This paper considers the following boundary value problems of fractional order differential equations

{ D c a α x ( t ) = f ( t , x ( t ) ) ,     for       t   ∈   J = [ a ,   b ] ,       n − 1   <   α   ≤   n   , x ( k ) ( a ) = x k ,       k = 0 ,   1 ,   2 ,   ⋯ , n − 2   ;         x ( n − 1 ) ( b ) = x b (1.1)

where D c a α is the Caputo fractional derivative, f   :   J × R   → R is continuous function and x 0 , x 1 , ⋯ , x n − 2 , x b are real constants.

Fractional order Differential equations have recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Applications can be found in fields of control, porous media, eletromagnetic, etc. (see [1] [2] [3] [4] [5] ). There has been a significant progress in the investigation of fractional differential equations in recent years, The readers are referred to the monographs of Oldham and Spanier [1] , Miller and Ross [2] , Podlubny [3] , Hilfer [5] and the papers of Agarwal et al. [6] , El-Sayed [7] [8] [9] [10] , Benchohra et al. [11] [12] , Yu and Gao [13] [14] , Zhang [15] , He [4] and the others references therein [16] - [23] .

Recently some basic theory for the initial value problems of fractional differential equations involving Riemann-Liouville differential operator ( 0 < α < 1 ) has been discussed by Lakshmikantham et al. [24] [25] [26] . In a series of papers (see [6] [11] ), the authors considered some classes of boundary value problems for differential equations involving Riemann-Liouville and Caputo fractional derivatives of order 0 < α < 1 and 2 < α < 3 .

This paper generalizes the results of the papers above [6] and presents some existence theorems for the boundary value problems (BVP) (1.1). Two theorems are based on the Banach fixed point theorem, and the others are based on Schau- der’s fixed point theorem and Leray-Schauder type nonlinear alternative. An example is given to demonstrate the application of our main results.

Some notions and Lemmas are important in order to state our results. Denote by C ( J , R ) the Banach space of all continuous functions from J into R with the norm

‖ x ‖ ∞   :   = sup t ∈ J { | x ( t ) | } ,       J = [ a , b ] .

Definition 2.1 ( [6] [11] ) The fractional order integral of the function h ( t )   ∈   L 1 ( [ a , b ] , R + ) is defined by

I a α h ( t ) = 1 Γ ( α ) ∫ a t ( t − s ) α − 1 h ( s ) d s (2.1)

where Γ is the gamma function.

Definition 2.2 ( [6] [11] ) For a function h given on the interval [a,b], the α - th Caputo fractional-order derivative of h is defined by

( D c a α h ) ( t ) = 1 Γ ( n − α ) ∫ a t ( t − s ) n − α − 1 h ( n ) ( s ) d s (2.2)

where n = [ α ] + 1 and [ α ] denotes the integer part of α .

A solution of the problem (1.1) is defined as follows.

Definition 2.3 A function x ( t )   ∈ C n − 1 ( J , R ) that satisfies (1.1) is called a solution of (1.1).

Lemma 2.1 ( [15] ) Let α > 0 , then the differential equation

D c a α h ( t ) = 0

has solutions

h ( t ) = c 0 + c 1 ( t − a ) + c 2 ( t − a ) 2 + ⋯ + c n − 1 ( t − a ) n − 1 ,     c i ∈ R ,     i = 0 , 1 , 2 , ⋯ , n − 1 ,   n = [ α ] + 1.

Lemma 2.2 Let α > 0 , then

I a α D c a α h ( t ) = h ( t ) + c 0 + c 1 ( t − a ) + c 2 ( t − a ) 2 + ⋯ + c n − 1 ( t − a ) n − 1 .

In particular, when a = 0 ,

I α D c α h ( t ) = h ( t ) + c 0 + c 1 t + c 2 t 2 + ⋯ + c n − 1 t n − 1 ,

for some c i ∈ R ,     i = 0 , 1 , 2 , ⋯ , n − 1 ,   n = [ α ] + 1 .

