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Existence of solutions for fractional differential equations with integral boundary conditions

Advances in Difference Equations20142014:25

https://doi.org/10.1186/1687-1847-2014-25

  • Received: 29 September 2013
  • Accepted: 23 December 2013
  • Published:

Abstract

In this paper, we study boundary-value problems for the following nonlinear fractional differential equations involving the Caputo fractional derivative: D 0 + α C x ( t ) = f ( t , x ( t ) , C D 0 + β x ( t ) ) , t [ 0 , 1 ] , x ( 0 ) + x ( 0 ) = y ( x ) , 0 1 x ( t ) d t = m , x ( 0 ) = x ( 0 ) = = x ( n 1 ) ( 0 ) = 0 , where D 0 + α C , D 0 + β C are the Caputo fractional derivatives, f : [ 0 , 1 ] × R × R R is a continuous function, y : C ( [ 0 , 1 ] , R ) R is a continuous function and m R , n 1 < α < n ( n 2 ), 0 < β < 1 is a real number. By means of the Banach fixed-point theorem and the Schauder fixed-point theorem, some solutions are obtained, respectively. As applications, some examples are presented to illustrate our main results.

MSC:34A08, 34B10.

Keywords

  • fractional differential equation
  • boundary-value problem
  • fixed-point theorem

1 Introduction

Fractional differential equations have been of increasing importance in the past decades due to their diverse applications in science and engineering, such as the memory of a variety of materials, signal identification and image processing, optical systems, thermal system materials and mechanical systems, control system, etc., see [1, 2]. Many interesting results on the existence of solutions of various classes of fractional differential equations have been obtained, see [317], and the references therein.

Recently, much attention has been focused on the study of the existence and multiplicity of solutions or positive solutions for boundary-value problems of fractional differential equations with local boundary-value problems by the use of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, the upper and lower solution method, etc.), see [717].

On the other hand, integer-order differential equations boundary-value problems with integral boundary conditions arise in a variety of different areas of applied mathematics and physics. For example, blood flow problems, chemical engineering, thermo-elasticity, underground water flow, population dynamics, and so forth can be reduced to nonlocal problems with integral boundary conditions. For a detailed description of the integral boundary conditions, we refer the reader to some recent papers [1820] and the references therein.

In fact, there we have the same requirements for fractional differential equations. Boundary-value problems for fractional-order differential equations with nonlocal boundary conditions constitute a very interesting and important class of problems. This type of boundary-value problems has been investigated in [2124]. Lately, Zhang et al. [25] investigated the existence of solutions for a fractional nonlinear integro-differential equation of mixed type on a semi-infinite interval in a Banach space E. Li et al. [26] studied the existence and uniqueness of a positive solution for nonlinear fractional differential equations. Anguraj et al. [27] obtained new existence results for fractional integro-differential equations with impulsive and integral conditions.

There were several definitions of fractional derivatives such as Riemann-Liouville, Caputo, Weyl, etc. Applied problems require those definitions of fractional derivatives that allow the utilization of physically interpretable initial and boundary conditions. The Caputo fractional derivative fulfills these requirements.

Cabada et al. investigated the existence of positive solutions of the following nonlinear fractional differential equations with integral boundary-value conditions [22]:
D α C u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 1 u ( s ) d s ,

where 2 < α < 3 , 0 < λ < 2 , D α C is the Caputo fractional derivative and f : [ 0 , 1 ] × [ 0 , ) [ 0 , ) is a continuous function.

In 2013, Vong et al. studied the existence of positive solutions of a class of singular fractional differential equations with nonlocal boundary conditions [28],
D 0 + α C u ( t ) + f ( t , u ( t ) ) = 0 , u ( 0 ) = = u ( n 1 ) = 0 , u ( 1 ) = 0 1 u ( s ) d μ ( s ) ,

where n 2 , α ( n 1 , n ) , and μ ( s ) is a function of bounded variation, f may have a singularity at t = 1 , and 0 1 d μ ( s ) < 1 .

