Open Access

Three-point boundary value problems of fractional functional differential equations with delay

Boundary Value Problems20132013:38

DOI: 10.1186/1687-2770-2013-38

Received: 6 December 2012

Accepted: 7 February 2013

Published: 22 February 2013

Abstract

In this paper, we study three-point boundary value problems of the following fractional functional differential equations involving the Caputo fractional derivative:

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equa_HTML.gif

where D α C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq1_HTML.gif, D β C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq2_HTML.gif denote Caputo fractional derivatives, 2 < α < 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq3_HTML.gif, 0 < β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq4_HTML.gif, η ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq5_HTML.gif, 1 < λ < 1 2 η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq6_HTML.gif. We use the Green function to reformulate boundary value problems into an abstract operator equation. By means of the Schauder fixed point theorem and the Banach contraction principle, some existence results of solutions are obtained, respectively. As an application, some examples are presented to illustrate the main results.

MSC:34A08, 34K37.

Keywords

fractional functional differential equation delay three-point boundary value problems fixed point theorem existence of solutions

1 Introduction

Fractional calculus is a branch of mathematics, it is an emerging field in the area of the applied mathematics that deals with derivatives and integrals of arbitrary orders as well as with their applications. The origins can be traced back to the end of the seventeenth century. During the history of fractional calculus, it was reported that the pure mathematical formulations of the investigated problems started to be addressed with more applications in various fields. With the help of fractional calculus, we can describe natural phenomena and mathematical models more accurately. Therefore, fractional differential equations have received much attention and the theory and its application have been greatly developed; see [16].

Recently, there have been many papers focused on boundary value problems of fractional ordinary differential equations [715] and an initial value problem of fractional functional differential equations [1628]. But the results dealing with the boundary value problems of fractional functional differential equations with delay are relatively scarce [2935]. It is well known that in practical problems, the behavior of systems not only depends on the status just at the present, but also on the status in the past. Thus, in many cases, we must consider fractional functional differential equations with delay in order to solve practical problems. Consequently, our aim in this paper is to study the existence of solutions for boundary value problems of fractional functional differential equations.

In 2011, Rehman [12] studied the existence and uniqueness of solutions to nonlinear three-point boundary value problems for the following fractional differential equation:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equb_HTML.gif

where 1 < δ < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq7_HTML.gif, 0 < σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq8_HTML.gif, α , β R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq9_HTML.gif, α η ( 1 β ) + ( 1 α ) ( t β η ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq10_HTML.gif and D 0 + δ C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq11_HTML.gif, D 0 + σ C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq12_HTML.gif denote Caputo fractional derivatives. By the Banach contraction principle and the Schauder fixed point theorem, they obtained some new existence and uniqueness results.

For 0 < r < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq13_HTML.gif, we denote by C r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq14_HTML.gif the Banach space of all continuous functions φ : [ r , 0 ] R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq15_HTML.gif endowed with the sup-norm
φ [ r , 0 ] : = sup { | φ ( s ) | : s [ r , 0 ] } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equc_HTML.gif
If u : [ r , 1 ] R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq16_HTML.gif, then for any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq17_HTML.gif, we denote by u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq18_HTML.gif the element of C r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq14_HTML.gif defined by
u t ( θ ) = u ( t + θ ) , for  θ [ r , 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equd_HTML.gif
Enlightened by literature [12], in this paper we study the following three-point boundary value problem for the fractional functional differential equation:
D α C u ( t ) = f ( t , u t , C D β u ( t ) ) , 0 < t < 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ1_HTML.gif
(1.1)
where 2 < α < 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq3_HTML.gif, 0 < β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq4_HTML.gif and D α C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq1_HTML.gif, D β C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq2_HTML.gif denote Caputo fractional derivatives, f ( t , u t , C D β u ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq19_HTML.gif is a continuous function associated with the boundary conditions
u ( 0 ) = 0 , u ( 1 ) = λ u ( η ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ2_HTML.gif
(1.2)
and u 0 = φ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq20_HTML.gif, where η ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq5_HTML.gif, 1 < λ < 1 2 η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq6_HTML.gif and φ is an element of the space
C r + ( 0 ) : = { ψ C r | ψ ( s ) 0 , s [ r , 0 ] , ψ ( 0 ) = 0 , C D β ψ ( s ) = 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Eque_HTML.gif