Proof. By (2.1), (2.2),

I a α D a α h ( t ) = 1 Γ ( α ) ∫ a t ( t − s ) α − 1 D a α h ( s ) d s = 1 Γ ( α ) ∫ a t ( t − s ) α − 1 [ 1 Γ ( n − α ) ∫ a s ( s − τ ) n − α − 1 h ( n ) ( τ ) d τ ] d s = 1 Γ ( α ) Γ ( n − α ) ∫ a t h ( n ) ( τ ) d τ ∫ τ t ( t − s ) α − 1 ( s − τ ) n − α − 1 d s = 1 Γ ( α ) Γ ( n − α ) ∫ a t h ( n ) ( τ ) d τ ( α − 1 ) ( n − α ) ∫ τ t ( t − s ) α − 2 ( s − τ ) n − α d s = 1 Γ ( α ) Γ ( n − α ) ∫ a t h ( n ) ( τ ) d τ ( α − 1 ) ( α − 2 ) ⋯ 2 × 1 ( n − α ) ( n − α + 1 ) ⋯ ( n − 2 ) ∫ τ t ( s − τ ) n − 2 d s = 1 Γ ( n ) ∫ a t ( t − τ ) n − 1 h ( n ) ( τ ) d τ = 1 Γ ( n ) ∫ a t ( t − τ ) n − 1 d h ( n − 1 ) ( τ ) = − h ( n − 1 ) ( a ) Γ ( n ) ( t − a ) n − 1 − h ( n − 2 ) ( a ) Γ ( n − 1 ) ( t − a ) n − 2 + 1 Γ ( n − 2 ) ∫ a t ( t − τ ) n − 3 h ( n − 2 ) ( τ ) d τ = − h ( n − 1 ) ( a ) Γ ( n ) ( t − a ) n − 1 − h ( n − 2 ) ( a ) Γ ( n − 1 ) ( t − a ) n − 2 − ⋯ − h ′ ( a ) Γ ( 2 ) ( t − a ) − h ( a ) Γ ( 1 ) + h ( t ) = h ( t ) + c 0 + c 1 ( t − a ) + c 2 ( t − a ) 2 + ⋯ + c n − 1 ( t − a ) n − 1

where c i = − h ( i ) ( a ) Γ ( i + 1 ) ,     i = 0 ,   1 ,   2 ,   ⋯ , n − 1 .

Lemma 2.3 ( [27] ) The relation

D c a α I a α h ( t ) = h ( t ) , I a α I a β h ( t ) = I a α + β h ( t ) (2.3)

is valid in following case

Re     α   > 0 ,       Re     β   > 0 ,       h ( t )   ∈   L 1 ( a , b ) .

As a consequence of Lemmas 2.1, Lemmas 2.2 and Lemmas 2.3, the following result is useful in what follows.

Lemma 2.4 Let n − 1 < α < n , n = [ α ] + 1 , and let h :   J → R be continuous. A function x ( t ) is a solution of the fractional BVP

{ D c a α x ( t ) = h ( t ) ,       t   ∈   J   ,                                                                           x ( k ) ( a ) = x k ,       k = 0 ,   1 ,   2 ,   ⋯ , n − 2   ;         x ( n − 1 ) ( b ) = x b , (2.4)

if and only if x ( t ) is a solution of the fractional integral equation

x ( t ) = 1 Γ ( α ) ∫ a t ( t − s ) α − 1 h ( s ) d s + ( x b ( n − 1 ) ! + h ( a ) ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ) ( t − a ) n − 1 − ( t − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n h ( s ) d s + ∑ k = 0 n − 2 x k k ! ( t − a ) k   . (2.5)

Proof. Assume x ( t ) satisfies (2.4), then Lemma 2.2 implies that

x ( t ) = c 0 + c 1 ( t − a ) + c 2 ( t − a ) 2 + ⋯ + c n − 1 ( t − a ) n − 1 + 1 Γ ( α ) ∫ a t ( t − s ) α − 1 h ( s ) d s .

