# On More General Fractional Differential Equations Involving Mixed Generalized Derivatives with Nonlocal Multipoint and Generalized Fractional Integral Boundary Conditions

Authors: Wafa Shammakh; Hadeel Z. Alzumi; Bushra A. AlQahtani
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

This paper deals with the existence of solutions for a new boundary value problem involving mixed generalized fractional derivatives of Riemann-Liouville and Caputo supplemented with nonlocal multipoint boundary conditions. The existence results for inclusion case are also discussed. The nonlinear term belongs to a general abstract space, and our results rely on modern theorems of fixed point. Ulam stability is also presented. We provide some examples that well-illustrate our main results.


---

## Body

## 1. Introduction and Preliminaries

In this paper, we study a new generalized boundary value problem (GBVP) that involves both generalized Riemann-Liouville and Caputo derivatives via nonlocal multipoint and generalized fractional integral boundary conditions given by(1)DRL0+α,ρCD0+βyt,0<α,β≤1,y0=α1∑i=1mσiI0+q,ρyεi+α2∑i=1pδiyηi,γy1=b1∑i=1mμiI0+q,ργyεi+b2∑i=1priγyηi,where the derivatives RLD0+α,ρ and CD0+β,ρ denote the generalized Riemann-Liouville fractional derivative of order 0<α≤1 and generalized Caputo fractional derivative of order 0<β≤1, respectively. I0+q,ρ is the generalized Riemann-Liouville fractional integral of order 0<q<1, and the nonlinear term g is given in the Orlicz space LG0,1.We also investigate the corresponding inclusion problem which is given by(2)DRL0+α,ρCD0+β,ρyt∈Gt,yt,0<α,α≤1,y0=α1∑i=1mIq,ρyεi+α2∑i=1pδiyηi,γy1=b1∑i=1mμiIq,ργyεi+b2∑i=1priγyηi,where ai, bi∈ℝi=1,2; δi,ri∈ℝ+i=1,⋯,p; σi,μi∈ℝ+i=1,⋯,m; γ=t1−ρd/dt; and ρ>0, and G:0,1×ℝ⟶Pℝ is a multivalued map. Pℝ is the family of all subsets of ℝ.The concept of generalized fractional derivatives and integrals was introduced by Katugampola in [1, 2]. This new form of fractional integral operator generalizes Riemann-Liouville and Hadamard fractional derivative into a single form and satisfy a semigroup property. Jarad et al. [3] proposed a new generalized fractional calculus based on a special case of proportional derivatives where the kernel of the fractional operator involves an exponential function. This generalization contains Riemann-Liouville and Caputo fractional derivatives and integrals as special cases. Also, there were works on general fractional derivative and integral containing a Mittag-Leffler function. For more new definitions of generalized fractional calculus, one can see [4–7]. These generalizations are helpful to reach a better form of the real models that appeared in various fields such as physics, chemistry, aerodynamics, and electrorheology employing fractional differential equations (see [8–17]). The existence and uniqueness results of fractional boundary value problems involving fractional derivatives and derivative of a generalized type were established by many authors (for example, see [18–21]).On the other hand, Birnbaum and Orlicz [22] generalized the classical Lebesgue space Lp,0<p<∞, into a general abstract space, which was called Orlicz space. This generalization is based on replacing the function xp with more general function f.In what follows, we recall some basic concepts of fractional calculus [1, 2, 4] and Orlicz space [22] that are needed throughout the paper.Consider the spaceXcpa,bc∈ℝ,1≤p≤∞ of those complex-valued Lebesgue measurable functions f on a,b for which fXcpa,b<∞, where the norm is defined by
(3)fXcp=∫abtcftpdtt1/p<∞.For0≤ε≤1, we denote by ACδna,b the space of all absolutely continuous real-valued functions on a,b, where
(4)ACδna,b=f:a,b⟶ℂandδn−1f∈ACa,b,δ=t1−ρddt,endowed with the norm fCδn=∑k=0nδkfC. Moreover, we introduce
(5)Cδ,εna,b=:a,b⟶ℂandδn−1f∈Ca,b,δnf∈Cε,ρa,b,δ=t1−ρddt,equipped with the norm fCδ,εn=∑k=0n−1δkfC+δnfCε,ρ, where Cε,ρa,b=f:a,b⟶ℝ:tρ−aρεft∈Ca,b,ℝ and Ca,b,ℝ is a Banach space of all continuous functions on a,b into ℝ endowed with the norms fCε,ρ=tρ−aρεftC and xC=sup∣xt∣:t∈a,b, respectively. Notice that Cδ,0na,b=Cδna,b.Definition 1.
For−∞<a<t<b<∞, the left-sided generalized fractional integral off∈Xcpa,bof orderα>0andρ>0is defined by(6)Ia+α,ρft=ρ1−αΓα∫atsρ−1tρ−sρ1−αfsds.Definition 2.
For0≤a<x<b<∞,the generalized Riemann-Liouville fractional derivative associated with generalized fractional integrals (2) is defined by(7)Da+α,ρRLft=t1−ρddtnIa+n−α,ρft=ρα−n+1Γn−αt1−ρddtn∫atsρ−1tρ−sρα−n+1fsds.Definition 3.
The generalized Caputo-type fractional derivative off∈ACδna,bof orderα≥0is defined by(8)Da+α,ρfCt=Da+α,ρft−∑k=0n−1δkfak!tρ−aρρkx,δ=x1−ρddx,where n=α+1.The following shows some important properties which are used in the sequel.Lemma 4.
Letf∈ACδna,borCδna,b,andα∈ℝ. Then,(9)Ia+α,ρCDa+α,ρft=ft−∑k=0n−1δkfak!tρ−aρρk.
In particular, if0<α≤1,we have(10)Ia+α,ρCDa+α,ρft=ft−fa.Lemma 5.
LetReα>0,n≤−−Reα,f∈La,bandIa+α,ρf∈ACδna,b. Then,(11)Ia+α,ρRLDa+α,ρft=ft−∑j=1n−1Da+α−j,ρfaΓα−j+1tρ−aρρα−j.
In particular, if0<α≤1,we have(12)Ia+α,ρRLDa+α,ρft=ft−Da+α−1,ρfaΓαtρ−aρρα−1.Lemma 6.
For a functionf∈Xcpa,b,ρ≥c, we have(13)Da+α,ρIa+β,ρft=Ia+β−α,ρft,β>α>0,which holds almost everywhere ona,b. If α=β, we have(14)Da+α,ρIa+α,ρft=ft,almost everywhere.Lemma 7.
For0<α<β, we have(15)I0+β,ρtρα−1=tρα+β−1ΓαρβΓα+β.Definition 8.
Letφ:0,∞⟶0,∞be right continuous, monotone, increasing function withφ0=0,limt⟶∞φt=∞,andφt>0whenevert>0.Then, the function defined by(16)Gx=∫0xφtdt,x≥0,is called theN-function.Alternatively, the functionG is an N-function iff G is continuous, even, and convex with limx⟶0Gx/x=0,limx⟶∞Gx/x=∞ and Gx>0 if x>0.Note that a functionG:0,∞⟶0,∞ is called a Young function if it is convex and satisfies the conditions
(17)G0=limx⟶0+Gx=0,limx⟶∞Gx=∞.For anN-function G, we define
(18)G∗x=∫0xφ−1tdt,x≥0,where φ−1 is the right inverse of the right derivative of φ of G and is called the complementary of G and it satisfies the condition
(19)G∗x=suptx−Gt:t≥0,∀x≥0.Notice that the functionG∗ is also an N-function and the complementary pairs G and G∗ satisfy the following Young inequality:
(20)xt≤Gx+G∗t,∀x,t≥0.Definition 9.
For anN-functionG, the Orlicz spaceLG0,1is the space of measurable functionsx:0,1⟶ℝsuch that∫01G∣xt∣dt<∞. This space is endowed with the Luxemburg norm, i.e.,(21)∥x∥G=infλ>0:∫01G∣xt∣λdt≤1,and the pairLG0,1,.G is a Banach space.For an Orlicz space, the Hölder inequality holds; that is,(22)∫01uvdt≤2uGvG∗,where u∈LG0,1 and v∈LG∗0,1.The existence and uniqueness of solutions of GBVP (1) rely on the following fixed point theorems [23].Theorem 10.
LetPbe a closed, convex, bounded, and nonempty subset of a Banach spaceX.LetT1,T2be operators such that(i)
T1u1+T2u2 belong to P whenever u1,u2∈P(ii)
T1 is a compact and T2 is a contraction mappingThen, there existsu0∈P such that u0=T1u0+T2u0.Theorem 11.
LetXbe a Banach space. Assume thatT:X⟶Xis a completely continuous operator and the setV=u∈X:u=εTu,0<ε<1is bounded. Then,Thas a fixed point inX.
For computational convenience,(23)ϕt=1ψ1Γαtρα+β−1ρβΓα+β+ψ3ψ2,ψ1=Γαρβ−1Γα+β−11−b1∑i=1mμiξiρα+β+q−2Γα+β−1ρqΓα+β+q−1−b2∑i=1priηiρα+β−2≠0,ψ2=1−a1∑i=1mσiξiρqρqΓq+1−a2∑i=1pδi≠0,ψ3=a1∑i=1mσiΓαξiρα+β+q−1ρβ+qΓα+β+q+a2∑i=1pδiΓαηiρα+β−1ρβΓα+β.Lemma 12.
Leth∈LG0,1. Then, the solution of the following GBVP:(24)DRL0+α,ρCD0+β,ρyt=ht,0<α,β≤1,y0=α1∑i=1mσiI0+q,ρyεi+α2∑i=1pδiyηi,γy1=b1∑i=1mμiI0+q,ργyεi+b2∑i=1priγyηi,is given by
(25)yt=I0+α+β,ρht+ϕtb1∑i=1mμiI0+α+β+q−1,ρhξi+b2∑i=1priI0+α+β−1,ρhηi−I0+α+β−1,ρh1+1ψ2a1∑i=1mσiI0+α+β+q,ρhξi+a2∑i=1pδiI0+α+β,ρhηi,where ϕ and ψ2 are given in (23).Proof.
To obtain the integral equation modeled by the GBVP(24), we apply the generalization Riemann-Liouville fractional integral of order α to both sides of (24), and we get
(26)CD0+β,ρyt=I0+α,ρht+c1tρα−1.
Next, applying the generalization Caputo fractional integral of orderβ to both sides of (26), we have
(27)yt=I0+α+β,ρht+c1I0+β,ρtρα−1+c2,where c1 and c2∈ℝ. By using Lemma 7, we get
(28)yt=I0+α+β,ρht+c1tρα+β−1ΓαρβΓα+β+c2.
So, the quantitiesγyt and Iq,ργyt are given, respectively, by
(29)γyt=Iα+β−1,ρht+c1tρα+β−2ΓαρβΓα+β−1,Iq,ργyt=Iα+β+q−1,ρht+c1ΓαρβΓα+β−1tρα+β+q−2Γα+βρqΓα+β+q.
In view of the boundary conditions given in (24), the constants c1 and c2 are defined by
(30)c2=1ψ2ψ3ψ1b1∑i=1mμiI0+α+β+q−1,ρhξi+b2∑i=1priI0+α+β−1,ρhηi−I0+α+β−1,ρh1+a1∑i=1mσiI0+α+β+q,ρhξi+a2∑i=1pδiI0+α+β,ρhηi,c1=1ψ1b1∑i=1mμiI0+α+β+q−1,ρhξi+b2∑i=1priI0+α+β−1,ρhηi−I0+α+β−1,ρh1,where ψ1,ψ2,ψ3,andϕt are given in (23), and by substituting of values of c1andc2, we obtain (25).


## 2. Main Results

In this section, we are looking to prove the existence and uniqueness of solutions of GBVP (1). We will utilize in our proofs the fixed point theorems. We shall assume that g is in the Orlicz space LG0,1. Let Y=y:y∈C0,1,ℝ denote the Banach space of all continuous functions on 0,1 into ℝ endowed with the norm y=supyt:t∈0,1. Now, we define an operator F:Y⟶Y associated with problem (1) by
(31)Fyt=I0+α+β,ρgt,yt+ϕtb1∑i=1mμiI0+α+β+q−1,ρgξi,yξi+b2∑i=1priI0+α+β−1,ρgηi,yηi−I0+α+β−1,ρg1,y1+1ψ2a1∑i=1mσiI0+α+β+q,ρgξi,gξi+a2∑i=1pδiI0+α+β,ρgηi,yηi.Therefore, the GBVP (1) has a solution if and only if the operator F has a fixed point.The following lemma plays a curial role in the sequel.Lemma 13.
Letα,β∈0,1,q∈0,1.LetGbe a Young function which has a Young complement functionG∗satisfying(32)∫0tG∗sγds<∞,where γ∈α+β,α+β−1,α+β+q,α+β+q−1. Then, the operator F exists and is well defined.Proof.
Letα,β∈0,1,q∈0,1, and y∈Y. Define a function
(33)Λ1s=tρ−sρα+βsρ−1ifs∈0,t,t>0,0,otherwise,
We show thatΛ1∈LG∗0,1. By using appropriate substitution and properties of the Young functions, one obtains
(34)∫01G∗Λ1skds=∫0tG∗tρ−sρα+βksρ−1ds=1ρk1/α+β∫01/k1/α+βα+β1/α+βtρG∗sα+βds.
By the assumption of the theorem, we getΛ1∈LG∗0,1. Similarly, set
(35)Λ2s=tρ−sρα+β−1sρ−1if∈0,t,t>0,0,otherwise,Λ3s=tρ−sρα+β+qsρ−1ifs∈0,t,t>0,0,otherwise,Λ4s=tρ−sρα+β+q−1sρ−1ifs∈0,t,t>0,0,otherwise.
Then, we can getΛ2,Λ3, and Λ4∈LG∗0,1. Next, we show that F is well defined, i.e., Fyt∈C0,1,ℝ. Let 0≤τ≤t≤1, then we have
(36)∣Fyt−Fyτ∣≤supt∈0,11ρα+β−1Γα+β∫0τtρ−sρα+β−1−τρ−sρα+β−1sρ−1gs,ysds+1ρα+β−1Γα+β∫τttρ−sρα+β−1sρ−1gs,ysds+1ψ1Γα∣tρα+β−1−τρα+β−1∣ρβΓα+βb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,ysds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,ysds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,ysds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,ysds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,ysds=1ρα+β−1Γα+β∫01χ1s+χ2sgs,ysds+1ψ1Γα∣tρα+β−1−τρα+β−1∣ρβΓα+βb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,ysds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,ysds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,ysds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,ysds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,ysds,where
(37)χ1s=∣tρ−sρα+β−1−τρ−sρα+β−1sρ−1∣ifs∈0,τ,0,otherwise,χ2s=∣tρ−sρα+β−1sρ−1∣ifs∈0,τ,0,otherwise.
The functionsχi,i=1,2 belong to LG∗0,1 with ∥χi∥G∗≤J∣tρ−τρ∣,i=1,2, where J:ℝ+⟶ℝ+ is a continuous, increasing function with J0=0. Using the Hölder inequality, we get
(38)∣Fyt−Fyτ∣≤∥g∥G∗12ρα+β−1Γα+β∥χ1∥G∗+∥χ2∥G∗+1ψ1Γα∣tρα+β−1−τρα+β−1∣ρβΓα+βb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1.
Hence, for0<∣tρ−τρ∣<ε and by continuity of J, we see that F is continuous, which completes the proof.For convenience, we define(39)ψ=1ρα+βΓα+β+1+ϕb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1.Now, we are in a position to state and prove the first existence result. The following result is based on Banach fixed point theorem.Theorem 14.
Assume that there exists a constantΩ>0such that
(S1)∣gt,x−gt,y∣≤Ω∣x−y∣forallt∈0,1andx,y∈ℝ.
IfΩψ<1 where ψ is given by (39), then the boundary value problem (1) has a unique solution on [0,1].Proof.
We defineBr=y∈Y:∥y∥≤r, with r≥g∗ψ/1−Ωψ, and ψ is defined by (39). Set
(40)g∗=supt∈0,1∣gt,0∣<∞.
We shall show thatFBr⊂Br. For any y∈Br, by assumption (S1), we obtain that
(41)∥Fy∥≤Ω∥y∥+g∗1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1ds+∥ϕ∥b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1ds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1ds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−spα+β+q−1sρ−1ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1ds≤Ωr+g∗1ρα+βΓα+β+1+∥ϕ∥b1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1.
Therefore,(42)∥Fy∥≤Ωr+g∗ψ≤r.
Now, fory,y~∈C0,1,ℝ, and for each t∈0,1, we get
(43)Fy−Fy~≤supt∈0,11ρα+β−1Γα+β∫0ttρ−sρα+β+1sρ−1gs,ys−gs,y~sds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,ys−gs,y~sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,ys−gs,y~sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,ys−gs,y~ssds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,ys−gs,y~sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,ys−gs,y~sds≤Ωy−y~1ρα+βΓα+β+1+ϕb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1ρδiηiρα+βρα+βΓα+β+1.
So,(44)Fy−Fy~≤ψΩy−y~.
AsΩψ<1, then F is a contraction. Hence, by the Banach fixed point theorem, F has a unique fixed point which is a unique solution of boundary value problem (1) on 0,1.
In the next result, we prove the existence of problem (1) by applying Schaefer’s fixed point theorem.Theorem 15.
Suppose that
(S2) there exist a constantν>0 such that ∣gt,y∣≤ν for each t∈0,1 and all y∈ℝ.
Then, the boundary value problem (1) has at least one solution in 0,1.Proof.
The proof is divided into three steps.Step 1.
LetK be a bounded subset of Y. By (S2), for all t∈0,1 and y∈K, we have
(45)∥Fy∥≤supt∈0,11ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gs,ysds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,ysds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,ysds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,ysds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,ysds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,ysds≤ν1ρα+βΓα+β+1+ϕb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1=νψ=ν∗.Then,∥Fy∥≤ν∗, and so F maps bounded sets into bounded sets in Y.Step 2.
Lett1,t2∈0,1,t1<t2, and K be a bounded set of Y. Thus, for y∈K and by (S2), we have
(46)∣Fyt2−Fyt1∣≤ν1ρα+β−1Γα+β∫0t2∣t2ρ−sρα+β−1sρ−1∣ds−∫0t1∣t1ρ−sρα+β−1sρ−1∣ds+1ψ1Γα∣tρα+β−1−τρα+β−1∣ρβΓα+βb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξi∣ξiρ−sρα+β+q−2sρ−1∣ds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηi∣ηiρ−sρα+β−2sρ−1∣ds−1ρα+β−2Γα+β−1∫01∣1−sρα+β−2sρ−1∣ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξi∣ξiρ−sρα+β+q−1sρ−1∣ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηi∣ηiρ−sρα+β−1sρ−1∣ds≤ν1ρα+βΓα+β+12t2ρ−t1ρα+β+∣t2ρα+β−t1ρα+β∣+1ψ1Γα∣tρα+β−1−τρα+β−1∣ρβΓα+βb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1.It follows that the right-hand side tends to zero ast2−t1⟶0 independent of y; by Arzela-Ascoli theorem, we conclude that F is completely continuous.Step 3.
It remains to show that the setM=y∈Y:y=àFyforsome0<à<1 is a bounded set. Let y∈M, and then, yt=àFyt,0<à<1,t∈0,1. Hence, by (S2), we have
(47)∥y∥≤supt∈0,11ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gs,ysds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,ysds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,ysds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,ysds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,ysds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,ysds≤ν1ρα+βΓα+β+1+ϕb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1≤νψ.This shows thatM is bounded. Schaefer’s fixed point theorem guarantees the existence of a fixed point of F. Hence, the integral equation (31) has a solution on [0,1].Now, we establish the existence of solutions of problem (1) via Krasnoselskii’s fixed point theorem.Theorem 16.
Assume thatg:0,1×ℝ⟶ℝis a continuous function such that it satisfies (S1). In addition, we suppose that
(S3)gt,x≤ζtfort∈0,1andζ∈C0,1,ℝwithζ=maxt∈0,1ζt.
Then, the boundary value problem (1) has at least one solution on [0,1], provided that
(48)Ωψ∗<1,where ψ∗=ψ−1/ρα+βΓα+β+1.Proof.
We define(49)supt∈0,1∣ζt∣=∥ζ∥and choose a suitable constant r∗ such that r∗≥∥ζ∥ψ, where ψ is given by (39). Furthermore, we split F into two operators F1 and F2 on Br∗=y∈Y:∥y∥≤r∗ by
(50)F1y=I0+α+β,ρgt,yt,forallt∈0,1,F2y=ϕtb1∑i=1mμiI0+α+β+q−1,ρgξi,yξi+b2∑i=1priI0+α+β−1,ρgηi,yηi−I0+α+β−1,ρg1,y1+1ψ2a1∑i=1mσiI0+α+β+q,ρgξi,yξi+a2∑i=1pδiI0+α+β,ρgηi,yηi.
Fory∈Br∗, we get
(51)∥F1y+F2y∥≤supt∈0,11ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gs,ysds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,ysds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,ysds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,ysds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,ysds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,ysds≤∥ζ∥ψ≤r∗,which shows that F1y+F2y∈Br∗. Now, we need to prove that F2 is a contraction. By assumption (S1), we have
(52)F2y−F2y~≤supt∈0,1ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,ys−gs,y~sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,ys−gs,y~sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,ys−gs,y~sds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,ys−gs,y~sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,ys−gs,y~sds.
Therefore,(53)∥F2y−F2y~∥≤Ω∥y−y~∥∥ϕ∥b1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1≤Ω∥y−y~∥ψ∗.
By using (S1) and (48), F2 is a contraction. Since g is continuous, we have F1 which is continuous. Now,
(54)F1y≤supt∈0,11ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gs,ysds≤supt∈0,11ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gs,ysds≤∥ζ∥1ρα+βΓα+β+1.
Thus,F1 is uniformly bounded on Br∗. Next, for t1,t2∈0,1 with 0<t1<t2<1, we obtain
(55)F1yt2−F1yt1≤1ρα+β−1Γα+β∫0t2t2ρ−sρα+β−1sρ−1∣gs,ys∣ds−∫0t1t1ρ−sρα+β−1sρ−1∣gs,ys∣ds≤νρ1−α+βρα+βΓα+β2t2ρ−t1ρα+β+t2ρα+β−t1ρα+β≤νρα+βΓα+β+12t2ρ−t1ρα+β+t2ρα+β−t1ρα+β,which tends to zero as t2−t1⟶0 independent of y. Thus, F1 is equicontinuous. Hence, F1 is relatively compact on Br∗. Then, all assumptions of Krasnoselskii’s fixed point theorem are satisfied, so the boundary value problem (1) has at least one solution on [0,1].The following example illustrates our main results.Example 1.
(56)D0+1/2,1/6RLD0+1/4,1/6ytC=gt,yt,0<α,β≤1,y0=17∑i=1mI3/4,1/6yξi+27∑i=11/6δiyηi,γy1=37∑i=12μiI3/4,1/6γyξi+47∑i=12riγyηi,where ξ1=1/5, ξ2=1/9, η1=1/8, η2=1/8, δ1=3/8, δ2=5/8, r1=3/10, r2=7/10, σ1=4/9, σ2=5/9, μ1=1/11, μ1=2/11, γ=t1−ρd/dt, and 0<q<1.Let the functiong be defined by
(57)gt,y=tan−1y10cost+16+2−t305∣y∣5∣y∣+1+e−t20cost+3.Using the given data, we findψ1=0.1089838156, ψ2=0.2461605410, ψ3=1.807429635, and ϕ=75.80569440. Choose Gu=uα/α,α>0, with α=2; then, G is an N-function and satisfies
(58)∫01G∣gs,us∣ds<∞,which shows that G belongs to an Orlicz space LG0,1. We have
(59)∣gt,y−gt,y~∣=tan−1y10cost+16+2−t305∣y∣5∣y∣+1+e−t20cost+3−tan−1y~10cost+16+2−t305∣y~∣5∣y~∣+1+e−t20cost+3≤110cost+16tan−1y−tan−1y~+2−t305y5y+1−5∣y~∣5∣y~∣+1.Consequently,(60)∣gt,y−gt,y~∣≤140∣y−y~∣+115∣y−y~∣≤140∣y−y~∣+140∣y−y~∣=120∣y−y~∣≤120∥y−y~∥,∣gt,y−gt,0∣≤120∥y∥,which shows the authenticity of (S1) with Ωψ=0.8902242605<1; then, Theorem 14 can be illustrated and problem (56) has a unique solution on [0,1]. Furthermore,
(61)gt,yt≤νt,with νt=Π/20cost+16+2−t/30+e−t/20cost+3 and ∥νt∥=0.2.Since the condition (S2) holds and by the conditions of Theorem15, problem (56) has at least one solution on [0,1].


## 3. Ulam Stability

In this section, we establish the criteria of Ulam stability of problem (1) by means of its equivalent integral equation
(62)wt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gs,wsds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gs,wsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gs,wsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gs,wsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gs,wsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gs,wsds,where w∈Y and g:0,1×ℝ⟶ℝ is a continuous function. We define a continuous nonlinear operator Λ:Y⟶Y(63)Λwt=RLD0+α,ρCD0+β,ρwt−gt,wt.Now, we recall some basic concepts of the Ulam stability. For more details, one can see [24–27].Definition 17.
GBVP is said to be Ulam-Hyers stable if there exists a real numberk>0such that for eachε>0and for each solutionw∈Yof ((1)) satisfying the inequality(64)∥Λwt∥≤ε,t∈0,1,there exists a solution y∈Y of GBVP satisfying the inequality
(65)∥y−w∥≤kε∗,t∈0,1,
whereε∗ is a positive real number depending on ε.Definition 18.
GBVP is generalized Ulam-Hyers stable if there existsτ∈Cℝ+,ℝ+such that for each solutionw∈Yof GBVP, there exists a solutiony∈Yof GBVP with(66)∣yt−wt∣≤τε,t∈0,1.Definition 19.
GBVP is Ulam-Hyers Rassias stable with respect toς∈C0,1,ℝ+if there exists a real numberk>0such that for eachε>0and for each solutionw∈Yof GBVP(67)∣Λwt∣≤εςt,t∈0,1,
we can find a solutiony∈Y of GBVP satisfying the inequality
(68)∣yt−wt∣≤kε∗ςt,t∈0,1,where ε∗ is a positive real number depending on ε.Theorem 20.
Assume that the condition (S1) andΩψ<1hold. Then, the GBVP satisfies both Ulam-Hyers and generalized Ulam-Hyers stability criteria.Proof.
We know thaty∈Y is a unique solution of GBVP as in Theorem 14. Let w∈Y be another solution of GBVP satisfying (64). Observe that the operator F and F−I are equivalent for every solution w∈Y (given by (62)) of GBVP. Therefore, by the fixed point property of operator F, we have
(69)∣wt−yt∣=∣wt−Fwt+Fwt−Fyt∣≤∣Fyt−Fwt∣+∣Fwt−wt∣≤Ωψ∣w−y∣+ε,where ε>0 and Ωψ<1. Taking the norm of (69) for t∈0,1 and solving for ∥w−y∥, we obtain
(70)∥w−y∥≤ε1−Ωψ.
If we letε∗=ε/1−Ωψ and take k=1, then the Ulam-Hyers stability condition is satisfied. More generally, defining τε=ε/1−mψ, the generalized Ulam-Hyers stability condition is also satisfied.Theorem 21.
Assume that (S1) andΩψ<1hold and there exists a functionς∈C0,1,ℝ+satisfying condition ((67)). Then, the problem GBVP is Ulam-Hyers Rassias stable with respect toς.Proof.
Following the idea of the proof of Theorem 19, we can obtain that(71)∥y−w∥≤ε∗ςt,with ε∗=ε/1−Ωψ.


## 4. Existence Results for Multivalued Maps

In this section, we extend the results to cover the multivalued case. We recall some basic definitions on multivalued maps. For more details, one can see [28].For a normed spaceX,., let PclX=y∈Px;yisclosed, PbX=y∈Px;yisbounded, PcpX=y∈Px;yiscompact, and Pcp,cX=y∈Px;yiscompactandconvex. A multivalued map G:X⟶PX is convex (closed) valued if Gx is convex (closed) for all x∈X. The map G is bounded on bounded sets if GB=∪x∈BGx is bounded in X for all B∈PbX, i.e.,
(72)supx∈Bsup∣y∣:y∈Gx<∞.G is called upper semicontinuous (u.s.c.) on X if for each x0∈X, the set Gx0 is a nonempty closed subset of X, and if for each open set N of X containing Gx0, there exists an open neighbourhood N0 of x0 such that GN0⊆N. G is said to be completely continuous if GB is relatively compact for every B∈PbX. If the multivalued map G is completely continuous with nonempty compact values, then G is u.s.c. if and only if G has a closed graph, i.e.,
(73)xn⟶x∗,yn⟶y∗,yn∈Gxnimplyy∗∈Gx∗.G has a fixed point if there is x∈X such that x∈Gx. The fixed point set of multivalued operator G will be denoted by FixG. A multivalued map G:0,1⟶Pclℝ is said to be measurable if for every y∈ℝ, the function
(74)t⟶y,Gt=inf∣y−z∣:z∈Gtis measurable.Definition 22.
A multivalued mapF:0,1×ℝ⟶Pℝis 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∈0,1. Further, a Carathéodory function F is called L1- Carathéodory if(iii)
for eachα>0, there exist φα∈L10,1,ℝ+ such that(75)∥Ft,x∥=supv:v∈Ft,x≤φαt,for all ∥x∥∞≤α and for a.e. t∈0,1
For eachy∈C0,1,ℝ, the set of selections of F is defined by
(76)SF,y≔v∈L10,1,ℝ+:vt∈Ft,ytfora.e.t∈0,1.Lemma 23.
LetXbe a Banach space. LetF:0,1×ℝ⟶Pcp,cXbe anL1- Carathéodory multivalued map and letΘbe a linear continuous mapping fromL10,1,XtoC0,1,X. Then, the operator(77)Θ∘SF:C0,1,X⟶Pcp,cC0,1,X,x↦Θ∘SFx=ΘSF,xis a closed graph operator in C0,1,X×C0,1,X.Definition 24.
A functiony∈C0,1,ℝis called a solution of problem ((2)) if we can find a functiong∈L10,1,ℝwithgt∈Gt,ya.e.on0,1such thaty0=a1∑i=1mσiI0+q,ρyξi+a2∑i=1pδiyηi,γy1=b1∑i=1mμiI0+q,ργyξi+b2∑i=1priγyηi,and(78)yt=I0+α+β,ρgt+ϕtb1∑i=1mμiI0+α+β+q−1,ρgξi+b2∑i=1priI0+α+β−1,ρgηi−I0+α+β−1,ρg1+1ψ2a1∑i=1mσiI0+α+β+q,ρgξi+a2∑i=1pδiI0+α+β,ρgηi.