Agarwal et al. investigated the existence of solutions for the singular fractional boundary-value problems [29]
D α u ( t ) + f ( t , u ( t ) , D μ u ( t ) ) = 0 , u ( 0 ) = u ( 1 ) = 0 ,

where 1 < α < 2 , 0 < μ α 1 are real numbers, D α is the standard Rieman-Liouville fractional derivative, f satisfies the Caratheodory conditions on [ 0 , 1 ] × B , B = ( 0 , ) × R ( f Car ( [ 0 , 1 ] × B ) ), f is positive, and f ( t , x , y ) is singular at x = 0 .

Benchohra et al. studied the boundary-value problem for the fractional differential equations with nonlocal conditions [30]
D 0 + α C y ( t ) = f ( t , y ( t ) ) , t J = [ 0 , T ] , 1 < α 2 , y ( 0 ) = g ( y ) , y ( T ) = y T ,

where D 0 + α C is the Caputo fractional derivative, f : [ 0 , T ] × R R is a continuous function, g : C ( J , R ) R is a continuous function and y T R .

Motivated by all the works above, this paper deals with the existence and uniqueness of solutions for the boundary-value problem of the fractional differential equations
D 0 + α C x ( t ) = f ( t , x ( t ) , C D 0 + β x ( t ) ) , t [ 0 , 1 ] ,
(1.1)
x ( 0 ) + x ( 0 ) = y ( x ) , 0 1 x ( t ) d t = m ,
(1.2)
x ( 0 ) = x ( 0 ) = = x ( n 1 ) ( 0 ) = 0 ,
(1.3)

where D 0 + α C , D 0 + β C are the Caputo fractional derivatives, f : [ 0 , 1 ] × R × R R is a continuous function, y : C ( [ 0 , 1 ] , R ) R is a continuous function, and m R , n 1 < α < n ( n 2 ), 0 < β < 1 is a real number.

The paper is organized as follows. In Section 2, we shall introduce some definitions and lemmas to prove our main results. In Section 3, we establish some criteria for the existence for the boundary problem (1.1) with nonlocal boundary conditions (1.2) and (1.3) by using the Banach fixed-point theorem and the Schauder fixed-point theorem. Finally, we present three examples to illustrate our main results.

2 Preliminaries

In this section, we introduce notations and definitions of fractional calculus, and we prove a lemma before stating our main results.

Let X = { x : x C [ 0 , 1 ] , D β x C [ 0 , 1 ] } . We define x = max t [ 0 , 1 ] | x ( t ) | , x 1 = max { x , D β x } ; then ( X , 1 ) is a Banach space.

Definition 2.1 ([2])

For a continuous function y : ( 0 , ) R , the Riemann-Liouville fractional integral of order α is defined as
I 0 + α y ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s , α > 0 ,

provided the right side integral is pointwise defined on [ 0 , ) .

Definition 2.2 ([2])

The Caputo fractional derivative of order α for a continuous function y ( t ) is defined by
D 0 + α C y ( t ) = 1 Γ ( n α ) 0 t ( t s ) n α 1 y ( n ) ( s ) d s , α > 0 ,

where Γ is the Gamma function, n = [ α ] + 1 , and [ α ] denotes the integer part of number α, and provided the right side integral is pointwise defined on [ 0 , ) .

Lemma 2.1 ([2])

