Some Results for a Four $ Point Boundary Value Problems for a coupled system Involving Caputo Derivatives

Some results for a four-point boundary value problems for coupled

system involving Caputo derivative

M. Houas,

M. Benbachir

and Z. Dahmani

a_{Faculty of Sciences and Technology Khemis Miliana University, Ain Defla, Algeria.} b_{LPAM, Faculty SEI, UMAB Mostaganem, Algeria.}

Abstract

In this paper, we prove the existence and uniqueness of solutions for a system for fractional differential equations with four point boundary conditions. The results are obtained using Banach contraction principle and Krasnoselkii’s fixed point theorem

         Dα_{x (t) + f t, y (t) , D}δ_{y (t)
= 0, t ∈ J,} Dβ_{y (t) + g (t, x (t) , D}σ_{x (t)) = 0, t ∈ J,} x (0) = y (0) = 0, x (1) − λ1x (η) = 0, y (1) − λ1y (η) = 0, x00(0) = y00(0) = 0, x00(1) − λ2x00(ξ) =0, y00(1) − λ2y 00 (ξ) =0,

where 3 < α, β ≤ 4, α − 2 < σ ≤ α − 1, β − 2 < δ ≤ β − 1, 0 < ξ, η < 1, and Dα_{, D}β_{, D}δ _{and D}σ_{, are the}

Caputo fractional derivatives, J = [0, 1] , λ1, λ2are real constants with λ1η 6= 1, λ2ξ 6= 1 and f , g continuous

functions on [0, 1] ×R2.

Keywords: Caputo derivative; Boundary Value Problem; fixed point theorem.

2010 MSC:26A33, 34B25, 34B15. c2012 MJM. All rights reserved.

1

Introduction

Differential equations of fractional order have been shown to be very useful in the study of models of many phenomena in various fields of science and engineering, such as electrochemistry, physics, chemistry, viscoelasticity, control, image and signal processing, biophysics. For more details, we refer the reader to [4, 6, 9, 11, 12, 14, 16, 17] and references therein. There has been a significant progress in the investigation of these equations in recent years, see [5, 7, 8, 14, 15, 26]. More recently, some basic theory for the initial bound-ary value problems of fractional differential equations has been discussed in [1, 13, 14]. Recently, existence and uniqueness of solutions to boundary value problems for fractional differential equations had attracted the attention of many authors, see for example, [4, 5, 7, 8, 14, 15, 18, 26] and the references therein. The study of coupled system of fractional order is also important as such systems occur in various problems of applied science [3, 10, 19, 20, 23, 25]. In the last decade, many authors have established the existence and uniqueness for solutions of some systems of nonlinear fractional differential equations, one can see [19, 22, 23, 24] and references cited therein. For example in [2, 20, 25] the authors established sufficient conditions for the exis-tence of solutions for a two-point and three-point boundary value problem for a coupled system of fractional differential equations.

In [2, 20, 21, 25], the existence and uniqueness of solutions was investigated for a nonlinear coupled system for fractional differential equations with two-point and three-point boundary conditions by using Schauder’s fixed point theorem.

∗_{Corresponding author.}

Motivated by the above mentioned work, this paper deals with the existence of solution for four point boundary value problems for a coupled system of fractional differential equations for the following problem

       Dα_{x (t) + f t, y (t) , D}δ_{y (t)
= 0, t ∈ J,} Dβ_{y (t) + g (t, x (t) , D}σ_{x (t)) = 0, t ∈ J,} x (0) = y (0) = 0, x (1) − λ1x (η) = 0, y (1) − λ1y (η) = 0, x00(0) = y00(0) = 0, x00(1) − λ2x00(ξ) =0, y00(1) − λ2y00(ξ) =0, (1.1)

where 3 < α, β ≤ 4, α − 2 < σ ≤ α − 1, β − 2 < δ ≤ β − 1, 0 < ξ, η < 1, and Dα_{, D}β_{, D}δ _{and D}σ_{, are}

the Caputo fractional derivatives, J = [0, 1] , λ1, λ2 are real constants with λ1η 6= 1, λ2ξ 6= 1 and f , g are

continuous functions on [0, 1] ×R2.

The rest of this paper is organized as follows. In section 2, we present some preliminaries and lemmas. Section 3 is devoted to existence of solution of problem (1.1). In section 4 examples are treated illustrating our results.

2

Preliminaries

The following notations, definitions, and preliminary facts will be used throughout this paper.

Definition 2.1. The Riemann-Liouville fractional integral operator of order α ≥ 0, for a continuous function f on [0,∞[ is defined as: Jα_{f (t) =} 1 Γ (α) Z t 0 (t − τ)α−1_{f (τ) dτ, α > 0, (2.2)} J0f (t) = f (t) , whereΓ (α) := R₀∞e−uuα−1_du.

Definition 2.2. The fractional derivative of f ∈ Cn([0,∞[) in the Caputo’s sense is defined as:

Dα_{f (t) =} 1

Γ (n − α) Z t

0

(t − τ)n−α−1f(n)(τ) dτ, n − 1 < α, n ∈ N∗. (2.3) For more details about fractional calculus, we refer the reader to [14, 17].

Let us now introduce the spaces

X = {x : x ∈ C ([0, 1]) , Dσ_{x ∈ C ([0, 1])},}

and

Y = {y : y ∈ C ([0, 1]) , Dδ_{y ∈ C ([0, 1])},}

endowed with the norm

kxk_X =kxk + kDσ_{xk , kxk = sup} t∈J |x (t)| , kDσ_{xk = sup} t∈J |Dσ_{x (t)| ,} and kyk_Y=kyk + D δ_y , kyk = sup t∈J |y (t)| , D δ_y =sup t∈J    D δ_{y (t)}  .

