1. Introduction
Impulsive differential equations and inclusions describe phenomena in which states are changing rapidly at certain moments. In [
1,
2,
3,
4,
5,
6,
7,
8], the authors examined whether a mild solution for different types of impulsive differential inclusions exist.
The study of neutral differential equations appears in many applied mathematical sciences, such as viscoelasticity and equations that describe the distribution of heat. The structure of neutral equations involve derivatives related to delay beside the function. Neutral differential equations and inclusions were studied in [
9,
10,
11,
12]. These papers examined the mild solutions and controllability of the system.
Because the set of mild solutions for a differential inclusion having the same initial point may not be a singleton, many authors are interested in investigating the structure of this set in a topological point of view. An important aspect of such structure is the 
-property, which means that the homology group of the set of mild solutions is the same as a one-point space. We list some studies in which the authors demonstrated the solution sets satisfying 
-property: Gabor [
13] considered impulsive semilinear differential inclusions with finite delay on the half-line of order one generated by a non-compact semi-group; Djebali et al. [
14] worked on impulsive differential inclusions on unbounded domains; Zhou et al. [
15] studied the neutral evolution inclusions of order one generated by a non-compact semi-group; Zhou et al. [
16] considered fractional stochastic evolution inclusions generated by a compact semi-group; Zhao et al. [
17] studied a stochastic differential equation of Sobolev-type which is semilinear with Poisson jumps of order 
Beddani [
18] examined a differential inclusion involving Riemann–Liouville fractional derivatives; Wang et al. [
19] worked on semilinear fractional differential inclusions with non-instantaneous impulses; Ouahab et al. [
20] considered fractional inclusions that are non-local and have impulses at different times; Zaine [
21] studied weighted fractional differential equations. Recently, Zhang et al. [
22] proved that the set of 
-solutions for impulsive evolution inclusions of order one is an 
-set and generated by 
m–dissipative operator. Wang et al. [
23] proved that the solution for evolution equations that have nonlinear delay and multivalued perturbation on a non-compact interval is an 
-set.
In [
6,
24,
25,
26], the authors studied different kinds of fractional differential inclusions, and, in all cases, they showed that the set of solutions is a compact set. For more work related to this, the reader can consult the book in [
27] about the topological properties for evolution inclusions.
However, up to now, proving that the solution set for fractional impulsive neutral semilinear differential inclusions involving delay and generated by a non-compact semigroup is an -set has not been considered in the literature. Thus, this topic is new and interesting and, hence, the question whether there exists a solution set carrying an -structure remains unsolved for fractional differential inclusions when there are impulses, delay (finite or infinite) and the operator families generated by the linear part lack compactness. Therefore, our main goal is to give an affirmative answer to this question. In fact, we study a neutral fractional impulsive differential inclusion with delay which is generated by a non-compact semigroup, and we show that the set of solutions is non-empty and equal to an intersection of a decreasing sequence of sets each of which is non-empty compact and has a homotopy equivalent to a point.
Let ,  a semigroup on E, which is Banach space, and the infinitesimal generator of  Let  be a multifunction, , and be given. For every , let , ; where and  are defined later.
The present paper shows the solution set of a fractional neutral impulsive semilinear differential inclusion with delay having details as follows:
      is not empty, compact and an 
-set, where 
, and
  are the limits of the function 
 evaluated at 
 from the right and the left. Furthermore, 
 denotes the Caputo derivative that has order 
 and lower limit at zero [
28].
In the following points, we clarify the originality, importance and the main contributions of this article:
Up to now, proving that the solution set is an -set for fractional impulsive neutral semilinear differential inclusions involving delay and generated by a non- compact semigroup has not been considered in the literature.
Demonstrating that the set of solutions is an -set for fractional neutral differential inclusions involving impulses and delay has not been considered yet.
We do not assume that the semi-group which generates the linear part is compact.
Proving that the set of solutions is an 
-set for neutral differential inclusions (without impulses) with a finite delay, 
, and generated by a non-compact semigroup, has been investigated in [
15], while stochastic neutral differential inclusions (without impulsive effects) with finite delay of order 
 and generated by a compact semigroup has been examined in [
16].
Gabor [
13] considered Problem 
1 on the half-line when 
and 
Problem 
1 is investigated in [
19] when 
 and in the absence of delay.
Our technique can be used to derive suitable conditions, which implies that the solution set is an 
-set for the problems studied in [
13,
14,
15,
16,
17,
18,
19,
20,
21,
22,
23] when they contain impulses and delay.
In order to clarify the difficulties encountered to achieve our aim, we point to the normed space 
,which consists of piecewise continuous bounded functions defined on 
 with a finite number of discontinuity points and is left continuous at the discontinuity points, and is not necessarily complete. Moreover, unlike the Banach spaces 
 and 
, the Hausdorff measure of noncompactness on 
 is not specific. Thus, when the problem involves delay and impulses, we cannot consider 
 as the space of solutions. To overcome these difficulties, a complete metric space 
H is introduced as the space of mild solutions (see the next section). In addition, the function 
is not necessarily measurable (see Remark 
1, and so, a norm different from the uniform convergence norm is introduced (see Equation (
2) below).
For recent contributions on neutral differential inclusions of fractional order, Burqan et al. [
29] give a numerical approach in solving fractional neutral pantograph equations via the ARA integral transform. Ma et al. [
30] studied the controllability for a neutral differential inclusion with Hilfer derivative, and Etmad et al. [
31] investigated a neutral fractional differential inclusion of Katugampola-type involving both retarded and advanced arguments.
For more recent papers we cite [
32,
33,
34].
The sections of the paper are organized as follows: We include some background materials in 
Section 2 as we need them in the main sections. 
Section 3 is assigned for proving that the solution set of Problem (1) is non-empty and compact. In 
Section 4, we show that this set is an 
-set in the complete metric space 
H. In 
Section 5, e give an example as an application of the obtained results. 
Section 6 and 
Section 7 are the discussion and conclusion sections.
  2. Preliminaries and Notation
In all the text we denote for the set of mild solutions for Problem 
1 by 
and by 
 to the quotient space consisting of 
valued Bohner integrable functions defined on 
J having the norm 
. Let 
 be non-empty, convex and compact}.
Definition 1. (Ref. [35]) Let ,   a semigroup and be the infinitesimal generator of it. A continuous function  is called a mild solution for the problem:ifwhere   and .
 Lemma 1. (Ref. [35] (lemma 3.1)) The properties stated below are held: -  (i) 
 For every fixed  are linear and bounded.
-  (ii) 
 Assuming  we have that for any and 
-  (iii) 
 If ; then for any  
 Consider the spaces:
The normed space
          
