Existence Results for a Riemann-Liouville-Type Fractional Multivalued Problem with Integral Boundary Conditions

Abstract

We discuss the existence of solutions for a boundary value problem of Riemann-Liouville fractional differential inclusions of order 𝛼∈(2,3] with integral boundary conditions. We establish our results by applying the standard tools of fixed point theory for multivalued maps when the right-hand side of the inclusion has convex as well as nonconvex values. An illustrative example is also presented.

1. Introduction

In the last few decades, fractional calculus is found to be an effective modeling tool in many branches of physics, economics, and technical sciences [1–3]. A fractional-order differential operator is nonlocal in its character in contrast to its counterpart in classical calculus. It means that the future state of a dynamical system or process based on fractional-order derivative depends on both its current and past states. Thus, the application of fractional calculus in various materials and processes enables an investigator to study the complete behavior (ranging from past to current states) of such stuff. This is indeed an important feature that makes fractional-order models more realistic and practical than the integer-order models and has accounted for the popularity of the subject. For some recent development on the topic, see [4–17] and the references therein.

Differential inclusions appear in the mathematical modeling of certain problems in economics, optimal control, and so forth and are widely studied by many authors. Examples and details can be found in a series of papers [18–23] and the references cited therein.

In this paper, we study the following boundary value problem: 𝐷^𝛼₀₊𝑥(𝑡)((𝑡𝑥(0)∈𝐹𝑡,𝑥)),0<𝑡<1,2<𝛼≤3,=𝑥^′(0)𝑥(1)==0,₁₀𝑔(𝑠)𝑥(𝑠)𝑑𝑠, (1) where 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative of order 2<𝛼≤3,𝐹∶[0,1]×ℝ→𝒫(ℝ)is a multivalued map, 𝒫(ℝ) is the family of all subsets of ℝ, and 𝑔 is a continuous function.

Here we remark that the present work is motivated by a recent paper [17], where problem (1) is considered with 𝐹 as single valued and the results on existence and nonexistence of positive solutions are obtained.

The main tools of our study include nonlinear alternative of Leray-Schauder type, a selection theorem due to Bressan and Colombo for lower semicontinuous multivalued maps, and Covitz and Nadler's fixed point theorem for contraction multivalued maps. The application of these results is new in the framework of the problem at hand. We recall some preliminaries in Section 2 while the main results are presented in Section 3.

2. Preliminaries

2.1. Fractional Calculus

Let us recall some basic definitions of fractional calculus [1, 2].

Definition 1. The Riemann-Liouville derivative of fractional order 𝑞 is defined as 𝐷^𝑞₀₊𝑔(𝑡)=1Γ𝑑𝑛−𝑞𝑑𝑡^𝑛∫^𝑡₀(𝑡−𝑠)^{𝑛−𝑞−1}𝑔(𝑠)𝑞𝑑𝑠,𝑛−1<𝑞<𝑛,𝑛=+1, (2) provided the integral exists, where [𝑞] denotes the integer part of the real number 𝑞.

Definition 2. The Riemann-Liouville fractional integral of order 𝑞 is defined as 𝐼^𝑞₀₊𝑔(𝑡)=1Γ𝑞^𝑡₀𝑔(𝑠)(𝑡−𝑠)^1−𝑞𝑑𝑠,𝑞>0, (3) provided the integral exists.

Definition 3. A function 𝑥∈𝐴𝐶²([0,1],ℝ) is called a solution of problem (1) if there exists a function 𝑣∈𝐿¹([0,1],ℝ) with 𝑣(𝑡)∈𝐹(𝑡,𝑥(𝑡)), a.e. [0,1] such that 𝐷^𝛼𝑥(𝑡)=𝑣(𝑡), a.e. [0,1] and 𝑥(0)=𝑥^′(0)=0, 𝑥(1)=∫¹₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡.