### 4.1. The Upper Semicontinuous Case

To prove the existence of solutions for (2), we will apply Leray-Schauder nonlinear alternative for multivalued maps [29].Theorem 25.
Assume that
(H1)G:0,1×ℝ⟶Pcp,cℝisL1−Carathéodory where
(79)Pcp,cℝ=γ∈Pℝ:γiscompactandconvex.
(H2) there exists a continuous nondecreasing functionϖ:0,∞⟶0,∞ and a function P∈C0,1,ℝ+ such that
(80)∥gt,y∥P=sup∣y∣:y∈gt,y≤Ptϖ∥y∥,for all t,y∈0,1×ℝ.
(H3) there exists a numberL>0 such that
(81)L∥p∥ϖLψ>1,where ψ is given in (39).Then, the B.V.P has at least one solution on [0,1].Proof.
Let the operatorF~:C0,1,ℝ⟶PC0,1,ℝ be defined by
(82)F~y=ht∈C0,1,ℝ:ht=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−21Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds,or g∈SG,y. It is clear the fixed points of F~ are the solutions of problem (2). So, we need to verify that the operator satisfies all the conditions of the Leray-Schauder nonlinear alternative. This will be done in five steps.Step 1.
F~y is convex for all y∈C0,1,ℝ. Indeed, if h1 and h2 belong to F~y, then there exist g1,g2∈SG,y such that
(83)SG,y=g∈L10,1,ℝ+:gt∈Gt,ytfora.e.t∈0,1.
For eacht∈0,1, we have
(84)hit=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gisds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gisds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gisds−1ρα+β−21Γα+β−1∫011−sρα+β−2sρ−1gisds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gisds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gisds,for i=1,2. Let t∈0,1 and θ∈0,1, then
(85)θh1+1−θh2t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1θg1s+1−θg2sds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1θg1s+1−θg2sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1θg1s+1−θg2sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1θg1s+1−θg2sds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1θg1s+1−θg2sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1θg1s+1−θg2sds.SinceG has convex values (SG,y is convex), therefore, θh1+1−θh2∈F~y.Step 2.
F~y maps bounded sets into bounded sets in C0,1,ℝ. Let Bε=y∈C0,1,ℝ:∥y∥≤ε, where ε is a positive number. Then, for each h∈F~y,y∈Bε, there exists g∈SG,y such that
(86)ht=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.
For eacht∈0,1, we find that
(87)∣ht∣≤1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1∣gs∣ds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1∣gs∣ds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1∣gs∣ds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1∣gs∣ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1∣gs∣ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1∣gs∣ds≤∥P∥ϖ∥y∥1ρα+βΓα+β+1+∥ϕ∥b1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1≤∥P∥ϖ∥y∥ψ,which gives ∥h∥≤∥P∥ϖεψ.Step 3.
F~y maps bounded sets into equicontinuous sets in C0,1,ℝ. Let y be any element in Bε and t1,t2∈0,1,t1<t2; then, by using (H2), we get that for each t∈0,1, we find that
(88)∣ht2−ht1∣≤ptϖε1ρα+βΓα+β+12t2ρ−t1ρα+β+t2ρα+β−t1ρα+β+1ψ1Γα∣tρα+β−1−τρα+β−1∣ρβΓα+βb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1.It follows that the right-hand side tends to zero independently ofy∈Bε as t2−t1⟶0. Combining the outcomes of Steps 1–3 with the Arzela-Ascoli theorem, we conclude that F~ is completely continuous.Step 4.
F~ has a closed graph. Suppose that there is yn⟶y∗,hn∈F~yn, and hn⟶h∗. Then, we have to establish that h∗∈F~y∗. Since hn∈F~yn, there exists gn∈SG,yn and therefore, for each t∈0,1, we get
(89)hnt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gnsds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gnsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gnsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gnsds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gnsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gnsds.Next, we show that there existsg∗∈SG,y∗ such that for each t∈0,1,
(90)h∗t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1g∗sds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1g∗sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1g∗sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1g∗sds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1g∗sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1g∗sds.Consider the continuous linear operator(91)Ξ:L10,1,ℝ⟶C0,1,ℝgiven by
(92)g⟶Ξgt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gsds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.Then, we can get(93)∥hnt−h∗t∥≤1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1∣gns−g∗s∣ds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1∣gns−g∗s∣ds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1∣gns−g∗s∣ds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1∣gns−g∗s∣ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1∣gns−g∗s∣ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1∣gns−g∗s∣ds,for some g∗∈SG,y∗. Notice that ∥hnt−h∗t∥⟶0 as n⟶∞. So, we conclude that Ξ∘SG,y is a closed graph operator. Then, hn∈ΞSG,yn⟶h∗∈SG,y∗.Step 5.
There exists an open setG⊆C0,1,ℝ with y∉θF~y for any θ∈0,1 and for each y∈∂G. Take θ∈0,1. Let y be a solution of (2); then, there exists g∈L10,1,ℝ with g∈SG,y such that for t∈0,1, we have
(94)yt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−1−1Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−1−1Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−1−1Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.
Similar to the computation in Step2, for each t∈0,1, we obtain
(95)∣yt∣≤∥p∥ϖ∥y∥ψ,which implies
(96)∥yt∥∥p∥ϖ∥y∥ψ<1.
By (H3), there existsL>0 such that ∥y∥≠L. Define a set
(97)G=y∈C0,1,ℝ:y<L.Note that the operatorF~:G¯⟶PC0,1,ℝ is a compact multivalued map, u.s.c. with convex closed values. With the given choice of G, it is not possible to find y∈∂G satisfying y∈θF~y for some θ∈0,1. Consequently, by the nonlinear alternative of the Leray-Schauder theorem, the operator F~y has a fixed point y∈G¯, which is a solution of problem (2).


### 4.2. The Lipschitz Case

The following result is concerned with the nonconvex valued case of problem (2) and is based on Covitz and Nadler’s fixed point theorem [30].Theorem 26.
Assume that
(H4)G:0,1×ℝ⟶Pcpℝ is such that G.,y:0,1⟶Pcpℝ is measurable for each x∈ℝ, where
(98)Pcpℝ=γ∈Pℝ:γiscompactandconvex;
(H5)H¯dGt,y,Gt,ỹ≤ϖt∣y−ỹ∣ for almost all t∈0,1 and y,y~∈ℝ with ϖ∈C0,1,ℝ+ and d0,G0,Gt,0≤ϖt for almost all t∈0,1.
Then, problem (81) has at least one solution on [0,1] if
(99)∥ϖ∥ψ<1.Proof.
Let us verify that the operatorF~:C0,1,ℝ⟶PC0,1,ℝ satisfies the hypothesis of the Covitz and Nadler fixed point theorem. We show that F~y is nonempty and closed for every g~∈SG,y. Since the set-valued map G.,y. is measurable (by Theorem III.6 in [28]), it admits a measurable selection g:0,1⟶ℝ, and we have
(100)∣gt∣≤ϖt1+∣yt∣,i.e., g∈L10,1,ℝ, and hence, G is integrably bounded. Therefor, SG,y≠ϕ.
To prove thatF~y is closed for each y∈C0,1,ℝ, let hnn≥0∈F~y be such that hn⟶h as n⟶∞ in C0,1,ℝ. Then, h∈C0,1,ℝ and we can find gn∈SG,yn such that for each t∈0,1,
(101)hnt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gnsds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gnsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gnsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gnsds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gnsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gnsds.
AsG has compact values, we can pass onto a subsequence (if necessary) to obtain that gn converge to g in L10,1,ℝ. So g∈SG,y. Then, for each t∈0,1, we have
(102)hnt⟶ht=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.which implies that h∈F~y. Now, we establish that there exists 0<θ^<1, where it satisfies
(103)HdF~y,F~y~≤θ^∥y−y~∥foreachy,y~∈C0,1,ℝ.
Let us takey,y~∈C0,1,ℝ and h1∈F~y. Then, there exists g~1∈Gt,yt such that for all t∈0,1,(104)h1t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1g1sds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1g1sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1g1sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1g1sds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1g1sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1g1sds.
By (H5), we have(105)HdGt,y,Gt,y~t≤ϖtyt−y~t.
So, there existsρ∈Gt,y~t satisfying
(106)g1t−ρ≤ϖtyt−y~t,t∈0,1.
DefineEt=ρ∈ℝ:∣g1t−ρ∣≤ϖt∣yt−y~t∣. As the multivalued operator Et∩Gt,y~t is measurable, we can find a measurable selection g2t for Et (by Theorem III.2 in [28]). Thus, g2t∈Gt,y~t and for each t∈0,1, we have
(107)g1t−g2t≤ϖty−y~.
For eacht∈0,1, define
(108)h2t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1g2sds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1g2sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1g2sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1g2sds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1g2sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1g2sds.
As a result, we get(109)∥h1t−h2t∥≤1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1∣g1s−g2s∣ds+∣ϕt∣b1∑i=1mμi1ρα+β+q−1−1Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1∣g1s−g2s∣ds+b2∑i=1pri1ρα+β−1−1Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1∣g1s−g2s∣ds−1ρα+β−1−1Γα+β−1∫011−sρα+β−2sρ−1∣g1s−g2s∣ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1∣g1s−g2s∣ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1∣g1s−g2s∣ds.
Hence,(110)∣h1t−h2t∣≤∥ϖ∥ψ∥y−y~∥.
By interchange the roles ofy and y~, we get
(111)HdFy−F~y~≤∥ϖ∥ψ∥y−y~∥,which implies that F is a contraction. Hence, by the conclusion of the Covitz and Nadler fixed point theorem, F has a fixed point.


## 4.1. The Upper Semicontinuous Case

To prove the existence of solutions for (2), we will apply Leray-Schauder nonlinear alternative for multivalued maps [29].Theorem 25.
Assume that
(H1)G:0,1×ℝ⟶Pcp,cℝisL1−Carathéodory where
(79)Pcp,cℝ=γ∈Pℝ:γiscompactandconvex.
(H2) there exists a continuous nondecreasing functionϖ:0,∞⟶0,∞ and a function P∈C0,1,ℝ+ such that
(80)∥gt,y∥P=sup∣y∣:y∈gt,y≤Ptϖ∥y∥,for all t,y∈0,1×ℝ.
(H3) there exists a numberL>0 such that
(81)L∥p∥ϖLψ>1,where ψ is given in (39).Then, the B.V.P has at least one solution on [0,1].Proof.
Let the operatorF~:C0,1,ℝ⟶PC0,1,ℝ be defined by
(82)F~y=ht∈C0,1,ℝ:ht=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−21Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds,or g∈SG,y. It is clear the fixed points of F~ are the solutions of problem (2). So, we need to verify that the operator satisfies all the conditions of the Leray-Schauder nonlinear alternative. This will be done in five steps.Step 1.
F~y is convex for all y∈C0,1,ℝ. Indeed, if h1 and h2 belong to F~y, then there exist g1,g2∈SG,y such that
(83)SG,y=g∈L10,1,ℝ+:gt∈Gt,ytfora.e.t∈0,1.
For eacht∈0,1, we have
(84)hit=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gisds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gisds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gisds−1ρα+β−21Γα+β−1∫011−sρα+β−2sρ−1gisds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gisds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gisds,for i=1,2. Let t∈0,1 and θ∈0,1, then
(85)θh1+1−θh2t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1θg1s+1−θg2sds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1θg1s+1−θg2sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1θg1s+1−θg2sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1θg1s+1−θg2sds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1θg1s+1−θg2sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1θg1s+1−θg2sds.SinceG has convex values (SG,y is convex), therefore, θh1+1−θh2∈F~y.Step 2.
F~y maps bounded sets into bounded sets in C0,1,ℝ. Let Bε=y∈C0,1,ℝ:∥y∥≤ε, where ε is a positive number. Then, for each h∈F~y,y∈Bε, there exists g∈SG,y such that
(86)ht=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.
For eacht∈0,1, we find that
(87)∣ht∣≤1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1∣gs∣ds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1∣gs∣ds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1∣gs∣ds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1∣gs∣ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1∣gs∣ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1∣gs∣ds≤∥P∥ϖ∥y∥1ρα+βΓα+β+1+∥ϕ∥b1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1≤∥P∥ϖ∥y∥ψ,which gives ∥h∥≤∥P∥ϖεψ.Step 3.
F~y maps bounded sets into equicontinuous sets in C0,1,ℝ. Let y be any element in Bε and t1,t2∈0,1,t1<t2; then, by using (H2), we get that for each t∈0,1, we find that
(88)∣ht2−ht1∣≤ptϖε1ρα+βΓα+β+12t2ρ−t1ρα+β+t2ρα+β−t1ρα+β+1ψ1Γα∣tρα+β−1−τρα+β−1∣ρβΓα+βb1∑i=1mμiξiρα+β+q−1ρα+β+q−1Γα+β+q+1ρα+β−1Γα+βb2∑i=1priηiρα+β−1−1+1∣ψ2∣a1∑i=1mσiξiρα+β+qρα+β+qΓα+β+q+1+a2∑i=1pδiηiρα+βρα+βΓα+β+1.It follows that the right-hand side tends to zero independently ofy∈Bε as t2−t1⟶0. Combining the outcomes of Steps 1–3 with the Arzela-Ascoli theorem, we conclude that F~ is completely continuous.Step 4.
F~ has a closed graph. Suppose that there is yn⟶y∗,hn∈F~yn, and hn⟶h∗. Then, we have to establish that h∗∈F~y∗. Since hn∈F~yn, there exists gn∈SG,yn and therefore, for each t∈0,1, we get
(89)hnt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gnsds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gnsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gnsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gnsds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gnsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gnsds.Next, we show that there existsg∗∈SG,y∗ such that for each t∈0,1,
(90)h∗t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1g∗sds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1g∗sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1g∗sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1g∗sds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1g∗sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1g∗sds.Consider the continuous linear operator(91)Ξ:L10,1,ℝ⟶C0,1,ℝgiven by
(92)g⟶Ξgt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gsds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.Then, we can get(93)∥hnt−h∗t∥≤1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1∣gns−g∗s∣ds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1∣gns−g∗s∣ds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1∣gns−g∗s∣ds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1∣gns−g∗s∣ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1∣gns−g∗s∣ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1∣gns−g∗s∣ds,for some g∗∈SG,y∗. Notice that ∥hnt−h∗t∥⟶0 as n⟶∞. So, we conclude that Ξ∘SG,y is a closed graph operator. Then, hn∈ΞSG,yn⟶h∗∈SG,y∗.Step 5.
There exists an open setG⊆C0,1,ℝ with y∉θF~y for any θ∈0,1 and for each y∈∂G. Take θ∈0,1. Let y be a solution of (2); then, there exists g∈L10,1,ℝ with g∈SG,y such that for t∈0,1, we have
(94)yt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−1−1Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−1−1Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−1−1Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.
Similar to the computation in Step2, for each t∈0,1, we obtain
(95)∣yt∣≤∥p∥ϖ∥y∥ψ,which implies
(96)∥yt∥∥p∥ϖ∥y∥ψ<1.
By (H3), there existsL>0 such that ∥y∥≠L. Define a set
(97)G=y∈C0,1,ℝ:y<L.Note that the operatorF~:G¯⟶PC0,1,ℝ is a compact multivalued map, u.s.c. with convex closed values. With the given choice of G, it is not possible to find y∈∂G satisfying y∈θF~y for some θ∈0,1. Consequently, by the nonlinear alternative of the Leray-Schauder theorem, the operator F~y has a fixed point y∈G¯, which is a solution of problem (2).


