Some results for a four-point boundary value problems for coupled system involving Caputo derivative

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Some Results for a Four-Point Boundary Value Problems for a Coupled

System Involving Caputo Derivatives

M. Houas

a,∗

, M. Benbachir

a

and Z. Dahmani

b

a_{Faculty of Sciences and Technology Khemis Miliana University, Ain Defla, Algeria.}

b_{LPAM, Faculty SEI, UMAB Mostaganem, Algeria.}

Abstract

Motivated by the problem (1.1) in [5], in this paper, we prove the existence and uniqueness of solutions for

the following system of fractional differential equations with four point boundary conditions:

       

      

Dα_x₍_t_{) +}_{f t,}_y₍_t₎_,_Dδ_y₍_t₎

=0,t∈J,

Dβ_y₍_t_{) +}_g₍_t,_x₍_t₎_,_Dσ_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, andDα,Dβ,Dδ andDσ, are the

Caputo fractional derivatives,J = [0, 1],λ1,λ2are real constants withλ1η 6= 1,λ2ξ 6= 1 and f,gcontinuous

functions on[0, 1]×R2_.

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

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, 7, 10, 12, 13, 15, 17, 18] and references therein. There has been a significant progress in the investigation of

these equations in recent years, see [6, 8, 9, 15, 16, 27]. More recently, some basic theory for the initial

bound-ary value problems of fractional differential equations has been discussed in [1, 14, 15]. Recently, existence

∗

Corresponding author.

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and uniqueness of solutions to boundary value problems for fractional differential equations had attracted

the attention of many authors, see for example, [4, 6, 8, 9, 15, 16, 19, 27] 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, 11, 20, 21, 24, 26]. 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 [20, 23, 24, 25] and

references cited therein. For example in [2, 5, 21, 26] 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, 5, 21, 22, 26], 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.

Motivated by the problem (1.1) in [5], this paper deals with the existence of solution for the following

fractional differential problem:

       

      

Dα_x₍_t_{) +}_{f t,}_y₍_t₎_,_Dδ_y₍_t₎

=0,t∈J,

Dβ_y₍_t_{) +}_g₍_t,_x₍_t₎_,_Dσ_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, andDα,Dβ,Dδ and Dσ, 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−1f(τ)dτ,α>0, (2.2)

J0f(t) = f(t), whereΓ(α):=R₀∞e−uuα−1du.

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

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For more details about fractional calculus, we refer the reader to [15, 18].

We will consider the following 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 norms:

kxk_X =kxk+kDσ_x_k_, _k_x_k₌_sup

t∈J

|x(t)|,kDσ_x_k₌_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₎ .

We know that(X,k.k_X)and (Y,k.k_Y), is a Banach space. The product space X×Y,k(x,y)k_X_×_Y

is also a

Banach space, with normk(x,y)k_X_×_Y=kxk_X+kyk_Y.

We recall the following important lemmas [13]:

Lemma 2.1. Forα>0,the general solution of the fractional differential equation Dαx(t) =0is given by

x(t) =c0+c1t+c2t2+...+cn−1tn−1, (2.4)

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

Lemma 2.2. Letα>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 prove the following 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 conditions

x(0) =0, x(1)−λ1x(η) =0, (2.7)

x00(0) =0, x00(1)−λ2x00(ξ) =0,

is given by:

x(t) =− 1

Γ(α)

Z t

0

(t−s)α−1g(s)ds (2.8)

+ λ1t (λ1η−1)Γ(α)

Z η 0

(η−s)α−1g(s)ds

− t

(λ1η−1)Γ(α)

Z 1 0

(1−s)α−1g(s)ds

+ λ2−λ2λ1η 3

t+ (λ2λ1η−λ2)t3

6(λ1η−1) (λ2ξ−1)Γ(α−2)

Z ξ 0

(ξ−s)α−3g(s)ds

− 1−λ1η

3

t+ (λ1η−1)t3

6(λ1η−1) (λ2ξ−1)Γ(α−2)

Z 1 0

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Proof. We use the same technics as in [5]. Forci∈R,i=0, 1, 2, 3, and by Lemmas 2.1, 2.2, we have

x(t) =− 1

Γ(α)

Z t

0

(t−s)α−1g(s)ds−c0−c1t−c2t2−c3t3 (2.9)

Using(2.7), we getc0=c2=0, and

c1=−

λ1 (λ1η−1)Γ(α)

Z η 0

(η−s)α−1g(s)ds (2.10)

+ 1

(λ1η−1)Γ(α)

Z 1 0

(1−s)α−1g(s)ds

− λ2 1−λ1η

3

6(λ1η−1) (λ2ξ−1)Γ(α−2)

Z ξ 0

(ξ−s)α−3g(s)ds

+ (1−λ1η)

6(λ1η−1) (λ2ξ−1)Γ(α−2)

Z 1 0

(1−s)α−3g(s)ds

and

c3=−

λ2

6(λ2ξ−1)Γ(α−2)

Z ξ 0

(ξ−s)α−3g(s)ds (2.11)

+ 1

6(λ2ξ−1)Γ(α−2)

Z 1 0

(1−s)α−3g(s)ds

Substituting the value ofc1andc3in(2.9), we obtain the desired quantity in Lemma.