Lemma 4 (see [17]). Given 𝑦∈𝐿([0,1],ℝ), then the unique solution of the problem 𝐷^𝛼𝑥(𝑡)(𝑡)𝑥(0)+𝑦=0,0<𝑡<1,2<𝛼≤3,=𝑥^′(0)𝑥(1)==0,₁₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡 (4) is given by 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (5) where 𝐺(𝑡,𝑠)=𝐺₁(𝑡,𝑠)+𝐺₂(),𝐺𝑡,𝑠₁()=1𝑡,𝑠Γ(𝛼)[𝑡(1−𝑠)]^𝛼−1[𝑡(,0≤𝑡≤𝑠≤1,1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1𝐺,0≤𝑠≤𝑡≤1,₂()=𝑡𝑡,𝑠^𝛼−11−𝜎¹₀𝐺₁(𝜏,𝑠)𝑔(𝜏)𝑑𝜏,𝜎=¹₀𝑠^𝛼−1𝑔(𝑠)𝑑𝑠. (6)(7)(8)

Lemma 5 (see [17]). The functions 𝐺₁(𝑡,𝑠),𝐺(𝑡,𝑠) have the following properties: (i)𝐺₁(𝑡,𝑠)≥0, for all 𝑡,𝑠∈[0,1];(ii)𝐺₁(𝑡,𝑠)≤((𝛼−1)𝑠(1−𝑠)^𝛼−1)/Γ(𝛼) for all 𝑡,𝑠∈[0,1];(iii)𝐺(𝑡,𝑠)≤(((𝛼−1)𝑠(1−𝑠)^𝛼−1)/Γ(𝛼))(1+(1/(1−𝜎))∫¹₀𝑔(𝜏)𝑑𝜏) for all 𝑡,𝑠∈[0,1].

2.2. Basic Concepts of Multivalued Maps

Let (𝒳,‖⋅‖) be a normed space and let 𝐻∶𝒳→𝒫(𝒳) be a multivalued map. 𝐻 is said to be(i) convex (closed) valued if 𝐻(𝑥) is convex (closed) for all 𝑥∈𝒳;(ii) bounded on bounded sets if 𝐻(𝔹)=∪_𝑥∈𝔹𝐻(𝑥) is bounded in 𝒳 for all 𝔹∈𝒫_𝑏(𝒳), where 𝒫_𝑏(𝒳)={𝒜∈𝒫(𝒳)∶𝒜isbounded}, (i.e., sup_𝑥∈𝔹{sup{|𝑤|∶𝑤∈𝐻(𝑥)}}<∞);(iii) upper semicontinuous (u.s.c.) on 𝒳 if the set 𝐻(𝑎₀) is a nonempty closed subset of 𝒳 for each 𝑎₀∈𝒳 and if for each open set 𝑀 of 𝒳 containing 𝐻(𝑎₀) there exists an open neighborhood 𝑀₀ of 𝑎₀ such that 𝐻(𝑀₀)⊆𝑀; (iv) completely continuous if 𝐻(𝔹) is relatively compact for every 𝔹∈𝒫_𝑏(𝒳).

Definition 6. If the multivalued map 𝐻 is completely continuous with nonempty compact values, then 𝐻 is u.s.c. if and only if 𝐻 has a closed graph; that is, 𝑤_𝑛∈𝐻(𝑥_𝑛) imply that 𝑤_∗∈𝐻(𝑥_∗) when 𝑥_𝑛→𝑥_∗,𝑤_𝑛→𝑤_∗.

Definition 7. 𝐻 has a fixed point if there is 𝑥∈𝒳 such that 𝑥∈𝐻(𝑥). The fixed point set of the multivalued operator 𝐻 will be denoted by Fix𝐻.

Definition 8. Let 𝒫_𝑐𝑙(ℝ)={𝑊∈𝒫(ℝ)∶𝑊isclosed}. A multivalued map 𝐻∶[0;1]→𝒫_𝑐𝑙(ℝ) is said to be measurable if the function 𝑡↦𝑑𝑦,𝐻(𝑡)=inf𝑦−𝑧∶𝑧∈𝐻(𝑡) (9) is measurable for every 𝑦∈ℝ.

