# Three-Point Boundary Value Problems for the Langevin Equation with the Hilfer Fractional Derivative

Authors: Athasit Wongcharoen; Bashir Ahmad; Sotiris K. Ntouyas; Jessada Tariboon
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

We discuss the existence and uniqueness of solutions for the Langevin fractional differential equation and its inclusion counterpart involving the Hilfer fractional derivatives, supplemented with three-point boundary conditions by means of standard tools of the fixed-point theorems for single and multivalued functions. We make use of Banach’s fixed-point theorem to obtain the uniqueness result, while the nonlinear alternative of the Leray-Schauder type and Krasnoselskii’s fixed-point theorem are applied to obtain the existence results for the single-valued problem. Existence results for the convex and nonconvex valued cases of the inclusion problem are derived via the nonlinear alternative for Kakutani’s maps and Covitz and Nadler’s fixed-point theorem respectively. Examples illustrating the obtained results are also constructed.(2010) Mathematics Subject Classifications. This study is classified under the following classification codes: 26A33; 34A08; 34A60; and 34B15.


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## Body

## 1. Introduction

Fractional calculus is an emerging field in applied mathematics that deals with derivatives and integrals of arbitrary orders. For details and applications, we refer the reader to the texts in [1–6]. In the literature, there exist several definitions of fractional integrals and derivatives, from the most popular Riemann-Liouville and Caputo-type fractional derivatives to others such as the Hadamard fractional derivative and the Erdeyl-Kober fractional derivative. A generalization of both the Riemann-Liouville and Caputo derivatives was given by Hilfer in [7], which is known as the Hilfer fractional derivative Dα,βxt of order α and type β∈0,1. One can observe that the Hilfer fractional derivative interpolates between the Riemann-Liouville and Caputo derivatives as it reduces to the Riemann-Liouville and Caputo fractional derivatives for β=0 and β=1, respectively. Some properties and applications of the Hilfer derivative can be found in [8, 9] and references cited therein.One of the important equations governing several phenomena occurring in physical sciences and electrical engineering is the Langevin differential equation, first formulated by Langevin in 1908 [10]. In recent years, several fractional variants of the Langevin equation have been introduced and studied; see, for example, [11–19] and the references cited therein.Initial value problems involving the Hilfer fractional derivatives were studied by several authors; see for example [20–22]. Nonlocal boundary value problems for the Hilfer fractional differential equation have been discussed in [23]. In [24], the authors proved some results for initial value problems of the Langevin equation with the Hilfer fractional derivative.Exploring the literature on fractional order boundary value problems, we find that there does not exist any work on boundary value problems of the Langevin equation with the Hilfer fractional derivative. Motivated by this observation, we fill this gap by introducing a new class of boundary value problems of the Hilfer-type Langevin fractional differential equation with three-point nonlocal boundary conditions. In precise terms, we investigate the existence and uniqueness criteria for the solutions of the following nonlocal boundary value problem:(1)DHα1,β1DHα2,β2+λxt=ft,xt,t∈a,b,(2)xa=0,xb=θxη,θ∈ℝ,η∈a,b,where DHαi,βi, i=1,2 is the Hilfer fractional derivative of order αi, 0<αi<1 and parameter βi, 0≤βi≤1, i=1,2, λ∈ℝ, a≥0, and f:a,b×ℝ→ℝ is a continuous function.In order to study problem (1)–(2), we convert it into an equivalent fixed-point problem and then use Banach’s fixed-point theorem to prove the uniqueness of its solutions. We also obtain two existence results for problem (1)–(2) by applying the nonlinear alternative of the Leray-Schauder type [25] and Krasnoselskii’s fixed-point theorem [26].As a second problem, we switch onto the multivalued analogue of (1) and (2) by considering the inclusion problem:
(3)DHα1,β1DHα2,β2+λxt∈Ft,xt,t∈a,b,(4)xa=0,xb=θxη,θ∈ℝ,η∈a,b,where F:a,b×ℝ⟶Pℝ is a multivalued map (Pℝ is the family of all nonempty subjects of ℝ).Existence results for problem (3)–(4) with convex and nonconvex valued maps are respectively derived by applying the nonlinear alternative for Kakutani’s maps and Covitz and Nadler’s fixed-point theorem for contractive maps.The rest of the paper is organized as follows: Section3 contains the main results for problem (1)–(2), while the existence results for problem (3)–(4) are presented in Section 4. We recall the related background material in Section 2.


## 2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus and multivalued analysis and present preliminary results needed in our proofs later [1].Definition 1.
The Riemann-Liouville fractional integral of orderα>0 for a continuous function u:a,∞⟶ℝ is defined by
(5)Iαut=1Γα∫att−sα−1usds,provided that the right-hand side exists on a,∞.Definition 2.
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function u is defined by
(6)DRLαut≔DnIn−αut=1Γn−αddtn∫att−sn−α−1usds,n−1<α<n,where n=α+1, α denotes the integer part of real number α, provided that the right-hand side is point-wise defined on a,∞.Definition 3.
The Caputo fractional derivative of orderα>0 of a continuous function u is defined by
(7)DCαut≔In−αDnut=1Γn−α∫att−sn−α−1ddsnusds,n−1<α<n,provided that the right-hand side is point-wise defined on a,∞.Definition 4 (Hilfer fractional derivative [7, 8]).
The Hilfer fractional derivative of orderα and parameter β of a function (also known as the generalized Riemann-Liouville fractional derivative) is defined by
(8)DHα,βut=Iβn−αDnI1−βn−αut,where n−1<α<n, 0≤β≤1, t>a, and D=d/dt.Remark 5.
Whenβ=0, the Hilfer fractional derivative corresponds to the Riemann-Liouville fractional derivative:
(9)DHα,0ut=DnIn−αut,while β=1 in the definition of the Hilfer fractional derivative corresponds to the Caputo fractional derivative:
(10)DHα,1ut=In−αDnut.In the following lemma, we present the compositional property of the Riemann-Liouville fractional integral operator with the Hilfer fractional derivative operator.Lemma 6 (see [8]).
Letf∈La,b, n−1<α≤n, n∈ℕ, 0≤β≤1, and In−α1−βf∈ACka,b. Then
(11)IαDHα,βft=ft−∑k=0n−1t−ak−n−α1−βΓk−n−α1−β+1limt→a+dkdtkI1−βn−αft.The following lemma deals with a linear variant of boundary value problem (1)–(2).Lemma 7.
Leta≥0, 0<αi<1, γi=αi+βi−αiβi, i=1,2, γ1+α2>1, and h∈Ca,b,ℝ. Then, the function x is a solution of the boundary value problem:
(12)DHα1,β1DHα2,β2+λxt=ht,(13)xa=0,xb=θxη,if and only if
(14)xt=Iα1+α2ht−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2hη−λθIα2xη−Iα1+α2hb+λIα2xb,where it is assumed that
(15)Λ=1Γγ1+α2b−aγ1+α2−1−θη−aγ1+α2−1≠0.Proof.
Applying the Riemann-Liouville fractional integral of orderα1 to both sides of (12), we obtain by using Lemma 6
(16)DHα2,β2xt+λxt=c0Γγ1t−aγ1−1+Iα1ht,where c0 is an arbitrary constant and 2−α11−β1=2−γ1. Applying the Riemann-Liouville fractional integral of order α2 to both sides of (16), we obtain
(17)Iα2DHα2,β2xt=Iα1+α2ht−λIα2xt+c0Γγ1Iα2t−aγ1−1=Iα1+α2ht−λIα2xt+c0Γγ1+α2t−aγ1+α2−1.
Applying Lemma6 to (17), we obtain
(18)xt=Iα1+α2ht−λIα2xt+c0Γγ1+α2t−aγ1+α2−1+c1Γγ2t−aγ2−1.
Usingxa=0 in (18), we obtain c1=0, and hence we get
(19)xt=Iα1+α2ht−λIα2xt+c0Γγ1+α2t−aγ1+α2−1.
Next, combining the conditionxb=θxη with (19), we have
(20)c0=1ΛθIα1+α2hη−λθIα2xη−Iα1+α2hb+λIα2xb.
Substituting the value ofc0 in (19) yields the solution (14). The converse follows by direct computation. This completes the proof.


## 3. Existence and Uniqueness Results for Single-Valued Problem (1)–(2)

In view of Lemma7, we define an operator A:C⟶C associated with problem (1)–(2) by
(21)Axt=Iα1+α2ft,xt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2fη,xη−λθIα2xη−Iα1+α2fb,xb+λIα2xb,where C=Ca,b,ℝ denotes the Banach space of all continuous functions from a,b into ℝ with the norm x≔supxt:t∈a,b. One can observe that the existence of a fixed point of operator A implies the existence of a solution for problem (1)–(2).For computational convenience, we introduce the following notations:(22)Ω1=b−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+θη−aα1+α2Γα1+α2+1,(23)Ω2=λb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα2Γα2+1+θη−aα2Γα2+1.Now, we present our main results for boundary value problem (1)–(2). Our first existence result is based on the well-known Krasnoselskii’s fixed-point theorem [26].Theorem 8.
Assume that the following conditions hold:(H1) f:a,b×ℝ⟶ℝ is a continuous function such that ft,x≤φt, ∀t,x∈a,b×ℝ, with φ∈Ca,b,ℝ+
(H2) Ω2<1, where Ω2 is given by (23)
Then, there exists at least one solution for problems (1) and (2) on a,b.Proof.
In order to verify the hypothesis of Krasnoselskii’s fixed-point theorem [26], we split operator A defined by (21) into the sum of two operators A1 and A2 on the closed ball Bρ=x∈C:x≤ρ with ρ≥φΩ1/1−Ω2,supt∈a,bφt=φ, where
(24)A1xt=Iα1+α2ft,xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2fη,xη−Iα1+α2fb,xb,t∈a,b,and
(25)A2xt=−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2−λθIα2xη+λIα2xb,t∈a,b.
For anyx,y∈Bρ, we have
(26)A1xt+A2yt≤supt∈a,bIα1+α2ft,xt+λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2fη,xη+λθIα2xη+Iα1+α2fb,xb+λIα2xb≤φΩ1+ρΩ2≤ρ.
This shows thatA1x+A2y∈Bρ. By using (H2), it is easy to establish that A2 is a contraction mapping.
Continuity of operatorA1 follows from that of f. Also, A1 is uniformly bounded on Bρ as
(27)A1x≤Ω1φ.
Now, we prove that operatorA1 is compact. Setting supt,x∈a,b×Bρft,x=f¯<∞, we obtain
(28)A1xt2−A1xt1=1Γα1+α2∫at1t2−sα1+α2−1−t1−sα1+α2−1fs,xsds+∫t1t2t2−sα1+α2−1fs,xsds+t2−aγ1+α2−1−t1−aγ1+α2−1ΛΓγ1+α2θf¯η−aα1+α2Γα1+α2+1+f¯b−aα1+α2Γα1+α2+1≤f¯Γα1+α2+12t2−t1α1+α2+t2−aα1+α2−t1−aα1+α2+t2−aγ1+α2−1−t1−aγ1+α2−1ΛΓγ1+α2θf¯η−aα1+α2Γα1+α2+1+f¯b−aα1+α2Γα1+α2+1⟶0ast2−t1⟶0,independently of x∈Bρ. Thus, A1 is equicontinuous and hence A1 is relatively compact on Bρ. In consequence, it follows from the Arzelá-Ascoli theorem that A1 is compact on Bρ. In view of the foregoing arguments, we deduce that the hypothesis of Krasnoselskii’s fixed-point theorem [26] holds true and consequently its conclusion implies that problem (1)–(2) has at least one solution on a,b.Example 9.
Consider the three-point boundary value problem of the Langevin equation with the Hilfer fractional derivative of the form:(29)DH4/5,3/4DH3/5,1/2+14xt=3sin2t5+3txt1+xt+14,t∈13,53,x13=0,x53=27x23.
