Đang tải... (xem toàn văn)
We consider a class of boundary value problem in a separable Banach space E, involving a nonlinear differential inclusion of fractional order with integral bounday conditions, of the form D αu(t) ∈ F(t, u(t), D α−1u (t)), a.e., t ∈ 0, 1, I β u(t) t=0 = 0, u(1) = R 1 0 u (t) d t, (1) where D α is the standard RiemannLiouville fractional derivative, F is a closed valued mapping. Under the suitable conditions we prove that the solutions set of (1) is nonempty and is a retract in W α,1 E (I). An application in control theory is also provided by using Young measures. Key word and phrases: Fractional differential inclusion; boundary value problem; Green’s function; contractive set valuedmap; retract; Young measures.
On a fractional differential inclusion with integral boundary conditions in Banach space P D PHUNG ∗, L X TRUONG † Abstract We consider a class of boundary value problem in a separable Banach space E, involving a nonlinear differential inclusion of fractional order with integral bounday conditions, of the form α α−1 D u(t) ∈ F (t, u(t), D u (t)), a.e., t ∈ [0, 1], (1) β I u(t) t=0 = 0, u(1) = u (t) d t, where Dα is the standard Riemann-Liouville fractional derivative, F is a closed valued mapping Under the suitable conditions we prove that the solutions set of (1) is nonempty and is a retract α,1 in WE (I) An application in control theory is also provided by using Young measures Key word and phrases: Fractional differential inclusion; boundary value problem; Green’s function; contractive set valued-map; retract; Young measures Introduction Differential equations of fractional order have recently showed to be strongly tools in the modelling of many physical phenomena (see [10, 16, 18, 19]) As a consequence there was an increasing interest in studying the initial value problems or boundary value problems for fractional differential equation ([3, 10, 11, 15] and references therein) El-Sayed and Ibrahim initiated the study of fractional differential inclusions in [12] Recently several qualitative results for fractional differential inclusion several results were obtained in [2, 8, 17] It should be noted that most of papers on fractional differential equations or fractional differential inclusions are devoted to the solvability in the cases wherein the nonlinear terms not depend on derivatives of unknown function Further, there are few works consider the such problems in the general context of Banach spaces In the present paper, with E is a separable Banach space, we consider the following problem Dα u(t) ∈ F (t, u(t), Dα−1 u (t)), a.e., t ∈ [0, 1] , t I β u(t) t=0 := lim t→0 (t − s)β−1 Γ (β) (1.1) u(s)ds = 0, u(1) = u (t) d t, (1.2) ∗ Nguyen Tat Thanh University, 300A, Nguyen Tat Thanh Str, District 4, HoChiMinh city, Vietnam, Email: pdphung@ntt.edu.vn † Department of Mathematics and Statistics, University of Economics, HoChiMinh city, 59C, Nguyen Dinh Chieu Str, District 3, HoChiMinh city, Vietnam, Email: lxuantruong@gmail.com where α ∈ (1, 2], β ∈ [0, − α] are given constants, Γ is the gamma function, Dα is the standard Riemann-Liouville fractional derivative and F : [0, 1] × E × E → E is a closed valued multifunction In the case of α = 2, (1.1) is a second order differential inclusion which has been studied by many authors We refer to [1, 5, 13] and references therein