# A New Result forψ-Hilfer Fractional Pantograph-Type Langevin Equation and Inclusions

Authors: Hamid Lmou; Khalid Hilal; Ahmed Kajouni
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we deal with the existence and uniqueness of solution forψ-Hilfer Langevin fractional pantograph differential equation and inclusion; these types of pantograph equations are a special class of delay differential equations. The existence and uniqueness results are obtained by making use of the Krasnoselskii fixed-point theorem and Banach contraction principle, and for the inclusion version, we use the Martelli fixed-point theorem to get the existence result. In the end, we are giving an example to illustrate our results.


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## Body

## 1. Introduction

Over the past few years, fractional differential equations have attracted the interest of many mathematicians due to their ability to describe several complex problems in different scientific and engineering fields such as physics, biology, chemistry, and control theory (for more details, see [1–14]).With the recent outstanding development in fractional differential equations, the Langevin equation has been considered a part of fractional calculus, and its form ismd2x/dt2=λdx/dt+ηt, introduced by Paul Langevin in 1908. The Langevin equation is an effective tool that can describe processes as regards time evolution of the velocity of the Brownian motion [15–19] and also describe the evolution of physical phenomena in fluctuating environments [8] (for more details see, for example, [2, 20, 21]).There are diverse definitions of fractional integrals and derivatives, the famous definitions are the Riemann-Liouville and the Caputo fractional derivatives. Hilfer [3] introduced the generalization of these derivatives under the name of Hilfer fractional derivative of order α and parameter β∈0,1.The authors in [14] have investigated the existence and uniqueness of solution for an initial value problem of Langevin equation involving two fractional orders, as follows:
(1)DβDα+λxt=ft,xt,0≤t≤1,0<α≤1,1≤β≤2,x0=x1=0,D2αx1+λDαx1=0.where Dα is is the Caputo fractional derivative of order α, f:0,1×ℝ⟶ℝ is a given continuous function, and λ is a real number.Motivated by the mentioned work and the new research going on in this direction, we study a new and a challenging case of fractional derivative called theψ-Hilfer derivative [22]; this brand of fractional derivative generalizes the well-known fractional derivatives (Riemman-Liouville, Caputo, ψ-Riemman-Liouville, Hilfer-Hadamard, and Katugampola derivative), for different values of function ψ and parameter β; these values are described in what follows.In this paper, we study the existence and uniqueness results of the solutions for the following problem:(2)DHα1,β1;ψHDα2,β2;ψ+μxt=ft,xt,xλt,a≤t≤bxa=0,xb=∑i=1nωiIσi;ψxηi.where HDαj,βj;ψ,j=1,2 are the ψ-Hilfer fractional derivative of order αj, 0<αj<1 and parameter βj, 0≤βj≤1, j=1,2, μ∈ℝ∗, 0<λ<1, a≥0, Iσi;ψ, are the ψ-Riemann-Liouville fractional integral of order σi>0, ωi∈ℝ∗, i=1,⋯,n, a<η1..⋯<ηn<b, and f:a,b×ℝ2⟶ℝ is a continuous function.We also will study the multivalued version of the problem (2) by considering the following problem:
(3)HDα1,β1;ψHDα2,β2;ψ+μxt∈Ft,xt,xλt,a≤t≤bxa=0,xb=∑i=1nωiIσi;ψxηi.where F:a,b×ℝ2⟶Pℝ is a multivalued map (Pℝ is the family of all nonempty subjects of ℝ).The novelty of this paper and its challenges are that it collects and generalizes the types of fractional derivative, for different values of functionψ and parameter βi,i=1,2 such as follows:
(i)
Ifψx=x and βi=1, then the problems (2) and (3) reduce to the Caputo-type(ii)
Ifψx=x and βi=0, then the problems (2) and (3) reduce to the Riemman-Liouville-type(iii)
Ifβi=0, then the problems (2) and (3) reduce to the ψ-Riemman-Liouville-type(iv)
Ifψx=x, then the problems (2) and (3) reduce to the Hilfer-type(v)
Ifψx=logx, then the problems (2) and (3) reduce to the Hilfer-Hadamard-type(vi)
Ifψx=xρ, then the problems (2) and (3) reduce to Katugampola-typeThis paper is structured as follows: In the second section, we will present some auxiliary lemmas, some basic definitions, and theorems which are needed throughout this paper. In the third section, we discuss the existence and uniqueness results for the first problem, by using Krasnoselskii’s fixed-point theorem and Banach’s contraction principle. In the fourth section, we deal with the existence results for the inclusion version, by making use of Martelli’s fixed-point theorem, which is applicable to completely continuous operators. Finally, in the last part, we give an example to support our study.