Here,α1=4/5, α2=3/5, β1=3/4, β2=1/2, λ=1/4, a=1/3, b=5/3, η=2/3, and θ=2/7. Using the given values, it is found that γ1=19/20, γ2=4/5, Λ≈1.142227093, Ω1≈2.650804917, and Ω2≈0.7638622147<1. Notice that
(30)ft,x≤3sin2t5+3t+14≔φt.Clearly the hypothesis of Theorem8 holds true and consequently its conclusion implies that the boundary value problem (29) has at least one solution on 1/3,5/3.The Leray-Schauder Nonlinear Alternative [25] is used for our next existence result.Theorem 10.
Suppose that (H2) and the following conditions hold:
(H3) ft,x≤ptψx for each t,x∈a,b×ℝ, where ψ:0,∞⟶0,∞ is a continuous nondecreasing function and p∈Ca,b,ℝ+
(H4) There exists a constant M>0, such that
(31)1−Ω2MψMpΩ1>1,where Ω1,Ω2 are respectively given by (22) and (23).Then, there exists at least one solution for problem (1)–(2) on a,b.Proof.
Let us verify that operatorA defined by (21) satisfies the hypothesis of the Leray-Schauder Nonlinear Alternative [25]. In our first step, we establish that operator A maps bounded sets (balls) into a bounded set in C. For a number r>0, let Br=x∈C:x≤r be a bounded ball in C. Then, for t∈a,b, we have
(32)Axt≤supt∈a,bIα1+α2ft,xt+λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2fη,xη+λθIα2xη+Iα1+α2fb,xb+λIα2xb≤pψxb−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+θη−aα1+α2Γα1+α2+1+xλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2λb−aα2Γα2+1+λθη−aα2Γα2+1=pψxΩ1+xΩ2,and consequently,
(33)Ax≤pψrΩ1+rΩ2.
Next, we will show thatA maps bounded sets into equicontinuous sets of C. Let τ1,τ2∈a,b with τ1<τ2 and x∈Br. Then we have
(34)Axτ2−Axτ1≤pψrΓα1+α2+1τ2−aα1+α2−τ1−aα1+α2+λrΓα2+1τ2−τ1α2+τ2−aγ1+α2−1−τ1−aγ1+α2−1ΛΓγ+α2θη−aα1+α2Γα1+α2+1pψr+λθη−aα2Γα2+1pψr+b−aα1+α2Γα1+α2+1pψr+λb−aα2Γα2+1.
Observe that the right-hand side of the above inequality tends to zero independently ofx∈Br as τ2−τ1⟶0. Thus, the set ABr is equicontinuous. Therefore, the Arzelá-Ascoli theorem applies and hence operator A is completely continuous.
Finally, we show that the set of all solutions to equationsx=λAx is bounded for λ∈0,1.
Following the computation in the first step, we obtain(35)xt≤ψxpΩ1+xΩ2,t∈a,b,which yields
(36)1−Ω2xψxpΩ1≤1.
According to (H4), there exists M>0 satisfying x≠M. Introduce a set
(37)U=x∈Ca,b,ℝ:x<M,and notice that A:U¯⟶C is continuous and completely continuous. Then, the choice of U implies that there is no x∈∂U, such that x=λAx for some λ∈0,1. In consequence, we deduce by the nonlinear alternative of the Leray-Schauder type [25] that A has a fixed-point x∈U¯, which corresponds to a solution of problem (1)–(2). This completes the proof.Example 11.
Consider the three-point boundary value problem of the Langevin equation with the Hilfer fractional derivative of the form:(38)DH4/7,1/3DH2/5,4/5+15xt=1804t+1x8tx6t+1+14,t∈14,34,x14=0,x34=23x12.
