Attainability to Solve Fractional Differential Inclusion on the Half Line at Resonance

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.

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. us, 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 [1][2][3]. 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 [4][5][6][7][8][9] 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: ] ″ (τ) + g(τ, ](τ)) � 0, τ ∈ [0, ∞), ](0) � 0, where α i > 0, m− 1 i�1 α i � 1, 0 � ζ 1 < , . . . , < ζ m− 1 < ∞ and g(t, 0) 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: R D α ](τ) ∈ g(τ, ](τ)), τ ∈ [0, 1], 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. e 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.

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

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

Fredholm Operator.
We recall some preliminaries about the operators specially Fredholm type where we can get them from [3,[22][23][24]. Let L: E ⟶ H be any considered operator. en, we define the following.
(1) e kernel of L by (2) e image (rang) of L by (3) e cokernel of L by (4) e index of F by Note that then L| do m(L)∩Ker(P) is invertible, and so we can define In fact, there is an isomorphic map J: Im(Q) ⟶ Ker(L).

Consider the inclusion
which is equivalent to Let C ⊂ E be a convex and a closed subset. en, C is called a cone if Lemma 3. Let E be a Banach space contains a cone C. en, Define μ: E ⟶ C to be a retraction (continuous mapping constrained with μ(η) � η, ∀η ∈ C). Define the map Θ as en, where

en, the inclusion L] ∈ N] can be solvable in the set
with the norm And let with the norm en, (E, ‖·‖ E ) and (H, ‖·‖ H ) are Banach spaces, see [11].
Expand the operator L, for which where en, 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]. (i) e subset Ω is uniformly bounded, i.e., (ii) Let I be compact subinterval of [0, ∞) and let for all

Acquired Upshots
In this section, we discuss some results for the sake of seeing the attainability of solution by apply (O'Regan and Zima, eorem 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.

Basic Requirements.
Here, by using the general form of solution, we prove several facts in order to explore that Also, we explain that this operator is invertible in the set do mL ∩ KerP to compute the formula of its inverse K p . Finally, under the version of Lemma 3 (i) and by the known maps J, K p , Q, and I, we calculate the operator JQ + K P (I − Q) via green functions and show some related inequalities.

Lemma 5. Consider h(τ) ∈ H. en, ] ∈ E is a solution of the boundary value problem:
if and only if Proof. e proof can be completely seen if we are backing to both Lemmas 1 and 2 and applying accompanying conditions. Now, using (33)-(35), we obtain which implies that By using the equation, en, P is a continuous projection on E.
Proof. Let ] ∈ E and P be defined as in (46). en, which completes the proof.
□ Remark 2. We can recall that Immediately, we have According to both integrals ∞ 0 e − t dt � 1 and ∞ 0 h(t)dt � 0 in (41), we can explain and prove that there exists a continuous projection Q on the space H. See the next lemma.

Lemma 7. Let Q: H ⟶ H be a mapping formed by
where τ ≥ 0, h(τ) ∈ H. en, Q is a continuous projection on H.
Remark 3. We can make sure that KerQ � ImL.
Claim that H � KerQ ⊕ ImQ � ImL ⊕ ImQ. For that, take h(τ) ∈ H. en, we have where Qh(τ) ∈ ImQ. Moreover, Hence, H � ImL ⊕ ImQ. It remains to prove that L is of index zero. Due to the previous results of Q, we find that Since ImL is closed in H, it implies that L is the Fredholm operator of index zero.
Furthermore, we can see clearly that KerL � ImQ if we define the isomorphism J: ImQ ⟶ KerL by as long as ImQ defined by (52).
Once we have L is the Fredholm operator of index zero, then it is invertible in the set domL ∩ KerP. Create (L | do mL∩KerP ) � L P : do mL∩KerP ⟶ ImL and L − 1 P � K P . Let ](τ) ∈ do mL∩KerP, which follows that ](τ) is defined by (40) and satisfying the condition in (48). By substituting (40) in the equation given in (48), the constant b will be subject to erefore, (60)

Lemma 8. Let L be the Fredholm operator of index zero. at is,
where do mL, P and KerP are defined by (32), (46), and (48), respectively. Let K P be elected such as in (60). en,

Complexity
Proof. Opting ](τ) ∈ do mL∩KerP drives And use (43) to verify that erefore, K P (L P ) � L P (K P ) � I. According to Lemma 3 (i), we need to compute the value: which can be introduced by where Proof. Basically, we have the following inequalities: In order to prove the L.H.S of (69), we note that, for all positive reals m < t ⟹ e − m > e − t , to imply that which leads to By analytical methods, we can see that 6 Complexity (73) Now, to make sure about the R.H.S of (69), use en,

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

Theorem 2.
Reckoning in addition of (H 1 ) that the following.

Proof. Let E and H be Banach spaces defined as in Remark.
Define the linear operator L and its property by (32), (33), (43), and (44), and the nonlinear operator N by (34) as well as (35) holds. en, 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, eorem 1-).
Step 1: the operator L will be the Fredholm operator of index zero by using (H 1 ).
Step 2: by (H 2 ), the operator (N) is upper semicontinuous mapping with nonempty compact convex values.
Step 3: according to (H 2 ), QN is bounded in the bounded set and is compact, where P, Q, and K P are all conformed by (46), (51), and (60), respectively. Define the needed sets Ω 1 , Ω 2 , and C as follows: , Clearly, Ω 1 ⊂ Ω 2 and C forms a cone in E.
To explore (H 3 ), we need to compute the values hereinafter: