Weak Solutions for Nonlinear Fractional Differential Equations in Banach Spaces

We discuss the existence of weak solutions for a nonlinear boundary value problem of fractional differential equations in Banach space. Our analysis relies on the Monch's fixed point theorem combined with the technique of measures of weak noncompactness.


Introduction
This paper is mainly concerned with the existence results for the following fractional differential equation: where 1 < α ≤ 2 is a real number, c D α 0 is the Caputo's fractional derivative, λ 1 , λ 2 , μ 1 , μ 2 ∈ R. f : J × E → E is a given function satisfying some assumptions that will be specified later, and E is a Banach space with norm u .
Recently, fractional differential equations have found numerous applications in various fields of physics and engineering 1, 2 .It should be noted that most of the books and papers on fractional calculus are devoted to the solvability of initial value problems for differential equations of fractional order.In contrast, the theory of boundary value problems for nonlinear fractional differential equations has received attention quite recently and many aspects of this theory need to be explored.For more details and examples, see 3-19 and the references therein.
To investigate the existence of solutions of the problem above, we use M önch's fixed point theorem combined with the technique of measures of weak noncompactness, which is an important method for seeking solutions of differential equations.This technique was mainly initiated in the monograph of Banaś and Goebel 20 and subsequently developed and used in many papers; see, for example, Banaś and Sadarangani 21 , Guo et al. 22 , Krzyśka and Kubiaczyk 23 , Lakshmikantham and Leela 24 , Mönch 25 , O'Regan 26, 27 , Szufla 28, 29 , and the references therein.As far as we know, there are very few results devoted to weak solutions of nonlinear fractional differential equations 30-32 .Motivated by the abovementioned papers 30-32 , the purpose of this paper is to establish the existence results for the boundary value problem 1.1 by virtue of the M önch's fixed point theorem combined with the technique of measures of weak noncompactness.Our results can be seen as a supplement of the results in 32 see Remark 3.8 .
The remainder of this is organized as follows.In Section 2, we provide some basic definitions, preliminaries facts, and various lemmas which are needed later.In Section 3, we give main results of problem 1.1 .In the end, we also give an example for the illustration of the theories established in this paper.

Preliminaries and Lemmas
In this section, we present some basic notations, definitions, and preliminary results which will be used throughout this paper.
Let J : 0, T and L 1 J, E denote the Banach space of real-valued Lebesgue integrable functions on the interval J, L ∞ J, E denote the Banach space of real-valued essentially bounded and measurable functions defined over J with the norm Let E be a real reflexive Banach space with norm • and dual E * , and let E, ω E, σ E, E * denote the space E with its weak topology.Here, C J, E is the Banach space of continuous functions x : J → E with the usual supremum norm x ∞ : sup{ x t : t ∈ J}.
Moreover, for a given set V of functions v : J → R, let us denote by V t {v t : v ∈ V }, t ∈ J, and V J {v t : v ∈ V, t ∈ J}.
The following result follows directly from the Hahn-Banach theorem.
For completeness, we recall the definitions of the Pettis-integral and the Caputo derivative of fractional order.Definition 2.7 see 26 .Let h : J → E be a function.The fractional Pettis integral of the function h of order α ∈ R is defined by where the sign " " denotes the Pettis integral and Γ is the Gamma function.
Definition 2.8 see 3 .For a function f : J → E, the Caputo fractional-order derivative of f is defined by where n α 1 and α denotes the integer part of α.
Lemma 2.9 see 28 .Let D be a closed convex and equicontinuous subset of a metrizable locally convex vector space C J, E such that 0 ∈ D. Assume that A : holds for every subset V of D, then A has a fixed point.

Main Results
Let us start by defining what we mean by a solution of the problem 1.1 .
Definition 3.1.A function x ∈ C J, E ω is said to be a solution of the problem 1.1 if x satisfies the equation c D α 0 u t f t, u t on J and satisfies the conditions u 0 For the existence results on the problem 1.1 , we need the following auxiliary lemmas.Lemma 3.2 see 3, 7 .For α > 0, the general solution of the fractional differential equation c D α 0 u t 0 is given by for some c i ∈ R, i 0, 1, 2, . . ., n − 1, where n α 1.We derive the corresponding Green's function for boundary value problem 1.1 which will play major role in our next analysis.

