On a fractional differential inclusion with integral boundary conditions in Banach space

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

non-linear differential inclusion of fractional order with integral bounday conditions, of the form







D α u (t) ∈ F(t, u(t), D α−1 u (t)), a.e., t ∈ [0, 1],

I β u (t)

t=0 = 0, u(1) =

1

R

0

where D α 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

in W E α,1 (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.

1 Introduction

Differential equations of fractional order have recently showed to be strongly tools in the mod-elling 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 differ-ential 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

D α u (t) ∈ F(t, u(t), D α−1 u (t)), a.e., t ∈ [0, 1] , (1.1)

I β u (t)

t=0 := lim

t→0

Z t

0

(t − s) β−1

Γ (β) u (s)ds = 0, u(1) =

1

Z

0

u (t) d t, (1.2)

∗ Nguyen Tat Thanh University, 300A, Nguyen Tat Thanh Str, District 4, HoChiMinh city, Vietnam, Email: pd-phung@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

whereα ∈ (1, 2], β ∈ [0, 2 − α] are given constants, Γ is the gamma function, D α is the standard

Riemann-Liouville fractional derivative and F : [0, 1]× E × E → 2 Eis a closed valued multifunction

In the case ofα = 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 2 we introduce some notions and recall some definitions and needed results, in particular on the fractional calculus In section 3 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 Bolza-type problem in optimal control for fractional order differential equation where the controls are Young measures

2 Some preliminaries

Let I be the interval [0, 1] Let E be a separable Banach space and E0 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,

- L (I) : the σ algebra of Lebesgue measurable sets on I,

- B(E) : the σ algebra of Borel subsets of E,

- L1E (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

k f k∞= sup

t ∈I k f (t)k.

- 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

- d H (A, B) : the Hausdorff distance between two subsets A and B of E, defined by

d H (A, B) = max

 sup

a ∈A

d (a, B), sup

b ∈B

d (b, A)



Definition 2.1 (Fractional Bochner integral) Let f : I → E The fractional Bochner-integral of order

α > 0 of the function f is defined by

I α f (t) := 1

Γ (α)

Z t

0

(t − s) α−1 f (s)ds, t > 0.

In the above definition, the sign ”R ” denotes the Bochner integral

Lemma 2.2 Let f ∈ L1E (I) We have

(i) If α ∈ (0, 1) then I α f (t) exists for almost every t ∈ I and I α f ∈ L1

E (I).

(ii) If α ≥ 1 then I α f (t) exists for all t ∈ I and I α f ∈ C E (I).

Proof. (i) Forα ∈ (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 ∈ L1E (I).

(ii) Let α ≥ 1 By using [14, Theorem 3.5.4], the function s 7→ (t − s) α−1 f (s) is strongly

measurable Morever we have

Z t

0

(t − s) α−1 f (s) ds ≤ t α−1 k f k L1E(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 ∈ L1E (I) We define the Riemann-Liouville fractional derivative of order α > 0 of

f by

D α f (t) := d

n

d t n I n −α f (t) = d

n

d t n

Z t

0

(t − s) n −α−1

Γ (n − α) f (s)ds, where n = [α] + 1.

In the case E≡ R, we have the following well-known results

Lemma 2.4. [3] Let α > 0 The general solution of the fractional differential equation D α x (t) = 0 is

given by

x (t) = c1t α−1 + c2t α−2 + · · · + c n t α−n, (2.3)

where c i ∈ R, i = 1, 2, , n (n = [α] + 1).

In view of Lemma 2.4, it follows that

x (t) = I α D α x (t) + c1t α−1 + · · · + c n t α−n, (2.4)

for some c i ∈ R, i = 1, 2, , n.

In the rest of this paper we denote by W E α,1 (I) the space of all continuous functions in C E (I) such

that their Riemann-Liouville fractional derivative of orderα − 1 are continuous and their

Riemann-Liouville fractional derivative of orderα belong to L1

E (I).

