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 , D β C denote Caputo fractional derivatives, 2 < α < 3 , 0 < β < 1 , η ( 0 , 1 ) , 1 < λ < 1 2 η . 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 , 0 < σ < 1 , α , β R , α η ( 1 β ) + ( 1 α ) ( t β η ) 0 and D 0 + δ C , D 0 + σ C 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 , we denote by C r the Banach space of all continuous functions φ : [ r , 0 ] R endowed with the sup-norm
φ [ r , 0 ] : = sup { | φ ( s ) | : s [ r , 0 ] } .
If u : [ r , 1 ] R , then for any t [ 0 , 1 ] , we denote by u t the element of C r defined by
u t ( θ ) = u ( t + θ ) , for  θ [ r , 0 ] .
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 ,
(1.1)
where 2 < α < 3 , 0 < β < 1 and D α C , D β C denote Caputo fractional derivatives, f ( t , u t , C D β u ( t ) ) is a continuous function associated with the boundary conditions
u ( 0 ) = 0 , u ( 1 ) = λ u ( η ) ,
(1.2)
and u 0 = φ , where η ( 0 , 1 ) , 1 < λ < 1 2 η 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 } .

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 ) in [12] to a fractional functional differential equation of α-order ( 2 < α < 3 ). 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 ) of a function f : ( t 0 , + ) R is given by
I α f ( t ) = 1 Γ ( α ) t 0 t f ( s ) ( t s ) 1 α d s , t > t 0 ,

where Γ ( ) is the gamma function, provided that the right-hand side is point-wise defined on ( t 0 , + ) .

Definition 2.2 ([1])

The Caputo fractional derivative of order α ( n 1 < α < n ) of a function f : ( t 0 , + ) R 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 ,

where Γ ( ) is the gamma function, provided that the right-hand side is point-wise defined on ( t 0 , + ) .

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 , ) . Then
D α C ( I α f ( t ) ) = f ( t ) , t > t 0 and 0 < α < 1 .

Lemma 2.2 ([2])

Let α > 0 . Then
I α C D α f ( t ) = f ( t ) c 1 c 2 t c n t n 1

for some c i R , i = 1 , 2 , , n , where n = [ α ] + 1 and [ α ] 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 , 0 < η < 1 , 1 < λ < 1 2 η and h : [ 0 , 1 ] R 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 ) ,
(2.1)
has a unique solution
u ( t ) = 0 1 G ( t , s ) h ( s ) d s ,
(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 ) .
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 ) , 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 ) .
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 , 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 .
For t η , 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 .
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 .

The proof is completed. □

Lemma 2.4 ([36] Schauder fixed point theorem)

Let ( D , d ) be a complete metric space, U be a closed convex subset of D, and T : D D be the map such that the set T u : u U is relatively compact in D. Then the operator T has at least one fixed point u U :
T u = u .

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 } . Also, if I is an interval of the real line , by C ( I ) and C 1 ( I ) we denote the set of continuous and continuously differentiable functions on I, respectively. Moreover, for u C ( I ) , we define
u I = max t I | u ( t ) | + max t I | C D β u ( t ) | .
(3.1)
For u 0 = φ , in view of the definitions of u t and φ, we have
u 0 = u ( θ ) = φ ( θ ) , for  θ [ r , 0 ] .
Thus, we have
u ( t ) = φ ( t ) , for  t [ r , 0 ] .
Since f : [ 0 , 1 ] × C r × R R is a continuous function, set f ( t , u t , C D β u ( t ) ) : = h ( t ) 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 ] .
We define an operator T : X X 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 ] ,
(3.2)
and
l = max 0 t 1 ( 0 1 | G ( t , s ) g ( s ) | d s ) ,
l = max 0 t 1 ( 0 1 | t G ( t , s ) g ( s ) | d s ) ,
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 .

Theorem 3.1 Assume the following:

(H1) There exists a nonnegative function g L [ 0 , 1 ] such that
| f ( t , v , w ) | g ( t ) + a | v | k 1 + b | w | k 2

for each v C r , w R , where a , b R are nonnegative constants and 0 < k 1 , k 2 < 1 ; or

(H2) There exists a nonnegative function g L [ 0 , 1 ] such that
| f ( t , v , w ) | g ( t ) + a | v | k 1 + b | w | k 2

for each v C r , w R , where a , b R are nonnegative constants and k 1 , k 2 > 1 .

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 }
(3.3)

and define the cone U = { u X | u ω , ω > 0 } .

For any u U , 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 ) .
(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 ) .
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 ,
(3.5)

which implies that T : U U . 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 }
(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 ,

which implies that T : U U .

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 . Then for u U and t 1 , t 2 [ r , 1 ] with t 1 < t 2 , in view of Lemma 2.3, if 0 t 1 < t 2 1 , 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 , then
| T u ( t 2 ) T u ( t 1 ) | = | φ ( t 2 ) φ ( t 1 ) | .
If r t 1 < 0 < t 2 1 , 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 ) .
Hence, if 0 t 1 < t 2 1 , 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 , 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 .
If r t 1 < 0 < t 2 1 , 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 , 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 , we have
T u ( t 2 ) T u ( t 1 ) = φ ( t 2 ) φ ( t 1 ) .
If r t 1 < 0 < t 2 1 , 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 ) .

In any case, it implies that T u ( t 2 ) T u ( t 1 ) 0 as t 2 t 1 , i.e., for any ϵ > 0 , there exists δ > 0 , independent of t 1 , t 2 and u, such that | T u ( t 2 ) T u ( t 1 ) | ϵ , whenever | t 2 t 1 | < δ . Therefore T : X X 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 such that | f ( t , μ , ν ) f ( t , μ ¯ , ν ¯ ) | p ( | μ μ ¯ | + | ν ν ¯ | ) for each μ , μ ¯ C r , ν , ν ¯ R . If
p < ( M + N ) 1 ,

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

Proof Consider the operator T : X X 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 ] ,
| 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 .
(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 .
(3.8)
In view of the definition of G ( t , s ) , 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 .
(3.9)
Clearly, for each t [ r , 0 ] , we have | T u ( t ) T u ¯ ( t ) | = 0 . 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 ¯

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 , D β C denote Caputo fractional derivatives, 2 < α < 3 , 0 < β < 1 , t ( 0 , 1 ) .

Choose λ = 6 5 , η = 1 3 , ϕ ( t ) = e t 1 36 , a = e t 110 , b = e t 143 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 .
Then, for t ( 0 , 1 ) , 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 .

For 0 < k 1 , k 2 < 1 , (H1) is satisfied and for k 1 , k 2 > 1 , (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 , D β C denote Caputo fractional derivatives, 2 < α < 3 , 0 < β < 1 , t ( 0 , 1 ) .

Choose λ = 7 6 , η = 3 8 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 ) | ) .
Set
f ( t , μ , ν ) = | μ | + | ν | ( 6 + 9 e t ) ( 1 + | μ | + | ν | ) .
Let μ , μ ¯ C r , ν , ν ¯ R . Then for each t [ 0 , 1 ] ,
| 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 ¯ | ) .
For each t [ 1 , 0 ] ,
| 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 ¯ | ) .
Thus the condition (H3) holds with p = 1 15 . For λ = 7 6 , η = 3 8 , 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 , 0 < β < 1 , we have
1 3 1 α < 1 2
and
M > 1 Γ ( α ) 131 54 , N > 32 9 Γ ( 2 β ) Γ ( α ) .
It implies that
p = 1 15 < 0.167 < ( M + N ) 1 .

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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