1. Introduction
Fractional calculus has emerged as an important area of investigation in view of the application of its tools in scientific and engineering disciplines. Examples include bio-medical sciences, ecology, finance, reaction-diffusion systems, wave propagation, electromagnetics, viscoelasticity, material sciences, and so forth. Fractional-order operators give rise to more informative and realistic mathematical models in contrast to their integer-order counterparts. It has been due to the non-local nature of fractional-order operators, which enables us to gain insight into the hereditary behavior (past history) of the associated phenomena. For examples and recent development of the topic, see References [
1,
2] and the references cited therein.
Differential inclusions—known as generalization of differential equations and inequalities—are found to be of great utility in the study of dynamical systems, stochastic processes, optimal control theory, and so forth. One can find a detailed account of the topic in Reference [
3]. In recent years, an overwhelming interest in the subject of fractional-order differential equations and inclusions has been shown, for instance, see References [
4,
5,
6,
7,
8,
9,
10,
11,
12,
13,
14] and the references cited therein.
In Reference [
15], the authors obtained some existence results for sequential neutral differential equations involving Hadamard derivatives:
	  where 
 are the Hadamard fractional derivatives of order 
 respectively and 
 are continuous functions, 
 and 
In this paper, we cover the multivalued case of problem (
1) and investigate the Hadamard type neutral fractional differential inclusions given by
      
      where 
 is a multivalued map, 
 represents the family of all nonempty subsets of 
 and the other quantities in (
2) are the same as taken in (
1). Here 
 is an element of the Banach space 
 equipped with norm 
 and is defined by 
 where 
y is a function defined on 
 and 
 The standard fixed point theorems for multivalued maps are applied to establish the existence results for the problem (
2).
The remaining content of the paper is composed as follows. In 
Section 2, we describe the necessary background material needed for our work. 
Section 3 deals with the main theorems. In 
Section 4, we construct illustrative examples for the obtained results.
  3. Existence Results
For a normed space , we define ,  and  In passing, we remark that a closed and bounded set in a metric space is not necessarily compact in general; however, it is true that a set in a metric space of real or complex numbers is compact if and only if it is closed and bounded.
For each 
, define the set of selections of 
F by
      
Denote by  the Banach space of all continuous functions from J into  endowed with the norm   represents the space of functions  such that 
Our first existence result deals with the case when 
F has convex values and is based on nonlinear alternative for Kakutani maps [
16] with the assumption that the multivalued map 
F is Carathéodory.
Definition 3. (Granas, Dugundji [
16]). 
A multivalued map  is said to be Carathéodory if-  is measurable for each ; 
-  is upper semicontinuous for almost all  
Further a Carathéodory function F is called  Carathéodory if
 for each , there exists  such thatfor all  with  and for almost everywhere   Theorem 1. Assume that:
- there exists a non-negative constant  such that 
-  is -Carathéodory; 
- there exists a continuous non-decreasing function  and a function  such that 
- there exists a constant  such thatwhere  
Then the problem (2) has at least one solution on   Proof.  Let us first transform the problem (
2) into a fixed point problem by introducing an operator 
 by
        
        for 
 It is obvious by Lemma 1 that the fixed points of the operator 
 are solutions of the problem (
2).
We verify the hypothesis of nonlinear alternative for Kakutani maps [
16] in several steps.
Step 1.  is convex for each  It directly follows from the fact that  is convex (F has convex values).
Step 2.  maps bounded sets (balls) into bounded sets in  Let 
 be a bounded set in 
. Then, for each 
, there exists 
 such that
            
 Then, for 
 we have
        
		Thus,
        
Step 3.  maps bounded sets into equicontinuous sets of 
Let 
 with 
 and 
 Then, for each 
 we obtain
        
        which tends to zero as 
 independently of 
 For the cases 
 and 
, the equicontinuity can be established in a similar manner. Thus, by Arzelá-Ascoli theorem [
17], we deduce that 
 is completely continuous.
Now we show that 
 has a closed graph. Then it will follow by the Proposition 1.2 in Reference [
18] that 
 is upper semi-continuous, as it is already proved to be completely continuous.
Step 4.  has a closed graph. We need to show that 
 when 
 and 
 Associated with 
 there exists 
 such that, for each 
			Thus it suffices to show that there exists 
 such that, for each 
 Let us introduce the linear operator 
 given by
        