## 4.2. The Lipschitz Case

The following result is concerned with the nonconvex valued case of problem (2) and is based on Covitz and Nadler’s fixed point theorem [30].Theorem 26.
Assume that
(H4)G:0,1×ℝ⟶Pcpℝ is such that G.,y:0,1⟶Pcpℝ is measurable for each x∈ℝ, where
(98)Pcpℝ=γ∈Pℝ:γiscompactandconvex;
(H5)H¯dGt,y,Gt,ỹ≤ϖt∣y−ỹ∣ for almost all t∈0,1 and y,y~∈ℝ with ϖ∈C0,1,ℝ+ and d0,G0,Gt,0≤ϖt for almost all t∈0,1.
Then, problem (81) has at least one solution on [0,1] if
(99)∥ϖ∥ψ<1.Proof.
Let us verify that the operatorF~:C0,1,ℝ⟶PC0,1,ℝ satisfies the hypothesis of the Covitz and Nadler fixed point theorem. We show that F~y is nonempty and closed for every g~∈SG,y. Since the set-valued map G.,y. is measurable (by Theorem III.6 in [28]), it admits a measurable selection g:0,1⟶ℝ, and we have
(100)∣gt∣≤ϖt1+∣yt∣,i.e., g∈L10,1,ℝ, and hence, G is integrably bounded. Therefor, SG,y≠ϕ.
To prove thatF~y is closed for each y∈C0,1,ℝ, let hnn≥0∈F~y be such that hn⟶h as n⟶∞ in C0,1,ℝ. Then, h∈C0,1,ℝ and we can find gn∈SG,yn such that for each t∈0,1,
(101)hnt=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gnsds+∣ϕt∣b1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gnsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gnsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gnsds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gnsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gnsds.
AsG has compact values, we can pass onto a subsequence (if necessary) to obtain that gn converge to g in L10,1,ℝ. So g∈SG,y. Then, for each t∈0,1, we have
(102)hnt⟶ht=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1gsds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1gsds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1gsds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1gsds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1gsds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1gsds.which implies that h∈F~y. Now, we establish that there exists 0<θ^<1, where it satisfies
(103)HdF~y,F~y~≤θ^∥y−y~∥foreachy,y~∈C0,1,ℝ.
Let us takey,y~∈C0,1,ℝ and h1∈F~y. Then, there exists g~1∈Gt,yt such that for all t∈0,1,(104)h1t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1g1sds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1g1sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1g1sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1g1sds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1g1sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1g1sds.
By (H5), we have(105)HdGt,y,Gt,y~t≤ϖtyt−y~t.
So, there existsρ∈Gt,y~t satisfying
(106)g1t−ρ≤ϖtyt−y~t,t∈0,1.
DefineEt=ρ∈ℝ:∣g1t−ρ∣≤ϖt∣yt−y~t∣. As the multivalued operator Et∩Gt,y~t is measurable, we can find a measurable selection g2t for Et (by Theorem III.2 in [28]). Thus, g2t∈Gt,y~t and for each t∈0,1, we have
(107)g1t−g2t≤ϖty−y~.
For eacht∈0,1, define
(108)h2t=1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1g2sds+ϕtb1∑i=1mμi1ρα+β+q−2Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1g2sds+b2∑i=1pri1ρα+β−2Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1g2sds−1ρα+β−2Γα+β−1∫011−sρα+β−2sρ−1g2sds+1ψ2a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1g2sds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1g2sds.
As a result, we get(109)∥h1t−h2t∥≤1ρα+β−1Γα+β∫0ttρ−sρα+β−1sρ−1∣g1s−g2s∣ds+∣ϕt∣b1∑i=1mμi1ρα+β+q−1−1Γα+β+q−1∫0ξiξiρ−sρα+β+q−2sρ−1∣g1s−g2s∣ds+b2∑i=1pri1ρα+β−1−1Γα+β−1∫0ηiηiρ−sρα+β−2sρ−1∣g1s−g2s∣ds−1ρα+β−1−1Γα+β−1∫011−sρα+β−2sρ−1∣g1s−g2s∣ds+1∣ψ2∣a1∑i=1mσi1ρα+β+q−1Γα+β+q∫0ξiξiρ−sρα+β+q−1sρ−1∣g1s−g2s∣ds+a2∑i=1pδi1ρα+β−1Γα+β∫0ηiηiρ−sρα+β−1sρ−1∣g1s−g2s∣ds.
Hence,(110)∣h1t−h2t∣≤∥ϖ∥ψ∥y−y~∥.
By interchange the roles ofy and y~, we get
(111)HdFy−F~y~≤∥ϖ∥ψ∥y−y~∥,which implies that F is a contraction. Hence, by the conclusion of the Covitz and Nadler fixed point theorem, F has a fixed point.



---

*Source: 3102142-2020-11-16.xml*