Existence and uniqueness results for a nonlinear differential equations of arbitrary order

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ISSN: 2008-6822 (electronic)

http://dx.doi.org/10.22075/ijnaa.2015.256

Existence and uniqueness results for a nonlinear

differential equations of arbitrary order

Mohamed Houas∗, Maamar Benbachir

Faculty of Sciences and Technology, Khemis-Milian University, Ain Defla, Algeria.

(Communicated by R. Memarbashi)

Abstract

This paper studies a fractional boundary value problem of nonlinear differential equations of arbitrary orders. New existence and uniqueness results are established using Banach contraction principle. Other existence results are obtained using Schaefer and Krasnoselskii fixed point theorems. In order to clarify our results, some illustrative examples are also presented.

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

2010 MSC: Primary 34A08; Secondary 47H10.

1. Introduction

Boundary value problems for fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of a complex medium, polymer rheology In consequence, fractional differential equations have been of great interest, see [3, 4, 5, 6, 7, 8, 10, 15, 16, 19, 20, 24] and refer-ences therein. Recently, there has been a significant progress in the investigation of these equations, (see [1, 2, 9, 11, 12, 13, 19, 24]). More recently, some basic theory for the initial boundary value problems of fractional differential equations has been discussed in [1, 4, 13, 17, 19, 23]. Moreover, existence and uniqueness of solutions to boundary value problems for fractional differential equations had attracted the attention of many authors, see for example,[1, 2, 3, 9, 11, 12, 19, 21, 25] and the references therein.

In [9, 13, 21, 22], the existence and uniqueness of solutions was investigated for a nonlinear frac-tional differential equations with integral boundary conditions by using Schauder and Krasnoselskii’s fixed point theorem.

∗_{Corresponding author}

Email addresses: [email protected] (Mohamed Houas),[email protected](Maamar Benbachir)

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This paper deals with the existence and uniqueness of solutions for the following problem:

Dα0_x₍_t_{) =} _f₍_{t, x}₍_t₎_{, D}α1_x₍_t₎_{, D}α2_x₍_t₎_{, ..., D}αn−1_x₍_t₎₎_{, t} _∈_[0_,_1]_,

x(0) =x∗, x0(0) =x00(0) =...=x(n−2)_{(0) = 0}_{, J}β_x_{(1) =}_λJβ_x₍_η₎_, (1.1)

where Dαi_{, i} _{= 0}_,₁_,₂_{, ..., n}₋₁_, _{denote the Caputo fractional derivatives, with} _n₋₁ _{< α}

n−1 < ... < α1 < α0 < n and n ∈ N∗, n 6= 1, J = [0,1], λ 6= 0 is real constant, x∗ ∈ _R,0< η < 1 are real numbers and f is a function which will be specified later.

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 the 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 orderα ≥0, for a continuous function f on [0,∞[ is defined as:

Jαf(t) = 1 Γ (α)

Z t

0

(t−τ)α−1f(τ)dτ, α >0, (2.1)

J0f(t) = f(t),

where Γ (α) :=R₀∞e−u_uα−1_du.

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

Dαf(t) = 1 Γ (n−α)

Z t

0

(t−τ)n−α−1f(n)(τ)dτ, n−1< α, n∈N∗. (2.2)

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

Let us now introduce the Banach space

X :={x:x∈C([0,1],_R) ;Dα1_{x, D}α2_{x, ..., D}αn−1_x_∈_C_([0_,_1]_,

R)},

endowed with the norm

kxk_X =kxk+kDα1_xk₊_kDα2_xk₊_...₊_kDαn−1_xk_;

kxk= sup

t∈J

|x(t)|, kDα1_xk_{= sup}

t∈J

|Dα1_x₍_t₎_|_, kDα2_xk_{= sup}

t∈J

|Dα2_x₍_t₎_|_{, ...,} _kDαn−1_xk_{= sup}

t∈J

|Dαn−1_x₍_t₎_|_.

We give the following lemmas [14, 15]:

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Lemma 2.4. Let s > r >0, f ∈L1([a, b]). Then DrIsf(t) =Is−rf(t), t∈[a, b].

To study the problem (1), we need the following two lemmas [14]:

Lemma 2.5. 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, (2.3)

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

Lemma 2.6. Let α >0.Then

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

for some ci ∈R, i= 0,1,2, ..., n−1, n= [α] + 1. We need also the following auxiliary result:

Lemma 2.7. Let g ∈C([0,1]), the solution of the equation

Dα0_x₍_t_{) =} _g₍_t₎_{, t}_∈_{J, n}₋₁_{< α}

0 < n, (2.5)

subject to the boundary condition

x(0) =x∗, x0(0) =x00(0) =...=x(n−2)(0) = 0,

and

Jβx(1) =λJβx(η),

is given by:

x(t) = 1 Γ (α0)

Z t

0

(t−s)α0−1

g(s)ds+x∗ (2.6)

− Γ (β+n)t

n−1

(1−ληβ+n−1_{) Γ (}_n_{) Γ (}_α 0 +β)

Z 1

0

(1−s)β+α0−1_g₍_s₎_ds

+ λΓ (β+n)t

n−1

(1−ληβ+n−1_{) Γ (}_n_{) Γ (}_α 0+β)

Z η

0

(η−s)β+α0−1_g₍_s₎_ds

+ x

∗ _ληβ₋₁

Γ (β+n)tn−1

(1−ληβ+n−1_{) Γ (}_n_{) Γ (}_β_{+ 1)}.

Proof . By lemmas 2.5 and 2.6, the general solution of (6) is given by

x(t) = 1 Γ (α0)

Z t

0

(t−s)α0−1_g₍_s₎_ds₋_c

0−c1t−c2t2 −....−cn−1tn−1. (2.7)

By x(0) = x∗, and x0(0) = .... = x(n−2)(0) = 0, we can obtain c0 = −x∗, and c1 = c2 = ....= cn−2 = 0.

Thanks to lemma 2.3, we get

Jβx(t) = 1 Γ (β+α0)

Z t

0

(t−s)β+α0−1_g₍_s₎_ds₊ x ∗_tβ

Γ (β+ 1) −cn−1

Γ (n) Γ (β+n)t

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Using the boundary condition

Jβx(1) =λJβx(η),

we get

cn−1 =

Γ (β+n)

(1−ληβ+n−1_{) Γ (}_n_{) Γ (}_β₊_α 0)

Z 1

0

(1−s)β+α0−1

g(s)ds (2.9)

− λΓ (β+n)

(1−ληβ+n−1_{) Γ (}_n_{) Γ (}_β₊_α 0)

Z η

0

(η−s)β+α0−1

g(s)ds

− x

∗ _ληβ ₋₁

Γ (β+n) (1−ληβ+n−1_{) Γ (}_n_{) Γ (}_β_{+ 1)}.

Substituting the values ofc0, c1, c2, ..., cn−2, cn−1in (2.7),we obtain the desired quantity in lemma.

3. Main Results

For convenience, we set:

L0 := _Γ(_α1₀₊₁₎ +

Γ(β+n)₍1+|λ|ηβ+α₀₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1), (3.1)

Lk:= _Γ(_α₀−1αk+1) +

Γ(β+n)₍1+|λ|ηβ+α₀₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1), k= 1, ..., n−1, ω:=ω0+ω1+...+ωn−1,

θ:= Γ(β+n)(1+|λ|η

β+_αk₎ |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1) +

n−1

X

k=1

Γ(β+n)₍1+|λ|ηβ+_αk₎ |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1), θ0 = |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎ +

n−1

X

k=1

|x∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1).

Now list the following hypotheses for convenience: (H1) : The function f : [0,1]×_Rn_→

R, is continuous.

(H2) : There exist non negative continuous functionsai ∈C([0,1]), i= 0,1,2, ..., n−1 such that

for all t∈[0,1] and (x0, x1, x2, ..., xn−1),(y0, y1, y2, ..., yn−1)∈Rn, we have

|f(t, x0, x1, x2, ..., xn−1)−f(t, y0, y1, y2, ..., yn−1)|

≤a0(t)|x0−y0|+a1(t)|x1 −y1|+a2(t)|x2−y2|+...+an−1(t)|xn−1−yn−1|.

where,

ω0 = sup

t∈J

a0(t), ω1 = sup

t∈J

a1(t), ω2 = sup

t∈J

a2(t), ..., ωn−1 = sup

t∈J

an−1(t).

(H3) : There exist positive constants m(t) such that

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with

M = sup

t∈J

m(t).