In the sequel, by 𝐶([0,1]) we mean a Banach space of continuous functions from [0,1] into ℝ with the norm ‖𝑥‖=sup_𝑡∈[0,1]|𝑥(𝑡)| whereas 𝐿¹([0,1],ℝ) is the Banach space of measurable functions 𝑥∶[0,1]→ℝ which are Lebesgue integrable and normed by ‖𝑥‖_𝐿¹=∫¹₀|𝑥(𝑡)|𝑑𝑡.

Definition 9. A multivalued map ℋ∶[0,1]×ℝ→𝒫(ℝ) is said to be Carathéodory if the map 𝑡↦ℋ(𝑡,𝑥) is measurable for each 𝑥∈ℝ and the map 𝑥↦ℋ(𝑡,𝑥) is upper semicontinuous for almost all 𝑡∈[0,1]. Further, a Carathéodory function ℋ is called 𝐿¹-Carathéodory if for each 𝛼>0, there exists 𝜑_𝛼∈𝐿¹([0,1],ℝ⁺) such that ‖ℋ(𝑡,𝑥)‖=sup{|𝑣|∶𝑣∈ℋ(𝑡,𝑥)}≤𝜑_𝛼(𝑡) for all ‖𝑥‖≤𝛼 and for a. e. 𝑡∈[0,1].

Definition 10. For each 𝑤∈𝐶([0,1],ℝ), the set of selections of ℋ is defined by 𝑆_ℋ,𝑤∶=𝑣∈𝐿¹])(𝑡)((𝑡[.([0,1,ℝ∶𝑣∈ℋ𝑡,𝑤))fora.e.𝑡∈0,1]} (10)

Definition 11. For a nonempty closed subset 𝑉 of a Banach space 𝐸, let ℋ_𝑐∶𝑉→𝒫(𝐸) be a nonempty multivalued operator with closed values. We call ℋ_𝑐 to be lower semi-continuous (l.s.c.) if the set {𝑣∈𝑉∶ℋ_𝑐(𝑣)∩𝐵≠∅} is open for any open set 𝑈 in 𝐸.

Definition 12. A subset 𝒜 of [0,1]×ℝ is ℒ⊗ℬ measurable if 𝒜 belongs to the 𝜎-algebra generated by all sets of the form 𝒥×𝒟, where 𝒥 is Lebesgue measurable in [0,1] and 𝒟 is Borel measurable in ℝ.

Definition 13. A subset 𝒜_𝑑 of 𝐿¹([0,1],ℝ) is said be decomposable if for all 𝑎,𝑏∈𝒜_𝑑 and measurable 𝒦⊂[0,1], the function 𝑎𝜙_𝒦+𝑏𝜙_[0,1]⧵𝒦∈𝒜_𝑑, where 𝜙_𝒦 stands for the characteristic function of 𝒦.

Definition 14. A multivalued operator ℋ_𝑏𝑐∶𝑊→𝒫(𝐿¹([0,1],ℝ)) has a property (BC) if ℋ_𝑏𝑐 is lower semi-continuous (l.s.c.) and has nonempty closed and decomposable values, where 𝑊 is a separable metric space.

Definition 15. Let 𝐹∶[0,1]×ℝ→𝒫(ℝ) be a multivalued map with nonempty compact values. We say that 𝐹 is of lower semi-continuous type (l.s.c. type) if its associated Nemytskii operator ℱ_𝑁 is lower semi-continuous and has nonempty closed and decomposable values, where ℱ_𝑁(𝑥)=𝑢∈𝐿¹])∶𝑢(𝑡)((𝑡[.([0,1,ℝ∈𝐹𝑡,𝑥))fora.e.𝑡∈0,1] (11)

Definition 16. Let (𝒳,𝑑) be a metric space induced from the normed space (𝒳;‖⋅‖) and let ℳ_𝑑∶𝒫(𝒳)×𝒫(𝒳)→ℝ∪{∞} be defined by ℳ_𝑑(𝐴,𝐵)=maxsup_𝑎∈𝐴𝑑(𝑎,𝐵),sup_𝑏∈𝐵𝑑(𝐴,𝑏), (12) where 𝑑(𝐴,𝑏)=inf_𝑎∈𝐴𝑑(𝑎;𝑏) and 𝑑(𝑎,𝐵)=inf_𝑏∈𝐵𝑑(𝑎;𝑏). Then (𝒫_{𝑏,𝑐𝑙}(𝒳),ℳ_𝑑) is a metric space and (𝒫_𝑐𝑙(𝒳),𝑀_𝑑) is a generalized metric space (see [24]).