		Notice that 
 as 
 Therefore, it follows from a result dealing with the closed graph operators derived in Reference [
19] that 
 is a closed graph operator. Further, we have 
 Since 
 we have
        
        for some 
.
Step 5. We can find an open set  with  for any  and all 
Let 
 and 
 Then there exists 
 with 
 such that for 
,
        
        which implies that
        
By 
, there exists a real number 
 such that 
. Let us consider an open set
        
        with 
 Notice that 
 is compact and upper semi-continuous multivalued map with convex closed values. The choice of 
U implies that there does not exist any 
 satisfying 
 for some 
. In consequence, we deduce from the nonlinear alternative for Kakutani maps [
16] that 
 has a fixed point 
 which corresponds to a solution to the problem (
2). This finishes the proof. ☐
 In the following result, we make use of the nonlinear alternative for contractive maps ([
20] Corollary 3.8) to show the existence of solutions for the problem (
2).
Lemma 2. (Nonlinear alternative [20]) Let D be a bounded neighborhood of  where X is a Banach space. Let  and  be multivalued operators such that (a)  is contraction, and (b)  is upper semi-continuous and compact. Then, if  either (i) G has a fixed point in  or (ii) there is a point  and  with .  Theorem 2. If the conditions  of Theorem 1 hold, then there exists at least one solution for the problem (2) on   Proof.  In order to verify the hypotheses of Lemma 2, we introduce the operator 
 by
        
        and the multivalued operator 
 by
        
        for 
 Observe that 
 where 
 is defined by (
5). In the first step, it will be established that the operators 
 and 
 define the multivalued operators 
 where 
 is a bounded set in 
. Let us show that 
 is compact-valued on 
. Observe that the operator 
 is equivalent to the composition 
, where 
 is the continuous linear operator on 
 into 
, defined by
        
		Let 
 be arbitrary and let 
 be a sequence in 
. Then it follows by the definition of 
 that 
 for almost all 
. As 
 is compact for all 
, we have a convergent subsequence of 
 (we denote it by 
 again) that converges in measure to some 
 for almost all 
. On the other hand, 
 is continuous, so 
 pointwise on 
J.
The convergence will be uniform once it is shown that 
 is an equicontinuous sequence. For 
 with 
, we have
        
        as 
 which shows that the sequence 
 is equicontinuous. As a consequence of the Arzelá-Ascoli theorem, there exists a uniformly convergent subsequence of 
 (we denote it again by 
) such that 
. Noting that 
, we deduce that 
 is compact for all 
. So 
 is compact.
Now, we show that 
 is convex for all 
. Let 
. We select 
 such that
        
        for almost all 
. Then
        
        where 
. Since 
 is convex (as 
F has convex values), 
. Thus 
 which shows that 
 is convex-valued.
On the other hand, it is easy to show that 
 is compact and convex-valued. Next we prove that 
 is a contraction on 
. For 
, we have
        
        which implies that 
 By the assumption 
 we conclude that 
 is a contraction.
As in the proof of Theorem 1, it can easily be shown that the operator  is compact and upper semi-continuous.
In view of the foregoing steps, we deduce that 
 and 
 satisfy the hypothesis of Lemma 2. So, from the conclusion of Lemma 2, either condition (i) or condition (ii) holds. We show that conclusion (ii) is not possible. If 
 for 
 then there exist 
 such that
        