Obviously, (X, k . kX)and (Y, k . kY),is a Banach space. The product space X × Y, k(x, y)kX×Y
is also

Ba-nach space with norm k(x, y)k_X×Y=kxk_X+kyk_Y. We give the following lemmas [12]:

Lemma 2.1. For α >0, the general solution of the fractional differential equation Dα_{x (t) = 0 is given by}

x (t) = c0+c1t + c2t2+... + cn−1tn−1, (2.4)

Lemma 2.2. Let α >0. Then

Jα_Dα_{x (t) = x (t) + c}

0+c1t + c2t2+... + cn−1tn−1, (2.5)

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

We need also the following auxiliary result:

Lemma 2.3. Let g ∈ C ([0, 1]) , the solution of the equation

Dα_{x (t) + g (t) = 0, t ∈ J, 3 < α ≤ 4, (2.6)}

subject to the boundary condition

x (0) = 0, x (1) − λ1x (η) = 0, (2.7) x00(0) = 0, x00(1) − λ2x00(ξ) =0, is given by: x (t) = −_{Γ (α)}1 Z t 0 (t − s)α−1_{g (s) ds (2.8)} + λ1t (λ1η −1)Γ (α) Z η 0 (η − s)α−1_{g (s) ds} − t (λ1η −1)Γ (α) Z 1 0 (1 − s)α−1_{g (s) ds} + λ2− λ2λ1η 3
_{t + (λ} 2λ1η − λ2)t3 6 (λ1η −1) (λ2ξ −1)Γ (α − 2) Z ξ 0 (ξ − s)α−3_{g (s) ds} − 1 − λ1η 3
_{t + (λ} 1η −1) t3 6 (λ1η −1) (λ2ξ −1)Γ (α − 2) Z 1 0 (1 − s)α−3_{g (s) ds.}

Proof. For ci∈R, i = 0, 1, 2, 3, and by Lemmas ((2.1), (2.2)), the general solution of (2.6) is given by

x (t) = − 1 Γ (α) Z t 0 (t − s)α−1_{g (s) ds − c} 0− c1t − c2t2− c3t3 (2.9)

Using the boundary condition (2.7), we have c0=c2=0, and

c1= − λ1 (λ1η −1)Γ (α) Z η 0 (η − s)α−1_{g (s) ds (2.10)} + 1 (λ1η −1)Γ (α) Z 1 0 (1 − s)α−1_{g (s) ds} − λ2 1 − λ1η 3
6 (λ1η −1) (λ2ξ −1)Γ (α − 2) Z ξ 0 (ξ − s)α−3_{g (s) ds} + (1 − λ1η) 6 (λ1η −1) (λ2ξ −1)Γ (α − 2) Z 1 0 (1 − s)α−3_{g (s) ds} and c3= − λ2 6 (λ2ξ −1)Γ (α − 2) Z ξ 0 (ξ − s)α−3_{g (s) ds (2.11)} + 1 6 (λ2ξ −1)Γ (α − 2) Z 1 0 (1 − s)α−3_{g (s) ds}

3

Main Results

Let us the take of convenience, we set:

M1= |λ1η−1|+|λ1|η α₊₁ |λ1η−1|Γ(α+1) (3.12) +(|λ2−λ2λ1η 3_|+|λ 2λ1η−λ2|)ξα−2+|1−λ1η3|+|λ1η−1| 6|λ₁η−1||λ2ξ−1|Γ(α−1) , M2= _{Γ(α−σ+1)}1 + |λ1|η α₊₁ |λ1η−1|Γ(α+1)Γ(2−σ) +_6|λ|λ2−λ2λ1η3|ξα−2+|1−λ1η3| 1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ) + |λ2λ1η−λ2|ξα−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ), M3= |λ1η−1|+|λ1|η β₊₁ |λ1η−1|Γ(β+1) + (|λ2−λ2λ1η3|+|λ2λ1η−λ2|)ξβ−2+|1−λ1η3|+|λ1η−1| 6|λ1η−1||λ2ξ−1|Γ(β−1) , M4= _{Γ(β−δ+1)}1 + |λ1|η β₊₁ |λ1η−1|Γ(β+1)Γ(2−δ) +_6|λ|λ2−λ2λ1η3|ξβ−2+|1−λ1η3| 1η−1||λ2ξ−1|Γ(β−1)Γ(2−δ) + |λ2λ1η−λ2|ξβ−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(β−1)Γ(4−δ), L1= (|λ2−λ2λ1η3|+|λ2λ1η−λ2|)ξα−2+|1−λ1η3|+|λ1η−1| 6|λ1η−1||λ2ξ−1|Γ(α−1) , L2= |λ2−λ2λ1η 3_|ξα−2_+|1−λ 1η3| 6|λ1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ)+ |λ2λ1η−λ2|ξα−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ), L3= (|λ2−λ2λ1η 3_|+|λ 2λ1η−λ2|)ξβ−2+|1−λ1η3|+|λ1η−1| 6|λ₁η−1||λ2ξ−1|Γ(β−1) , L4= |λ2−λ2λ1η 3_|ξβ−2_+|1−λ 1η3| 6|λ1η−1||λ2ξ−1|Γ(β−1)Γ(2−δ)+ |λ2λ1η−λ2|ξβ−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(β−1)Γ(4−δ).

