On Higher-Order Sequential Fractional Differential Inclusions with Nonlocal Three-Point Boundary Conditions

and Applied Analysis 3 Definition 9. AmultivaluedmapG : [0; 1] → Pcl(R) is said to be measurable if for every y ∈ R, the function t 󳨃󳨀→ d (y,G (t)) = inf {󵄨󵄨󵄨y − z 󵄨󵄨󵄨 : z ∈ G (t)} (10) is measurable. 3.1. The Carathéodory Case Definition 10. A multivalued map F : [0, 1] × R → P(R) is said to be Carathéodory if (i) t 󳨃→ F(t, x) is measurable for each x ∈ R; (ii) x 󳨃→ F(t, x) is upper semicontinuous for almost all t ∈ [0, 1]. Further a Carathéodory functionF is calledL-Carathéodory if (iii) for each ρ > 0, there exists φ ρ ∈ L1([0, 1],R+) such that ‖F (t, x)‖ = sup {|V| : V ∈ F (t, x)} ≤ φ ρ (t) (11) for all ‖x‖ ≤ ρ and for a.e. t ∈ [0, 1]. For each y ∈ C([0, 1],R), define the set of selections of F by S F,y := {V ∈ L1 ([0, 1] , R) : V (t) ∈ F (t, y (t)) for a.e. t ∈ [0, 1] } . (12) For the forthcoming analysis, we need the following lemmas. Lemma11 (nonlinear alternative forKakutanimaps [26]). Let E be a Banach space, C a closed convex subset of E, U an open subset of C, and 0 ∈ U. Suppose that F : U → P cp,c (C) is an upper semicontinuous compact map. Then either (i) F has a fixed point in U, or (ii) there is a u ∈ ∂U and λ ∈ (0, 1) with u ∈ λF(u). Lemma 12 (see [27]). LetX be a Banach space. Let F : [0, 1]× R → P cp,c (X) be an L-Carathéodory multivalued map and let Θ be a linear continuous mapping from L1([0, 1], X) to C([0, 1], X). Then the operator Θ ∘ S F : C ([0, 1] , X) 󳨀→ P cp,c (C ([0, 1] , X)) , x 󳨃󳨀→ (Θ ∘ S F ) (x) = Θ (S F,x ) (13) is a closed graph operator in C([0, 1], X) × C([0, 1], X). Now we are in a position to prove the existence of the solutions for the boundary value problem (1) when the righthand side is convex-valued. Theorem 13. Assume that the nonresonance condition (6) holds. In addition, we suppose that (H1) F : [0, 1] × R → P(R) is Carathéodory and has nonempty compact and convex values; (H2) there exist a continuous nondecreasing function ψ : [0,∞) → (0,∞) and a function p ∈ L1([0, 1],R+) such that ‖F (t, x)‖P := sup { 󵄨󵄨󵄨y 󵄨󵄨󵄨 : y ∈ F (t, x)} ≤ p (t) ψ (‖x‖) for each (t, x) ∈ [0, 1] ×R; (14) (H3) there exists a constantM > 0 such that M( ψ (M) Γ (ξ) { (1 + P 1 ) ∫ 1 0 e −λ(1−s) p (s) ds

In this paper, we study the following boundary value problem: ( + )  () ∈  (,  ()) , 0 <  < 1,  − 1 <  ≤ ,  (0) = 0,   (0) = 0,   (0) = 0, . . .,  (−1) (0) = 0,  (1) =  () , where   is the Caputo fractional derivative,  is the ordinary derivative,  : [0, 1] × R → P(R) is a multivalued map, P(R) is the family of all subsets of R, 0 <  < 1,  is a positive real number, and  is a real number.The present work is motivated by a recent paper of the authors [14], where the problem (1) was considered for a single-valued case.The existence of solutions for the given multivalued problem is discussed for three cases: (a) convexvalued maps; (b) not necessarily convex-valued maps; (c) nonconvex-valued maps.To establish the existence results, we make use of nonlinear alternative for Kakutani maps, nonlinear alternative of Leray-Schauder type for singlevalued maps, selection theorem due to Bressan and Colombo for lower semicontinuous multivalued maps with nonempty closed and decomposable values, and a fixed point theorem for contractive multivalued maps due to Covitz and Nadler.The tools employed in this paper are standard; however, their exposition in the framework of the problem at hand is new.
The paper is organized as follows: in Section 2 we recall some preliminary facts that we used in the sequel.Section 3 contains the main results and an example.In Section 4, we summarize the work obtained in this paper and discuss some special cases.