## 2. Preliminaries and Notations

### 2.1. Fractional Calculus

In this section, we introduce some definitions, lemmas, and useful notations that will be used throughout this paper.LetC=Ca,b,ℝ denote the Banach space of all continuous functions from a,b into ℝ with the norm defined by f=supt∈a,bft. We denote by ACna,b,ℝ the n-times absolutely continuous functions given by
(4)ACna,b,ℝ=f:a,b⟶ℝ;fn−1∈ACa,b,ℝ.Definition 2.1 [23].
Leta,b, −∞≤a<b≤+∞, be a finite or infinite interval of the half-axis 0,+∞ and α>0. In addition, let ψt be a positive increasing function on a,b, which has a continuous derivative ψ′t on a,b. The ψ-Riemann-Liouville fractional integral of a function f with respect to another function ψ on a,b is defined by
(5)Ia+α;ψft=1Γα∫atψ′tψt−ψsα−1fsds,where Γ. represents the Gamma function.Definition 2.2 [23].
Letψ′t≠0 and α>0, n∈ℕ. The Riemann-Liouville derivative of a function f with respect to another function ψ of order α, correspondent to the Riemann-Liouville is defined by
(6)Da+α;ψft=1ψ′tddtnIa+n−α;ψft,=1Γn−α1ψ′tddtn∫atψ′tψt−ψsn−α−1fsds,where n−1<α<n,n=α+1, and α denote the integer part of the real number α.Definition 2.3 [23].
Letn−1<α<n with n∈ℕ, a,b is the interval such that −∞≤a<b≤+∞ and f,ψ∈Cna,b,ℝ are two functions such that ψ is increasing and ψ′t≠0 for all t∈a,b. The ψ-Hilfer fractional derivative of a function f of order α and type 0≤β≤1 is defined by
(7)HDa+α,β;ψft=Ia+βn−α;ψ1ψ′tddtnIa+1−βn−α;ψft=Ia+γ−α;ψDa+γ;ψft,where n−1<α<n,n=α+1, and α denote the integer part of the real number α, with γ=α+βn−α.Lemma 2.4 [23].
Letα,β>0. Then we have the following semigroup property given by
(8)Ia+α;ψIa+β;ψft=Ia+α+β;ψft,t>a.Proposition 2.5 [23, 24].
Leta≥0, υ>0, and t>a. Then, ψ-fractional integral and derivative of a power function are given by
(i)
Ia+α;ψψs−ψaυ−1t=Γυ/Γυ+αψs−ψaυ+α−1t(ii)
HDa+α;ψψs−ψaυ−1t=Γυ/Γυ+αψs−ψaυ−α−1t, n−1<α<n,υ>nLemma 2.6 [23].
Iff∈Cna,b,ℝ, n−1<α<n,0≤β≤1, and γ=α+βn−α, then
(9)Ia+α;ψDa+α,β;ψHft=ft−∑k=1nψt−ψaγ−kΓγ−k+1fψn−kIa+1−βn−α;ψfa,for all t∈a,b, where fψn−kft≔1/ψ′td/dtnft.Lemma 2.7.
Leta≥0, 0<αi<1, i=1,2, and h∈Ca,b,ℝ. The function x is a solution of the problem:
(10)HDα1,β1;ψHDα2,β2;ψ+μxt=ht,a≤t≤b,xa=0,xb=∑i=1nωiIσi;ψxηi,a<ηi<b,i=1,⋯,n,if and only if
(11)xt=Iα1+α2;ψht−μIα2;ψxt+ψt−ψaγ1+α2−1/ΔΓγ1+α2Iα1+α2;ψhb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψhηi+μ∑i=1nωiIα2+σi;ψxηi,where
(12)Δ=∑i=1nωiψηi−ψaγ1+α2+σi−1Γγ1+α2+σi−ψb−ψaγ1+α2−1Γγ1+α2≠0.Proof.
The problem (10) can be written as
(13)Iγ1−α1;ψHDγ1;ψDα2,β2;ψH+μxt=ht.
Applying theψ-Riemann-Liouville fractional integral of order α1 to both sides, we obtain the following by using Lemma 2.6(14)HDα2,β2;ψxt+μxt=Iα1;ψht+d0Γγ1ψt−ψaγ1−1,where d0 is constant and γ1=α1+β1−α1β1. Next, applying the ψ-Riemann-Liouville fractional integral of order α2 to both sides of (14), we get by using Lemma 2.6:
(15)xt=Iα1+α2;ψht−μIα2;ψxt+d0Γγ1+α2ψt−ψaγ1+α2−1+d1Γγ2ψt−ψaγ2−1.
From using the boundary conditionxa=0 in (15), we obtain that d1=0. We get
(16)xt=Iα1+α2;ψht−μIα2;ψxt+d0Γγ1+α2ψt−ψaγ1+α2−1.
From using the boundary conditionxb=∑i=1nωiIσi;ψxηi, in (15), we find
(17)d0=1ΔIα1+α2;ψhb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψhηi+μ∑i=1nωiIα2+σi;ψxηi.
Substituting the value ofd0 in (16), we obtain the solution (11). The converse follows by direct computation.