3 The solutions in WE α,1(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) , 0≤ s ≤ t ≤ 1

0, 0≤ t ≤ s ≤ 1 + t α−1

(α − 1) Γ (α) (1 − s) α − α (1 − s) α−1
(3.1)

Then the following assertions hold:

(i) G (., ) satisfies the following estimate

|G(t, s)| ≤ 2α

(α − 1)Γ (α)

(ii) If u ∈ W E α,1 (I) with I β u (t)

t=0 = 0 and u(1) =R1

0

u (t) d t, then

u (t) =

1

Z

0

G (t, s) D α u (s) ds, ∀t ∈ I.

(iii) Let f ∈ L1E (I) and let u f : I → E be the function defined by

u f (t) =

1

Z

0

G (t, s) f (s) ds, ∀t ∈ I.

Then I β u

f (t)

t=0 = 0 and u f(1) =

1

R

0

u f (t) d t Furthermore u f ∈ W E α,1 (I) and we have

D α−1 u

f (t) =

t

Z

0

f (s) ds + 1

α − 1

1

Z

0

(1 − s) α − α (1 − s) α−1
f (s) ds, ∀t ∈ I, (3.2)

D α 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 ∈ E0 For all t ∈ I, we have

*

y,

1

Z

0

G (t, s) D α u (s) ds

+

=

1

Z

0

G (t, s) D α y , u (s)

ds

= I α D α y , u (t)
+ αt α−1

α − 1



I α+1 D α y , u(1) − Iα D α y , u(1) (3.4)

Using the assumption limt→0+I β u (t) = 0 it follows from (2.4) that

y , u (t) = I α D α y , u (t) + c1t α−1, (3.5)

for some c1∈ R So we have

y , u(1) = Iα D α y , u(1) + c1, (3.6) and

®

y,

Z 1 0

u (t)d t

¸

=

Z 1 0

y , u (t)

d t

=

Z 1 0

I α D α y , u (t)

d t+c1

α

= I α+1 D α y , u(1) + c1

As u(1) =R1

0 u (t)d t it follows from (3.6) and (3.7) that

c1= α

α − 1



I α+1 D α y , u(1) − Iα D α y , u(1) (3.8) Combining (3.4), (3.5) and (3.8) we get

*

y,

1

Z

0

G (t, s) D α u (s) ds

+

= y , u (t)

Since this equality holds for every y ∈ E0so we have u (t) =R1

0

G (t, s) D α u (s) ds, ∀t ∈ I.

(iii) Let f ∈ L1E (I) and u f (t) =

1

R

0

G (t, s) f (s) ds, ∀t ∈ I By the definition of G(·, ·) we have

u f (t) = I α f (t) + αt α−1

α − 1



I α+1 f (1) − I α f(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) = 1

α − 1 αI α+1 f (1) − I α f(1) ,

and

1

Z

0

u f (t) d t =

Z 1 0

I α f (t)d t + 1

α − 1



I α+1 f (1) − I α f(1)

= I α+1 f(1) + 1

α − 1



I α+1 f (1) − I α f(1)

= 1

α − 1 αI α+1 f (1) − I α f(1)

Hence u f(1) =R1

0

u f (t) d t Now, let y ∈ E0be arbitrary

y , I β u f (t)

= I β y , u f (t) = I β





1

Z

0

G (t, s) y , f (s)

ds





= I α+β y , f (t) + I β



αt α−1

α − 1 y , I α+1 f (1) − I α f(1)



= I α+β y , f (t) + αΓ (α)

(α − 1)Γ α + β
y , I α+1 f (1) − I α f(1)

t α+β−1 (3.10)

Letting t→ 0+in (3.10) we get limt→0+ y , I β u f (t) = 0, ∀y ∈ E0 This show that I β u f (t)

t=0 = 0

It’s remains to check the equalities (3.2) - (3.3) Indeed, since the function I α f(·) has Riemann-Liouville fractional derivatives of orderγ, for all γ ∈ (0, α], so is the function u f(·) by using (3.9)

On the other hand, for each y ∈ E0, we have

y , D γ u f (t)

= D γ y , u f (t) = D γ

1

Z

0

G (t, s) y , f (s)

ds

= D γ I α y , f (t) + α

(α − 1) I α+1 y , f(1) − Iα y , f(1)



D γ (t α−1) (3.11)

Since D γ I α y , f (t) = I α−γ y , f (t) and D γ t α−1
=

¨ Γ (α)

Γ(α−γ)t α−γ−1, 0< γ < α,

from (3.11) that

〈 y, D α−1 u f (t)〉 =

Z t

0

y , f (s)

ds+ αΓ (α)

(α − 1) I α+1 y , f(1) − Iα y , f(1)
, ∀t ∈ I,

and

y , D α u f (t)

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) =R1

0 G (t, s)f (s)ds, f ∈ L1

E (I), then

u f (t) ≤ M G f L1E(I) and D α−1 u f (t) ≤ M G f L1E(I), (3.12)

for all t ∈ I, where

(α − 1)Γ (α).

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 L (I) ⊗ B(E) ⊗ B(E)-measurable,

(A2) There exists positive functions `1,`2∈ LR1(I) with M G `1+ `2 1< 1 such that

d H F t , x1, y1
, F t, x2, y2

≤ `1(t) x1− x2 + `2(t) y1− y2 ,

for all t , x1, y1
, t, x2, y2
∈ I × E × E.

(A3) The function t 7→ sup {kzk : z ∈ F (t, 0, 0)} is integrable.

Then the problem (1.1)-(1.2) has at least one solution in W α,1

E (I).

Proof. We defined the following set valued map

S : L1E (I) → c L1

E (I)
defined by

S (h) =

f ∈ L1E (I) : f (t) ∈ F t, u h (t) , D α−1 u h (t)
, a.e t ∈ I , h ∈ L1

E (I) , where c L1E (I)
denotes the set of all nonempty closed subsets of L1

E (I) and u h ∈ W E α,1 (I),

u h (t) =

Z 1 0

G (t, s)h(s)ds.

It is clear that u (·) is a solution of (1.1) − (1.2) if and only if D α u (·) is a fixed point of S We shall show that S is a contraction The proof will be given in two steps

Step 1 S (h) is nonempty and closed for every h ∈ L1

E (I) It’s note that, by the assumptions, the

multifunction

F ·, u h (·) , D α−1 u h(·)

is closed valued and measurable on I Using the standard measurable selections theorem we infer that F ·, u h (·) , D α−1 u h(·)
admits a measurable selection z (·) One has

kz (t)k ≤ sup {kak : a ∈ F (t, 0, 0)} + d H F (t, 0, 0) , F t, u h (t) , D α−1 u h (t)

≤ sup {kak : a ∈ F (t, 0, 0)} + `1(t) u h (t) + `2(t) D α−1 u

h (t)

≤ sup {kak : a ∈ F (t, 0, 0)} + M G `1(t) + `2(t)
khk L1E(I),

for almost every t ∈ I, which shows that z ∈ L1E (I) and then S (h) is nonempty On the other hand,

it is easy to see that, for each h ∈ L1E (I), S (h) is closed in L1

E (I).

Step 2 The multi-valued map S is a contraction.