Preliminaries
Let us recall some basic definitions of fractional calculus [2,4,6].Definition 1.For (−1)-times absolutely continuous function  : [0, ∞) → R, the Caputo derivative of fractional order  is defined as where [] denotes the integer part of the real number .
Definition 2. The Riemann-Liouville fractional integral of order  is defined as provided the integral exists.

Existence Results
We begin this section with some preliminary material on multivalued maps [24,25] that we need in the sequel.
Lemma 11 (nonlinear alternative for Kakutanimaps [26]).Let  be a Banach space,  a closed convex subset of ,  an open subset of , and 0 ∈ .Suppose that  :  → P , () is an upper semicontinuous compact map.Then either (i)  has a fixed point in , or (ii) there is a  ∈  and  ∈ (0, 1) with  ∈ ().
Lemma 12 (see [27]).Let  be a Banach space.Let  : [0, 1]× R → P , () be an  1 -Carathéodory multivalued map and let Θ be a linear continuous mapping from  1 ([0, 1], ) to ([0, 1], ).Then the operator Now we are in a position to prove the existence of the solutions for the boundary value problem (1) when the righthand side is convex-valued.
Theorem 13.Assume that the nonresonance condition (6) holds.In addition, we suppose that

is Carathéodory and has nonempty compact and convex values;
(H 2 ) there exist a continuous nondecreasing function  : (H 3 ) there exists a constant  > 0 such that where ) and ( 5)).

Then the boundary value problem (1) has at least one solution on [0, 1].
Proof.Define the operator for V ∈  , .We will show that Ω  satisfies the assumptions of the nonlinear alternative of Leray-Schauder type.The proof consists of several steps.As a first step, we show that Ω  is convex for each  ∈ ([0, 1], R).This step is obvious since  , is convex ( has convex values), and therefore we omit the proof.
Remark 14.The condition ( 3 ) in the statement of Theorem 13 may be replaced with the following one.
( 3 )  There exists a constant  > 0 such that where  1 is the same as defined in ( 3 ).

The Lower Semicontinuous Case.
As a next result, we study the case when  is not necessarily convex-valued.
Our strategy to deal with this problem is based on the nonlinear alternative of Leray-Schauder type together with the selection theorem of Bressan and Colombo [28] for lower semicontinuous maps with decomposable values.
Let  be a nonempty closed subset of a Banach space  and let  :  → P() be a multivalued operator with nonempty closed values. is lower semicontinuous (l.s.c.) if the set { ∈  : () ∩  ̸ = 0} is open for any open set  in .Let  be a subset of [0, 1] × R.  is L ⊗ B measurable if  belongs to the -algebra generated by all sets of the form J × D, where J is Lebesgue measurable in [0, 1] and D is Borel measurable in R. A subset A of  1 ([0, 1], R) is decomposable if, for all , V ∈ A and measurable J ⊂ [0, 1] = , the function  J + V −J ∈ A, where  J stands for the characteristic function of J. Definition 15.Let  be a separable metric space and let  :  → P( 1 ([0, 1], R)) be a multivalued operator.We say  has a property (BC) if  is lower semicontinuous (l.s.c.) and has nonempty closed and decomposable values.
Definition 16.Let  : [0, 1] × R → P(R) be a multivalued function with nonempty compact values.We say  is of lower semicontinuous type (l.s.c.type) if its associated Nemytskii operator F is lower semicontinuous and has nonempty closed and decomposable values.
Proof.It follows from ( 2 ) and ( 4 ) that  is of l.s.c.type.Then from Lemma 17, there exists a continuous function  : Consider the problem Observe that if  ∈  1 ([0, 1], R) is a solution of (32), then  is a solution to the problem (1).In order to transform the problem (32) into a fixed point problem, we define the operator Ω  as It can easily be shown that Ω  is continuous and completely continuous.The remaining part of the proof is similar to that of Theorem 13.So we omit it.This completes the proof.
Theorem 21.Assume that the nonresonance condition (6) holds.In addition, suppose that the following conditions hold: (36) Proof.Observe that the set  , is nonempty for each  ∈ ([0, 1], R) by the assumption ( 5 ), so  has a measurable selection (see Theorem III.6 [32]).Now we show that the operator Ω  , defined in the beginning of proof of Theorem 13, satisfies the assumptions of Lemma 20.To show that Ω  () ∈ As  has compact values, we pass onto a subsequence (if necessary) to obtain that V  converges to V in  1 ([0, 1], R).Thus, V ∈  , and, for each  ∈ [0, 1], we have Hence,  ∈ Ω().