Our first result is based on Banach contraction principle:

Theorem 3.1. Suppose that ηβ+n−1 ₆₌ 1

λ, and assume that the hypothesis(H2)holds.

If

L0+

n−1

X

k=1 Lk

!

ω <1, (3.2)

then the fractional problem (1.1)has a unique solution.

Proof . Consider the operator φ:X →X defined by:

φx(t) := 1 Γ(α0)

Z t

0

(t−s)α0−1_f₍_{s, x}₍_s₎_{, D}α1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_ds₊_x∗ _(3.3)

− Γ(β+n)tn−1

(1−ληβ+n−1₎_Γ(_n_)Γ(_β₊_α 0)

Z 1 0

(1−s)β+α0−1

f(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_ds

+ λΓ(β+n)tn−1 (1−ληβ+n−1₎_Γ(_n_)Γ(_β₊_α

0) Z η

0

(η−s)β+α0−1

+ x

∗₍_ληβ₋₁₎_Γ(_β₊_n₎_tn−1

(1−ληβ+n−1₎_Γ(_n_)Γ(_β₊₁₎.

We shall prove that φ is contraction mapping : For x, y ∈X and for each t∈J, we have:

|φx(t)−φy(t)| ≤ 1 Γ(α0)

Z t

0

(t−s)α0−1

f(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎ −f(s, y(s), Dα1_y₍_s₎_{, ..., D}αn−1_y₍_s₎₎

ds (3.4)

+ Γ(β+n)tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z 1

0

(1−s)β+α0−1

ds

+ |λ|Γ(β+n)tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z η

0

(η−s)β+α0−1

ds.

Using the (H2), we can write:

|φx(t)−φy(t)| ≤ (ω0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα₁_y_k₊_...₊_k_Dαn−1_x₋_Dαn−1_y_k_]

Γ(α0+1) (3.5)

+Γ(β+n)(ω0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα₁_y_k₊_...₊_k_Dαn−1_x₋_Dαn−1_y_k_] |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1)

+|λ|Γ(β+n)ηβ+α0(ω0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk] |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1) .

Thus,

|φx(t)−φy(t)| ≤ (ω0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk]

Γ(α0+1) (3.6)

+ Γ(β+n)[1+|λ|η

β+α₀_]₍_ω

0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα₁_y_k₊_...₊_k_Dαn−1_x₋_Dαn−1_y_k_] |1−ληβ+n−1_|_Γ(_n_)Γ(_α

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Consequently,we have

|φx(t)−φy(t)| ≤L0ω(kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk), (3.7)

which implies that

kφ(x)−φ(y)k ≤L0ω(kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk). (3.8)

On the other hand, for all k = 1, ..., n−1 and for each t ∈[0,1], we have

|Dαk_φx₍_t₎₋_Dαk_φy₍_t₎_{| ≤} 1

Γ(α0−αk) Z t

0

(t−s)α0−αk−1

ds

+ Γ(β+n)

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0) Z 1

0

(1−s)β+α0−1

ds (3.9)

+ |λ|Γ(β+n)

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0) Z η

0

(η−s)β+α0−1

ds.

By (H2) we obtain

|Dαk_φx₍_t₎₋_Dαk_φy₍_t₎_{| ≤} (ω0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα₁_y_k₊_...₊_k_Dαn−1_x₋_Dαn−1_y_k_]

Γ(α0−αk+1) (3.10)

+Γ(β+n)(ω0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk] |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1)

+|λ|Γ(β+n)ηβ+α0(ω0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα₁_y_k₊_...₊_k_Dαn−1_x₋_Dαn−1_y_k_] |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1) .

Hence, we have

|Dαk_φ

1y(t)−Dαkφ1y1(t)| ≤ (ω0+ω1+...+ωn−1)[

kx−yk+kDα₁_x₋_Dα₁_y_k₊_...₊_k_Dαn−1_x₋_Dαn−1_y_k_] Γ(α0−αk+1)

+Γ(β+n)[1+|λ|η

β+α₀_]₍_ω

0+ω1+...+ωn−1)[kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk] |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1) . (3.11)

Therefore,

|Dαk_φx₍_t₎₋_Dαk_φy₍_t₎_{| ≤}_L

kω(kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk), (3.12)

which implies that

kDαk_φ₍_x₎₋_Dαk_φ₍_y₎_{k ≤}_L

kω(kx−yk+kDα1x−Dα1yk+...+kDαn−1x−Dαn−1yk). (3.13)

Thanks to (3.8) and (3.13), we obtain

kφ(x)−φ(y)k_X ≤ L0+

n−1

X

k=1 Lk

!

