# Boundary Value Problem for the Langevin Equation and Inclusion with the Hilfer Fractional Derivative

Authors: Khalid Hilal; Ahmed Kajouni; Hamid Lmou
Journal: International Journal of Differential Equations
 (2022)

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## Abstract

In this work, we discuss the existence and uniqueness of solution for a boundary value problem for the Langevin equation and inclusion with the Hilfer fractional derivative. First of all, we give some definitions, theorems, and lemmas that are necessary for the understanding of the manuscript. Second of all, we give our first existence result, based on Krasnoselskii’s fixed point, and to deal with the uniqueness result, we use Banach’s contraction principle. Third of all, in the inclusion case, to obtain the existence result, we use the Leray–Schauder alternative. Last but not least, we give an illustrative example.


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## Body

## 1. Introduction

Fractional derivatives give an excellent description of memory and hereditary properties of different processes. Properties of the fractional derivatives make the fractional-order models more useful and practical than the classical integral-order models. Several researchers in the recent years have employed the fractional calculus as a way of describing natural phenomena in different fields such as physics, biology, finance, economics, and bioengineering (for more details see [1–10] and many other references).With the recent outstanding development in fractional differential equations, the Langevin equation has been considered a part of fractional calculus, and thus, important results have been elaborated [11–15].An equation of the formmd2x/dt2=λdx/dt+ηt is called Langevin equation, introduced by Paul Langevin in 1908. The Langevin equation is found to be an effective tool to describe the evolution of physical phenomena in fluctuating environments [8]. For some new developments on the fractional Langevin equation, see, for example, [16–18].In this study, we investigate the existence and uniqueness criteria for the solutions of the following nonlocal boundary value problem:(1)Dα1,β1HDα2,β2H+λxt=ft,xt,a≤t≤b,xa=0,xb=∑i=1nμiIνixη,a<η<b,where Dαi,βiH,i=1,2 is the Hilfer fractional derivative of order αi, 0<αi<1, and parameter βi, 0≤βi≤1, i=1,2, 1<α1+α2≤2, λ∈ℝ, a≥0, Iνi is the Riemann–Liouville fractional integral of order νi>0, μi∈ℝ, i=1,2, and f:a,b×ℝ⟶ℝ is a continuous function.In order to study problem (1), we transform it into a fixed-point problem and then use Krasnoselskii’s fixed-point theorem to prove the existence results.As a second problem, we study the multivalued case of (1) by considering the inclusion problem:(2)Dα1,β1HDα2,β2H+λxt∈Ft,xt,a≤t≤b,xa=0,xb=∑i=1nμiIνixη,a<η<b,where F:a,b×ℝ⟶Aℝ is a multivalued map (Aℝ is the family of all nonempty subjects of ℝ).We prove the existence of solution for problem (2) by applying the nonlinear alternative of the Leray–Schauder [19]. Finally, in the last part, we give an example to support our study.


## 2. Preliminaries

Let us recall some basic definitions and notations of fractional calculus and multivalued analysis which are needed throughout this study.