Now list the following hypotheses for convenience:

(H1) : There exist two constants k1and k2such that for all t ∈ [0, 1] and (x1, y1), (x2, y2) ∈R2, we have

| f (t, x1, y1) −f (t, x2, y2)| ≤ k1(|x1− x2| + |y1− y2|) , (3.13)

|g (t, x1, y1) −g (t, x2, y2)| ≤ k2(|x1− x2| + |y1− y2|) .

(H2)) : The functions f , g : [0, 1] ×R2→R are continuous (H3) : There exists positive constants N1and N2such that

| f (t, x, y)| ≤ N1, |g (t, x, y)| ≤ N2for each t ∈ J and all x, y ∈R.

Our first result is based on Banach contraction principle:

Theorem 3.1. Assume that the hypothesis (H1)holds. If

k1(M1+M2) +k2(M3+M4) <1, (3.14)

then the boundary value problem (1.1) has a unique solution. Proof. Consider the operator φ : X × Y → X × Y defined by:

φ (x, y) (t) := (φ1y (t) , φ2x (t)) , (3.15) where φ1y (t) := − 1 Γ (α) Z t 0 (t − s)α−1_f_{s, y (s) , D}δ_{y (s)}_{ds (3.16)} + λ1t (λ1η −1)Γ (α) Z η 0 (η − s)α−1_f_{s, y (s) , D}δ_{y (s)}_ds − t (λ1η −1)Γ (α) Z 1 0 (1 − s)α−1_f s, y (s) , Dδ_{y (s)}_ds + λ2− λ2λ1η 3
_{t + (λ} 2λ1η − λ2)t3 6 (λ1η −1) (λ2ξ −1)Γ (α − 2) Z ξ 0 (ξ − s)α−3_f_{s, y (s) , D}δ_{y (s)}_ds − 1 − λ1η 3
_{t + (λ} 1η −1) t3 6 (λ1η −1) (λ2ξ −1)Γ (α − 2) Z 1 0 (1 − s)α−3_f_{s, y (s) , D}δ_{y (s)}_ds,

and φ2x (t) := − 1 Γ (β) Z t 0 (t − s)β−1_{g (s, x (s) , D}σ_{x (s)) ds (3.17)} + λ1t (λ1η −1)Γ (β) Z η 0 (η − s)α−1_{g (s, x (s) , D}σ_{x (s)) ds} − t (λ1η −1)Γ (β) Z 1 0 (1 − s)α−1_{g (s, x (s) , D}σ_{x (s)) ds} + λ2− λ2λ1η 3
_{t + (λ} 2λ1η − λ2)t3 6 (λ1η −1) (λ2ξ −1)Γ (β − 2) Z ξ 0 (ξ − s)α−3_{g (s, x (s) , D}σ_{x (s)) ds} − 1 − λ1η 3
_{t + (λ} 1η −1) t3 6 (λ1η −1) (λ2ξ −1)Γ (β − 2) Z 1 0 (1 − s)α−3_{g (s, x (s) , D}σ_{x (s)) ds.}

We shall prove that φ is a contraction mapping :

For (x, y) , (x1, y1) ∈X × Y and for each t ∈ J, we have

|φ1y (t) − φ1y1(t)| ≤ _Γ(α)1 Z t 0 (t − s)α−1  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds + |λ1|t |λ1η−1|Γ(α) Z η 0 (η − s)α−1  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds +_|λ t 1η−1|Γ(α) Z 1 0 (1 − s)α−1  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds +|λ2_6|λ−λ2λ1η3|t+|λ2λ1η−λ2|t3 1η−1||λ2ξ−1|Γ(α−2) Z ξ 0 (ξ − s)α−3  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds + |1−λ1η 3_|t+|λ 1η−1|t3 6|λ₁η−1||λ2ξ−1|Γ(α−2) Z 1 0 (1 − s)α−3  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds. Using the (H1) ,we obtain

|φ1y (t) − φ1y1(t)| ≤ k1(|λ1η−1|+|λ1|ηα+1)(ky−y1k+kDδy−Dδy1k) |λ1η−1|Γ(α+1) +k1[(|λ2−λ2λ1η3|+|λ2λ1η−λ2_6|λ|)ξα−2+|1−λ1η3|+|λ1η−1|](ky−y1k+kDδy−Dδy1k) 1η−1||λ2ξ−1|Γ(α−1) . Consequently,we have |φ1x (t) − φ1y (t)| ≤ k1M1  ky − y1k + D δ_{y − D}δ_y 1  . (3.18)

Which implies that

kφ1(x) − φ1(y)k ≤ k1M1  ky − y1k + D δ_{y − D}δ_y 1  , (3.19) and |Dσ φ1y (t) − Dσφ1y1(t)| ≤ _Γ(α−σ)1 Z t 0 (t − s)α−σ−1  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds + |λ1|t1−σ |λ1η−1|Γ(α)Γ(2−σ) Z η 0 (η − s)α−1  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds + t1−σ |(λ1η−1)|Γ(α)Γ(2−σ) Z 1 0 (1 − s)α−1  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds +   |λ2−λ2λ1η3|t1−σ 6|λ1η−1||λ2ξ−1|Γ(α−2)Γ(2−σ) + |λ2λ1η−λ2|t3−σ |λ1η−1||λ2ξ−1|Γ(α−2)Γ(4−σ)   Z ξ 0 (ξ − s)α−3  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds

+   |1−λ1η3|t1−σ 6|λ₁η−1||λ2ξ−1|Γ(α−2)Γ(2−σ) + |λ1η−1|t3−σ |λ1η−1||λ2ξ−1|Γ(α−2)Γ(4−σ)   Z 1 0 (1 − s)α−3  f  s, y (s) , Dδ_{y (s)}_{− f}_{s, y} 1(s) , Dδy1(s)    ds. By (H1), yields |Dσ_φ 1y (t) − Dσφ1y1(t)| ≤ k1(ky−y1k+kDδy−Dδy1k) Γ(α−σ+1) + k1[|λ1|ηα+1](ky−y1k+kDδy−Dδy1k) |λ1η−1|Γ(α+1)Γ(2−σ) +k1[|λ2−λ2λ1η_6|λ3|ξα−2+|1−λ1η3|](ky−y1k+kDδy−Dδy1k) 1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ) +k1[|λ2λ1η−λ2|ξα−2+|λ1η−1|](ky−y1k+kDδy−Dδy1k) |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ) . Hence,we have |Dσ_φ 1y (t) − Dσφ1y1(t)| ≤ k1 h 1 Γ(α−σ+1)+ |λ1|η α₊₁ |λ1η−1|Γ(α+1)Γ(2−σ) i  ky − y1k + D δ_{y − D}δ_y 1  +k1   |λ2−λ2λ1η3|ξα−2+|1−λ1η3| 6|λ1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ) + |λ2λ1η−λ2|ξα−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)    ky − y1k + D δ_{y − D}δ_y 1  . Therefore, |Dσ φ1y (t) − Dσφ1y1(t)| ≤ k1M2  ky − y1k + D δ_{y − D}δ_y 1  . (3.20) And consequently, kDσ_φ 1(y) − Dσφ1(y1)k ≤ k1M2  ky − y1k + D δ_{y − D}δ_y 1  . (3.21)

By (3.19) and (3.21),we can write

kφ1(y) − φ1(y1)kX ≤ k1(M1+M2)  ky − y1k + D δ_{y − D}δ_y 1  . (3.22)

With the same arguments as before, we have

kφ2(x) − φ2(x1)kY≤ k2(M3+M4) (kx − x1k + kDσx − Dσx1k) . (3.23)

And by,(3.22) and (3.23) we obtain

kφ (x, y) − φ (x1, y1)kX×Y≤  k1(M1+M2) +k2(M3+M4)  k(x − x1, y − y1)kX×Y. (3.24)

Consequently by (3.14) , we conclude that φ is contraction. As a consequence of Banach fixed point theorem, we deduce that φ has a fixed point which is a solution of the boundary value problem (1.1) .

Now, we use Krasnselskii’s fixed point theorem to prove the following result:

Theorem 3.2. Assume that the hypotheses (H1) − (H2) and (H3) are satisfied,such that

k1θ1+k2θ2<1, (3.25) where θ1= |λ1η−1|+|λ1|η α₊₁ |λ1η−1|Γ(α+1) + 1 Γ(α−σ+1)+ |λ1|η α₊₁ |λ1η−1|Γ(α+1)Γ(2−σ), θ2= |λ1η−1|+|λ1|η β₊₁ |λ1η−1|Γ(β+1) + 1 Γ(β−δ+1)+ |λ1|η β₊₁ |λ1η−1|Γ(β+1)Γ(2−δ),

if there exist µ ∈R such that

N1(M1+M2) +N2(M3+M4) ≤ µ, (3.26)

Proof. We shall use Krasnselskii’s fixed point theorem to prove that φ has at least a fixed point on X × Y. Suppose that (3.26) holds and let us take

φ (x, y) (t) := T (x, y) (t) + R (x, y) (t) , (3.27) where T (x, y) (t) := (T1y (t) , T2x (t)) , (3.28) T1y (t) = −_Γ(α)1 Z t 0 (t − s)α−1_f_{s, y (s) , D}δ_{y (s)}_{ds (3.29)} + λ1t (λ1η−1)Γ(α) Z η 0 (η − s)α−1_f_{s, y (s) , D}δ_{y (s)}_ds −_(λ t 1η−1)Γ(α) Z 1 0 (1 − s)α−1_f_{s, y (s) , D}δ_{y (s)}_ds, T2x (t) = −_Γ(β)1 Z t 0 (t − s)β−1_{g (s, x (s) , D}σ_{x (s)) ds (3.30)} + λ1t (λ1η−1)Γ(β) Z η 0 (η − s)α−1_{g (s, x (s) , D}σ_{x (s)) ds} − t (λ1η−1)Γ(β) Z 1 0 (1 − s)α−1_{g (s, x (s) , D}σ_{x (s)) ds,} and R (x, y) (t) := (R1y (t) , R2x (t)) , (3.31) where R1y (t) = (λ2−λ2λ1η 3_)t+(λ 2λ1η−λ2)t3 6(λ1η−1)(λ2ξ−1)Γ(α−2) Z ξ 0 (ξ − s)α−3_f_{s, y (s) , D}δ_{y (s)}_ds − (1−λ1η3)t+(λ1η−1)t3 6(λ1η−1)(λ2ξ−1)Γ(α−2) Z 1 0 (1 − s)α−3_f_{s, y (s) , D}δ_{y (s)}_{ds, (3.32)} R2x (t) = (λ2−λ2λ1η3)t+(λ2λ1η−λ2)t3 6(λ1η−1)(λ2ξ−1)Γ(β−2) Z ξ 0 (ξ − s)α−3_{g (s, x (s) , D}σ_{x (s)) ds} − (1−λ1η3)t+(λ1η−1)t3 6(λ1η−1)(λ2ξ−1)Γ(β−2) Z 1 0 (1 − s)α−3_{g (s, x (s) , D}σ_{x (s)) ds. (3.33)}

The proof will be given in several steps.

