Weak Solutions for Partial Random Hadamard Fractional Integral Equations with Multiple Delays

Abstract

We present some results concerning the existence of weak solutions for some functional integral equations of Hadamard fractional order with random effects and multiple delays by applying Mönch’s and Engl’s fixed point theorems associated with the technique of measure of weak noncompactness.

1. Introduction

Random differential equations arise in many applications and have been studied in the literature on bounded as well as unbounded internals of the real line for different aspects of the solution. See, for example, [1]. We refer the reader to the monograph [2] and the papers [1, 3] and the references therein. There are real-world phenomena with anomalous dynamics such as signals transmissions through strong magnetic fields, atmospheric diffusion of pollution, network traffic, and the effect of speculations on the profitability of stocks in financial markets, where the classical models are not sufficiently good to describe these features. In this case, the theory of fractional differential equations is a good tool for modeling such phenomena. Therefore, the study of the fractional differential equations with random parameters seems to be a natural one. For some fundamental results in the theory of fractional calculus and fractional differential equations, we refer the reader to the monographs of Abbas et al. [4, 5], Baleanu et al. [6], and Kilbas et al. [7] and a series of recent research articles [8–12] and the references therein.

The measure of weak noncompactness is introduced by De Blasi [13]. The strong measure of noncompactness was developed first by Banaś and Goebel [14] and subsequently developed and used in many papers; see, for example, Akhmerov et al. [15], Alvàrez [16], and Guo et al. [17] and the references therein. In [18], the authors considered some existence results by applying the techniques of the measure of noncompactness. Recently, several researchers obtained other results by application of the technique of measure of weak noncompactness; see [5]. Existence of random solutions for functional differential and integral equations has extensively been studied in various papers; see [19, 20] and the references therein.

In this paper, we discuss the existence of random solutions for the partial Hadamard fractional integral equation of the following form:where , , , (), , , , , (), and are given continuous functions, is a measurable space, and is a real (or complex) Banach space with norm and dual , such that is the dual of a weakly compactly generated Banach space , is the left-sided mixed Hadamard integral of order , and is a given continuous and measurable function such that

2. Preliminaries

Let be the Banach space of all continuous functions from into with the supremum (uniform) norm . By , we denote the Banach space of measurable functions which are essentially bounded equipped with the norm Denote by the Banach space with its weak topology.

Definition 1. A Banach space is called weakly compactly generated (WCG, in short) if it contains a weakly compact set whose linear span is dense in

Definition 2. A function is said to be weakly sequentially continuous if takes each weakly convergent sequence in to a weakly convergent sequence in (i.e., for any in with in , one has in ).

Definition 3 (see [21]). The function is said to be Pettis integrable on if and only if there is an element corresponding to each such that for all , where the integral on the right-hand side is assumed to exist in the sense of Lebesgue (by definition,

Let be the space of all -valued Pettis integrable functions on , and let be the Banach space of Lebesgue measurable functions Define the class by The space is normed by where stands for a Lebesgue measure on

The following result is due to Pettis (see [21], Theorem  3.4 and Corollary  3.41).

Proposition 4 (see [21]). If and is a measurable and essentially bounded -valued function, then

For all what follows, the sign “” denotes the Pettis integral.

Let us recall the definitions of Pettis integral and Hadamard integral of fractional order.

Definition 5 (see [7]). The left-sided mixed Pettis Hadamard integral of order , for a function , is defined as

Remark 6. Let For every , one has

Definition 7. Let , , and For , define the left-sided mixed Pettis Hadamard partial fractional integral of order by the expression

Let be the -algebra of Borel subsets of . A mapping is said to be measurable if, for any , one has

To define integrals of sample paths of random process, it is necessary to define a jointly measurable map.

Definition 8. A function is called jointly measurable if, for any , one has where is the direct product of the -algebras and , those defined in and , respectively.

Lemma 9 (see [22]). A function is jointly measurable if is measurable for all and is continuous for all

Definition 10. A function is called random Carathéodory if the following conditions are satisfied: (i)The map is jointly measurable for all .(ii)The map is continuous for all and

Let be a mapping. Then is called a random operator if is measurable in for all and it is expressed as In this case, we also say that is a random operator on A random operator on is called continuous (resp., compact, totally bounded, and completely continuous) if is continuous (resp., compact, totally bounded, and completely continuous) in for all The details of completely continuous random operators in Banach spaces and their properties appear in Itoh [23].

Definition 11 (see [24]). Let be the family of all nonempty subsets of and let be a mapping from into A mapping is called random operator with stochastic domain if is measurable (i.e., for all closed , is measurable) and for all open and all , is measurable. will be called continuous if every is continuous. For a random operator , a mapping is called random (stochastic) fixed point of if for almost all , , and and for all open , is measurable.

