Existence Results for Fractional Differential Inclusions with Multivalued Term Depending on Lower-Order Derivative

and Applied Analysis 3 The results of this paper can easily to be generalized to the boundary value problems of fractional differential inclusions 1.4 with the following integral boundary conditions: a1x 0 b1 cDγx 0 ) c1 ∫T


Introduction
Recently, the subject of fractional differential equations has emerged as an important area of investigation.Indeed, we can find numerous applications of fractional-order derivatives in the mathematical modeling of physical and biological phenomena in various fields of science and engineering 1-3 .A variety of results on initial and boundary value problems of fractional differential equations and inclusions can easily be found in the literature on this topic.For some recent results, we can refer to, for instance, 4-20 equations 21-27 inclusions and the references therein.
Ahmad and Ntouyas 22 considered a boundary value problem of fractional differential inclusions with fractional separated boundary conditions given by where c D q denotes the Caputo fractional derivative of order q, F : 0, 1 × R → 2 R is a multivalued map, and α i , β i , γ i i 1, 2 are real constants, with α 1 / 0.
In Cernea 24 , the following multipoint boundary value problem for a fractionalorder differential inclusion was studied D α x t ∈ F t, x t , x t a.e.t ∈ 0, 1 , 2 < α ≤ 3, where D α is the standard Riemann-Liouville fractional derivative, m ≥ 1, 0 < 1, λ > 0, a i > 0, i 1, 2, . . ., m, and In Khan et al. 11 , the authors studied the existence and uniqueness results of nonlinear fractional differential equation of the type c D q x t f t, x t , c D σ x t , t ∈ 0, T , αx 0 − βx 0 where 0 < σ < 1, 1 < q < 2, α, δ > 0, β, γ ≥ 0 or α, δ ≥ 0, β, γ > 0 and c D q , c D σ are the Caputo fractional derivatives.The results in 11, 22, 24 are obtained by using appropriate standard fixed point theorems.Motivated by the papers cited above, in this paper, we consider the existence results for a new class of fractional differential inclusions of the form c D α x t ∈ F t, x t , c D β x t , a.e.t ∈ 0, T , 1.4 where c D α denotes the Caputo fractional derivative of order α, F : 0, 1 × R × R → 2 R is a multivalued map, 1 < α ≤ 2, 0 < β ≤ 1, and T > 0. We study 1.4 subject to two families of boundary conditions: 1 separated boundary conditions 2 Nonseparated boundary conditions The results of this paper can easily to be generalized to the boundary value problems of fractional differential inclusions 1.4 with the following integral boundary conditions: where g, h : 0, T × R → R are given functions.We remark that when the third variable of the multifunction F in 1.4 vanishes, the problem 1.4 , 1.5 reduces to the case considered in 22 .When a 1 b 1 1, a 2 b 2 1, and c 1 c 2 0, the problem 1.4 , 1.6 reduces to an antiperiodic fractional boundary value problem the case of F f a given continuous function was studied in 4, 15 .Our results generalize some results from the literature cited above and constitute a contribution to this emerging field of research.
The rest of the paper is organized as follows: in Section 2 we present the notations and definitions and give some preliminary results that we need in the sequel, Section 3 is dedicated to the existence results of the fractional differential inclusion 1.4 with boundary conditions 1.5 and 1.6 , in Section 4 we indicate a possible generalization for the inclusion problem 1.4 with integral boundary conditions 1.7 and 1.8 , and two illustrative examples are given in Section 5.

Preliminaries
In this section, we introduce notations, definitions, and preliminary facts that will be used in the remainder of this paper.
Let X, • be a normed space.We use the notations Let X be a Banach space.Let G : 0, T × X → P cp,c X be an L 1 -Carathéodory multivalued map and Γ a linear continuous map from L 1 0, T , X to C 0, T , X , then the operator Here S G,y {v ∈ L 1 0, T , X : v t ∈ G t, y t for a.e.t ∈ 0, T }.
Definition 2.4 see 30 .The Riemann-Liouville fractional integral of order q for a function f is defined as provided the integral exists.
Abstract and Applied Analysis 5 Definition 2.5 see 30 .For at least n-times differentiable function f, the Caputo derivative of order q is defined as where q denotes the integer part of the real number q.
Lemma 2.6 see 20 .Let α > 0; then the differential equation The following lemma obtained in 6 is useful in the rest of the paper.
Lemma 2.7 see 6 .For a given y ∈ C 0, T , R , the unique solution of the fractional separated boundary value problem 2.9 is given by where We notice that the solution 2.10 of the problem 2.9 does not depend on the parameter b 1 , that is to say, the parameter b 1 is of arbitrary nature for this problem.And by 2.10 , we should assume that a 1 / 0 and a 2 T γ Γ 2 − γ / − b 2 .

