Open Access

A nonlocal multi-point multi-term fractional boundary value problem with Riemann-Liouville type integral boundary conditions involving two indices

  • Ahmed Alsaedi1,
  • Sotiris K Ntouyas2,
  • Ravi P Agarwal1, 3 and
  • Bashir Ahmad1Email author
Advances in Difference Equations20132013:369

DOI: 10.1186/1687-1847-2013-369

Received: 6 November 2013

Accepted: 1 December 2013

Published: 13 December 2013

Abstract

In this paper, we study the existence of solutions for fractional differential equations of arbitrary order with multi-point multi-term Riemann-Liouville type integral boundary conditions involving two indices. The Riemann-Liouville type integral boundary conditions considered in the problem address a more general situation in contrast to the case of a single index. Our results are based on standard fixed point theorems. Some illustrative examples are also presented.

MSC:26A33, 34A08.

Keywords

fractional differential equations nonlocal boundary conditions fixed point theorems

1 Introduction

In the last few decades, the subject of fractional differential equations has become a hot topic for the researchers due to its intensive development and applications in the field of physics, mechanics, chemistry, engineering, etc. For a reader interested in the systematic development of the topic, we refer the books [16]. A fractional-order differential operator distinguishes itself from the integer-order differential operator in the sense that it is nonlocal in nature, that is, the future state of a dynamical system or process involving fractional derivative depends on its current state as well its past states. In other words, differential equations of arbitrary order describe memory and hereditary properties of various materials and processes. As a matter of fact, this characteristic of fractional calculus makes the fractional-order models more realistic and practical than the classical integer-order models. There has been a great surge in developing the theoretical aspects such as periodicity, asymptotic behavior and numerical methods for fractional equations. For some recent work on the topic, see [725] and the references therein. In particular, the authors studied nonlinear fractional differential equations and inclusions of arbitrary order with multi-strip boundary conditions in [21], while a boundary value of nonlinear fractional differential equations of arbitrary order with Riemann-Liouville type multi-strip boundary conditions was investigated in [22]. Sudsutad and Tariboon [26] obtained some existence results for an integro-differential equation of fractional order q ( 1 , 2 ] with m-point multi-term fractional-order integral boundary conditions.

In this paper, we study a boundary value problem of fractional differential equations of arbitrary order q ( n 1 , n ] , n 2 with m-point multi-term Riemann-Liouville type integral boundary conditions involving two indices given by
{ D q c x ( t ) = f ( t , x ( t ) ) , t [ 0 , T ] , x ( 0 ) = 0 , x ( 0 ) = 0 , , x ( n 2 ) ( 0 ) = 0 , x ( T ) = i = 1 m 1 j = 1 ν i γ j , i [ I k j , i x ] | η i 1 η i ,
(1.1)
where D q c denotes the Caputo fractional derivative of order q, f is a given continuous function, [ I k j , i x ] | η i 1 η i = [ I k j , i x ] ( η i ) [ I k j , i x ] ( η i 1 ) , I k j , i is the Riemann-Liouville fractional integral of order k j , i > 0 , j = 1 , 2 , , ν i , ν i N = { 1 , 2 , } , i = 1 , 2 , , m 1 , 0 = η 0 < η 1 < η 2 < < η m 2 < η m 1 = T , and γ j , i R is such that
( T n 1 i = 1 m 1 j = 1 ν i γ j , i ( η i k j , i + n 1 η i 1 k j , i + n 1 ) Γ ( n ) Γ ( k j , i + n ) ) 0 .

Here we emphasize that Riemann-Liouville type integral boundary conditions involving two indices give rise to a more general situation in contrast to the case of a single index [22]. Furthermore, the present work dealing with an arbitrary-order problem generalizes the results for the problem of order q ( 1 , 2 ] obtained in [26]. Several examples are considered to show the worth of the results established in this paper.

We develop some existence results for problem (1.1) by using standard techniques of fixed point theory. The paper is organized as follows. In Section 2 we recall some preliminary facts that we need in the sequel, and Section 3 contains the main results. Section 4 provides some examples for the illustration of the main results.

2 Preliminaries from fractional calculus

Let us recall some basic definitions of fractional calculus [24].