Step1: We shall prove that for any (x, y) , (x1, y1) ∈ Bµ, then T (x, y) + R(x1, y1) ∈ Bµ, Such that Bµ =

{(x, y) ∈ X × Y; k(x, y)k_X×Y≤ µ}.

For any (x, y) , (x1, y1) ∈Bµand for each t ∈ J we have:

|T1y (t) + R1y1(t)| =| −_Γ(α)1 Z t 0 (t − s)α−1_f_{s, y (s) , D}δ_{y (s)}_ds + λ1t (λ1η−1)Γ(α) Z η 0 (η − s)α−1_f_{s, y (s) , D}δ_{y (s)}_ds −_(λ t 1η−1)Γ(α) Z 1 0 (1 − s)α−1_f_{s, y (s) , D}δ_{y (s)}_ds +(λ2−λ2λ1η 3_)t+(λ 2λ1η−λ2)t3 6(λ1η−1)(λ2ξ−1)Γ(α−2) Z ξ 0 (ξ − s)α−3_f_{s, y (s) , D}δ_{y (s)}_ds − (1−λ1η3)t+(λ1η−1)t3 6(λ1η−1)(λ2ξ−1)Γ(β−2) Z 1 0 (1 − s)α−3_f_{s, y (s) , D}δ_{y (s)}_{ds |}

then, |T1y (t) + R1y1(t)| ≤ _Γ(α)1 Z t 0 (t − s)α−1  f  s, y (s) , Dδ_{y (s)}  ds + |λ1| |λ1η−1|Γ(α) Z η 0 (η − s)α−1  f  s, y (s) , Dδ_{y (s)}  ds + 1 |λ1η−1|Γ(α) Z 1 0 (1 − s)α−1  f  s, y (s) , Dδ_{y (s)}  ds +|λ_6|λ2−λ2λ1η3|+|λ2λ1η−λ2| 1η−1||λ2ξ−1|Γ(α−2) Z ξ 0 (ξ − s)α−3  f  s, y1(s) , Dδy1(s)    ds + |1−λ1η 3_|+|λ 1η−1| 6|λ₁η−1||λ2ξ−1|Γ(α−2) Z 1 0 (1 − s)α−3  f  s, y1(s) , Dδy1(s)    ds. Using the (H3), we obtain

|T1y (t) + R1y1(t)| ≤ N1 h_|λ 1η−1|+|λ1|ηα+1 |λ1η−1|Γ(α+1) i +N1  [(|λ2−λ2λ1η3|+|λ2λ1η−λ2|)ξα−2+|1−λ1η3|+|λ1η−1|] 6|λ₁η−1||λ2ξ−1|Γ(α−1)  . Consequently, |T1y (t) + R1y1(t)| ≤ N1M1. Thus, kT1(y) + R1(y1)k ≤ N1M1, (3.34) and |Dσ_T 1y (t) + DσR1y1(t)| ≤ _Γ(α−σ)1 Z t 0 (t − s)α−σ−1  f  s, y (s) , Dδ_{y (s)}  ds + |λ1| |λ1η−1|Γ(α)Γ(2−σ) Z η 0 (η − s)α−1  f  s, y (s) , Dδ_{y (s)}  ds +_|λ 1 1η−1|Γ(α)Γ(2−σ) Z 1 0 (1 − s)α−1  f  s, y (s) , Dδ_{y (s)}  ds +   |λ2−λ2λ1η3| 6|λ1η−1||λ2ξ−1|Γ(α−2)Γ(2−σ) + |λ2λ1η−λ2| |λ1η−1||λ2ξ−1|Γ(α−2)Γ(4−σ)   Z ξ 0 (ξ − s)α−3  f  s, y1(s) , Dδy1(s)    ds +   |1−λ1η3| 6|λ1η−1||λ2ξ−1|Γ(α−2)Γ(2−σ) + |λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−2)Γ(4−σ)   Z 1 0 (1 − s)α−3  f  s, y1(s) , Dδy1(s)    ds. By (H3),we have |Dσ_T 1y (t) + DσR1y1(t)| ≤ N1 h 1 Γ(α−σ+1)+ |λ1|η α₊₁ |λ1η−1|Γ(α+1)Γ(2−σ) i +N1  |λ2−λ2λ1η3|ξα−2+|1−λ1η3| 6|λ1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ) + |λ2λ1η−λ2|ξα−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)  . Consequently we obtain |Dσ_T 1y (t) + DσR1y1(t)| ≤ N1M2. Hence, kDσ_T 1(y) + DσR1(y1)k ≤ N1M2. (3.35)

Combining (3.34) and (3.35) yields

kT1(y) + R1(y1)kX ≤ N1(M1+M2). (3.36)

Analogously, we have

Hence, it follows from (3.36) and (3.37) that

kT (x, y) + R(x1, y1)kX×Y ≤ N1(M1+M2) +N2(M3+M4) < µ. (3.38)

Step2:We shall prove that R is continuous and compact.