Lemma 3.4. Let ρ ∈ C J, E be a given function, then the boundary-value problem
where G t, s is defined by the formula

3.5
Here G t, s is called the Green's function of boundary value problem 3.3 .
Proof.By the Lemma 3.3, we can reduce the equation of problem 3.3 to an equivalent integral equation for some constants c 1 , c 2 ∈ R. On the other hand, by relations D α 0 I α 0 u t u t and I m 0 I n 0 u t I m n 0 u t , for m, n > 0, u ∈ L 0, 1 , we have Applying the boundary conditions 3.3 , we have

3.8
Therefore, the unique solution of problem 3.3 is which completes the proof.
Remark 3.5.From the expression of G t, s , it is obvious that G t, s is continuous on J × J. Denote To prove the main results, we need the following assumptions: H1 for each t ∈ J, the function f t, • is weakly sequentially continuous; H2 for each x ∈ C J, E , the function f and consider the set

3.16
Clearly, the subset D is closed, convex, and equicontinuous.We shall show that A satisfies the assumptions of Lemma 2.9.The proof will be given in three steps.
Step 1.We will show that the operator A maps D into itself.Take x ∈ D, t ∈ J and assume that Ax t / 0. Then there exists ψ ∈ E * such that Ax t ψ Ax t .Thus

3.17
Let Step 2. We will show that the operator A is weakly sequentially continuous.Let x n be a sequence in D and let x n t → x t in E, w for each t ∈ J. Fix t ∈ J. Since f satisfies assumptions H1 , we have f t, x n t converge weakly uniformly to f t, x t .Hence, the Lebesgue Dominated Convergence Theorem for Pettis integrals implies that Ax n t converges weakly uniformly to Ax t in E ω .Repeating this for each t ∈ J shows Ax n → Ax.Then A : D → D is weakly sequentially continuous.
Step 3. The implication 2.3 holds.Now let V be a subset of D such that V ⊂ conv A V ∪{0} .Clearly, V t ⊂ conv A V ∪{0} for all t ∈ J. Hence, AV t ⊂ AD t , t ∈ J, is bounded in E. Thus, AV t is weakly relatively compact since a subset of a reflexive Banach space is weakly relatively compact if and only if it is bounded in the norm topology.Therefore, thus, V is relatively weakly compact in E. In view of Lemma 2.9, we deduce that A has a fixed point which is obviously a solution of the problem 1.1 .This completes the proof.Remark 3.8.In the Theorem 3.7, we presented an existence result for weak solutions of the problem 1.1 in the case where the Banach space E is reflexive.However, in the nonreflexive case, conditions H1 -H3 are not sufficient for the application of Lemma 2.9; the difficulty is with condition 2.3 .Our results can be seen as a supplement of the results in 32 see Remark 3.8 .Theorem 3.9.Let E be a Banach space, and assume assumptions (H1), (H2), (H3), (H4) are satisfied.If 3.13 holds, then the problem 1.1 has at least one solution on J. Theorem 3.10.Let E be a Banach space, and assume assumptions (H1), (H2), (H3)', (H4), (H5) are satisfied.If 3.13 holds, then the problem 1.1 has at least one solution on J.
Proof.Assume that the operator A : C J, E → C J, E is defined by the formula 3.14 .It is well known the fixed points of the operator A are solutions of the problem 1.1 .
First notice that, for x ∈ C J, E , we have f •, x • ∈ P J, E assumption H2 .Since, s → G t, s ∈ L ∞ J , then G t, • f •, x • for all t ∈ J is Pettis integrable Lemma 2.3 and thus the operator A makes sense.
Let R > 0, and consider the set

3.20
Clearly the subset D is closed, convex and equicontinuous.We shall show that A satisfies the assumptions of Lemma 2.9.The proof will be given in three steps.
Step 1.We will show that the operator A maps D into itself.