Definition 2.1 For an at least n-times continuously differentiable function g : [ 0 , ) R , the Caputo derivative of fractional order q is defined as
D q c g ( t ) = 1 Γ ( n q ) 0 t ( t s ) n q 1 g ( n ) ( s ) d s , n 1 < q < n , n = [ q ] + 1 ,

where [ q ] denotes the integer part of the real number q.

Definition 2.2 The Riemann-Liouville fractional integral of order q is defined as
I q g ( t ) = 1 Γ ( q ) 0 t g ( s ) ( t s ) 1 q d s , q > 0 ,

provided the integral exists.

Lemma 2.3 For y C [ 0 , T ] , the fractional boundary value problem
{ D q c x ( t ) = y ( t ) , t [ 0 , T ] , q ( n 1 , n ] , x ( 0 ) = 0 , x ( 0 ) = 0 , , x ( n 2 ) = 0 , x ( T ) = i = 1 m 1 j = 1 ν i γ j , i [ I k j , i x ] | η i 1 η i
(2.1)
has a unique solution given by
x ( t ) = 1 Γ ( q ) 0 t ( t s ) q 1 y ( s ) d s t n 1 δ Γ ( q ) 0 T ( T s ) q 1 y ( s ) d s + t n 1 δ Γ ( q ) i = 1 m 1 j = 1 ν i γ j , i Γ ( k j , i ) [ 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 y ( u ) d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 y ( u ) d u d s ] ,
(2.2)
where
δ = ( T n 1 i = 1 m 1 j = 1 ν i γ j , i ( η i k j , i + n 1 η i 1 k j , i + n 1 ) Γ ( n ) Γ ( k j , i + n ) ) 0 .
(2.3)
Proof The general solution of fractional differential equations in (2.1) can be written as [4]
x ( t ) = 1 Γ ( q ) 0 t ( t s ) q 1 y ( s ) d s c 0 c 1 t c n 1 t n 1 .
(2.4)
Using the given boundary conditions, it is found that c 0 = 0 , c 1 = 0 , …, c n 2 = 0 . Applying the Riemann-Liouville integral operator I k j , i on (2.4), we get
I k j , i x ( t ) = 1 Γ ( k j , i ) 0 t ( t s ) k j , i 1 ( 1 Γ ( q ) 0 s ( s u ) q 1 y ( u ) d u c n 1 s n 1 ) d s = 1 Γ ( k j , i ) Γ ( q ) 0 t 0 s ( t s ) k j , i 1 ( s u ) q 1 y ( u ) d u d s c n 1 1 Γ ( k j , i ) 0 t ( t s ) k j , i 1 s n 1 d s .
Using the concept of beta function, we find that
1 Γ ( k j , i ) 0 t ( t s ) k j , i 1 s n 1 d s = t k j , i + n 1 Γ ( n ) Γ ( k j , i + n ) .
Now using the condition
x ( T ) = i = 1 m 1 j = 1 ν i γ j , i [ I k j , i x ] | η i 1 η i = [ I k j , i x ] ( η i ) [ I k j , i x ] ( η i 1 ) ,
we obtain
1 Γ ( q ) 0 T ( T s ) q 1 y ( s ) d s c n 1 T n 1 = i = 1 m 1 j = 1 ν i γ j , i Γ ( q ) Γ ( k j , i ) [ 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 y ( u ) d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 y ( u ) d u d s ] c n 1 i = 1 m 1 j = 1 ν i γ j , i ( η i k j , i + n 1 η i 1 k j , i + n 1 ) Γ ( n ) Γ ( k j , i + n ) ,
which yields
c n 1 = 1 δ Γ ( q ) 0 T ( T s ) q 1 y ( s ) d s 1 δ Γ ( q ) i = 1 m 1 j = 1 ν i γ j , i Γ ( k j , i ) [ 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 y ( u ) d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 y ( u ) d u d s ] ,

where δ is given by (2.3). Substituting the values of c 0 , c 1 , , c n 2 , c n 1 in (2.4), we obtain (2.2). This completes the proof. □

3 Main results

Let C : = C ( [ 0 , T ] , R ) denote the Banach space of all continuous functions defined on [ 0 , T ] × R endowed with a topology of uniform convergence with the norm x = sup t [ 0 , T ] | x ( t ) | .

To prove the existence results for problem (1.1), we need the following known results.

Theorem 3.1 (Leray-Schauder alternative [[27], p.4])

Let X be a Banach space. Assume that T : X X is a completely continuous operator and the set
V = { u X u = μ T u , 0 < μ < 1 }

is bounded. Then T has a fixed point in X.