And the following simple calculation can be obtained by (2.4)

  c k = − x k k ! ,     k = 0 , 1 , 2 , ⋯ , n − 2 ,

c n − 1 = x b ( n − 1 ) ! + h ( a ) ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ( t − a ) n − 1 − ( t − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n h ( s ) d s .

Hence Equation (2.5). Conversely, it is clear that if x ( t ) satisfies Equation (2.5), then Equations (2.4) hold.

In this section, Our first result is based on the Banach fixed point theorem (see [28] ).

Theorem 3.1 Assume that

(H1) There exists a function λ ( t )   ∈   C ( J , R ) such that

| f ( t , u ( t ) ) − f ( t , v ( t ) ) | ≤   λ ( t ) | u ( t ) − v ( t ) | ,     ∀ t ∈   J = [ a , b ] ;   u ( t ) , v ( t )   ∈ R .

If

θ   =   I α λ ( t ) + ( b − a ) α λ ( a ) ( n − 2 ) ! Γ ( α − n + 2 ) + ( b − a ) n − 1 ( n − 1 ) ! I α − n + 1 λ ( b ) < 1. (3.1)

Then the BVP (1.1) has a unique solution on J.

Proof. Transform the problem (1.1) into a fixed point problem. Consider the operator

T   :   C n − 1 ( J , R )   →   C n − 1 ( J , R )

defined by

T x ( t ) = 1 Γ ( α ) ∫ a t ( t − s ) α − 1 f ( s , x ( s ) ) d s + ( x b ( n − 1 ) ! + f ( a , x ( a ) ) ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ) ( t − a ) n − 1 − ( t − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n f ( s , x ( s ) ) d s + ∑ k = 0 n − 2 x k k ! ( t − a ) k   . (3.2)

The Banach contraction principle is used to prove that T has afixed point.

Let x ( t ) ,   y ( t )   ∈   C n − 1 ( J , R ) . Then ∀ t ∈ J ,

| T x ( t ) − T y ( t ) |   ≤   1 Γ ( α ) ∫ a t ( t − s ) α − 1 | f ( s , x ( s ) ) − f ( s , y ( s ) ) | d s + ( t − a ) n − 1 ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) | f ( a , x ( a ) ) − f ( a , y ( a ) ) | + ( t − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n | f ( s , x ( s ) ) − f ( s , y ( s ) ) | d s ≤ ‖ x − y ‖ ∞ Γ ( α ) ∫ a t ( t − s ) α − 1 λ ( s ) d s + ( b − a ) α − n + 1 λ ( a ) ‖ x − y ‖ ∞ ( n − 2 ) ! Γ ( α − n + 2 ) + ( b − a ) n − 1 ‖ x − y ‖ ∞ ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n λ ( s ) d s =   ( I α λ ( t )   +   ( b − a ) α λ ( a ) ( n − 2 ) ! Γ ( α − n + 2 ) + ( b − a ) n − 1 ( n − 1 ) ! I α − n + 1 λ ( b ) )   ‖ x − y ‖ ∞ .

Thus

‖ T x − T y ‖ ∞   ≤   ( I α λ ( t )   +   ( b − a ) α λ ( a ) ( n − 2 ) ! Γ ( α − n + 2 ) + ( b − a ) n − 1 ( n − 1 ) ! I α − n + 1 λ ( b ) )   ‖ x − y ‖ ∞ .

Consequently, by (3.1) T is a contraction operator. As a consequence of the Banach Fixed point theorem, T has a fixed point which is the unique solution of the problem (1.1). The proof is completed.

In Theorem 3.1, if the function λ ( t ) is replaced by a constant L > 0, the second result follows.