dealing with boundary value problem in interger order differential inclusion This paper is organized as follows In section we introduce some notions and recall some definitions and needed results, in particular on the fractional calculus In section we provide the results for existence of W α,1 (I)-solutions and properties of solutions set of the problem (1.1)−(1.2) via some classical tools such as fixed points theorem and retract property for the set of all fixed points of a contractive multivalued mapping In section 4, as an application we present a Bolzatype problem in optimal control for fractional order differential equation where the controls are Young measures Some preliminaries Let I be the interval [0, 1] Let E be a separable Banach space and E be its topological dual For the convenience of the reader, we fisrt state here several notations that will used in the sequel - B E : the closed unit ball of E, - (I) : the σ algebra of Lebesgue measurable sets on I, - (E) : the σ algebra of Borel subsets of E, - L 1E (I) : the Banach space of all Lebesgue-Bochner integrable E-valued functions defined on I, - C E (I) : the Banach space of all continuous functions f from [0, 1] into E endowed with the norm f ∞ = sup f (t) t∈I - c(E) : the set of all nonempty and closed subsets of E, - cc(E) : the set of all nonempty and closed and convex subsets of E, - ck(E) : the set of all nonempty and compact and convex subsets of E, - cwk(E) : the set of all nonempty and weakly compact and convex subsets of E, - bc(E) : the set of all nonempty bounded closed subsets of E, - d(x, A) : the distance of a point x of E to a subset A of E, that is d(x, A) = inf x− y : y ∈A - dH (A, B) : the Hausdorff distance between two subsets A and B of E, defined by dH (A, B) = ma x sup d(a, B), sup d(b, A) a∈A b∈B Definition 2.1 (Fractional Bochner integral) Let f : I → E The fractional Bochner-integral of order α > of the function f is defined by I α f (t) := In the above definition, the sign ” t Γ (α) (t − s)α−1 f (s)ds, t > 0 ” denotes the Bochner integral Lemma 2.2 Let f ∈ L 1E (I) We have (i) If α ∈ (0, 1) then I α f (t) exists for almost every t ∈ I and I α f ∈ L 1E (I) (ii) If α ≥ then I α f (t) exists for all t ∈ I and I α f ∈ C E (I) Proof (i) For α ∈ (0, 1), the existence of I α f (t), for a.e t ∈ [0, 1] has been proved in [20, Theorem 2.4] It is not difficult to check that I α f ∈ L 1E (I) (ii) Let α ≥ By using [14, Theorem 3.5.4], the function s → (t − s)α−1 f (s) is strongly measurable Morever we have t (t − s)α−1 f (s) ds ≤ t α−1 f L 1E (I) , ∀t ∈ I So I α f (t) exists for all t ∈ I It’s clear that I α f ∈ C E (I) Definition 2.3 Let f ∈ L 1E (I) We define the Riemann-Liouville fractional derivative of order α > of f by t d n n−α dn (t − s)n−α−1 α D f (t) := n I f (t) = n f (s)ds, dt d t Γ (n − α) where n = [α] + In the case E ≡ R, we have the following well-known results Lemma 2.4 [3] Let α > The general solution of the fractional differential equation Dα x(t) = is given by x(t) = c1 t α−1 + c2 t α−2 + · · · + cn t α−n , (2.3) where ci ∈ R, i = 1, 2, , n (n = [α] + 1) In view of Lemma 2.4, it follows that x(t) = I α Dα x(t) + c1 t α−1 + · · · + cn t α−n , (2.4) for some ci ∈ R, i = 1, 2, , n α,1 In the rest of this paper we denote by WE (I) the space of all continuous functions in C E (I) such that their Riemann-Liouville