Let u C m [ 0 , T ] and q ( m 1 , m ] , m N . Then for t [ 0 , T ] ,
I q C D 0 + q u ( t ) = u ( t ) k = 0 m 1 t k k ! u ( k ) ( 0 ) .
Lemma 2.2 Let n 1 < α < n ; if x C n [ 0 , 1 ] is a solution of the following fractional differential equations:
D 0 + α C x ( t ) = h ( s ) , t [ 0 , 1 ] , x ( 0 ) + x ( 0 ) = y ( x ) , 0 1 x ( t ) d t = m , x ( 0 ) = x ( 0 ) = = x ( n 1 ) ( 0 ) = 0 ,
then x ( t ) can be represented by
x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 h ( s ) d s + 2 ( 1 t ) m + ( 2 t 1 ) y ( x ) + 2 ( t 1 ) Γ ( α + 1 ) 0 1 ( 1 s ) α h ( s ) d s .
(2.1)
Proof By Lemma 2.1 and the boundary conditions x ( 0 ) = x ( 0 ) = = x ( n 1 ) ( 0 ) = 0 , we have
x ( t ) = I 0 + α h ( t ) + x ( 0 ) + x ( 0 ) t + x ( 0 ) 2 ! t 2 + + x ( n 1 ) ( 0 ) ( n 1 ) ! t n 1 = 1 Γ ( α ) 0 t ( t s ) α 1 h ( s ) d s + x ( 0 ) + x ( 0 ) t .
Hence
0 1 x ( t ) d t = 0 1 0 t ( t s ) α 1 Γ ( α ) h ( s ) d s d t + x ( 0 ) + 1 2 x ( 0 ) .
By the boundary conditions x ( 0 ) + x ( 0 ) = y ( x ) , 0 1 x ( t ) d t = m , we obtain
x ( 0 ) = 2 m y ( x ) 2 Γ ( α + 1 ) 0 1 ( 1 s ) α h ( s ) d s
and
x ( 0 ) = 2 Γ ( α + 1 ) 0 1 ( 1 s ) α h ( s ) d s + 2 y ( x ) 2 m .
Consequently
x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 h ( s ) d s + 2 ( 1 t ) m + ( 2 t 1 ) y ( x ) + 2 ( t 1 ) Γ ( α + 1 ) 0 1 ( 1 s ) α h ( s ) d s .

 □

Lemma 2.3 (Schauder’s fixed point theorem)

Let ( E , d ) be a complete metric space, let U be a closed convex subset of E, and let A : U U be a mapping such that the set { A u : u U } is relatively compact in E. Then A has at least one fixed point.

3 Main results

Now we are in the position to establish the main results.

Theorem 3.1 Assume that:
  1. (H1)

    There exists a constant l > 0 such that | f ( t , x 1 , z 1 ) f ( t , x 2 , z 2 ) | l ( | x 1 x 2 | + | z 1 z 2 | ) , for each t [ 0 , T ] and all x 1 , x 2 , z 1 , z 2 R .

     
  2. (H2)

    There exists a constant l 1 > 0 such that | y ( x 1 ) y ( x 2 ) | l 1 x 1 x 2 , for each x 1 , x 2 C ( [ 0 , T ] , R ) .

     
  3. (H3)

    θ = max { ( 2 Γ ( α + 1 ) + 4 Γ ( α + 2 ) ) l + l 1 , 1 Γ ( 2 β ) [ ( 2 Γ ( α ) + 4 Γ ( α + 2 ) ) l + 2 l 1 ] } < 1 .

    Then the BVP (1.1)-(1.3) has a unique solution.

     
Proof Transform the BVP (1.1)-(1.3) into a fixed-point problem. Consider the operator
F : X X ,
defined by
F ( x ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 f ( s , x ( s ) , C D β x ( s ) ) d s + 2 ( 1 t ) m + ( 2 t 1 ) y ( x ) + 2 ( t 1 ) Γ ( α + 1 ) 0 1 ( 1 s ) α f ( s , x ( s ) , C D β x ( s ) ) d s .

Clearly, the fixed points of the operator F are solutions of the problem (1.1)-(1.3).

Let x 1 , x 2 X . Then
| F ( x 1 ) ( t ) F ( x 2 ) ( t ) | 1 Γ ( α ) 0 t ( t s ) α 1 | f ( s , x 1 ( s ) , C D β x 1 ( s ) ) f ( s , x 2 ( s ) , C D β x 2 ( s ) ) | d s + 2 ( 1 t ) Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x 1 ( s ) , C D β x 1 ( s ) ) f ( s , x 2 ( s ) , C D β x 2 ( s ) ) | d s + | y ( x 1 ) y ( x 2 ) | .
Consider the conditions (H1) and (H2), implying that
| f ( t , x 1 ( t ) , C D β x 1 ( t ) ) f ( t , x 2 ( t ) , C D β x 2 ( t ) ) | l ( | x 1 ( t ) x 2 ( t ) | + | C D β x 1 ( t ) C D β x 2 ( t ) | ) l ( x 1 x 2 + C D β x 1 C D β x 2 ) 2 l x 1 x 2 1
and
| y ( x 1 ) y ( x 2 ) | l 1 x 1 x 2 l 1 x 1 x 2 1 ,
thus, we have
F ( x 1 ) F ( x 2 ) 1 Γ ( α ) 0 t ( t s ) α 1 d s 2 l x 1 x 2 1 + l 1 x 1 x 2 1 + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α d s 2 l x 1 x 2 1 [ ( 2 Γ ( α + 1 ) + 4 Γ ( α + 2 ) ) l + l 1 ] x 1 x 2 1 .
As
| ( F x 1 ) ( t ) ( F x 2 ) ( t ) | 1 Γ ( α 1 ) 0 t ( t s ) α 2 | f ( s , x 1 ( s ) , C D β x 1 ( s ) ) f ( s , x 2 ( s ) , C D β x 2 ( s ) ) | d s + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x 1 ( s ) , C D β x 1 ( s ) ) f ( s , x 2 ( s ) , C D β x 2 ( s ) ) | d s + 2 | y ( x 1 ) y ( x 2 ) | 1 Γ ( α 1 ) 0 t ( t s ) α 2 d s 2 l x 1 x 2 1 + 2 l 1 x 1 x 2 1 + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α d s 2 l x 1 x 2 1 [ ( 2 Γ ( α ) + 4 Γ ( α + 2 ) ) l + 2 l 1 ] x 1 x 2 1 ,
we obtain
| C D β ( F x 1 ) ( t ) C D β ( F x 2 ) ( t ) | 1 Γ ( 1 β ) 0 t ( t s ) β | ( F x 1 ) ( s ) ( F x 2 ) ( s ) | d s 1 Γ ( 2 β ) [ ( 2 Γ ( α ) + 4 Γ ( α + 2 ) ) l + 2 l 1 ] x 1 x 2 1 .
Consequently
F x 1 F x 2 1 = max { F ( x 1 ) F ( x 2 ) , C D β ( F x 1 ) C D β ( F x 2 ) } θ x 1 x 2 1 ,

then F is a contraction with θ < 1 . As a consequence of the Banach fixed-point theorem, we deduce that F has a fixed point which is the unique solution of the problem (1.1)-(1.3). The proof is complete. □

Next, we will use the Schauder’ fixed-point theorem to prove our result. For the sake of convenience, we set
M = max t [ 0 , 1 ] | f ( t , x ( t ) , C D β x ( t ) | , M 1 = max t [ 0 , 1 ] | y ( x ( t ) ) | , A = ( 1 Γ ( α + 1 ) + 2 Γ ( α + 2 ) ) M + M 1 + 2 | m | , B = ( 1 Γ ( α β + 1 ) + 2 Γ ( 2 β ) Γ ( α + 2 ) ) M + 2 Γ ( 2 β ) ( M 1 + | m | ) .

Theorem 3.2 Assume f : [ 0 , 1 ] × R × R R is continuous, y : ( C [ 0 , 1 ] , R ) R is continuous. Then the BVP (1.1)-(1.3) has a solution.

Proof Let E = { x : x X , x 1 r } , where r = max { A , B } . First, we prove that F : E E .

