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

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
(1)
β
 I u(t) t=0 = 0, u(1) = u (t) d t,
0

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.

1

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

0

(t − s)β−1
Γ (β)

(1.1)

1

u(s)ds = 0,

u(1) =

u (t) d t,

(1.2)

0

∗

Nguyen Tat Thanh University, 300A, Nguyen Tat Thanh Str, District 4, HoChiMinh city, Vietnam, Email:
†
Department of Mathematics and Statistics, University of Economics, HoChiMinh city, 59C, Nguyen Dinh Chieu Str,
District 3, HoChiMinh city, Vietnam, Email:

1


where α ∈ (1, 2], β ∈ [0, 2 − α] are given constants, Γ is the gamma function, Dα is the standard
Riemann-Liouville fractional derivative and F : [0, 1] × E × E → 2 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 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 Bolzatype 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 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

2

b∈B


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) :=
In the above definition, the sign ”

t

1
Γ (α)

(t − s)α−1 f (s)ds, t > 0.
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 ≥ 1 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 ≥ 1. 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
0

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 > 0 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 0 Γ (n − α)
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 Dα x(t) = 0 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 − 1 are continuous and their RiemannLiouville fractional derivative of order α belong to L 1E (I).

3


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)
1

(I) with I β u(t)

t=0

= 0 and u(1) =

u (t) d t, then
0

1

G (t, s) Dα u (s) ds, ∀t ∈ I.

u (t) =
0

(iii) Let f ∈ L 1E (I) and let u f : I → E be the function defined by
1

G (t, s) f (s) ds, ∀t ∈ I.

u f (t) =
0
β

Then I u f (t)

1
t=0

= 0 and u f (1) =
0

(I) and we have

1

t

Dα−1 u f (t) =

α,1

u f (t) d t. Furthermore u f ∈ WE

f (s) ds +

1

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

α−1

(3.2)

0

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

1

G (t, s) Dα u (s) ds

y,
0

G (t, s) Dα y, u (s) ds

=
0

= I α Dα y, u (t)

+

αt α−1
α−1
4

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) = 0 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

1

y,

=

u(t)d t
0

y, u (t) d t
0
1

I α Dα y, u (t) d t +

=
0

= I α+1 Dα y, u (1) +
As u(1) =

1
u(t)d t
0

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
1

G (t, s) Dα u (s) ds

y,

= y, u (t) .

0
1

Since this equality holds for every y ∈ E so we have u (t) =

G (t, s) Dα u (s) ds,

∀t ∈ I.

0
1

(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
0
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
1
u f (1) =
αI α+1 f (1) − I α f (1) ,
α−1
and
1

1

I α f (t)d t +

u f (t) d t =
0

0

= I α+1 f (1) +
=

1
α−1

1

α−1

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) .

5


1

Hence u f (1) =

u f (t) d t. Now, let y ∈ E be arbitrary.
0
1


y, I β u f (t)

= I β y, u f (t) = I β 


G (t, s) y, f (s) ds

0

= 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

= 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
1

y, Dγ u f (t)

= Dγ y, u f (t) = Dγ

G (t, s) y, f (s) ds
0

= 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)

0 < γ < α,
γ = α,

we deduce

from (3.11) that
t

〈 y, D

α−1

y, f (s) ds +

u f (t)〉 =
0

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 =

1
G(t, s) f
0

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,
6


(A2 ) There exists positive functions

1, 2

∈ LR1 (I) with MG
≤

dH F t, x 1 , y1 , F t, x 2 , y2

1 (t)

1

+

2 1

< 1 such that

x1 − x2 +

2 (t)

y1 − y2 ,

for all t, x 1 , y1 , t, x 2 , 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),
1

uh(t) =

G(t, s)h(s)ds.
0

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 1. 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)} +

1 (t)

≤ sup { a : a ∈ F (t, 0, 0)} + MG

uh (t) +

2 (t)

1 (t) + 2 (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 2. 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 > 0. 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)
7

+ ,


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)
1 (t)

≤

u g (t) − uh (t) +

+

Dα−1 u g (t) − Dα−1 uh (t) + ,

2 (t)

for all t ∈ I. This follows that
φ− f

L 1E (I)

≤ MG

1

+

g −h

2 L 1 (I)
R

L 1E (I)

+ , ∀ f ∈ S (h) .

Hence φ ∈ S(g) and
≤ MG

sup d f , S g

1

f ∈S(h)

+

g −h

2 1

L 1E (I)

+ .

Whence we get
sup d f , S g

≤ MG

f ∈S(h)

since

1

+

g −h

2 1

L 1E (I)

,

is arbitrary. By interchanging the variables g, h we obtain
dH S g , S (h) ≤ MG

1

+

g −h

2 1

L 1E (I)

, ∀g, h ∈ L 1E (I) .

Since k := MG 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
(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 1 + 2 1 < 1 such that
f t, x 1 , y1 − f t, x 2 , y2

≤

1 (t)

x1 − x2 +

2 (t)

1, 2

∈ LR1 (I) with

y1 − y2 ,

for all t, x 1 , y1 , t, x 2 , 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,

1

t=0

= 0, u(1) =

(3.13)

u (t) d t,

0

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)

≤

1 (t)

u1 (t) − u2 (t) +
8

2 (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

1

+

2 L 1 (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),

1

uh(t) =

G(t, s)h(s)ds,
0

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
1

G(t, s)ψ (Dα u) (s)ds, t ∈ I.

Φ(u)(t) =

(3.18)

0

Using Lemma 3.1 we have
1

I β (Φ(u)) (t)

= 0,
t=0

t

Dα−1 (Φ(u)) (t) =

ψ (Dα u) (s)ds +
0

1
α−1

Φ(u)(1) =

Φ(u)(s)ds,

(3.19)

0
1

(1 − s)α − α(1 − s)α−1 ψ (Dα u) (s)ds,
0

9

(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 > 0. As ψ is continuous on L 1E (I), there exists δ > 0 such that
h − Dα u

L 1E (I)

<δ

ψ(h) − ψ (Dα u)

=⇒

L 1E (I)

< ,

(3.22)
α,1

for all h ∈ L 1E (I). Let us consider the ball BW α,1 (I) (u, δ) of center u with radius δ in WE (I) , · W .
E
Then, for v ∈ BW α,1 (I) (u, δ), one have Dα v − Dα u L 1 (I) < δ using the definition of the norm · W .
E
E
So it follows from (3.21) and(3.22) that
Dα (Φ (v)) − Dα (Φ (u))

L 1E (I)

= ψ(Dα v) − ψ (Dα u)

L 1E (I)

< .

(3.23)

< MG ,

(3.24)

Using again Lemma 3.1 we deduce, from (3.18), (3.20) and (3.23),
≤ MG Dα (Φ (v)) − Dα (Φ (u))

Φ (v) (t) − Φ (u) (t)

L 1E (I)

Dα−1 (Φ (v)) (t) − Dα−1 (Φ (u)) (t) ≤ MG Γ (α) Dα (Φ (v)) − Dα (Φ (u))

L 1E (I)

< MG Γ (α) , (3.25)

for all t ∈ I. Combining (3.23) − (3.25) we obtain the continuity of Φ. Finally, for u ∈
Dα (u) ∈ F i x(S). So
ψ(Dα (u)) = Dα (u) ,

, we have

by the property of ψ. It follows that
1

G(t, s)ψ (Dα u) (s)ds

Φ(u)(t) =
0
1

G(t, s)Dα u(s)ds = u(t),

=
0

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 order α considered in the preceding section. Let E ≡ Rd be a finite dimensional space and Z be
a compact metric space. By +1 (Z) we denote the space of all probability Radon measures on Z. It
is well-known that +1 (Z) is a compact metrizable space for the vague topology. We also denote by
I; +1 (Z) the space of all Young measures defined on I endowed with the stable topology so
that
I; +1 (Z) is a compact metrizable space with respect to this topology. For the convenience
of the reader we recall that a sequence (νn ) in
I; +1 (Z) stably converges to ν ∈
I; +1 (Z)
if
1

lim

n→∞

1

h t (z)dνnt (z) =

dt
0

h t (z)dν t (z),

dt
0

Z

Z

for all h ∈ L 1 (Z) ([0, 1]), here (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.
10


Theorem 4.1. ([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(., un (.), vn (.)) is uniformly integrable. Then the following holds

h(ω, un (ω), v n (ω)) d P(ω) =

lim

n→∞

[

Ω

Ω

Z

h(ω, u∞ (ω), z) dν∞
ω (z)] d P(ω).

Let ∆ be a measurable multifunction defined on I with nonempty compact values in Z. We
consider the following control system
 α
α−1
 D u(t) = f t, u(t), D u(t), z(t) , a.e., t ∈ I,
1
(4.1)
β
 I u(t) t=0 = 0, u(1) = u (t) d t,
0

where z belongs to the set S∆ of all original controls, which mean that z is a Lebesgue-measurable
mapping from I to Z with z(t) ∈ ∆(t) for almost every t ∈ I, and the function f : I × E × E × Z → E

satisfying the following conditions
(B1 ) for every (x, y, z) ∈ E × E × Z, f (·, x, y, z) is Lebesgue-measurable on I,
(B2 ) for every fixed t ∈ I, f (t, ·, ·, ·) is continuous on E × E × Z,
(B3 ) there exist

1, 2

∈ LR1 (I) with MG

1

+

2 L1
R

< 1 such that

f (t, x 1 , y1 , z) − f (t, x 2 , y2 , z) ≤

1 (t)

x1 − x2 +

2 (t)

y1 − y2 ,

for all (t, x 1 , y1 , z), (t, x 2 , y2 , z) ∈ I × E × E × Z.
(B4 ) there is a positive function ρ ∈ LR1 (I) such that f (t, x, y, z) ∈ Λ(t) := ρ(t)B E , for all (t, x, y, z) ∈

I × E × E × Z.
α,1

By Corollary 3.4, for each z ∈ S∆ , the system (4.1) admits an unique solution uz ∈ WE
we have

(I). Morever

1

G(t, s)Dα uz (s)ds,

uz (t) =

(4.2)

0
t

D

α−1

α

uz (t) =

D uz (s)ds +
0

for all t ∈ I and

1

1
α−1

(1 − s)α − α(1 − s)α−1 Dα uz (s)ds,

(4.3)

0

Dα uz (t) = f (t, uz (t), Dα−1 uz (t), z(t)), a.e., t ∈ I.

(4.4)

Let L(t, x, y, z) be real-valued function defined on I ×E×E×Z. We consider the following Bolza-type
optimal control problem associated with the system (4.1)
1

minimize

L t, uz (t), Dα−1 uz (t), z(t) d t,

J(z) =
0

11

subject to z ∈ S∆ and the corresponding state trajectory uz . We would like to describe the relaxation
for above optimal control problem. For this purpose, we first introduce the following relaxed system
 α
α−1
 D u(t) = Z f t, u(t), D u(t), z ν t (dz), a.e., t ∈ I,
1
(4.5)
 I β u(t) t=0 = 0, u(1) = u (t) d t,
0

where ν belongs to the set
:= SΣ of all relaxed controls, which means that ν is a Lebesguemeasurable selection of the multifunction Σ defined by
1
+ (Z)

Σ(t) := σ ∈

: σ(∆(t)) = 1
α,1

for all t ∈ I. It is note that, for each ν ∈ , the existence of WE (I)-solutions for the relaxed
systems (4.5) again follows from Corollary 3.4, because the function
f (t, x, y, ν) =

f (t, x, y, z)ν(dz),
Z

(t, x, y, ν) ∈ I × E × E ×

1
+ (Z),

inherits the properties of the function f . Namly we have

Theorem 4.2. Let (B1 ) − (B4 ) hold. Then, for each ν ∈
α,1
(4.5) has a unique solution uν ∈ WE (I). Further,

, the fractional boundary value problem

1

G(t, s)Dα uν (s)ds,

uν (t) =

(4.6)

0

t

Dα uν (s)ds +

Dα−1 uν (t) =
0

1

1

(1 − s)α − α(1 − s)α−1 Dα uν (s)ds,

α−1

(4.7)

0

for all t ∈ I and
f (t, uν (t), Dα−1 uν (t), z)ν t (dz), a.e., t ∈ I.

Dα uν (t) =

(4.8)

Z
α,1

At this point we will state some properties of the WE (I)-solutions sets of the problems (4.5)
and (4.1).
α,1

Lemma 4.3. Let (B1 ) − (B4 ) hold. Then the set
of all WE (I)-solutions of the problem (4.5) is
α,1
compact with respect to the topology of uniform convergence. Morever the WE (I)-solutions set
of the problem (4.1) is dense in
with respect to this topology.

Proof. The proof consists two steps.
Step 1. Let (νn ) be a sequence in
≡ SΣ which stably converges to ν∞ ∈
n ∈ N ∪ {∞}, let uνn be the unique solution to the problem
 α
α−1
n
 D uνn (t) = Z f t, uνn (t), D uνn (t), z ν t (dz), a.e., t ∈ I,
 I β uνn (t)

1

t=0

= 0, uνn (1) =

uνn (t) d t.
0

12

, and, for each


It follows from Theorem 4.2 that
1

G(t, s)Dα uνn (s)ds,

uνn (t) =

(4.9)

0

t

D

α−1

α

uνn (t) =

D uνn (s)ds +
0

1

1

(1 − s)α − α(1 − s)α−1 Dα uνn (s)ds,

α−1

(4.10)

0

for all t ∈ I and
Dα uνn (t) =

f t, uνn (t), Dα−1 uνn (t), z νnt (dz) ∈ Λ(t), a.e., t ∈ I.

(4.11)

Z

Let t 1 , t 2 ∈ I, t 1 < t 2 . It follows from (4.9) − (4.11) that
1

uνn (t 1 ) − uνn (t 2 )

G(t 1 , s) − G(t 2 , s) ρ(s)ds

≤
0

≤

t1

1
Γ (α)
+

α−1

α−1

(t 1 − s)

− (t 2 − s)

t2

ρ(s)ds +

0

(t 2 − s)α−1

t1

t 1α−1 − t 2α−1
(α − 1)Γ (α)

Γ (α)

ρ(s)ds

1

(1 − s)α − α(1 − s)α−1 ρ(s)ds.

(4.12)

0

By using the inequality |a p − b p | ≤ |a − b| p , for all a, b ≥ 0 and p ∈ (0, 1], we deduce from (4.12)
that
uνn (t 1 ) − uνn (t 2 )

≤
≤

α−1

t1 − t2

t2

t1

ρ(s)ds +

ρ(s)ds +

Γ (α)

0

2α
(α − 1)Γ (α)

ρ

t1
LR1 (I)

t1 − t2

α−1

1

1+α
α−1

ρ(s)ds
0

.

(4.13)

The above inequality shows that uνn : n ∈ N is equicontinuous in C E (I). Further, for each t ∈ I,
the set uνn (t) : n ∈ N is relatively compact in E because it is included in the norm compact set
MG ρ L 1 (I) B E (recall that E is finite dimensional). So, by using Arzela-Ascoli theorem, uνn : n ∈ N
R

is relatively compact in C E (I). Hence, it’s not difficult to show that, by extractting subsequences,
α,1
uνn converges uniformly to a function u∞ ∈ WE (I) such that
1

u∞ (1) =

u∞ (s)ds,

I β u∞ (t)

0

t=0

= 0.

Morever the sequence Dα−1 un converges pointwisely to Dα−1 u∞ and Dα un weakly converges in
L 1E (I) to Dα u∞ .
α,1
Step 2. We claim that u∞ coincides with the WE (I)-solution uν∞ associated with ν∞ ∈ , that is,
 α
 D uν∞ (t) =

Z

 I β uν∞ (t)

= 0, uν∞ (1) =

t=0

f t, uν∞ (t), Dα−1 uν∞ (t), z ν∞
t (dz), a.e., t ∈ I,
1

uν∞ (t) d t.
0

13


Indeed, multiplying scalarly the equation
Dα uνn (t) =

f t, uνn (t), Dα−1 uνn (t), z νnt (dz)
Z

by w ∈ L ∞
E (I) and integrating on I we get
1

1

w(t), Dα uνn (t) d t =
0

f t, uνn (t), Dα−1 uνn (t), z νnt (dz) d t.

w(t),
0

(4.14)

Z

It is note that the fiber product δuνn ⊗ δ Dα−1 uνn ⊗ νn stably converges to δuν∞ ⊗ δ Dα−1 uν∞ ⊗ ν∞ and
the Carathéodory integrand f w (t, x, y, z) = w(t), f (t, x, y, z) satisfies the following estimate

f w (t, x, y, z) =

w(t), f (t, x, y, z)

≤ ρ(t) w(t) ,

with t → ρ(t) w(t) ∈ LR1 + (I). Applying the Theorem 4.1 we deduce from (4.14) that
1

lim

n→∞

1
α

f w t, uνn (t), D uνn (t), z

dt
0

νnt (dz)