### 2.2. Multivalued Analysis

For a normed spaceX,., we define
(18)PX=Y⊂X:Y≠∅,Pb,c,clX=Y⊂X:Yisbounded,convexandclosed.For more details of multivalued analysis, see ([2, 3]).Definition 2.8.
A multivalued mapF:a,b×ℝ⟶Pℝ 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,b
Furthermore, a Carathéodory functionF is called L1-Carathéodory if:
(iii)
For eachρ>0, there exists φρ∈L1a,b;ℝ such that(19)Ft,x=supv:v∈Ft,x≤φρt,for all x∈ℝ with x≤ρ and for a.e. t∈a,b.Theorem 2.9 (Krasnoselskii fixed-point theorem [25]).
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 2.10 (Martelli fixed-point theorem [26]).
LetX be a Banach space and T:X⟶Pb,c,clX be a completely continuous multivalued map. If the set Λ=x∈X:θx∈Tx,θ>1 is bounded, then T has a fixed point.


## 2.1. Fractional Calculus

In this section, we introduce some definitions, lemmas, and useful notations that will be used throughout this paper.LetC=Ca,b,ℝ denote the Banach space of all continuous functions from a,b into ℝ with the norm defined by f=supt∈a,bft. We denote by ACna,b,ℝ the n-times absolutely continuous functions given by
(4)ACna,b,ℝ=f:a,b⟶ℝ;fn−1∈ACa,b,ℝ.Definition 2.1 [23].
Leta,b, −∞≤a<b≤+∞, be a finite or infinite interval of the half-axis 0,+∞ and α>0. In addition, let ψt be a positive increasing function on a,b, which has a continuous derivative ψ′t on a,b. The ψ-Riemann-Liouville fractional integral of a function f with respect to another function ψ on a,b is defined by
(5)Ia+α;ψft=1Γα∫atψ′tψt−ψsα−1fsds,where Γ. represents the Gamma function.Definition 2.2 [23].
Letψ′t≠0 and α>0, n∈ℕ. The Riemann-Liouville derivative of a function f with respect to another function ψ of order α, correspondent to the Riemann-Liouville is defined by
(6)Da+α;ψft=1ψ′tddtnIa+n−α;ψft,=1Γn−α1ψ′tddtn∫atψ′tψt−ψsn−α−1fsds,where n−1<α<n,n=α+1, and α denote the integer part of the real number α.Definition 2.3 [23].
Letn−1<α<n with n∈ℕ, a,b is the interval such that −∞≤a<b≤+∞ and f,ψ∈Cna,b,ℝ are two functions such that ψ is increasing and ψ′t≠0 for all t∈a,b. The ψ-Hilfer fractional derivative of a function f of order α and type 0≤β≤1 is defined by
(7)HDa+α,β;ψft=Ia+βn−α;ψ1ψ′tddtnIa+1−βn−α;ψft=Ia+γ−α;ψDa+γ;ψft,where n−1<α<n,n=α+1, and α denote the integer part of the real number α, with γ=α+βn−α.Lemma 2.4 [23].
Letα,β>0. Then we have the following semigroup property given by
(8)Ia+α;ψIa+β;ψft=Ia+α+β;ψft,t>a.Proposition 2.5 [23, 24].
Leta≥0, υ>0, and t>a. Then, ψ-fractional integral and derivative of a power function are given by
(i)
Ia+α;ψψs−ψaυ−1t=Γυ/Γυ+αψs−ψaυ+α−1t(ii)
HDa+α;ψψs−ψaυ−1t=Γυ/Γυ+αψs−ψaυ−α−1t, n−1<α<n,υ>nLemma 2.6 [23].
Iff∈Cna,b,ℝ, n−1<α<n,0≤β≤1, and γ=α+βn−α, then
(9)Ia+α;ψDa+α,β;ψHft=ft−∑k=1nψt−ψaγ−kΓγ−k+1fψn−kIa+1−βn−α;ψfa,for all t∈a,b, where fψn−kft≔1/ψ′td/dtnft.Lemma 2.7.
Leta≥0, 0<αi<1, i=1,2, and h∈Ca,b,ℝ. The function x is a solution of the problem:
(10)HDα1,β1;ψHDα2,β2;ψ+μxt=ht,a≤t≤b,xa=0,xb=∑i=1nωiIσi;ψxηi,a<ηi<b,i=1,⋯,n,if and only if
(11)xt=Iα1+α2;ψht−μIα2;ψxt+ψt−ψaγ1+α2−1/ΔΓγ1+α2Iα1+α2;ψhb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψhηi+μ∑i=1nωiIα2+σi;ψxηi,where
(12)Δ=∑i=1nωiψηi−ψaγ1+α2+σi−1Γγ1+α2+σi−ψb−ψaγ1+α2−1Γγ1+α2≠0.Proof.
The problem (10) can be written as
(13)Iγ1−α1;ψHDγ1;ψDα2,β2;ψH+μxt=ht.
Applying theψ-Riemann-Liouville fractional integral of order α1 to both sides, we obtain the following by using Lemma 2.6(14)HDα2,β2;ψxt+μxt=Iα1;ψht+d0Γγ1ψt−ψaγ1−1,where d0 is constant and γ1=α1+β1−α1β1. Next, applying the ψ-Riemann-Liouville fractional integral of order α2 to both sides of (14), we get by using Lemma 2.6:
(15)xt=Iα1+α2;ψht−μIα2;ψxt+d0Γγ1+α2ψt−ψaγ1+α2−1+d1Γγ2ψt−ψaγ2−1.
From using the boundary conditionxa=0 in (15), we obtain that d1=0. We get
(16)xt=Iα1+α2;ψht−μIα2;ψxt+d0Γγ1+α2ψt−ψaγ1+α2−1.
From using the boundary conditionxb=∑i=1nωiIσi;ψxηi, in (15), we find
(17)d0=1ΔIα1+α2;ψhb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψhηi+μ∑i=1nωiIα2+σi;ψxηi.
Substituting the value ofd0 in (16), we obtain the solution (11). The converse follows by direct computation.