6
Abstract and Applied Analysis Lemma 2.8.For any y ∈ C 0, T , R , the unique solution of the fractional nonseparated boundary value problem 12 is given by

2.13
Proof.For 1 < α ≤ 2, by Lemma 2.6, we know that the general solution of the equation c D α x t y t can be written as where Using the boundary conditions, we obtain

2.16
Therefore, we have

2.17
Substituting the values of k 1 , k 2 in 2.14 , we obtain 2.13 .This completes the proof.
From the proof of the above lemma, we notice that the solution 2.13 of the problem 2.12 does not depend on the parameter a 2 , that is to say, the parameter a 2 is of arbitrary nature for this problem.In this situation, we need to assume that a 1 b 1 / 0 and b 2 / 0.
Let us define what we mean by a solution of the problem 1.4 , 1.5 and the problem 1.4 , 1.6 .Definition 2.9.A function x ∈ AC 1 0, T , R is a solution of the problem 1.4 , 1.5 if it satisfies the boundary conditions 1.5 and there exists a function

2.18
Definition 2.10.A function x ∈ AC 1 0, T , R is a solution of the problem 1.4 , 1.6 if it satisfies the boundary conditions 1.6 and there exists a function f ∈ L 1 0, T , R such that f t ∈ F t, x t , c D β x t a.e. on t ∈ 0, T and

2.19
Let C 0, T , R be the space of all continuous functions defined on 0, T .Define the space X {x : We know that X, • is a Banach space see 14 .
We end this section with two fixed point theorems, which will be used in the sequel.

Existence Results
In this section, we will give some existence results for the problems 1.4 , 1.5 and 1.4 , 1.6 .
For each x ∈ X, define the set of selections of F by In view of Lemmas 2.7 and 2.8, we define operators N, M : X → P X as

3.4
It is clear that if x ∈ X is a fixed point of the operator N the operator M , then x is a solution of the problem 1.4 , 1.5 the problem 1.4 , 1.6 .Now we are in a position to present our main results.The methods used to prove the existence results are standard; however, their exposition in the framework of problems 1.4 , 1.5 and 1.4 , 1.6 is new.

Convex Case
We consider first the case when F is convex valued. where

3.7
Then the problem 1.4 , 1.5 has at least one solution on 0, T .
Proof.Consider the operator N : X → P X defined by 3.2 .From H1 , we have for each x ∈ X, the set S F,x is nonempty 29 .For x ∈ X, let u ∈ S F,x and h Su, that is, h ∈ N x , we have where k is a constant given by 3.9 Hence we know that the operator N : X → P X is well defined.We put Su S 1 u S 2 u where

3.10
Here k u means that the constant k defined by 3.9 is related to u.
We will show that N satisfies the requirements of the nonlinear alternative of Leray-Schauder type.The proof will be given in five steps.
Step 1 N x is convex valued .Since F is convex valued, we know that S F,x is convex and therefore it is obvious that N x is convex for each x ∈ X.
Step 2 N maps bounded sets into bounded sets in X .Let B r be a bounded subset of X such that for any x ∈ B r , x ≤ r, r > 0. We prove that there exists a constant l > 0 such that for each x ∈ B r , one has h ≤ l for each h ∈ N x .Let x ∈ B r and h ∈ N x , then there exists u ∈ S F,x such that h t Su t for t ∈ 0, T .

3.11
By simple calculations, we have

3.12
Hence we obtain

3.13
Step 3 N maps bounded sets into equicontinuous sets in X .Let B r be a bounded set of X as in Step 2. Let 0 ≤ t 1 < t 2 ≤ T and x ∈ B r .For each h ∈ N x , then there is u ∈ S F,x such that h t Su t .Since and the limits are independent of x ∈ B r and h ∈ N x .
Step 4 N has a closed graph .Let x n → x * , h n ∈ N x n , and h n → h * ; we need to show h * ∈ N x * .Now h n ∈ N x n implies that there exists u n ∈ S F,x n such that h n t Su n t for t ∈ 0, T .Let us consider the continuous linear operator Γ : and denote w t v 2 t c 1 /a 1 .Then h n t − w t Γu n t and

3.17
We apply Lemma 2.3 to find that Γ • S F has closed graph and from the definition of Γ we get Step 5 a priori bounds on solutions .Let x ∈ λN x for some λ ∈ 0, 1 .Then there exists u ∈ S F,x such that x t λ Su t for t ∈ 0, T .With the same arguments as in Step 2 of our proof, for each t ∈ 0, T , we obtain Thus Now we set U {x ∈ X : x < L}.

