FILIPPOV’S THEOREM FOR IMPULSIVE DIFFERENTIAL INCLUSIONS WITH FRACTIONAL ORDER

In this paper, we present an impulsive version of Filippov’s Theorem for fractional differential inclusions of the form: D ∗ y(t) ∈ F (t, y(t)), a.e. t ∈ J\{t1, . . . , tm}, α ∈ (1, 2], y(t+k )− y(t − k ) = Ik(y(t − k )), k = 1, . . . ,m, y(t+k )− y (t−k ) = Ik(y (t−k )), k = 1, . . . ,m, y(0) = a, y′(0) = c, where J = [0, b], D ∗ denotes the Caputo fractional derivative and F is a setvalued map. The functions Ik, Ik characterize the jump of the solutions at impulse points tk (k = 1, . . . ,m).


Introduction
Differential equations with impulses were considered for the first time in the 1960's by Milman and Myshkis [47,46].A period of active research, primarily in Eastern Europe from 1960-1970, culminated with the monograph by Halanay and Wexler [32].
The dynamics of many evolving processes are subject to abrupt changes, such as shocks, harvesting and natural disasters.These phenomena involve short-term perturbations from continuous and smooth dynamics, whose duration is negligible in comparison with the duration of an entire evolution.In models involving such perturbations, it is natural to assume these perturbations act instantaneously or in the form of "impulses".As a consequence, impulsive differential equations have been developed in Benchohra et al [8,10,11], Henderson and Ouahab [34], and Ouahab [52].
In this paper, we shall be concerned with Filippov's theorem and global existence of solutions for impulsive fractional differential inclusions with fractional order.More precisely, we will consider the following problem, where D α * is the Caputo fractional derivative, F : J × R → P(R) is a multivalued map with compact values (P(R) is the family of all nonempty subsets of R), ) and y(t − k ) = lim h→0 + y(t k − h) stand for the right and the left limits of y(t) at t = t k , respectively.
The paper is organized as follows.We first collect some background material and basic results from multi-valued analysis and fractional calculus in Sections 2 and 3, respectively.Then, we shall be concerned with Filippov's theorem for impulsive differential inclusions with fractional order in Section 4.

Preliminaries
In this section, we introduce notations, definitions, and preliminary facts that will be used in the remainder of this paper.Let AC i ([0, b], R n ) be the space of functions y : [0, b] → R n , i-differentiable, and whose i th derivative, y (i) , is absolutely continuous.
We take C(J, R) to be the Banach space of all continuous functions from J into R with the norm y ∞ = sup{|y(t)| : 0 ≤ t ≤ b}.

A. Ouahab
Let (X, • ) be a Banach space, and denote: We say that a multivalued mapping G : X → P(X) has a fixed point if there exists [25], Theorem 19.7)Let E be a separable metric space and G a multi-valued map with nonempty closed values.Then G has a measurable selection.
Let (X, d) be a metric space induced from the normed space (X, | • |).Consider ).Then (P b,cl (X), H d ) is a metric space and (P cl (X), H d ) is a generalized metric space; see [38].