Definition 12 (see [13]). Let be a Banach space, let be the bounded subsets of , and let be the unit ball of De Blasi measure of weak noncompactness is the map defined by

De Blasi measure of weak noncompactness satisfies the following properties:(a).(b) is weakly relatively compact.(c).(d) ( denotes the weak closure of ).(e).(f).(g).(h).

The next result follows directly from the Hahn-Banach theorem.

Proposition 13. Let be a normed space, and with Then, there exists with and .

For a given set of functions , let us denote that

Lemma 14 (see [17]). Let be bounded and equicontinuous. Then the function is continuous on , and where ; , and is De Blasi measure of weak noncompactness defined on the bounded sets of

We will need the following fixed point theorems.

Theorem 15 (see [25]). Let be a nonempty, closed, convex, and equicontinuous subset of a metrizable locally convex vector space such that . Suppose that is weakly, sequentially continuous. If the implicationholds for every subset , then the operator has a fixed point.

Theorem 16 (see [24]). Let be measurable with closed, convex, and solid (i.e., ) for all We assume that there exists measurable with for all Let be a continuous random operator with stochastic domain such that, for every . Then has a stochastic fixed point.

3. Existence Results

Let us start by defining what we mean by a random solution of problem (1).

Definition 17. By a random solution of problem (1), we mean a measurable function that satisfies the integral equation on , as well as on

The following hypotheses will be used in the sequel:The functions , , are bounded for a.e. , and The function is random Carathéodory on for each For a.e. , as well as all , the function is weakly, sequentially continuous.There exist functions with , , for each , such that, for all , one has  for all and a.e. For all , there exists a continuous function with , such that, for each and any , one has There exists a function with for each such that, for any bounded ,  where There exists a random function such that  where Set

Theorem 18. Assume that hypotheses hold. Ifthen problem (1) has a random solution defined on

Proof. Define the operator by The functions , , are continuous for all Again, as the function is continuous on , defines a mapping Thus, is a solution for problem (1) if and only if . We shall show that the operator satisfies all conditions of Theorem 16. The proof will be given in several steps.
Step  1 ( is a random operator with stochastic domain on ). Since is random Carathéodory, the map is measurable in view of Definition 10. Therefore, the map is measurable. As a result, is a random operator on into
Let be defined by Clearly, the subset is closed, convex, end equicontinuous for all Then is measurable by Lemma  17 in [24]. Therefore, is a random operator with stochastic domain
Step  2 ( is continuous). Let be a sequence such that in Then, there exists such that , and
Thus, for each and , one has Using the Lebesgue Dominated Convergence Theorem, we get As a consequence of Steps 1 and 2, we can conclude that is a continuous random operator with stochastic domain , and is bounded.
Step  3 (for every ). For this, we apply Theorem 15. The proof will be given in several claims.
Claim  1 ( maps into itself). Let be fixed, and let Assume that Then there exists such that Thus, we get Next, for any fixed , let such that and , and let , where Then there exists such that and Thus, one has Hence Therefore, maps into itself.
Claim  2 ( is weakly, sequentially continuous). Let be a sequence in and let in for any , and each Fix , and since satisfies assumption , one has that converges weakly uniformly to Hence, the Lebesgue Dominated Convergence Theorem for Pettis integral implies that converges weakly uniformly to in We do it for any , and each , so Then is weakly, sequentially continuous.
Claim  3 (implication (14) holds). Let be a subset of such that Obviously, ). Further, as is bounded and equicontinuous, by Lemma  3 in [26], the function is continuous on Since the function is continuous on , the set is compact. From Lemma 14 and the properties of the measure , for any , as well as each , one has Thus, From (21), we get , that is; , for any , and each Hence, Theorem  2 in [27] shows that is weakly relatively compact in .
As consequence of Claims 1–3 and from Theorem 15, it follows that, for every , Apply now Theorem 16; Steps 1–3 show that, for each has at least one fixed point in Since , and a measurable selector of exists, the operator has a stochastic fixed point; that is, problem (1) has at least one random solution defined on

4. An Example

Let be the Banach space with norm and let be equipped with the usual -algebra consisting of Lebesgue measurable subsets of . Given a measurable function , consider the functional random integral problem of the following form:where , , and ; , , , and , , , and is a continuous and measurable function such that , where Set Then, For each , , and , one has Hence condition is satisfied with and

A simple computation shows that all other conditions of Theorem 18 are satisfied. Consequently, Theorem 18 implies that problem (33) has a random solution defined on .

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This research is supported by the National Natural Science Foundation of China (11671339).