3.21
Let τ 1 , τ 2 ∈ J, τ 1 < τ 2 and ∀x ∈ D, so Ax τ 2 − Ax τ 1 / 0. Then there exist ψ ∈ E * such that Step 2. We will show that the operator A is weakly sequentially continuous.Let x n be a sequence in D and let x n t → x t in E, w for each t ∈ J. Fix t ∈ J. Since f satisfies assumptions H1 , we have f t, x n t , converging weakly uniformly to f t, x t .Hence the Lebesgue Dominated Convergence theorem for Pettis integral implies Ax n t converging weakly uniformly to Ax t in E ω .We do it for each t ∈ J so Ax n → Ax.Then A : D → D is weakly sequentially continuous.
Step 3. The implication 2.3 holds.Now let V be a subset of D such that V ⊂ conv A V ∪{0} .Clearly, V t ⊂ conv A V ∪{0} for all t ∈ J. Hence, AV t ⊂ AD t , t ∈ J, is bounded in E. Since function g is continuous on J, the set {g t , t ∈ J} ⊂ E is compact, so β g t 0. Using this fact, assumption H4 , Lemma 2.5 and the properties of the measure β, we have for each By 3.13 it follows that v ∞ 0, that is v t 0 for each t ∈ J, and then V t is relatively weakly compact in E. In view of Lemma 2.9, we deduce that A has a fixed point which is obviously a solution of the problem 1.1 .This completes the proof.

An Example
In this section we give an example to illustrate the usefulness of our main result.

4.2
We have

4.3
A simple computation gives We shall check that condition 3.13 is satisfied.Indeed which is satisfied for some α ∈ 1, 2 .Then by Theorem 3.7, the problem 4.1 has at least one solution on J for values of α satisfying 4.5 .

Lemma 2 . 5
see 34 .The De sBlasi measure of noncompactness satisfies the following properties: a S ⊂ T ⇒ β S ≤ β T ; b β S 0 ⇔ S is relatively weakly compact; c β S ∪ T max{β S , β T }; d β S ω β S , where S ω denotes the weak closure of S; e β S T ≤ β S β T ; f β aS |a|α S ;

Example 4 . 1 .
Let us consider the following fractional boundary value problem: J : 0, T , 1 < α ≤ 2, Banach space, Ω E the set of all bounded subsets of E, and B 1 the unit ball in E. The De Blasi measure of weak noncompactness is the map β : Ω E → 0, ∞ defined by β X inf{ > 0 : there exists a weakly compact subset Ω of E such that X ⊂ B 1 Ω}.
• is Pettis integrable for all t ∈ J by Lemma 2.3, and so the operator A is well defined.
•, x • is Pettis integrable on J; H3 there exists p f ∈ L ∞ J, R such that f t, u ≤ p f t u , for a.e.t ∈ J and each u ∈ E; H3 ' there exists p f ∈ L ∞ J, E and a continuous nondecreasing function ψ : 0, ∞ → 0, ∞ such that f t, u ≤ p f t ψ u , for a.e.t ∈ J and each u ∈ E; H4 for each bounded set D ⊂ E, and each t ∈ J, the following inequality holds ∞ sup{p f t : t ∈ J}.Theorem 3.7.Let E be a reflexive Banach space and assume that (H1)-(H3) are satisfied.If p f L ∞ G * < 1, 3.13 then the problem 1.1 has at least one solution on J. Proof.Let the operator A : C J, E → C J, E defined by the formula Au t : T 0 G t, s f s, u s ds g t , 3.14 where G •, • is the Green's function defined by 3.5 .It is well known the fixed points of the operator A are solutions of the problem 1.1 .First notice that, for x ∈ C J, E , we have f •, x • ∈ P J, E assumption H2 .Since, s → G t, s ∈ L ∞ J , then G t, • f •, x Take x ∈ D, t ∈ J and assume that Ax t / 0. Then there exists ψ ∈ E * such that