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.