Abstract

In the present research work, we investigate the existence of a solution for new boundary value problems involving fractional differential equations with -Caputo fractional derivative supplemented with nonlocal multipoint, Riemann–Stieltjes integral and -Riemann–Liouville fractional integral operator of order boundary conditions. Also, we study the existence result for the inclusion case. Our results are based on the modern tools of the fixed-point theory. To illustrate our results, we provide examples.

1. Introduction

Fractional calculus has played a very important role in different areas of research (see [1, 2] and the references cited therein). Consequently, fractional differential equations have grasped the interest of many researchers working in diverse applications [36]. Recently, several researchers have tried to propose different types of fractional operators that deal with derivatives and integrals of arbitrary orders and their applications. For instance, Kilbas et al. in [2] introduced the properties of fractional integrals and fractional derivatives concerning another function. Some generalized fractional integral and differential operators and their properties were introduced by Agrawal in [7]. Very recently, Almeida in [8] presented a new type of fractional differentiation operator, the so-called -Caputo fractional operator, and extended work of Caputo [2, 9]. Almeida et al. in [10, 11] investigated the existence and uniqueness of the results of nonlinear fractional differential equations involving a Caputo-type fractional derivative with respect to another function, employing the fixed-point theorem and Picard iteration method. Numerous interesting results concerning the existence, uniqueness, and stability of initial value problems and boundary value problems for fractional differential equations with -Caputo fractional derivatives by applying different types of fixed-point techniques were obtained by Abdo et al. [12, 13], Vivek et al. [14], and Wahash et al. [15]. An important application that is controlled by the theory of -fractional differentiation can be found in [16].

In this paper, we investigate a new boundary value problem of fractional differential equations supplemented with nonlocal multipoint, Riemann–Stieltjes integral fractional boundary conditions involving Riemann–Liouville fractional integral operator of order with respect to function given by the formwhere denotes the -Caputo fractional derivatives of orders and , respectively, is the -Riemann–Liouville fractional integral operator of order . The functions and are continuous, whereas and are Riemann–Stieltjes integral and is a function of bounded variation. is a real constant, and are positive constants.

We also study the corresponding inclusion problem that is given bywhere is a multivalued function, where is the family of all subsets of and the other quantities are the same as defined in problem (1).

Notice that this Riemann–Stieltjes integral fractional boundary conditions arise in manifold applications of computational fluid dynamics, distribution methods, and so forth (for example, see [17, 18]).

This paper is organized as follows. In Section 2, we recall some preliminary results and some related definitions. In Section 3, we discuss the existence results of solutions by relying on Krasnoselskii fixed-point theorem and Leray–Schauder nonlinear alternative. Also, we present an example. Finally, we describe the inclusion case and deduce the existence of solutions by applying Krasnoselskii’s multivalued fixed-point theorem in Section 4.

2. Preliminaries

For the convenience of the reader, we present here some necessary basic definitions, lemmas, and results which are used throughout this paper [2, 8, 10, 12, 19, 20].

Definition 1. Let be a finite interval and and be the set of absolute continuous functions on . Then, we defineendowed with the norm. The convention endowed with the norm is used.

Definition 2. Let , be an integrable function defined on , and be an increasing differentiable function such that for all . The left-sided -Riemann–Liouville fractional integral of order of a function is given by

Definition 3. Let be an integrable function and be an increasing differentiable function such that for all . The left-sided -Riemann–Liouville fractional derivative of order of a function is defined bywhere and denote the integer part of the real number .

Definition 4. Let , and be an increasing differentiable function such that for all . The left-sided -Caputo fractional derivative of order of a function is defined bywhere and for . Furthermore, if and , thenThus, if , one has

Lemma 1. Given a function and , then

Lemma 2. Let and ; then,(i)(ii)(iii)(iv)The existence of solutions of problem (1) relies on the following fixed-point theorems [21, 22].

Theorem 1. (Krasnoselskii’s fixed-point theorem). Let be a closed, convex, bounded, and nonempty subset of a Banach space . Let be operators such that(i) belong to whenever (ii) is compact and is a contraction mappingThen, there exist such that .

Theorem 2. (Leray–Schauder fixed-point theorem). Let be a closed and convex subset of a Banach space and be an open subset of with . Suppose that is a continuous, compact (that is, is a relatively compact subset of ) map. Then, either(i) has a fixed point in or(ii)there are (the boundary of in ) and with For computational convenience, we set the following:

Lemma 3. Let ; then, the linear fractional differential equationhas a solution on given by

Proof. We apply -Riemann–Liouville fractional integral of order to both sides of the linear fractional differential equation:We obtainNext, applying -Riemann–Liouville fractional integral of order to both sides of (14), we obtainwhere are arbitrary constants. By Definition 1, general solution (15) can be written asUsing the boundary condition , we obtain . Thus, (16) takes the formApplying the operator , on equation (17), we obtainUsing the boundary condition , we find thatInserting the value of in (17) yields solution (12). The converse follows by direct computation.

3. Main Results

In this section, we prove the existence of solutions of problem (1). We shall assume that are in the Banach space . Let denote the Banach space of all continuous functions on into endowed with the norm . Here, we define an operator associated with problem (1) by

Therefore, problem (1) has a solution if and only if the operator has a fixed point.

For computational convenience, we introduce the notations

Now, we will state and prove the existence result via Krasnoselskii’s fixed-point theorem.

Theorem 3. Let be continuous functions satisfying the conditions:(H1) for all , every and .(H2) There exist continuous nonnegative functions such thatand . Then, problem (1) has at least one solution on .