In fact, for each t [ 0 , 1 ] , we have
| F ( x ) ( t ) | 1 Γ ( α ) 0 t ( t s ) α 1 | f ( s , x ( s ) , C D β x ( s ) ) | d s + 2 | m | + | y ( x ) | | F ( x ) ( t ) | + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x ( s ) , C D β x ( s ) ) | d s | F ( x ) ( t ) | 1 Γ ( α ) 0 t ( t s ) α 1 M d s + 2 | m | + M 1 + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α M d s | F ( x ) ( t ) | ( 1 Γ ( α + 1 ) + 2 Γ ( α + 2 ) ) M + M 1 + 2 | m | = A , F ( x ) = max t [ 0 , 1 ] | F ( x ) ( t ) | ( 1 Γ ( α + 1 ) + 2 Γ ( α + 2 ) ) M + M 1 + 2 | m | = A .
Considering
F ( x ) ( t ) = 1 Γ ( α 1 ) 0 t ( t s ) α 2 f ( s , x ( s ) , C D β x ( s ) ) d s 2 m + 2 y ( x ) + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α f ( s , x ( s ) , C D β x ( s ) ) d s ,
and
0 t ( t s ) β s α 1 d s = t α β B ( α , 1 β ) , B ( α , n β ) = Γ ( α ) Γ ( n β ) ( α β + n 1 ) Γ ( α β + n 1 ) ,
we obtain
D β C ( F x ) ( t ) = 1 Γ ( 1 β ) 0 t ( t s ) β ( 1 Γ ( α 1 ) 0 s ( s τ ) α 2 f ( τ , x ( τ ) , C D β x ( τ ) ) d τ 2 m + 2 y ( x ) + 2 Γ ( α + 1 ) 0 1 ( 1 τ ) α f ( τ , x ( τ ) , C D β x ( τ ) ) d τ ) d s = 1 Γ ( α β ) 0 t ( t s ) α 1 β f ( s , x ( s ) , C D β x ( s ) ) d s + 2 t 1 β Γ ( 2 β ) ( y ( x ) m ) + 2 t 1 β Γ ( 2 β ) Γ ( α + 1 ) 0 1 ( 1 s ) α f ( s , x ( s ) , C D β x ( s ) ) d s ,
thus
C D β ( F x ) 1 Γ ( α β ) 0 t ( t s ) α 1 β | f ( s , x ( s ) , C D β x ( s ) ) | d s + 2 Γ ( 2 β ) ( | y ( x ) | + | m | ) + 2 Γ ( 2 β ) Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x ( s ) , C D β x ( s ) ) | d s ( 1 Γ ( α β + 1 ) + 2 Γ ( 2 β ) Γ ( α + 2 ) ) M + 2 Γ ( 2 β ) ( M 1 + | m | ) = B .
Hence, we can conclude that
F x 1 = max { F x , C D β ( F x ) } r .

From the expression of ( F x ) ( t ) and D β C ( F x ) ( t ) , it is easy to see that ( F x ) ( t ) C [ 0 , 1 ] , D β C ( F x ) ( t ) C [ 0 , 1 ] . Consequently F : E E .

In what follows we show that F is completely continuous.
  1. (a)
    For each t [ 0 , 1 ] , we have
    F x ( 1 Γ ( α + 1 ) + 2 Γ ( α + 2 ) ) M + M 1 + 2 | m | = A , C D β ( F x ) ( 1 Γ ( α β + 1 ) + 2 Γ ( 2 β ) Γ ( α + 2 ) ) M + 2 Γ ( 2 β ) ( M 1 + | m | ) = B ,
     
which shows that F is uniform bounded.
  1. (b)
    For each t 1 , t 2 [ 0 , 1 ] , t 1 < t 2 , and it implies that
    | F ( x ) ( t 2 ) F ( x ) ( t 1 ) | | 1 Γ ( α ) 0 t 1 [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] f ( s , x ( s ) , C D β x ( s ) ) d s + 1 Γ ( α ) t 1 t 2 ( t 2 s ) α 1 f ( s , x ( s ) , C D β x ( s ) ) d s | + 2 ( | m | + | y ( x ) | + 1 Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x ( s ) , C D β x ( s ) ) | d s ) ( t 2 t 1 ) M Γ ( α ) 0 t 1 [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] d s + M Γ ( α ) t 1 t 2 ( t 2 s ) α 1 d s + ( 2 | m | + 2 M 1 + 2 M Γ ( α + 2 ) ) ( t 2 t 1 ) = M Γ ( α + 1 ) ( t 2 α t 1 α ) + ( 2 | m | + 2 M 1 + 2 M Γ ( α + 2 ) ) ( t 2 t 1 )
    (3.1)
     