Definition 17. A multivalued operator ℋ∶𝒳→𝒫_𝑐𝑙(𝒳) is called 𝛾-Lipschitz if and only if there exists 𝛾>0 such that ℳ_𝑑𝑁(𝑥),𝑁𝑦≤𝛾𝑑𝑥,𝑦foreach𝑥,𝑦∈𝒳, (13) and a contraction if and only if it is 𝛾-Lipschitz with 𝛾<1.

For further details on multi-valued maps, we refer the reader to the books [25, 26].

3. Existence Results for the Multivalued Problem

In this section, we present some existence results for the problem (1). Our first result deals with the case when 𝐹 is Carathéodory. We make use of the following known results to establish the proof.

Lemma 18 (see [27]). Let 𝑋 be a Banach space. Let 𝐹∶[0,𝑇]×ℝ→𝒫_{𝑐𝑝,𝑐}(𝑋) be an 𝐿¹-Carathéodory multivalued map with compact and convex values and let 𝜃 be a linear continuous mapping from 𝐿¹([0,1],𝑋) to 𝐶([0,1],𝑋). Then the operator 𝜃∘𝑆_𝐹∶𝐶([0,1],𝑋)→𝒫_{𝑐𝑝,𝑐}(𝐶([0,1],𝑋)),𝑥↦𝜃∘𝑆_𝐹(𝑥)=𝜃𝑆_𝐹,𝑥 (14) is a closed graph operator in 𝐶([0,1],𝑋)×𝐶([0,1],𝑋).

Lemma 19 (nonlinear alternative for Kakutani maps [28]). Let 𝐸 be a Banach space, 𝐶 a closed convex subset of 𝐸,𝑈 an open subset of 𝐶, and 0∈𝑈. Suppose that 𝐹∶𝑈→𝑃_{𝑐,𝑐𝑣}(𝐶) is an upper semicontinuous compact map; here 𝑃_{𝑐,𝑐𝑣}(𝐶) denotes the family of nonempty, compact convex subsets of 𝐶. Then either (i)𝐹 has a fixed point in 𝑈, or (ii)there is a 𝑢∈𝜕𝑈 and 𝜆∈(0,1) with 𝑢∈𝜆𝐹(𝑢).

Theorem 20. Assume that (H₁)𝐹∶[0,1]×ℝ→𝒫(ℝ) is Carathéodory and has nonempty compact and convex values; (H₂)there exists a continuous nondecreasing function 𝜓∶[0,∞)→(0,∞) and a function 𝑝∈𝐿¹([0,1],ℝ⁺) such that ‖𝐹(𝑡,𝑥)‖_𝒫𝑦((𝑡)𝜓𝑥()∈[]∶=sup∶𝑦∈𝐹𝑡,𝑥)≤𝑝(‖‖)𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑡,𝑥0,1×ℝ; (15)(H₃)there exists a constant 𝑀>0 such that 𝑀𝜓𝑀(𝛼−1)Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑑𝑠⁻¹>1. (16)Then the boundary value problem (1) has at least one solution on [0,1].

Proof. In view of Lemma 4, we define an operator 𝒬∶𝐶([0,1],ℝ)→𝒫(𝐶([0,1],ℝ)) by 𝒬(𝑥)=ℎ∈𝐶([0,1],ℝ)∶ℎ(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣(𝑠)𝑑𝑠,𝑣∈𝑆_𝐹,𝑥 (17) and show that it satisfies the hypotheses of Lemma 19. Since 𝑆_𝐹,𝑥 is convex (𝐹 has convex values), therefore, it can be easily shown that 𝒬 is convex for each 𝑥∈𝐶([0,1],ℝ).
As a next step, we prove that 𝒬 maps the bounded sets (balls) into bounded sets in 𝐶([0,1],ℝ). For a positive number 𝜉, let 𝐵_𝜉={𝑥∈𝐶([0,1],ℝ)∶‖𝑥‖≤𝜉} be a bounded ball in 𝐶([0,1],ℝ). Then, for each ℎ∈𝒬(𝑥), 𝑥∈𝐵_𝜉, there exists 𝑣∈𝑆_𝐹,𝑥 such that ℎ(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣(𝑠)𝑑𝑠. (18) Then, for 𝑡∈[0, 1], we have |ℎ(𝑡)|≤¹₀𝐺(𝑣(𝑠𝑥()𝑡,𝑠)|)|𝑑𝑠≤𝜓(‖‖)𝛼−1Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)(𝜉)()𝑑𝑠≤𝜓𝛼−1Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑑𝑠. (19) Thus, ‖ℎ‖(𝜉)()≤𝜓𝛼−1Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑑𝑠. (20) Now we show that 𝒬 maps the bounded sets into equicontinuous sets of 𝐶([0,1],ℝ).
Let 𝑡^′, 𝑡^′′∈[0,1] with 𝑡^′<𝑡^′′ and 𝑥∈𝐵_𝜉. For each ℎ∈𝒬, we obtain ℎ𝑡^′′−ℎ𝑡^′≤¹₀𝐺𝑡^′,𝑠−𝐺𝑡^′′(𝑠≤,𝑠∣∣𝑣)𝑑𝑠₁₀𝐺𝑡^′,𝑠−𝐺𝑡^′′𝑝(𝑠)𝜓‖𝑥‖(𝜉),𝑠𝑑𝑠≤𝜓¹₀𝐺𝑡^′,𝑠−𝐺𝑡^′′,𝑠𝑝(𝑠)𝑑𝑠. (21) Obviously the right-hand side of the above inequality tends to zero independently of 𝑥∈𝐵_𝜉 as 𝑡^′′−𝑡^′→0. As 𝒬 satisfies the above three assumptions, therefore, by the Ascoli-Arzelá theorem, it follows that 𝒬∶𝐶([0,1],ℝ)→𝒫(𝐶([0,1],ℝ)) is completely continuous.
In our next step, we show that 𝒬 has a closed graph. Let 𝑥_𝑛→𝑥_∗,ℎ_𝑛∈𝒬(𝑥_𝑛), and ℎ_𝑛→ℎ_∗. Then we need to show that ℎ_∗∈𝒬(𝑥_∗). Associated with ℎ_𝑛∈𝒬(𝑥_𝑛), there exists 𝑣_𝑛∈𝑆_{𝐹,𝑥𝑛} such that for each 𝑡∈[0,1], ℎ_𝑛(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣_𝑛(𝑠)𝑑𝑠. (22) Thus, it suffices to show that there exists 𝑣_∗∈𝑆_𝐹,𝑥∗ such that for each 𝑡∈[0,1], ℎ_∗(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣_∗(𝑠)𝑑𝑠. (23) Define a linear operator 𝜑∶𝐿¹([0,1],ℝ)→𝐶([0,1],ℝ) as 𝑓↦𝜑(𝑣)(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣(𝑠)𝑑𝑠. (24) Notice that ℎ_𝑛(𝑡)−ℎ_∗(𝑡=₁₀)𝐺(𝑡,𝑠)𝑣_𝑛(𝑠)−𝑣_∗(𝑠)𝑑𝑠→0,as𝑛→∞. (25) Thus, it follows from Lemma 18 that 𝜑∘𝑆_𝐹 is a closed graph operator. Further, we have ℎ_𝑛(𝑡)∈𝜑(𝑆_{𝐹,𝑥𝑛}), since 𝑥_𝑛→𝑥_∗. Thus, for some 𝑣_∗∈𝑆_𝐹,𝑥∗, we have ℎ_∗(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣_∗(𝑠)𝑑s. (26) In the last step, we show that there exists an open set 𝒱⊆𝐶([0,1],ℝ) with 𝑥∉𝒬(𝑥) for any 𝜇∈(0,1) and all 𝑥∈𝜕𝒱. Let 𝜇∈(0,1) and 𝑥∈𝜇𝒬(𝑥). Then there exists 𝑣∈𝐿¹([0,1],ℝ) with 𝑣∈𝑆_𝐹,𝑥 such that for 𝑡∈[0,1], we have 𝑥(𝑡)=𝜇¹₀𝐺(𝑡,𝑠)𝑣(𝑠)𝑑𝑠, (27) and using the computations used in the second step, we obtain |𝑥(𝑡𝑥())|≤𝜓(‖‖)𝛼−1Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑑𝑠. (28) In consequence, we have ‖𝑥‖𝜓𝑥()(‖‖)𝛼−1Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑑𝑠⁻¹≤1. (29) By the assumption (H₃), there exists 𝑀 such that ‖𝑥‖≠𝑀. Let us set 𝒱=𝑥∈𝐶([0,1],ℝ)∶‖𝑥‖<𝑀. (30) Observe that the operator 𝒬∶𝒱→𝒫(𝐶([0,1],ℝ)) is upper semicontinuous and completely continuous. From the choice of 𝒱, there is no 𝑥∈𝜕𝒱 such that 𝑥∈𝜇𝒬(𝑥) for some 𝜇∈(0,1). Consequently, by Lemma 19, we have that 𝒬 has a fixed point 𝑥∈𝒱 which is a solution of the problem (1). This completes the proof.

Example 21. Consider the following boundary value problem: ^𝑐𝐷^5/2𝑥(𝑡)((𝑡[],𝑥(0)∈𝐹𝑡,𝑥)),𝑡∈0,1=𝑥^′(0)𝑥(1)==0,₁₀√𝑡𝑥(𝑡)𝑑𝑡, (31) where 𝐹∶[0,1]×ℝ→𝒫(ℝ) is a multivalued map given by ()=𝑡|𝑥|𝑥→𝐹𝑡,𝑥³|𝑥|³,𝑡|𝑥|+32𝑥|).(|+1 (32) For 𝑓∈𝐹, we have 𝑓𝑡|𝑥|≤max³|𝑥|³,𝑡|𝑥|+32𝑥|)≤√(|+1𝑡,𝑥∈ℝ. (33) Here 𝜎=∫¹₀𝑠^𝛼−1𝑔(𝑠)𝑑𝑠=∫¹₀𝑠^3/2𝑠^1/2𝑑𝑠=(1/3)<1, ‖𝐹(𝑡,𝑥)‖_𝒫∶=sup{|𝑦|∶𝑦∈𝐹(𝑡,𝑥)}≤𝑡=𝑝(𝑡)𝜓(‖𝑥‖), 𝑥∈ℝ, with 𝑝(𝑡)=𝑡,𝜓(‖𝑥‖)=1, and ∫¹₀𝑠(1−𝑠)^3/2𝑠𝑑𝑠=16/315. Using the given values in the condition (H₃): 𝑀𝜓(𝑀)(𝛼−1)Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑝(𝑠)𝑑𝑠⁻¹>1, (34) we find that 𝑀>96315√𝜋. (35) Clearly, all the conditions of Theorem 20 are satisfied. Hence, the conclusion of Theorem 20 applies to the problem (31).

In our next result, we assume that 𝐹 is not necessarily convex valued. We complete the proof of this result by applying the nonlinear alternative of Leray-Schauder type together with the selection theorem of Bressan and Colombo [29] for lower semi-continuous maps with decomposable values, which is stated below.

Lemma 22 (see [29, 30]). Let 𝑌 be a separable metric space and let 𝑁∶𝑌→𝒫(𝐿¹([0,1],ℝ)) be a multivalued operator satisfying the property (BC). Then 𝑁 has a continuous selection; that is, there exists a continuous function (single-valued) 𝑔∶𝑌→𝐿¹([0,1],ℝ) such that 𝑔(𝑥)∈𝑁(𝑥) for every 𝑥∈𝑌.

Theorem 23. Suppose that (H₂) and (H₃) hold. In addition, we assume the following condition: (H₄)𝐹∶[0,1]×ℝ→𝒫(ℝ) is a nonempty compact-valued multivalued map such that (a)(𝑡,𝑥)↦𝐹(𝑡,𝑥) is ℒ⊗ℬ measurable, (b)𝑥↦𝐹(𝑡,𝑥) is lower semicontinuous for each 𝑡∈[0,1]. Then the problem (1) has at least one solution on [0,1].

Proof. Observe that the assumptions (H₂) and (H₄) imply that 𝐹 is of l.s.c. type. Then, by Lemma 22, there exists a continuous function 𝜈∶𝐴𝐶²([0,1],ℝ)→𝐿¹([0,1],ℝ) such that 𝜈(𝑥)∈ℱ(𝑥) for all 𝑥∈𝐶([0,1],ℝ).
Let us consider the problem 𝐷^𝛼𝑥(𝑡)(𝑥(𝑡[],𝑥(0)=𝜈)),𝑡∈0,1=𝑥^′(0)𝑥(1)==0,₁₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡. (36) One can note that if 𝑥∈𝐴𝐶²([0,1],ℝ) is a solution of (36), then 𝑥 is a solution to the problem (1). To convert the problem (36) to a fixed point problem, we define an operator 𝒬 as 𝒬𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝜈(𝑥(𝑠))𝑑𝑠. (37) It is easy to show that the operator 𝒬 is continuous and completely continuous. The rest of the proof is similar to that of Theorem 20. So we omit it. This completes the proof.

Finally we show the existence of solutions for the problem (1) with a nonconvex valued right-hand side by applying a fixed point theorem for multivalued maps according to Covitz and Nadler [31].

Lemma 24 ([31]). Let (𝑋,𝑑) be a complete metric space. If 𝑁∶𝑋→𝒫_𝑐𝑙(𝑋) is a contraction, then Fix𝑁≠∅.

Theorem 25. Assume that (H₅)𝐹∶[0,1]×ℝ→𝒫_𝑐𝑝(ℝ) is such that 𝐹(⋅,𝑥)∶[0,1]→𝒫_𝑐𝑝(ℝ) is measurable for each 𝑥∈ℝ; (H₆)ℳ_𝑑(𝐹(𝑡,𝑥),𝐹(𝑡,𝑦))≤𝑚(𝑡)|𝑥−𝑦| for almost all 𝑡∈[0,1] and 𝑥,𝑦∈ℝ with 𝑚∈𝐿¹([0,1],ℝ⁺) and 𝑑(0,𝐹(𝑡,0))≤𝑚(𝑡) for almost all 𝑡∈[0,1]. Then the boundary value problem (1) has at least one solution on [0,1] if (𝛼−1)Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)𝑑𝑠¹₀𝑠(1−𝑠)^𝛼−1𝑚(𝑠)𝑑𝑠<1. (38)