## 2.2. Multivalued Analysis

For a normed spaceX,., we define
(18)PX=Y⊂X:Y≠∅,Pb,c,clX=Y⊂X:Yisbounded,convexandclosed.For more details of multivalued analysis, see ([2, 3]).Definition 2.8.
A multivalued mapF:a,b×ℝ⟶Pℝ 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,b
Furthermore, a Carathéodory functionF is called L1-Carathéodory if:
(iii)
For eachρ>0, there exists φρ∈L1a,b;ℝ such that(19)Ft,x=supv:v∈Ft,x≤φρt,for all x∈ℝ with x≤ρ and for a.e. t∈a,b.Theorem 2.9 (Krasnoselskii fixed-point theorem [25]).
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 2.10 (Martelli fixed-point theorem [26]).
LetX be a Banach space and T:X⟶Pb,c,clX be a completely continuous multivalued map. If the set Λ=x∈X:θx∈Tx,θ>1 is bounded, then T has a fixed point.


## 3. Existence and Uniqueness Results for Problem (2)

In this section, we investigate the existence and uniqueness results for the problem (2), for this to simplify the computations, we use the following notations:
(20)Φ1=ψb−ψaα1+α2Γα1+α2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1,(21)Φ2=μψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1.In view of Lemma2.7, we define the operator K:C⟶C by
(22)Kxt=Iα1+α2;ψft,xt,xλt−μIα2;ψxt+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψfb,xb,xλb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψfηi,xηi,xληi+μ∑i=1nωiIα2+σi;ψxηi,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.


### 3.1. Existence Result via Krasnoselskii’s Fixed-Point Theorem

Theorem 3.1.
Assume that
H1f:a,b×ℝ2⟶ℝ is a continuous function such that ft,x,y≤ϕt, ∀t,x,y∈a,b×ℝ2, with ϕ∈Ca,b;ℝ+
H2Φ2<1, where Φ2 is given by (21)
Then there exists at least one solution for the problem (2) on a,b.Proof.
Letsupt∈a,bϕt=ϕ and Br=x∈C;x≤r, where r≥ϕΦ1/1−Φ2, we will show that the operator K defined by (22) satisfies the conditions of Krasnoselskii’s fixed-point theorem, for that we split the operator K into the sum of two operators K1 and K2 defined, on the closed ball, by
(23)K1xt=Iα1+α2ft,xt,xλt+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2fb,xb,xλb−∑i=1nωiIα1+α2+σifηi,xηi,xληi,K2xt=−μIα2xt+ψt−ψaγ1+α2−1ΔΓγ1+α2−μIα2xb+μ∑i=1nωiIα2+σixηi.
For everyx,y∈Br, we have
(24)K1xt+K2yt≤Iα1+α2ft,xt,xλt+μIα2yt+ψt−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2fb,xb,xλb+∑i=1nωiIα1+α2+σifηi,xηi,xληi+μIα2yb+μ∑i=1nωiIα2+σiyηi,≤ϕψb−ψaα1+α2Γα1+α2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1+yμψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1,≤ϕΦ1+rΦ2,≤r,then K1x+K2y≤r, which implies that K1x+K2y∈Br.
Sincef is continuous, then the operator K1 is continuous, and it is uniformly bounded on Br as
(25)K1x≤Φ1ϕ.
Next, we prove that the operatorK1 is compact, for that setting supt,x,y∈a,b×Brft,x,y=f¯<∞, and let τ1,τ2∈a,b, τ1<τ2; we obtain
(26)K1xτ2−K1xτ1=Iα1+α2fτ2,xτ2,xλτ2+ψτ2−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2fb,xb,xλb−∑i=1nωiIα1+α2+σifηi,xηi,xληi−Iα1+α2fτ1,xτ1,xλτ1−ψτ1−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2fb,xb,xλb−∑i=1nωiIα1+α2+σifηi,xηi,xληi≤f¯Γα1+α2∫aτ1ψ′sψτ2−ψsα1+α2−1−ψτ1−ψsα1+α2−1ds+∫τ1τ2ψ′sψτ2−ψsα1+α2−1ds+ψτ2−ψaγ1+α2−1−ψτ1−ψaγ1+α2−1ΔΓγ1+α2×f¯ψb−ψaα1+α2Γα1+α2+1+f¯∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1.
The right-hand side tends to zero asτ2−τ1⟶0, independently of x∈Br. Then, K1 is equicontinuous; hence, K1 is relatively compact on Br. By the Arzelà-Ascoli theorem, it implies that K1 is compact on Br.
In the next step, we will show thatK2 is a contraction mapping; for that, let x,y∈C, and for t∈a,b, we have
(27)K2xt+K2yt≤μIα2xt−yt+ψb−ψaγ1+α2−1ΔΓγ1+α2×μIα2xb−yb+μ∑i=1nωiIα2+σixηi−yηi,≤x−yμψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2,×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1,≤Φ2x−y.
This implies thatK2x+K2y≤Φ2x−y, by using H2; we deduce that K2 is a contraction mapping. It follows by using Krasnoselskii’s fixed-point theorem, the problem (2) has at least one solution on a,b.


