# A New Class ofψ-Caputo Fractional Differential Equations and Inclusion

Authors: Wafa Shammakh; Hadeel Z. Alzumi; Bushra A. AlQahtani
Journal: Journal of Mathematics
 (2021)

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## Abstract

In the present research work, we investigate the existence of a solution for new boundary value problems involving fractional differential equations withψ-Caputo fractional derivative supplemented with nonlocal multipoint, Riemann–Stieltjes integral and ψ-Riemann–Liouville fractional integral operator of order γ boundary conditions. Also, we study the existence result for the inclusion case. Our results are based on the modern tools of the fixed-point theory. To illustrate our results, we provide examples.


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## Body

## 1. Introduction

Fractional calculus has played a very important role in different areas of research (see [1, 2] and the references cited therein). Consequently, fractional differential equations have grasped the interest of many researchers working in diverse applications [3–6]. Recently, several researchers have tried to propose different types of fractional operators that deal with derivatives and integrals of arbitrary orders and their applications. For instance, Kilbas et al. in [2] introduced the properties of fractional integrals and fractional derivatives concerning another function. Some generalized fractional integral and differential operators and their properties were introduced by Agrawal in [7]. Very recently, Almeida in [8] presented a new type of fractional differentiation operator, the so-called ψ-Caputo fractional operator, and extended work of Caputo [2, 9]. Almeida et al. in [10, 11] investigated the existence and uniqueness of the results of nonlinear fractional differential equations involving a Caputo-type fractional derivative with respect to another function, employing the fixed-point theorem and Picard iteration method. Numerous interesting results concerning the existence, uniqueness, and stability of initial value problems and boundary value problems for fractional differential equations with ψ-Caputo fractional derivatives by applying different types of fixed-point techniques were obtained by Abdo et al. [12, 13], Vivek et al. [14], and Wahash et al. [15]. An important application that is controlled by the theory of ψ-fractional differentiation can be found in [16].In this paper, we investigate a new boundary value problem of fractional differential equations supplemented with nonlocal multipoint, Riemann–Stieltjes integral fractional boundary conditions involving Riemann–Liouville fractional integral operator of orderγ>0 with respect to function ψ given by the form(1)D0+,ψαcD0+,ψβcut+ht,ut=ft,ut,0<α,β≤1,u0=a1Ωu,u1=a2θu+a3∑i=1mμiI0+,ψγuηi,where D0+,ψαcandD0+,ψβc denotes the ψ-Caputo fractional derivatives of orders 0<α⩽1 and 0<β⩽1, respectively, I0+,ψγ is the ψ-Riemann–Liouville fractional integral operator of order γ>0. The functions h:0,1×ℝ⟶×ℝ and f:0,1×ℝ⟶×ℝ are continuous, whereas Ωu=∫01usdA1s and θu=∫01usdA2s are Riemann–Stieltjes integral and Ai⋅,i=1,2 is a function of bounded variation. aii=1,2,3 is a real constant, and μi,ηii=1,…,m are positive constants.We also study the corresponding inclusion problem that is given by(2)D0+,ψαcD0+,ψβcut+ht,ut∈Ft,ut,0<α,β≤1,u0=a1Ωu,u1=a2θu+a3∑i=1mμiI0+,ψγuηi,where F:C0,1,ℝ⟶Pℝ is a multivalued function, where Pℝ is the family of all subsets of ℝ and the other quantities are the same as defined in problem (1).Notice that this Riemann–Stieltjes integral fractional boundary conditions arise in manifold applications of computational fluid dynamics, distribution methods, and so forth (for example, see [17, 18]).This paper is organized as follows. In Section2, we recall some preliminary results and some related definitions. In Section 3, we discuss the existence results of solutions by relying on Krasnoselskii fixed-point theorem and Leray–Schauder nonlinear alternative. Also, we present an example. Finally, we describe the inclusion case and deduce the existence of solutions by applying Krasnoselskii’s multivalued fixed-point theorem in Section 4.


## 2. Preliminaries

For the convenience of the reader, we present here some necessary basic definitions, lemmas, and results which are used throughout this paper [2, 8, 10, 12, 19, 20].Definition 1.
Leta,b be a finite interval and 0≤ε<1 and ACa,b be the set of absolute continuous functions on a,b. Then, we define(3)ACγna,b=f:a,b⟶C andγn−1f∈ACa,b,γ=x1−ρddx,ACγ1a,b=ACa,b,Cγ,εna,b=f:a,b⟶C andγn−1f∈Ca,b,γnf∈Cε,ρ,γ=x1−ρddx,endowed with the normfCγ,εn=∑k=0n−1γkfC+γnfCε,ρ. The convention Cγ,0na,b=Cγna,b endowed with the norm fCγn=∑k=0nγkfC is used.Definition 2.
