On a hybrid inclusion problem via hybrid boundary value conditions

*Correspondence: rezapourshahram@yahoo.ca; sh.rezapour@azaruniv.ac.ir 1Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran 3Department of Medical Research, China Medical University Hospital, China Medical University, Taichung, Taiwan Full list of author information is available at the end of the article Abstract In this manuscript, the existence of solutions for a novel category of the fractional differential equation of hybrid type with hybrid boundary value conditions is studied. Also, we review the existence result for its related hybrid inclusion problem with hybrid conditions. In the end of the paper, two illustrative examples are given to demonstrate the consistency to our key results.


Introduction
As you know, we can introduce different kinds of modeling made by means of powerful and logical tools in mathematics. One of these tools is formulas and relations defined in the fractional calculus. Indeed, the importance and efficiency of this branch of mathematics has caused recent developments of FDEs or inclusions. Therefore, many researchers have studied some results on the properties of solutions for fractional BVPs under general boundary conditions. In this way, the authors use different approaches for their special goals. In other words, some recent published works show the importance of fractional differential equations in modeling of a variety of applied sciences (see, for example, [1][2][3][4][5][6][7][8][9]), numerical computations (see [10][11][12][13][14][15]), and different views on this field (see, for example, ).
By mixing and generalizing the above ideas, we are going to derive a new existence theorem for solution functions (t) of the following hybrid differential equation of order ν: with hybrid boundary conditions where ν ∈ (2, 3], θ ∈ (1, 2], D = d dt , D 2 = d 2 dt 2 , and c D β is the Caputo derivative of order β ∈ {ν, θ }. Moreover, Φ is a real-valued continuous function defined on [0, 1] × R and w ∈ C R ([0, 1] × R) is a nonzero function. In the following, we intend to study the related hybrid inclusion problem with hybrid boundary value conditions where Ψ : [0, 1] × R → P(R) is a given set-valued map via some properties.
To reach the main purposes of this manuscript, the techniques of the fixed point theory are employed to prove the theoretical results. Our investigations are two fold: we first deal with a hybrid differential equation and then with its related hybrid differential inclusion. It is worth mentioning that the proposed hybrid problems (1)-(2) and (3)-(4) differ from the newly defined ones. In both hybrid problems (1)-(2) and (3)-(4), we consider three boundary conditions of hybrid type in terminal points. We believe that our hybrid problems involve some particular cases, which can be extended to more general hybrid problems. The fractional hybrid modeling is of great significance in different engineering fields, and it can be a unique idea for the future combined research between various applied sciences (see [57]). As a practical example of applicability of our results, we can point out to our newly published work [58]. The fractional hybrid problem given in [58] is a particular case of the proposed hybrid problem (1)- (2) in this work, in which a fractional hybrid modeling of a thermostat is simulated.
The content of this article is arranged as follows: In Sect. 2, some required concepts in this regard are recalled. Section 3 is devoted to proving the main theorems relying on some mathematical inequalities and two versions of fixed point theorems due to Dhage. At the end of the paper, we give two illustrative examples to support the applicability of our findings.

Preliminaries
Prior to proceeding to the main objectives, we here recall some essential auxiliary concepts which are needed in the sequel. Let ν ∈ R + so that ν ∈ (n -1, n] and assume that the realvalued function is integrable on [a, b]. In this case, the Riemann-Liouville integral of the function of order ν is given by provided that the integral is finite-valued [59,60]. With the same assumptions, the Caputo derivative of order ν for a function ∈ C (n) R ([a, b]) is given by provided that the integral is finite-valued and n = 1 + [ν] [59,60]. According to the existing propositions, if we solve the fractional homogeneous differential equation c D ν 0 (t) = 0, then its general solution is obtained as (t) = b 0 + b 1 t + b 2 t 2 + · · · + b n-1 t n-1 , where b 0 , . . . , b n-1 ∈ R and n = 1 + [ν] [59,60]. For every ∈ C R ([0, T]) with T > 0, the linear equation holds, where b 0 , . . . , b n-1 ∈ R and n = 1 + [ν] [59,60].
Here, consider the normed space (X , · X ). Then the collection of all subsets of X , the collection of closed subsets of X , the collection of bounded subsets of X , the collection of convex subsets of X , and the collection of compact subsets of X are denoted by P(X ), P cl (X ), P b (X ), P cv (X ), and P cp (X ), respectively. A set-valued map Ψ is convex-valued if for each ∈ X the set Ψ ( ) is convex. The set-valued map Ψ has an upper semi-continuity property whenever, for every * ∈ X , Ψ (ρ * ) belongs to P cl (X ) and for each open . Moreover, * ∈ X is a fixed point for the set-valued map Ψ : X → P(X ) whenever * ∈ Ψ ( * ) [61]. The notation Fix(Ψ ) denotes the set of all fixed points of set-valued Ψ [61].
A set-valued operator Ψ has the complete continuity property if the set Ψ (W) has the relative compactness property for all W ∈ P b (X ). Let Ψ : X → P cl (Q) have the upper semi-continuity property. Then Graph(Ψ ) ⊆ X × Q is a closed set. On the other hand, assume that Ψ has a closed graph with the complete continuity property. Then Ψ has the upper semi-continuity property [61]. We say that Ψ : [0, 1] × R → P(R) is a Caratheodory set-valued map if the mapping → Ψ (t, ) is upper semi-continuous for almost all t ∈ [0, 1] and t → Ψ (t, ) is measurable for each ∈ R [61,62]. In addition, a Caratheodory set-valued map Ψ : for almost all t ∈ [0, 1] and for each | | ≤ r [61,62]. The collection of all selections of Ψ at 61,62]. As it has been proved before in [61], (SEL) Ψ , = ∅ for all ∈ C X ([0, 1]) whenever dim X < ∞. We need the following results.

Theorem 3 ([65]) Let X be a Banach algebra. Assume that there are a set-valued map
there exists a solution belonging to X for the inclusion ∈ Υ 1 Υ 2 .

Main results
In this part of the paper, we intend to state our main theoretical findings on the existence results. To reach this aim, we consider X = { (t) : (t) ∈ C R ([0, 1])} equipped with the supremum norm X = sup t∈[0,1] | (t)| and the multiplication action on the space X defined by ( · )(t) = (t) (t) for all , ∈ X . Then an ordered triple (X , · X , ·) is a Banach algebra. In this moment, we present an essential lemma which converts fractional BVP (1)-(2) into integral equation.

Lemma 4 Assume that g belongs to X . Then 0 is a solution function for the hybrid equation of fractional order
with hybrid boundary value conditions iff the function 0 is a solution for the following integral equation of fractional order: Proof Let 0 be a solution for hybrid equation (5). Then the general solution of homogeneous equation (5) is obtained by the equality 0 (t) That is, Now, we employ the Caputo derivative of arbitrary orders on both sides of equation (8), and we get where 1 < θ ≤ 2. Corresponding to the boundary value conditions, we obtain b * 0 = 0 and b * 1 = - This means that 0 is a solution for integral equation (7). On the contrary, it is easy to check that 0 satisfies the fractional hybrid BVP (5)-(6) if 0 is a solution for the integral equation of fractional order (7).

