0 ( t ) | K Open image in new window for all t [ ν 3 , ν + b ] Z ν 3 Open image in new window.

Proof Consider the Banach space B : = { y R b + 4 : y K } Open image in new window. T is defined as (3.1). It is obvious that T is a continuous operator. Therefore, our main objective is to show that T : B B Open image in new window. That is, whenever y K Open image in new window, it follows that T y K Open image in new window. Once this is established, we use the Brouwer theorem to deduce the conclusion.

Assume that inequalities (3.7) and (3.8) hold for given f and g. For convenience, we let
Φ : = K 2 Γ ( ν + b + 1 ) Γ ( ν + 1 ) Γ ( b + 1 ) + 1 , Open image in new window
(3.9)
which is a strictly positive constant. Then we have
T y max t [ ν 3 , ν + b ] Z ν 3 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ | f ( s + ν 1 , y ( s + ν 1 ) ) | + max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ Γ ( ν ) ( ν + b ) ν 1 s = 0 b ( ν + b s 1 ) ν 1 ̲ | f ( s + ν 1 , y ( s + ν 1 ) ) | + max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 | g ( y ) | Φ max t [ ν 3 , ν + b ] Z ν 3 [ 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ + t ν 1 ̲ Γ ( ν ) ( ν + b ) ν 1 s = 0 b ( ν + b s 1 ) ν 1 ̲ ] + Φ max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 . Open image in new window
(3.10)
As in the proof of Theorem 3.1, we can simplify the expression on the right-hand side of inequality (3.10). Indeed, we know that
1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ + t ν 1 ̲ Γ ( ν ) ( ν + b ) ν 1 s = 0 b ( ν + b s 1 ) ν 1 ̲ 1 Γ ( ν ) s = 0 t ν ( t s 1 ) ν 1 ̲ + 1 Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ 1 Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ + 1 Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ = 2 Γ ( ν ) s = 0 b ( ν + b s 1 ) ν 1 ̲ . Open image in new window
(3.11)
On the one hand, from Lemma 2.1 we know t ν 1 ̲ Open image in new window is increasing in t, thus we have
s = 0 b ( ν + b s 1 ) ν 1 ̲ = [ 1 ν ( ν + b s ) ν ̲ ] s = 0 b + 1 = Γ ( ν + b + 1 ) ν Γ ( b + 1 ) . Open image in new window
(3.12)
On the other hand,
t ν 1 ̲ ( ν + b ) ν 1 1 . Open image in new window
(3.13)
Inserting (3.11)-(3.13) into (3.10), we can obtain
T y Φ [ 2 Γ ( ν + b + 1 ) Γ ( ν + 1 ) Γ ( b + 1 ) ] + Φ = Φ [ 2 Γ ( ν + b + 1 ) Γ ( ν + 1 ) Γ ( b + 1 ) + 1 ] . Open image in new window
(3.14)
By substituting (3.9) into (3.14), we have
T y Φ [ 2 Γ ( ν + b + 1 ) Γ ( ν + 1 ) Γ ( b + 1 ) + 1 ] = K . Open image in new window
(3.15)

Thus, from (3.15) we deduce that T : B B Open image in new window. Consequently, it follows at once by the Brouwer theorem that there exists a fixed point of the map T, say T y 0 = y 0 Open image in new window with y 0 B Open image in new window. So, this function y 0 Open image in new window is a solution of (1.1) and y 0 Open image in new window satisfies the bound | y 0 ( t ) | K Open image in new window for each t [ ν 3 , ν + b ] Z ν 3 Open image in new window. And this completes the proof of the theorem. □

4 Existence of a positive solution

In this section, we show the existence of positive solutions for boundary value problem (1.2).

Lemma 4.1 [26]

Letbe a Banach space, and let K B Open image in new window be a cone. Assume that Ω 1 Open image in new window and Ω 2 Open image in new window are two bounded open subsets contained insuch that 0 Ω 1 Open image in new window and Ω ¯ 1 Ω 2 Open image in new window. Assume further that T : K ( Ω ¯ 2 Ω 1 ) K Open image in new window is a completely continuous operator. If either
  1. (1)

    T y y Open image in new window for y K Ω 1 Open image in new window and T y y Open image in new window for y K Ω 2 Open image in new window, or

     
  2. (2)

    T y y Open image in new window for y K Ω 1 Open image in new window and T y y Open image in new window for y K Ω 2 Open image in new window,

     

then T has at least one fixed point in K ( Ω ¯ 2 Ω 1 ) Open image in new window.