Theorem 3.2 Assume that

(H2) There exists a constant L > 0 (i.e. λ ( t )   =   L > 0 ), such that

| f ( t , u ( t ) ) − f ( t , v ( t ) ) | ≤   L | u ( t ) − v ( t ) | , ∀ t   ∈   J = [ a , b ] ;     u ( t ) ,   v ( t ) ∈ R .

If

θ   =   L ( b − a ) α ( 1 Γ ( α + 1 ) + n ( n − 1 ) ! Γ ( α − n + 2 ) ) < 1. (3.3)

Then the BVP (1.1) has a unique solution on J.

The third result is based on Schauder’s Fixed point theorem.

Theorem 3.3 Assume that

(H3) The function f   :   J × R   → R is continuous.

(H4) There exists a constant M > 0, such that

| f ( t , u ( t ) ) |   ≤   M   for   each     t   ∈   J = [ a , b ]       and     ∀ u ( t ) ∈ R . (3.4)

Then the BVP (1.1) has at least one solution on J.

Proof. Schauder’s Fixed point theorem is used to prove that T defined by (3.2) has a fixed point. The proof will be given in several steps.

Step 1: T is continuous.

Let { x m } be a sequence such that x m → x in C ( J , R ) . Then for each t   ∈   J

| T x m ( t ) − T x ( t ) | ≤   1 Γ ( α ) ∫ a t ( t − s ) α − 1 | f ( s , x m ( s ) ) − f ( s , x ( s ) ) | d s + ( t − a ) n − 1 ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) | f ( a , x m ( a ) ) − f ( a , x ( a ) ) | + ( t − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n | f ( s , x m ( s ) ) − f ( s , x ( s ) ) | d s ≤ 1 Γ ( α ) ∫ a t ( t − s ) α − 1 sup s   ∈   J | f ( s , x m ( s ) ) − f ( s , x ( s ) ) | d s + ( b − a ) α ( n − 2 ) ! Γ ( α − n + 2 ) sup s   ∈   J | f ( s , x m ( s ) ) − f ( s , x ( s ) ) | + ( b − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n sup s   ∈   J | f ( s , x m ( s ) ) − f ( s , x ( s ) ) | d s

then

| T x m ( t ) − T x ( t ) |   ≤   ‖ f ( • , x m ( • ) ) − f ( • , x ( • ) ) ‖ ∞   ( 1 Γ ( α ) ∫ a t ( t − s ) α − 1 d s + ( b − a ) α ( n − 2 ) ! Γ ( α − n + 2 ) + ( b − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n d s ≤   ‖ f ( • , x m ( • ) ) − f ( • , x ( • ) ) ‖ ∞ ( b − a ) α ( 1 Γ ( α + 1 ) + n ( n − 1 ) ! Γ ( α − n + 2 ) ) .

Since f is a continuous function, it can be shown that

‖ T x m − T x ‖ ∞ ≤ ( b − a ) α ( 1 Γ ( α + 1 ) + n ( n − 1 ) ! Γ ( α − n + 2 ) ) ‖ f ( · , x m ( · ) ) − f ( · , x ( · ) ) ‖ ∞ .

And hence

‖ T x m − T x ‖ ∞   → 0   ,         m   →   ∞ .

Step 2: T maps the bounded sets into the bounded sets in C ( J , R ) .

For any η * > 0 , it can be shown that there exists a positive constant l such that ∀ x ∈ B η * =   { x   ∈   C ( J , R )   :   ‖ x ‖ ∞ ≤   η * } , ‖ T x ‖ ∞   ≤   l .

In fact, ∀ t   ∈   J , by (3.2) and (H4)

| T x ( t ) | ≤   ∑ k = 0 n − 2 | x k | k ! ( b − a ) k + ( | x b | ( n − 1 ) ! + | f ( a , x ( a ) ) | ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ) ( b − a ) n − 1 + 1 Γ ( α ) ∫ a t ( t − s ) α − 1 | f ( s , x ( s ) ) | d s + ( b − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n | f ( s , x ( s ) ) | d s   ≤   ∑ k = 0 n − 2 | x k | k ! ( b − a ) k + | x b | ( b − a ) n − 1 ( n − 1 ) ! + M ( b − a ) α ( 1 Γ ( α + 1 ) + n ( n − 1 ) ! Γ ( α − n + 2 ) ) .