Theorem 5 Suppose that w is a nonzero continuous real-valued function on
. Furthermore, assume that the following statements hold: There exist a nondecreasing and continuous map ξ : R ≥0 → R + and a continuous map h : where where ε satisfies inequality (9). Consider the operators Υ 1 , Υ 2 : V (0) → X given by (Υ 1 )(t) = w(t, (t)) and Obviously, ∈ X as a solution for hybrid BVP (1)-(2) satisfies equation First, we want to show that the operator Υ 1 is Lipschitzian on X with Lipschitz constant s * = sup t∈[0,1] |s(t)|. To check this, let 1 , 2 ∈ V ε (0). By using assumption (A1), we have . This inequality shows that Υ 1 is Lipschitzian on V ε (0) with constant s * . In the sequel, we have to show that Υ 2 on V ε (0) is completely continuous. In this way, we need to show the continuity of Υ 2 on V ε (0). Consider the convergent sequence { n } in the ball V ε (0) with n → , where is an arbitrary member belonging to V ε (0). By the hypothesis, we know that Φ is continuous. Thus we have lim n→∞ Φ(t, n (t)) = Φ(t, (t)). In view of the dominated convergence theorem due to Lebesgue, we have for any t ∈ [0, 1]. Therefore, Υ 2 n → Υ 2 , and so we deduce that Υ 2 is continuous on V ε (0). Now, we check the uniform boundedness of Υ 2 on V ε (0). By using assumption (A2), we can write for any t ∈ [0, 1] and ∈ V ε (0). Hence Υ 2 X ≤ H * ξ ( ) * , where * is defined in (10). This means that Υ 2 (V ε (0)) is a uniformly bounded subset of X . Here, we proceed to proving that the operator Υ 2 is equicontinuous. Without loss of generality, let us assume that 0 ≤ t 1 , t 2 ≤ 1 provided that t 1 < t 2 , and let ∈ V ε (0). Then Letting t 1 → t 2 , we observe that the right-hand side of the above inequality converges to zero independently of ∈ V ε (0). Thus |(Υ 2 )(t 2 )-(Υ 2 )(t 1 )| → 0 as t 1 → t 2 . Consequently, the operator Υ 2 is equicontinuous. By utilizing the Arzela-Ascoli theorem, we arrive at the desired aim, which is the complete continuity of Υ 2 on V ε (0). In addition, because of hypothesis (A3), we get Setting l * = s * , we get l * H * < 1. Now, to complete the proof, we claim that one of conditions (b1) or (b2) in Theorem 1 is valid. Let = μΥ 1 Υ 2 for some constant 0 < μ < 1 and = ε. Then we have the following estimate: and so ε ≤ W * * H * ξ ( ) 1-s * * H * ξ ( ) , which is a contradiction to inequality (9). It follows that condition (b1) of Theorem 1 is not possible. Hence, condition (b2) is valid and so the mentioned fractional hybrid problem (1)-(2) has a solution.
In what follows, we are going to provide another essential result for the fractional hybrid inclusion problem (3)-(4). Existence results herein are carried out in light of the assumptions of Theorem 3.

Definition 6 We say that the function ∈ AC R ([0, 1]) is a solution for the hybrid inclusion BVP of fractional order (3)-(4) whenever there exists an integrable function
w(t, (t)) )| t=0 = 0 and Here, we can formulate desired theorem on the existence of solution function of the above form.