Hereα1=4/7, α2=2/5, β1=1/3, β2=4/5, λ=1/5, a=1/4, b=3/4, η=1/2, and θ=2/3. Using the given values, we obtain γ1=5/7, γ2=22/25, Λ≈0.3752096355, Ω1≈2.316402992, and Ω2≈0.8402809682. Next, the nonlinear function is bounded as
(39)ft,x=1804t+1x8x6+1+14≤1804t+1x2+14≔ptψx,which satisfies condition H4 with p=1/20 and ψM=M2+1/4. Furthermore, we find that M∈0.2147201605,1.164306134, satisfying H4 of Theorem 10. Therefore, by applying Theorem 10, the boundary value problem (38) has at least one solution on 1/4,3/4.In the following result, we apply Banach’s fixed-point theorem to prove the existence of a unique solution of the problem at hand.Theorem 12.
Assume that(H5) ft,x−ft,y≤Lx−y, L>0 for each t∈a,b and x,y∈ℝ.
If the constantsΩ1,Ω2 defined by (22) and (23), respectively, are such that
(40)LΩ1+Ω2<1,then problem (1)–(2) has a unique solution on a,b.Proof.
Let us first show thatA defined by (21) satisfies ABr⊂Br, where Br=x∈C:x≤r with r≥MΩ1/1−LΩ1−Ω2 and supt∈a,bft,0=M<∞. For any x∈Br, we have
(41)Axt≤supt∈a,bIα1+α2ft,xt+λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2fη,xη+λθIα2xη+Iα1+α2fb,xb+λIα2xb≤Lx+Mb−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+θη−aα1+α2Γα1+α2+1+xλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2λb−aα2Γα2+1+λθη−aα2Γα2+1≤Lr+MΩ1+rΩ2<r,which implies that ABr⊂Br.
Next, we letx,y∈C. Then for t∈a,b, we have
(42)Axt−Ayt≤b−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+θη−aα1+α2Γα1+α2+1x−y+λb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2λb−aα2Γα2+1+λθη−aα2Γα2+1x−y=LΩ1+Ω2x−y,which implies that Ax−Ay≤LΩ1+Ω2x−y. As LΩ1+Ω2<1, A is a contraction. Therefore, by Banach’s fixed-point theorem, operator A has a fixed point which is indeed a unique solution of problem (1)–(2). The proof is finished.Example 13.
Consider the three-point boundary value problem of the Langevin equation with the Hilfer fractional derivative of the form:(43)DH1/2,3/5DH3/4,2/5+217xt=2e1/2−t159+2tx2t+2xt1+xt+23,t∈12,52,x12=0,x52=58x32.
Here,α1=1/2, α2=3/4, β1=3/5, β2=2/5, λ=2/17, a=1/2, b=5/2, η=3/2, θ=5/8, and
(44)ft,x=2e1/2−t159+2tx2+2x1+x+23.
Using the given data, we obtainγ1=4/5, γ2=17/20, Λ≈0.9439932223, Ω1≈6.724559445, Ω2≈0.7305167191, and L=1/40 as
(45)ft,x−ft,y≤4e1/2−t159+2tx−y≤140x−y.
Moreover, we haveLΩ1+Ω2≈0.8986307052<1. As all the assumptions of Theorem 12 are satisfied, we therefore deduce by its conclusion that problem (43) has a unique solution on 1/2,5/2.


## 4. Existence Results for Multivalued Problems (3) and (4)

Definition 14.
A continuous functionx is said to be a solution of problem (3)–(4) if xa=0,xb=θxη and there exists a function v∈L1a,b,ℝ with v∈Ft,x, a.e., on a,b such that
(46)xt=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vη−λθIα2xη−Iα1+α2vb+λIα2xb.
For eachy∈Ca,b,ℝ, define the set of selections of F by
(47)SF,x≔v∈L1a,b,ℝ:v∈Ft,xtona,b.Lemma 15 (see [27]).
LetX be a separable Banach space. Let F:a,b×ℝ⟶Pcp,cX be an L1-multivalued map and let Θ be a linear continuous mapping from L1a,b,X to Ca,b,X. Then the operator
(48)Θ∘SF:Ca,b,X→Pcp,cCa,b,X,x↦Θ∘SFx=ΘSF,x,is a closed graph operator in Ca,b,X×Ca,b,X.Our first existence result, dealing with the convex valuedF, is based on the nonlinear alternative of the Leray-Schauder type for (Kakutani) multivalued maps [25] with the assumption that F is Carathéodory.Theorem 16.