=
0

Z

f w t, u∞ (t), Dα u∞ (t), z ν∞
(4.15)

t (dz).

dt
Z

Whence passing to the limit in when n goes to ∞ in the equality (4.14) we get
1

1

w(t), Dα u∞ (t) d t =
0

0

f t, u∞ (t), Dα−1 u∞ (t), z ν∞
t (dz) d t,

w(t),

(4.16)

Z

by using (4.15) and by Dα uνn weakly converges to Dα u∞ in L 1E (I). This implies that
Dα u∞ (t) =

f t, u∞ (t), Dα−1 u∞ (t), z ν∞
t (dz),
Z

for almost every t ∈ I. Finally, by the uniqueness of the solutions we conclude that u∞ = uν∞ . This
proves first part of the theorem, while the second part follows by the continuity and the density (S∆
is dense in SΣ with respect to the stable topology ([6], Lemma 7.1.1)).
Now we define the cost function J(ν) by
1

L t, uν (t), Dα−1 uν (t), z ν t (dz),

J(ν) =
0

Z

where as usual uν (t) is the solution of (4.5). Then we have the following theorem
Theorem 4.4. Assume that E is a finite dimensional space. Let L : I × E × E × Z → R is a Carathéodory
integrand (that is, L(t, ·, ·, ·) is continuous on E × E × Z for every t ∈ I and I(·, x, y, z) is Lebesguemeasurable on I, for every (x, y, z) ∈ E × E × Z ) which satisfies the condition: there exists a positive
Lebesgue-integrable function ζ defined on I such that
L(t, x, y, z) ≤ ζ(t), ∀(t, x, y, z) ∈ I × E × E × Z,
for all (t, x, y, z) ∈ I × E × E × Z. Then we have infz∈S∆ J(z) = minν∈ J(ν).

14


Proof. We are given ν ∈ SΣ . By S∆ is dense in SΣ with respect to the stable topology, there exists a
sequence (zn ) in S∆ such that the associated Young measures δzn converges stably to ν. In view
of Lemma 4.3, the sequence uzn converges uniformly to uν , Dα−1 uzn converges pointwisely to
Dα−1 uν and Dα uzn converges weakly to Dα uν in L 1E (I), where uzn is the unique solution of the
problem
 α

α−1
 D uzn (t) = f t, uzn (t), D uzn (t), zn (t) , a.e., t ∈ I,
β
 I uzn (t)

1

t=0

and uν is the unique solution of
 α
 D uν (t) =
β

 I uν (t)

t=0

Z

= 0, uzn (1) =

uzn (t) d t,

0

f t, uν (t), Dα−1 uν (t), z ν t (dz), a.e., t ∈ I,
1

= 0, uν (1) =

uν (t) d t.
0

Using the fiber product for the stable convergence in the space of Young measures, see ([6], Theorem
3.3.1), we conclude that δuz ⊗δ Dα−1 uz ⊗δzn stably converges to δuν ⊗δ Dα−1 uν ⊗ν. By the assumptions
n

n

of L, the sequence of functions t → L(t, uνn (t), Dα−1 uνn (t), ), n ∈ N is integrable, so that, in view
of Theorem 4.1, we get
1

1

L t, uzn (t), Dα−1 uzn (t), zn (t) d t =

lim

n→∞

L t, uν (t), Dα−1 uν (t), z ν t (dz).

dt
0

0

Z

This shows that
1

1
0

L t, uz (t), Dα−1 uz (t), z(t) d t = inf J(z),

L t, uν (t), Dα−1 uν (t), z ν t (dz) ≥ inf

dt

z∈S∆

Z

as

z∈S∆

0

1

L t, uzn (t), Dα−1 uzn (t), zn (t) 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 ν ∈ SΣ such that δu j converges stably to ν. As before we have
1
z∈S∆

j→∞

1

L t, uz j (t), Dα−1 uz j (t), z j (t) d t =

inf J(z) = lim

0

L t, uν (t), Dα−1 uν (t), z ν t (dz) = J(ν).

dt
0

Z

Therefore ν is a minimizer for J. The proof of Theorem is completed.

15


Acknowledgements
The authors wish to express their sincere thanks to the support given by Vietnam’s National
Foundation for Science and Technology Development (NAFOSTED) under Project 101.01-2012.12.
The authors are also grateful for the support given by Vietnam Institute for Advanced Study in
Mathematics.

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17