          endowed with the norm:
          
The Banach space
          
          where 
, 
,  and 
.
The complete metric space
          
          where the metric function is given by:
          
The Banach space
          
          together with the norm 
The Hausdorff measure of noncompactness on a Banach space 
 is given by
      
      where
is a bounded subset of 
is the Hausdorff measure of noncompactness on the Banach space 
 and
      
The Hausdorff measure of noncompactness on 
 is defined by:
      where
is a bounded subset of
Remark 1. Since the function is not necessarily measurablewe do not consider the uniform convergence norm to be the norm defined on the space Θ (see Example 3.1, [36]). Therefore, the multivalued superposition operatorwould not be well defined. Therefore, we consider a norm defined by Equation (2).  Definition 2. A function  is said to be a mild solution for  if
      where 
.
  We assume the following conditions:
is the infinitesimal generator of T, 0 is an element of the resolvent of  and , where
 where:
For any  the multifunctionhas a measurable selection, and for the multifunctionis upper semicontinuous.
 There exists a 
 satisfying
      
 There is a 
such that, for any 
 that is bounded, we have
      
 For any 
, the function 
is continuous, and there are 
 and 
 satisfying 
and for any bounded subset 
Lemma 2. (Ref. [37]) Under condition for any, the fractional power  can be defined, and it is linear and closed on its domain . In addition, the following properties are satisfied: -  (i) 
 is a Banach space with the norm -  (ii) 
 For any , , we have  and, assuming , we get .
-  (iii) 
 For every ,  is bounded on and there is a constant  such that -  (iv) 
  is a bounded linear operator on E.
-  (v) 
 
We need the next lemmas in order to prove our main results.
 Lemma 3. Assume  to be bounded, closed and convex,  is a single-valued function,  is a multifunction, and for any  Suppose that
-  (a) 
  is a contraction with the contraction constant 
-  (b) 
  is a closed and completely continuous multifunction.
 Then, the fixed point set of is not empty. Moreover, the set of fixed points for  is compact if it is bounded.
Proof.   is continuous on 
W since it is a contraction and, hence, it follows by the closeness of 
 that the multifunction 
 is closed. We show that 
R is 
condensing, where 
 is the Hausdorff measure of noncompactness on 
E. Let 
be a bounded set of 
W. Since 
 is a contraction with the contraction constant 
k, we get 
, where 
 is the Kuratowski measure of noncompactness on 
E. Because 
 is compact, 
. Therefore,
        