3

Main Results

Let us set:

M1= |λ1η−1|+|λ1|η α₊₁

|λ1η−1|Γ(α+1) (3.12) +(|λ2−λ2λ1η3|+|λ_6|λ2λ1η−λ2|)ξα−2+|1−λ1η3|+|λ1η−1|

1η−1||λ2ξ−1|Γ(α−1) ,

M2= _Γ_(α−σ+1)1 + |λ1|η α₊₁ |λ₁η−1|Γ(α+1)Γ(2−σ) + |λ2−λ2λ1η3|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),

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|ηβ+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

−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),

L3=

(|λ2−λ2λ1η3|+|λ2λ1η−λ2|)ξβ−2+|1−λ1η3|+|λ1η−1|

6|λ₁η−1||λ₂ξ−1|Γ(β−1) ,

L4= |λ2

−λ2λ1η3|ξβ−2+|1−λ1η3| 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).

Let us also consider the following hypotheses:

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|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→Rare continuous

(H3): There exists positive constantsN1andN2such that

|f(t,x,y)| ≤ N1, |g(t,x,y)| ≤N2for eacht∈ Jand allx,y∈R.

We prove the following theorem:

Theorem 3.1. Assume that(H1)holds.

If

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

then the problem(1.1)has a unique solution.

Proof. The proof is similarly to that of Theorem 3.1 in [5] by takingk1=ω1+ω2andk2=v1+v2.

Now, we prove the following result:

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

k1θ1+k2θ2<1, (3.15)

where

θ1= |λ1η−1|+|λ1|η α₊₁ |λ1η−1|Γ(α+1) +

1 Γ(α−σ+1) +

|λ₁|ηα₊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|ηβ+1

|λ1η−1|Γ(β+1)Γ(2−δ).

If there existsµ∈Rsuch that

N1(M1+M2) +N2(M3+M4)≤µ, (3.16)

then, the problem(1.1)has at least a solution.

Proof. We shall use Krasnselskii’s fixed point theorem to prove thatφhas at least a fixed point onX×Y.

Suppose that(3.16)holds and let us take

φ(x,y) (t):=T(x,y) (t) +R(x,y) (t), (3.17)

where

T(x,y) (t):= (T1y(t),T2x(t)), (3.18)

T1y(t) =−_Γ(α)1 Z t

0

(t−s)α−1fs,y(s),Dδ_y₍_s₎_ds _(3.19)

+ λ1t

(λ₁η−1)Γ(α) Z η

0

(η−s)α−1f

s,y(s),Dδ_y₍_s₎_ds

− t

(λ1η−1)Γ(α) Z 1

0

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T2x(t) =−_Γ(1β) Z t

0

(t−s)β−1g(s,x(s),Dσ_x₍_s₎₎_ds _(3.20)

+ λ1t

(λ1η−1)Γ(β) Z η

0

(η−s)α−1g(s,x(s),Dσx(s))ds

−_(λ t

1η−1)Γ(β)

Z 1 0

(1−s)α−1g(s,x(s),Dσ_x₍_s₎₎_ds,

and

R(x,y) (t):= (R1y(t),R2x(t)), (3.21)

where,

R1y(t) =

(λ2−λ2λ1η3)t+(λ₂λ1η−λ2)t3

6(λ1η−1)(λ2ξ−1)Γ(α−2)

Z ξ 0

(ξ−s)α−3f

− (1−λ1η3)t+(λ1η−1)t3

6(λ1η−1)(λ2ξ−1)Γ(α−2)

Z 1 0

(1−s)α−3fs,y(s),Dδ_y₍_s₎_ds, _(3.22)

R2x(t) = (λ2

−λ2λ1η3)t+(λ2λ1η−λ2)t3

6(λ1η−1)(λ2ξ−1)Γ(β−2) Z ξ

0

(ξ−s)α−3g(s,x(s),Dσx(s))ds

−_6(λ(1−λ1η3)t+(λ1η−1)t3 1η−1)(λ2ξ−1)Γ(β−2)

Z 1 0

(1−s)α−3g(s,x(s),Dσ_x₍_s₎₎_ds. _(3.23)

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 eacht∈Jwe have:

|T1y(t) +R1y1(t)|=| −_Γ(1α) Z t

0

(t−s)α−1fs,y(s),Dδ_y₍_s₎_ds

+ λ1t

(λ1η−1)Γ(α) Z η

0

(η−s)α−1f

−_(λ t

1η−1)Γ(α)

Z 1 0

(1−s)α−1fs,y(s),Dδ_y₍_s₎_ds

+(λ_6(λ2−λ2λ1η3)t+(λ2λ1η−λ2)t3

1η−1)(λ2ξ−1)Γ(α−2)

Z ξ 0

(ξ−s)α−3f

− (1−λ1η3)t+(λ1η−1)t3

6(λ1η−1)(λ2ξ−1)Γ(β−2)

Z 1 0

(1−s)α−3fs,y(s),Dδ_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

+_|λ 1

1η−1|Γ(α) Z 1

0

(1−s)α−1 f

+|λ2−λ2λ1η

3_|_+|λ 2λ1η−λ2|

6|λ₁η−1||λ₂ξ−1|Γ(α−2) Z ξ

0

(ξ−s)α−3

f

s,y1(s),Dδy1(s)

ds

+_6|λ|1−λ1η3|+|λ1η−1|

1η−1||λ2ξ−1|Γ(α−2)

Z 1 0

(1−s)α−3 f

s,y1(s),Dδy1(s)

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Using(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, (3.24)

and

|Dσ_T

1y(t) +DσR1y1(t)| ≤ _Γ(α−σ1 ) Z t

0

(t−s)α−σ−1 f

+ |λ1|

|λ1η−1|Γ(α)Γ(2−σ) Z η

0

(η−s)α−1

f

+_|λ 1

1η−1|Γ(α)Γ(2−σ)

Z 1 0

(1−s)α−1 f

+ 



|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|ηα+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. (3.25)

Combining(3.24)and(3.25), yields

kT1(y) +R1(y1)kX ≤N1(M1+M2). (3.26)

Analogously, we have

kT2(x) +R2(x1)kY≤N2(M3+M4). (3.27)

Hence, it follows from(3.26)and(3.27)that

kT(x,y) +R(x1,y1)kX×Y≤N1(M1+M2) +N2(M3+M4)<µ. (3.28)

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[1∗]: The continuity of f andgimplies that the operatorRis continuous.

[2∗]: Now, we prove thatRmaps bounded sets into bounded sets ofX×Y.

For(x,y)∈Bµand for eacht∈ J, we have:

|R1y(t)| ≤ |

λ2−λ2λ1η3|t+|λ₂λ1η−λ2|t3

6|λ₁η−1||λ₂ξ−1|Γ(α−2) Z ξ

0

(ξ−s)α−3

f

+_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.

|R1y(t)| ≤

N1[(|λ2−λ2λ1η3|+|λ2λ1η−λ2|)ξα−2+|1−λ1η3|+|λ1η−1|]

6|λ1η−1||λ2ξ−1|Γ(α−1) ≤N1

(|λ2−λ2λ1η3|+|λ2λ1η−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. (3.29)

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

+_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

By(H3), we have

|Dσ_φ

1y(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. (3.30)

Hence, from(3.29)and(3.30), we have

kR1(y)kX≤N1(L1+L2). (3.31)

Similarly, it can be shown that

kR2(x)kY≤N2(L3+L4). (3.32)

It follows from(3.31)and(3.32)that

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Consequently,

kR(x,y)k_X_×_Y<∞.

[3∗]: We show thatRis equi-continuous:

Lett1,t2∈ J, such thatt1<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

+|1−λ1η

3_|₍_t

1−t2)+|λ1η−1|(t31−t32)

6|λ₁η−1||λ2ξ−1|Γ(α−2)

Z 1 0

(1−s)α−3 f

|R1y(t2)−R1y(t1)| ≤

N1|λ2−λ2λ1η3|ξα−2

6|λ₁η−1||λ₂ξ−1|Γ(α−1)(t2−t1) (3.34) +_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−σ

2 −t1−1 σ)

6|λ₁η−1||λ₂ξ−1|Γ(α−2)Γ(2−σ) + |λ2λ1η−λ2|(t

3−σ

2 −t3

−σ

1 )

|λ₁η−1||λ₂ξ−1|Γ(α−2)Γ(4−σ) 

 Z ξ

0

(ξ−s)α−3

f

+ 



|1−λ₁η3|(t1−σ

1 −t1

−σ

2 )

6|λ₁η−1||λ2ξ−1|Γ(α−2)Γ(2−σ)

+ |λ1η−1|(t31−σ−t3 −σ

2 )

|λ1η−1||λ2ξ−1|Γ(α−2)Γ(4−σ)



 Z 1

0

(1−s)α−3 f

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 (3.35)

+ 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−σ 2 −t

3−σ 1

+ N1|λ1η−1|

|λ1η−1||λ2ξ−1|Γ(α−1)Γ(4−σ)

t3−σ₁ −t3−σ₂ .