Theorem 3.2 [28]

Let X be a Banach space. Assume that Ω is an open bounded subset of X with θ Ω , and let T : Ω ¯ X be a completely continuous operator such that
T u u , u Ω .

Then T has a fixed point in Ω ¯ .

By Lemma 2.3, we define an operator P : C C as
( P x ) ( t ) = 1 Γ ( q ) 0 t ( t s ) q 1 f ( s , x ( s ) ) d s t n 1 δ Γ ( q ) 0 T ( T s ) q 1 f ( s , x ( s ) ) d s + t n 1 δ Γ ( q ) i = 1 m 1 j = 1 ν i γ j , i Γ ( k j , i ) [ 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 f ( u , x ( u ) ) d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 f ( u , x ( u ) ) d u d s ] , t [ 0 , T ] .
(3.1)

Observe that problem (1.1) has a solution if and only if the associated fixed point problem P x = x has a fixed point.

For the sake of convenience, we set
ϑ = T q Γ ( q + 1 ) + T q + n 1 | δ | Γ ( q + 1 ) + T n 1 | δ | i = 1 m 1 j = 1 ν i | γ j , i | η i q + k j , i η i 1 q + k j , i Γ ( q + k j , i + 1 ) .
(3.2)

Theorem 3.3 Assume that there exists a positive constant L 1 such that | f ( t , x ) | L 1 for t [ 0 , T ] , x R . Then problem (1.1) has at least one solution.

Proof First of all, we show that the operator P is completely continuous. Note that the operator P is continuous in view of the continuity of f. Let B C be a bounded set. By the assumption that | f ( t , x ) | L 1 , for x B , we have
| ( P x ) ( t ) | 1 Γ ( q ) 0 t ( t s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) 0 T ( T s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) ( 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s ) L 1 [ 1 Γ ( q ) 0 t ( t s ) q 1 d s + t n 1 | δ | Γ ( q ) 0 T ( T s ) q 1 d s + t n 1 | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) ( 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 d u d s ) ] L 1 { T q Γ ( q + 1 ) + T q + n 1 | δ | Γ ( q + 1 ) + T n 1 | δ | i = 1 m 1 j = 1 ν i | γ j , i | η i q + k j , i η i 1 q + k j , i Γ ( q + k j , i + 1 ) } = L 2 ,
which implies that ( P x ) L 2 . Further, we find that
| ( P x ) ( t ) | = 1 Γ ( q 1 ) 0 t ( t s ) q 2 | f ( s , x ( s ) ) | d s + ( n 1 ) t n 2 | δ | Γ ( q ) 0 T ( T s ) q 1 | f ( s , x ( s ) ) | d s + ( n 1 ) t n 2 | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) ( 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s ) L 1 [ 1 Γ ( q 1 ) 0 t ( t s ) q 2 d s + ( n 1 ) t n 2 | δ | Γ ( q ) 0 T ( T s ) q 1 d s + ( n 1 ) t n 2 | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) ( 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 d u d s ) ] L 1 { T q 1 Γ ( q ) + ( n 1 ) T q + n 2 | δ | Γ ( q + 1 ) + ( n 1 ) T n 2 | δ | i = 1 m 1 j = 1 ν i | γ j , i | η i q + k j , i η i 1 q + k j , i Γ ( q + k j , i + 1 ) } = L 3 .
Hence, for t 1 , t 2 [ 0 , T ] , we have
| ( P x ) ( t 2 ) ( P x ) ( t 1 ) | t 1 t 2 | ( P x ) ( s ) | d s L 3 ( t 2 t 1 ) .

This implies that P is equicontinuous on [ 0 , T ] . Thus, by the Arzela-Ascoli theorem, the operator P : C C is completely continuous.

Next, we consider the set
V = { x C x = μ P x , 0 < μ < 1 } ,
and show that the set V is bounded. Let x V , then x = μ P x , 0 < μ < 1 . For any t [ 0 , T ] , we have
| x ( t ) | = μ | ( P x ) ( t ) | 1 Γ ( q ) 0 t ( t s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) 0 T ( T s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) ( 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s ) L 1 { T q Γ ( q + 1 ) + T q + n 1 | δ | Γ ( q + 1 ) + T n 1 | δ | i = 1 m 1 j = 1 ν i | γ j , i | η i q + k j , i η i 1 q + k j , i Γ ( q + k j , i + 1 ) } = M 1 .

Thus, x M 1 for any t [ 0 , T ] . So, the set V is bounded. Thus, by the conclusion of Theorem 3.1, the operator P has at least one fixed point, which implies that (1.1) has at least one solution. □

Theorem 3.4 Let there exist a small positive number τ such that | f ( t , x ) | ϵ | x | for 0 < | x | < τ , with 0 < ϵ 1 / ϑ , where ϑ is given by (3.2). Then problem (1.1) has at least one solution.

Proof Let us define B τ = { x C x < τ } and take x C such that x = τ , that is, x B τ . As before, it can be shown that P is completely continuous and
P x sup t [ 0 , t ] { 1 Γ ( q ) 0 t ( t s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) 0 T ( T s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) ( 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s ) } ϑ ϵ x ,

which, in view of the given condition ( ϵ ϑ 1 ) , gives P x x , x B τ . Therefore, by Theorem 3.2, the operator P has at least one fixed point, which in turn implies that problem (1.1) has at least one solution. □

Our next result is based on Leray-Schauder nonlinear alternative.

Lemma 3.5 (Nonlinear alternative for single-valued maps [[27], p.135])

Let E be a Banach space, C be a closed, convex subset of E, U be an open subset of C and 0 U . Suppose that F : U ¯ C is a continuous, compact (that is, F ( U ¯ ) is a relatively compact subset of C) map. Then either
  1. (i)

    F has a fixed point in U ¯ , or

     
  2. (ii)

    there are u U (the boundary of U in C) and λ ( 0 , 1 ) with u = λ F ( u ) .

     

Theorem 3.6 Assume that

(A1) there exist a function σ C ( [ 0 , 1 ] , R + ) and a nondecreasing function ψ : R + R + such that | f ( t , x ) | σ ( t ) ψ ( x ) , ( t , x ) [ 0 , T ] × R ;

(A2) there exists a constant M > 0 such that
M ψ ( M ) ϑ σ > 1 .

Then boundary value problem (1.1) has at least one solution on [ 0 , T ] .

Proof Consider the operator P : C C defined by (3.1). We show that P maps bounded sets into bounded sets in C ( [ 0 , T ] , R ) . For a positive number r, let B r = { x C ( [ 0 , T ] , R ) : x r } be a bounded set in C ( [ 0 , T ] , R ) . Then
P x sup t [ 0 , T ] { 1 Γ ( q ) 0 t ( t s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) 0 T ( T s ) q 1 | f ( s , x ( s ) ) | d s + t n 1 | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) ( 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) | d u d s ) } ψ ( r ) { T q Γ ( q + 1 ) + T q + n 1 | δ | Γ ( q + 1 ) + T n 1 | δ | i = 1 m 1 j = 1 ν i | γ j , i | η i q + k j , i η i 1 q + k j , i Γ ( q + k j , i + 1 ) } σ .
Next we show that F maps bounded sets into equicontinuous sets of C ( [ 0 , 1 ] , R ) . Let t , t [ 0 , 1 ] with t < t and x B r , where B r is a bounded set of C ( [ 0 , 1 ] , R ) . Then we obtain
| ( P x ) ( t ) ( P x ) ( t ) | = | 1 Γ ( q ) 0 t ( t s ) q 1 f ( s , x ( s ) ) d s 1 Γ ( q ) 0 t ( t s ) q 1 f ( s , x ( s ) ) d s [ ( t ) n 1 ( t ) n 1 ] δ Γ ( q ) 0 T ( T s ) q 1 f ( s , x ( s ) ) d s + [ ( t ) n 1 ( t ) n 1 ] δ Γ ( q ) i = 1 m 1 j = 1 ν i γ j , i Γ ( k j , i ) [ 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 f ( u , x ( u ) ) d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 f ( u , x ( u ) ) d u d s ] | 1 Γ ( q ) 0 t | ( t s ) q 1 ( t s ) q 1 | ψ ( r ) σ ( s ) d s + 1 Γ ( q ) t t | t s | q 1 ψ ( r ) σ ( s ) d s + | ( t ) n 1 ( t ) n 1 | | δ | Γ ( q ) 0 T | T s | q 1 ψ ( r ) σ ( s ) d s + | ( t ) n 1 ( t ) n 1 | | δ | Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) [ 0 η i 0 s | ( η i s ) k j , i 1 ( s u ) q 1 | ψ ( r ) σ ( u ) d u d s + 0 η i 1 0 s | ( η i 1 s ) k j , i 1 ( s u ) q 1 | ψ ( r ) σ ( u ) d u d s ] .

Obviously the right-hand side of the above inequality tends to zero independently of x B r as t t 0 . As P : C ( [ 0 , T ] , R ) C ( [ 0 , T ] , R ) satisfies the above assumptions, therefore it follows by the Arzelá-Ascoli theorem that P is completely continuous.

Let x be a solution. Then, for t [ 0 , T ] , and following the similar computations as before, we find that
| x ( t ) | = | μ ( P x ) ( t ) | ψ ( x ) { T q Γ ( q + 1 ) + T q + n 1 | δ | Γ ( q + 1 ) + T n 1 | δ | i = 1 m 1 j = 1 ν i | γ j , i | η i q + k j , i η i 1 q + k j , i Γ ( q + k j , i + 1 ) } σ = ψ ( x ) ϑ σ .
In consequence, we have
x ψ ( x ) ϑ σ 1 .
Thus, by (A2), there exists M such that x M . Let us set
V = { x C ( [ 0 , T ] , R ) : x < M + 1 } .

Note that the operator P : V ¯ C ( [ 0 , T ] , R ) is continuous and completely continuous. From the choice of V, there is no x V such that x = μ P ( x ) for some μ ( 0 , 1 ) . Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 3.5), we deduce that P has a fixed point x V ¯ which is a solution of problem (1.1). This completes the proof. □

Finally we prove an existence and uniqueness result by means of Banach’s contraction mapping principle.

Theorem 3.7 Suppose that f : [ 0 , T ] × R R is a continuous function and satisfies the following assumption:

(A3) | f ( t , x ) f ( t , y ) | L | x y | , t [ 0 , 1 ] , L > 0 , x , y R .

Then boundary value problem (1.1) has a unique solution provided
L < 1 / ϑ ,
(3.3)

where ϑ is given by (3.2).

Proof With r M ϑ / ( 1 L ϑ ) , we define B r = { x C : x r } , where M = sup t [ 0 , T ] | f ( t , 0 ) | < and ϑ is given by (2.3). Then we show that P B r B r . For x B r , by means of the inequality | f ( s , x ( s ) ) | | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | L x + M L r + M , it can easily be shown that
P x = ( L r + M ) ϑ r .
Now, for x , y C and for each t [ 0 , T ] , we obtain
( P x ) ( P y ) sup t [ 0 , T ] { 1 Γ ( q ) 0 t ( t s ) q 1 | f ( s , x ( s ) ) f ( s , y ( s ) ) | d s + t n 1 δ Γ ( q ) 0 T ( T s ) q 1 | f ( s , x ( s ) ) f ( s , y ( s ) ) | d s + t n 1 δ Γ ( q ) i = 1 m 1 j = 1 ν i | γ j , i | Γ ( k j , i ) [ 0 η i 0 s ( η i s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) f ( u , y ( u ) ) | d u d s 0 η i 1 0 s ( η i 1 s ) k j , i 1 ( s u ) q 1 | f ( u , x ( u ) ) f ( u , y ( u ) ) | d u d s ] } L ϑ x y .

Note that ϑ depends only on the parameters involved in the problem. As L ϑ < 1 , therefore P is a contraction. Hence, by Banach’s contraction mapping principle, problem (1.1) has a unique solution on [ 0 , T ] . □

4 Examples

In this section, we present some examples for the illustration of the results established in Section 3 by choosing the nonlinear function f ( t , x ( t ) ) appropriately. Let us consider the following nonlocal boundary value problem:
{ D 11 / 2 c x ( t ) = f ( t , x ( t ) ) , t [ 0 , 2 ] , x ( 0 ) = 0 , x ( 0 ) = 0 , , x ( 4 ) ( 0 ) = 0 , x ( T ) = i = 1 m 1 j = 1 ν i γ j , i [ I k j , i x ] | η i 1 η i ,
(4.1)
where q = 11 / 2 , n = 6 , m = 5 , η 0 = 0 , η 1 = 1 / 2 , η 2 = 1 , η 3 = 3 / 2 , η 4 = T = 2 , ν 1 = 1 , ν 2 = 2 , ν 3 = 3 , ν 4 = 4 , γ 1 , 1 = 4 , γ 1 , 2 = 7 , γ 2 , 2 = 8 , γ 1 , 3 = 10 , γ 2 , 3 = 15 , γ 3 , 3 = 21 , γ 1 , 4 = 3 , γ 2 , 4 = 9 , γ 3 , 4 = 15 , γ 4 , 4 = 21 , k 1 , 1 = 3 / 2 , k 1 , 2 = 5 / 4 , k 2 , 2 = 7 / 4 , k 1 , 3 = 1 / 2 , k 2 , 3 = 5 / 4 , k 3 , 3 = 9 / 4 , k 1 , 4 = 3 / 4 , k 2 , 4 = 7 / 4 , k 3 , 4 = 11 / 4 , k 4 , 4 = 15 / 4 . Using the given data, we find that
δ = T n 1 i = 1 m 1 j = 1 ν i γ j , i ( η i k j , i + n 1 η i 1 k j , i + n 1 ) Γ ( n ) Γ ( k j , i + n ) 113.395760 ,
and
ϑ = T q Γ ( q + 1 ) + T q + n 1 | δ | Γ ( q + 1 ) + T n 1 | δ | i = 1 m 1 j = 1 ν i | γ j , i | η i q + k j , i η i 1 q + k j , i Γ ( q + k j , i + 1 ) 0.37757 .
  1. (a)
    As a first example, let us take
    f ( t , x ( t ) ) = e ( 2 cos 2 x ( t ) ) 2 [ 5 sin 2 t + 10 ln ( 1 + 4 cos 2 x ( t ) ) ] ( 26 + sin x ( t ) ) .
    (4.2)
     
Observe that | f ( t , x ) | L 1 with L 1 = e 4 ( 1 + 2 ln 5 ) . Thus the hypothesis of Theorem 3.3 is satisfied. Hence, by the conclusion of Theorem 3.3, problem (4.1) with f ( t , x ( t ) ) given by (4.2) has at least one solution.
  1. (b)
    Let us consider
    f ( t , x ( t ) ) = x ( b 4 + x 4 ( t ) ) 1 4 + 2 ( 1 + sin ( t 4 + 1 ) ) 3 ( 1 cos x ( t ) ) , x 0 , b > 0 .
    (4.3)
     
For sufficiently small x (ignoring x 2 and higher powers of x), we have
| x ( b 4 + x 4 ( t ) ) 1 4 + 2 ( 1 + sin ( t 4 + 1 ) ) 3 ( 1 cos x ( t ) ) | b | x | .
Choosing b 1 / ϑ , all the assumptions of Theorem 3.4 hold. Therefore, the conclusion of Theorem 3.4 implies that problem (4.1) with f ( t , x ( t ) ) given by (4.3) has at least one solution.
  1. (c)
    Consider
    f ( t , x ( t ) ) = t + 14 ( 1 + | x | 1 + | x | ) σ ( t ) ψ ( x ) ,
    (4.4)
     
with σ ( t ) = t + 14 and ψ ( x ) = 2 . Using σ = 4 , ϑ 0.37757 , we find by condition (A2) that M > M 1 , where M 1 3.02056 . Thus all the assumptions of Theorem 3.6 are satisfied. Hence, it follows by Theorem 3.6 that problem (4.1) with f ( t , x ( t ) ) defined by (4.4) has at least one solution.
  1. (d)
    For the illustration of the existence-uniqueness result, we choose
    f ( t , x ( t ) ) = 1 ( t + 16 ) ( | x | 1 + | x | + tan 1 x ) + 4 + sin t .
    (4.5)
     

Clearly, L = 1 / 2 as | f ( t , x ) f ( t , y ) | 1 2 | x y | and L < 1 ϑ 2.64852 . Therefore all the conditions of Theorem 3.7 hold, and consequently there exists a unique solution for problem (4.1) with f ( t , x ( t ) ) given by (4.5).

Declarations

Acknowledgements

This research was partially supported by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Abdulaziz University
(2)
Department of Mathematics, University of Ioannina
(3)
Department of Mathematics, Texas A&M University

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© Alsaedi et al.; licensee Springer. 2013

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.