### 3.2. Uniqueness Result via Banach’s Fixed-Point Theorem

To deal with the uniqueness of solution for our problem (2), we use Banach’s fixed-point theorem.Theorem 3.2.
Assume thatft,x,y−ft,z,w≤Lx−z+y−w; L>0, for each t∈a,b and x,y,z,w∈ℝ.
If2LΦ1+Φ2<1, where Φ1,Φ2 are, respectively, given by (20) and (21); then the problem (2) has a unique solution on a,b.Proof.
By considering the operatorK defined in (22), we transform the problem (2) into a fixed-point problem x=Kx. By using Banach contraction principle, we will show that K has a unique fixed point.
We setsupt∈a,b=ft,0,0=M<∞ and choose ρ>0 such that
(28)ρ≥MΦ11−2LΦ1−Φ2.
Bρ=x∈Ca,b,ℝ;x≤ρ, where Φ1,Φ2 are, respectively, given by (20) and (21).
Step 1: we show thatKBρ⊂Bρ.
For anyx∈Bρ we have
(29)ft,xt,xλt≤ft,xt,xλt−ft,0,0+ft,0,0≤Lxt+xλt+M≤2Lx+M.
Then we have(30)Kxt≤Iα1+α2ft,xt,xλt+μIα2xt+ψt−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2fb,xb,xλb+∑i=1nωiIα1+α2+σifηi,xηi,xληi+μIα2xb+μ∑i=1nωiIα2+σixηi,≤2Lx+Mψb−ψaα1+α2Γα1+α2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1+xμψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1,≤2Lx+MΦ1+xΦ2≤2Lρ+MΦ1+ρΦ2,≤ρ,which implies that KBρ⊂Bρ.
Step 2: next we show that the operatorK is a contraction.
For anyx,y∈C, and for t∈a,b, we have
(31)Kxt−Kyt≤ψb−ψaα1+α2Γα1+α2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+12Lx−y+μψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1x−y≤2LΦ1+Φ2x−y,which implies Kxt−Kyt≤2LΦ1+Φ2x−y. As 2LΦ1+Φ2<1, then K is a contraction, and by applying Banach’s fixed-point theorem, we get that the operator K has a unique fixed point which is the unique solution of our problem (2).


## 3.1. Existence Result via Krasnoselskii’s Fixed-Point Theorem

Theorem 3.1.
Assume that
H1f:a,b×ℝ2⟶ℝ is a continuous function such that ft,x,y≤ϕt, ∀t,x,y∈a,b×ℝ2, with ϕ∈Ca,b;ℝ+
H2Φ2<1, where Φ2 is given by (21)
Then there exists at least one solution for the problem (2) on a,b.Proof.
Letsupt∈a,bϕt=ϕ and Br=x∈C;x≤r, where r≥ϕΦ1/1−Φ2, we will show that the operator K defined by (22) satisfies the conditions of Krasnoselskii’s fixed-point theorem, for that we split the operator K into the sum of two operators K1 and K2 defined, on the closed ball, by
(23)K1xt=Iα1+α2ft,xt,xλt+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2fb,xb,xλb−∑i=1nωiIα1+α2+σifηi,xηi,xληi,K2xt=−μIα2xt+ψt−ψaγ1+α2−1ΔΓγ1+α2−μIα2xb+μ∑i=1nωiIα2+σixηi.
For everyx,y∈Br, we have
(24)K1xt+K2yt≤Iα1+α2ft,xt,xλt+μIα2yt+ψt−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2fb,xb,xλb+∑i=1nωiIα1+α2+σifηi,xηi,xληi+μIα2yb+μ∑i=1nωiIα2+σiyηi,≤ϕψb−ψaα1+α2Γα1+α2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1+yμψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1,≤ϕΦ1+rΦ2,≤r,then K1x+K2y≤r, which implies that K1x+K2y∈Br.
Sincef is continuous, then the operator K1 is continuous, and it is uniformly bounded on Br as
(25)K1x≤Φ1ϕ.
Next, we prove that the operatorK1 is compact, for that setting supt,x,y∈a,b×Brft,x,y=f¯<∞, and let τ1,τ2∈a,b, τ1<τ2; we obtain
(26)K1xτ2−K1xτ1=Iα1+α2fτ2,xτ2,xλτ2+ψτ2−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2fb,xb,xλb−∑i=1nωiIα1+α2+σifηi,xηi,xληi−Iα1+α2fτ1,xτ1,xλτ1−ψτ1−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2fb,xb,xλb−∑i=1nωiIα1+α2+σifηi,xηi,xληi≤f¯Γα1+α2∫aτ1ψ′sψτ2−ψsα1+α2−1−ψτ1−ψsα1+α2−1ds+∫τ1τ2ψ′sψτ2−ψsα1+α2−1ds+ψτ2−ψaγ1+α2−1−ψτ1−ψaγ1+α2−1ΔΓγ1+α2×f¯ψb−ψaα1+α2Γα1+α2+1+f¯∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1.
The right-hand side tends to zero asτ2−τ1⟶0, independently of x∈Br. Then, K1 is equicontinuous; hence, K1 is relatively compact on Br. By the Arzelà-Ascoli theorem, it implies that K1 is compact on Br.
In the next step, we will show thatK2 is a contraction mapping; for that, let x,y∈C, and for t∈a,b, we have
(27)K2xt+K2yt≤μIα2xt−yt+ψb−ψaγ1+α2−1ΔΓγ1+α2×μIα2xb−yb+μ∑i=1nωiIα2+σixηi−yηi,≤x−yμψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2,×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1,≤Φ2x−y.
This implies thatK2x+K2y≤Φ2x−y, by using H2; we deduce that K2 is a contraction mapping. It follows by using Krasnoselskii’s fixed-point theorem, the problem (2) has at least one solution on a,b.