[1∗]: The continuity of f and g implies that the operator R is continuous. [2∗]: Now, we prove that R maps bounded sets into bounded sets of X × Y. For (x, y) ∈ Bµand for each t ∈ J,we have:

|R1y (t)| ≤ |λ2−λ2λ1η 3_|t+|λ 2λ1η−λ2|t3 6|λ1η−1||λ2ξ−1|Γ(α−2) Z ξ 0 (ξ − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds +_6|λ|1−λ1η3|t+|λ1η−1|t3 1η−1||λ2ξ−1|Γ(α−2) Z 1 0 (1 − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds. Using the (H3), we obtain

|R1y (t)| ≤ N1[(|λ2−λ2λ1η3|+|λ2λ1η−λ2|)ξα−2+|1−λ1η3|+|λ1η−1|] 6|λ1η−1||λ2ξ−1|Γ(α−1) ≤ N1 (|λ2−λ2λ1η 3_|+|λ 2λ1η−λ2|)ξα−2+|1−λ1η3|+|λ1η−1| 6|λ1η−1||λ2ξ−1|Γ(α−1)  . Thus, |R1y (t)| ≤ N1L1, t ∈ J, Therefore, kR1(y)k ≤ N1L1. (3.39)

On the other hand,

|Dσ_R 1y (t)| ≤ _Γ(α−2)1   |λ2−λ2λ1η3|t1−σ 6|λ1η−1||λ2ξ−1|Γ(2−σ)+ |λ2λ1η−λ2|t3−σ |λ1η−1||λ2ξ−1|Γ(4−σ)   Z ξ 0 (ξ − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds +_Γ(α−2)1   |1−λ1η3|t1−σ 6Γ(2−σ)|λ1η−1||λ2ξ−1|+ |λ1η−1|t3−σ |λ1η−1||λ2ξ−1|Γ(4−σ)   Z 1 0 (1 − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds. By (H3), we have, |Dσ φ1y (t)| ≤ N1  |λ2−λ2λ1η3|ξα−2+|1−λ1η3| 6|λ1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ) + |λ2λ1η−λ2|ξα−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)  ≤ N1  |λ2−λ2λ1η3|ξα−2+|1−λ1η3| 6|λ1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ) + |λ2λ1η−λ2|ξα−2+|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)  . Consequently we obtain, |Dσ_R 1y (t)| ≤ N1L2, t ∈ J. Therefore, kDσ_R 1(y)k ≤ N1L2. (3.40)

Hence, from (3.39) and (3.40), we have

kR1(y)kX ≤ N1(L1+L2). (3.41)

Similarly, it can be shown that,

kR2(x)kY≤ N2(L3+L4). (3.42)

It follows from (3.41) and (3.42) that

Consequently

kR (x, y)k_X×Y <∞.

[3∗]: In the end we show that R is equicontinuous on J: Let t1, t2∈ J, such that t1<t2and (x, y) ∈ Bµ, .Then, we have:

|R1y (t2) −R1y (t1)| ≤ |λ2−λ2λ1η3|(t2−t1)+|λ2λ1η−λ2|(t32−t31) 6|λ1η−1||λ2ξ−1|Γ(α−2) Z ξ 0 (ξ − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds +|1−λ1η3|(t1−t2)+|λ1η−1|(t31−t32) 6|λ1η−1||λ2ξ−1|Γ(α−2) Z 1 0 (1 − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds. Using the (H3),we obtain

|R1y (t2) −R1y (t1)| ≤ N1|λ2−λ2λ1η3|ξα−2 6|λ1η−1||λ2ξ−1|Γ(α−1)(t2− t1) (3.44) +_6|λ N1|1−λ1η3| 1η−1||λ2ξ−1|Γ(α−1)(t1− t2) + N1|λ2λ1η−λ2|ξα−2 6|λ1η−1||λ2ξ−1|Γ(α−1)  t32− t31  + N1|λ1η−1| 6|λ1η−1||λ2ξ−1|Γ(α−1)  t3₁− t3₂, and |Dσ_R 1y (t2) −DσR1y (t1)| ≤   |λ2−λ2λ1η3|(t1−σ₂ −t1−σ₁ ) 6|λ1η−1||λ2ξ−1|Γ(α−2)Γ(2−σ) + |λ2λ1η−λ2|(t 3−σ 2 −t3−σ1 ) |λ1η−1||λ2ξ−1|Γ(α−2)Γ(4−σ)   Z ξ 0 (ξ − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds +   |1−λ1η3|(t1−σ₁ −t1−σ₂ ) 6|λ1η−1||λ2ξ−1|Γ(α−2)Γ(2−σ) + |λ1η−1|(t 3−σ 1 −t3−σ2 ) |λ1η−1||λ2ξ−1|Γ(α−2)Γ(4−σ)   Z 1 0 (1 − s)α−3  f  s, y (s) , Dδ_{y (s)}  ds, by (H3), we have: |Dσ_R 1y (t2) −DσR1y (t1)| ≤ N1|λ2−λ2λ1η3|ξα−2 6|λ1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ)  t1−σ₂ − t1−σ₁  (3.45) +_6|λ N1|1−λ1η3| 1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ)  t1−σ₁ − t1−σ₂  + N1|λ2λ1η−λ2|ξα−2 |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)  t3−σ₂ − t3−σ₁  + N1|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)  t3−σ₁ − t3−σ₂ .

Hence,by (3.44) and (3.45), we obtain

kR1y (t2) −R1y (t1)kX≤ N1|λ2−λ2λ1η3|ξα−2 6|λ1η−1||λ2ξ−1|Γ(α−1)(t2− t1) + N1|1−λ1η3| 6|λ1η−1||λ2ξ−1|Γ(α−1)(t1− t2) + N1|λ2λ1η−λ2|ξα−2 6|λ1η−1||λ2ξ−1|Γ(α−1)  t3₂− t3₁+ N1|λ1η−1| 6|λ1η−1||λ2ξ−1|Γ(α−1)  t3₁− t3₂ +_6|λ N1|λ2−λ2λ1η3|ξα−2 1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ)  t1−σ₂ − t1−σ₁  (3.46) + N1|1−λ1η 3_| 6|λ1η−1||λ2ξ−1|Γ(α−1)Γ(2−σ)  t1−σ₁ − t1−σ₂  + N1|λ2λ1η−λ2|ξα−2 |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)  t3−σ₂ − t3−σ₁  + N1|λ1η−1| |λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)  t3−σ₁ − t3−σ₂ .