We need to prove that there exists k∈ (0, 1) satisfying

d H S (h), S(g)
≤ k h − g L1E(I),

for any h, g ∈ L1

E (I), where d H denotes the Hausdorff distance on closed subsets in the Banach

space L1E (I) Let f ∈ S (h) and " > 0 By a standard measurable selections theorem, there exists a

Lebesgue-measurableφ : I → E such that

φ (t) ∈ F t, u g (t) , D α−1 u g (t)
, and

φ (t) − f (t) ≤ d f (t), F t, u g (t), D α−1 u

g (t)

+ ",

for all t ∈ I As f ∈ S(h) we have

φ (t) − f (t) ≤ d H F t , u h (t), D α−1 u

h (t)
, F t, u g (t), D α−1 u

g (t)

+ "

≤ `1(t) u g (t) − u h (t) + `2(t) D α−1 u

g (t) − D α−1 u h (t) + ", for all t ∈ I This follows that

φ − f L1E(I) ≤ M G `1+ `2 L1

R(I) g − h L1E(I) + ", ∀f ∈ S (h)

Henceφ ∈ S(g) and

sup

f ∈S(h)

d f , S g

≤ M G `1+ `2 1 g − h L1

E (I) + ".

Whence we get

sup

f ∈S(h)

d f , S g

≤ M G `1+ `2 1 g − h L1

E (I),

since" is arbitrary By interchanging the variables g, h we obtain

d H S g
, S (h)
≤ M G `1+ `2 1 g − h L1E(I), ∀g, h ∈ L1E (I) Since k : = M G `1+ `2 1< 1 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

(A0

1) for every x, y
∈ E × E, the function f ·, x, y
is measurable on I,

(A0

2) for every t ∈ I, f (t, ·, ·) is continuous and there exists positive functions `1,`2 ∈ L1R(I) with

M G `1+ `2 1< 1 such that

f t , x1, y1
− f t, x2, y2 ≤ `1(t) x1− x2 + `2(t) y1− y2 ,

for all t , x1, y1
, t, x2, y2
∈ I × E × E,

(A0

3) the function t 7→ f (t, 0, 0) is Lebesgue-integrable on I.

Then the fractional differential equation







D α u (t) = f t, u(t), D α−1 u (t)
, a.e., t ∈ I,

I β u (t)

t=0 = 0, u(1) =R1

0

has a unique solution u ∈ W E α,1 (I).

Proof The existence of solution u is guaranteed by Theorem 3.3 Let u1, u2be two W E α,1 (I)-solutions

to the problem (3.13) For each t ∈ I, we have

D α u

1(t) − D α u2(t) = f t , u1(t), D α−1 u1(t)
− f t, u2(t), D α−1 u2(t)

≤ `1(t) u1(t) − u2(t) + `2(t) D α−1 u

1(t) − D α−1 u

2(t) (3.14)

On the other hand, it follows from Lemma 3.1 that

u1(t) − u2(t) ≤ M G D α u

1− D α u2 L1

E (I), (3.15)

and

D α−1 u

1(t) − D α−1 u2(t) ≤ M G D α u

1− D α u2 L1

E (I). (3.16)

Combining (3.14), (3.15) and (3.16) we deduce that

D α u

1− D α u2 L1

E (I) ≤ M G `1+ `2 L1

R(I) D α u1− D α u2 L1

E (I),

which ensures D α u1= D α u

2and 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

conditions (A1)−(A3) in Theorem 3.3 Then the W E α,1 (I)-solutions set, S , of the problem (1.1)−(1.2)

is retract in W α,1

E (I), here the space W E α,1 (I) is endowed with the norm

kuk W = kuk∞+ D α−1 u

∞+ kD α ukL1E(I).

Proof. According to Theorem 3.3 and our assumptions, the multifunction

S : L1E (I) → c L1

E (I)
defined by

S (h) =

f ∈ L1E (I) : f (t) ∈ F t, u h (t) , D α−1 u h (t)
, a.e t ∈ I , h ∈ L1

E (I) , where c L1E (I)
denotes the set of all nonempty closed subsets of L1

E (I) and u h ∈ W E α,1 (I),

u h (t) =

Z 1 0

G (t, s)h(s)ds,

is a contraction with the nonempty, bounded, closed and decomposable values in L1E (I) So by a

result of Bressan-Cellina-Fryszkowski ([4]), the set F ix(S) of all fixed points of S is a retract in