This means that 
R is 
condensing. By Proposition 3.5.1 in [
38], the fixed point set of 
is not empty. The second part follows from Proposition 3.5.1 in [
38].    □
   3. The Compactness of 
In this section, we show that the set of mild solutions for Problem 
1 is nonempty and compact.
For any 
 with 
 let 
be defined by
      
Lemma 4. For any , the function  is continuous from to Θ.
 Proof.  Assume 
. Then,
        
Because  is continuous on except for a finite number of points, it follows that , a.e. Since , , and the proof is completed.      □
 Theorem 1. Assume that  are held and that  is equicontinuous. Assume also that the following conditions are satisfied.
 The function  is continuous and there exists a  satisfying  and
- (i)
 For anyis strongly measurable.
- (ii)
 There are 
 and 
 with
          
          and
          
Then, 
is not empty and a compact subset of 
H provided that
      
      and
      
      where 
 and 
.
Proof.  A multioperator 
is defined as the following: let 
, hence, as a consequence of 
, the multifunction 
admits a measurable selection which, by 
, belongs to 
, and, therefore, 
 can be defined by
        
        where 
fand 
is defined by (
8).
We show that a point 
is a fixed point for 
 if and only if 
. Assume 
 is a fixed point to 
. Hence,
        
Therefore,
        
        which means that 
satisfies (
3), and, thus, it is a mild solution for problem (
1). In a similar way, it can be seen that if 
satisfies (
3), then 
is a fixed point for 
. Let 
  and 
 be such that
        
        and a function 
if and only if
        
        where 
f Notice that 
. Let 
        and 
 be a positive real number satisfying
        
Put 
. Due to (
12), 
is well defined. The rest of the proof is divided in the following steps:
Step 1. This step shows that 
. Let 
 and 
 There exists 
 where
        
Let 
For every 
, we get
        
        which implies that 
. So, by (ii) of Lemma 1, and the Holder inequality, it follows that
        
Then, from (
6), (
7), (
10)
one has, for 
This equation with (
12) leads to
        
Then, .
Step 2.  is a contraction with a contraction constant 
Let 
 and 
. Then, 
. From (
6), (
7) and (
11), for every 
and any 
, we have that
        
        which yields with (9) that 
 is a contraction with a contraction constant 
Step 3. has a closed graph and  is compact.
Assume 
 and 
 are sequences in 
 where 
 and 
. Then,
        
        where 
. Using 
, it yields that
        
So, 
 is bounded in 
 and, hence, there exists a subsequence of 
. We denote them by 
 where 
. From Mazur’s Lemma, there exists a sequence of convex combination, 
 of 
 that converges almost everywhere to 
Note that by
, again, for any
 and any 
Note that by (
18), 
, 
Moreover, since 
, 
 is upper semicontinuous, it yields 
 Therefore, from the continuity of 
, and by taking the limit of (
19) as 
, one gets 
To prove that the values of  are compact, assume  and  Using similar arguments to the above, we get that  has a convergent subsequence . So, is relatively compact. Since the graph of  is closed its values are closed and, hence, is relatively compact in 
Step 4. We claim that the subsets 
 are equicontinuous, where
        
Assume 
. Then, there exists 
and 
, where, for 
        and 
.
Case 1. Let 
be two points in 
.Then,
        
The hypothesis 
implies 
 and, hence, by Lemma 
1, we get
        
Note that 
 then, for 
we have 
. As an application of Lemma 3 in [
8] and considering 
, we get
        
Observe that for every 
,
        
Moreover, since 
is equicontinuous, and, using the Lebesgue-dominated convergence theorem, one gets
        
Case 2. 
, 
. Assume 
, 
 and 
 where 
 Hence, as above, it can be shown that
        
Then,  are equicontinuous.
Step 5. Set 
 and 
,
. Then, the sequence 
 is a decreasing sequence of not empty, closed and bounded subsets of 
. So, the set 
 is bounded, closed, convex and 
. Next, we show that 
is compact. According to the generalized Cantor’s intersection property, we only need to prove that
        
        where 
 is the Hausdorff measure of noncompactness on 
. Assume 
 and 
 are fixed. From the fact that
 is a contraction with a contraction constant 
, it follows that
        