Analogously, we can obtain kR2x (t2) −R2x (t1)kY≤ N2|λ2−λ2λ1η3|ξβ−2 6|λ1η−1||λ2ξ−1|Γ(β−1)(t2− t1) + N2|1−λ1η3| 6|λ1η−1||λ2ξ−1|Γ(β−1)(t1− t2) + N2|λ2λ1η−λ2|ξβ−2 6|λ1η−1||λ2ξ−1|Γ(β−1)  t3₂− t3₁+ N2|λ1η−1| 6|λ1η−1||λ2ξ−1|Γ(β−1)  t3₁− t3₂ +_6|λ N2|λ2−λ2λ1η3|ξβ−2 1η−1||λ2ξ−1|Γ(β−1)Γ(2−δ)  t1−δ₂ − t1−δ₁  (3.47) + N2|1−λ1η 3_| 6|λ1η−1||λ2ξ−1|Γ(β−1)Γ(2−δ)  t1−δ₁ − t1−δ₂  + N2|λ2λ1η−λ2|ξβ−2 |λ1η−1||λ2ξ−1|Γ(β−1)Γ(4−δ)  t3−δ₂ − t3−δ₁  + N2|λ1η−1| |λ1η−1||λ2ξ−1|Γ(β−1)Γ(4−δ)  t3−δ₁ − t3−δ₂ .

Thanks to (3.46) and (3.47), can state that kφ (x, y) (t2) − φ (x, y) (t1)k → 0 as t1 → t2. Then, as a

conse-quence of steps ([1∗], [2∗], [3∗]); we can conclude that R is continuous and compact.

Step3:Now, we prove that T is contraction mapping. Let (x, y) , (x1, y1) ∈X × Y. Then, for each t ∈ J, we have

|T1y (t) − T1y1(t)| ≤ _Γ(α)1 Z t 0 (t − s)α−1     f s, y (s) , Dδ_{y (s)}
− f s, y1(s) , Dδy1(s)
    ds + λ1t (λ1η−1)Γ(α) Z η 0 (η − s)α−1     f s, y (s) , Dδ_{y (s)}
− f s, y1(s) , Dδy1(s)
   ds +_(λ t 1η−1)Γ(α) Z 1 0 (1 − s)α−1     f s, y (s) , Dδ_{y (s)}
− f s, y1(s) , Dδy1(s)
   ds.

Thanks to (H1), we can write

|T1y (t) − T1y1(t)| ≤ _Γ(α+1)k1  ky − y1k + D δ_{y − D}δ_y 1  + k1(|λ1|ηα+1) |λ1η−1|Γ(α+1)  ky − y1k + D δ_{y − D}δ_y 1  . Consequently, kT1(y) − T1(y1)k ≤ k1[|λ1η−1|+|λ1|η α_+1] |λ1η−1|Γ(α+1)  ky − y1k + D δ_{y − D}δ_y 1  , (3.48) and |Dσ_T 1y (t) − DσT1y1(t)| ≤ _Γ(α−σ)1 Z t 0 (t − s)α−σ−1     f s, y (s) , Dδ_{y (s)}
− f s, y1(s) , Dδy1(s)
   ds + |λ1|t1−σ |λ1η−1|Γ(α)Γ(2−σ) Z η 0 (η − s)α−1     f s, y (s) , Dδ_{y (s)}
− f s, y1(s) , Dδy1(s)
   ds + t1−σ |λ1η−1|Γ(α)Γ(2−σ) Z 1 0 (1 − s)α−1     f s, y (s) , Dδ_{y (s)}
− f s, y1(s) , Dδy1(s)
   ds. By (H1) ,yields |Dσ_T 1y (t) − DσT1y1(t)| ≤ _{Γ(α−σ+1)}k1  ky − y1k + D δ_{y − D}δ_y 1  + k1|λ1|ηα |λ1η−1|Γ(α+1)Γ(2−σ)  ky − y1k + D δ_{y − D}δ_y 1  + k1 |λ1η−1|Γ(α+1)Γ(2−σ)  ky − y1k + D δ_{y − D}δ_y 1  .

Hence, kDσ_T 1(y) − DσT1(y1)k ≤ k1 " ₁ Γ(α−σ+1) + |λ1|ηα+1 |λ1η−1|Γ(α+1)Γ(2−σ) #  ky − y1k + D δ_{y − D}δ_y 1  . (3.49)

By (3.48) and (3.49) we can write

kT1(y) − T1(y1)kX ≤ k1   |λ1η−1|+|λ1|ηα+1 |λ1η−1|Γ(α+1) + 1 Γ(α−σ+1) + |λ1|ηα+1 |λ1η−1|Γ(α+1)Γ(2−σ)    ky − y1k + D δ_{y − D}δ_y 1  . Thus, kT1(y) − T1(y1)kX≤ k1θ1  ky − y1k + D δ_{y − D}δ_y 1  . (3.50)

Analogously, we can get

kT2(x) − T2(x1)kY≤ k2θ2(kx − x1k + kDσx − Dσx1k) . (3.51)

It follows from (3.50) and (3.51) that

kT (x, y) − T (x1, y1)kX×Y ≤ [k1θ1+k2θ2] k(x − x1, y − y1)kX×Y
.