Proof. For a positive number , consider where , and we split into two operators andwhere, on the bounded set byFor any , by using (H2), we haveHence, . Next, we show that is a contraction mapping. Let and , so by using (H1), we haveBy assumption , we obtain that is a contraction mapping. Since are continuous functions, we have as continuous. Also, is uniformly bounded on asFinally, we prove the compactness of the operator . To show this, we define and and take . Thus, we haveindependent of . Thus, is equicontinuous. So, is relatively compact on . Hence, by the Arzela–Ascoli theorem, is compact on . Thus, the hypotheses of Theorem 6 are satisfied which leads problem (1) to have at least one solution on .Now, we apply Leray–Schauder nonlinear alternative fixed-point theorem to establish an existence result for problem (1).

Theorem 4. Let be continuous functions and the following conditions are satisfied:(H3) There exist functions with and nondecreasing functions such that(H4) There exists a constant such thatThen, problem (1) has at least one solution on .

Proof. Let us show that the operator maps bounded sets into bounded sets in . For a positive , let be a bounded set in . For each , by (H3), we haveNext, we show that maps bounded sets into equicontinuous sets of . Let ; then, for , with , we havewhich tends to zero independent of as . So, we deduce that the operator is completely continuous (by Arzela–Ascoli theorem). It remains to show the boundedness of the set of all solutions for . Let be a solution of problem (1). So, for , we obtainwhich, on taking the norm for , yieldsand thenFrom (H4), we can find such that . Take a set and notice that the operator is continuous and completely continuous. By choice of , we cannot find a such that for some . Hence, by Theorem 2, the operator has a fixed point which is a solution of problem (1).

Example 1. Consider the following boundary value problem:where , , and . Clearly, is an increasing function on and is a continuous function on .
To illustrate the application of Theorem 3, we takewhere satisfy the assumption of Theorem 3. By using the given data, we find , , , and . In addition, . Therefore, the result of Theorem 3 applies to problem (38) with and given above.

4. Existence Results for Inclusion Case

In this section, we extend the results to cover the inclusion problem and prove the existence of solutions for problem (2) by applying the fixed-point theorem [23]. We recall some basic notations for the inclusion case [2430].

For a normed space , let

A multivalued map is said to be carathodory if(i) is measurable for each (ii) is upper semicontinuous for almost all Furthermore, a carathodory function is called -Carathodory if(iii)for each , there exist such thatfor all and for a.e. .

For each , define the set of selections of by

Let be a metric space induced from the normed space . We have given bywhere and . So, is a metric space [31].

Lemma 4. Let be a Banach space. Let be an -Carathodory multivalued map, and let be a linear continuous mapping from . Then, the operatoris a closed graph operator in .

Lemma 5. If is u.s.c., then is a closed subset of ; i.e., for every sequence and , if , , and , then . Conversely, if is completely continuous and has a closed graph, then it is upper semicontinuous.

Definition 5. A function is called a solution of problem (2) if we can find a function with such that andFor convenience, we denoteOur result is based on the following fixed-point theorem.

Theorem 5. Let and be, respectively, the open and closed subsets of Banach space , such that ; let be multivalued and be single-valued such that is bounded. Suppose that(a) is a contraction with a contraction (b) is u.s.c and compactThen, either(i)the operator inclusion has a solution for or(ii)there is an element such that for some , where is the boundary of

Theorem 6. Assume that(N1) .(N2) There exists a continuous function and such that(N3) Let be a continuous functions satisfying(N4) There exists a number such thatwhere are defined in (21) and (22), respectively. Then, problem (2) has at least one solution on if .

Proof. Let be an open set in U. Define the multivalued operator byand define the single-valued operator byObserve that , and it is given byIndeed, if , then there exists , such thatWe will show that the maps and satisfy the hypotheses of Theorem 5. This will be done in several steps.

Step 1. is a contraction. Let , by (N3), we havewhich proves that is a contraction map.

Step 2. is convex for all . Let . We select such that, for each , we obtainfor .
Let and. So, we haveSince is convex, it follows that and then is convex-valued.

Step 3. is compact and upper semicontinuous. This will be done in various statements. First, we show that maps bounded sets into bounded sets in .
For a positive number , let be a bounded ball in . So, for all , , there exists such thatBy using (N2), for each , we haveConsequently,where and are defined in (22) and (47), respectively. Second, we prove that maps bounded sets into equicontinuous sets. Let with and ; we haveIn the above inequality, the right hand side tends to zero independent of as . Consequently, by the Arzela–Ascoli theorem, we conclude that is completely continuous and then is completely continuous. Finally, we show has a closed graph. Let , and . Then, we show that . Since , there exists such that for each , we find thatNow, we have to show that there exists such that for each ,Consider the continuous linear operator given byNote thatwhich goes to 0, as .
It follows by Lemma 5 that is a closed graph operator. Furthermore, we obtain . Since , we havefor some . Hence, has a closed graph (and therefore has closed values). Hence, we conclude that is a compact multivalued map, upper semicontinuous with convex closed values.

Step 4. There exists an open set with for any and for each . Take . Let be a solution of (2); then, there exists with such that for , we havewhich impliesBy (N4), there exists such that . Define a setNote that the operator is a compact multivalued map, u.s.c. with convex closed values. With the given choice of , it is not possible to find satisfying for some . Consequently, the operator has a fixed point , which is a solution of problem (2).

Example 2. Consider the following boundary value problem:where is a multivalued map given byFor , we obtainHere, ,
, and . Clearly, is an increasing function on and is a continuous function on .
Clearly,with .
Next, we takewhere the function satisfies the assumption of Theorem 6. By using the given data, we find , , and . Thus,and we haveTherefore, all the conditions of Theorem 6 are satisfied. Then, there exists at least one solution of problem (70) on .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare no conflicts of interest.