L1E (I) Hence there exists a continuous mapping ψ : L1

E (I) → F ix(S) such that

For each u ∈ W E α,1 (I), let us set

Φ(u)(t) =

Z 1 0

G (t, s)ψ (D α u ) (s)ds, t ∈ I. (3.18) Using Lemma 3.1 we have

I β (Φ(u)) (t)

t=0= 0, Φ(u)(1) =

Z 1 0

D α−1 (Φ(u)) (t) =

Z t

0

ψ (D α u ) (s)ds + 1

α − 1

Z 1 0

(1 − s) α − α(1 − s) α−1
ψ (D α u ) (s)ds, (3.20)

D α (Φ(u)) (t) = ψ(D α u )(t), a.e t ∈ I. (3.21)

This shows that D α (Φ(u)) ∈ F ix(S) So Φ(u) is a W E α,1 (I)-solution of problem (1.1) − (1.2), that is

Φ(u) ∈ S It remains to prove that Φ is continuous mapping from W E α,1 (I) in to S Let u ∈ W E α,1 (I)

and" > 0 As ψ is continuous on L1

E (I), there exists δ > 0 such that

kh − D α ukL1E(I) < δ =⇒ ψ(h) − ψ (D α u) L1

E (I) < ", (3.22)

for all h ∈ L1E (I) Let us consider the ball B W α,1

E (I) (u, δ) of center u with radius δ in€W α,1

E (I) , k·k W

Š

Then, for v ∈ B W α,1

E (I) (u, δ), one have kD α v − D α ukL1E(I) < δ using the definition of the norm k·k W

So it follows from (3.21) and(3.22) that

kD α (Φ (v)) − D α (Φ (u))k L1

E (I)= ψ(D α v ) − ψ (D α u) L1E(I) < ". (3.23) Using again Lemma 3.1 we deduce, from (3.18), (3.20) and (3.23),

kΦ (v) (t) − Φ (u) (t)k ≤ M G kD α (Φ (v)) − D α (Φ (u))k L1E(I) < M G ", (3.24)

D α−1 (Φ (v)) (t) − D α−1 (Φ (u)) (t) ≤ M G Γ (α) kD α (Φ (v)) − D α (Φ (u))k L1E (I) < M G Γ (α)", (3.25)

for all t ∈ I Combining (3.23) − (3.25) we obtain the continuity of Φ Finally, for u ∈ S , we have

D α (u) ∈ F ix(S) So

ψ(D α (u)) = D α (u) ,

by the property ofψ It follows that

Φ(u)(t) =

Z 1 0

G (t, s)ψ (D α u ) (s)ds

=

Z 1 0

G (t, s)D α u (s)ds = u(t), for all t ∈ I The proof is therefore complete.

4 Application to control theory

In this section we present a relaxation problem in control theory related to differential inclusion

of orderα considered in the preceding section Let E ≡ R d be a finite dimensional space and Z be

a compact metric space ByM+1(Z) we denote the space of all probability Radon measures on Z It

is well-known thatM+1(Z) is a compact metrizable space for the vague topology We also denote by

Y I; M+1(Z)
the space of all Young measures defined on I endowed with the stable topology so

thatY I; M+1(Z)
is a compact metrizable space with respect to this topology For the convenience

of the reader we recall that a sequence(ν n ) in Y I; M1

+(Z)
stably converges to ν ∈ Y I; M1

+(Z)
if

lim

n→∞

Z1 0

d t

Z

Z

h t (z)dν n

t (z) =

Z 1 0

d t

Z

Z

h t (z)dν t (z),

for all h ∈ L C (Z)1 ([0, 1]), here C (Z) denotes the space of all continuous real valued functions defined

on Z endowed with the norm of uniform convergence.

The following theorem is useful for our main result

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,... noted that most of papers on fractional differential equations or fractional differ-ential inclusions are devoted to the solvability in the cases wherein the nonlinear terms not depend

on derivatives... 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)