Using the condition (3.25) we conclude that T is a contraction mapping.

As a consequence of Krasnoselskii’s fixed point theorem we deduce that φ has a fixed point which is a solution of (1.1) .

4

Examples

In this section we give an example to illustrate the usefulness of our main results.

Example 4.1. Let us consider the following system of fractional boundary value problem:

D72x (t) + √ πe−πt2cos(πt)  y(t)+D52y(t)  (5√π+7et₎  1+y(t)+D52y(t)  +ln 1 + t2
= 0, t ∈ J, D113_{y (t) +} √ πe−πt2cos(πt)  x(t)+D94x(t)  (5√π+7et)  1+x(t)+D94x(t)  +ln 1 + t2
= 0, t ∈ J, x (0) = 0, x (1) −3₄x1₃=0, y (0) = 0, y (1) −3₄y1₃=0, x00(0) = 0, x00(1) −4₅x00 2₃
= 0, y00(0) = 0, y00(1) −4₅y00 2₃
= 0, Set f (t, x, y) = g (t, x, y) = √

πe−πt|cos (πt)| (|x| + |y|)

5√π +7et
2(1 + |x| + |y|)

+ln1 + t2, t ∈ [0, 1] , x, y ∈ [0,∞) , For t ∈ J = [0, 1] and x1, y1, x2, y2∈ [0,∞) ,we have:

| f (t, x, y) − f (t, x1, y1)| = √ πe−πt|cos (πt)| 5√π +7et
2     x + y (1 + |x| + |y|) − x1+y1 (1 + |x1| + |y1|)     ≤ √ πe−πt|cos (πt)| (|x − x1| + |y − y1|) 5√π +7et
2(1 + |x| + |y|) (1 + |x1| + |y1|) ≤ √ πe−πt|cos (πt)| (|x − x1| + |y − y1|) 5√π +7et
2 ≤ √ π 5√π +7
2 (|x − x1| + |y − y1|) .

Hence the condition (H1) holds with k1=k2= √

π

(5√π+7)2

For α = 7₂, β = 11₃, σ = 9₄, δ = 5₂and λ1= 34, λ2= 45 = η =13, ξ =23, we have: M1=1, 089, M2=3, 503, M3=0, 909, M4=3, 089, and, k1(M1+M2) +k2(M3+M4) =0, 0605075. Therefore, k1(M1+M2) +k2(M3+M4) <1.

Hence, the condition (3.14) of Theorem (3.1) is satisfied. Therefore the boundary value problem (1.1) has a unique solution. So, a simple computation shows that

θ1=1, 283, θ2=1, 058,

and, we have

k1θ1+k2θ2=0, 0164904.

Using the condition (3.25), we get,

k1θ1+k2θ2<1.

Therefore it follow from Theorem (3.2) that the boundary value problem (1.1) has a solution.

Example 4.2. Consider the following system of fractional Bounded value problem:

                       D72_{x (t) +}    D 7 3y(t)     5π(√π+2et₎+ e−t2_|y(t)| 5π(√πet+2)2(1+|y(t)|) =0, t ∈ J, D113y (t) + |x(t)| 14√π(1+|x(t)|)+ |cos(πt)|     D52x(t)     7√π(t+1)2 =0, t ∈ J, x (0) = 0, x (1) − 2₃x1₅=0, y (0) = 0, y (1) −2₃y1₅ x00(0) = 0, x00(1) −1₂x001₄=0, y00(0) = 0, y00(1) −1₂y001₄=0. For this example, we have

f (t, x, y) = |x| 5π √π +2et
+ e−t2|y| 5π √πet+2
2(1 + |y|) , t ∈ [0, 1] , x, y ∈ [0,∞) , g (t, x, y) = |x| 14√π (1 + |x|) + |cos (πt)| |y| 7√π (t +1)2 , t ∈ [0, 1] , x, y ∈ [0,∞) .

For t ∈ J = [0, 1] and x, y, x1, y1∈ [0,∞) . Then we have:

| f (t, x, y) − f (t, x1, y1)| = e−t2|x − x1| 5 √πet+2
2(1 + |x|) (1 + |x1|) + |y − y1| 5π √π +2et
≤ e −t2 5π √πet+2
2 |x − x1| + 1 5π √π +2et
|y − y1 | ≤ 1 5π √π +2
2 (|x − x1| + |y − y1|) , and |g (t, x, y) − g (t, x1, y1)| = |x − x1| 14√π (1 + |x|) (1 + |x1|) +|cos (πt)| |y − y1| 7√π (t +1)2 ≤ 1 14√π|x − x1 | + |cos (πt)| 7√π (t +1)2 |y − y1| ≤ 1 14√π(|x − x1| + |y − y1|) .

So,we can take: k1= 1 5π(√π+2)2

and k2= ₁₄1√_π.

It, follows then that

M1=0, 51254, M2=1, 9531, M3=0, 43113, M4=1, 32719, and k1(M1+M2) +k2(M3+M4) = 2, 46564 5π √π +2
2 +1, 75832 14√π =0, 081914 < 1. Hence by Theorem (3.1) the boundary value problem (1.1) has a unique solution.

Now, using the condition (3.25), we get:

k1θ1+k2θ2= 1, 15722 5π √π +2
2 +0, 96759 14√π =0.044183. Therefore, k1θ1+k2θ2<1.

By Theorem (3.2) the boundary value problem (1.1) has a solution.

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Received: July 12, 2014; Accepted: October 15, 2014

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