Define the Banach space ℬ by
B = { y : [ ν 3 , ν + b ] Z ν 3 R : y ( ν 2 ) = y ( ν 3 ) = 0 , y ( ν + b ) = g ( y ) } , Open image in new window

with the norm y = max { | y ( t ) | , t [ ν 3 , ν + b ] Z ν 3 } Open image in new window.

For t ν 1 ̲ Open image in new window is increasing, we get max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ = 1 Open image in new window. Thus, there exists a positive constant γ 0 Open image in new window such that
min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ γ 0 max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ . Open image in new window
In fact, for t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 Open image in new window, t ν 1 ̲ ( ν + b ) ν 1 ̲ Open image in new window is strictly positive, then we let
k = min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ . Open image in new window
Take 0 < γ 0 k Open image in new window. Then
min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ γ 0 max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ . Open image in new window
Denote γ ˜ = min { γ , γ 0 } Open image in new window. Then we have
min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ γ ˜ max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ . Open image in new window
(4.1)
Define the cone
P = { y B : y ( t ) 0 , min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 y ( t ) γ ˜ y , t [ ν 3 , ν + b ] Z ν 3 } . Open image in new window
We define an operator T λ : P B Open image in new window as follows:
( T λ y ) ( t ) = λ s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + λ t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) , t [ ν 3 , ν + b ] Z ν 3 . Open image in new window

It is easy to see from Lemma 2.4 that y is a solution of (1.2) if and only if y is a fixed point of T λ Open image in new window.

Suppose that f is a nonnegative function. Then, from Theorem 2.1 and (4.1), we have
min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 ( T λ y ) ( t ) = min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 λ s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 λ t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) γ λ max t [ ν 3 , ν + b ] Z ν 3 s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + γ ˜ λ max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) γ ˜ λ max t [ ν 3 , ν + b ] Z ν 3 s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + γ ˜ λ max t [ ν 3 , ν + b ] Z ν 3 t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) = γ ˜ T λ y . Open image in new window
(4.2)

Thus, T λ ( P ) P Open image in new window.

Lemma 4.2 T λ : P P Open image in new window is completely continuous.

Proof Note that T λ Open image in new window is a summation operator on a discrete finite set, so T λ Open image in new window is trivially completely continuous. □

For convenience, we define:

(F1) f ( t , y ) = h ( t ) g ( y ) Open image in new window, where h is a positive function, g is a nonnegative functional;

(F2)
lim y 0 + g ( y ) y = 0 , lim y + g ( y ) y = + ; Open image in new window
(F3)
lim y 0 + g ( y ) y = + , lim y + g ( y ) y = 0 . Open image in new window

Set l = 3 ( ν + b ) 4 ν + 1 + 1 ν + b 4 ν + 1 Open image in new window, M = max G ( t , s ) Open image in new window for ( t , s ) [ ν 3 , ν + b ] Z ν 3 × [ 0 , b ] Open image in new window, m = s = ν + b 4 ν + 1 3 ( ν + b ) 4 ν + 1 G ( s + ν 1 , s ) Open image in new window, h = min h ( t ) Open image in new window, H = max h ( t ) Open image in new window for t [ ν 3 , ν + b ] Z ν 3 Open image in new window.

Theorem 4.1 Suppose that conditions (F1) and (F2) hold. If there exist a sufficiently small positive constant δ and a sufficiently large constant L 1 Open image in new window such that ( M H ( b + 1 ) + 1 ) δ < m h L 1 Open image in new window holds, then for each
λ ( ( m h L 1 ) 1 , ( ( M H ( b + 1 ) + 1 ) δ ) 1 ) , Open image in new window
(4.3)

boundary value problem (1.2) has at least one positive solution.

Proof By condition (F2), there exists r 1 > 0 Open image in new window such that
g ( y ) δ r 1 , 0 < y r 1 . Open image in new window
(4.4)
So, for y P Open image in new window with y = r 1 Open image in new window, by (4.3) and (4.4), we have, for all t [ ν 3 , ν + b ] Z ν 3 Open image in new window,
( T λ y ) ( t ) = λ s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + λ t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) λ M s = 0 b h ( s + ν 1 ) g ( y ( s + ν 1 ) ) + λ g ( y ) λ M H ( b + 1 ) δ r 1 + λ δ r 1 = λ ( M H ( b + 1 ) + 1 ) δ r 1 r 1 = y . Open image in new window
(4.5)
Thus, if we choose Ω 1 = { y B : y < r 1 } Open image in new window, then (4.5) implies that
T λ y y , y P Ω 1 . Open image in new window
(4.6)
Similarly, by condition (F2), we can find 0 < r 1 < r 2 Open image in new window and a sufficiently large constant L 1 Open image in new window such that
g ( y ) L 1 γ 2 r 2 , y r 2 . Open image in new window
(4.7)
And then we set r 2 = r 2 / γ > r 2 Open image in new window and Ω 2 = { y B : y < r 2 } Open image in new window. Then y P Open image in new window and y = r 2 Open image in new window imply
min t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 2 y ( t ) γ y = γ r 2 = r 2 , Open image in new window
thus
y ( t ) r 2 for all  t [ ν + b 4 , 3 ( ν + b ) 4 ] Z ν 3 . Open image in new window
Therefore, for given t 0 [ b + ν 4 , 3 ( b + ν ) 4 ] Open image in new window, by Theorem 2.1, (4.3) and (4.7), we have
( T λ y ) ( t 0 ) = λ s = 0 b G ( t 0 , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + λ t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) λ s = ν + b 4 ν + 1 3 ( ν + b ) 4 ν + 1 G ( t 0 , s ) h ( s + ν 1 ) g ( y ( s + ν 1 ) ) λ γ h L 1 γ 2 r 2 s = ν + b 4 ν + 1 3 ( ν + b ) 4 ν + 1 G ( s + ν 1 , s ) λ m h L 1 r 2 r 2 = y . Open image in new window
(4.8)
Hence, from (4.8) we have
T λ y y , y P Ω 2 . Open image in new window
(4.9)

Now, from (4.6), (4.9) and Lemma 4.1, we have T λ Open image in new window has a fixed point y P ( Ω ¯ 2 Ω 1 ) Open image in new window with r 1 y r 2 Open image in new window. Then the theorem is proved. □

Theorem 4.2 Assume that conditions (F1) and (F3) hold. If there exists a sufficiently large constant L 2 Open image in new window such that ( M b H + 1 ) < m h L 2 Open image in new window holds, then for each
λ ( ( m h L 2 ) 1 , ( M H ( b + 1 ) + 1 ) 1 ) , Open image in new window
(4.10)

boundary value problem (1.2) has at least one positive solution.

Proof By condition (F3), there exist r 3 > 0 Open image in new window and a sufficiently large constant L 2 > 0 Open image in new window such that
g ( y ) > L γ r 3 for  0 < y < r 3 . Open image in new window
(4.11)
Let Ω 1 = { y B , y < r 3 } Open image in new window. Then, for y Ω 1 Open image in new window,
( T λ y ) ( t ) = λ s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + λ t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) λ s = ν + b 4 ν + 1 3 ( ν + b ) 4 ν + 1 G ( t , s ) h ( s + ν 1 ) g ( y ( s + ν 1 ) ) λ γ m h L 2 γ r 3 = λ m h L 2 r 3 r 3 = y . Open image in new window
(4.12)
Then (4.12) implies that
T λ y y , y P Ω 1 . Open image in new window
(4.13)

Next, we consider two cases for the construction of Ω 2 Open image in new window.