Thus

‖ T x ‖ ∞ ≤ l

where

l = ∑ k = 0 n − 2 | x k | k ! ( b − a ) k + | x b | ( b − a ) n − 1 ( n − 1 ) !           + M ( b − a ) α ( 1 Γ ( α + 1 ) + n ( n − 1 ) ! Γ ( α − n + 2 ) ) .

Step 3: Tmaps the bounded sets into the equicontinuous sets of C ( J , R ) .

Let t 1 ,   t 2   ∈   J ,   t 1 <   t 2 , B η * be abounded set of C ( J , R ) as above, and x   ∈   B η * .

| T x ( t 2 ) − T x ( t 1 ) | ≤ | 1 Γ ( α ) ( ∫ a t 2 ( t 2 − s ) α − 1 f ( s , x ( s ) ) d s − ∫ a t 1 ( t 1 − s ) α − 1 f ( s , x ( s ) ) d s ) + ( x b ( n − 1 ) ! + f ( a , x ( a ) ) ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ) ( ( t 2 − a ) n − 1 − ( t 1 − a ) n − 1 ) − ( t 2 − a ) n − 1 − ( t 1 − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n f ( s , x ( s ) ) d s + ∑ k = 0 n − 2 x k k ! ( ( t 2 − a ) k − ( t 1 − a ) k ) |   .

Then

| T x ( t 2 ) − T x ( t 1 ) |   ≤ M Γ ( α ) ∫ a 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 + ( | x b | ( n − 1 ) ! + M ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ) ( ( t 2 − a ) n − 1 − ( t 1 − a ) n − 1 ) + M ( ( t 2 − a ) n − 1 − ( t 1 − a ) n − 1 ) ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n d s + ∑ k = 0 n − 2 | x k | k ! ( ( t 2 − a ) k − ( t 1 − a ) k ) ≤ M Γ ( α + 1 )   ( ( t 2 − a ) α − ( t 1 − a ) α ) + ∑ k = 0 n − 2 | x k | k ! ( ( t 2 − a ) k − ( t 1 − a ) k ) + ( | x b | ( n − 1 ) ! + n M ( b − a ) α − n + 1 ( n − 1 ) ! Γ ( α − n + 2 ) ) ( ( t 2 − a ) n − 1 − ( t 1 − a ) n − 1 ) .

As t 1   →   t 2 , the right-hand side of the aboveinequality tends to zero. As a consequence of Steps 1 to 3 together with the Arzelá-Ascoli theorem, T   :   C ( J , R )   →   C ( J , R )   is completely continuous.

Step 4: A priori bounds.

Let   ε =   { x   ∈   C ( J ,   R )   :   x   =   λ T x     for     some     0   <   λ   <   1 } , it shall be shown that the set is bounded.

Let x   ∈   ε , then x =   λ T x for some 0   <   λ   <   1 . Thus ∀ t   ∈   J ,

x =   λ T x = λ Γ ( α ) ∫ a t ( t − s ) α − 1 f ( s , x ( s ) ) d s + λ ( x b ( n − 1 ) ! + f ( a , x ( a ) ) ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ) ( t − a ) n − 1 − λ ( t − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n f ( s , x ( s ) ) d s + λ ∑ k = 0 n − 2 x k k ! ( t − a ) k   .

By the condition (H4) and Step 2,

  | x ( t ) | ≤   ∑ k = 0 n − 2 | x k | k ! ( b − a ) k + | x b | ( b − a ) n − 1 ( n − 1 ) ! + M ( b − a ) α ( 1 Γ ( α + 1 ) + n ( n − 1 ) ! Γ ( α − n + 2 ) ) .