ω ky−y1k+

Dβ1y−Dβ1y₁

+...+

Dβn−1y−Dβn−1y₁

.

(3.14) Consequently by (3.2), we conclude thatφ is contraction. As a consequence of Banach contrac-tion, we deduce thatφ has a fixed point which is a solution of the boundary

value problem (1.1).

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Theorem 3.2. Suppose that for all ηβ+n−1 6= 1_λ and assume that the hypotheses (H1) and (H3) are satisfied.

Then, the problem (1.1)has at least a solution on J.

Proof . We shall use Scheafer’s fixed point theorem to prove that φ has at least a fixed point onX. Step1: φ is continuous on X : By (H1) we conclude that the operator φ is continuous.

Step2: The operator φ maps bounded sets into bounded sets in X : For σ > 0, we take x ∈ Bσ ={x∈X;kxkX ≤σ}. For each t∈J, we have:

|φx(t)| ≤ 1 Γ(α0)

Z t

0

(t−s)α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds₊_|x∗_| _(3.15)

+ Γ(β+n)tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z 1

0

(1−s)β+α0−1_|f₍_{s, x}₍_s₎_{, D}α1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z η

0

(η−s)β+α0−1_|f₍_{s, x}₍_s₎_{, D}α1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds |x∗||ληβ−1|Γ(β+n)tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎.

Using the (H3), we obtain

|φx(t)| ≤sup

t∈J

m(t)

1 Γ(α0+1) +

Γ(β+n)(1+|λ|ηβ+α0) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1)

+|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎. (3.16)

Thus,

|φx(t)| ≤L0sup

t∈J

m(t) +|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎, t ∈J, (3.17)

which implies that

kφ(x)k ≤M L0+|x∗|+

|x∗||ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎, (3.18)

On the other hand, for all k = 1,2, ..., n−1, we have

|Dαk_φx₍_t₎_{| ≤} 1

Γ(α0−αk) Z t

0

(t−s)α0−αk−1_|f₍_{s, x}₍_s₎_{, D}α1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds _(3.19)

+ Γ(β+n)tn−αk−1

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0) Z 1

0

(1−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds

+ |λ|Γ(β+n)tn−αk−1

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0) Z η

0

(η−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds |x∗||ληβ₋₁_|_Γ(_β₊_n₎_tn−_αk−1

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1).

By (H3) we have,

kDαk_φ₍_x₎_{k ≤}_M

1

Γ(α0−αk+1)+

Γ(β+n)₍1+L|λ|ηβ+α₀₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1)

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

. (3.20)

Hence,

kDαk_φ₍_x₎_{k ≤}_{M L}

k+

|x∗||ληβ−1|Γ(β+n) |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

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Combining (3.18) and (3.21), yields

kφ(x)k_X ≤M L0+

n−1

X

k=1 Lk

!

+|x∗_|₊ |x∗||ληβ−1|Γ(β+n)

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎ +

|x∗||ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

. (3.22)

Consequently

kφ(x)k_X <∞. (3.23)

Step3: In the end we show that φ is equicontinuous on J:

Let us take (x, y)∈Bσ, t1, t2 ∈J, such that t1 < t2 and Thanks to (H3). We can write:

|φx(t2)−φx(t1)| ≤

supt∈Jm(t)

Γ(α0) Z t1

0

(t2−s)

α0−1₋

(t1−s)

α0−1

ds+supt∈Jm(t)

Γ(α0) Z t2

t1

(t2−s)

α0−1 ds

(3.24)

+supt∈Jm(t)Γ(β+n)(tn −1 1 −t

n−1 2 ) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0) Z 1

0

(1−s)β+α0−1 ds

+supt∈Jm(t)|λ|Γ(β+n)(tn −1 2 −t

n−1 1 ) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0)

Z η

0

(η−s)β+α0−1 ds

|x∗||ληβ−1|Γ(β+n)(tn2−1−t n−1 1 ) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎ .

Thus,

|φx(t2)−φx(t1)| ≤ _Γ(_αM₀₊₁₎(tα10 −t

α0

2 ) + Γ(α2M0+1)(t2−t1)

α0

(3.25)

+ MΓ(β+n)

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0+1)

tn₁−1−tn₂−1

+

M|λ|Γ(β+n)ηβ+α0 |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1) +

|x∗||ληβ−1|Γ(β+n) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎

tn₂−1−tn₁−1,

on the other hand, for all k= 1,2..., n−1,

Dkφ1y(t2)−Dkφ1y(t1)

≤ _Γ(_α M 0−αk+1) t

α0−αk

1 −t

α0−αk

2

+ _Γ(_α 2M

0−αk+1)(t2−t1)

α0−αk

+ MΓ(β+n)

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1)

tn−αk−1

1 −t

n−αk−1

2

(3.26)

+ M|λ|Γ(β+n)ηβ+α0

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1) t

n−αk−1

2 −t

n−αk−1

1

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1) t

n−αk−1

2 −t

n−αk−1

1

.

Thanks to (3.25) and (3.26), we can state that kφ(x) (t2)−φ(x) (t1)k → 0 as t2 → t1. By

Arzela-Ascoli theorem, we conclude thatφ is completely continuous operator.

Step4: Finally, we discuss a priory bounds on solutions. We shall show that the set Ω defined by

Ω ={x∈X, x=σφ(x),0< σ <1}, (3.27)

is bounded:

Let x∈Ω, then x=σφ(x), for some 0< σ <1. Thus, for each t∈J, we have:

(9)

Then

1

σ|x(t)| ≤ 1 Γ(α0)

Z t

0

(t−s)α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds₊_|x∗_| _(3.29)

+ Γ(β+n)tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z 1

0

(1−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z η

0

(η−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎_tn−1 |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎.

Thanks to (H3), we can write

1

σ |x(t)| ≤supt∈J

m(t)

1 Γ(α0+1) +

Γ(β+n)₍1+|λ|ηβ+α₀₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

+|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎. (3.30)

Therefore,

|x(t)| ≤σsup

t∈J

m(t)

1 Γ(α0+1) +

Γ(β+n)₍1+|λ|ηβ+α₀₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

+|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎. (3.31)

Hence,

|x(t)| ≤σL0sup

t∈J

m(t) +|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎, t∈J, (3.32)

which implies that,

kxk ≤σM L0+|x∗|+

|x∗||ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎. (3.33)

On the other hand, for all k,= 1,2, ..., n−1,we have

|Dαk_φx₍_t₎_{| ≤}_σ_sup

t∈J

m(t)

1

Γ(α0−αk+1) +

Γ(β+n)(1+|λ|ηβ+α₀₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1)

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

, t∈J.

(3.34) Thus,

kDαk_φ₍_x₎_{k ≤}_{σM L}

k+

|x∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

. (3.35)

From (3.33) and (3.35) we get,

kxk_X ≤σM L0+

n−1

X

k=1 Lk

!

+|x∗|+|x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n_)[Γ(_n₋_α k)+Γ(n)]

|1−ληβ+n−1_|_Γ(_n_)Γ(_n₋_α

k)Γ(β+1)

. (3.36)

Hence

kφ(x)k_X <∞. (3.37)

This shows that the set Ω is bounded.

As consequence of Schaefer’s fixed point theorem, we deduce that φ at least a fixed point, which is a solution of the fractional differential problem (1.1).

(10)

Theorem 3.3. Let ηβ+n−1 6= 1_λ. Suppose that (H1), (H2)and (H3) are satisfied, such that

1 Γ(α0+1) +

n−1

X

k=1 1 Γ(α0−αk+1)

!

ω <1, (3.38)

if there exist δ∈_R such that

M L0+

n−1

X

k=1 Lk

!

+|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n_)[Γ(_n₋_α k)+Γ(n)]

|1−ληβ+n−1_|_Γ(_n_)Γ(_n₋_α

k)Γ(β+1)

≤δ. (3.39)

Then, the fractional problem (1.1) has at least one solution on [0,1].

Proof . We shall use Krasnoselskii fixed point theorem to prove that φ has at least a fixed point on

X.