Fractional Calculus
According to the Riemann-Liouville approach to fractional calculus, the notation of fractional integral of order α (α > 0) is a natural consequence of the well known formula (usually attributed to Cauchy), that reduces the calculation of the n−fold primitive of a function f (t) to a single integral of convolution type.In our notation, the Cauchy formula reads where Γ is the gamma function.When a = 0, we write for t > 0, and φ α (t) = 0 for t ≤ 0, and φ α → δ(t) as α → 0, where δ is the delta function and Γ is the Euler gamma function defined by Also J 0 = I (Identity operator), i.e.J 0 f (t) = f (t).Furtheremore, by J α f (0 + ) we mean the limit (if it exists) of J α f (t) for t → 0 + ; this limit may be infinite.) After the notion of fractional integral, that of fractional derivative of order α (α > 0) becomes a natural requirement and one is attempted to substitute α with −α in the above formulas.However, this generalization needs some care in order to guarantee the convergence of the integral and preserve the well known properties of the ordinary derivative of integer order.Denoting by D n with n ∈ N, the operator of the derivative of order n, we first note that i.e., D n is the left-inverse (and not the right-inverse) to the corresponding integral operator J n .We can easily prove that As consequence, we expect that D α is defined as the left-inverse to J α .For this purpose, introducing the positive integer n such that n − 1 < α ≤ n, one defines the fractional derivative of order α > 0: where n = [α] + 1 and [α] is the integer part of α.
Also, we define D 0 = J 0 = I.Then we easily recognize that and Of course, the properties ( 5) and ( 6) are a natural generalization of those known when the order is a positive integer.
Note the remarkable fact that the fractional derivative D α f is not zero for the constant function f (t) = 1 if α ∈ N. In fact, ( 6) with γ = 0 teaches us that It is clear that D α 1 = 0 for α ∈ N, due to the poles of the gamma function at the points 0, −1, −2, . . . .We now observe an alternative definition of fractional derivative, originally introduced by Caputo [12,13] in the late sixties and adopted by Caputo and Mainardi [14] in the framework of the theory of "linear viscoelasticity" (see a review in [43]).
This definition is of course more restrictive than Riemann-Liouville definition, in that it requires the absolute integrability of the derivative of order m.Whenever we use the operator D α * we (tacitly) assume that this condition is met.We easily recognize that in general unless the function f (t), along with its first m − 1 derivatives, vanishes at t = a + .In fact, assuming that the passage of the m−derivative under the integral is legitimate, one recognizes that, m − 1 < α < m and t > 0, EJQTDE Spec.Ed.I, 2009 No. 23 and therefore, recalling the fractional derivative of the power function (6), The alternative definition, that is, Definition 3.3, for the fractional derivative thus incorporates the initial values of the function and of order lower than α.The subtraction of the Taylor polynomial of degree m − 1 at t = a + from f (t) means a sort of regularization of the fractional derivative.In particular, according to this definition, a relevant property is that the fractional derivative of a constant is sill zero, i.e., We now explore the most relevant differences between Definition 3.2 and Definition 3.3 for the two fractional derivatives.From the Riemann-Liouville fractional derivative, we have From ( 11) and ( 12) we thus recognize the following statements about functions, which for t > 0 admit the same fractional derivative of order α, and In these formulas the coefficients c j are arbitrary constants.For proving all mains results we present the following auxiliary lemmas.
For further readings and details on fractional calculus we refer to the books and papers by Kilbas [36], Podlubny [54], Samko [56], Captuo [12,13,14].Let J k = (t k , t k+1 ], k = 0, . . ., m, and let y k be the restriction of a function y to J k .In order to define mild solutions for problem (1)-( 4), consider the space this is a Banach space.Definition 4.1 A function y ∈ P C is said to be a solution of ( 1)-( 4) if there exists v ∈ L 1 (J, R) with v(t) ∈ F (t, y(t)) for a.e.t ∈ J such that y satisfies the fractional differential equation D α * y(t) = v(t) a.e. on J, and the conditions ( 2)-( 4).Let a, c ∈ R, g ∈ L 1 (J, R) and let x ∈ P C be a solution of the impulsive differential problem with fractional order: We will need the following two assumptions: ) and (H 2 ), it follows that the multi-function t → F (t, x t ) is measurable, and by Lemmas 1.4 and 1.5 from [22], we deduce that γ(t) = d(g(t), F (t, x(t)) is measurable (see also the Remark on p. 400 in [4]).
Let P (t) = t 0 p(s)ds.Define the functions η 0 and H 0 by where H 0 (t) = δ 0 M exp Me P (t) + M and and for k = 1, . . ., m, where and where where Before proving the theorem, we present a lemma.
Then from Lemma 2.2, there exists a measurable selection g ǫ of G such that Using the fact that G is integrable bounded, then Proof of Theorem 4.1.We are going to study Problem ( 1)-( 4) in the respective intervals [0, t 1 ], (t 1 , t 2 ], . . ., (t m , b].The proof will be given in three steps and then continued by induction, and then summarized in a fourth step. Step 1.In this first step, we construct a sequence of functions (y n ) n∈N which will be shown to converge to some solution of Problem ( 1)-( 4) on the interval [0, t 1 ], namely to Let f 0 = g on [0, t 1 ] and y 0 (t) = x(t), t ∈ [0, t 1 ], i.e., Then define the multi-valued map Then u ∈ U 1 (t), proving our claim.We deduce that the intersection multivalued operator U 1 (t) is measurable (see [4,15,25]).By Lemma 2.1 (Kuratowski-Ryll-Nardzewski selection theorem), there exists a function t → f 1 (t) which is a measurable selection for U 1 .Consider For each t ∈ [0, t 1 ], we have Hence Then using (H 2 ), we get i.e. u ∈ U 2 (t), proving our claim.Now, since the intersection multi-valued operator U 2 defined above is measurable (see [4,15,25]), there exists a measurable selection Define Using ( 17) and ( 18), a simple integration by parts yields the following estimates, valid for every t ∈ [0, t 1 ], For t ∈ [0, t 1 ], we have Performing an integration by parts, we obtain, since P is a nondecreasing function, the following estimates Repeating the process for n = 1, 2, 3, . . ., we arrive at the following bound By induction, suppose that ( 19) holds for some n and check (19) ).Since U n+1 is a nonempty measurable set, there exists a measurable selection f n+1 (t) ∈ U n+1 (t), which allows us to define for n ∈ N Therefore, for a.e.t ∈ [0, t 1 ], we have Again, an integration by parts leads to Consequently, (19) holds true for all n ∈ N. We infer that {y n } is a Cauchy sequence in P C 1 , converging uniformly to a limit function y ∈ P C 1 , where Moreover, from the definition of {U n }, we have

A. Ouahab
Hence, for almost every t ∈ [0, t 1 ], {f n (t)} is also a Cauchy sequence in R and then converges almost everywhere to some measurable function f (•) in R. In addition, since f 0 = g, we have for a.e.t ∈ [0, t 1 ] where From the Lebesgue dominated convergence theorem, we deduce that {f n } converges to f in L 1 ([0, t 1 ], R).Passing to the limit in (20), we find that the function is solution to Problem (1)-( 4) on [0, t 1 ]; thus y ∈ S [0,t 1 ] (a, c).Moreover, for a.e.t ∈ [0, t 1 ], we have Arguing as in (21) and passing to the limit as n → +∞, we deduce that The obtained solution is denoted by Step 2. Consider now Problem ( 1)-( 4) on the second interval (t Let f 0 = g and set Notice that (22) allows us to use Assumption (H 2 ), apply again Lemma 1.4 in [22] and argue as in Step 1 to prove that the multi-valued map U 1 : [t 1 , t 2 ] → P(R) defined by U 1 (t) = F (t, y 0 (t)) ∩ (g(t) + γ(t)B(0, 1)) is U 1 (t) is measurable.Hence, there exists a function t → f 1 (t) which is a measurable selection for U 1 .Define Next define the measurable multi-valued map U 2 (t) = F (t, y 1 (t)) ∩ (f 1 (t) + p(t)|y 1 (t) − y 0 (t)|B(0, 1)).It has a measurable selection f 2 (t) ∈ U 2 (t) by the Kuratowski-Ryll-Nardzewski selection theorem.Repeating the process of selection as in Step 1, we can A. Ouahab define by induction a sequence of multi-valued maps U n (t) = F (t, y n−1 (t)) ∩ (f n−1 (t) + p(t)|y n−1 (t) − y n−2 (t)|B(0, 1)) where {f n } ∈ U n and (y n ) n∈N is as defined by and we can easily prove that Step 3. We continue this process until we arrive at the function y m+1 := y Then, for a.e.t ∈ (t m , b], the following estimates are easily derived:  1)-( 4) can be defined as follows where γ k := γ | J k .The proof of Theorem 4.1 is complete.

Filippov's Theorem on the Half-Line
We may consider Filippov's Problem on the half-line as given by, where Then we can extend Filippov's Theorem to the half-line.Proof.The solution will be sought in the space P C = {y: [0, ∞) → R, y k ∈ C(J k , R), k = 0, . . ., such that y(t − k ) and y(t + k ) exist and satisfy y(t − k ) = y(t k ) for k = 1, . ..},where y k is the restriction of y to J k = (t k , t k+1 ], k ≥ 0. Theorem 4.1 yields estimates of y k on each one of the bounded intervals J 0 = [0, t 1 ], and J k = (t k−1 , t k ], k = 2, . . . .Let y 0 be solution of Problem ( 1)-(4) on J 0 . Then