To the best of our knowledge, no one has studied the existence of positive solutions for problem (1.1)-(1.2). The aim of this paper is to fill the gap in the relevant literatures. In this paper, we firstly give the fractional Green function and some properties of the Green function. Consequently, boundary value problem (1.1) and (1.2) is reduced to an equivalent Fredholm integral equation. Then we extend the existence results for boundary value problems of an ordinary fractional differential equation of δ-order ( 1 < δ < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq7_HTML.gif) in [12] to a fractional functional differential equation of α-order ( 2 < α < 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq3_HTML.gif). As an application, some examples are presented to illustrate the main results.

2 Preliminaries

For the convenience of the reader, we give the following background material from fractional calculus theory to facilitate the analysis of boundary value problem (1.1) and (1.2). This material can be found in the recent literature; see [1, 2, 36].

Definition 2.1 ([1])

The fractional integral of order α ( α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq21_HTML.gif) of a function f : ( t 0 , + ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq22_HTML.gif is given by
I α f ( t ) = 1 Γ ( α ) t 0 t f ( s ) ( t s ) 1 α d s , t > t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equf_HTML.gif

where Γ ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq23_HTML.gif is the gamma function, provided that the right-hand side is point-wise defined on ( t 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq24_HTML.gif.

Definition 2.2 ([1])

The Caputo fractional derivative of order α ( n 1 < α < n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq25_HTML.gif) of a function f : ( t 0 , + ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq22_HTML.gif is given by
D α C f ( t ) = 1 Γ ( n α ) t 0 t f ( n ) ( s ) ( t s ) α + 1 n d s , t > t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equg_HTML.gif

where Γ ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq23_HTML.gif is the gamma function, provided that the right-hand side is point-wise defined on ( t 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq24_HTML.gif.

Obviously, the Caputo derivative for every constant function is equal to zero.

From the definition of the Caputo derivative, we can acquire the following statement.

Lemma 2.1 ([2])

Let f ( t ) L 1 [ t 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq26_HTML.gif. Then
D α C ( I α f ( t ) ) = f ( t ) , t > t 0 and 0 < α < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equh_HTML.gif

Lemma 2.2 ([2])

Let α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq21_HTML.gif. Then
I α C D α f ( t ) = f ( t ) c 1 c 2 t c n t n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equi_HTML.gif

for some c i R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq27_HTML.gif, i = 1 , 2 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq28_HTML.gif, where n = [ α ] + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq29_HTML.gif and [ α ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq30_HTML.gif denotes the integer part of α.

Next, we introduce the Green function of fractional functional differential equations boundary value problems.