## 3.2. Uniqueness Result via Banach’s Fixed-Point Theorem

To deal with the uniqueness of solution for our problem (2), we use Banach’s fixed-point theorem.Theorem 3.2.
Assume thatft,x,y−ft,z,w≤Lx−z+y−w; L>0, for each t∈a,b and x,y,z,w∈ℝ.
If2LΦ1+Φ2<1, where Φ1,Φ2 are, respectively, given by (20) and (21); then the problem (2) has a unique solution on a,b.Proof.
By considering the operatorK defined in (22), we transform the problem (2) into a fixed-point problem x=Kx. By using Banach contraction principle, we will show that K has a unique fixed point.
We setsupt∈a,b=ft,0,0=M<∞ and choose ρ>0 such that
(28)ρ≥MΦ11−2LΦ1−Φ2.
Bρ=x∈Ca,b,ℝ;x≤ρ, where Φ1,Φ2 are, respectively, given by (20) and (21).
Step 1: we show thatKBρ⊂Bρ.
For anyx∈Bρ we have
(29)ft,xt,xλt≤ft,xt,xλt−ft,0,0+ft,0,0≤Lxt+xλt+M≤2Lx+M.
Then we have(30)Kxt≤Iα1+α2ft,xt,xλt+μIα2xt+ψt−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2fb,xb,xλb+∑i=1nωiIα1+α2+σifηi,xηi,xληi+μIα2xb+μ∑i=1nωiIα2+σixηi,≤2Lx+Mψb−ψaα1+α2Γα1+α2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1+xμψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1,≤2Lx+MΦ1+xΦ2≤2Lρ+MΦ1+ρΦ2,≤ρ,which implies that KBρ⊂Bρ.
Step 2: next we show that the operatorK is a contraction.
For anyx,y∈C, and for t∈a,b, we have
(31)Kxt−Kyt≤ψb−ψaα1+α2Γα1+α2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+12Lx−y+μψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1x−y≤2LΦ1+Φ2x−y,which implies Kxt−Kyt≤2LΦ1+Φ2x−y. As 2LΦ1+Φ2<1, then K is a contraction, and by applying Banach’s fixed-point theorem, we get that the operator K has a unique fixed point which is the unique solution of our problem (2).