If s * * σ L 1 < 1 2 , then the hybrid boundary value problem of fractional order (3)-(4) has a solution.
Proof To transform the hybrid inclusion problem (3)-(4) into a fixed point problem, we define K : It is clear that each fixed point of K is a solution for hybrid inclusion BVP (3)-(4). We split the operator K into two parts as Υ 1 : X → X and Υ 2 : Note that K( ) = Υ 1 Υ 2 . We are going to prove that operators Υ 1 and Υ 2 satisfy all the assumptions of Theorem 3. Obviously, in view of hypothesis (A 1) and in a similar way used in Theorem 5, one can easily find that Υ 1 is Lipschitzian on X . In the following, we need to show that Υ 2 is convex-valued. To do this, let 1 , 2 ∈ Υ 2 . Choose ϑ 1 , ϑ 2 ∈ (SEL) Ψ , so that for all t ∈ [0, 1] (a.e.) and j = 1, 2. For any 0 < γ < 1, we obtain for almost all t ∈ [0, 1]. By the hypothesis of theorem, Ψ has convex values. Thus (SEL) Ψ , is convex-valued and γ ϑ 1 (t) + (1γ )ϑ 2 (t) ∈ (SEL) Ψ , for all t ∈ [0, 1] and so Υ 2 is a convex set for all ∈ X . Now, to check the complete continuity of Υ 2 , we have to prove that Υ 2 (X ) is uniformly bounded and equicontinuous. To reach this purpose, we show that Υ 2 maps all bounded sets into bounded subsets of the space X . For a number ε * ∈ R + , we construct a bounded ball V ε * = { ∈ X : X ≤ ε * }. For every ∈ V ε * and ζ ∈ Υ 2 , there exists ϑ ∈ (SEL) Ψ , so that for each t ∈ [0, 1]. Then we have the following estimate for function ζ : where * is given in (10). Thus, ζ ≤ * σ L 1 and this means that the set Υ 2 (X ) is uniformly bounded. The next step in this part of the proof is to show that Υ 2 maps bounded sets into equicontinuous sets. Let ∈ V ε * and ζ ∈ Υ 2 . Choose ϑ ∈ (SEL) Ψ , provided that for all t ∈ [0, 1]. We may assume that 0 ≤ t 1 , t 2 ≤ 1 with t 1 < t 2 . Then we have Letting t 1 → t 2 , one can see that the above inequality converges to 0 independently of ∈ V ε * . It follows from the Arzela-Ascoli theorem that Υ 2 : C R ([0, 1]) → P(C R ([0, 1])) is a completely continuous operator. Here, we claim that Υ 2 has a closed graph. Then, because of the complete continuity of Υ 2 , we find that Υ 2 is upper semi-continuous. For this aim, suppose that n ∈ V ε * and ζ n ∈ Υ 2 n are such that n → * and ζ n → ζ * . We can verify that ζ * ∈ Υ 2 * . Indeed, for each n ≥ 1 and ζ n ∈ Υ 2 n , select ϑ n ∈ (SEL) Ψ , n such that for any t ∈ [0, 1]. In this case, we have to show that there is v * ∈ (SEL) Ψ , * such that for all t ∈ [0, 1]. Define the continuous linear operator Ξ : for each t ∈ [0, 1]. Then we have Letting n → ∞, the above estimate yields ζ n (t)ζ * (t) → 0. Hence with due attention to Theorem 2, one can deduce that Ξ •(SEL) Ψ is an operator having a closed graph property. As ζ n ∈ Ξ ((SEL) Ψ , n ) and n → * , so there exists ϑ * ∈ (SEL) Ψ , * such that for all t ∈ [0, 1]. Hence, ζ * ∈ Υ 2 * and so Υ 2 has a closed graph. Therefore, the upper semicontinuity of the operator Υ 2 is fulfilled. By utilizing the assumption of theorem, we know that Υ 2 has compact values. Consequently, Υ 2 is an upper semi-continuous and compact operator. Now, under assumption (A 3), we have By putting l * = s * , we obtain l * K * 0 < 1 2 . Hence, all the assumptions of Theorem 3 hold for Υ 1 . Here, we claim that only one of conditions (b 1) or (b 2) is valid. By applying Theorem 3 and assumption (A 4), consider an arbitrary element of Y * with =ε. Then μ (t) ∈ Υ 1 (t)Υ 2 (t) for each μ > 1. Select the related function ϑ ∈ (SEL) Ψ , . Then, for each μ > 1, we obtain for all t ∈ [0, 1]. Thus, we have for any t ∈ [0, 1]. By simple computations, we getε ≤ W * * σ L 1 1-s * * σ L 1 . According to condition (11), we find that condition (b 1) of Theorem 3 is not possible and condition (b 2) is valid. Therefore, the operator inclusion ∈ Υ 1 Υ 2 has a solution, and so the hybrid inclusion BVP (3)-(4) has a solution. This ends the proof.
To demonstrate the consistency and the applicability of the obtained results, two illustrative numerical examples are provided herein.

Conclusion
Nowadays, it is a vital goal that we could model most phenomena in the real world. For example, modeling of chemical reactions using some modern software to reduce the use of materials in chemical laboratories. This will contribute to environmental protection. Thus, we should endeavor to increase our creativity to study the complicated modeling of differential equations and inclusions. In the present research work, we design a novel fractional hybrid differential equation and its related inclusion version with hybrid conditions. To reach the desired findings, some analytical techniques are adopted from the concepts of nonlinear analysis. Finally, to demonstrate the consistency and applicability of the obtained results, two illustrative numerical examples are provided. We believe that our hybrid problems involve some particular cases, which can extend to more general hybrid problems. The fractional hybrid modeling is of great significance in different engineering fields, and it can be a unique idea for the future combined research between various applied sciences. So we leave the new abstract idea for interested researchers as future projects.