### 2.1. Fractional Calculus

Definition 1 (see [4]).
The Riemann–Liouville fractional integral of orderα>0 for a continuous function f:a,∞⟶ℝ can be defined as(3)Iαft=1Γα∫0tt−sα−1fsds,provided that the right-hand side exists on a,∞.Definition 2 (see [4]).
The Riemann–Liouville fractional derivative of orderα>0 of a continuous function f is defined by(4)DαRLft≔DnIn−αft=1Γn−αddtn∫0tt−sn−α−1fsds,where n−1<α<n,n=α+1, and α denotes the integer part of the real number α .Definition 3 (see [4]).
The Caputo fractional derivative of orderα>0 of a continuous function f is defined by(5)DαCft≔In−αDnft=1Γn−α∫0tt−sn−α−1ddtnfsds,where n−1<α<n,n=α+1, and α denotes the integer part of the real number α.Definition 4.
Hilfer fractional derivative [3].
The Hilfer fractional derivative of orderα and parameter β of a function (also known as the generalized Riemann–Liouville fractional derivative) is defined by(6)Dα,βHft=Iβn−αDnI1−βn−αft,where n−1<α<n, 0≤β≤1, t>a, and D=d/dt.Remark 1.
Whenβ=0, the Hilfer fractional derivative corresponds to the Riemann–Liouville fractional derivative:(7)Dα,0Hft=DnIn−αft.
Whenβ=1, the Hilfer fractional derivative corresponds to the Caputo fractional derivative:(8)Dα,1Hft=In−αDnft.
The following lemma plays a fundamental role in establishing the existence results for the given problem.Lemma 1 (see [3]).
Letf∈La,b, n−1<α≤n, n∈ℕ, 0≤β≤1, and I1−βn−αf∈ACka,b; then,(9)IαHDα,βft=ft−∑k=1n−1t−ak−n−α1−βΓk−n−α1−β+1.limt⟶+adkdtkI1−βn−αft.The following lemma deals with a linear variant of the boundary value problem (1).Lemma 2.
Leta≥0, 0<αi<1, γi=αi+βi−αiβi, i=1,2, 1<α1+α2≤2, and h∈Ca,b,ℝ. Then, the function x is a solution of the boundary value problem:(10)Dα1,β1HDα2,β2H+λxt=ht,a≤t≤b,xa=0,xb=∑i=1nμiIνixηds,a<η<b,if and only if(11)xt=Iα1+α2ht−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη,where(12)Λ=∑i=1nμiη−aγ1+α2+νi−1Γγ1+α2+νi−b−aγ1+α2−1Γγ1+α2≠0.Proof.
Applying the Riemann–Liouville fractional integral of orderα1 to both sides of (10), we obtain by using Lemma 1(13)Dα2,β2Hxt+λxt=Iα1ht+c0Γγ1t−aγ1−1,where c0 is the constant and γ1=α1+β1−α1β1. Applying the Riemann–Liouville fractional integral of order α2 to both sides of (13), we obtain by using Lemma 1(14)xt=Iα1+α2ht−λIα2xt+c0Γγ1+α2t−aγ1+α2−1+c1Γγ2t−aγ2−1.
From using the boundary conditionxa=0 in (14), we obtain that c1=0. Then, we get(15)xt=Iα1+α2ht−λIα2xt+c0Γγ1+α2t−aγ1+α2−1.
From using the boundary conditionxb=∑i=1nμiIνixη, in (15), we find(16)c0=1ΛIα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη.
Substituting the value ofc0 in (13), we obtain the solution (11).
Conversely, suppose thatx is the solution of the fractional integral (11). By applying fractional derivetive Dα2,β2H on both sides of (11) and then applying fractional derivative, we obtain that(17)Dα1,β1HDα2,β2Hxt=ht−λHDα1,β1xt.
It follows that(18)Dα1,β1HDα2,β2H+λxt=ht,a≤t≤b.
Now, we will prove thatx satisfies the boundary conditions; for that, we have xa=0, and from (11), we have(19)∑i=1nμiIνixη=∑i=1nμiIα1+α2+νihη−λ∑i=1nμiIα2+νixη+∑i=1nμiη−aγ1+α2+νi−1Γγ1+α2+νi×1ΛIα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη.
By (12), we get(20)∑i=1nμiIνixη=∑i=1nμiIα1+α2+νihη−λ∑i=1nμiIα2+νixη+1+b−aγ1+α2−1ΛΓγ1+α2×Iα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη=Iα1+α2hb−λIα2xb+b−aγ1+α2−1ΛΓγ1+α2Iα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη=xb.
This completes the proof.


### 2.2. Multivalued Analysis

For a normed spaceX,., we define(21)AX=Y⊂X:Y≠∅,Ac,cpX=Y⊂X:Yis convex and compact.For the basic concepts of multivalued analysis, we refer to ([2, 3]).Definition 5.
A multivalued mapF:a,b×ℝ⟶Aℝ is said to be Carathéodory if(i)
t⟶Ft,x is measurable for each x∈ℝ(ii)
x⟶Ft,x is upper semicontinuous for almost all t∈a,bFurthermore, a Carathéodory functionF is called L1-Carathéodory if(iii)
For eachρ>0, there exists φρ∈L1a,b;ℝ, such that(22)Ft,x=supv:v∈Ft,x≤φρt,for all x∈ℝ with x≤ρ and for a.e. t∈a,b.
Fixed-point theorems play a major role in establishing the existence theory for problem (1) and problem (2). We collect here some well-known fixed-point theorems used in this study.Theorem 1 Krasnoselskii’s fixed-point theorem [19].
Letℳ be a closed, bounded, convex, and nonempty subset of a Banach space. Let A,ℬ be the operators such that(i)
Ax+ℬy∈ℳ whenever x,y∈ℳ(ii)
A is compact and continuous(iii)
ℬ is contraction mapping
Then, there existsz∈ℳ, such that z=Az+ℬz.Theorem 2 (Leray–Schauder nonlinear alternative [19]).
LetX be a Banach space, C a closed, convex subset of X, U an open subset of C, and 0∈U. Suppose that F:U¯⟶C is a continuous, compact (FU¯ is a relatively compact subset of C) map. Then, either(i)
F has a fixed point in U¯ or(ii)
There existsx∈∂U (the boundary of U in C) and θ∈0,1 with x=θFx.


## 2.1. Fractional Calculus

Definition 1 (see [4]).
The Riemann–Liouville fractional integral of orderα>0 for a continuous function f:a,∞⟶ℝ can be defined as(3)Iαft=1Γα∫0tt−sα−1fsds,provided that the right-hand side exists on a,∞.Definition 2 (see [4]).
The Riemann–Liouville fractional derivative of orderα>0 of a continuous function f is defined by(4)DαRLft≔DnIn−αft=1Γn−αddtn∫0tt−sn−α−1fsds,where n−1<α<n,n=α+1, and α denotes the integer part of the real number α .Definition 3 (see [4]).
The Caputo fractional derivative of orderα>0 of a continuous function f is defined by(5)DαCft≔In−αDnft=1Γn−α∫0tt−sn−α−1ddtnfsds,where n−1<α<n,n=α+1, and α denotes the integer part of the real number α.Definition 4.
Hilfer fractional derivative [3].
The Hilfer fractional derivative of orderα and parameter β of a function (also known as the generalized Riemann–Liouville fractional derivative) is defined by(6)Dα,βHft=Iβn−αDnI1−βn−αft,where n−1<α<n, 0≤β≤1, t>a, and D=d/dt.Remark 1.
Whenβ=0, the Hilfer fractional derivative corresponds to the Riemann–Liouville fractional derivative:(7)Dα,0Hft=DnIn−αft.
Whenβ=1, the Hilfer fractional derivative corresponds to the Caputo fractional derivative:(8)Dα,1Hft=In−αDnft.
The following lemma plays a fundamental role in establishing the existence results for the given problem.Lemma 1 (see [3]).
Letf∈La,b, n−1<α≤n, n∈ℕ, 0≤β≤1, and I1−βn−αf∈ACka,b; then,(9)IαHDα,βft=ft−∑k=1n−1t−ak−n−α1−βΓk−n−α1−β+1.limt⟶+adkdtkI1−βn−αft.The following lemma deals with a linear variant of the boundary value problem (1).Lemma 2.
Leta≥0, 0<αi<1, γi=αi+βi−αiβi, i=1,2, 1<α1+α2≤2, and h∈Ca,b,ℝ. Then, the function x is a solution of the boundary value problem:(10)Dα1,β1HDα2,β2H+λxt=ht,a≤t≤b,xa=0,xb=∑i=1nμiIνixηds,a<η<b,if and only if(11)xt=Iα1+α2ht−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη,where(12)Λ=∑i=1nμiη−aγ1+α2+νi−1Γγ1+α2+νi−b−aγ1+α2−1Γγ1+α2≠0.Proof.
Applying the Riemann–Liouville fractional integral of orderα1 to both sides of (10), we obtain by using Lemma 1(13)Dα2,β2Hxt+λxt=Iα1ht+c0Γγ1t−aγ1−1,where c0 is the constant and γ1=α1+β1−α1β1. Applying the Riemann–Liouville fractional integral of order α2 to both sides of (13), we obtain by using Lemma 1(14)xt=Iα1+α2ht−λIα2xt+c0Γγ1+α2t−aγ1+α2−1+c1Γγ2t−aγ2−1.
From using the boundary conditionxa=0 in (14), we obtain that c1=0. Then, we get(15)xt=Iα1+α2ht−λIα2xt+c0Γγ1+α2t−aγ1+α2−1.
From using the boundary conditionxb=∑i=1nμiIνixη, in (15), we find(16)c0=1ΛIα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη.
Substituting the value ofc0 in (13), we obtain the solution (11).
Conversely, suppose thatx is the solution of the fractional integral (11). By applying fractional derivetive Dα2,β2H on both sides of (11) and then applying fractional derivative, we obtain that(17)Dα1,β1HDα2,β2Hxt=ht−λHDα1,β1xt.
It follows that(18)Dα1,β1HDα2,β2H+λxt=ht,a≤t≤b.
Now, we will prove thatx satisfies the boundary conditions; for that, we have xa=0, and from (11), we have(19)∑i=1nμiIνixη=∑i=1nμiIα1+α2+νihη−λ∑i=1nμiIα2+νixη+∑i=1nμiη−aγ1+α2+νi−1Γγ1+α2+νi×1ΛIα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη.
By (12), we get(20)∑i=1nμiIνixη=∑i=1nμiIα1+α2+νihη−λ∑i=1nμiIα2+νixη+1+b−aγ1+α2−1ΛΓγ1+α2×Iα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη=Iα1+α2hb−λIα2xb+b−aγ1+α2−1ΛΓγ1+α2Iα1+α2hb−λIα2xb−∑i=1nμiIα1+α2+νihη+λ∑i=1nμiIα2+νixη=xb.
This completes the proof.


## 2.2. Multivalued Analysis

For a normed spaceX,., we define(21)AX=Y⊂X:Y≠∅,Ac,cpX=Y⊂X:Yis convex and compact.For the basic concepts of multivalued analysis, we refer to ([2, 3]).Definition 5.
A multivalued mapF:a,b×ℝ⟶Aℝ is said to be Carathéodory if(i)
t⟶Ft,x is measurable for each x∈ℝ(ii)
x⟶Ft,x is upper semicontinuous for almost all t∈a,bFurthermore, a Carathéodory functionF is called L1-Carathéodory if(iii)
For eachρ>0, there exists φρ∈L1a,b;ℝ, such that(22)Ft,x=supv:v∈Ft,x≤φρt,for all x∈ℝ with x≤ρ and for a.e. t∈a,b.
Fixed-point theorems play a major role in establishing the existence theory for problem (1) and problem (2). We collect here some well-known fixed-point theorems used in this study.Theorem 1 Krasnoselskii’s fixed-point theorem [19].
Letℳ be a closed, bounded, convex, and nonempty subset of a Banach space. Let A,ℬ be the operators such that(i)
Ax+ℬy∈ℳ whenever x,y∈ℳ(ii)
A is compact and continuous(iii)
ℬ is contraction mapping
Then, there existsz∈ℳ, such that z=Az+ℬz.Theorem 2 (Leray–Schauder nonlinear alternative [19]).
LetX be a Banach space, C a closed, convex subset of X, U an open subset of C, and 0∈U. Suppose that F:U¯⟶C is a continuous, compact (FU¯ is a relatively compact subset of C) map. Then, either(i)
F has a fixed point in U¯ or(ii)
There existsx∈∂U (the boundary of U in C) and θ∈0,1 with x=θFx.


## 3. Existence and Uniqueness Results for Problem (1)

In this section, we deal with the existence and uniqueness of solution for the boundary value problem (1).By Lemma2, we define an operator A:C⟶C by(23)Axt=Iα1+α2ft,xt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2fb,xb−λIα2xb−∑i=1nμiIα1+α2+νifη,xη+λ∑i=1nμiIα2+νixη,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. We will show that the boundary value problem (1) has a solution if and only if the operator A has a fixed point.To simplify the computations, we use the following notations:(24)Ω1=b−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα1+α2Γα1+α2+1+∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1,(25)Ω2=λb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2b−aα2Γα2+1+∑i=1nμiη−aα2+νiΓα2+νi+1.Our first result is an existence result, based on well-known Krasnoselskii’s fixed-point theorem.Theorem 3.
Assume that(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 (19)Then, there exists at least one solution for the boundary value problem (1) on a,b.Proof.
We will show that the operatorA defined by (17) satisfies the assumptions of Krasnoselskii’s fixed-point theorem. We split the operator A 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(26)A1xt=Iα1+α2ft,xt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2fb,xb−∑i=1nμiIα1+α2+νifη,xη,(27)A1xt=−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2−λIα2xb+λ∑i=1nμiIα2+νixη.
For anyx,y∈ℬρ, we have(28)A1xt+A2yt≤supt∈a,bIα1+α2ft,xt+λIα2yt+t−aγ1+α2−1ΛΓγ1+α2×Iα1+α2fb,xb+∑i=1nμiIα1+α2+νifη,xη+λIα2yb+λ∑i=1nμiIα2+νiyη≤φb−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα1+α2Γα1+α2+1+∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1+yλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα2Γα2+1+∑i=1nμiη−aα2+νiΓα2+νi+1,≤φΩ1+ρΩ2,≤ρ,and hence, A1x+A2y≤ρ, which implies that A1x+A2y∈ℬρ.
Next, by usingH2, we show that A2 is a contraction mapping. Let x,y∈C, for t∈a,b,(29)A2xt+A2yt≤λIα2xt−yt+b−aγ1+α2−1ΛΓγ1+α2×λIα2xb−yb+λ∑i=1nμiIα2+νixη−yη,≤x−yλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα2Γα2+1+∑i=1nμiη−aα2+νiΓα2+νi+1≤Ω2x−y.
This shows thatA2x+A2y≤Ω2x−y; then, by using H2, A2 is a contraction mapping.
The operatorA1 is continuous, since f is continuous. It is uniformly bounded on ℬρ as(30)A1x≤Ω1φ.
Now, we prove that the operatorA1 is compact. Setting supt,x∈a,b×ℬρft,x=f¯<∞, and let t1,t2∈a,b, t1<t2, we obtain(31)A1xt2−A1xt1=Iα1+α2ft2,xt2+t2−aγ1+α2−1ΛΓγ1+α2Iα1+α2fb,xb−∑i=1nμiIα1+α2+νifη,xη−Iα1+α2ft1,xt1−t1−aγ1+α2−1ΛΓγ1+α2Iα1+α2fb,xb−∑i=1nμiIα1+α2+νifη,xη,≤f¯Γα1+α2∫at1t2−sα1+α2−1−t1−sα1+α2−1ds+∫t1t2t2−sα1+α2−1ds+t2−aγ1+α2−1−t1−aγ1+α2−1ΛΓγ1+α2×f¯b−aα1+α2Γα1+α2+1+f¯∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1,where the right-hand side tends to zero as t2−t1⟶0, independently of x∈ℬρ. Then, A1 is equicontinuous, and hence, A1 is relatively compact on ℬρ. By the Arzelá–Ascoli theorem, A1 is compact on ℬρ. It follows by Krasnoselskii’s fixed-point theorem that problem (1) has at least one solution on a,b.
To deal with the uniqueness of solution for our problem (1), we use Banach’s contraction principle.Theorem 4.
Assume thatft,x−ft,y≤Lx−y, L>0, for each t∈a,b and x,y∈ℝ.IfLΩ1+Ω2<1, where Ω1,Ω2 are, respectively, given by (24) and (25), then problem (1) has a unique solution on a,b.Proof.
Consider the operatorA defined in (17). The problem (1) is then transformed into a fixed-point problem x=Ax. By using Banach contraction principle, we will show that A has a unique fixed point.
We setsupt∈a,b=ft,0=M<∞ and choose r>0, such that(32)r≥MΩ11−LΩ2−Ω1.
Now, we show thatAℬr⊂ℬr, where ℬr=x∈Ca,b,ℝ;x≤r. For any x∈ℬr, we have(33)Axt≤supt∈a,bIα1+α2ft,xt+λIα2xt|+t−aγ1+α2−1ΛΓγ1+α2×Iα1+α2fb,xb+∑i=1nμiIα1+α2+νifη,xη+λxb+λ∑i=1nμiIα2+νixη≤Lx+Mb−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα1+α2Γα1+α2+1+∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1+xλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα2Γα2+1+∑i=1nμiη−aα2+νiΓα2+νi+1≤Lx+MΩ1+xΩ2,≤Lr+MΩ1+rΩ2,which implies that Aℬr⊂ℬr.
Next, letx,y∈Ca,b,ℝ. Then, for t∈a,b, we have(34)Axt−Ayt≤b−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα1+α2Γα1+α2+1+∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1Lx−y+λb−aα2Γα2+1+b−aγ1+α2−1Λγ1+α2×b−aα2Γα2+1+∑i=1nμiη−aα2+νiΓα2+νi+1x−y,≤LΩ1+Ω2x−y,which implies Axt−Ayt≤LΩ1+Ω2x−y. As LΩ1+Ω2<1, A is a contraction. Therefore, by Banach’s fixed-point theorem, the operator A has a fixed point which is indeed a unique solution of problem (1).


## 4. Existence Results for Problem (2)

Definition 6.
A continuous functionx is said to be a solution of problem (2) if xa=0,xb=∑i=1nμiIνixη, and there exists a function v∈L1a,b,ℝ with v∈Ft,xt, a.e., on a,b, such that(35)xt=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2vb−λIα2xb−∑i=1nμiIα1+α2+νivη+λ∑i=1nμiIα2+νixη.
For eachx∈Ca,b,ℝ, define the set of selections of F by(36)SF,x≔v∈L1a,b,ℝ:v∈Ft,xtona,b.Lemma 3 (see [5]).
Let X be a Banach space, andF:a,b×ℝ⟶Ac,cp be a L1-Carathéodory multivalued map. Let Υ be a linear continuous mapping from L1a,b,X to Ca,b,X. Then, the operator isϒ∘SF:Ca,b,X⟶Ac,cpCa,b,X;x⟶ϒ∘SFx=ϒSF,x is a closed graph operator in Ca,b,X×Ca,b,X.Our second existence result is based on the nonlinear alternative of the Leray–Schauder for multivalued maps [19].Theorem 5.
Suppose thatH2 and the following assumptions hold:(H3)F:a,b×ℝ⟶Ac,cpℝ is L1-Carathéodory and has nonempty compact and convex values, and for each fixed x∈Ca,b,ℝ, the setSF,x=v∈L1a,b,X:vt∈Ft,xt;t∈a,b is nonempty.(H4)Ft,x≔supv:v∈Ft,x≤ptΨx for all t∈a,b and all x∈Ca,b,X, where p∈L1a,b,ℝ+, and Ψ:ℝ+⟶0,+∞ is a continuous and nondecreasing function.(H5) There exists a constantM>0, such that(37)1−Ω2MpΨMΩ1>1,where Ω1,Ω2 are, respectively, given by (24) and (25).Then, there exists at least one solution for problem (2) on a,b.Proof.
Let us introduce the operatorA:Ca,b,ℝ⟶Ac,cpa,b,ℝ, in order to transform problem (2) into a fixed-point problem:(38)Ax≔h∈Ca,b,ℝ:ht=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2×Iα1+α2vb−λIα2xb−∑i=1nμiIα1+α2+νivη+λ∑i=1nμiIα2+νixη,t∈a,b,v∈SF,x.
We will show that the operatorA satisfies all conditions of the Leray–Schauder nonlinear alternative [19], and we give the proof in several steps:Step 1.
Ax is convex for each x∈Ca,b,ℝ.
Indeed, ifh1,h2 belong to Ax, then there exist v1,v2∈SF,x, such that for each t∈a,b, we have(39)hjt=Iα1+α2vjt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2vjb−λIα2xb−∑i=1nμiIα1+α2+νivjη+λ∑i=1nμiIα2+νixη,for j=1,2. Let 0≤k≤1; then, for each t∈0,1, we have(40)kh1t+1−kh2t=Iα1+α2kv1s+1−kv2s−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2×Iα1+α2kv1s+1−kv2s−λIα2xb−∑i=1nμiIα1+α2+νikv1η+1−kv2η+λ∑i=1nμiIα2+νixη.
Thus,kv1+1−kv2∈Axbecause SF,xis convex; then, Ax is convex for each x∈Ca,b,ℝ.Step 2.
Ax maps bounded set into bounded set in Ca,b,ℝ.
Indeed, it is enough to show that there exists a positive constantl, such that for each h∈Ax, x∈ℬρ=x∈Ca,b,ℝ:x≤ρ, and we have h≤l.
Ifh∈Ax, then there exist v∈SF,x, such that(41)ht=Iα1+α2vt−λIα2xt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2vb−λIα2xb−∑i=1nμiIα1+α2+νivη+λ∑i=1nμiIα2+νixη.
Then, for everyt∈a,b, we have(42)hxt≤supt∈a,bIα1+α2vt+λIα2xt+t−aγ1+α2−1ΛΓγ1+α2×Iα1+α2vb+∑i=1nμiIα1+α2+νivη+λIα2xb+λ∑i=1nμiIα2+νixη≤pΨxb−aα1+α2Γα1+α2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα1+α2Γα1+α2+1+∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1+xλb−aα2Γα2+1+b−aγ1+α2−1ΛΓγ1+α2×b−aα2Γα2+1+∑i=1nμiη−aα2+νiΓα2+νi+1≤pΨxΩ1+xΩ2,≤pΨρΩ1+ρΩ2.
Then,(43)h≤pΨρΩ1+ρΩ2≔l,where Ω1,Ω2 are, respectively, given by (24) and (25).Step 3.
A maps bounded set into equicontinuous sets of Ca,b,ℝ.
Lett1,t2∈a,b;t1<t2, and x∈ℬρ, where ℬρ, as above, is a bounded set of Ca,b,ℝ; for each x∈ℬρ and h∈Ax, there exist v∈SF,x; then, we obtain(44)ht2−ht1≤1Γα1+α2∫at1t2−sα1+α2−1−t1−sα1+α2−1vsds+∫t1t2t2−sα1+α2−1vsds+λΓα2∫at1t2−sα2−1−t1−sα2−1xsds+∫t1t2t2−sα2−1xsds+t2−aγ1+α2−1−t1−aγ1+α2−1ΛΓγ1+α2×vsb−aα1+α2Γα1+α2+1+vs∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1+xbλb−aα2Γα2+1+xηλ∑i=1nμiη−aα2+νiΓα2+νi+1≤pΨρΓα1+α2∫at1t2−sα1+α2−1−t2−sα1+α2−1ds+∫t1t2t2−sα1+α2−1ds+ρλΓα2∫at1t2−sα2−1−t2−sα2−1ds+∫t1t2t2−sα2−1ds+t2−aγ1+α2−1−t2−aγ1+α2−1ΛΓγ1+α2×pΨρb−aα1+α2Γα1+α2+1+pΨρ∑i=1nμiη−aα1+α2+νiΓα1+α2+νi+1+ρλb−aα2Γα2+1+ρλ∑i=1nμiη−aα2+νiΓα2+νi+1,as t2⟶t1, the right-hand side of the above inequality tends to zero, implies that Ax is equicontinuous. Therefore, it follows by Arzelà-Ascoli theorem that A:Ca,b,ℝ⟶Ac,cpCa,b,ℝ is relatively compact; then, A is completely continuous.
Now, we show that the operatorA is upper semicontinuous. To prove this, it is enough to show that A has a closed graph.Step 4.
N¯ has a closed graph.
Letxn⟶x∗, hn∈Axn, and hn⟶h∗; we shall prove that h∗∈Ax∗. hn∈Axn, and then, there exists vn∈SF,xn, such that for each t∈a,b,(45)hnt=Iα1+α2vnt−λIα2xnt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2vnb−λIα2xnb−∑i=1nμiIα1+α2+νivnη+λ∑i=1nμiIα2+νixnη.
We should prove thatv∗∈SF,x∗, such that for each t∈a,b,(46)h∗t=Iα1+α2v∗t−λIα2x∗t+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2v∗b−λIα2x∗b−∑i=1nμiIα1+α2+νiv∗η+λ∑i=1nμiIα2+νix∗η,and we have that(47)hnt+λIα2xnt−t−aγ1+α2−1ΛΓγ1+α2−λIα2xnb+λ∑i=1nμiIα2+νixnη−h∗t+λIα2x∗t−t−aγ1+α2−1ΛΓγ1+α2−λIα2x∗b+λ∑i=1nμiIα2+νix∗η⟶0,as n⟶∞.
Consider the linear operator:(48)v⟶ϒvt,ϒ:L1a,b,ℝ⟶Ca,b,ℝ,with(49)ϒvt=Iα1+α2vt+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2vb−∑i=1nμiIα1+α2+νivη.
From Lemma3, ϒ∘SF is a closed graph operator; then, we have that(50)hnt+λIα2xnt−t−aγ1+α2−1ΛΓγ1+α2−λIα2xnb+λ∑i=1nμiIα2+νixnη∈ϒSF,xn.
Sincexn⟶x∗ and hn⟶h∗ then(51)h∗t+λIα2x∗t−t−aγ1+α2−1ΛΓγ1+α2−λIα2x∗b+λ∑i=1nμiIα2+νix∗η=ϒv∗∈ϒSF,x∗.
It follows thatv∗∈SF,x∗, such that(52)h∗t=Iα1+α2v∗t−λIα2x∗t+t−aγ1+α2−1ΛΓγ1+α2Iα1+α2v∗b−λIα2x∗b−∑i=1nμiIα1+α2+νiv∗η+λ∑i=1nμiIα2+νix∗η.Step 5.
A has a fixed point in ℬρ.
We show thatii from Theorem 2 is not possible. Then, if x∈θAx for θ∈0,1, there exist v∈SF,x, such that xt=θht implies xt≤ht; then,(53)x≤pΨxΩ1+xΩ2,and then,(54)1−Ω2x≤pΨx.
Ifii from Theorem 2 hold, then there exist θ∈0,1 and x∈∂ℬM avec x=θA which means that x is solution to (2) with x=M; then, we have from (32) that(55)1−Ω2M≤pΨM,and then,(56)1−Ω2MpΨM≤1,which contradicts H5. Consequently, A has a fixed point in a,b.
By the nonlinear alternative of Leray–Schauder, we deduce that our problem (2) has at least one solution. This completes the proof.


## 5. Example

Consider the following problem:(57)D3/7,2/7D5/7,4/7+113xt=24t+31x2t+2xt1+xt+14,14≤t≤54,x14=0,x54=38I2/3x34+58I4/3x34+78I5/3x34,where α1=3/7, α2=5/7, β1=2/7, β2=4/7, λ=1/13, a=1/4, b=5/4, n=3, μ1=3/8, μ2=5/8, μ3=7/8, ν1=2/3, ν2=4/3, ν3=5/3, and η=3/4, with α1+α2=8/7∈1,2.With these values, we getγ1=α1+β1−α1β1=29/49, γ2=α2+β2−α2β2=43/49, Λ≃0⋅6572399879, Ω1=4⋅725942806, Ω2=0⋅3825052846<1, and L=1/8, while(58)ft,x−ft,y≤18x−y,x,y∈ℝ.SinceLΩ1+Ω2≃0⋅9732481354<1, then the conditions of Theorem 4 are satisfied; finally, problem (1) has a unique solution on 1/4,5/4.


## 6. Conclusion

In the current article, we study and investigate the existence and uniqueness of solution for a boundary value problem for the Langevin equation and inclusion with the Hilfer fractional derivative. The novelty of this work is that it is more general than the works based on the derivative of Caputo and Riemann–Liouville because whenβ=0, we find the Riemann–Liouville fractional derivative, and when β=1, we find the Caputo fractional derivative. In this study, we established the existence and uniqueness results for the first problem, by using the fixed-point theorems (Banach’ fixed-point theorem and Krasnoselskii’s fixed-point theorem), and for the inclusion case, we use Leray–Schauder alternative to prove the existence of solution. In the end, we give an example to illustrate our results.



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*Source: 3386198-2022-03-11.xml*