Thus for every ∀ t   ∈   J ,

  ‖ x ‖ ∞   ≤   ∑ k = 0 n − 2 | x k | k ! ( b − a ) k + | x b | ( b − a ) n − 1 ( n − 1 ) ! + M ( b − a ) α ( 1 Γ ( α + 1 ) + n ( n − 1 ) ! Γ ( α − n + 2 ) )   :   =   R .

This shows that the set ε is bounded. As a consequence of Schauder’s fixed point theorem, T has a fixed point which is a solution of the problem (1.1).

In Theorem 3.3, if the condition (H4) is weakened, the fourth result can be obtained, which is a more general existence result (see [6] ).

Theorem 3.4 Assume that (H3) and the following conditionshold.

(H5) There exist a functional ψ f   ∈   L 1 ( J , R + ) and a continuous and nondecreasing   φ   :   [ 0 ,   ∞ )   →   ( 0 ,   ∞ ) , such that

| f ( t , x ( t ) ) | ≤ ψ f ( t ) φ ( | x ( t ) | ) for   each   t ∈ J = [ a ,   b ]   and   ∀ x ( t ) ∈ R .

(H6) There exists a number K > 0, such that

θ   =   K − 1 ( φ ( K ) ‖ I α ψ f ‖ L 1 + λ ( b − a ) n − 1 φ ( K ) ( n − 1 ) ! I α − n + 1 ψ f ( b ) + ∑ k = 0 n − 2 | x k | k ! ( b − a ) k + | x b | ( n − 1 ) ! ( b − a ) n − 1 + ψ f ( a ) φ ( | x ( ( a ) | ) ( n − 2 ) ! Γ ( α − n + 2 ) ( b − a ) α ) < 1. (3.5)

Then the BVP (1.1) has at least one solution on J.

Proof. Consider the operator T defined by (3.2), ∀ λ   ∈   [ 0 , 1 ] ,     t   ∈   J =   [ a , b ] , let x ( t ) meets x ( t ) =   λ ( T x ) ( t ) , then from (H5) and (H6),

| x ( t )   | =   | λ ( T x ) ( t ) |   ≤   | T x ( t ) |   ≤   1 Γ ( α ) ∫ a t ( t − s ) α − 1 f ( s , x ( s ) ) d s + | ( x b ( n − 1 ) ! + f ( a , x ( a ) ) ( b − a ) α − n + 1 ( n − 2 ) ! Γ ( α − n + 2 ) ) ( t − a ) n − 1 | + | ( t − a ) n − 1 | ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n | f ( s , x ( s ) ) | d s + ∑ k = 0 n − 2 | x k | k ! ( t − a ) k ≤   φ ( ‖ x ‖ ∞ ) Γ ( α ) ∫ a t ( t − s ) α − 1 ψ f ( s ) d s + | x b | ( n − 1 ) ! ( b − a ) n − 1 + ψ f ( a ) φ ( | x ( a ) | ) ( b − a ) α ( n − 2 ) ! Γ ( α − n + 2 ) + φ ( ‖ x ‖ ∞ ) ( b − a ) n − 1 ( n − 1 ) ! Γ ( α − n + 1 ) ∫ a b ( b − s ) α − n ψ f ( s ) d s + ∑ k = 0 n − 2 | x k | k ! ( b − a ) k ≤ φ ( ‖ x ‖ ∞ ) ‖ I α ψ f ‖ L 1 + | x b | ( n − 1 ) ! ( b − a ) n − 1 + ψ f ( a ) φ ( | x ( a ) | ) ( b − a ) α ( n − 2 ) ! Γ ( α − n + 2 ) + φ ( ‖ x ‖ ∞ ) ( b − a ) n − 1 ( n − 1 ) ! I α − n + 1 ψ f ( b ) + ∑ k = 0 n − 2 | x k | k ! ( b − a ) k .

By (H6), there exists K such that ‖ x ‖ ∞ ≠   K . Let D = { x   ∈   C ( J , R ) : ‖ x ‖ ∞ < K } , the operator T   :   D ¯ → C ( J , R ) is completely continuous. Through proper selection of D, there exists no x ( t )   ∈   ∂ D such that x ( t ) = λ ( T x ) ( t ) for some λ   ∈   ( 0 , 1 ) .

Therefore, T is Leray-Schauder type operator (see [6] ), so that it has a fixed point x ( t ) in U ¯ , which is a solution of the BVP (1.1).

For the boundary value problem

{ D c a α x ( t ) = | x ( t ) | t 1 + x ( t ) ,           t   ∈   J = [ 0 , 1 ] ,       n − 1   <   α   ≤   n   ,   x ( k ) ( 0 ) = 0 ,       k = 0 ,   1 ,   2 ,   ⋯ , n − 2   ;         x ( n − 1 ) ( 1 ) = 1     .         (4.1)

Take

f ( t ,   u ( t ) )   =   u ( t ) t 1 + u ( t )     , ( t ,   u ( t ) )   ∈   J ×   [ 0 ,   ∞ ) .

Let x ( t ) ,   y ( t )     ∈   [ 0 ,   ∞ ) ,     t   ∈   J . Then

| f ( t ,   x ( t ) )   − f ( t ,   y ( t ) ) | = t   | x ( t ) 1 + x ( t )     − y ( t ) 1 + y ( t ) | = t | x ( t ) − y ( t ) | ( 1 + x ( t ) ) ( 1 + y ( t ) )     ≤ t | x ( t ) − y ( t ) | . (4.2)

Hence the condition (H1) holds with λ ( t )   =   t   ∈   C ( J , R ) . It can be checked that condition (3.2) is satisfied with b = 1 . In fact,

θ   =   I α λ ( t )   +   λ ( 0 ) ( n − 2 ) ! Γ ( α − n + 2 ) + 1 ( n − 1 ) ! I α − n + 1 λ ( 1 )   =   1 Γ ( α + 2 ) t α + 1   +   1 ( n − 1 ) ! Γ ( α − n + 3 ) < 1   ,       ( t   ≤   1 ,     λ ( 0 )   =   0 ) (4.3)

only if

  1 Γ ( α + 2 )   + 1 ( n − 1 ) ! Γ ( α − n + 3 ) < 1   . (4.4)

For example, α   =   5 2 , then n   =   [ α ]   + 1   =   3 , Γ ( α + 2 ) =   Γ ( 9 2 )   =   105 π 16 ,

Γ ( α − n + 3 )   =   Γ ( α )   =   Γ ( 5 2 )   =   3 π 4 , ( n − 1 ) !   =   2 !   = 2 . Then

θ   ≤ 1 Γ ( α + 2 ) + 1 ( n − 1 ) ! Γ ( α − n + 3 )   = 1 Γ ( α + 2 ) + 1 2 Γ ( α ) = ( 16 105   + 2 3 ) 1 π = 0.4621 < 1   . (4.5)

Then by Theorem 3.1 the boundary value problem (4.1) has a uniquesolution on   J = [ 0 ,   1 ] for the values of α ∈ ( 2 ,   3 ] .

The authors would like to thank the reviewers for their valuable suggestions and comments, which improved the completeness of the paper. Research of J. Song is funded by the High-level Talents Project of Guangdong Province Colleges and Universities (2013-178). Research of Z. Zeng is funded by the Natural Science Foundation of Guangdong Province of China (S2012010010069). These supports are greatly appreciated.

Sun, Y.F., Zeng, Z. and Song, J. (2017) Existence and Uniqueness for the Boundary Value Problems of Nonlinear Fractional Differential Equation. Applied Mathematics, 8, 312-323. https://doi.org/10.4236/am.2017.83026