Suppose that (H2), (H4), and the following conditions hold:
(A1) F:a,b×ℝ⟶Pcp,cℝ is L1-Carathéodory, where Pcp,cℝ=Y∈Pℝ:Yiscompactandconvex
(A2) Ft,xP≔supy:y∈Ft,x≤ptψx for each t,x∈a,b×ℝ, where ψ:0,∞⟶0,∞ is a continuous nondecreasing function and p∈Ca,b,ℝ+
Then, there exists at least one solution for problem (3)–(4) on a,b.Proof.
Let us transform problem (3)–(4) into a fixed-point problem by introducing an operator ℱ:Ca,b,ℝ⟶PCa,b,ℝ as
(49)ℱx=h∈Ca,b,ℝ:ht=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vη−λθIα2xη−Iα1+α2vb+λIα2xb,for t∈a,b and v∈SF,x. Notice that the existence of a fixed point of ℱ ensures the existence of a solution of problems (1) and (2). This will be achieved by establishing that operator ℱ satisfies the hypothesis of the Leray-Schauder nonlinear alternative for the Kakutani maps [25]. We do it in several steps.Step 1.
SinceSF,x is convex (F has convex values), therefore ℱx is convex for each x∈Ca,b,ℝ.Step 2.
ℱ maps bounded sets (balls) into bounded sets in Ca,b,ℝ.LetBr=x∈Ca,b,ℝ:x≤r be a bounded set in Ca,b,ℝ. Then, for each h∈ℬx,x∈Br, there exists v∈SF,x such that
(50)ht=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vη−λθIα2xη−Iα1+α2vb+λIα2xb.Then, fort∈a,b, we have
(51)ht≤supt∈a,bIα1+α2vt+λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vη+λθIα2xη+Iα1+α2vb+λIα2xb≤pψxb−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+θη−aα1+α2Γα1+α2+1+xλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2λb−aα2Γα2+1+λθη−aα2Γα2+1=pψxΩ1+xΩ2.Thus,(52)h≤pψrΩ1+rΩ2.Step 3.
ℱ maps bounded sets into equicontinuous sets of Ca,b,ℝ.Letτ1,τ2∈a,b with τ1<τ2 and x∈Br. Then, for each h∈ℬx, we obtain
(53)ht2−ht1≤1Γα1+α2∫aτ1τ2−sα1+α2−1−τ1−sα1+α2−1vsds+∫τ1t2τ2−sα1+α2−1vsds+1Γα2∫aτ1τ2−sα2−1−τ1−sα2−1xsds+1Γα2∫τ1τ2τ2−sα2−1xsds+τ2−aγ1+α2−1−τ1−aγ1+α2−1Γγ+α2λθIα2xη+λIα2xb+τ2−aγ1+α2−1−τ1−aγ1+α2−1ΛΓγ1+α2θη−aα1+α2Γα1+α2+1+b−aα1+α2Γα1+α2+1pψr≤pψrΓα1+α22τ2−τ1α1+α2+τ2−aα1+α2−τ1−aα1+α2+rΓα22τ2−τ1α2+τ2−aα2−τ1−aα2+τ2−aγ1+α2−1−τ1−aγ1+α2−1Γγ+α2λθIα2xη+λIα2xbr+τ2−aγ1+α2−1−τ1−aγ1+α2−1ΛΓγ1+α2θη−aα1+α2Γα1+α2+1+b−aα1+α2Γα1+α2+1pψr⟶0asτ2−τ1⟶0independentlyofx∈Br.Therefore,ℱ:Ca,b,ℝ⟶PCa,b,ℝ is completely continuous by the application of the Arzelá-Ascoli theorem.Next, we show thatℱ is upper semicontinuous by proving that it has a closed graph ([28], Proposition 1.2) as ℱ is already shown to be completely continuous.Step 4.