and
| C D β ( F x ) ( t 2 ) C D β ( F x ) ( t 1 ) | | 1 Γ ( α β ) 0 t 2 ( t 2 s ) α 1 β f ( s , x ( s ) , C D β x ( s ) ) d s 1 Γ ( α β ) 0 t 1 ( t 1 s ) α 1 β f ( s , x ( s ) , C D β x ( s ) ) d s | + 2 Γ ( 2 β ) ( | y ( x ) | + | m | ) ( t 2 1 β t 1 1 β ) + 2 ( t 2 1 β t 1 1 β ) Γ ( 2 β ) Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x ( s ) , C D β x ( s ) ) | d s M Γ ( α β + 1 ) ( t 2 α β t 1 α β ) + 2 ( M 1 + | m | ) Γ ( 2 β ) ( t 2 1 β t 1 1 β ) + 2 M Γ ( α + 2 ) Γ ( 2 β ) ( t 2 1 β t 1 1 β ) .
(3.2)

The right-hand sides of equations (3.1) and (3.2) tend to zero when t 1 t 2 , so F is compact as consequence of the Arzela-Ascoli theorem, and F is continuous. We claim that F is completely continuous. Combing the two steps above with lemma 2.3, we deduce that the problem (1.1)-(1.3) has a solution on E. □

Theorem 3.3 Assume that f : [ 0 , 1 ] × R × R R is continuous, y : ( C [ 0 , 1 ] , R ) R is continuous, and they satisfy
  1. (H4)

    | f ( t , x , z ) | k ( t ) + c 1 | x | ρ 1 + c 2 | z | ρ 2 , x , z R ;

     
  2. (H5)

    | y ( x ) | c 3 x ρ 3 , x C ( [ 0 , 1 ] , R ) ;

     

where k ( t ) 0 L [ 0 , 1 ] , 0 < ρ i < 1 , and c i 0 for i = 1 , 2 , 3 . Then the BVP (1.1)-(1.3) has a solution.