Letα>0, h be an integrable function defined on a,b, and ψ∈Cna,b be an increasing differentiable function such that ψ´t≠0 for all t∈a,b. The left-sided ψ-Riemann–Liouville fractional integral of order α of a function h is given by(4)I0+,ψαht=1Γα∫0tψt−ψsα−1ψ´shsds.Definition 3.
Letn−1<α<n,h:a,b⟶ℝ be an integrable function and ψ∈Cna,b be an increasing differentiable function such that ψ´t≠0 for all t∈a,b. The left-sided ψ-Riemann–Liouville fractional derivative of order α of a function h is defined by(5)D0+,ψαht=1ψ´tddtnI0+,ψn−αht,where n=α+1 and α denote the integer part of the real number α.Definition 4.
Letn−1<α<n,h∈Cn−1a,b, and ψ∈Cna,b be an increasing differentiable function such that ψ´t≠0 for all t∈a,b. The left-sided ψ-Caputo fractional derivative of order α of a function h is defined by(6)D0+,ψαcht=D0+,ψαht−∑k=0n−1hψk0k!ψt−ψ0k,where hψkt=1/ψ´td/dtkht and n=α+1 for α∉ℕandn=αforα∈ℕ. Furthermore, if h∈Cna,b and α∉ℕ, then(7)D0+,ψαcht=I0+,ψn−α1ψ´tddtnht=1Γn−α∫0tψt−ψsn−α−1ψ´shψnsds.
Thus, ifn=α∈ℕ, one has(8)D0+,ψαcht=hψnt.Lemma 1.
Given a functionh∈ACψn0,1 and α∈ℝ+, then(9)I0+,ψαD0+,ψαcht=ht−∑k=0n−1δψkh0k!ψt−ψ0k.Lemma 2.
Letα,β>0 and h:a,b⟶ℝ; then,(i)
I0+,ψαψt−ψ0β−1=Γβ/Γα+βψt−ψ0α+β−1(ii)
D0+,ψαcψt−ψ0β−1=Γβ/Γβ−αψt−ψ0β−α−1(iii)
D0+,ψαcψt−ψ0k=0,∀k∈0,1,…,n−1,n∈ℕ(iv)
I0+,ψαI0+,ψβht=I0+,ψα+βht
The existence of solutions of problem (1) relies on the following fixed-point theorems [21, 22].Theorem 1.
(Krasnoselskii’s fixed-point theorem). Letp be a closed, convex, bounded, and nonempty subset of a Banach space X. Let T1andT2 be operators such that(i)
T1u1+T2u2 belong to p whenever u1,u2∈p(ii)
T1 is compact and T2 is a contraction mapping
Then, there existu0∈p such that u0=T1u0+T2u0.Theorem 2.
(Leray–Schauder fixed-point theorem). LetC be a closed and convex subset of a Banach space E and U be an open subset of C with 0∈U. Suppose that V:U¯⟶C is a continuous, compact (that is, VU¯ is a relatively compact subset of C) map. Then, either(i)
V has a fixed point in U¯ or(ii)
there areu∈∂U (the boundary of U in C) and λ∈0,1 with u=λVu
For computational convenience, we set the following:(10)φ=1−a3∑i=1mμiψηi−ψ0γΓγ+1,ζ=a3∑i=1mμiψηi−ψ0β+γΓβ+γ+1−ψ1−ψ0βΓβ+1.Lemma 3.
Lety,g∈C0,1,ℝ; then, the linear ψ−fractional differential equation(11)D0+,ψαcD0+,ψαcut+gt=yt,u0=a1Ωu,u1=a2θu+a3∑i=1mμiI0+,ψγuηi,has a solution ut on 0,1 given by(12)ut=I0+,ψα+βyt−I0+,ψβgt+ψt−ψ0βζΓβ+1I0+,ψα+βy1−I0+,ψβg1+a3∑i=1mμiI0+,ψα+β+γyηi−I0+,ψβ+γgηi+a1φΩu−a2θu+a1Ωu.Proof.
We applyψ-Riemann–Liouville fractional integral of order α to both sides of the linear ψ−fractional differential equation:(13)D0+,ψαcD0+,ψβcut+gt=yt,0<α,β≤1.We obtain(14)D0+,ψβcut+gt=I0+,ψαyt+c1.
Next, applyingψ-Riemann–Liouville fractional integral of order β to both sides of (14), we obtain(15)ut=−I0+,ψβgt+I0+,ψα+βyt+I0+,ψβc1+c2,where c1andc2 are arbitrary constants. By Definition 1, general solution (15) can be written as(16)ut=I0+,ψα+βyt−I0+,ψβgt+c1Γβ+1ψt−ψ0β+c2.
Using the boundary conditionu0=a1Ωu, we obtain c2=a1Ωu. Thus, (16) takes the form(17)ut=I0+,ψα+βyt−I0+,ψβgt+c1Γβ+1ψt−ψ0β+a1Ωu.
Applying the operatorI0+,ψγ,γ>0, on equation (17), we obtain(18)I0+,ψγut=I0+,ψα+β+γyt−I0+,ψβ+γgt+c1ψt−ψ0β+γΓβ+γ+1+a1Ωuψt−ψ0γΓγ+1.
Using the boundary conditionu1=a2θu+a3∑i=1mμiI0+,ψγuηi, we find that(19)c1=1ζI0+,ψα+βy1−I0+,ψβg1−a2θu+a1Ωu1−a3∑i=1mμiψηi−ψ0γΓγ+1+a3∑i=1mμiI0+,ψα+β+γyηi−I0+,ψβ+γgηi.
Inserting the value ofc2 in (17) yields solution (12). The converse follows by direct computation.


## 3. Main Results

In this section, we prove the existence of solutions of problem (1). We shall assume that fandh are in the Banach space C0,1,ℝ. Let U=u:u∈C0,1,ℝ denote the Banach space of all continuous functions on 0,1 into ℝ endowed with the norm u=suput:t∈0,1. Here, we define an operator T:U⟶U associated with problem (1) by(20)Tut=I0+,ψα+βft,ut−I0+,ψβht,ut+ψt−ψ0βζΓβ+1I0+,ψα+βf1,u1−I0+,ψβh1,u1+a3∑i=1mμiI0+,ψα+β+γfηi,uηi−I0+,ψβ+γhηi,uηi+a1φΩu−a2θu+a1Ωu.Therefore, problem (1) has a solution if and only if the operator T has a fixed point.For computational convenience, we introduce the notations(21)G=ψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+ψ1−ψ0βζΓα+β⋅ψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1−ψηi−ψ0β+γΓβ+γ+1,(22)G1=a1Ωu1+ψ1−ψ0βζΓα+β,(23)G2=ψ1−ψ0βζΓα+βψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1−ψηi−ψ0β+γΓβ+γ+1,(24)G3=a11+ψ1−ψ0βζΓα+β∫01dA1s−a2ψ1−ψ0βζΓα+β∫01dA2s.Now, we will state and prove the existence result via Krasnoselskii’s fixed-point theorem.Theorem 3.
Letf,h:0,1×ℝ⟶ℝ be continuous functions satisfying the conditions:(H1)ft,u−ft,u˜≤ρ1u−u˜,andht,u−ht,u˜≤ρ2u−u˜ for all t∈0,1, every u,u˜∈ℝ and ρ>0,withρG2+G3<1andρ=maxρ1,ρ2.(H2) There exist continuous nonnegative functionsM1,M2∈C0,1,ℝ such that(25)ft,u≤M1tandht,u≤M2t,∀t,u∈0,1×ℝ,andM=maxM1,M2. Then, problem (1) has at least one solution on 0,1.Proof.
For a positive numberϵ, consider Bϵ=u∈U:u⩽ϵ,where ϵ≥MG+G1, and we split T into two operators T1 andT2whereT=T1+T2, on the bounded set Bϵ by(26)T1ut=I0+,ψα+βft,ut−I0+,ψβht,ut,T2ut=ψt−ψ0βζΓβ+1I0+,ψα+βf1,u1−I0+,ψβh1,u1+a3∑i=1mμi⋅I0+,ψα+β+γfηi,uηi−I0+,ψβ+γhηi,uηi+a1φΩu−a2θu+a1Ωu.For anyu∈Bϵ, by using (H2), we have(27)T1u+T2u≤supt∈0,1∫0tψt−ψsα+β−1Γα+βψ´sfs,usds−∫0tψt−ψsβ−1Γβψ´shs,usds+ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´sfs,usds−∫01ψ1−ψsβ−1Γβψ´shs,usds+a3∑i=1mμi∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfs,usds−∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shs,uss+a1φΩu−a2θu+a1Ωu≤Msupt∈0,1ψt−ψ0α+βΓα+β+1−ψt−ψ0βΓβ+1+ψt−ψ0βζΓα+β⋅ψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1−ψηi−ψ0β+γΓβ+γ+1+a1Ωu1+ψ1−ψ0βζΓα+β≤MG+G1≤ϵ.Hence,T1u+T2u∈Bϵ. Next, we show that T2 is a contraction mapping. Let u,u˜∈ℝ and t∈0,1, so by using (H1), we have(28)T2u−T2u˜≤supt∈0,1ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´sfs,us−fs,u˜sds−∫01ψ1−ψsβ−1Γβψ´shs,us−hs,u˜sds+a3∑i=1mμi∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´s⋅fs,us−fs,u˜sds−∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shs,us−hs,u˜sds+a1φΩu−Ωu˜−a2θu−θu˜+a1Ωu−Ωu˜≤ρu−u˜supt∈0,1ψt−ψ0βζΓα+βψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1−ψηi−ψ0β+γΓβ+γ+1+a11+ψ1−ψ0βζΓα+β⋅∫01u−u˜dA1s−a2ψ1−ψ0βζΓα+β⋅∫01u−u˜dA2s≤ρu−u˜ψ1−ψ0βζΓα+βψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1−ψηi−ψ0β+γΓβ+γ+1+a11+ψ1−ψ0βζΓα+β⋅∫01u−u˜dA1s−a2ψ1−ψ0βζΓα+β∫01u−u˜dA2s≤ρG2+a11+ψ1−ψ0βζ|Γα+β∫01dA1s−a2ψ1−ψ0βζΓα+β∫01dA2su−u˜≤ρG2+G3u−u˜.By assumptionρG2+G3<1, we obtain that T2 is a contraction mapping. Since fandh are continuous functions, we have T1 as continuous. Also, T1 is uniformly bounded on Bϵ as(29)T1u=supt∈0,1I0+,ψα+βft,ut−I0+,ψβht,ut≤supt∈0,1∫0tψt−ψsα+β−1Γα+βψ´sfs,usds−∫0tψt−ψsβ−1Γβψ´shs,usds≤Mψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1.Finally, we prove the compactness of the operatorT1. To show this, we define supt,u∈0,1×Bϵft,u=f¯<∞ and supt,u∈0,1×Bϵht,u=h¯<∞ and take 0≤t1<t2≤1. Thus, we have(30)T1ut2−T1ut1≤∫0t1ψt2−ψsα+β−1−ψt1−ψsα+β−1Γα+β⋅ψ´sfs,usds+∫t1t2ψt2−ψsα+β−1Γα+βψ´sfs,usds−∫0t1ψt2−ψsβ−1−ψt1−ψsβ−1Γβ⋅ψ´shs,usds−∫t1t2ψt2−ψsβ−1Γβψ´shs,usds≤f¯Γα+β+12ψt2−ψt1α+β+ψt2−ψ0α+β−ψt1−ψ0α+β+h¯Γβ+12ψt2−ψt1β−ψt2−ψ0β+ψt1−ψ0β⟶0,ast2−t1⟶0,independent ofu∈Bϵ. Thus, T1 is equicontinuous. So, T1 is relatively compact on Bϵ. Hence, by the Arzela–Ascoli theorem, T1 is compact on Bϵ. Thus, the hypotheses of Theorem 6 are satisfied which leads problem (1) to have at least one solution on 0,1.Now, we apply Leray–Schauder nonlinear alternative fixed-point theorem to establish an existence result for problem (1).Theorem 4.
Letf,h:0,1×ℝ⟶ℝ be continuous functions and the following conditions are satisfied:(H3) There exist functionsl1,l2∈C0,1,ℝ with l=maxl1,l2 and nondecreasing functions q1,q2:ℝ+⟶ℝ+,q=maxq1,q2 such that(31)ft,u≤l1q1u,ht,u≤l2q2u,for allt,u∈0,1×ℝ.(H4) There exists a constantN>0 such that(32)NlqNG+G1>1.Then, problem (1) has at least one solution on 0,1.Proof.
Let us show that the operatorT maps bounded sets into bounded sets in C0,1,ℝ. For a positive r, let Br=u∈C0,1,ℝ:u≤r be a bounded set in C0,1,ℝ. For each t∈0,1, by (H3), we have(33)Tut≤supt∈0,1∫0tψt−ψsα+β−1Γα+βψ´sfs,usds−∫0tψt−ψsβ−1Γβψ´shs,usds+ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´sfs,usds−∫01ψ1−ψsβ−1Γβψ´shs,usds+a3∑i=1mμi⋅∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfs,usds−∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shs,usds+a1φΩu−a2θu+a1Ωu≤l1q1usupt∈0,1ψt−ψ0α+βΓα+β+1+ψt−ψ0βζΓα+β⋅ψ1−ψ0α+βΓα+β+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1+l2q2usupt∈0,1ψt−ψ0βζΓβ+1−ψt−ψ0βζΓα+β.ψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0β+γΓβ+γ+1+a1Ωu1+ψt−ψ0βζΓα+β≤lqrψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+ψ1−ψ0βζΓα+β⋅ψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+a3∑i=1mμi.ψηi−ψ0α+β+γΓα+β+γ+1−ψηi−ψ0β+γΓβ+γ+1+a1Ωu1+ψ1−ψ0βζΓα+β≤lqrG+G1.
Next, we show thatT maps bounded sets into equicontinuous sets of C0,1,ℝ. Let u∈Br; then, for t1,t2∈0,1, with t1<t2, we have(34)Tut2−Tut1≤∫0t1ψt2−ψsα+β−1−ψt1−ψsα+β−1Γα+β⋅ψ´sfs,usds+∫t1t2ψt2−ψsα+β−1Γα+βψ´sfs,usds−∫0t1ψt2−ψsβ−1−ψt1−ψsβ−1Γβ⋅ψ´shs,usds−∫t1t2ψt2−ψsβ−1Γβψ´shs,usds+ψt2−ψ0β−ψt1−ψ0βζΓα+β⋅∫01ψ1−ψsα+β−1Γα+βψ´sfs,usds−∫01ψ1−ψsβ−1Γβψ´shs,usds+a3∑i=1mμi∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfs,usds−∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shs,usds+a1φΩu−a2θu≤l1q1r2ψt2−ψt1α+β+ψt2−ψ0α+β−ψt1−ψ0α+βΓα+β+1+ψt2−ψ0β−ψt1−ψ0βζΓα+βψ1−ψ0α+βΓα+β+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1+l2q2r⋅2ψt2−ψt1β−ψt2−ψ0β+ψt1−ψ0βΓβ+1−ψt2−ψ0β−ψt1−ψ0βζΓα+βψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0β+γΓβ+γ+1,which tends to zero independent of u∈Br as t2−t1⟶0. So, we deduce that the operator T:C0,1,ℝ⟶C0,1,ℝ is completely continuous (by Arzela–Ascoli theorem). It remains to show the boundedness of the set of all solutions u=λTu for λ∈0,1. Let u∈U be a solution of problem (1). So, for t∈0,1, we obtain(35)ut=λTut≤Tut,which, on taking the norm for t∈0,1, yields(36)u≤lqrG+G1,and then(37)ulqrG+G1≤1.
From (H4), we can findN>0 such that u≠N. Take a set U∗=u∈C0,1,ℝ:u<N and notice that the operator T:U¯⟶C0,1,ℝ is continuous and completely continuous. By choice of U, we cannot find a u∈∂U such that u=λTu for some λ∈0,1. Hence, by Theorem 2, the operator T has a fixed point u∈U¯ which is a solution of problem (1).Example 1.
Consider the following boundary value problem:(38)D0+,ψ1/4cD0+,ψ1/2cut+ht,ut=ft,ut,0<α,β≤1,u0=a1Ωu,u1=a2θu+a3∑i=12μiI0+,ψγuηi,where α=1/4,β=1/2,A1s=A2s=s,a1=a2=3/4,a3=1,m=2, μ1=1/5,μ2=2/5,η1=1/6,η2=5/6,γ=1/3, and ψt=2t2+1. Clearly, ψ is an increasing function on 0,1 and ψ´t=4t is a continuous function on 0,1.
To illustrate the application of Theorem3, we take(39)ft,u=cosu2274+e−t4t2+20,ht,u=148sinu+5t22,where fandh satisfy the assumption of Theorem 3. By using the given data, we find ρ=0.0208,ζ=0.589473482, M=0.072916, G2=0.3951268038, and G3=0. In addition, ρG2+G3≈0.008218637519<1. Therefore, the result of Theorem 3 applies to problem (38) with ft,u and ht,u given above.


## 4. Existence Results for Inclusion Case

In this section, we extend the results to cover the inclusion problem and prove the existence of solutions for problem (2) by applying the fixed-point theorem [23]. We recall some basic notations for the inclusion case [24–30].For a normed spaceX,⋅, let(40)PclX=y∈Px:yis closed,Pcp,cvℝ=γ∈Pℝ:γis compact and convex.A multivalued mapF:0,1×ℝ⟶Pℝ is said to be carathe´odory if(i)
t⟶Ft,x is measurable for each x∈ℝ(ii)
x⟶Ft,x is upper semicontinuous for almost all t∈0,1Furthermore, a carathe´odory function F is called L1-Carathe´odory if(iii)
for eachα>0, there exist φα∈L10,1,ℝ+ such that(41)Ft,x=supv:v∈Ft,x≤φαt,for allx∞≤α and for a.e. t∈0,1.For eachy∈C0,1,ℝ, define the set of selections of F by(42)SF,y≔v∈L10,1,ℝ+:vt∈Ft,yt,for a.e.t∈0,1.LetX,d be a metric space induced from the normed space X;⋅. We have Hd:PX×PX⟶ℝ∪∞ given by(43)HdA,B=maxsupa∈Ada,B,supb∈BdA,b,where dA,b=infa∈Ada;b and da,B=infb∈Bda;b. So, PclX,Hd is a metric space [31].Lemma 4.
LetX be a Banach space. Let F:0,T×ℝ⟶Pcp,cvX be an L1-Carathe´odory multivalued map, and let Θ be a linear continuous mapping from L10,1,XtoC0,1,X. Then, the operator(44)Θ°SF:C0,1,X⟶Pcp,cvC0,1,X,x⟶Θ°SFx=ΘSF,xis a closed graph operator in C0,1,X×C0,1,X.Lemma 5.
IfF:X⟶PclY is u.s.c., then FrF is a closed subset of X×Y; i.e., for every sequence xnn∈ℕ⊂X and ynn∈ℕ⊂Y, if n⟶∞, xn⟶x∗,yn⟶y∗, and yn∈Fxn, then y∗∈Fx∗. Conversely, if F is completely continuous and has a closed graph, then it is upper semicontinuous.Definition 5.
A functionu∈C0,1,ℝ is called a solution of problem (2) if we can find a function f∈L10,1,ℝ with ft∈Ft,ua.e.on0,1 such that u0=a1Ωu,u1=a2θu+a3∑i=1mμiI0+,ψγuηi and(45)ut=I0+,ψα+βft−I0+,ψβht+ψt−ψ0βζΓβ+1I0+,ψα+βf1−I0+,ψβh1+a3∑i=1mμiI0+,ψα+β+γfηi−I0+,ψβ+γhηi+a1φΩu−a2θu+a1Ωu.
For convenience, we denote(46)Δ1=ψ1−ψ0βΓβ+1+ψ1−ψ0βζΓα+βψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0β+γΓβ+γ+1,(47)Δ2=ψ1−ψ0α+βΓα+β+1+ψt−ψ0βζΓα+βψ1−ψ0α+βΓα+β+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1.
Our result is based on the following fixed-point theorem.Theorem 5.
LetU and U¯ be, respectively, the open and closed subsets of Banach space X, such that 0∈U; let χ1u:U¯⟶Pcp,cvX be multivalued and χ2u:U¯⟶X be single-valued such that χ1U¯+χ2U¯ is bounded. Suppose that(a)
χ2 is a contraction with a contraction k<1/2(b)
χ1 is u.s.c and compact
Then, either(i)
the operator inclusionλx∈χ1x+χ2x has a solution for λ=1 or(ii)
there is an elementu∈∂U such that λu∈χ1u+χ2u for some λ>1, where ∂U is the boundary of UTheorem 6.
Assume that(N1)F:0,1×ℝ⟶Pcp,cℝisL1−Carathe´odory.(N2) There exists a continuous functionϖ∈C0,1,ℝ+ and Λ∈C0,1,ℝ+ such that(48)Ft,uP=supu:u∈Ft,u≤Λtϖu;for allt,u∈0,1×ℝ.(N3) Leth:0,1×ℝ⟶ℝ be a continuous functions satisfying(49)hs,u−hs,u˜≤Eu−u˜,∀u,u˜∈ℝandE>0.(N4) There exists a numberτ>0 such that(50)τΛϖτG+G1<1,whereGandG1 are defined in (21) and (22), respectively. Then, problem (2) has at least one solution on 0,1 if EΔ1<1/2.Proof.
LetD=u∈U:u<ε be an open set in U. Define the multivalued operator χ1:D¯⟶PU by(51)χ1u=z∈U:zt=I0+,ψα+βft+ψt−ψ0βζΓβ+1I0+,ψα+βf1+a3∑i=1mμiI0+,ψα+β+γfηi+a1φΩu−a2θu+a1Ωu,and define the single-valued operator χ2:D¯⟶U by(52)χ2u=z∈U:zt=−I0+,ψβht−ψt−ψ0βζΓβ+1I0+,ψβh1+a3∑i=1mμiI0+,ψβ+γhηi.
Observe thatχ=χ1+χ2, and it is given by(53)χu=zt∈C0,1,ℝ:zt=∫0tψt−ψsα+β−1Γα+βψ´sfsds−∫0tψt−ψsβ−1Γβψ´shsds+ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´sfsds−∫01ψ1−ψsβ−1Γβψ´shsds+a3∑i=1mμi∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfsds−∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shsds+a1φΩu−a2θu+a1Ωu,f∈SF,u.
Indeed, ifz∈χu, then there exists f∈SF,u, such that(54)SF,u=f∈L10,1,ℝ+:ft∈Ft,ut,for a.e.t∈0,1.
We will show that the mapsχ1 and χ2 satisfy the hypotheses of Theorem 5. This will be done in several steps.Step 1.
χ2 is a contraction. Let u,u˜∈ℝ, by (N3), we have(55)χ2u−χ2u˜⩽supt∈0,1−∫0tψt−ψsβ−1Γβψ´shs,us−hs,u˜sds−ψt−ψ0βζΓα+β∫01ψ1−ψsβ−1Γβψ´shs,us−hs,u˜sds+a3∑i=1mμi∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shs,us−hs,u˜sds≤Eu−u˜supt∈0,1−ψt−ψ0βΓβ+1−ψt−ψ0βζΓα+βψt−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0β+γΓβ+γ+1≤Eu−u˜ψ1−ψ0βΓβ+1+ψ1−ψ0βζΓα+βψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0β+γΓβ+γ+1≤EΔ1u−u˜,which proves that χ2 is a contraction map.Step 2.
χ1u is convex for all u∈D¯. Let z1,z2∈χ1u. We select f1,f2∈SF,u such that, for each t∈0,1, we obtain(56)zit=I0+,ψα+βfit+ψt−ψ0βζΓβ+1I0+,ψα+βfi1+a3∑i=1mμiI0+,ψα+β+γfiηi+a1φΩu−a2θu+a1Ωu.for i=1,2.
Lett∈0,1andϕ∈0,1. So, we have(57)ϕz1+1−ϕz2t=∫0tψt−ψsα+β−1Γα+βψ´sϕf1s+1−ϕf2sds+ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´s⋅ϕf1s+1−ϕf2sds+a3∑i=1mμi⋅∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´s⋅ϕf1s+1−ϕf2sds+a1ϕ+1−ϕφΩu−a2ϕ+1−ϕθu+a1ϕ+1−ϕΩu.Since SF,u is convex, it follows that ϕz1+1−ϕz2∈χ1u and then χ1u is convex-valued.Step 3.
χ1 is compact and upper semicontinuous. This will be done in various statements. First, we show that χ1 maps bounded sets into bounded sets in U.
For a positive numberς, let Bς=u∈U:u≤ς be a bounded ball in D¯. So, for all z∈χ1, u∈Bς, there exists f∈SF,u such that(58)zt=I0+,ψα+βft+ψt−ψ0βζΓβ+1I0+,ψα+βf1+a3∑i=1mμiI0+,ψα+β+γfηi+a1φΩu−a2θu+a1Ωu.
By using (N2), for eacht∈0,1, we have(59)zt≤∫0tψt−ψsα+β−1Γα+βψ´sfs,usds+ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´sfs,usds+a3∑i=1mμi∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfs,usds+a1φΩu−a2θu+a1Ωu≤Λϖuψ1−ψ0α+βΓα+β+1+ψt−ψ0βζΓα+βψ1−ψ0α+βΓα+β+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1+a1Ωu1+ψ1−ψ0βζΓα+β.
Consequently,(60)z≤ΛϖuΔ2+G1,where G1 and Δ2 are defined in (22) and (47), respectively. Second, we prove that χ1 maps bounded sets into equicontinuous sets. Let τ1,τ2∈0,1 with τ1<τ2 and u∈Bς; we have(61)zτ2−zτ1≤∫0τ1ψτ2−ψsα+β−1−ψτ1−ψsα+β−1Γα+βψ´sfs,usds+∫τ1τ2ψτ2−ψsα+β−1Γα+βψ´sfs,usds+ψτ2−ψ0β−ψτ1−ψ0βζΓα+β⋅∫01ψ1−ψsα+β−1Γα+βψ´sfs,usds+a3∑i=1mμi⋅∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfs,usds≤2ψτ2−ψτ1α+β+ψτ2−ψ0α+β−ψτ1−ψ0α+βΓα+β+1+ψτ2−ψ0β−ψτ1−ψ0βζΓα+βψ1−ψ0α+βΓα+β+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1.
In the above inequality, the right hand side tends to zero independent ofu∈Be as τ2−τ1⟶0. Consequently, by the Arzela–Ascoli theorem, we conclude that χ1:D¯⟶PU is completely continuous and then χ1 is completely continuous. Finally, we show χ1 has a closed graph. Let un⟶u∗,zn∈χ1un, and zn⟶z∗. Then, we show that z∗∈χ1u∗. Since zn∈χ1un, there exists zn∈SF,un such that for each t∈0,1, we find that(62)znt=I0+,ψα+βfnt+ψt−ψ0βζΓβ+1I0+,ψα+βfn1+a3∑i=1mμiI0+,ψα+β+γfnηi+a1φΩu−a2θu+a1Ωu.
Now, we have to show that there existsz∗∈SF,u∗ such that for each t∈0,1,(63)z∗t=I0+,ψα+βf∗t+ψt−ψ0βζΓβ+1I0+,ψα+βf∗1+a3∑i=1mμiI0+,ψα+β+γf∗ηi+a1φΩu−a2θu+a1Ωu.