Case 1. Suppose that g is bounded. Then there exists some R 1 > r 3 Open image in new window such that
g ( y ) R 1 for  y P . Open image in new window
(4.14)
From (4.10) we know
λ < ( M H ( b + 1 ) + 1 ) 1 . Open image in new window
(4.15)
Thus, from (4.14) and (4.15), we get
( T λ y ) ( t ) = λ s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + λ t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) λ M s = 0 b h ( s + ν 1 ) g ( y ( s + ν 1 ) ) + λ g ( y ) λ M H R 1 ( b + 1 ) + λ R 1 = λ ( M H ( b + 1 ) + 1 ) R 1 R 1 = y . Open image in new window
(4.16)
Case 2. Suppose that g is unbounded. From (4.10) we know λ < ( M H ( b + 1 ) + 1 ) 1 Open image in new window, so λ < ( ( M H ( b + 1 ) + 1 ) δ 2 ) 1 Open image in new window for a sufficiently small constant δ 2 Open image in new window. Then, by condition (F3), there exists some R 2 Open image in new window such that
g ( y ) δ 2 y , y R 2 . Open image in new window
(4.17)
Choose R 3 Open image in new window such that R 3 > r 3 Open image in new window and for 0 < y R 3 Open image in new window, g ( y ) g ( R 3 ) Open image in new window. Define R = max { R 2 , R 3 } Open image in new window. Now we set Ω 2 = { y B : y < R } Open image in new window, then g ( R ) δ 2 R Open image in new window. Thus, for y Ω 2 Open image in new window, we have
( T λ y ) ( t ) = λ s = 0 b G ( t , s ) f ( s + ν 1 , y ( s + ν 1 ) ) + λ t ν 1 ̲ ( ν + b ) ν 1 ̲ g ( y ) λ M s = 0 b h ( s + ν 1 ) g ( y ( s + ν 1 ) ) + λ g ( y ) λ M H ( b + 1 ) δ 2 R + λ δ 2 R = λ ( M H ( b + 1 ) + 1 ) δ 2 R R = y . Open image in new window
(4.18)
Then, in both Case 1 and Case 2, we have
T λ y y , y P Ω 2 . Open image in new window
(4.19)

From (4.13), (4.19) and Lemma 4.1, we get T λ Open image in new window has a fixed point y P ( Ω ¯ 2 Ω 1 ) Open image in new window with r 3 y R Open image in new window. This completes the proof. □

5 Nonexistence

In this section, we give some sufficient conditions for the nonexistence of a positive solution to boundary value problem (1.2).

We state the following hypotheses that will be used in what follows.

(F4)
g 0 = lim sup y 0 + g ( y ) y , g = lim sup y + g ( y ) y ; Open image in new window
(F5)
g 0 = lim inf y 0 + g ( y ) y , g = lim inf y + g ( y ) y . Open image in new window

Theorem 5.1 Assume that (F1) and (F4) hold. If g 0 < + Open image in new window and g < + Open image in new window, then there exists λ 0 Open image in new window such that for all 0 < λ < λ 0 Open image in new window, boundary value problem (1.2) has no positive solution.

Proof Since g 0 < + Open image in new window and g < + Open image in new window, there exist positive numbers c 1 Open image in new window, c 2 Open image in new window, r 1 Open image in new window and r 2 Open image in new window such that r 1 < r 2 Open image in new window and
g ( y ) c 1 y for  y [ 0 , r 1 ] , g ( y ) c 2 y for  y [ r 2 , + ) . Open image in new window
Let c = max { c 1 , c 2 , max r 1 y r 2 g ( y ) y } Open image in new window. Then we have
g ( y ) c y for all  y [ 0 , + ) . Open image in new window
Suppose that y 0 ( t ) Open image in new window is a positive solution of (1.2). Then we show that this leads to a contradiction for 0 < λ < λ 0 : = ( ( M H ( b + 1 ) + 1 ) c ) 1 Open image in new window. Since T λ y 0 ( t ) = y 0 ( t ) Open image in new window, for t [ ν 3 , ν + b ] Z ν 3 Open image in new window,
y 0 = T λ y 0 λ M H s = 1 b g ( y 0 ( s + ν 1 ) ) + λ g ( y 0 ) λ ( M H ( b + 1 ) + 1 ) c y 0 < y 0 , Open image in new window

which is a contradiction. Therefore, (1.2) has no positive solution. This completes the proof. □

Theorem 5.2 Assume that (F1) and (F5) hold. If g 0 > 0 Open image in new window and g > 0 Open image in new window, then there exists λ 0 Open image in new window such that for all λ > λ 0 Open image in new window, boundary value problem (1.2) has no positive solution.