Hence, by(3.34)and(3.35), 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

+ N1|λ2−λ2λ1η

3_|_ξα−2

6|λ₁η−1||λ₂ξ−1|Γ(α−1)Γ(2−σ)

t1−σ₂ −t1−σ₁ (3.36)

+ 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)

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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

+ N2|λ2−λ2λ1η3|ξβ

−2

6|λ1η−1||λ2ξ−1|Γ(β−1)Γ(2−δ)

t1−δ₂ −t1−δ₁ (3.37)

+_6|λ N2|1−λ1η3|

1η−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(3.36) and (3.37), we can state that kφ(x,y) (t2)−φ(x,y) (t1)k → 0 as t1 → t2. Then, as a

consequence of steps([1∗],[2∗],[3∗]), we can conclude thatRis continuous and compact.

Step3:Now, we prove thatTis contractive.

Let(x,y),(x1,y1)∈X×Y. Then, for eacht∈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

+_(λ t

1η−1)Γ(α)

Z 1 0

(1−s)α−1

−f s,y1(s),Dδy1(s) ds.

Thanks to(H1), we can write

|T1y(t)−T1y1(t)| ≤ _Γ(αk+1)1

ky−y1k+ D

δ_y₋_Dδ_y 1

+ k1(|λ1|ηα+1) |λ1η−1|Γ(α+1)

ky−y1k+ D

δ_y₋_Dδ_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₋_Dδ_y 1 , (3.38) and

|Dσ_T

1y(t)−DσT1y1(t)| ≤ _Γ(α−σ)1 Z t

0

(t−s)α−σ−1

+ |λ1|t1−σ

|λ1η−1|Γ(α)Γ(2−σ) Z η

0

(η−s)α−1

+_|λ t1−σ

1η−1|Γ(α)Γ(2−σ)

Z 1 0

(1−s)α−1

(11)

By(H1), yields

|Dσ_T

1y(t)−DσT1y1(t)| ≤ _Γ(α−σ+1)k1

ky−y1k+ D

δ_y₋_Dδ_y 1

+ k1|λ1|ηα

|λ₁η−1|Γ(α+1)Γ(2−σ)

ky−y1k+ D

δ_y₋_Dδ_y 1

+ k1

|λ₁η−1|Γ(α+1)Γ(2−σ)

ky−y1k+ D

δ_y₋_Dδ_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₋_Dδ_y 1 . (3.39)

By(3.38)and(3.39), 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₋_Dδ_y 1 . Thus,

kT1(y)−T1(y1)kX≤k1θ1

ky−y1k+ D

δ_y₋_Dδ_y 1 . (3.40)

Analogously, we can get

kT2(x)−T2(x1)kY≤k2θ2(kx−x1k+kDσx−Dσx1k). (3.41)

It follows from(3.40)and(3.41)that

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

.

Using(3.15), we conclude thatTis a contraction mapping.

As a consequence of Krasnoselskii’s fixed point theorem we deduce that φhas a fixed point which is a

solution of(1.1).

4

Examples

In this section we give some examples to illustrate our main results.

Example 4.1. Let us consider the following system:

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+7et2(1+|x|+|y|)

(12)

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+7et2

x+y

(1+|x|+|y|)−

x1+y1 (1+|x1|+|y1|)

≤ √

πe−πt|cos(πt)|(|x−x1|+|y−y1|)

5√π+7et2(1+|x|+|y|) (1+|x1|+|y1|)

≤ √

πe−πt|cos(πt)|(|x−x1|+|y−y1|)

5√π+7et2

≤ √

π

5√π+72

(|x−x1|+|y−y1|).

The condition(H1)holds with k1=k2= √

π (5√π+7)2

.

Forα= 7₂,β=11₃,σ= 9₄,δ= 5₂andλ1= 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, by Theorem3.1, the problem(1.1)has a unique solution.

Example 4.2. Consider the following system:

           

          

D72x(t) +

D73y(t) 5π(√π+2et)+

e−t2_|_y₍_t_)|

5π(√πet+2)2(1+|y(t)|)

=0,t∈J,

D113 _y(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+22(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,∞),we have:

|f(t,x,y)−f(t,x1,y1)|=

e−t2|x−x1|

5 √πet+22(1+|x|) (1+|x1|)

+ |y−y1|

5π

√

π+2et

≤ e

−t2

5π

√

πet+22

|x−x1|+ 1

5π

√

π+2et

|y−y1|

≤ 1

5π

√

π+22

(13)

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|).

By Theorem3.2,the problem(1.1)has at leat one solution.

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

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