Consider the continuous linear operatorΘ:L10,1,ℝ⟶U given by(64)f↦Θft=I0+,ψα+βft+ψt−ψ0βζΓβ+1I0+,ψα+βf1+a3∑i=1mμiI0+,ψα+β+γfηi+a1φΩu−a2θu+a1Ωu.
Note that(65)znt−z∗t=supt∈0,1I0+,ψα+βfnt−f∗t+ψt−ψ0βζΓβ+1I0+,ψα+βfn1−f∗1+a3∑i=1mμiI0+,ψα+β+γfnηi−f∗ηi+a1φΩu−a2θu,which goes to 0, as n⟶∞.
It follows by Lemma5 that Θ°SF is a closed graph operator. Furthermore, we obtain znt∈ΘSF,un. Since un⟶u∗, we have(66)z∗t=I0+,ψα+βf∗t+ψt−ψ0βζΓβ+1I0+,ψα+βf∗1+a3∑i=1mμiI0+,ψα+β+γf∗ηi+a1φΩu−a2θu+a1Ωu,for some z∗∈SF,u∗. Hence, χ1 has a closed graph (and therefore has closed values). Hence, we conclude that χ1 is a compact multivalued map, upper semicontinuous with convex closed values.Step 4.
There exists an open setQ⊆C0,1,ℝ with u∉λχ2u for any λ>1 and for each u∈∂Q. Take λ>1. Let u be a solution of (2); then, there exists f∈L10,1,ℝ with f∈Sf,u such that for t∈0,1, we have(67)ut=ψt−ψsα+β−1Γα+βψ´sfsds−∫0tψt−ψsβ−1Γβψ´shsds+ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´sfsds−∫01ψ1−ψsβ−1Γβψ´shsds+a3∑i=1mμi∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfsds−∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shsds+a1φΩu−a2θu+a1Ωuut≤∫0tψt−ψsα+β−1Γα+βψ´sfsds−∫0tψt−ψsβ−1Γβψ´shsds+ψt−ψ0βζΓα+β∫01ψ1−ψsα+β−1Γα+βψ´sfsds−∫01ψ1−ψsβ−1Γβψ´shsds+a3∑i=1mμi∫0ηiψηi−ψsα+β+γ−1Γα+β+γψ´sfsds−∫0ηiψηi−ψsβ+γ−1Γβ+γψ´shsds+a1φΩu−a2θu+a1Ωu≤Λϖuψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+ψ1−ψ0βζΓα+β⋅ψ1−ψ0α+βΓα+β+1−ψ1−ψ0βΓβ+1+a3∑i=1mμiψηi−ψ0α+β+γΓα+β+γ+1−ψηi−ψ0β+γΓβ+γ+1+a1Ωu1+ψ1−ψ0βζΓα+β≤ΛϖuG+G1,which implies(68)uΛϖuG+G1<1.
By (N4), there existsτ>0 such that u≠τ. Define a set(69)Q=u∈C0,1,ℝ:u<τ.
Note that the operatorχ1:Q¯⟶PU is a compact multivalued map, u.s.c. with convex closed values. With the given choice of Q, it is not possible to find u∈∂Q satisfying u∈λχu for some λ>1. Consequently, the operator χu has a fixed point u∈Q¯, which is a solution of problem (2).Example 2.
Consider the following boundary value problem:(70)D0+,ψ1/3cD0+,ψ1/4cut+ht,ut∈Ft,ut,0<α,β≤1,u0=a1Ωu,u1=a2θu+a3∑i=12μiI0+,ψγuηi,where F:C0,1,ℝ⟶Pℝ is a multivalued map given by(71)u⟶Ft,u=uu+11+1+2t9,cosu+1+t3.
Forf∈F, we obtain(72)f≤maxuu+11+1+2t9,cosu+1+t3≤53.
Here,α=1/3,β=1/4,A1s=A2s=s,a1=0,a2=3/4,a3=1,m=2,
μ1=1/7,μ2=3/7,η1=1/5,η2=4/5,γ=1/2, and ψt=3t2+2. Clearly, ψ is an increasing function on 0,1 and ψ´t=6t is a continuous function on 0,1.
Clearly,(73)Ft,uP=supu:u∈Ft,u≤Λtϖu;for allt,u∈0,1×ℝ,with Λ=1andϖu=5/3.
Next, we take(74)ht,u=4217sinu+t4+et26+3t,where the function h satisfies the assumption of Theorem 6. By using the given data, we find E=0.0184331797,ζ=−0.378649448,Δ1=6.552924874, G=2.343253024, and G1=0. Thus,(75)u>ΛϖuG+G1≈3.905421707,and we have(76)EΔ1≈0.1207912419<12.
Therefore, all the conditions of Theorem6 are satisfied. Then, there exists at least one solution of problem (70) on 0,1.



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