Suppose that M L0+

Pn−1

k=1Lk

+|x∗_|₊|x∗||ληβ−1|Γ(β+n)[Γ(n−αk)+Γ(n)]

|1−ληβ+n−1_|_Γ(_n_)Γ(_n₋_α

k)Γ(β+1)

≤δ and let us take

φ(x) (t) :=T (x) (t) +R(x) (t), (3.40)

where

T x(t) := _Γ(1_α

0) Z t

0

(t−s)α0−1

f(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1₍_s₎₎_ds₊_x∗_, _(3.41)

and

Rx(t) :=− Γ(β+n)tn−1

(1−ληβ+n−1₎_Γ(_n_)Γ(_β₊_α 0)

Z 1

0

(1−s)β+α0−1

f(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1₍_s₎₎_ds

+ λΓ(β+n)tn−1 (1−ληβ+n−1₎_Γ(_n_)Γ(_β₊_α

0) Z η

0

(η−s)β+α0−1

+ x

∗₍_ληβ₋₁₎_Γ(_β₊_n₎_tn−1

(1−ληβ+n−1₎_Γ(_n_)Γ(_β₊₁₎. (3.42)

The proof will be given in several steps.

(1∗ :) We shall prove that for any x, y ∈ Bδ, then T(x) (t) +R(y) (t) ∈ Bδ, such that Bδ =

{x∈X;kxk_X ≤δ}.

For any x, y ∈Bδ and for eacht ∈J we have:

|T x(t) +Ry(t)| ≤ 1 Γ(α0)

Z t

0

(t−s)α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1₍_s₎₎_|_ds₊_|x∗_| _(3.43)

+ Γ(β+n)tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z 1

0

(1−s)β+α0−1_|f₍_{s, x}₍_s₎_{, D}α1_x₍_s₎_{, ..., D}αn−1₍_s₎₎_|_ds

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z η

0

(η−s)β+α0−1_|f₍_{s, x}₍_s₎_{, D}α1_x₍_s₎_{, ..., D}αn−1₍_s₎₎_|_ds |x∗||ληβ₋₁_|_Γ(_β₊_n₎_tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎.

|T x(t) +Ry(t)| ≤sup

t∈J

m(t)

1 Γ(α0+1) +

Γ(β+n)(1+|λ|ηβ+α0) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1)

(3.44)

+|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

(11)

Consequently,

|T x(t) +Ry(t)| ≤L0sup

t∈J

m(t) +|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎, t ∈J, (3.45)

Thus,

kT (x) +R(y)k ≤M L0+|x∗|+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎. (3.46)

|Dαk_{T x}₍_t_{) +}_Dαk_Ry₍_t₎_{| ≤}_sup

t∈J

m(t)

1

Γ(α0−αk+1) +

Γ(β+n)₍1+|λ|ηβ+α₀₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1)

(3.47)

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

,

Thus,

kDαk_{T x}₍_t_{) +}_Dαk_Ry₍_t₎_{k ≤}_{M L}

k+

|x∗||ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

. (3.48)

Combining (3.58) and (3.60) yields

kT (x) +R(y)k_X ≤M L0+

n−1

X

k=1 Lk

!

+|x∗|+|x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n_)[Γ(_n₋_α k)+Γ(n)]

|1−ληβ+n−1_|_Γ(_n_)Γ(_n₋_α

k)Γ(β+1)

. (3.49)

kT (x) +R(y)k_X ∈Bδ. (3.50)

(2∗ :) We shall prove thatR is continuous and compact. Note that R is continuous on X in view of the continuity of f (hypothesis (H1)).

(a∗) : Now, we prove that R maps bounded sets into bounded sets of X.

For x∈Bδ and for each t∈J, we have:

|Rx(t)| ≤ Γ(β+n)tn−1 |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0) Z 1

0

(1−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1₍_s₎₎_|_ds

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z η

0

(η−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎_tn−1

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎. (3.51)

|Rx(t)| ≤ supt∈Jm(t)Γ(β+n)(1+|λ|ηβ+αk)

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0+1)

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎, t∈J, (3.52)

Thus,

kR(x)k ≤ MΓ(β+n)(1+|λ|η

β+_αk₎ |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1)

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎. (3.53)

kDαk_R₍_x₎_{k ≤} MΓ(β+n)(1+|λ|η β+_αk₎ |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1) +

|x∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

(12)

From (3.53) and (3.54), we have

kR(x)k_X ≤M Γ(β+n)(1+|λ|η

β+_αk₎ |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1) +

n−1

X

k=1

Γ(β+n)₍1+|λ|ηβ+_αk₎ |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1) !

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎ +

n−1

X

k=1

|x∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1).

Consequently,

kR(x)k_X ≤M θ+θ0 <∞. (3.55)

(b∗) : In the end we show that R is equicontinuous on J : Let t1, t2 ∈J, such that t2 < t1 and (x, y)∈Bδ, Then, we have:

|Rx(t1)−Rx(t2)| ≤

Γ(β+n)₍tn₂−1−tn₁−1₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z 1

0

(1−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1₍_s₎₎_|_ds

+ |λ|Γ(β+n)(t

n−1 1 −t

n−1 2 ) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0)

Z η

0

(η−s)β+α0−1_|f

(s, x(s), Dα1_x₍_s₎_{, ..., D}αn−1_x₍_s₎₎_|_ds

+ |x

∗_|_|_ληβ₋₁_|_Γ(_β₊_n₎₍_tn−1 1 −t

n−1 2 )

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎ . (3.56)

|Rx(t1)−Rx(t2)| ≤

MΓ(β+n) |1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α

0+1) t

n−1 2 −t

n−1 1

(3.57)

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊_α 0+1)

+ tn₁−1−tn₂−1

|x∗||ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n_)Γ(_β₊₁₎ t

n−1 1 −t

n−1 2

.

On the other hand, for all k = 1,2..., n−1,

|Dαk_Rx₍_t

1)−DαkRx(t2)| ≤

MΓ(β+n) |1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1) t

n−1 2 −tn

−1 1

(3.58)

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+α0+1)

tn−αk−1

1 −t

n−αk−1

2

|x∗||ληβ₋₁_|_Γ(_β₊_n₎

|1−ληβ+n−1_|_Γ(_n₋_α

k)Γ(β+1)

tn−αk−1

1 −t

n−αk−1

2

.

As t2 → t1, the right-hand sides of the inequalities (3.57) and (3.58) tend to zero. Then, as a

consequence of the steps (a∗) and (b∗) and by Arzela-Ascoli theorem, we conclude that R is completely continuous.

(3∗ :) Finally, we prove that T is contraction mapping Let x, y ∈X.Then, for each t ∈J and by (H2) we have

|T x(t)−T y(t)| ≤ (ω0+ω1+...+ωn−1)

Γ(α0+1) (kx−yk+kD

α1_x₋_Dα1_yk₊_...₊_kDαn−1_x₋_Dαn−1_yk₎_. _(3.59)

|Dαk_{T x}₍_t₎₋_Dαk_{T y}₍_t₎_{| ≤} (ω0+ω1+...+ωn−1)

Γ(α0−αk+1) (kx−yk+kD

α1_x₋_Dα1_yk₊_...₊_kDαn−1_x₋_Dαn−1_yk₎_.

(13)

By (3.59) and (3.60), we obtain

|T x(t)−T y(t)| ≤

"

1 Γ(α0+1) +

n−1

X

k=1 1 Γ(α0−αk+1)

#

ω(kx−yk+kDα1_x₋_Dα1_yk₊_...₊_kDαn−1

x−Dαn−1_yk

).

(3.61) Using the condition (3.51) we conclude that T is a contraction mapping.

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

Corollary 3.4. Assume thatηβ+n−1 ₆₌ 1

λ and there exist non negative real numbersθi, i= 0,1, ..., n−

1 such that for allt ∈[0,1] and (x0, x1, x2, ..., xn−1),(y0, y1, y2, ..., yn−1)∈Rn,we have

|f1(t, x0, x1, x2, ..., xn−1)−f1(t, y0, y1, y2, ..., yn−1)| ≤θ0|x0−y0|+...+θn−1|xn−1−yn−1|,

If

L0+

n−1

X

k=1 Lk

!

(θ0+...+θn−1)<1, (3.62)

then the fractional problem (1.1) has a unique solution on J.

Corollary 3.5. Assume that (H1) holds and ηβ+n−1 ₆₌ 1

λ. If There exist positive constants k1 and

k2 such that

f1 ≤k1, f2 ≤k2 onJ×Rn,then, the problem (1.1) has at least a solution on J. Example 3.6. Consider the following fractional differential problem:

  

D72x(t) =

|x(t)|+

D

9 4x(t)

+

D

3 2x(t)

+

D

1 2x(t)

(t2₊₃₂_π₎_et₊_|_x₍_t₎_|₊ D

9 4x(t)

+

D

3 2x(t)

+

D

1 2x(t)

+ cosh (2 +t

2₎_{, t} _∈_[0_,_1]_,

x(0) =√2, x0(0) =x00(0) = 0, J12x(1) = 4 5J

1 2x 3

4

,

(3.63)

For this example, we have

f(t, x1, x2, x3, x4) = |x1|+|x2|+|x3|+|x4| (t2_{+ 32}_π_{) (}_et₊_|x

1|+|x2|+|x3|+|x4|)

+cosh 2 +t2, t∈[0,1], x1, x2, x3, x4 ∈_R.

For t∈[0,1] and (x0, x1, x2, x3),(y0, y1, y2, y3)∈R4, we have:

|f(t, x0, x1, x2, x3)−f(t, y0, y1, y2, y3)| ≤

1

(t2_{+ 32}_π₎(|x0−y0|+|x1−y1|+|x2 −y2|+|x3−y3|),

So,we have

ai(t) =

1

(t2 _{+ 32}_π₎, i= 0, ...,3.

Then,

ωi = sup t∈J

ai(t) =

1

32π, i= 0, ...,3,

and then,

L0 = 0,228998, L1 = 2,203434, L2 = 1,145589, L3 = 0,424889, ω=

(14)

and

L0+

n−1

X

k=1 Lk

!

ω = 1,799476.0,039808 = 0,071633<1.

Hence by Theorem 3.1 then the problem (3.63) has a unique solution on [0,1].

Example 3.7. Let us consider the following fractional boundary value problem:

  

D144 x(t) =

sinx(t)+cos

D52_x₍_t₎₊_D43_x₍_t₎

(5√π+et2

)2 , t∈[0,1],

x(0) = 3, x0(0) =x00(0) = 0, D45x(1) = 3 4D

4 5x 1

3

.

(3.64)

Then, we have

f(t, x, y, z) = sin (x) + cos (y+z) (5√π+et2

)2 , t∈[0,1],(x, y, z)∈R

3 .

Let x, y, z∈_R and t ∈[0,1]. Then

|f(t, x, y, z)| ≤ 2

(5√π+et2

)2.

So, we can take

m(t) = 2 (5√π+et2

)2. Then,

M = 2 (5√π+ 1)2.

Thanks to Theorem 3.2, the problem (3.64) has at least one solution on [0,1].

Example 3.8. Our third example is the following:

  

D175 x(t) =

e−t2_e−|x(t)|_+sin_D15₆ _x₍_t_)+sin2_D11₇ _x₍_t₎_+cos_D2₃_x₍_t₎

20π+et2 + ln (2 +t

2₎_{, t}_∈_[0_,_1]_, x(0) = 2, x0(0) =x00(0) = 0, D54x(1) = 5

6D 5 4x 2

5

,

(3.65)

where,

f(x1, x2, x3, x4) =

e−t2 e−|x1|_{+ sin (}_x

2) + sin2(x3) + cos (x4)

20π+et2 +ln 2 +t

2

, t∈[0,1], x1, x2, x3, x4 ∈R.

Taking x0, x1, x2, x3, y0, y1, y2, y3 ∈R, t∈[0,1],then

|f(t, x0, x1, x2, x3)−f(t, y0, y1, y2, y3)| ≤ e−t2

(|x0−y0|+|x1−y1|+|x2 −y2|+|x3−y3|)

20π+et2 .

So,we can take:

ai(t) =

e−t2

(15)

then,

ωi = sup t∈[0,1]

ai(t) =

1

20π+ 1, i= 0, ...,3. For k= 1,2,3 we have

1 Γ(α0+1) +

n−1

X

k=1 1

Γ(α0−αk+1) = 1,951303.

We have also

1 Γ(α0+1) +

n−1

X

k=1 1 Γ(α0−αk+1)

!

ω= 0,1223387.

By Theorem 3.3, we can state that the problem (3.65) has at least one solution on [0,1].

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