ℱ has a closed graph.Letxn⟶x∗,hn∈ℱxn and hn⟶h∗. Then, we need to show that h∗∈ℱx∗. Associated with hn∈ℱxn, there exists vn∈SF,xn such that for each t∈a,b,
(54)ht=Iα1+α2vnt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vnη−λθIα2xη−Iα1+α2vnb+λIα2xb.Thus, it suffices to show that there existsv∗∈SF,x∗, such that for each t∈a,b,
(55)h∗t=Iα1+α2v∗t−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2v∗η−λθIα2xη−Iα1+α2v∗b+λIα2xb.Let us introduce the linear operatorΘ:L1a,b,ℝ⟶Ca,b,ℝ as by
(56)v↦Θvt=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vη−λθIα2xη−Iα1+α2vb+λIα2xb.Observe thathn−h∗→0, as n→∞. So, by Lemma 15, Θ∘SF is a closed graph operator. Furthermore, we have hnt∈ΘSF,xn. Since xn→x∗, we have
(57)h∗t=Iα1+α2v∗t−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2v∗η−λθIα2xη−Iα1+α2v∗b+λIα2xb,for some v∗∈SF,x∗.Step 5.
We show that there exists an open setU⊆Ca,b,ℝ with x∉νℱx for any ν∈0,1 and all x∈∂U.Letν∈0,1 and x∈νℱx. Then, there exists v∈L1a,b,ℝ with v∈SF,x, such that for t∈a,b, we have
(58)xt=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vη−λθIα2xη−Iα1+α2vb+λIα2xb.As in the second step, it can be shown that(59)xt≤pψxb−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+θη−aα1+α2Γα1+α2+1+xλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2λb−aα2Γα2+1+λθη−aα2Γα2+1=pψxΩ1+xΩ2,which implies that
(60)x≤pψxΩ1+xΩ2.Consequently(61)1−Ω2xpψxΩ1≤1.By H4, we can find a number M with x≠M. Define
(62)U=x∈Ca,b,ℝ:x<M.Notice that operatorℱ:U¯⟶PCa,b,ℝ is compact, upper semicontinuous, and convex valued. By the choice of U, we cannot find x∈∂U satisfying x∈νℱx for some ν∈0,1. Therefore, by the Leray-Schauder nonlinear alternative for the Kakutani maps [25], ℱ has a fixed point y∈U¯ which is a solution of problem (3)–(4). This completes the proof.In our next result, we show the existence of solutions for the nonconvex valued case of problem (3)–(4). For that, we need the following assumptions.
(B1) F:a,b×ℝ⟶Pcpℝ is such that F⋅,x:a,b⟶Pcpℝ is measurable for each x∈ℝ, where Pcpℝ=Y∈Pℝ:Yiscompact
(B2) HdFt,x,Ft,x¯≤mtx−x¯ for almost all t∈a,b and x,x¯∈ℝ with m∈Ca,b,ℝ+ and d0,Ft,0≤mt for almost all t∈a,bRecall thatHd:PX×PX→ℝ∪∞ is defined by
(63)HdV,W=maxsupv∈Vdv,W,supw∈WdV,w,where dV,w=infv∈Vdv;w, dv,W=infw∈Wdv;w, and X,d is a metric space induced from the normed space X;⋅.We apply a fixed-point theorem for multivalued maps due to Covitz and Nadler [29]: if N:X⟶PclX is a contraction, then FixN≠∅, where PclX=Y∈PX:Yisclosed.Theorem 17.
Suppose that conditions (B1) and (B2) hold and that
(64)δ≔mΩ1+Ω2<1.
Then, problem (3)–(4) has at least one solution on a,b.Proof.
In view of (B1), the set SF,x is nonempty for each x∈Ca,b,ℝ, and hence F has a measurable selection (see Theorem III.6 in [30]). Now, we verify that operator ℱ defined by (49) satisfies the hypothesis of Covitz and Nadler’s fixed theorem [29]. In order to establish that ℱx∈PclCa,b,ℝ for each x∈Ca,b,ℝ, let unn≥0∈ℱx be such that un⟶un⟶∞ in Ca,b,ℝ. Then, u∈Ca,b,ℝ and there exists vn∈SF,xn such that, for each t∈a,b
(65)unt=Iα1+α2vnt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vnη−λθIα2xη−Iα1+α2vnb+λIα2xb.
SinceF is compact valued, we pass onto a subsequence (if necessary) to obtain that vn converges to v in L1a,b,ℝ. Therefore, v∈SF,x, and for each t∈a,b, we have
(66)unt⟶ut=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2vη−λθIα2xη−Iα1+α2vb+λIα2xb.which implies that u∈ℱx.
Next, we show that we can find aδ<1 (defined by (64)) satisfying
(67)Hdℱx,ℱx¯≤δx−x¯foreachx,x¯∈C2a,b,ℝ.
Letx,x¯∈C2a,b,ℝ and h1∈ℱx. Then, there exists v1t∈Ft,xt such that, for each t∈a,b,
(68)h1t=Iα1+α2v1t−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2v1η−λθIα2xη−Iα1+α2v1b+λIα2xb.
By assumption (B2), we have that HdFt,x,Ft,x¯≤mtxt−x¯t. So we can find ϱ∈Ft,x¯t, such that
(69)v1t−ϱ≤mtxt−x¯t,t∈a,b.
DefineY:a,b⟶Pℝ by
(70)Yt=ϱ∈ℝ:v1t−ϱ≤mtxt−x¯t.
As the multivalued operatorYt∩Ft,x¯t is measurable (Proposition III.4 in [30]), we can find a function v2t which is a measurable selection for Y. So v2t∈Ft,x¯t, and for each t∈a,b, we have v1t−v2t≤mtxt−x¯t.
For eacht∈a,b, let us define
(71)h2t=Iα1+α2v2t−λIα2x¯t+t−aγ1+α2−1ΛΓγ1+α2θIα1+α2v2η−λθIα2x¯η−Iα1+α2v2b+λIα2x¯b.
Thus,(72)h1t−h2t=Iα1+α2v1t−v2t+λIα2xt−x¯t+b−aγ1+α2−1ΛΓγ1+α2θIα1+α2v1η−v2η+λθIα2xη−x¯η+Iα1+α2v1b−v2b+λIα2xb−x¯b≤b−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+θη−aα1+α2Γα1+α2+1mx−x¯+λb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2λb−aα2Γα2+1+λθη−aα2Γα2+1x−x¯=mΩ1+Ω2x−x¯.
Hence(73)h1−h2≤mΩ1+Ω2x−x¯.
By interchanging the roles ofx and x¯, one can obtain in a similar manner that
(74)Hdℱx,ℱx¯≤mΩ1+Ω2x−x¯,
This shows thatℱ is a contraction. So, by Covitz and Nadler’s fixed-point theorem [29], operator ℱ has a fixed-point x which corresponds to a solution of problem (3)–(4). This completes the proof.Example 18.
Consider the three-point boundary value problem of the Langevin inclusion with the Hilfer fractional derivative of the form:(75)DH3/7,2/3DH5/7,1/4+110xt∈Ft,xt,t∈35,95,x35=0,x95=34x75.where
(76)Ft,x=1+sinx15+t2,t+1411+2x1+x.Here,α1=3/7, α2=5/7, β1=2/3, β2=1/4, λ=1/10, a=3/5, b=9/5, η=7/5, and θ=3/4. We can find that γ1=17/21, γ2=11/14, Λ≈0.4879734002, Ω1≈5.461283890, and Ω2≈0.6208844245. It is clear that F is measurable for all x∈ℝ. Now, we see that
(77)HdFt,x,Ft,y≤t+141x−y,x,y∈ℝ,t∈35,95.By choosingmt=t+1/41, we have m=14/205 and also we obtain d0,Ft,0≤mt, t∈3/5,9/5. Then we find that
(78)mΩ1+Ω2≈0.9938501536<1.Hence, by using Theorem17, we get that the boundary value problem (75) has at least one solution on 3/5,9/5.



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