		By our assumptions, we can obtain
        
		Thus
        
		If condition (ii) of Lemma 2 is satisfied, then there exists 
 and 
 with 
 Then, 
y is a solution of (
2) with 
 Now, by the inequality (
8), we get
        
        which contradicts 
 Hence, 
 has a fixed point on 
 by Lemma 2, which implies that the problem (
2) has a solution. The proof is complete. ☐
 Our next result deals with the non-convex valued map in the problem (
2) and is based on Covitz and Nadler’s fixed point theorem [
21] (
If  is a contraction, then  where X is a metric space).
For a metric space 
 induced from the normed space 
 it is argued in Reference [
22] that 
 is a metric space, where 
 is defined by 
  and 
.
Definition 4. (Granas, Dugundji [
16]). 
A multivalued operator  is called- (a) 
-  Lipschitz if and only if there exists  such that 
- (b) 
- a contraction if and only if it is  Lipschitz with . 
 Theorem 3. Assume that  and the following conditions hold:
-  is such that  is measurable for each . 
-  for almost all  and  with  and  for almost all . 
Then there exists at least one solution for the problem (2) on  provided that  Proof.  Observe that the set 
 is nonempty for each 
 by the assumption 
 Therefore 
F has a measurable selection (see Theorem III.6 [
23]). Next we consider the operator 
 given by (
5) and verify that it satisfies the hypothesis of the Covitz and Nadler theorem [
21]. We show that 
 for each 
 Let 
 be such that 
 in 
 Then 
 and we can find 
 such that, for each 
,
        
		Since 
F has compact values, we pass onto a subsequence (if necessary) such that 
 converges to 
 in 
 So 
 and for each 
, we have
        
		Hence, 
Next we prove that there exists 
 (
 is defined by (
9)) such that
        
		Let 
 and 
. Then there exists 
 such that, for each 
,
        
		By 
, we have
        
		So, there exists 
 such that
        
Define 
 by
        
		By Proposition III.4 in Reference [
23], it follows that the multivalued operator 
 is measurable. So we can find a measurable selection 
 for 
V. So 
 and satisfying 
 for each 
For each 
, we define
        
		Thus,
        
		Hence
        
		On the other hand, interchanging the roles of 
y and 
 leads to
        
		So 
 is a contraction. Therefore, from the conclusion of Covitz and Nadler theorem [
21], the operator 
 has a fixed point 
y which is indeed a solution of the problem (
2). This finishes the proof. ☐
 Finally, we prove an existence result by applying the multivalued version of Krasnoselskii’s fixed point theorem [
24], which is stated below.
Lemma 3. (Krasnoselskii [
24]). 
Let X be a Banach space,  and  be multivalued operators satisfying the conditions: (i)  for all ; (ii)  is contraction; and (iii)  is upper semicontinuous and compact. Then there exists  such that . Theorem 4. Suppose that ,  and the following assumption are satisfied
 there exists a function  such thatThen there exists at least one solution for the problem (2) on   Proof.  Let us consider the operators 
 and 
 defined by (
6) and (
7) respectively. As in Theorem 2, one can show that 
 are indeed multivalued operators, where 
 is a bounded set in 
 Moreover, 
 is a contraction on 
 and 
 is upper semi-continuous and compact.
Next we show that 
 for all 
 Let 
 and suppose that
        
		For 
 and 
, we have
        
		With the given assumptions, one can obtain
        
		Thus
        
        which means that 
 for all 
Thus, the operators 
 and 
 satisfy the hypothesis of Lemma 3 and hence its conclusion implies that 
 in 
 Therefore the problem (
2) has a solution in 
 and the proof is finished. ☐
   4. Examples
In this section, we demonstrate the application of our main results by considering the following Hadamard type neutral fractional differential inclusions:
	  Here 
	  With the given data, it is easy to see that 
 is satisfied with 
, 
 is satisfied with 
 and 
 holds true for 
 with a particular choice of 
 Thus all the conditions of Theorem 1 hold true. Hence the problem (
10) and (
11) has at least one solution on 
 by the conclusion of Theorem 1. In a similar manner, one can check that the hypotheses of Theorem 2 hold with 
 and consequently the conclusion of Theorem 2 applies to the problem (
10) and (
11).
In order to illustrate Theorem 3, let us take
      
      in (
10). Then 
 and from the condition (
9), 
. Clearly the hypothesis of Theorem 3 is satisfied. Therefore, there exists at least one solution for the problem (
10) and (
11) with 
 given by (
12) on