The presented article is deduced about the positive solutions of the fractional differential inclusion at resonance on the half line. The fractional derivative used is in the sense of Riemann–Liouville and the problem is supplemented by unseparated conditions. The existence results are illustrated in view of Leggett–Williams theorem due to O’Regan and Zima on unbounded domain.

1. Introduction

In this paper, we study the possibility of finding a positive solution on unbounded domain with unseparated conditions for the following fractional differential inclusion:which is contingent with for which is multivalued map, and is the symbol of the Riemann–Liouville fractional derivative.

Riemann–Liouville fractional derivative and corresponding integral and many other types of fractional derivatives and integrals with singular and nonsingular kernels are used in fractional calculus as generalizations of the ordinary differentiations and integrations. Thus, many scientific teams are attracted to link their works with fractional calculus, which has been effectively applied in the other scientific fields. As a result, there are some books talking about fractional calculus and some of its applications with fractional equations and inclusions, see [13]. In the existence and uniqueness fields, there are many articles concerned with different types of fractional differential equations or inclusions. For fractional differential equations, there are a huge number of contributions (see [49] and the references given therein). For fractional differential inclusions, Salem and Al-Dosari [10] have studied the existence results of solution for fractional SturmLiouville inclusion associated with composition with multimaps. Chen and Tang [11] have investigated probability of the positive solution for the fractional equation below on unbounded domain at resonance:

Jiang and Yang [12] have studied the positive solution for the second ordinary boundary value problems:where and is not always equal to zero.

On a bounded domain, Chen et al. [13] have explored sufficient conditions to obtain a positive solution for the problem:

Furthermore, Hu [14] has given some results of positive solution at resonance for the boundary value problem formulated as follows:

By reducing nonperturbed boundary value problems at resonance, Wang and Liu [15] have presented their results about positive solution of the problem:

About an overall summary of resonance and its applications, Haller [16] has mentioned that “resonances are regions in the phase space of a dynamical system in which the frequencies of some angular variables become nearly commensurate. Such regions have a profound effect on the dynamics of the system, since they are rich sources of highly complex motions. In molecular dynamics, resonances are known to give rise to chaotic patterns, multiple time scales, and apparent irreversibility in the transfer of energy between different oscillatory states of molecules. In engineering structures, interactions among resonant modes are responsible for most complicated dynamical phenomena, which again include energy transfer, multitime-scale behavior, and chaotic motions. It is of great practical importance to understand the common mechanism behind these irregular features, both qualitatively and quantitatively. The circle of further applications ranges from nonlinear optics through celestial and fluid mechanics to electromagnetism.”

Chaotic patterns, oscillatory states of molecules, complicated dynamical phenomena, and multitime-scale behavior are all given a strong cause to study differential problems at resonance associated with multivalued maps, which are not presented until now. So, our aim is to show the possibility of having at least one positive solution at resonance. For the sake of that, in Section 2, we give some basic notations and results of fractional differentiations and integrations, some facts of multimaps and their arguments, and some needed lemmas and fixed point theorems. Section 3 contains some required results to the existence of positive solution for the given problem. Before the conclusion, Section 4 shows a quintessence as an application of our outcomes.

2. Precursory Notations

This section is devoted to provide some notations and definitions for the fractional calculus, multivalued functions, and Fredholm operator.

2.1. Fractional Calculus

We introduce some definitions and fundamental rules on the fractional calculus that are needed for the new results [1, 17, 18].

Definition 1. For , the Riemann–Liouville fractional integral of a continuous function is defined by

Definition 2. For and , the Riemann–Liouville fractional derivative of an absolutely continuous function is given by

Lemma 1. Let and . Then,

Lemma 2. Let . Then,for some .

2.2. Multivalued Function

In this section, we state some necessary facts on multivalued maps [2, 1921].

Let be Banach spaces and be a multivalued map. The mapping is(i)Convex (closed) if is convex (closed), for every .(ii)Completely continuous if is relatively compact, for every .(iii)Upper semicontinuous if is closed subset of , for each closed subset . That is, the set is open for all open sets .(iv)Lower semicontinuous if is open subset of , for each open subset . In the other words, the set is open for all open sets .(v)Measurable multivalued if for every , the function is measurable function.

Definition 3. A multivalued map is known as a Caratheodory if(1)For all is measurable.(2)For a.e is upper semicontinuous.In addition to assumptions (1) and (2), the map is -Caratheodory if for each satisfying and , that is,for all .

2.3. Fredholm Operator

We recall some preliminaries about the operators specially Fredholm type where we can get them from [3, 2224].

Let be any considered operator. Then, we define the following.(1)The kernel of by(2)The image (rang) of by(3)The cokernel of by(4)The index of by

Note that

The operator is called Fredholm operator if it is a bounded linear operator with finite dimensional kernel and cokernel and closed rang. It is said to be Fredholm of index zero if and only if .

Let be Fredholm of index zero and be continuous projections. In the case ofthen is invertible, and so we can defineIn fact, there is an isomorphic map .

Consider the inclusionwhich is equivalent toLet be a convex and a closed subset. Then, is called a cone if(i)(ii)

Consequently, there is an induced partial order in subject to

Lemma 3. Let be a Banach space contains a cone . Then, for every , there exists a positive number which satisfies

Define to be a retraction (continuous mapping constrained with ). Define the map asThen,(i)(ii)

Theorem 1. (O’Regan and Zima). Let be a cone, and be open bounded subsets of which satisfy , and . Assume that(1) is a Fredholm operator of index zero.(2) is an upper-semicontinuous mapping with nonempty compact convex values.(3) is bounded on bounded subsets of , and is compact on every bounded subset of .(4) and .(5) subsets of into bounded subsets of .(6).(7)There is such thatwhereand is such that (8).(9).(10).Then, the inclusion can be solvable in the set .

Remark 1. Letwith the normAnd letwith the normThen, are Banach spaces, see [11].
Expand the operator , for whichwhereThen, the multivalue problems (1)–(3) are equivalent to the inclusion:We can understand the proof of the following lemma if we see the compactness hypothesis in [11, 25, 26].

Lemma 4. Let be subset of ; then, is a relatively compact subset if and only if the next conditions hold.(i)The subset is uniformly bounded, i.e.,(ii)Let be compact subinterval of and let for all there exists such that for . Then,(iii)For be given, there exists in since

3. Acquired Upshots

In this section, we discuss some results for the sake of seeing the attainability of solution by apply (O’Regan and Zima, Theorem 1) on the half line. Consequently, in view of the arguments of Fredholm operator, we first need to explore some lemmas and facts as follows.

3.1. Basic Requirements

Here, by using the general form of solution, we prove several facts in order to explore that() is Fredholm operator of index zero.

Also, we explain that this operator is invertible in the set to compute the formula of its inverse . Finally, under the version of Lemma 3 (i) and by the known maps , we calculate the operator via green functions and show some related inequalities.

Lemma 5. Consider . Then, is a solution of the boundary value problem:if and only ifwhere and

Proof. The proof can be completely seen if we are backing to both Lemmas 1 and 2 and applying accompanying conditions.
Now, using (33)–(35), we obtainwhich implies thatBy using the equation,we can define a continuous projection on as in the next Lemma.

Lemma 6. Let be a map defined as

Then, is a continuous projection on .

Proof. Let and be defined as in (46). Then,which completes the proof.

Remark 2. We can recall thatImmediately, we haveAccording to both integrals and in (41), we can explain and prove that there exists a continuous projection on the space . See the next lemma.

Lemma 7. Let be a mapping formed bywhere . Then, is a continuous projection on .

Remark 3. We can make sure thatClaim that . For that, take . Then, we havewhere . Moreover,Thus, . Assume that . Then,Hence, . It remains to prove that is of index zero. Due to the previous results of , we find thatSince is closed in , it implies that is the Fredholm operator of index zero.
Furthermore, we can see clearly that if we define the isomorphism byas long as defined by (52).
Once we have is the Fredholm operator of index zero, then it is invertible in the set . Create and . Let , which follows that is defined by (40) and satisfying the condition in (48). By substituting (40) in the equation given in (48), the constant will be subject toTherefore,

Lemma 8. Let be the Fredholm operator of index zero. That is,where and are defined by (32), (46), and (48), respectively. Let be elected such as in (60). Then,

Proof. Opting drivesAnd use (43) to verify thatTherefore,
According to Lemma 3 (i), we need to compute the value:which can be introduced bywhere

Lemma 9. We have

Proof. Basically, we have the following inequalities:In order to prove the L.H.S of (69), we note that, for all positive reals , to imply thatwhich leads toBy analytical methods, we can see thatNow, to make sure about the R.H.S of (69), useThen,

3.2. Essential Theorem

Apply all results in the first part to illustrate the main theorem with some needed hypothesis as follows.

Theorem 2. Reckoning in addition of () that the following.() Let be Caratheodory (Definition 3) with which has nonempty compact convex values conditional on which there exist nonnegative functions and satisfyingfor all and () For , we haveand there exists in which() There exist positive constants , where aswhere , and

Then, it is distinctly possible to find positive solution for problems (1)–(3) in .

Proof. Let and be Banach spaces defined as in Remark. Define the linear operator and its property by (32), (33), (43), and (44), and the nonlinear operator by (34) as well as (35) holds. Then, to show that if it is possible for the given problem to has a solution, we will prove all points of (O’Regan and Zima, Theorem 1-).Step 1: the operator will be the Fredholm operator of index zero by using ().Step 2: by (), the operator (N) is upper semicontinuous mapping with nonempty compact convex values.Step 3: according to (), is bounded in the bounded set andis compact, where are all conformed by (46), (51), and (60), respectively.Define the needed sets as follows:Clearly, and forms a cone in .Step 4: claim that .Assume by contradiction that such that . Sincewe have, by using Hölder’s inequality with and , (66) and (68) thatIn casewe can see by ( (76) and (77)) thatNow, we haveRelations (89), (90), and (91) give the ensuing results:that are equivalent, respectively, toFrom (95), it can be seen thatThe relations (81) and (94)–(96) together with () show that (87) will be equivalent towhich is contradiction to Step 5: to recall that maps subsets of into bounded sets of , let . Since , thenStep 6: this step will show thatDefine the map byLet , and consider by contradiction that . Since , then it is given by the formConsequently, we havewhere . Now, expound a mapsuch asThen, (