Proof Since g 0 > 0 Open image in new window and g > 0 Open image in new window, we can get that there exist positive numbers η 1 Open image in new window, η 2 Open image in new window, r 1 Open image in new window, r 2 Open image in new window such that r 1 < r 2 Open image in new window, and
g ( y ) η 1 y , y [ 0 , r 1 ] , g ( y ) η 2 y , y [ r 2 , + ) . Open image in new window
Let η = min { η 1 , η 2 , min r 1 y r 2 g ( y ) y } > 0 Open image in new window. Then
g ( y ) η y for  y [ 0 , + ) . Open image in new window
Assume that y 1 ( t ) Open image in new window is a positive solution of (1.2). We show that this leads to a contradiction for λ > λ 0 : = ( γ m H l η ) 1 Open image in new window. Since T λ y 1 ( t ) = y 1 ( t ) Open image in new window, for t [ ν 3 , ν + b ] Z ν 3 Open image in new window, thus
y 1 = T λ y 1 λ s = ν + b 4 ν + 1 3 ( ν + b ) 4 ν + 1 G ( t , s ) h ( s + ν 1 ) g ( y ( s + ν 1 ) ) λ γ m h η y 1 > y 1 , Open image in new window

which is a contradiction. Thus, (1.2) has no positive solution. The proof is completed. □

6 Example

In this section, we present some examples to illustrate the main results.

Example 6.1 Suppose that ν = 5 2 Open image in new window, b = 8 Open image in new window. Let f ( t , y ( t ) ) : = | cos y ( t ) | 1 , 000 + t 2 Open image in new window and g ( y ) : = 1 20 sin y Open image in new window. Then (1.1) becomes
{ Δ 5 2 y ( t ) = | cos y ( t ) | 1 , 000 + t 2 , t = 0 , 1 , 2 , , 8 , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = 0 , y ( 21 2 ) = 1 20 sin y . Open image in new window
(6.1)
In this case, let α = 1 500 Open image in new window, β = 1 20 Open image in new window. Inequality (3.2) is
2 α j = 1 b ( ν + j j ) + β 165 250 + 1 20 < 1 . Open image in new window

Therefore, from Theorem 3.1 we deduce that problem (6.1) has a unique solution.

Example 6.2 Suppose that ν = 5 2 Open image in new window, b = 8 Open image in new window and K = 1 , 000 Open image in new window. Let f ( t , y ) : = 1 21 t exp { 1 10 y 2 } Open image in new window and g ( y ) : = 10 cos y Open image in new window. Then problem (1.1) is
{ Δ 5 2 y ( t ) = 5 t exp { 1 10 y 2 ( t ) } , t = 0 , 1 , 2 , , 8 , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = 0 , y ( 21 2 ) = 10 cos y . Open image in new window
(6.2)

The Banach space B : = { y R 12 : y 1 , 000 } Open image in new window.

We note that
K 2 Γ ( ν + b + 1 ) Γ ( ν + 1 ) Γ ( b + 1 ) + 1 = 1 , 000 2 Γ ( 23 2 ) Γ ( 7 2 ) Γ ( 9 ) + 1 > 1 , 000 970 10.3 . Open image in new window

It is clear that | f ( t , y ) | 0.5 < 10.3 Open image in new window, | g ( y ) | 10 < 10.3 Open image in new window. So, f and g satisfy the conditions. Thus, by Theorem 3.2 we deduce that problem (6.2) has at least one solution.

Example 6.3 Suppose that ν = 5 2 Open image in new window, b = 8 Open image in new window. Let h ( t ) = 2 t Open image in new window for t [ ν 1 , ν + b ] Z ν 1 Open image in new window, g ( y ) = y 3 Open image in new window for y ( 0 , + ) Open image in new window. Take L 1 = 400 Open image in new window, δ = 1 50 , 000 Open image in new window. Then f ( t , y ) = 2 t y 3 Open image in new window, and problem (1.2) becomes
{ Δ 5 2 y ( t ) = 2 λ t y 3 ( t ) , t [ 0 , 8 ] Z , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = 0 , y ( 21 2 ) = y 3 . Open image in new window
(6.3)
By routine numerical calculations, we have
M = max ( t , s ) [ ν 3 , ν + b ] Z ν 3 × [ 0 , b ] G ( t , s ) 5.1454 , H = 2 ( ν + b ) = 2 ( 5 2 + 8 ) = 21 , h = 2 ( ν 1 ) = 2 × 3 2 = 3 , m = s = ν + b 4 ν + 1 3 ( ν + b ) 4 ν + 1 G ( s + ν 1 , s ) = s = 2 6 G ( s + ν 1 , s ) 8.5318 Open image in new window
and
lim y 0 + g ( y ) y = 0 , lim y + g ( y ) y = + . Open image in new window
Then
( M H ( b + 1 ) + 1 ) δ = ( 5.1454 × 21 × 9 + 1 ) × 1 50 , 000 = 0.0195 , m h L 1 = 8.5318 × 3 × 400 = 10 , 238.16 , Open image in new window
thus ( M H b + 1 ) δ < m h L 1 Open image in new window. So, the conditions of Theorem 4.1 are satisfied. Since
( ( M H ( b + 1 ) + 1 ) δ ) 1 = 1 0.0195 51.2821 , ( m h L 1 ) 1 0.0001 , Open image in new window

thus by Theorem 4.1 we have that boundary value problem (6.3) has at least one positive solution for each λ ( 0.0001 , 51.2821 ) Open image in new window.

Example 6.4 Suppose that ν = 5 2 Open image in new window, b = 8 Open image in new window. Let h ( t ) = 1 500 t Open image in new window for t [ ν 1 , ν + b ] Z ν 1 Open image in new window, g ( y ) = e y Open image in new window for y ( 0 , + ) Open image in new window. Take L 2 = 1 , 000 Open image in new window. Then f ( t , y ) = 1 500 t e y Open image in new window, and problem (1.2) becomes
{ Δ 5 2 y ( t ) = 1 500 λ t e y ( t ) , t [ 0 , 8 ] Z , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = 0 , y ( 21 2 ) = e y . Open image in new window
(6.4)
By routine numerical calculations, we have
M = max ( t , s ) [ ν 3 , ν + b ] Z ν 3 × [ 0 , b ] G ( t , s ) 5.1454 , H = 1 500 ( ν + b ) = 1 500 ( 5 2 + 8 ) = 21 1 , 000 , h = 1 500 ( ν 1 ) = 1 500 × 3 2 = 3 1 , 000 Open image in new window
and
lim y 0 + g ( y ) y = lim y 0 + e y y = + , lim y + g ( y ) y = lim y + e y y = 0 . Open image in new window
Then
M H ( b + 1 ) + 1 = ( 5.1454 × 21 1 , 000 × 9 + 1 ) 1.9724 , m h L 2 = 8.5318 × 3 1 , 000 × 1 , 000 = 25.5954 , Open image in new window
thus M H ( b + 1 ) + 1 < m h L 2 Open image in new window. So, the conditions of Theorem 4.2 are satisfied. Since
( M H ( b + 1 ) + 1 ) 1 = 1 1.9724 0.507 , ( m h L 2 ) 1 0.0391 , Open image in new window

then by Theorem 4.2 we deduce that boundary value problem (6.4) has at least one positive solution for each λ ( 0.0391 , 0.507 ) Open image in new window.

Example 6.5 Suppose that ν = 5 2 Open image in new window, b = 8 Open image in new window. Let h ( t ) = 2 t Open image in new window for t [ ν 1 , ν + b ] Z ν 1 Open image in new window, g ( y ) = y Open image in new window for y ( 0 , + ) Open image in new window. Take c 1 = c 2 = 2 Open image in new window. Then f ( t , y ) = 2 t y Open image in new window, and problem (1.2) becomes
{ Δ 5 2 y ( t ) = 2 λ t y ( t ) , t [ 0 , 8 ] Z , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = 0 , y ( 21 2 ) = y . Open image in new window
(6.5)
Thus, we have
g 0 = lim y 0 + sup g ( y ) y = 1 < + , g = lim y + sup g ( y ) y = 1 < + . Open image in new window
By calculation,
M = max ( t , s ) [ ν 3 , ν + b ] Z ν 3 × [ 0 , b ] G ( t , s ) 5.1454 , H = 2 ( ν + b ) = 2 ( 5 2 + 8 ) = 21 , c = max { c 1 , c 2 , max r 1 y r 2 g ( y ) y } = 2 , Open image in new window
so
λ 0 = ( ( M H ( b + 1 ) + 1 ) c ) 1 = ( ( 5.1454 × 21 × 9 + 1 ) × 2 ) 1 0.0005 . Open image in new window

Therefore, by Theorem 5.1 we deduce that (6.5) has no positive solution for 0 < λ < λ 0 Open image in new window.

Example 6.6 Suppose that ν = 5 2 Open image in new window, b = 8 Open image in new window. Let h ( t ) = 1 1 , 050 t Open image in new window for t [ ν 1 , ν + b ] Z ν 1 Open image in new window, g ( y ) = y Open image in new window for y ( 0 , + ) Open image in new window. Take η 1 = η 2 = 1 2 Open image in new window. Then f ( t , y ) = 1 1 , 050 t y Open image in new window, and problem (1.2) becomes
{ Δ 5 2 y ( t ) = 2 λ t y ( t ) , t [ 0 , 8 ] Z , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = 0 , y ( 21 2 ) = y . Open image in new window
(6.6)
Thus, we have
g 0 = lim y 0 + sup g ( y ) y = 1 < + , g = lim y + sup g ( y ) y = 1 < + . Open image in new window
By calculation,
M = max ( t , s ) [ ν 3 , ν + b ] Z ν 3 × [ 0 , b ] G ( t , s ) 5.1454 , H = 1 1 , 050 ( ν + b ) = 1 1 , 050 ( 5 2 + 8 ) = 1 100 , η = min { η 1 , η 2 , min r 1 y r 2 g ( y ) y } = 1 2 , l = 3 ( ν + b ) 4 ν + 1 + 1 ν + b 4 ν + 1 = 25 4 . Open image in new window
By the definition of [14], we know
γ = min { 1 ( 3 ( ν + b ) 4 ) ν 1 ̲ [ ( 3 ( ν + b ) 4 ) ν 1 ̲ ( 3 ( ν + b ) 4 ( 1 ) ) ν 1 ̲ ( ν + b + 1 ) ν 1 ̲ ( ν + b + 1 ( 1 ) ) ν 1 ̲ ] , ( ν + b 4 ) ν 1 ( b + ν ) ν 1 ̲ } 0.518 , Open image in new window
and then
λ 0 : = ( γ M H l η ) 1 = ( γ M × 1 100 × 25 4 × 1 2 ) 1 12.01 . Open image in new window

Hence by Theorem 5.2 we deduce that (6.6) has no positive solution for λ > λ 0 Open image in new window.

7 Conclusion

This paper is an extension of [14] and [18]. The main contributions of this paper include:

  • The existence and uniqueness of a solution to a class of boundary value problems for a fractional difference equation with 3 < α 4 Open image in new window are studied by the contraction mapping theorem.

  • The existence of a solution to a class of boundary value problems for a fractional difference equation with 3 < α 4 Open image in new window is studied by the Brouwer fixed point theorem.

  • The eigenvalue intervals of a boundary value problem for a class nonlinear fractional difference equations with 3 < α 4 Open image in new window are investigated by the Guo-Krasnosel’skii fixed point theorem.

  • The nonexistence of a positive solution boundary value problem for a class nonlinear fractional difference equations with 3 < α 4 Open image in new window is considered in terms of parameter.

In contrast to [14] and [18], the similarities and differences are as follows:

  • The methods used to prove the existence results are standard and the same; however, their exposition in the framework of problems (1.1) and (1.2) is new.

  • The major difference is that the equations have different fractional order. The order is 3 < α 4 Open image in new window in this paper and 1 < α 2 Open image in new window in [14] and [18]. The higher order leads the comparable process to being more difficult and complex.

  • Nonlocal boundary conditions are considered in this paper and [18], Dirichlet boundary conditions are considered in [14].

  • Both the existence and nonexistence are considered in this paper, but only the existence is considered in [14] and [18].

Notes

Acknowledgements

The authors sincerely thank the reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original manuscript. This research is supported by the Natural Science Foundation of China (11071143), Natural Science Outstanding Youth Foundation of Shandong Province (JQ201119) and supported by Shandong Provincial Natural Science Foundation (ZR2012AM009), also supported by the Natural Science Foundation of Educational Department of Shandong Province (J11LA01).

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© Pan et al.; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  • Yuanyuan Pan
    • 1
  • Zhenlai Han
    • 1
    Email author
  • Shurong Sun
    • 1
    • 2
  • Chuanxia Hou
    • 1
  1. 1.School of Mathematical SciencesUniversity of JinanJinanChina
  2. 2.Department of Mathematics and StatisticsMissouri University of Science and TechnologyRollaUSA

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