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Existence of positive solutions for a discrete fractional boundary value problem

Abstract

This paper is concerned with the existence of positive solutions to a discrete fractional boundary value problem. By using the Krasnosel’skii and Schaefer fixed point theorems, the existence results are established. Additionally, examples are provided to illustrate the effectiveness of the main results.

MSC:26A33, 39A10, 47H07.

1 Introduction

Fractional differential equations have received increasing attention within the last ten years or so. The theory of fractional differential equations has been a new important mathematical branch due to its wide applications in different research areas and engineering, such as physics, chemistry, economics, control of dynamical etc. For more details, see [19] and the references therein. On the other hand, accompanied with the development of the theory for fractional calculus, fractional difference equations have attracted increasing attention slowly but steadily in the past three years or so. Some research papers have appeared, see [1019]. For example, Atici and Eloe [10] analyzed the conjugate discrete fractional boundary value problem (FBVP) with delta derivative:

{ Δ v y ( t ) = f ( t + v 1 , y ( t + v 1 ) ) , t [ 0 , b ] N 0 , y ( v 2 ) = y ( v + b + 1 ) = 0 , 1 < v 2 .

Goodrich [11] studied the discrete fractional boundary value problems:

{ Δ v y ( t ) = λ f ( t + v 1 , y ( t + v 1 ) ) , t [ 0 , T ] Z , y ( v 1 ) = y ( v + T ) + i = 1 N F ( t i , y ( t i ) ) , 0 < v < 1 .

In [12], Lv discussed the existence of solutions for discrete fractional boundary value problems with a p-Laplacian operator:

{ Δ c β [ ϕ p ( Δ c α u ) ] ( t ) = f ( t + α + β 1 , u ( t + α + β 1 ) ) , t [ 0 , b ] N 0 , Δ c α u ( t ) | t = β 1 + Δ c α u ( t ) | t = β + b = 0 , u ( α + β 2 ) + u ( α + β + b ) = 0 , 0 < α , β 1 , 1 < α + β 2 .

They obtained a series of excellent results of discrete fractional boundary value problems. Motivated by the aforementioned works, in this paper we consider a discrete fractional boundary value problem (FBVP):

{ Δ v y ( t ) = f ( t + v 1 , y ( t + v 1 ) ) , t [ 0 , b ] N 0 , y ( v 2 ) = 0 , Δ y ( v 2 ) = Δ y ( v + b 1 ) ,
(1.1)

where 1<v2, Δ v denotes the Riemann-Liouville fractional difference operator, N a ={a,a+1,a+2,} and I N a =I N a for any number aR and each interval I of R, b N 1 . We appeal to the convention that s = k k 1 y(s)=0 for any k N a , where y is a function defined on N a . By using the Krasnosel’skii and Schaefer fixed point theorems, the existence results are established and two examples are also provided to illustrate the effectiveness of the main results.

The rest of the paper is organized as follows. In Section 2, we introduce some lemmas and definitions which will be used later. In Section 3, the existence of positive solutions for the boundary value problem (1.1) is investigated. In Section 4, two examples are provided to illustrate the effectiveness of the main results.

2 Basic definitions and preliminaries

Firstly we present here some necessary definitions and lemmas which are used throughout this paper.

Definition 2.1 [13, 14]

Define t v ̲ := Γ ( t + 1 ) Γ ( t + 1 v ) for any t and v for which the right-hand side is defined. If t+1v is a pole of the gamma function and t+1 is not a pole, then t v ̲ =0.

Definition 2.2 [15]

The v th fractional sum of a function f, for v>0, is defined to be

Δ v f(t)= Δ v f(t;a):= 1 Γ ( v ) s = a t v ( t s 1 ) v 1 ̲ f(s)
(2.1)

for t{a+v,a+v+1,}:= N a + v . Define the v th fractional difference for v>0 by Δ v f(t):= Δ N Δ v N f(t), t N a + v and NN satisfies 0N1<vN.

Lemma 2.3 [15]

Let t and v be any numbers for which t v ̲ and t v 1 ̲ are defined. Then Δ t v ̲ =v t v 1 ̲ .

Lemma 2.4 [15]

Assume that 0N1<vN. Then

Δ v Δ v y(t)=y(t)+ C 1 t v 1 ̲ + C 2 t v 2 ̲ ++ C N t v N ̲
(2.2)

for some C i R, with 1iN.

Lemma 2.5 (The nonlinear alternative of Leray and Schauder [20])

Let E be a Banach space with CE closed and convex. Let U be a relatively open subset of C with 0U and T: U ¯ C be a continuous and compact mapping. Then either

  1. (a)

    the mapping T has a fixed point in U ¯ ; or

  2. (b)

    there exist uU and λ(0,1) with u=λTu.

Lemma 2.6 [13]

Let B be a Banach space and let KB be a cone. Assume that Ω 1 and Ω 2 are bounded open sets contained in B such that 0 Ω 1 and Ω ¯ 1 Ω 2 . Assume further that T:K( Ω ¯ 2 Ω 1 )K is a completely continuous operator. If either

  1. (i)

    Tyy for yK Ω 1 and Tyy for yK Ω 2 ; or

  2. (ii)

    Tyy for yK Ω 1 and Tyy for yK Ω 2 ;

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

We state next the structural assumptions that we impose on (1.1).

(H1) Assume that the nonlinearity function f: [ v 1 , v + b 1 ] N v 1 ×R[0,+) is continuous.

(H2) Assume that there exist nonnegative continuous functions a 1 (t), a 2 (t), t [ v 1 , v + b 1 ] N v 1 such that |f(t,y)| a 1 (t)+ a 2 (t)|y|, t [ v 1 , v + b 1 ] N v 1 , yR.

(H3) Assume that lim y 0 + f ( t , y ) y =0 uniformly for t [ v 1 , v + b 1 ] N v 1 .

(H4) Assume that lim y + f ( t , y ) y =+ uniformly for t [ v 1 , v + b 1 ] N v 1 .

3 Existence results

In this section, we will establish the existence of at least one positive solution for problem (1.1). At first, we state and prove some preliminary lemmas.

Lemma 3.1 Let h: [ v 1 , v + b 1 ] N v 1 R be given. Then the unique solution of the discrete fractional boundary value problem

{ Δ v y ( t ) = h ( t + v 1 ) , t [ 0 , b ] N 0 , y ( v 2 ) = 0 , Δ y ( v 2 ) = Δ y ( v + b 1 ) ,
(3.1)

is

y(t)= 1 Γ ( v ) s = 0 b G(t,s)h(s+v1).
(3.2)

Here, for (t,s) [ v 2 , v + b ] N v 2 × [ 0 , b ] N 0 , G(t,s) is defined by

G(t,s)= { ( b + v s 2 ) v 2 ̲ t v 1 ̲ Γ ( v 1 ) ( b + v 1 ) v 2 ̲ + ( t s 1 ) v 1 ̲ , 0 s t v b , ( b + v s 2 ) v 2 ̲ t v 1 ̲ Γ ( v 1 ) ( b + v 1 ) v 2 ̲ , 0 t v < s b .
(3.3)

Proof Suppose that y(t) defined on [ v 2 , v + b ] N v 2 is a solution of (3.1). Using Lemma 2.4, for some constants C 1 , C 2 R, we have

y(t)= 1 Γ ( v ) s = 0 t v ( t s 1 ) v 1 ̲ h(s+v1)+ C 1 t v 1 ̲ + C 2 t v 2 ̲ ,t [ v 2 , v + b ] N v 2 .
(3.4)

By y(v2)=0 and Definition 2.1, we obtain C 2 =0.

Then, for all t [ v 2 , v + b 1 ] N v 2 , we obtain [21]

Δy(t)= 1 Γ ( v 1 ) s = 0 t ( v 1 ) ( t s 1 ) v 2 ̲ h(s+v1)+ C 1 (v1) t v 2 ̲ .
(3.5)

In view of Δy(v2)=Δy(v+b1), we have

C 1 = 1 Γ ( v 1 ) [ Γ ( v ) ( v 1 ) ( b + v 1 ) v 2 ̲ ] s = 0 b ( b + v s 2 ) v 2 ̲ h(s+v1).

Substituting the values of C 1 and C 2 in (3.4), we have

y ( t ) = 1 Γ ( v ) s = 0 t v ( t s 1 ) v 1 ̲ h ( s + v 1 ) + s = 0 b ( b + v s 2 ) v 2 ̲ t v 1 ̲ Γ ( v 1 ) [ Γ ( v ) ( v 1 ) ( b + v 1 ) v 2 ̲ ] h ( s + v 1 ) = 1 Γ ( v ) s = 0 b G ( t , s ) h ( s + v 1 ) , t [ v 2 , v + b 1 ] N v 2 .

 □

Lemma 3.2 The function G(t,s) given in (3.3) satisfies the following:

  1. (1)

    0G(t,s) D ( v + b ) v 1 ̲ ( s + v 1 ) v 1 ̲ G(s+v1,s), (t,s) [ v 2 , v + b ] N v 2 × [ 0 , b ] N 0 ;

  2. (2)

    min t [ v 1 , v + b ] N v 1 G(t,s) Γ ( v ) ( s + v 1 ) v 1 ̲ G(s+v1,s)>0.

Here,

D = max s [ 0 , b ] { 1 + Γ ( v 1 ) ( v + b 1 ) v 2 ̲ ( v + b s 2 ) v 2 ̲ } = 1 + Γ ( v 1 ) ( v + b 1 ) v 2 ̲ ( v + b 2 ) v 2 ̲ .
(3.6)

Proof First of all, (3.3) implies that G(s+v1,s)= ( v + b s 2 ) v 2 ̲ ( s + v 1 ) v 1 ̲ Γ ( v 1 ) ( v + b 1 ) v 2 ̲ . Note that Γ(v1) ( v + b 1 ) v 2 ̲ >0, we know G(s+v1,s)>0.

Second of all, by (3.3) and the definition of D in (3.6), we obtain

0 G ( t , s ) ( b + v s 2 ) v 2 ̲ ( v + b ) v 1 ̲ Γ ( v 1 ) ( b + v 1 ) v 2 ̲ + ( v + b ) v 1 ̲ D ( b + v ) v 1 ̲ ( s + v 1 ) v 1 ̲ G ( s + v 1 , s ) .
(3.7)

On the other hand,

min t [ v 1 , v + b ] N v 1 G ( t , s ) = min t [ v 1 , v + b ] N v 1 ( b + v s 2 ) v 2 ̲ t v 1 ̲ Γ ( v 1 ) ( b + v 1 ) v 2 ̲ ( v 1 ) v 1 ̲ ( s + v 1 ) v 1 ̲ ( b + v s 2 ) v 2 ̲ ( s + v 1 ) v 1 ̲ Γ ( v 1 ) ( v + b 1 ) v 2 ̲ = Γ ( v ) ( s + v 1 ) v 1 ̲ G ( s + v 1 , s ) > 0 .
(3.8)

The proof of Lemma 3.2 is completed. □

Let B be the collection of all functions y: [ v 2 , v + b ] N v 2 R with the norm y=max{|y(t)|:t [ v 2 , v + b ] N v 2 }.

Define the operator T:BB by

Ty(t)= 1 Γ ( v ) s = 0 b G(t,s)f ( s + v 1 , y ( s + v 1 ) ) .
(3.9)

In view of the continuity of f, it is easy to know that T is continuous. Furthermore, it is not difficult to verify that T maps bounded sets into bounded sets and equi-continuous sets. Therefore, in the light of the well-known Arzelá-Ascoli theorem, we know that T is a compact operator (see [11, 12]).

Let E={yB|y(t)0,t [ v 2 , v + b ] N v 2 } and set

A = s = 0 b D ( v + b ) v 1 ̲ G ( s + v 1 , s ) Γ ( v ) ( s + v 1 ) v 1 ̲ a 2 , B = s = 0 b D ( v + b ) v 1 ̲ G ( s + v 1 , s ) Γ ( v ) ( s + v 1 ) v 1 ̲ a 1 .

We have the following theorem.

Theorem 3.3 Assume that (H1) and (H2) hold. Then system (1.1) has at least one positive solution provided that

s = 0 b D ( v + b ) v 1 ̲ G ( s + v 1 , s ) Γ ( v ) ( s + v 1 ) v 1 ̲ a 2 <1.
(3.10)

Proof Let Ω={yE|y<r} with r= B 1 A >0. If y Ω ¯ , that is, yr. From (H1), (H2) and (3.9), we have

T y ( t ) = max t [ v 2 , v + b ] N v 2 | 1 Γ ( v ) s = 0 b G ( t , s ) f ( s + v 1 , y ( s + v 1 ) ) | 1 Γ ( v ) s = 0 b D ( v + b ) v 1 ̲ ( s + v 1 ) v 1 ̲ G ( s + v 1 , s ) ( | a 1 ( t ) | + | a 2 ( t ) | | y ( t ) | ) s = 0 b ( v + b ) v 1 ̲ Γ ( v ) D G ( s + v 1 , s ) ( s + v 1 ) v 1 ̲ a 1 + s = 0 b ( v + b ) v 1 ̲ Γ ( v ) D G ( s + v 1 , s ) ( s + v 1 ) v 1 ̲ a 2 y = B + A y r ,

which shows that Ty Ω ¯ .

Consider the eigenvalue problem

y=λTy,λ(0,1).
(3.11)

Assume that y is a solution of (3.11), we obtain

y=λTy<Tyr.
(3.12)

It shows that yΩ. By Lemma 2.5, T has a fixed point in Ω ¯ . The proof is completed. □

We define the cone KB by

K= { y E : min t [ v 1 , v + b ] N v 1 y ( t ) Γ ( v ) D ( v + b ) v 1 ̲ y } .
(3.13)

Lemma 3.4 Let T be the operator defined in (3.9) and K be the cone defined in (3.13). Then T:KK.

Proof Note that for each t [ v 2 , v + b ] N v 2 , we have

T y ( t ) = max t [ v 2 , v + b ] N v 2 | 1 Γ ( v ) s = 0 b G ( t , s ) f ( s + v 1 , y ( s + v 1 ) ) | 1 Γ ( v ) s = 0 b D ( v + b ) v 1 ̲ ( s + v 1 ) v 1 ̲ G ( s + v 1 , s ) | f ( s + v 1 , y ( s + v 1 ) ) | = s = 0 b D ( v + b ) v 1 ̲ G ( s + v 1 , s ) Γ ( v ) ( s + v 1 ) v 1 ̲ | f ( s + v 1 , y ( s + v 1 ) ) | .

Therefore, it holds that

min t [ v 1 , v + b ] N v 1 ( T y ) ( t ) = min t [ v 1 , v + b ] N v 1 1 Γ ( v ) s = 0 b G ( t , s ) f ( s + v 1 , y ( s + v 1 ) ) 1 Γ ( v ) s = 0 b Γ ( v ) ( s + v 1 ) v 1 ̲ G ( s + v 1 , s ) f ( s + v 1 , y ( s + v 1 ) ) = Γ ( v ) D ( v + b ) v 1 ̲ s = 0 b D ( v + b ) v 1 ̲ G ( s + v 1 , s ) Γ ( v ) ( s + v 1 ) v 1 ̲ f ( s + v 1 , y ( s + v 1 ) ) Γ ( v ) D ( v + b ) v 1 ̲ T y .

The conclusion of Lemma 3.4 holds. □

Theorem 3.5 Suppose that conditions (H1), (H3) and (H4) hold. Then problem (1.1) has at least one positive solution.

Proof We have already shown T(K)K in Lemma 3.4. By condition (H3), we can select η 1 >0 sufficiently small so that both |f(t,y)| η 1 y and η 1 s = 0 b ( v + b ) v 1 ̲ Γ ( v ) D G ( s + v 1 , s ) ( s + v 1 ) v 1 ̲ <1 hold for all t [ v 1 , v + b 1 ] N v 1 and 0<y< r 1 , where r 1 := r 1 ( η 1 ).

Let Ω 1 ={yB:y< r 1 }. Then, for y Ω 1 K, we have

T y ( t ) = max t [ v 2 , v + b ] N v 2 | 1 Γ ( v ) s = 0 b G ( t , s ) f ( s + v 1 , y ( s + v 1 ) ) | 1 Γ ( v ) s = 0 b D ( v + b ) v 1 ̲ ( s + v 1 ) v 1 ̲ G ( s + v 1 , s ) | f ( s + v 1 , y ( s + v 1 ) ) | s = 0 b D ( v + b ) v 1 ̲ G ( s + v 1 , s ) Γ ( v ) ( s + v 1 ) v 1 ̲ η 1 y < y .

It implies that T is a cone contraction on y Ω 1 K.

On the other hand, from condition (H4), we may select a number η 2 >0 such that both |f(t,y)|> η 2 y and η 2 s = 0 b G ( s + v 1 , s ) ( s + v 1 ) v 1 ̲ >1 hold for all t [ v 1 , v + b 1 ] N v 1 and 0<y< r 2 , where r 2 := r 2 ( η 2 ) and r 2 > r 1 >0. Define Ω 2 ={yB:y< r 2 }, we obtain

T y ( t ) = max t [ v 2 , v + b ] N v 2 1 Γ ( v ) s = 0 b G ( t , s ) | f ( s + v 1 , y ( s + v 1 ) ) | 1 Γ ( v ) s = 0 b min t [ v 1 , v + b ] N v 1 G ( t , s ) | f ( s + v 1 , y ( s + v 1 ) ) | > s = 0 b G ( s + v 1 , s ) ( s + v 1 ) v 1 ̲ η 2 y > y ,

whenever y Ω 2 K, so that T is a cone expansion on Ω 2 K.

In summary, we may invoke Lemma 2.6 to deduce the existence of a function y 0 K( Ω ¯ 2 Ω 1 ) such that T y 0 = y 0 , where y 0 is a positive solution to problem (1.1). The proof is completed. □

4 Example

Example 4.1 Consider the fractional difference boundary value problem

{ Δ 3 2 y ( t ) = f ( t + 1 2 , y ( t + 1 2 ) ) , t [ 0 , 3 ] N 0 , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = Δ y ( 7 2 ) .
(4.1)

Set a 1 (t)=1, a 2 (t)= t 30 , f(t,y)= t | y | 30 +|sint|, t [ 1 2 , 7 2 ] N 1 2 . We have

| f ( t , y ) | 1+ t 30 |y|.

By a simple computation, we can obtain D1.7750, s = 0 3 ( 9 2 ) 1 2 ̲ G ( s + 1 2 , s ) Γ ( 3 2 ) ( s + 1 2 ) 1 2 ̲ 4.7688, a 2 = 7 60 . Therefore, A0.9875<1. The conditions of Theorem 3.3 hold, the boundary value problem (4.1) has at least one positive solution.

Example 4.2 Consider the fractional difference boundary value problem

{ Δ 3 2 y ( t ) = f ( t + 1 2 , y ( t + 1 2 ) ) , t [ 0 , 11 ] N 0 , y ( 1 2 ) = 0 , Δ y ( 1 2 ) = Δ y ( 23 2 ) .
(4.2)

Set f(t,y)=t y 2 , t [ 1 2 , 23 2 ] N 1 2 . We have

(1) lim y 0 + f ( t , y ) y = lim y 0 + t y 2 y =0,(2) lim y + f ( t , y ) y = lim y + t y 2 y =+.

The conditions of Theorem 3.5 hold, the boundary value problem (4.2) has at least one positive solution.

Misc

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Acknowledgements

This work was jointly supported by the Natural Science Foundation of China under Grants 11471278, the Natural Science Foundation of Hunan Province under Grants 13JJ3120 and 14JJ2133, and the Construct Program of the Key Discipline in Hunan Province. We would like to show our great thanks to the anonymous referee for his/her valuable suggestions and comments, which improve the former version of this paper and make us rewrite the paper in a more clear way.

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Wang, J., Xiang, H. & Chen, F. Existence of positive solutions for a discrete fractional boundary value problem. Adv Differ Equ 2014, 253 (2014). https://doi.org/10.1186/1687-1847-2014-253

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Keywords

  • existence
  • positive solution
  • discrete
  • fractional boundary value problem