Proof First, we define
U = { x : x X , x 1 R } ,
where
R max { ( 4 P c 1 ) 1 1 ρ 1 , ( 4 P c 2 ) 1 1 ρ 2 , ( 4 c 3 ) 1 1 ρ 3 , ( 4 Q c 1 ) 1 1 ρ 1 , ( 4 Q c 2 ) 1 1 ρ 2 , ( 8 c 3 Γ ( 2 β ) ) 1 1 ρ 3 , 4 δ , 4 η }
and
P = 1 Γ ( α + 1 ) + 2 Γ ( α + 2 ) , Q = 1 Γ ( α β + 1 ) + 2 Γ ( 2 β ) Γ ( α + 2 ) , δ = 1 Γ ( α ) 0 t ( t s ) α 1 k ( s ) d s + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α k ( s ) d s + 2 | m | , η = 1 Γ ( α β ) 0 t ( t s ) α 1 β k ( s ) d s + 2 Γ ( 2 β ) Γ ( α + 1 ) 0 1 ( 1 s ) α k ( s ) d s + 2 | m | Γ ( 2 β ) .
Now we prove that F : U U . For any x U , we have
F x 1 Γ ( α ) 0 t ( t s ) α 1 | f ( s , x ( s ) , C D β x ( s ) ) | d s + 2 | m | + | y ( x ) | F x + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x ( s ) , C D β x ( s ) ) | d s F x δ + 1 Γ ( α ) 0 t ( t s ) α 1 d s ( c 1 R ρ 1 + c 2 R ρ 2 ) + c 3 R ρ 3 F x + 2 Γ ( α + 1 ) 0 1 ( 1 s ) α d s ( c 1 R ρ 1 + c 2 R ρ 2 ) F x δ + ( c 1 R ρ 1 + c 2 R ρ 2 ) P + c 3 R ρ 3 , C D β ( F x ) 1 Γ ( α β ) 0 t ( t s ) α 1 β | f ( s , x ( s ) , C D β x ( s ) ) | d s + 2 Γ ( 2 β ) ( | y ( x ) | + | m | ) C D β ( F x ) + 2 Γ ( 2 β ) Γ ( α + 1 ) 0 1 ( 1 s ) α | f ( s , x ( s ) , C D β x ( s ) ) | d s C D β ( F x ) η + 1 Γ ( α β ) 0 t ( t s ) α 1 β d s ( c 1 R ρ 1 + c 2 R ρ 2 ) + 2 c 3 R ρ 3 Γ ( 2 β ) C D β ( F x ) + 2 Γ ( 2 β ) Γ ( α + 1 ) 0 1 ( 1 s ) α d s ( c 1 R ρ 1 + c 2 R ρ 2 ) C D β ( F x ) η + ( c 1 R ρ 1 + c 2 R ρ 2 ) Q + 2 c 3 R ρ 3 Γ ( 2 β ) .
Considering F x 1 = max { F x , C D β ( F x ) } , we can conclude that
F x 1 max { δ + ( c 1 R ρ 1 + c 2 R ρ 2 ) P + c 3 R ρ 3 , η + ( c 1 R ρ 1 + c 2 R ρ 2 ) Q + 2 c 3 R ρ 3 Γ ( 2 β ) } R 4 + R 4 + R 4 + R 4 R .

Considering that f, y are continuous functions, we take M = max t [ 0 , 1 ] | f ( t , x ( t ) , C D β x ( t ) ) | , M 1 = max t [ 0 , 1 ] | y ( x ( t ) ) | , and we can see that F is completely continuous by considering the second step of Theorem 3.2.

As a consequence of Schauder’s fixed-point theorem, we claim that the problem (1.1)-(1.3) has a solution on U. □

4 Examples

In this section, in order to illustrate our results, we consider three examples.

Example 4.1 Consider the following boundary-value problem:
D 0 + 1.5 C x ( t ) = 0.01 t x ( t ) + 0.01 t 2 C D 0 + 0.5 x ( t ) + t ,
(4.1)
x ( 0 ) + x ( 0 ) = i = 1 n c i x ( t i ) , 0 1 x ( t ) d t = 1 ,
(4.2)
x ( 0 ) = x ( 0 ) = = x ( n 1 ) ( 0 ) = 0 ,
(4.3)

where 0 < t 1 < t 2 < < t n < 1 , c i , i = 1 , 2 , , n , are given positive constants with i = 1 n c i < 1 2 .

Set α = 1.5 ( n = 2 ), β = 0.5 , f ( t , x ( t ) , C D 0 + 0.5 x ( t ) ) = 0.01 t x ( t ) + 0.01 t 2 C D 0 + 0.5 x ( t ) + t , y ( x ) = i = 1 n c i x ( t i ) , m = 1 . Let t [ 0 , 1 ] and x 1 , x 2 , y 1 , y 2 R ,
| f ( t , x 1 , z 1 ) f ( t , x 2 , z 2 ) | = | 0.01 t x 1 0.01 t x 2 + 0.01 t 2 z 1 + 0.01 t 2 z 2 | 0.01 ( | x 1 x 2 | + | z 1 z 2 | ) .
Hence the condition (H1) holds with l = 0.01 . Also we have
| y ( x 1 ) y ( x 2 ) | = | i = 1 n c i x 1 ( t i ) i = 1 n c i x 2 ( t i ) | i = 1 n c i x 1 x 2 .
Hence (H2) is satisfied with l 1 = i = 1 n c i < 1 2 . As to (H3), we can show that
( 2 Γ ( α + 1 ) + 4 Γ ( α + 2 ) ) l + l 1 = ( 2 Γ ( 2.5 ) + 4 Γ ( 3.5 ) ) × 0.01 + 0.5 0.5096 < 1 , 1 Γ ( 2 β ) [ ( 2 Γ ( α ) + 4 Γ ( α + 2 ) ) l + 2 l 1 ] 0.76 < 1 .

Then by Theorem 3.1, the problem (4.1)-(4.3) has a unique solution.

Example 4.2 Consider the following boundary-value problem:
D 0 + 1.5 C x ( t ) = ( t 0.5 ) 4 ( x ( t ) ρ 1 + ( C D 0.25 x ( t ) ) ρ 2 ) ,
(4.4)
x ( 0 ) + x ( 0 ) = ( t 0 0.5 ) 4 x ( t 0 ) ρ 3 , 0 1 x ( t ) d t = 1 ,
(4.5)
x ( 0 ) = x ( 0 ) = = x ( n 1 ) ( 0 ) = 0 ,
(4.6)

where ρ i , t 0 are given positive constants with 0 < ρ i , t 0 < 1 .

Set α = 1.5 ( n = 2 ), β = 0.25 , f ( t , x ( t ) , C D 0 + 0.25 x ( t ) ) = ( t 0.5 ) 4 ( x ( t ) ρ 1 + ( C D 0.25 x ( t ) ) ρ 2 ) , y ( x ) = ( t 0 0.5 ) 4 x ( t 0 ) ρ 3 , m = 1 .

Note that
| f ( t , x , z ) | = | ( t 0.5 ) 4 ( x ρ 1 + z ρ 2 ) | 1 16 ( | x | ρ 1 + | z | ρ 2 ) , | y ( x ) | = | ( t 0 0.5 ) 4 x ( t 0 ) ρ 3 | 1 16 x ρ 3 .

Hence the conditions (H4) and (H5) hold with k ( t ) = 0 , c 1 = c 2 = c 3 = 1 16 . Then by Theorem 3.3, the problem (4.4)-(4.6) has a solution.

Example 4.3 Consider the following boundary-value problem:
D 0 + 1.5 C x ( t ) = t + 0.01 t x ( t ) ρ 1 + 0.01 t 2 ( C D 0.25 x ( t ) ) ρ 2 ,
(4.7)
x ( 0 ) = ( t 0 0.5 ) 4 x ( t 0 ) ρ 3 , 0 1 x ( t ) d t = 1 ,
(4.8)
x ( 0 ) = x ( 0 ) = = x ( n 1 ) ( 0 ) = 0 ,
(4.9)

where ρ i , t 0 are given positive constants with 0 < ρ i , t 0 < 1 .

Set α = 1.5 ( n = 2 ), β = 0.25 , f ( t , x ( t ) , C D 0 + 0.25 x ( t ) ) = t + 0.01 t x ( t ) ρ 1 + 0.01 t 2 ( C D 0.25 x ( t ) ) ρ 2 , y ( x ) = ( t 0 0.5 ) 4 x ( t 0 ) ρ 3 , m = 1 .

Note that
| f ( t , x , z ) | 1 + 0.01 ( | x | ρ 1 + | z | ρ 2 ) , | y ( x ) | 1 16 x ρ 3 .

Hence the conditions (H4) and (H5) hold with k ( t ) = 1 , c 1 = c 2 = 0.01 , c 3 = 1 16 . Then by Theorem 3.3, the problem (4.7)-(4.9) has a 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, 61374074), Natural Science Outstanding Youth Foundation of Shandong Province (JQ201119) and supported by Shandong Provincial Natural Science Foundation (ZR2012AM009), also supported by Natural Science Foundation of Educational Department of Shandong Province (J11LA01).

Authors’ Affiliations

(1)
School of Mathematical Sciences, University of Jinan, Jinan, Shandong, 250022, PR China

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© Yan et al.; licensee Springer. 2014

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.

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