3.20
Clearly, U is an open subset of X and 0 ∈ U.As a consequence of Steps 1-4, together with the Arzela-Ascoli theorem, we can conclude that N : U → P cp,c X is upper semicontinuous and completely continuous.From the choice of the U, there is no x ∈ ∂U such that x ∈ λN x for some λ ∈ 0, 1 .Therefore, by the nonlinear alternative of Leary-Schauder type Theorem 2.11 , we deduce that N has a fixed point x ∈ U, which is a solution of the problem 1.4 , 1.5 .This completes the proof.
Theorem 3.2.Assume that (H1) is satisfied and there exists L 1 > 0 such that where

3.22
Then the problem 1.4 , 1.6 has at least one solution on 0, T .
Proof.To obtain the result, the main aim is to study the properties of the operator M defined in 3.3 .The proof of them is similar to those of Theorem 3.1, so we omit the details.Here we just give some estimations, which are needed in the following theorems.Let x ∈ X and h ∈ M x ; then there exists u ∈ S F,x such that h t Ku t , for t ∈ 0, T .

3.23
We put Ku K 1 u K 2 u and 3.24 here k u 1 and k u 2 are constants given by

3.25
By simple calculations, we have

3.26
Hence we obtain

3.27
This is the end of the proof.

Nonconvex Case
Now we study the case when F is not necessarily convex valued.
A subset A of L 1 0, T , R is decomposable if for all u, v ∈ A and J ⊆ 0, T Lebesgue measurable, then uχ J vχ 0,T −J ∈ A, where χ stands for the characteristic function.The proof of this theorem is similar to that of Theorem 3.3.

H2 :
H3 : F : 0, T × R × R → P cp R is a multivalued map such that 1 F is integrably bounded and the map t → F t, x, y is measurable for all x, y ∈ R; 2 there exists m ∈ L ∞ 0, T , R such that for a.e.t ∈ 0, T and all

3.32
then the problem 1.4 , 1.5 has at least one solution on 0, T .
Proof.From H3 , we have that the multivalued map t → F t, x t , c D β x t is measurable 28, Proposition 2.7.9 and closed valued for each x ∈ X. Hence it has measurable selection 28, Theorem 2.2.1 and the set S F,x is nonempty.Let N be defined in 3.2 .We will show that, under this situation, N satisfies the requirements of Theorem 2.12.
Step 1.For each x ∈ X, N x ∈ P cl X .Let h n ∈ N x , n ≥ 1 such that h n → h in X.Then h ∈ X and there exists u n ∈ S F,x , n ≥ 1 such that h n t Su n t , t ∈ 0, T .

3.33
By H3 , the sequence u n is integrable bounded.Since F has compact values, we may pass to a subsequence if necessary to get that u n converges to u in L 1 0, T , R .Thus u ∈ S F,x and for each t ∈ 0, T h n t −→ h t Su t .

3.34
This means that h ∈ N x and N x is closed.

3.35
Let x, y ∈ X and h 1 ∈ N y ; then there exists u 1 ∈ S F,y such that

3.37
Hence, for a.e.t ∈ 0, T , there exists v ∈ F t, x t , c D β x t such that

3.38
Consider the multivalued map V : 0, T → P R given by , Theorem III.41 implies that V is measurable.It follows from H3 that the map t → F t, x t , c D β x t is measurable.Hence by 3.38 and 28, Proposition 2.1.43, the multivalued map t → V t ∩ F t, x t , c D β x t is measurable and nonempty closed valued.Therefore, we can find u 2 t ∈ F t, x t , c D β x t such that for a.e.t ∈ 0, T ,

3.41
we obtain x − y .

3.43
By using an analogous relation obtained by interchanging the roles of x and y, we get h N x , N y ≤ ρ x − y .

3.44
Therefore, from condition 3.32 , Theorem 2.12 implies that N has a fixed point, which is a solution of the problem 1.4 , 1.5 .This completes the proof.

3.45
then the problem 1.4 , 1.6 has at least one solution on 0, T .
Using the arguments employed in the proof of Theorem 3.5, we can prove this theorem similarly.Hence the details are omitted here.

Integral Boundary Conditions
In this section, the existence results of the problems 1.4 , 1.5 and 1.4 , 1.6 obtained in the previous section will be extended to the ones of the problems of fractional differential inclusions 1.4 subject to the integral boundary conditions 1.7 and 1.8 .Lemma 4.1.For any y, ξ, χ ∈ C 0, T , R , the unique solution of the fractional separated integral boundary value problem,

4.4
To obtain the existence results of the problems 1.4 , 1.7 and 1.4 , 1.8 , in view of Lemmas 4.1 and 4.2, we define two operators Π, Ω : X → P X as

4.7
Observe that if x ∈ X is a fixed point of the operator Π the operator Ω , that is, x ∈ Π x x ∈ Ω x , then x is a solution of the problem 1.4 , 1.7 the problem 1.4 , 1.8 .
From the definitions of the operators N, Π see 3.2 , 4.5 , we know that the difference between them is very apparent, that is, c 1 , c 2 in 3.2 were replaced by c 1 T 0 g s, x s ds and c 2 T 0 h s, x s ds in 4.5 .This fact is also true for the operators M, Ω see 3.3 , 4.6 .

Abstract and Applied Analysis
In the following, we state some existence results for the problems 1.4 , 1.7 and 1.4 , 1.8 .We omit the proofs as these are similar to the ones given in Section 3. A1 : The functions g, h : 0, T × R → R are continuous.There exist functions m 2 , m 3 ∈ L 1 0, T , R and ϕ 2 , ϕ 3 : 0, ∞ → 0, ∞ continuous, nondecreasing such that here Q is defined by 3.7 and Then the boundary value problem 1.4 , 1.7 has at least one solution on 0, T .
Theorem 4.4.Assume that (H1) and (A1) hold.If there exists a constant I 1 > 0 such that here Q 1 is defined by 3.22 and Then the boundary value problem 1.
here Q 1 is defined by 3.22 and R 1 , W 1 are defined by 4.12 , then the boundary value problem 1.4 , 1.8 has at least one solution on 0, T .

Examples
In this section, we give two simple examples to show the applicability of our results.In the context of this problem, we have

Example 5 . 1 .where α 3 / 2 ,
Consider the following fractional boundary value problem:c D 3/2 x t ∈ F t, x t , c D 3/4 x t , t ∈ 0, 1 , β 3/4, γ 1/2, a 1 1, b 1 −1/2, c 1 2.5, a 2 2, b 2 1/3, c 2 −1/3, T 1, and F : 0, 1 × R × R → P R is a multivalued map given by F t, x, y u ∈ R : e −|x| − y 1 y sin t ≤ u ≤ 5 |x| 1 x 2 6t 3 cos y .5.2 X, and for every open set O of X containing F x 0 , there exists an open neighborhood U 0 of x 0 such that F U 0 ⊆ O. Equivalently, F is upper semicontinuous if the set {x ∈ X : F x ⊆ O} is open for any open set O of X. F is called lower semicontinuous if the set {x ∈ X : F x ∩ O / ∅} is open for each open set O in X.If a multivalued map F is completely continuous with nonempty compact values, then F is upper semicontinuous if and only if F has a closed graph, that is, if x n → x * and y n → y * , then y n ∈ F x n implies y * ∈ F x * 28 .A multivalued map F : 0, T → P cl X is said to be measurable if, for every x ∈ X, the function t → d x, F t inf{d x, y : y ∈ F t } is a measurable function.Definition 2.1.A multivalued map F : X → P cl X is called 1 γ-Lipschitz if there exists γ > 0 such that h F x , F y ≤ γd x, y , for each x, y ∈ X, 2.2 2 a contraction if it is γ-Lipschitz with γ < 1.
It can easily be shown that N 1 is continuous and completely continuous and satisfies all conditions of the Leray-Schauder nonlinear alternative for single-valued maps 31 .The remaining part of the proof is similar to that of Theorem 3.1, so we omit it.This completes the proof.
x, y is lower semicontinuous for a.e.t ∈ 0, T .Theorem 3.3.Let (H1) 2 , (H2), and relation 3.6 hold; then the problem 1.4 , 1.5 has at least one solution on 0, T .islower semicontinuous and has nonempty closed and decomposable values.Then from a selection theorem due to Bressan and Colombo 34 , there exists a continuous function f : X → L 1 0, T , R such that f x ∈ F x for all x ∈ X.That is to say, we have f x t ∈ F t, x t , c D β x t for a.e.t ∈ 0, T .Now consider the problem 4 , 1.8 has at least one solution on 0, T .Assume that (H3) and (A2) hold.If, in addition,m L ∞ Q m 3 L 1 R m 2 L 1 W < 1, 4.14here Q is defined by 3.7 and R, W are defined by 4.10 , then the boundary value problem 1.4 , 1.7 has at least one solution on 0, T .