Proof. By the assumption (H₅), it follows that the set 𝑆_𝐹,𝑥 is nonempty for each 𝑥∈𝐶([0,1],ℝ). So 𝐹 has a measurable selection (see [32, Theorem III.6]). Now it will be shown that the operator 𝒬 defined by (17) satisfies the hypotheses of Lemma 24. To show that 𝒬(𝑥)∈𝒫_𝑐𝑙((𝐶[0,1],ℝ)) for each 𝑥∈𝐶([0,1],ℝ), let {𝑢_𝑛}_𝑛≥0∈𝒬(𝑥) be such that 𝑢_𝑛→𝑢(𝑛→∞) in 𝐶([0,1],ℝ). Then 𝑢∈𝐶([0,1],ℝ) and there exists 𝑣_𝑛∈𝑆_{𝐹,𝑥𝑛} such that, for each 𝑡∈[0,1], 𝑢_𝑛(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣_𝑛(𝑠)𝑑𝑠. (39) As 𝐹 has compact values, we pass onto a subsequence (if necessary) to obtain that 𝑣_𝑛 converges to 𝑣 in 𝐿¹([0,1],ℝ). Thus, 𝑣∈𝑆_𝐹,𝑥 and for each 𝑡∈[0,1], we have 𝑣_𝑛(𝑡)→𝑣(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣(𝑠)𝑑𝑠. (40) Hence, 𝑢∈𝒬.
Next, we show that there exists 𝛿<1 such that ℳ_𝑑𝒬(𝑥),𝒬𝑦≤𝛿𝑥−𝑦foreach𝑥,𝑦∈𝐴𝐶²([0,1],ℝ). (41) Let 𝑥,𝑦∈𝐴𝐶²([0,1],ℝ) and ℎ₁∈𝒬(𝑥). Then there exists 𝑣₁(𝑡)∈𝐹(𝑡,𝑥(𝑡)) such that, for each 𝑡∈[0,1], ℎ₁(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣₁(𝑠)𝑑𝑠. (42) By (H₆), we have that ℳ_𝑑𝐹(𝑡,𝑥),𝐹𝑡,𝑦≤𝑚(𝑡)𝑥(𝑡)−𝑦(𝑡). (43) So, there exists 𝑤∈𝐹(𝑡,𝑦(𝑡)) such that 𝑣₁(𝑡)−𝑤≤𝑚(𝑡)𝑥(𝑡)−𝑦(𝑡),𝑡∈[0,1]. (44) Define 𝒲∶[0,1]→𝒫(ℝ) by 𝒲(𝑡)=𝑤∈ℝ∶𝑣₁(𝑡)−𝑤≤𝑚(𝑡)𝑥(𝑡)−𝑦(𝑡). (45) Since the multivalued operator 𝒲(𝑡)∩𝐹(𝑡,𝑦(𝑡)) is measurable ([32, Proposition III.4]), there exists a function 𝑣₂(𝑡) which is a measurable selection for 𝒲. So 𝑣₂(𝑡)∈𝐹(𝑡,𝑦(𝑡)) and for each 𝑡∈[0,1], we have that |𝑣₁(𝑡)−𝑣₂(𝑡)|≤𝑚(𝑡)|𝑥(𝑡)−𝑦(𝑡)|.
For each 𝑡∈[0,1], let us define ℎ₂(𝑡)=¹₀𝐺(𝑡,𝑠)𝑣₂(𝑠)𝑑𝑠. (46) Thus, ℎ₁(𝑡)−ℎ₂(𝑡)≤¹₀𝐺(𝑡,𝑠)𝑣₁(𝑠)−𝑣₂(𝑠≤())𝑑𝑠𝛼−1Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑚(𝑠)𝑑𝑠𝑣₁(𝑠)−𝑣₂(𝑠)𝑑𝑠. (47) Hence, ℎ₁−ℎ₂≤(𝛼−1)Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)×𝑑𝑠₁₀𝑠(1−𝑠)^𝛼−1𝑚(𝑠)𝑑𝑠𝑥−𝑦. (48) Analogously, interchanging the roles of 𝑥 and 𝑥, we obtain ℳ_𝑑𝒬(𝑥)𝑦≤(),𝒬≤𝛿𝑥−𝑦𝛼−1Γ(𝛼)11+1−𝜎¹₀𝑔(𝑠)𝑑𝑠×¹₀𝑠(1−𝑠)^𝛼−1𝑚(𝑠)𝑑𝑠𝑥−𝑦. (49) Since 𝒬 is a contraction, it follows from Lemma 24 that 𝒬 has a fixed point 𝑥 which is a solution of (1). This completes the proof.

Acknowledgments

The authors thank the editor and the reviewer for their constructive comments that led to the improvement of the paper. The research of H. H. Alsulami and B. Ahmad was partially supported by Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia.