Existence of Solutions for Fractional Differential Inclusions with Separated Boundary Conditions in Banach Space

Abstract

We discuss the existence of solutions for a class of some separated boundary differential inclusions of fractional orders 2<𝛼<3 involving the Caputo derivative. In order to obtain necessary conditions for the existence result, we apply the fixed point technique, fractional calculus, and multivalued analysis.

1. Introduction

In recent years a great interest was devoted to the study of (singular and Neumann) boundary-value problems of fractional order [1–8], see also [9–12]. In the literature of fractional calculus, there are several definitions of fractional derivative that can be used. However, the most popular senses are the Riemann-Liouville and Caputo fractional derivatives; see, for instance, [13–19].

In this paper, we use the Caputo's fractional derivative since mathematical modeling of many physical problems requires initial and boundary conditions. These demands are satisfied using the Caputo fractional derivative. For more details we refer the reader to [20, 21] and references therein.

The importance of fractional boundary-value problems stems from the fact that they model various applications in fluid mechanics, viscoelasticity, physics, biology, and economics which cannot be modeled by differential equations with integer derivatives [21–23].

Delbosco and Rodino [24] considered the existence of a solution for the nonlinear fractional differential equation 𝑑^𝛼𝑥(𝑡)/𝑑𝑡^𝛼=𝑓(𝑡,𝑥), where 0<𝛼<1 and 𝑓∶[0,𝑎]×ℝ→ℝ,0<𝑎≤+∞, is a given function, continuous in(0,𝑎)×ℝ. They obtained results for solutions by using the Schauder fixed point theorem and the Banach contraction principle. Qiu and Bai [25] considered the existence of positive solution for equation: 𝑑^𝛼𝑥(𝑡)𝑑𝑡^𝛼((𝑡𝑥(0)+𝑓𝑡,𝑥))=0,0<𝑡<1,=𝑥^′(1)=𝑥^′′(0)=0, (1) where 2<𝛼≤3 and 𝑓∶(0,1]×[0,∞)→[0,∞) with 𝑓 being singular at 𝑡=0 (i.e., lim_𝑡→0⁺𝑓(𝑡,⋅)=+∞), by using Krasnoselskii's fixed point theorem and nonlinear alternative of Leray-Schauder type in a cone. Recently, Aitaliobrahim [26] considered the existence of solutions to the boundary-value problem: 𝑥^′′(𝑡)∈𝐹𝑡,𝑥(𝑡),𝑥^′(𝑡)[],𝑥,a.e.on0,1_′(0)=𝑟,𝑥^′(1)=𝑠, (2) where 𝐹 is a closed multifunction, measurable in the first argument, and Lipschitz continuous in the second argument, by using fixed point theory for multivalued maps.

Motivated by the previous results, in this work we establish the existence result of a new version of fractional separated boundary-value problem: 𝑑^𝛼𝑥(𝑡)𝑑𝑡^𝛼∈𝐹𝑡,𝑥(𝑡),𝑥^′(𝑡),𝑥^′′(𝑡)[],𝑥(0),a.e.on0,1=0,𝑥^′(1)=𝑠,𝑥^′′(0)=𝑟, (3) where 2<𝛼≤3,𝐹 is nonconvex, closed multifunction, measurable in the first argument, and Lipschitz continuous in the second argument, and 𝑟,𝑠 are in a Banach space 𝑋. The work is organized as follows. In Section 2, we recall some preliminary facts that we need in the sequel while in Section 3, we give the main result. Finally in Section 4 we give example to illustrate the application of our results.

2. Preliminaries

In this section, we present basic definitions of fractional calculus and some essential facts from multivalued analysis that will be used in this work to obtain our main results.

Definition 1. A real function 𝑓(𝑡),𝑡>0, is said to be in the space 𝐶_𝜇,𝜇∈ℝ, if there exists a real number 𝑝>𝜇, such that 𝑓(𝑡)=𝑡^𝑝𝑓₁(𝑡), where 𝑓₁(𝑡)∈𝐶[0,∞), and it is said to be in the space 𝐶^𝑚_𝜇 if 𝑓^(𝑚)∈𝐶_𝜇, 𝑚∈ℕ.

Definition 2. The left Riemann-Liouville fractional integral of order 𝛼>0, of a function 𝑓∈𝐶_𝜇,𝜇≥−1, is defined by 𝐼^𝛼𝑓(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠,𝑡>0. (4)

Definition 3. For 𝛼>0,𝑚−1<𝛼<𝑚,𝑚∈ℕ,𝑡>0, and 𝑓∈𝐶^𝑚₋₁, the left Caputo fractional derivative is defined by 𝐷^𝛼𝑓(𝑡)=1Γ(𝑚−𝛼)^𝑡₀(𝑡−𝑠)^{𝑚−1−𝛼}𝑓^(𝑚)(𝑠)𝑑𝑠, (5) where Γ is the well-known Gamma function.

The Caputo derivative defined in (5) is related to the Riemann-Liouville fractional integral, 𝐼^𝛼, of order 𝛼∈ℝ⁺, by 𝐷^𝛼𝑓(𝑡)=𝐼^𝑚−𝛼𝑓^(𝑚)(𝑡). (6) It is known (see [20]) that 𝐼^𝛼𝐷^𝛼𝑓(𝑡)=𝑓(𝑡)−^𝑚−1_𝑘=0𝑐_𝑘𝑡^𝑘,𝐷^𝛼𝐼^𝛼𝑓(𝑡)=𝑓(𝑡), (7)(8) where in (7), 𝑐_𝑘=𝑓^(𝑘)(0+)/𝑘!,0≤𝑘≤𝑚−1.

Now, let 𝑋 be a real separable Banach space with the norm ‖⋅‖. We denote by 𝒞([0,1],𝑋) the Banach space of continuous functions from [0,1] to 𝐸 equipped with the norm ‖𝑥(⋅)‖_∞∶=sup{‖𝑥(𝑡)‖;𝑡∈[0,1]}. For 𝑥∈𝑋 and for nonempty sets 𝐴,𝐵 of 𝑋 we denote 𝑑(𝑥,𝐴)=inf{𝑑(𝑥,𝑦);𝑦∈𝐴},𝑒(𝐴,𝐵)∶=sup{𝑑(𝑥,𝐵);𝑥∈𝐴}, and 𝐻(𝐴,𝐵)∶=max{𝑒(𝐴,𝐵),𝑒(𝐵,𝐴)}. A multifunction is said to be measurable if its graph is measurable.

Also, we recall the following results that will be used in this paper.

Definition 4 (see [26]). Let 𝑇∶𝑋→2^𝑋 be a multifunction with closed values:(i)𝑇is 𝑘-Lipschitz if 𝐻(𝑇(𝑥),𝑇(𝑦))≤𝑘‖𝑥−𝑦‖ for each 𝑥,𝑦∈𝑋,(ii)𝑇 is a contraction if it is 𝑘-Lipschitz with 𝑘<1,(iii)𝑇 has a fixed point if there exists 𝑥∈𝑋 such that 𝑥∈𝑇(𝑥).

Lemma 5 (see [26]). If 𝑇∶𝑋→2^𝑋 is a contraction with nonempty closed values, then it has a fixed point.

Lemma 6 (see [26]). Assume that 𝐹∶[𝑎,𝑏]×𝑋→2^𝑋 is a multifunction with nonempty closed values satisfying the following:(i)for every 𝑥∈𝑋,𝐹(⋅,𝑥) is measurable on [𝑎,𝑏];(ii)for every 𝑡∈[𝑎,𝑏],𝐹(𝑡,⋅) is (Hausdorff) continuous on 𝐸.Then for any measurable function 𝑥(⋅)∶[𝑎,𝑏]→𝑋, the multifunction 𝐹(⋅,𝑥(⋅)) is measurable on [𝑎,𝑏].

Definition 7 (see [26]). A measurable multivalued function 𝐹∶[0,1]→2^𝑋 is said to be integrably bounded if there exists a function ℎ∈𝐿¹([0,1],𝑋) such that for all 𝑣∈𝐹(𝑡),‖𝑣‖≤ℎ(𝑡) for almost every 𝑡∈[0,1].

Definition 8 (see [26]). A function 𝑥(⋅)∶[0,1]→𝑋 is said to be a solution of (3) if 𝑥(⋅) is absolutely continuous on [0,1] and satisfies (3).

Lemma 9 (see [25]). Given 𝜑∈𝐶[0,1] and 2<𝛼≤3, the unique solution of 𝑑^𝛼𝑦(𝑡)𝑑𝑡^𝛼(𝑡)𝑦(0)+𝜑=0,0<𝑡<1,=𝑦^′(1)=𝑦^′′(0)=0 (9) is 𝑦(𝑡)=¹₀𝐺(𝑡,𝑠)𝜑(𝑠)𝑑𝑠, (10) where 𝐺()=()𝑡()𝑡,𝑠𝛼−11−𝑠^𝛼−2−(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑡(),0≤𝑠≤𝑡≤1,1−𝑠^𝛼−2Γ(𝛼−1),0≤𝑡≤𝑠≤1. (11)

Obviously, 𝐺(𝑡,𝑠) is continuous on [0,1]×[0,1] and 0<𝐺(𝑡,𝑠)≤𝜆, for each 𝑡,𝑠∈[0,1] and some 𝜆.

3. Main Results

Now we are in a position to state and prove the main results of this paper.

Theorem 10. Let 𝐹∶[0,1]×𝑋×𝑋×𝑋→2^𝑋 be a set-valued map with nonempty closed values satisfying the following:(i)for each (𝑥,𝑦,𝑧)∈𝑋×𝑋×𝑋,𝑡↦𝐹(𝑡,𝑥,𝑦,𝑧) is measurable and integrably bounded;(ii)there exists a function 𝑚(⋅)∈𝐿¹([0,1],ℝ⁺) such that for all 𝑡∈[0,1] and for all 𝑥₁,𝑥₂,𝑦,𝑧∈𝑋𝐻𝐹𝑡,𝑥₁,𝑦,𝑧,𝐹𝑡,𝑥₂,𝑦,𝑧≤𝑚(𝑡)𝑥₁−𝑥₂. (12)Then, if ∫¹₀(1+𝑚(𝑠))𝑑𝑠<1/𝜆, for all 𝑟,𝑠∈𝑋, the problem (3) has at least one solution on [0,1].

Proof. For the proof of this theorem, we use the similar steps as those of [26, Theorem 2.6] together with the theory of fractional calculus. Let 𝑟,𝑠 be in 𝑋. We introduce first the function 𝜌∶[0,1]→𝑋 defined by 𝜌(𝑡)=()1𝑟+𝑠𝑡−2𝑟𝑡²,∀𝑡∈[0,1], (13) and the multifunction 𝐻∶[0,1]×𝒞([0,1],𝑋)→2^𝑋 defined by 𝐻(⋅)(𝑡)(𝑡)(𝑡)𝑡,𝑦=𝜑−𝐹𝑡,𝑦+𝜌,𝑦^′(𝑡)+𝜌^′(𝑡),𝑦^′′(𝑡)+𝜌^′′(𝑡), (14) for all (𝑡,𝑦(⋅))∈[0,1]×𝒞([0,1],𝑋). Consider the following problem: −𝑑^𝛼𝑦(𝑡)𝑑𝑡^𝛼(𝑡)(⋅)[],𝑦(0)+𝜑∈𝐻𝑡,𝑦,a.e.on0,1=𝑦^′(1)=𝑦^′′(0)=0. (15)
We should note that the function 𝑦(⋅) is a solution of (15), if and only if the function 𝑥(𝑡)=𝑦(𝑡)+𝜌(𝑡) is a solution of (3), for all 𝑡∈[0,1].
Next, by Lemma 6, for 𝑦(⋅)∈𝒞([0,1],𝑋),𝐹(⋅,𝑦(⋅),𝑦^′(⋅),𝑦^′′(⋅)) is closed and measurable; then it has a measurable selection which, by hypothesis (i), belongs to 𝐿¹([0,1],𝑋). Thus the set 𝑆_{𝐹,𝑦(⋅)}∶=𝑓∈𝐿¹])∶𝑓(𝑡)(𝑡)([0,1,𝑋∈𝐹𝑡,𝑦,𝑦^′(𝑡),𝑦^′′(𝑡)fora.e.𝑡∈[0,1] (16) is nonempty. Let us transform problem (15) into a fixed point problem. Consider the multivalued map 𝑇∶𝒞([0,1],𝑋)⟶2^{𝒞([0,1],𝑋)} (17) defined as follows, for 𝑦(⋅)∈𝒞([0,1],𝑋): 𝑇𝑦(⋅)=𝑧(⋅)])∶𝑧(𝑡)=∈𝒞([0,1,𝑋₁₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠,∀𝑡∈[0,1],ℎ∈𝑆_{𝐻,𝑦(⋅)}, (18) where 𝑆_{𝐻,𝑦(⋅)}∶=ℎ∈𝐿¹])∶ℎ(𝑡)(⋅)[.([0,1,𝑋∈𝐻𝑡,𝑦fora.e.𝑡∈0,1] (19)
We will show that 𝑇 satisfies the assumptions of Lemma 5. The proof will be given in two steps.
Step??1?? (T has nonempty closed values). Indeed, let (𝑦_𝑝(⋅))_𝑝≥0∈𝑇(𝑦(⋅)) such that (𝑦_𝑝(⋅))_𝑝≥0 converges to 𝑦(⋅) in 𝒞([0,1],𝑋). Then 𝑦(⋅)∈𝒞([0,1],𝑋) and for each 𝑡∈[0,1], 𝑦_𝑝(𝑡)∈¹₀𝐺(𝑡,𝑠)𝐻𝑠,𝑦(⋅)𝑑𝑠, (20) where ∫¹₀𝐺(𝑡,𝑠)𝐻(𝑠,𝑦(⋅))𝑑𝑠 is the Aumann integral of 𝐺(𝑡,⋅)𝐻(⋅,𝑦), which is defined as ¹₀𝐺()𝐻(⋅)=𝑡,𝑠𝑠,𝑦𝑑𝑠₁₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠,ℎ∈𝑆_{𝐻,𝑦(⋅)}. (21) Using the fact that the set-valued map 𝐹 is closed and by (14), we conclude that the set ¹₀𝐺(𝑡,𝑠)𝐻𝑠,𝑦(⋅)𝑑𝑠 (22) is closed for all 𝑡∈[0,1]. Then 𝑦(𝑡)∈¹₀𝐺(𝑡,𝑠)𝐻𝑠,𝑦(⋅)𝑑𝑠. (23) So, there exists ℎ∈𝑆_{𝐻,𝑦(⋅)} such that 𝑦(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠. (24) Hence 𝑦(⋅)∈𝑇(𝑦(⋅)). So 𝑇(𝑦(⋅)) is closed for each 𝑦(⋅)∈𝒞([0,1],𝑋).
Step??2?? (T is a contraction). Indeed, let 𝑦₁(⋅),𝑦₂(⋅)∈𝒞([0,1],𝑋) and consider 𝑧₁(⋅)∈𝑇(𝑦₁(⋅)). Then there exists ℎ₁∈𝑆_{𝐻,𝑦1(⋅)} such that 𝑧₁(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ₁(𝑠)𝑑𝑠,∀𝑡∈[0,1]. (25) Using (14), there exists 𝑓₁∈𝑆_{𝐹,𝑦1(⋅)} such that ℎ₁(𝑡)=𝑦₁(𝑡)−𝑓₁(𝑡),∀𝑡∈[0,1]. (26) On the other hand, let 𝜀>0 and consider the valued map 𝑈_𝜀∶[0,1]→2^𝑋, given by 𝑈_𝜀(𝑡)=𝑥∈𝑋∶𝑓₁(𝑡)−𝑥≤𝑚(𝑡)𝑦₁(𝑡)−𝑦₂(𝑡)+𝜀. (27) We claim that 𝑈_𝜀(𝑡) is nonempty, for each 𝑡∈[0,1]. Indeed, let 𝑡∈[0,1]; then we have 𝐻𝐹𝑡,𝑦₁(𝑡),𝑦^′₁(𝑡),𝑦₁^′′(𝑡),𝐹𝑡,𝑦₂(𝑡),𝑦^′₂(𝑡),𝑦₂^′′(𝑡)≤𝑚(𝑡)𝑦₁(𝑡)−𝑦₂(𝑡). (28) Hence, there exists 𝑥∈𝐹(𝑡,𝑦₂(𝑡),𝑦^′₂(𝑡),𝑦₂^′′(𝑡)), such that 𝑓₁(𝑡)−𝑥≤𝑚(𝑡)𝑦₁(𝑡)−𝑦₂(𝑡)+𝜀. (29) The multifunction 𝑉∶𝑡⟶𝑈_𝜀(𝑡)∩𝐹𝑡,𝑦₂(𝑡),𝑦^′₂(𝑡),𝑦₂^′′(𝑡)ismeasurable. (30) Then there exists a measurable selection for 𝑉 denoted by 𝑓₂ such that, for all 𝑡∈[0,1], 𝑓₂(𝑡)∈𝐹𝑡,𝑦₂(𝑡),𝑦^′₂(𝑡),𝑦₂^′′(𝑡) (31) and 𝑓₁(𝑡)−𝑓₂(𝑡)≤𝑚(𝑡)𝑦₁(𝑡)−𝑦₂(𝑡)+𝜀. (32) Now, for all 𝑡∈[0,1], set ℎ₂(𝑡)=𝑦₂(𝑡)−𝑓₂(𝑡) and 𝑧₂(𝑡)=¹₀𝐺(𝑡,𝑠)ℎ₂(𝑠)𝑑𝑠. (33) We have 𝑧₁(𝑡)−𝑧₂(𝑡≤₁₀)‖𝐺(𝑡,𝑠)‖ℎ₁(𝑠)−ℎ₂(𝑠)𝑑𝑠≤𝜆¹₀𝑦₁(𝑠)−𝑦₂(𝑠)𝑑𝑠+𝜆¹₀𝑓₁(𝑠)−𝑓₂(𝑠)𝑑𝑠≤𝜆¹₀𝑦₁(𝑠)−𝑦₂(𝑠)𝑑𝑠+𝜆¹₀𝑚(𝑠)𝑦₁(𝑠)−𝑦₂(𝑠𝑦)𝑑𝑠+𝜆𝜀≤𝜆₁(⋅)−𝑦₂(⋅)_∞¹₀(1+𝑚(𝑠))𝑑𝑠+𝜆𝜀. (34) So, we conclude that 𝑧₁(⋅)−𝑧₂(⋅)_∞≤𝜆𝑦₁(⋅)−𝑦₂(⋅)_∞¹₀(1+𝑚(𝑠))𝑑𝑠+𝜆𝜀. (35) By an analogous relation, obtained by interchanging the roles of 𝑦₁(⋅) and 𝑦₂(⋅), it follows that 𝐻𝑇𝑦₁(⋅),𝑇𝑦₂(⋅)≤𝜆𝑦₁(⋅)−𝑦₂(⋅)_∞¹₀(1+𝑚(𝑠))𝑑𝑠+𝜆𝜀. (36) By letting 𝜀→0, we obtain 𝐻𝑇𝑦₁(⋅),𝑇𝑦₂(⋅)≤𝜆𝑦₁(⋅)−𝑦₂(⋅)_∞¹₀(1+𝑚(𝑠))𝑑𝑠. (37) Consequently, if ∫¹₀(1+𝑚(𝑠))𝑑𝑠<1/𝜆, 𝑇 is a contraction. By Lemma 5, 𝑇 has a fixed point which is a solution of (15).

4. Example

In this section we present an example to illustrate the applications of our main results, and we consider the following fractional inclusion boundary-value problem: 𝑑^𝛼𝑥(𝑡)𝑑𝑡^𝛼∈𝐹𝑡,𝑥(𝑡),𝑥^′(𝑡),𝑥^′′(𝑡)[]𝑥(0),𝑡∈0,1,2<𝛼≤3,=0,𝑥^′(1)=𝑠,𝑥^′′(0)=𝑟, (38) where 𝑟 and 𝑠 are arbitrary real numbers and 𝐹∶[0,1]×ℝ×ℝ×ℝ→𝒫(ℝ) is a multivalued map given by 𝐹𝑡,𝑥,𝑦,𝑧=0,sin𝑥(1+𝑡)²+cos𝑦𝑧. (39) Then we have 𝑢|1sup|∶𝑢∈𝐹𝑡,𝑥,𝑦,𝑧≤1+(1+𝑡)²,𝐻𝐹𝑡,𝑥,𝑦,𝑧,𝐹𝑡,≤1𝑥,𝑦,𝑧(1+𝑡)²|𝑥−𝑥|. (40)

Let 𝑚(𝑡)=1/(1+𝑡)². Then ∫¹₀(1+𝑚(𝑠))𝑑𝑠=3/2<1/𝜆 for some 𝜆>2/3.

By Theorem 10 the problem (38) has at least one solution on [0,1].