Lemma 2.3 Let 2 < α < 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq3_HTML.gif, 0 < η < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq31_HTML.gif, 1 < λ < 1 2 η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq6_HTML.gif and h : [ 0 , 1 ] R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq32_HTML.gif be continuous. Then the boundary value problem
D α C u ( t ) = h ( t ) , 0 < t < 1 , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ u ( η ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ3_HTML.gif
(2.1)
has a unique solution
u ( t ) = 0 1 G ( t , s ) h ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ4_HTML.gif
(2.2)
where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ5_HTML.gif
(2.3)
Proof From equation (2.1), we know
I α C D α u ( t ) = I α h ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equj_HTML.gif
From Lemma 2.2, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ6_HTML.gif
(2.4)
According to (2.1), we know that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equk_HTML.gif
By u ( 1 ) = λ u ( η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq33_HTML.gif, we have
c 3 = α 1 ( 2 2 λ η ) Γ ( α ) ( 0 1 ( 1 s ) α 2 h ( s ) d s λ 0 η ( η s ) α 2 h ( s ) d s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equl_HTML.gif
Therefore,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equm_HTML.gif
Now, for t η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq34_HTML.gif, we have
u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 h ( s ) d s + ( α 1 ) t 2 ( 2 2 λ η ) Γ ( α ) ( ( 0 t + t η + η 1 ) ( 1 s ) α 2 h ( s ) d s λ ( 0 t + t η ) ( η s ) α 2 h ( s ) d s ) = 1 Γ ( α ) 0 t ( ( t s ) α 1 + ( α 1 ) t 2 2 2 λ η ( ( 1 s ) α 2 λ ( η s ) α 2 ) ) h ( s ) d s + 1 Γ ( α ) t η ( α 1 ) t 2 2 2 λ η ( ( 1 s ) α 2 λ ( η s ) α 2 ) h ( s ) d s + 1 Γ ( α ) η 1 ( α 1 ) t 2 2 2 λ η ( 1 s ) α 2 h ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equn_HTML.gif
For t η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq35_HTML.gif, we have
u ( t ) = 1 Γ ( α ) ( 0 η + η t ) ( t s ) α 1 h ( s ) d s + ( α 1 ) t 2 ( 2 2 λ η ) Γ ( α ) ( ( 0 η + η t + t 1 ) ( 1 s ) α 2 h ( s ) d s λ 0 η ( η s ) α 2 h ( s ) d s ) = 1 Γ ( α ) 0 η ( ( t s ) α 1 + ( α 1 ) t 2 2 2 λ η ( ( 1 s ) α 2 λ ( η s ) α 2 ) ) h ( s ) d s + 1 Γ ( α ) η t ( ( t s ) α 1 + ( α 1 ) t 2 2 2 λ η ( 1 s ) α 2 ) h ( s ) d s + 1 Γ ( α ) t 1 ( α 1 ) t 2 2 2 λ η ( 1 s ) α 2 h ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equo_HTML.gif
Hence, we can conclude (2.2) holds, where
G ( t , s ) = 1 Γ ( α ) { ( t s ) α 1 + ( α 1 ) t 2 2 2 λ η ( ( 1 s ) α 2 λ ( η s ) α 2 ) , 0 s t 1 , s η , ( t s ) α 1 + ( α 1 ) t 2 2 2 λ η ( 1 s ) α 2 , 0 s t 1 , η s , ( α 1 ) t 2 2 2 λ η ( ( 1 s ) α 2 λ ( η s ) α 2 ) , 0 t s 1 , s η , ( α 1 ) t 2 2 2 λ η ( 1 s ) α 2 , 0 t s 1 , η s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equp_HTML.gif

The proof is completed. □

Lemma 2.4 ([36] Schauder fixed point theorem)

Let ( D , d ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq36_HTML.gif be a complete metric space, U be a closed convex subset of D, and T : D D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq37_HTML.gif be the map such that the set T u : u U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq38_HTML.gif is relatively compact in D. Then the operator T has at least one fixed point u U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq39_HTML.gif:
T u = u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equq_HTML.gif

3 Main results

In this section, we discuss the existence and uniqueness of solutions for boundary value problem (1.1) and (1.2) by the Schauder fixed point theorem and the Banach contraction principle.

For convenience, we define the Banach space X = { u | u C [ r , 1 ] , C D β u C [ r , 1 ] , 0 < β < 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq40_HTML.gif. Also, if I is an interval of the real line , by C ( I ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq41_HTML.gif and C 1 ( I ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq42_HTML.gif we denote the set of continuous and continuously differentiable functions on I, respectively. Moreover, for u C ( I ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq43_HTML.gif, we define
u I = max t I | u ( t ) | + max t I | C D β u ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ7_HTML.gif
(3.1)
For u 0 = φ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq20_HTML.gif, in view of the definitions of u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq18_HTML.gif and φ, we have
u 0 = u ( θ ) = φ ( θ ) , for  θ [ r , 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equr_HTML.gif
Thus, we have
u ( t ) = φ ( t ) , for  t [ r , 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equs_HTML.gif
Since f : [ 0 , 1 ] × C r × R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq44_HTML.gif is a continuous function, set f ( t , u t , C D β u ( t ) ) : = h ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq45_HTML.gif in Lemma 2.3. We have by Lemma 2.3 that a function u is a solution of boundary value problem (1.1) and (1.2) if and only if it satisfies
u ( t ) = { 0 1 G ( t , s ) f ( s , u s , C D β u ( s ) ) d s , t ( 0 , 1 ) , φ ( t ) , t [ r , 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equt_HTML.gif
We define an operator T : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq46_HTML.gif as follows:
T u ( t ) = u ( t ) = { 0 1 G ( t , s ) f ( s , u s , C D β u ( s ) ) d s , t ( 0 , 1 ) , φ ( t ) , t [ r , 0 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ8_HTML.gif
(3.2)
and
l = max 0 t 1 ( 0 1 | G ( t , s ) g ( s ) | d s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equu_HTML.gif
l = max 0 t 1 ( 0 1 | t G ( t , s ) g ( s ) | d s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equv_HTML.gif
Q = 1 Γ ( 2 β ) + 1 + λ η α 1 Γ ( 2 β ) ( 1 λ η ) + 1 α + λ η α 1 + 1 2 2 λ η . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equw_HTML.gif

Theorem 3.1 Assume the following:

(H1) There exists a nonnegative function g L [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq47_HTML.gif such that
| f ( t , v , w ) | g ( t ) + a | v | k 1 + b | w | k 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equx_HTML.gif

for each v C r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq48_HTML.gif, w R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq49_HTML.gif, where a , b R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq50_HTML.gif are nonnegative constants and 0 < k 1 , k 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq51_HTML.gif; or

(H2) There exists a nonnegative function g L [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq47_HTML.gif such that
| f ( t , v , w ) | g ( t ) + a | v | k 1 + b | w | k 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equy_HTML.gif

for each v C r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq48_HTML.gif, w R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq49_HTML.gif, where a , b R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq50_HTML.gif are nonnegative constants and k 1 , k 2 > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq52_HTML.gif.

Then boundary value problem (1.1) and (1.2) has a solution.

Proof Suppose (H1) holds. Choose
ω max { 3 ( l + l Γ ( 2 β ) ) , ( 3 a Q ) 1 1 k 1 , ( 3 b Q ) 1 1 k 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ9_HTML.gif
(3.3)

and define the cone U = { u X | u ω , ω > 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq53_HTML.gif.

For any u U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq54_HTML.gif, we have
| T u ( t ) | = | 0 1 G ( t , s ) f ( s , u s , C D β u ( s ) ) d s | 0 1 | G ( t , s ) g ( s ) | d s + ( a | ω | k 1 + b | ω | k 2 ) ( 0 t ( t s ) α 1 Γ ( α ) d s + ( α 1 ) t 2 2 2 λ η 0 1 ( 1 s ) α 2 Γ ( α ) d s + ( α 1 ) λ t 2 2 2 λ η 0 η ( η s ) α 2 Γ ( α ) d s ) l + ( a | ω | k 1 + b | ω | k 2 ) ( t α α Γ ( α ) + t 2 ( 2 2 λ η ) Γ ( α ) + λ t 2 η α 1 ( 2 2 λ η ) Γ ( α ) ) l + a | ω | k 1 + b | ω | k 2 Γ ( α ) ( 1 α + λ η α 1 + 1 2 2 λ η ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ10_HTML.gif
(3.4)
Also,
| T u ( t ) | 0 1 | t G ( t , s ) | | f ( s , u s , C D β u ( s ) ) | d s 0 1 | t G ( t , s ) g ( s ) | d s + ( a | ω | k 1 + b | ω | k 2 ) ( 0 t ( t s ) α 2 Γ ( α 1 ) d s + 2 ( α 1 ) t 2 2 λ η 0 1 ( 1 s ) α 2 Γ ( α ) d s + 2 ( α 1 ) λ t 2 2 λ η 0 η ( η s ) α 2 Γ ( α ) d s ) l + ( a | ω | k 1 + b | ω | k 2 ) ( t α 1 Γ ( α ) + t ( 1 λ η ) Γ ( α ) + λ t η α 1 ( 1 λ η ) Γ ( α ) ) l + a | ω | k 1 + b | ω | k 2 Γ ( α ) ( 1 + 1 + λ η α 1 1 λ η ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equz_HTML.gif
Hence,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equaa_HTML.gif
In view of (3.1) and (3.3), we obtain
T u ( t ) l + l Γ ( 2 β ) + a | ω | k 1 + b | ω | k 2 Γ ( α ) ( 1 Γ ( 2 β ) + 1 + λ η α 1 Γ ( 2 β ) ( 1 λ η ) + 1 α + λ η α 1 + 1 2 2 λ η ) ω 3 + ( a | ω | k 1 + b | ω | k 2 ) Q ω 3 + ω 3 + ω 3 ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ11_HTML.gif
(3.5)

which implies that T : U U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq55_HTML.gif. The continuity of the operator T follows from the continuity of f and G.

Now, if (H2) holds, we choose
0 < ω min { 3 ( l + l Γ ( 2 β ) ) , ( 1 3 a Q ) 1 1 k 1 , ( 1 3 b Q ) 1 1 k 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ12_HTML.gif
(3.6)
and by the same process as above, we obtain
T u ( t ) l + l Γ ( 2 β ) + a | ω | k 1 + b | ω | k 2 Γ ( α ) ( 1 Γ ( 2 β ) + 1 + λ η α 1 Γ ( 2 β ) ( 1 λ η ) + 1 α + λ η α 1 + 1 2 2 λ η ) ω 3 + ( a | ω | k 1 + b | ω | k 2 ) Q ω 3 + ω 3 + ω 3 ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equab_HTML.gif

which implies that T : U U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq55_HTML.gif.

Now, we show that T is a completely continuous operator.

Let L = max 0 t 1 | f ( t , u t , C D β u ( t ) ) | + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq56_HTML.gif. Then for u U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq54_HTML.gif and t 1 , t 2 [ r , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq57_HTML.gif with t 1 < t 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq58_HTML.gif, in view of Lemma 2.3, if 0 t 1 < t 2 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq59_HTML.gif, then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equac_HTML.gif
If r t 1 < t 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq60_HTML.gif, then
| T u ( t 2 ) T u ( t 1 ) | = | φ ( t 2 ) φ ( t 1 ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equad_HTML.gif
If r t 1 < 0 < t 2 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq61_HTML.gif, then
| T u ( t 2 ) T u ( t 1 ) | | T u ( t 2 ) T u ( 0 ) | + | T u ( 0 ) T u ( t 1 ) | 0 1 | G ( t 2 , s ) G ( 0 , s ) | | f ( s , u s , C D β u ( s ) ) | d s + | φ ( 0 ) φ ( t 1 ) | L Γ ( α ) | t 2 α α + t 2 2 ( 1 λ η α 1 ) 2 2 λ η | + φ ( t 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equae_HTML.gif
Hence, if 0 t 1 < t 2 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq59_HTML.gif, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equaf_HTML.gif
If r t 1 < t 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq60_HTML.gif, in view of the definition of φ, we have
| C D β T u ( t 2 ) C D β T u ( t 1 ) | = | C D β φ ( t 2 ) C D β φ ( t 1 ) | = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equag_HTML.gif
If r t 1 < 0 < t 2 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq61_HTML.gif, then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equah_HTML.gif
Hence, if 0 t 1 < t 2 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq59_HTML.gif, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equai_HTML.gif
If r t 1 < t 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq60_HTML.gif, we have
T u ( t 2 ) T u ( t 1 ) = φ ( t 2 ) φ ( t 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equaj_HTML.gif
If r t 1 < 0 < t 2 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq61_HTML.gif, then
T u ( t 2 ) T u ( t 1 ) 1 Γ ( α ) | t 2 α α + t 2 2 ( 1 λ η α 1 ) 2 2 λ η | + φ ( t 1 ) + L t 2 1 β Γ ( 2 β ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equak_HTML.gif

In any case, it implies that T u ( t 2 ) T u ( t 1 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq62_HTML.gif as t 2 t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq63_HTML.gif, i.e., for any ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq64_HTML.gif, there exists δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq65_HTML.gif, independent of t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq66_HTML.gif, t 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq67_HTML.gif and u, such that | T u ( t 2 ) T u ( t 1 ) | ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq68_HTML.gif, whenever | t 2 t 1 | < δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq69_HTML.gif. Therefore T : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq46_HTML.gif is completely continuous. The proof is completed. □

For convenience, we denote
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equal_HTML.gif

Theorem 3.2 Assume that

(H3) There exists a constant p > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq70_HTML.gif such that | f ( t , μ , ν ) f ( t , μ ¯ , ν ¯ ) | p ( | μ μ ¯ | + | ν ν ¯ | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq71_HTML.gif for each μ , μ ¯ C r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq72_HTML.gif, ν , ν ¯ R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq73_HTML.gif. If
p < ( M + N ) 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equam_HTML.gif

then boundary value problem (1.1) and (1.2) has a unique solution.

Proof Consider the operator T : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq46_HTML.gif defined by (3.2). Clearly, the fixed point of the operator T is the solution of boundary value problem (1.1) and (1.2). We will use the Banach contraction principle to prove that T has a fixed point. We first show that T is a contraction. For each t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq17_HTML.gif,
| T u ( t ) T u ¯ ( t ) | 0 1 | G ( t , s ) | | f ( s , u s , C D β u ( s ) ) f ( s , u ¯ s , C D β u ¯ ( s ) ) | d s p u u ¯ Γ ( α ) ( 0 t ( t s ) α 1 d s + ( α 1 ) t 2 2 2 λ η 0 1 ( 1 s ) α 2 d s + λ ( α 1 ) t 2 2 2 λ η 0 η ( η s ) α 2 d s ) p u u ¯ Γ ( α ) ( t α α + t 2 2 2 λ η + t 2 λ η α 2 2 2 λ η ) p u u ¯ Γ ( α ) ( 1 α + 1 2 2 λ η + λ η α 2 2 2 λ η ) p u u ¯ M . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ13_HTML.gif
(3.7)
By a similar method, we get
| C D β T u ( t ) C D β T u ¯ ( t ) | = | 1 Γ ( 1 β ) 0 t ( t s ) β ( T u ( s ) T u ¯ ( s ) ) d s | 1 Γ ( 1 β ) 0 t ( t s ) β ( 0 1 | s G ( s , z ) | | f ( z , u z , C D β u ( z ) ) f ( z , u ¯ z , C D β u ¯ ( z ) ) | d z ) d s p u u ¯ Γ ( 1 β ) 0 t ( t s ) β ( 0 1 | s G ( s , z ) | d z ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ14_HTML.gif
(3.8)
In view of the definition of G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq74_HTML.gif, we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equan_HTML.gif
Hence,
| C D β T u ( t ) C D β T u ¯ ( t ) | p u u ¯ Γ ( 1 β ) 0 t ( t s ) β Γ ( α ) ( 1 + 1 + λ η α 1 1 λ η ) d s p u u ¯ Γ ( 1 β ) 1 Γ ( α ) 1 1 β ( 1 + 1 + λ η α 1 1 λ η ) p u u ¯ Γ ( 2 β ) Γ ( α ) ( 1 + 1 + λ η α 1 1 λ η ) p u u ¯ N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ15_HTML.gif
(3.9)
Clearly, for each t [ r , 0 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq75_HTML.gif, we have | T u ( t ) T u ¯ ( t ) | = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq76_HTML.gif. Therefore, by (3.7) and (3.9), we get
T u T u ¯ p u u ¯ M + p u u ¯ N p u u ¯ ( N + M ) u u ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equao_HTML.gif

and T is a contraction. As a consequence of the Banach contraction principle, we get that T has a fixed point which is a solution of boundary value problem (1.1) and (1.2). □

4 Example

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

Example 4.1

Consider boundary value problems of the following fractional functional differential equations:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ16_HTML.gif
(4.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ17_HTML.gif
(4.2)

where D α C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq1_HTML.gif, D β C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq2_HTML.gif denote Caputo fractional derivatives, 2 < α < 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq3_HTML.gif, 0 < β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq4_HTML.gif, t ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq77_HTML.gif.

Choose λ = 6 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq78_HTML.gif, η = 1 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq79_HTML.gif, ϕ ( t ) = e t 1 36 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq80_HTML.gif, a = e t 110 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq81_HTML.gif, b = e t 143 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq82_HTML.gif and
f ( t , u t , C D β u ( t ) ) = e t 1 36 + e t 110 | u t | k 1 + e t 143 | C D β u ( t ) | k 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equap_HTML.gif
Then, for t ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq77_HTML.gif, we have
| f ( t , u t , C D β u ( t ) ) | ϕ ( t ) + a | u t | k 1 + b | C D β u ( t ) | k 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equaq_HTML.gif

For 0 < k 1 , k 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq51_HTML.gif, (H1) is satisfied and for k 1 , k 2 > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq52_HTML.gif, (H2) is satisfied. Therefore, by Theorem 3.1, boundary value problem (4.1) and (4.2) has a solution.

Example 4.2

Consider boundary value problems of the following fractional functional differential equations:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ18_HTML.gif
(4.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equ19_HTML.gif
(4.4)

where D α C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq1_HTML.gif, D β C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq2_HTML.gif denote Caputo fractional derivatives, 2 < α < 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq3_HTML.gif, 0 < β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq4_HTML.gif, t ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq77_HTML.gif.

Choose λ = 7 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq83_HTML.gif, η = 3 8 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq84_HTML.gif and
f ( t , u t , C D β u ( t ) ) = | u t | + | C D β u ( t ) | ( 6 + 9 e t ) ( 1 + | u t | + | C D β u ( t ) | ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equar_HTML.gif
Set
f ( t , μ , ν ) = | μ | + | ν | ( 6 + 9 e t ) ( 1 + | μ | + | ν | ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equas_HTML.gif
Let μ , μ ¯ C r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq72_HTML.gif, ν , ν ¯ R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq73_HTML.gif. Then for each t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq17_HTML.gif,
| f ( t , μ , ν ) f ( t , μ ¯ , ν ¯ ) | = 1 6 + 9 e t | | μ | + | ν | 1 + | μ | + | ν | | μ ¯ | + | ν ¯ | 1 + | μ ¯ | + | ν ¯ | | = | μ + μ ¯ | | ν ν ¯ | ( 6 + 9 e t ) ( 1 + | μ ¯ | + | ν ¯ | ) ( 1 + | μ | + | ν | ) 1 6 + 9 e t ( | μ + μ ¯ | | ν ν ¯ | ) 1 15 ( | μ + μ ¯ | | ν ν ¯ | ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equat_HTML.gif
For each t [ 1 , 0 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq85_HTML.gif,
| f ( t , μ , ν ) f ( t , μ ¯ , ν ¯ ) | = 1 6 + 9 e t | | μ | + | ν | 1 + | μ | + | ν | | μ ¯ | + | ν ¯ | 1 + | μ ¯ | + | ν ¯ | | = | μ + μ ¯ | | ν ν ¯ | ( 6 + 9 e t ) ( 1 + | μ ¯ | + | ν ¯ | ) ( 1 + | μ | + | ν | ) 1 6 + 9 e t ( | μ + μ ¯ | | ν ν ¯ | ) 1 6 ( | μ + μ ¯ | | ν ν ¯ | ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equau_HTML.gif
Thus the condition (H3) holds with p = 1 15 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq86_HTML.gif. For λ = 7 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq83_HTML.gif, η = 3 8 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq84_HTML.gif, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equav_HTML.gif
By 2 < α < 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq3_HTML.gif, 0 < β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_IEq4_HTML.gif, we have
1 3 1 α < 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equaw_HTML.gif
and
M > 1 Γ ( α ) 131 54 , N > 32 9 Γ ( 2 β ) Γ ( α ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equax_HTML.gif
It implies that
p = 1 15 < 0.167 < ( M + N ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-38/MediaObjects/13661_2012_Article_292_Equay_HTML.gif

Then by Theorem 3.2, boundary value problem (4.3) and (4.4) has a unique solution.

Declarations

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, 60904024, 61174217), Natural Science Outstanding Youth Foundation of Shandong Province (JQ201119) and supported by Shandong Provincial Natural Science Foundation (ZR2012AM009, ZR2010AL002, ZR2011AL007), also supported by Natural Science Foundation of Educational Department of Shandong Province (J11LA01).

Authors’ Affiliations

(1)
School of Mathematical Sciences, University of Jinan

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