Let 
. Using Lemma 5 in [
39], there is a 
 in
with
        
From the fact that the subsets 
 are equicontinuous, one obtains
        
Now, let 
 and 
, 
.Then, for every 
 there is a 
such that, for any 
,
        
Note that the assumption 
 implies that for 
Moreover, from (
4), we have that for 
Again, by 
 for every 
and for almost 
and, hence, 
 is integrably bounded. As a consequence of Lemma 4 in [
40], there is a compact set 
 a measurable set 
 having a measure less than 
and 
 such that for every 
and
        
Then, by (
24) and (
25) and Minkowski’s inequality, it follows that for 
        and
        
Moreover, from the fact that 
is contained in a compact subset, we get
        
Combining this relation with (
26) and (
27),it follows that
        
        where 
. Using the fact that 
 is chosen arbitrary, relation (
28) becomes
        
Using the above inequality and (
21)–(
23), in addition to the fact that 
 is arbitrary, it follows that
        
The above inequality holds for any natural number 
n, and by (
13) together with taking the limit as 
 we get (
20). Then, 
is not empty and a compact subset of 
. So, 
 is completely continuous. By applying Lemma 3, we conclude that the fixed points set of 
 is not an empty subset of 
. Furthermore, by arguing as in Step 1, we can prove that the set of fixed points of 
is bounded and, hence, by Lemma 3, it is compact in 
. Therefore, the set 
 is not empty and a compact subset of 
H.      □
   4. The Structure Topological of 
In the section we prove that  is an -set
Definition 3 ([
41])
. A topological space X, which is homotopy equivalent to a point, is called contractible. In other words, there is a continuous map  and  Lemma 5 ([
41])
. Let , where A is not empty and X is a complete metric space. Then, A is said to be -set if and only if it is an intersection of a decreasing sequence of contractible sets and  as  Now, consider the multi-valued function 
 that is given by:
      where 
is defined by (
17). Since 
on 
the set of solutions consisting of mild solutions for Problem (1) is equal to the set of solutions consisting of mild solutions for the problem:
Obviously, 
verifies 
 and, for 
Then, we can assume that verifies the next condition:
 There exists a function 
, where for every 
We recall the next Lemma. For its proof, we refer the reader to the second step in the proof of Theorem 3.5 in [
13].
Lemma 6. Assume that  and  are satisfied. Then, there exists a sequence of multifunctions  with  such that:
-  (i) 
 Every is continuous for almost .
-  (ii) 
 , for each and 
-  (iii) 
 -  (iv) 
 For all there is a selection of such that  is measurable for each  and for  is locally Lipschitz.
 Remark 2. (Ref. [19]) The property (iv) in Lemma 6 implies that, for almost ,,  is continuous.  Assume 
is the mild solutions set of the following fractional neutral impulsive semilinear differential inclusions with delay:
Theorem 2. Assume that the conditions in Theorem 1 after substituting  by are held. Then, there exists  such that, for , the set  is compact and not empty in 
 Proof.  Let 
be a fixed natural number. We define a multioperator 
as the following: 
if and only if
          
          where 
f. Due to Lemma 5, 
verifies 
, 
. As a result of Theorem 1, 
 is closed, 
 and 
 is equicontinuous. Set 
 and 
,
. As in Theorem 1, the sequence 
 is a decreasing sequence of non-empty, closed and bounded subsets of 
We show that
          
Let 
. Choose a natural number 
with 
and let 
 be a fixed natural number. Using a similar argument as the one used in the proof of Theorem 1, one gets
          
where 
and 
 Next, due to Remark 4.2 in [
7], it follows that for any bounded subset 
,
          
Then, it yields from (ii) in Lemma 5 and (
31), for
As in (28) but by using (
32) instead of (
24), we get
          
Similarly, as in the proof of Theorem 1, we confirm the validity of (
30).Therefore, by the generalized Cantor’s intersection property, the set 
is not empty and compact in 
. As in Theorem 1, the fixed points set of the multivalued function 
is not empty and a compact subset in 
 Consequently, the set 
is not empty and a compact subset of H.      □
 Theorem 3. Under the conditions of Theorem 2, 
 Proof.  In view of (iii) in Lemma 8, it can be seen that 
. Let 
. Then, there is 
 such that
          
          It follows from 
 that
          
		  This means that the sequence 
 is weakly relatively compact in 
, so we can assume 
 weakly, where 
As in the proof of Theorem 1, there is a sequence of convex combinations 
 of 
that converges almost everywhere to f. Note that
          
         and 
. It yields, from (ii) of Lemma 8, that for almost 
          which implies that 
 for 
Moreover, using the fact that 
 is continuous, and taking the limit as 
in (
34), one gets
          
         This means that 
      □
 To prove our main results, we need the next lemma.
Lemma 7 ([
19], Lemma 4.5). 
Assume that andare two metric spaces.Then, if is locally Lipschitz, then it is Lipschitz on all subsets of X that are compact. Theorem 4. Under the assumptions of Theorem 2the set is an -set in H provided that 
 Proof.  Using Lemma 4 and Theorems 1–3, we only need to prove that 
 is contractible. Assume that 
 and 
. Consider the following fractional neutral impulsive semilinear:
        
Using Lemma 6 and Remark 3, 
 is measurable, and for 
 is continuous. Since the multi-valued 
F satisfies 
and 
, then, following the arguments employed in the proof of Theorem 2, the fractional differential Equation (
35) has a mild solution 
∈
satisfying the following integral equation:
        
Next, we show that the solution is unique. Assume that 
 is another mild solution for (
35). Then,
        
Let 
 be fixed. Due to (
6), (
7), (
11) (
36) and (
37), it yields
        
Now, from Lemma 5, the function 
 is continuous from 
to 
 and, hence, the subset 
 is compact in
. Similarly, the set 
 is compact in
 and, therefore, the set 
 is compact in
, and consequently, 
 is compact in 
. Thus, by (iv) in Lemma 6 and Lemma 7, there exists 
, for which the estimate
        
        holds for 
. Therefore, from (
38), it yields
        
Note that when 
, we have
        
Since 
 and 
 are continuous on 
, there is 
with 
 Then,
        
Since 
, the last relations lead to
        
Using the generalized Gronwall inequality [
42], one has 
. Since 
 is arbitrary, we conclude that 
 on 
.
Next, let 
 be fixed. Note that 
. Then,
        
By repeating the arguments employed above, we get  on  Continuing with the same processes, we arrive to  on 
Next, we prove that 
is homotopically equivalent to 
 To this end, we define a continuous function 
, where 
and 
. Assume 
 is fixed. Then, there exists a 
 such that
        
Consider the partition { for . We consider the following cases:
(i) 
. Put 
 The following fractional neutral differential inclusion is a result of the above discussion:
        
        has a unique mild solution 
 satisfying the next integral equation:
        
Note that 
(ii) 
Put 
. Again, the following fractional neutral differential inclusion:
        
        has a unique mild solution 
 and
        
We continue up to 
step. That is 
 and put 
Let 
 be the unique mild solution for the impulsive fractional neutral differential inclusion:
        
Note that 
 and
. Now, we define 
at 
 as
        
Therefore,  and 
It remains to clarify the continuity of 
 Let 
. Let 
. Then, by (
42), 
. Let 
. So, 
and 
, where
        
        and
        
 and 
. Obviously, 
 and, hence, by (
43) and (
44), and by arguing as above, we get
        
        which implies the continuity of 
 when 
Similarly, we can show the continuity of 
 and consequently, 
is contractible. This completes the proof.      □
   5. Example
Example 1. Assume that  and . For any , we denote by  the value of  at ω. Let ,  and domain  be defined as  Using [
37], there is a compact analytic semi-group 
 generated by 
A and
      
      where 
 is the orthonormal set of eigenvalues of 
In addition, for all 
 one gets
      
So, 
 Furthermore, for each 
      and 
. The domain of 
 is defined as
      
Let 
be such that
      
      where 
. We have
      
      and
      
Then, (
10) and (
11) are satisfied with
.
Let
be a convex compact subset in 
E, 
 and 
. Define 
by
      
Moreover, for any bounded subset  we have, where  and, hence, . Then, satisfies and with 
Next, let
      
      where 
 is a positive number. Obviously, 
verifies 
 with 
Therefore, by applying Theorems 1 and 4, the set of solutions for the following fractional neutral impulsive semilinear differential inclusions with delay:
      is a not empty, compact and an 
-set provided that
      
      and
      
      where 
are given by (
45)–(
47). By choosing 
 and 
 small enough and 
 large enough, we arrive to (
50) and (
51).
Example 2. Let  and ϱ be as in Example (1) and be a fixed element.  Let 
be such that
      
      where 
is measurable,
,
 is measurable, 
 and 
.
Next, let 
, 
 where 
 is a continuous function. Then, by choosing 
 and 
 small enough, one can show that 
and 
satisfy all assumptions of Theorems 2 (see [
15,
43]) and, hence, the set of mild solutions for the partial differential inclusions of impulsive neutral type with delay:
      is an 
-set.