## 4. Existence Results for the Inclusion Version

In this section, we will investigate the existence result for the inclusion version defined as problem (3).Definition 4.1.
A continuous functionx is said to be a solution of problem (3) if xa=0;xb=∑i=1nωiIσi;ψxηi, and there exists a function v∈L1a,b,ℝ with v∈Ft,xt,xλt, i.e., on a,b such that
(32)xt=Iα1+α2;ψvt−μIα2;ψxt+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψvb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψvηi+μ∑i=1nωiIα2+σi;ψxηi,for each x∈Ca,b,ℝ, define the set of selections of F by
(33)SF,x≔v∈L1a,b,ℝ:v∈Ft,xt,xλtona,b.Lemma 4.2 ([5]).
LetX be a Banach space and F:a,b×ℝ2⟶Pb,c,cl be a L1-Carathéodory multivalued map. And let Ξ be a linear continuous mapping from L1a,b,X to Ca,b,X. Then the operator
(34)Ξ∘SF:Ca,b,X⟶Pb,c,clCa,b,X;x⟶Ξ∘SFx=ΞSF,xis a closed graph operator in Ca,b,X×Ca,b,X.In what follows, we deal with the upper semicontinuous case, and for the existence results, we use Martelli’s fixed-point theorem.Theorem 4.3.
Suppose thatH2 holds and the following assumptions hold:
H3F:a,b×ℝ2⟶Pb,c,clℝ is L1-Carathéodory and has nonempty convex values, and for each fixed x∈Ca,b,ℝ, the set
(35)SF,x=v∈L1a,b,X:vt∈Ft,xt,xλt;t∈a,bis nonempty and convex.
H4Ft,x,y≔supv:v∈Ft,x,y≤ptΨx+y for all t∈a,b and all x,y∈Ca,b,X, where p∈L1a,b,ℝ+ and Ψ:ℝ+⟶0,+∞ is continuous and nondecreasing function.
Then the problem (3) has at least one solution on a,b.Proof.
In order to transform the problem (3) into a fixed-point problem. Let K:Ca,b,ℝ⟶Pa,b,ℝ be defined by
(36)Kx≔k∈Ca,b,ℝ:kt=Iα1+α2;ψvt−μIα2;ψxt+ψt−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2;ψvb−μIα2xb−∑i=1nωiIα1+α2+σi;ψvηi+μ∑i=1nωiIα2+σi;ψxηi;t∈a,b,v∈SF,x.
We will show thatK satisfies the conditions of the Theorem 2.10; for the poof, we give it in steps:
Step 1:Kx is convex for each x∈Ca,b,ℝ.
Ifk1,k2 belong to Kx, then there exist v1,v2∈SF,x such that for each t∈a,b we have:
(37)kjt=Iα1+α2;ψvjt−μIα2xt+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψvjb−μIα2ψxb−∑i=1nωiIα1+α2+σi;ψvjηi+μ∑i=1nωiIα2+σi;ψxηi,for j=1,2. Let 0≤ξ≤1; then, for each t∈a,b, we have
(38)ξk1t+1−ξk2t=Iα1+α2;ψξv1s+1−ξv2s−μIα2;ψxt+ψt−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2;ψξv1s+1−ξv2s−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψξv1ηi+1−ξv2ηi+μ∑i=1nωiIα2+σi;ψxηi,
Thus,ξv1+1−ξv2∈Kx (because SF,x is convex); then Kx is convex for each x∈Ca,b,ℝ
Step 2:K is bounded.
For a positive numberρ, let Bρ=x∈Ca,b,ℝ:x≤ρ be bounded ball in Ca,b,ℝ; then for each k∈Kx and x∈Bρ, there exists v∈SF,x, such that
(39)kt=Iα1+α2;ψvt−μIα2;ψxt+ψt−aγ1+α2−1ΔΓγ1+α2Iα1+α2;ψvb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψvηi+μ∑i=1nωiIα2+σi;ψxηi,then for every t∈a,b, we have
(40)kt≤supt∈a,bIα1+α2;ψvt+μIα2;ψxt+ψt−ψaγ1+α2−1ΔΓγ1+α2×Iα1+α2;ψvb+∑i=1nωiIα1+α2+σi;ψvηi+μIα2;ψxb+μ∑i=1nωiIα2+σi;ψxηi≤pΨ2xψb−ψaα1+α2Γα1+α2+1+ψb−aγ1+α2−1ΔΓγ1+α2×ψb−ψaα1+α2Γα1+α2+1+∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1+xμψb−ψaα2Γα2+1+ψb−ψaγ1+α2−1ΔΓγ1+α2×ψb−ψaα2Γα2+1+∑i=1nωiψηi−ψaα2+σiΓα2+σi+1≤pΨ2xΦ1+xΦ2≤pΨ2ρΦ1+ρΦ2.
Then(41)Kx≤pΨ2ρΦ1+ρΦ2≔l,where Φ1,Φ2 are, respectively, given by (20) and (21).
Step 3:K is equicontinuous.
Lett1,t2∈a,b;t1<t2, and x∈Bρ where Bρ, as above then for each x∈Bρ and k∈Kx; there exist v∈SF,x; then we obtain
(42)kt2−kt1≤1Γα1+α2∫at1ψ′sψt2−ψsα1+α2−1−ψt1−ψsα1+α2−1vsds+∫t1t2ψ′sψt2−ψsα1+α2−1vsds+μΓα2∫at1ψ′sψt2−ψsα2−1−ψt1−ψsα2−1xsds+∫t1t2ψ′sψt2−ψsα2−1xsds+ψt2−ψaγ1+α2−1−ψt1−ψaγ1+α2−1ΔΓγ1+α2,×vsψb−ψaα1+α2Γα1+α2+1+vs∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1+xbμψb−ψaα2Γα2+1+xηiμ∑i=1nωiψηi−ψaα2+σiΓα2+σi+1,≤pΨ2ρΓα1+α2∫at1ψ′sψt2−ψsα1+α2−1−ψt1−ψsα1+α2−1ds+∫t1t2ψ′sψt2−ψsα1+α2−1ds+ρμΓα2∫at1ψ′sψt2−ψsα2−1−ψt1−ψsα2−1ds+∫t1t2ψ′sψt2−ψsα2−1ds+ψt2−ψaγ1+α2−1−ψt1−ψaγ1+α2−1ΔΓγ1+α2×pΨ2ρψb−ψaα1+α2Γα1+α2+1+pΨ2ρ∑i=1nωiψηi−ψaα1+α2+σiΓα1+α2+σi+1+ρμψb−ψaα2Γα2+1+ρμ∑i=1nωiψηi−ψaα2+σiΓα2+σi+1.
Ast2⟶t1, the right-hand side of the above inequality tends to zero, implying that Kx is equicontinuous. By using the Arzelà-Ascoli theorem, we get that K is relatively compact; then K is completely continuous.
To prove that the operatorK is upper semicontinuous, it is enough to show that K has a closed graph.
Step 4:K has a closed graph.
Letxn⟶x∗, kn∈Kxn and kn⟶k∗; we will prove that k∗∈Kx∗.
Forkn∈Kxn, then there exists vn∈SF,xn such that for each t∈a,b:
(43)knt=Iα1+α2;ψvnt−μIα2;ψxnt+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψvnb−μIα2;ψxnb−∑i=1nωiIα1+α2+σi;ψvnηi+μ∑i=1nωiIα2+σi;ψxnηi.
We should prove thatv∗∈SF,x∗ such that for each t∈a,b(44)k∗t=Iα1+α2;ψv∗t−μIα2;ψx∗t+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψv∗b−μIα2;ψx∗b−∑i=1nωiIα1+α2+σi;ψv∗ηi+μ∑i=1nωiIα2+σi;ψx∗ηi.
We have that(45)knt+λIα2;ψxnt−ψt−ψaγ1+α2−1ΔΓγ1+α2−μIα2;ψxnb+μ∑i=1nωiIα2+σi;ψxnηi−k∗t+μIα2;ψx∗t−ψt−ψaγ1+α2−1ΔΓγ1+α2−λIα2;ψx∗b+μ∑i=1nωiIα2+σi;ψx∗ηi⟶0,as n⟶∞.
Consider the operator defined by(46)Ξ:L1a,b,ℝ⟶Ca,b,ℝ,v⟶Ξvt.with
(47)Ξvt=Iα1+α2;ψvt+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψvb−∑i=1nωiIα1+α2+σi;ψvηi,
By using Lemma4.2, Ξ∘SF is a closed graph operator; then we get
(48)knt+μIα2;ψxnt−ψt−ψaγ1+α2−1ΔΓγ1+α2−μIα2;ψxnb+μ∑i=1nωiIα2+σi;ψxnηi∈YSF,xn.
Sincexn⟶x∗, and kn⟶k∗, then
(49)k∗t+μIα2;ψx∗t−ψt−ψaγ1+α2−1ΔΓγ1+α2−μIα2;ψx∗b+μ∑i=1nωiIα2+σi;ψx∗ηi=Ξv∗∈ΞSF,x∗.
It follows thatv∗∈SF,x∗ such that
(50)k∗t=Iα1+α2;ψv∗t−μIα2;ψx∗t+ψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψv∗b−μIα2;ψx∗b−∑i=1nωiIα1+α2+σi;ψv∗ηi+μ∑i=1nωiIα2+σi;ψx∗ηi.
We deduce thatK is an upper semicontinuous multivalued map, with convex closed values.
Step 5:Λ=x∈X:θx∈Kx,θ>1 is bounded.
Letx∈Λ, and then θx∈Kx for some θ>1; thus, there exists a function v∈SF,x such that
(51)xt=1θIα1+α2;ψvt−1θμIα2;ψxt+1θψt−ψaγ1+α2−1ΔΓγ1+α2Iα1+α2;ψvb−μIα2;ψxb−∑i=1nωiIα1+α2+σi;ψvηi+μ∑i=1nωiIα2+σi;ψxηi.
From Step 2, and for everyt∈a,b, we have
(52)x≤pΨ2xΦ1+xΦ2.Then, by using H2, we get
(53)x≤pΨ2ρΦ11−Φ2.
Finally the setΛ is bounded; then from Theorem 2.10, we deduce that the problem (3) has at least one solution.


## 5. Example

Consider the followingψ-Hilfer fractional pantograph Langevin equation given by
(54)HD2/5,4/5;et/3HD1/5,3/5;et/3+19xt=15+17xt+17sinxt3,0≤t≤1x0=0,x1=58I3/2;et/3x13+78I3/2;et/3x12.where α1=2/5, α2=1/5, β1=4/5, β2=3/5, μ=1/9, λ=1/3, a=0, b=1, n=2, σ1=3/2, σ2=5/2, ω1=5/8, ω2=7/8, η1=1/3, η2=1/2, and ψt=et/3. With this given data, we get γ1=α1+β1−α1β1=22/25, Δ≃1,7825119, Φ1=2,1294613, and Φ2=0,2195946<1.Setft,x,y=1/5+1/7xt+1/7sinyt/3, x,y∈ℝ, t∈0,1.Letx1,x2,y1,y2∈ℝ and t∈0,1; then we get
(55)ft,x1,y1−ft,x2,y21/7x1−x2+1/7y1−y21/7x1−x2+y1−y2.This implies that the assumption of Theorem3.2 holds with L=1/7.It follows that2LΦ1+Φ2≃0.8280121<1; then by applying Theorem 3.2, our problem has a unique solution on 0,1.


## 6. Conclusion

The present paper examined theψ-Hilfer fractional pantograph Langevin equation and inclusion. The challenges and the novelty of this work generalize the types of fractional derivatives. With the assistance of Krasnoselskii and Banach fixed-point theorems, we investigate the existence and uniqueness results for the single valued problem, and by making use of the Martelli fixed-point theorem, we study the existence result for the multivalued problem. In the end, we illustrate our result with an example.



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