fractional derivative of order α − are continuous and their RiemannLiouville fractional derivative of order α belong to L 1E (I) 3 α,1 The solutions in WE (I) Lemma 3.1 Let E be a Banach space and let G (·, ·) : I × I → R be a function defined by G (t, s) = (t−s)α−1 , Γ (α) t α−1 0≤s≤t ≤1 + (1 − s)α − α (1 − s)α−1 (α − 1) Γ (α) 0≤t ≤s≤1 0, (3.1) Then the following assertions hold: (i) G(., ) satisfies the following estimate |G(t, s)| ≤ α,1 (ii) If u ∈ WE 2α (α − 1)Γ (α) (I) with I β u(t) t=0 = and u(1) = u (t) d t, then G (t, s) Dα u (s) ds, ∀t ∈ I u (t) = (iii) Let f ∈ L 1E (I) and let u f : I → E be the function defined by G (t, s) f (s) ds, ∀t ∈ I u f (t) = β Then I u f (t) t=0 = and u f (1) = (I) and we have t Dα−1 u f (t) = α,1 u f (t) d t Furthermore u f ∈ WE f (s) ds + (1 − s)α − α (1 − s)α−1 f (s) ds, ∀t ∈ I, α−1 (3.2) 0 Dα u f (t) = f (t) , a.e t ∈ I (3.3) Proof (i) From the definition of G it is easy to see that, for all s, t ∈ [0, 1], |G(t, s)| ≤ 2α (α − 1)Γ (α) (ii) Let y ∈ E For all t ∈ I, we have 1 G (t, s) Dα u (s) ds y, G (t, s) Dα y, u (s) ds = = I α Dα y, u (t) + αt α−1 α−1 I α+1 Dα y, u (1) − I α Dα y, u (1) (3.4) Using the assumption lim t→0+ I β u(t) = it follows from (2.4) that y, u (t) = I α Dα y, u (t) + c1 t α−1 , (3.5) y, u (1) = I α Dα y, u (1) + c1 , (3.6) for some c1 ∈ R So we have and 1 y, = u(t)d t y, u (t) d t I α Dα y, u (t) d t + = = I α+1 Dα y, u (1) + As u(1) = u(t)d t c1 α c1 α (3.7) it follows from (3.6) and (3.7) that c1 = α I α+1 Dα y, u (1) − I α Dα y, u (1) α−1 (3.8) Combining (3.4), (3.5) and (3.8) we get G (t, s) Dα u (s) ds y, = y, u (t) Since this equality holds for every y ∈ E so we have u (t) = G (t, s) Dα u (s) ds, ∀t ∈ I (iii) Let f ∈ L 1E (I) and u f (t) = G (t, s) f (s) ds, ∀t ∈ I By the definition of G(·, ·) we have α u f (t) = I f (t) + αt α−1 I α+1 f (1) − I α f (1) α−1 (3.9) Using Lemma 2.2 it’s clear that I α f ∈ C E (I) So u f is continuous on I On the other hand, from (3.9), it follows that u f (1) = αI α+1 f (1) − I α f (1) , α−1 and 1 I α f (t)d t + u f (t) d t = 0 = I α+1 f (1) + = α−1 α−1 α−1 I α+1 f (1) − I α f (1) I α+1 f (1) − I α f (1) αI α+1 f (1) − I α f (1) Hence u f (1) = u f (t) d t Now, let y ∈ E be arbitrary y, I β u f (t) = I β y, u f (t) = I β G (t, s) y, f (s) ds = I α+β y, f (t) + I = I α+β y, f (t) + αt α−1 β α−1 αΓ (α) y, I α+1 f (1) − I α f (1) y, I α+1 f (1) − I α f (1) t α+β−1 (3.10) (α − 1)Γ α + β Letting t → 0+ in (3.10) we get lim t→0+ y, I β u f (t) = 0, ∀ y ∈ E This show that I β u f (t) t=0 = α It’s remains to check the equalities (3.2) - (3.3) Indeed, since the function I f (·) has RiemannLiouville fractional derivatives of order γ, for all γ ∈ (0, α], so is the function u f (·) by using (3.9) On the other hand, for each y ∈ E , we have y, Dγ u f (t) = Dγ y, u f (t) = Dγ G (t, s) y, f (s) ds = Dγ I α y, f (t) + α (α − 1) I α+1 y, f (1) − I α y, f (1) Γ (α) t α−γ−1 , Γ (α−γ) Since Dγ I α y, f (t) = I α−γ y, f (t) and Dγ t α−1 = 0, Dγ (t α−1 ) (3.11) < γ < α, γ = α, we deduce from (3.11) that t 〈 y, D α−1 y, f (s) ds + u f (t)〉 = and αΓ (α) (α − 1) I α+1 y, f (1) − I α y, f (1) , ∀t ∈ I, y, Dα u f (t) = y, f (t) , a.e t ∈ I These imply that (3.2) and (3.3) hold The proof of this Lemma is completed Remark 3.2 From Lemma 3.1, it’s easy to see that if u f (t) = u f (t) ≤ MG f L 1E (I) (s)ds, f ∈ L 1E (I), then Dα−1 u f (t) ≤ MG f and for all t ∈ I, where MG = G(t, s) f 2α (α − 1)Γ (α) L 1E (I) , (3.12) Now we will establish the theorem for existence of the solutions of problem (1.1) − (1.2) by applying the Covitz - Nadler fixed point theorem (see [9]) Theorem 3.3 Let F : [0, 1] × E × E → c (E) be a closed valued multifunction satisfying the following conditions (A1 ) F is (I) ⊗ (E) ⊗ (E)-measurable, (A2 ) There exists positive functions 1, ∈ LR1 (I) with MG ≤ dH F t, x , y1 , F t, x , y2 (t) + < such that x1 − x2 + (t) y1 − y2 , for all t, x , y1 , t, x , y2 ∈ I × E × E (A3 ) The function t → sup { z : z ∈ F (t, 0, 0)} is integrable α,1 Then the problem (1.1)-(1.2) has at least one solution in WE (I) Proof We defined the following set valued map S : L 1E (I) → c L 1E (I) defined by S (h) = f ∈ L 1E (I) : f (t) ∈ F t, uh (t) , Dα−1 uh (t) , a.e t ∈ I , h ∈ L 1E (I) , α,1 where c L 1E (I) denotes the set of all nonempty closed subsets of L 1E (I) and uh ∈ WE (I), uh(t) = G(t, s)h(s)ds It is clear that u (·) is a solution of (1.1) − (1.2) if and only if Dα u (·) is a fixed point of S We shall show that S is a contraction The proof will be given in two steps Step S (h) is nonempty and closed for every h ∈ L 1E (I) It’s note that, by the assumptions, the multifunction F ·, uh (·) , Dα−1 uh (·) is closed valued and measurable on I Using the standard measurable selections theorem we infer that F ·, uh (·) , Dα−1 uh (·) admits a measurable selection z (·) One has z (t) ≤ sup { a : a ∈ F (t, 0, 0)} + dH F (t, 0, 0) , F t, uh (t) , Dα−1 uh (t) ≤ sup { a : a ∈ F (t, 0, 0)} + (t) ≤ sup { a : a ∈ F (t, 0, 0)} + MG uh (t) + (t) (t) + (t) h Dα−1 uh (t) L 1E (I) , for almost every t ∈ I, which shows that z ∈ L 1E (I) and then S (h) is nonempty On the other hand, it is easy to see that, for each h ∈ L 1E (I), S (h) is closed in L 1E (I) Step The multi-valued map S is a contraction We need to prove that there exists k ∈ (0, 1) satisfying dH S(h), S(g) ≤ k h − g L 1E (I) , for any h, g ∈ L 1E (I), where dH denotes the Hausdorff distance on closed subsets in the Banach space L 1E (I) Let f ∈ S (h) and > By a standard measurable selections theorem, there exists a Lebesgue-measurable φ : I → E such that φ (t) ∈ F t, u g (t) , Dα−1 u g (t) , and φ (t) − f (t) ≤ d f (t), F t, u g (t), Dα−1 u g (t) + , for all t ∈ I As f ∈ S(h) we have φ (t) − f (t) ≤ dH F t, uh(t), Dα−1 uh(t) , F t, u g (t), Dα−1 u g (t) (t) ≤ u g (t) − uh (t) + + Dα−1 u g (t) − Dα−1 uh (t) + , (t) for all t ∈ I This follows that φ− f L 1E (I) ≤ MG + g −h L (I) R L 1E (I) + , ∀ f ∈ S (h) Hence φ ∈ S(g) and ≤ MG sup d f , S g f ∈S(h) + g −h L 1E (I) + Whence we get sup d f , S g ≤ MG f ∈S(h) since + g −h L 1E (I) , is arbitrary By interchanging the variables g, h we obtain dH S g , S (h) ≤ MG + g −h L 1E (I) , ∀g, h ∈ L 1E (I) Since k := MG + < by our assumpsion, this prove that S is a contractive map Apply the Covitz-Nadler’s theorem ([9]) to the contractive multivalued map S shows that S has a fixed point The theorem is proved Corollary 3.4 Let f : I × E × E → E be a mapping satisfying the following conditions (A1 ) for every x, y ∈ E × E, the function f ·, x, y is measurable on I, (A2 ) for every t ∈ I, f (t, ·, ·) is continuous and there exists positive functions MG + < such that f t, x , y1 − f t, x , y2 ≤ (t) x1 − x2 + (t) 1, ∈ LR1 (I) with y1 − y2 , for all t, x , y1 , t, x , y2 ∈ I × E × E, (A3 ) the function t → f (t, 0, 0) is Lebesgue-integrable on I Then the fractional differential equation α α−1 D u(t) = f t, u(t), D u(t) , β I u(t) a.e., t ∈ I, t=0 = 0, u(1) = (3.13) u (t) d t, α,1 has a unique solution u ∈ WE (I) α,1 Proof The existence of solution u is guaranteed by Theorem 3.3 Let u1 , u2 be two WE (I)-solutions to the problem (3.13) For each t ∈ I, we have Dα u1 (t) − Dα u2 (t) = f t, u1 (t), Dα−1 u1 (t) − f t, u2 (t), Dα−1 u2 (t) ≤ (t) u1 (t) − u2 (t) + (t) Dα−1 u1 (t) − Dα−1 u2 (t) (3.14) On the other hand, it follows from Lemma 3.1 that u1 (t) − u2 (t) ≤ MG Dα u1 − Dα u2 and L 1E (I) Dα−1 u1 (t) − Dα−1 u2 (t) ≤ MG Dα u1 − Dα u2 , (3.15) L 1E (I) (3.16) Combining (3.14), (3.15) and (3.16) we deduce that D α u1 − D α u2 L 1E (I) ≤ MG + L (I) R D α u1 − D α u2 L 1E (I) , which ensures Dα u1 = Dα u2 and hence, by (3.15), we get u1 = u2 Theorem 3.5 Let F : [0, 1] × E × E → bc(E) be a bounded closed valued multifunction satisfying the α,1 conditions (A1 )−(A3 ) in Theorem 3.3 Then the WE (I)-solutions set, , of the problem (1.1)−(1.2) α,1 α,1 is retract in WE (I), here the space WE (I) is endowed with the norm u W = u ∞ + Dα−1 u ∞ + Dα u L 1E (I) Proof According to Theorem 3.3 and our assumptions, the multifunction S : L 1E (I) → c L 1E (I) defined by S (h) = f ∈ L 1E (I) : f (t) ∈ F t, uh (t) , Dα−1 uh (t) , a.e t ∈ I , h ∈ L 1E (I) , α,1 where c L 1E (I) denotes the set of all nonempty closed subsets of L 1E (I) and uh ∈ WE (I), uh(t) = G(t, s)h(s)ds, is a contraction with the nonempty, bounded, closed and decomposable values in L 1E (I) So by a result of Bressan-Cellina-Fryszkowski ([4]), the set F i x(S) of all fixed points of S is a retract in L 1E (I) Hence there exists a continuous mapping ψ : L 1E (I) → F i x(S) such that ψ(h) = h, ∀h ∈ F i x(S) α,1 For each u ∈ WE (3.17) (I), let us set G(t, s)ψ (Dα u) (s)ds, t ∈ I Φ(u)(t) = (3.18) Using Lemma 3.1 we have I β (Φ(u)) (t) = 0, t=0 t Dα−1 (Φ(u)) (t) = ψ (Dα u) (s)ds + α−1 Φ(u)(1) = Φ(u)(s)ds, (3.19) (1 − s)α − α(1 − s)α−1 ψ (Dα u) (s)ds, (3.20) and Dα (Φ(u)) (t) = ψ(Dα u)(t), a.e t ∈ I (3.21) α,1 WE (I)-solution α This shows that D (Φ(u)) ∈ F i x(S) So Φ(u) is a of problem (1.1) − (1.2), that is α,1 α,1 Φ(u) ∈ It remains to prove that Φ is continuous mapping from WE (I) in to Let u ∈ WE (I) and > As ψ is continuous on L 1E (I), there exists δ > such that h − Dα u L 1E (I) [...]... 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([6], Theorem 6.3.5) Assume that X and Z are Polish spaces Let (un ) be sequence of -measurable mappings from Ω into X such that (un ) converges in probability to a -measurable mapping u∞ from Ω into X and (v n ) be a sequence of -measurable mappings from Ω into Z such that (v n ) stably converges to ν∞ ∈ (Ω, , P; +1 (Z)) Let h : Ω × X × Z → R be a Carathéodory integrand such that the sequence (h(.,... d t ≥ inf J(z), z∈S∆ 0 for all n ∈ N Because ν is an arbitrary element in SΣ we get infν∈SΣ J(ν) ≥ infz∈S∆ J(z) On the other hand, it is evident that infν∈SΣ J(ν) ≤ infz∈S∆ J(z) So we can conclude that inf J(ν) = inf J(z) ν∈SΣ z∈S∆ In order to show that there is a minimizer in SΣ , let z j be minimizing for J(z), that is 1 L t, uz j (t), Dα−1 uz j (t), z j (t) d t = inf J(z), lim j→∞ z∈S∆ 0 and let ... 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