SECOND-ORDER VIABILITY RESULT IN BANACH SPACES

SECOND-ORDER VIABILITY RESULT IN BANACH SPACES

Myelkebir Aitalioubrahim and Said Sajid University Hassan II-Mohammedia

U.F.R Mathematics and Applications F.S.T, BP 146, Mohammedia, Morocco e-mail: saidsajid@hotmail.com, aitalifr@yahoo.fr

Abstract

We show the existence result of viable solutions to the second-order differential inclusion

¨

x(t) ∈ F (t, x(t), ˙x(t)),

x(0) = x

⁰

, ˙x(0) = y

⁰

, x(t) ∈ K on [0, T ],

where K is a closed subset of a separable Banach space E and F (·, ·, ·) is a closed multifunction, integrably bounded, measurable with respect to the first argument and Lipschitz continuous with respect to the third argument.

Keywords: differential inclusion,viability, measurability, selection.

2000 Mathematics Subject Classification: 34A60.

1. Introduction

The aim of this paper is to establish the existence of local solutions of the second-order viability problem:



 

 

¨

x(t) ∈ F (t, x(t), ˙x(t)) a.e.;

(x(0), ˙x(0)) = (x

₀

, y

₀

);

(x(t), ˙x(t)) ∈ K × Ω;

(1.1)

where K (resp. Ω) is a closed subset (resp. an open subset) of a separable

Banach space E and F : [0, 1] × K × Ω → 2

^E

is a measurable multifunction

with respect to the first argument and Lipschitz continuous with respect to the third argument.

Second-order viability problem was first introduced by Cornet and Had- dad [5]. The purpose was to study a problem of type:



 

 

¨

x(t) ∈ Q(x(t), ˙x(t)) a.e;

(x(0), ˙x(0)) = (x

0

, y

₀

);

(x(t), ˙x(t)) ∈ Gr(T

K

), K ⊂ R

ⁿ

; (1.2)

where Q is a convex multifunction and Gr(T

_K

) is the graph of the multi- function x 7→ T

K

(x): the contingent cone at x. In order to obtain solutions of the problem (1.2), Cornet and Haddad imposed stronger conditions on the viability set K and the tangent vector y

₀

, namely K = L∩M, y

₀

belongs to T

L

(x

0

) ∩ T I

M

(x

0

) and Gr(T

K

) is locally compact. Here T I

M

(x

0

) is the interior tangent cone introduced by Dubovitskij and Muljitin [6].

Another line of research appeared later, it considers the second order viability problems without convexity. Two kinds of problems are studied in this topic. In the first one, the space of state constraints is finite-dimensional (see [3, 9, 10]). In this work, the right hand-side is upper semi-continuous and cyclically monotone. Proof technics are based on Ascoli’s theorem and the basic relation

d

dt V (x(t)) = k ˙x(t)k

²

,

where V is a convex and lower semi-continuous function whose subdifferen- tial ∂V contains the right-hand side. The non-convex case in Hilbert space has been studied by Morchadi and Sajid [11]. The authors proved the exis- tence of solution for a second-order viability problem without convexity and compactness of the right-hand side of the inclusion. However, the viability set is convex. Their approach is based on the Baire category theorem. See also the work of Ibrahim and Alkulaibi [2].

First-order viability problem with the non-convex Carath´eodory Lips- chitzean right-hand side in Banach space has been studied recently by Duc Ha [7]. The author establishes a multi-valued version of Larrieu’s work [8], assuming the following tangential condition:

lim inf

h7→0⁺

1 h e x +

Z

t+h t

F (s, x)ds, K

!

= 0,

where K is the viability set and e(., .) denotes the Hausdorff’s excess.

In this paper we extend this result to the second-order case with the following tangential condition:

(1.3) lim inf

h7→0⁺

1

h

²

e x + hy + h 2

Z

t+h t

F (s, x, y)ds, K

!

= 0.

The case deserves mentioning: when F is compact values in finite-dimensional space and does not depend on time s. The condition (1.3) becomes

lim inf

h7→0⁺

1

h

²

e x + hy + h

²

2 F (x, y), K

!

= 0, which is equivalent to the following relation:

∀v ∈ F (x, y) lim inf

h7→0⁺

1

h

²

d

_K

x + hy + h

²

2 v

!

= 0,

i.e., F (x, y) is contained in the set second-order tangent of K at (x, y) in- troduced by Ben-Tal and defined by:

A

_K

(x, y) = (

z ∈ E : lim inf

h7→0⁺

1

h² 2

d

_K

x + hy + h

²

2 z

!

= 0 )

.

This condition was used in [1, 9, 10].

2. Preliminaries and statement of the main result

Let E be a separable Banach space with the norm k · k. For measurability

purposes, E (resp. Ω ⊂ E) is endowed with the σ-algebra B(E) (resp. B(Ω))

of Borel subsets for the strong topology and [0, 1] is endowed with Lebesgue

measure and the σ-algebra of Lebesgue measurable subsets. For x ∈ E and

r > 0 let B(x, r) := {y ∈ E; k y−x k< r} be the open ball centered at x with

radius r and ¯ B(x, r) be its closure and put B = B(0, 1). For x ∈ E and for

nonempty sets A, B of E we denote d

_A

(x) or d(x, A) := inf{k y−x k; y ∈ A},

e(A, B) := sup{d

B

(x); x ∈ A} and H(A, B) := max{e(A, B), e(B, A)}. A

multifunction is said to be measurable if its graph is measurable. For more

details on measurability theory, we refer the reader to the book of Castaing-

Valadier [4].

Let us recall the following Lemmas that will be used in the paper. For the proofs, we refer the reader to [12].

Lemma 2.1. Let Ω be a nonempty set in E. Assume that F : [a, b]×Ω → 2

^E

is a multifunction with nonempty closed values satisfying:

• For every x ∈ Ω, F (·, x) is measurable on [a, b];

• For every t ∈ [a, b], F (t, ·) is (Hausdorff) continuous on Ω.

Then for any measurable function x(·) : [a, b] → Ω, the multifunction F (·, x(·)) is measurable on [a, b].

Lemma 2.2. Let G : [a, b] → 2

^E

be a measurable multifunction and y(·) : [a, b] → E a measurable function. Then for any positive measurable function r (·) : [a, b] → R

⁺

, there exists a measurable selection g(·) of G such that for almost all t ∈ [a, b]

kg(t) − y(t)k ≤ d y(t), G(t)
+ r(t).

Assume that the following hypotheses hold:

(H1) K is a nonempty closed subset in E and Ω is a nonempty open subset in E;

(H2) F : [0, 1] × K × Ω → 2

^E

is a set valued map with nonempty closed values satisfying

(i) For each (x, y) ∈ K × Ω, t 7→ F (t, x, y) is measurable;

(ii) There exists a function m ∈ L

¹

([0, 1], R

⁺

) such that for all t ∈ [0, 1] and for all (x

1

, y

₁

), (x

2

, y

₂

) ∈ K × Ω

H 

F (t, x

1

, y

₁

), F (t, x

2

, y

₂

) 

≤ m(t)ky

1

− y

2

k;

(iii) For each bounded subset S of K × Ω, there exists a function g

_S

∈ L

¹

([0, 1], R

⁺

) such that for all t ∈ [0, 1] and for all (x, y) ∈ S

kF (t, x, y)k := sup

z∈F(t,x,y)

kzk ≤ g

S

(t);

(H3) (Tangential condition) For every (t, x, y) fixed in [0, 1] × K × Ω, lim inf

h7→0⁺

1

h

²

e x + hy + h 2

Z

t+h t

F (s, x, y)ds, K

!

= 0.

For any (x

₀

, y

₀

) ∈ K × Ω, consider the problem:



 

 

¨

x(t) ∈ F (t, x(t), ˙x(t)) a.e;

x(0) = x

0

, ˙x(0) = y

0

; x(t) ∈ K.

(2.4)

Theorem 2.3. If assumptions (H1), (H2) and (H3) are satisfied, then there exist T > 0 and an absolutely continuous function x(·) : [0, T ] → E, for which ˙x(·) : [0, T ] → E is also absolutely continuous such that x(·) is a solution of (2.4).

3. Proof of the main result

Let r > 0 such that ¯ B(y

0

, r) ⊂ Ω and g ∈ L

¹

([0, 1], R

⁺

) such that

(3.1) k F (t, x, y) k ≤ g(t) ∀(t, x, y) ∈ [0, 1] × (K ∩ B(x

₀

, r)) × ¯ B(y

₀

, r).

Let T

1

> 0 and T

2

> 0 such that (3.2)

Z

T1

0

m(t)dt < 1 and Z

T2

0

 g(t) + r+ k y

₀

k +1  dt < r

2 . For ε > 0 there exists η(ε) > 0 such that

(3.3)

    

Z

t2

t1

g(τ )dτ     

< ε whenever |t

₁

− t

₂

| < η(ε).

Set

(3.4) T = min n

T

₁

, T

₂

, 1 o

and α = min (

T, 1 2 η( ε

4 ), ε 4

) .

We will use the following approximation Lemma to prove the main result.

Lemma 3.1. If assumptions (H1)–(H3) are satisfied, then for all ε > 0 and for all y ∈ L

¹

([0, T ], E), there exist f ∈ L

¹

([0, T ], E), z : [0, T ] → E differentiable and a step function θ : [0, T ] → [0, T ] such that

• f (t) ∈ F t, z(θ(t)), ˙z(θ(t))
for all t ∈ [0, T ];

•

f (t) − y(t) ≤ d 

y(t), F t, z(θ(t)), ˙z(θ(t)


+ ε for all t ∈ [0, T ];

•

˙ z(t) − y

₀

− R

t

0

f (τ )dτ

≤ ε for all t ∈ [0, T ].

P roof. By (H3), for (0, x

₀

, y

₀

), we have lim inf

h7→0⁺

1

h

²

e x

₀

+ hy

₀

+ h 2

Z

h 0

F (s, x

₀

, y

₀

)ds, K

!

= 0.

Hence, there exists 0 < h ≤ α such that e x

₀

+ hy

₀

+ h

2 Z

h

0

F (s, x

₀

, y

₀

)ds, K

!

< αh

²

4 . Put

h

₀

:= max (

h ∈]0, α] : e x

₀

+ hy

₀

+ h 2

Z

h 0

F (s, x

₀

, y

₀

)ds, K

!

≤ αh

²

4

) .

In view of Lemma 2.2, there exists a function f

0

∈ L

¹

([0, h

0

], E) such that for all t ∈ [0, h

₀

], one has

f

₀

(t) ∈ F (t, x

₀

, y

₀

) and kf

₀

(t) − y(t)k ≤ d y(t), F (t, x

₀

, y

₀

)
+ ε.

Moreover, it is clear that

d

_K

x

₀

+ h

₀

y

₀

+ h

₀

2

Z

h0

0

f

₀

(s)ds

!

≤ αh

²₀

4 . So, there exists x

₁

∈ K such that

2 h

²₀

x

₁

− x

₀

+ h

0

y

₀

+ h

₀

2

Z

h0

0

f

₀

(s)ds

!

≤ 2

h

²₀

d

_K

x

₀

+ h

₀

y

₀

+ h

₀

2

Z

h0

0

f

₀

(s)ds

! + α

4 ,

hence

x

₁

− x

₀

− h

₀

y

₀

h²₀ 2

− 1 h

₀

Z

h0

0

f

₀

(s)ds

< α.

Set

u

₀

= x

₁

− x

₀

− h

₀

y

₀

h²₀ 2

,

then

(3.5) x

₁

= x

₀

+ h

₀

y

₀

+ h

²₀

2 u

₀

∈ K and u

₀

∈ 1 h

₀

Z

h0

0

f

₀

(s)ds + αB.

Put y

₁

= y

₀

+ h

₀

u

₀

. Since f

₀

(t) ∈ F (t, x

₀

, y

₀

) for all t ∈ [0, h

₀

], by (3.1), (3.2) and (3.4), we have

kx

1

− x

0

k =

h

₀

y

₀

+ h

²₀

2 u

₀

≤ h

0

ky

0

k + h

₀

2

Z

h0

0

g(s)ds + h

²₀

2 α

≤ h

₀

ky

₀

k + Z

h0

0

g(s)ds + h

₀

= Z

h0

0

g(s) + ky

₀

k + 1
ds < r 2 . Then x

₁

∈ B(x

₀

, r). Also by (3.5), we have

ky

₁

− y

₀

k = kh

₀

u

₀

k ≤ Z

h0

0

g(s)ds + h

₀

α ≤ Z

h0

0

g(s) + 1
ds < r 2 . Then y

₁

∈ B(y

₀

, r).

We reiterate this process for constructing sequences h

_q

, t

_q

, x

_q

, y

_q

, f

_q

and u

_q

satisfying for some rank m ≥ 1 the following assertions:

(a) For all q ∈ {0, . . . , m − 1}

h

_q

:= max (

h ∈]0, α] : e x

_q

+ hy

_q

+ h 2

Z

tq+1

tq

F (s, x

_q

, y

_q

)ds, K

!

≤ αh

²

4

)

(b) t

₀

= 0, t

_m−1

< T ≤ t

m

with t

q

= P

q−1

j=0

h

j

for all q ∈ {1, . . . , m};

(c) For all q ∈ {1, . . . , m}

x

q

= x

₀

+

q−1

X

j=0

h

j

y

j

+

q−1

X

j=0

h

²_j

2 u

j

!

∈ K ∩ B(x

₀

, r)

and

y

q

= y

₀

+

q−1

X

j=0

h

j

u

j

!

∈ B(y

₀

, r);

(d) For all q ∈ {0, . . . , m − 1}, for every t ∈ [t

q

, t

_q+1

] u

q

∈ 1

h

_q

Z

tq+1

tq

f

q

(s)ds + αB, f

q

(t) ∈ F (t, x

q

, y

q

) and

kf

q

(t) − y(t)k ≤ d 

y(t), F (t, x

q

, y

q

)  + ε;

(e) For all q ∈ {0, . . . , m − 1}

y

_q+1

− y

q

− Z

tq+1

tq

f

_q

(t)dt

< h

_q

α.

It is easy to see that for q = 1, the assertions (a)–(e) are fulfilled. Let now q ≥ 2. Assume that (a)–(e) are satisfied for any j = 1, . . . , q. If T ≤ t

_q+1

, then we take m = q + 1 and so the process of iterations is stopped and we get (a)–(e) satisfied with t

_m−1

< T ≤ t

m

. In the other case: t

_q+1

< T, we define x

_q+1

and y

_q+1

as follows

x

_q+1

:= x

q

+ h

q

y

_q

+ h

²_q

2 u

_q

= x

₀

+

q

X

j=0

h

_j

y

_j

+

q

X

j=0

h

²_j

2 u

_j

∈ K, and

y

_q+1

:= y

q

+ h

q

u

_q

= y

₀

+

q

X

j=0

h

_j

u

_j

. By (3.1), (3.2) and (3.4), we have

kx

_q+1

− x

₀

k ≤

q

X

j=0

h

_j

ky

j

k +

q

X

j=0

h

²_j

2 ku

j

k ≤

q

X

j=0

h

_j

ky

j

− y

₀

k + h

j

ky

₀

k + kh

j

u

_j

k

≤

q

X

j=0

h

_j

r + ky

0

k
+

q

X

j=0

Z

tj+1

tj

kf

j

(t)kdt + αh

j

≤ Z

tq+1

0

g(t) + r + ky

₀

k + 1
dt < r,

which ensures that x

_q+1

∈ K ∩ B(x

₀

, r). Also we have

ky

_q+1

− y

₀

k ≤

q

X

j=0

kh

_j

u

_j

k ≤

q

X

j=0

Z

t_j+1 tj

kf

_j

(t)kdt + αh

_j

≤ Z

t_q+1

0

g(t) + 1
dt < r,

so that y

_q+1

∈ B(y

₀

, r). Thus the conditions (a)–(e) are satisfied for q + 1.

Now, we have to prove that this iterative process is finite, i.e., there exists a positive integer m such that t

_m−1

< T ≤ t

_m

. Suppose the contrary:

t

_q

≤ T for all q ≥ 1. Then the bounded increasing sequence (t

q

)

q

converges to some ¯ t such that ¯ t ≤ T . By (c) and (d), for q > p we get

kx

_q

− x

_p

k ≤

q−1

X

j=p

h

_j

ky

_j

k +

q−1

X

j=p

h

²_j

2 ku

_j

k

≤

q−1

X

j=p

h

_j

ky

j

− y

0

k + ky

0

k
+

q−1

X

j=p

h

²_j

2

1 h

_j

Z

tj+1

t_j

kf

j

(s)kds + α

q−1

X

j=p

h

²_j

2

≤ r + ky

₀

k + α

q−1

X

j=p

h

_j

+

q−1

X

j=p

Z

tj+1

t_j

g(s)ds

≤ Z

tq

tp

g(t)dt + (t

_q

− t

_p

) k y

₀

k +r + 1
,

and

ky

_q

− y

_p

k ≤

q−1

X

j=p

h

_j

ku

_j

k ≤

q−1

X

j=p

h

_j

1 h

_j

Z

tj+1

tj

kf

_j

(s)kds + α

q−1

X

j=p

h

_j

≤

q−1

X

j=p

Z

t_j+1 tj

g(s)ds + α

q−1

X

j=p

h

j

≤ Z

tq

tp

g(t)dt + t

q

− t

p

.

The last terms of the above tow inequalities converge to 0 when p, q → +∞, then (x

q

)

q

and (y

q

)

q

are Cauchy sequences, and hence they converge to some

¯

x ∈ K and ¯ y ∈ ¯ B (y

₀

, r) respectively. As (t, ¯ x, y) ∈ [0, T ] × K × Ω, by (H3), ¯

there exist h ∈]0, α] and an integer q

0

≥ 1 such that for all q ≥ q

0



 

 

 

 

 

 

 



 

 

 

 

 

 

 

 e



x + h¯ y +

^h₂

R

¯t+h

t¯

F (s, ¯ x, y)ds, K ¯



≤

^h₃₀^2α

; kx

q

− xk ≤

^h₃₀^2α

;

ky

_q

− yk ≤

^hα₃₀

;

¯ t − t

_q

< min{η(

^2hα₃₀

), h};

k y

_q

− ¯ y k R

¯t+h

¯t

m(t)dt ≤

^2hα₃₀

. (3.6)

Let q > q

0

be given. For an arbitrary measurable selection φ

q

of F (t, x

q

, y

_q

) on [0, ¯ t + h], there exists a measurable selection φ of F (t, ¯ x, y) on [0, ¯ ¯ t + h]

such that

(3.7) kφ

q

(t) − φ(t)k ≤ d 

φ

q

(t), F (t, ¯ x, y) ¯  + 2α

30 ≤ m(t)ky

q

− ¯ yk + 2α 30 . Relations (3.6) and (3.7) ensure

d

_K

x

_q

+ hy

q

+ h 2

Z

tq+h tq

φ

_q

(s)ds

!

≤ kx

q

− xk + hky

q

− yk + h 2

Z

t¯ tq

kφ

q

(s)kds + d

K

x + h¯ y + h 2

Z

¯t+h

¯t

φ(s)ds

!

+ h 2

Z

tq+h

¯t

kφ

q

(s) − φ(s)kds + h 2

Z

¯t+h tq+h

kφ(s)kds

≤ kx

_q

− xk + hky

_q

− yk + h 2

Z

t¯ tq

g(s)ds + d

_K

x + h¯ y + h 2

Z

¯t+h

¯t

φ(s)ds

!

+ h 2

Z

¯t+h

¯t

m(s) k y

_q

− ¯ y k ds + h

²

α 30 + h

2 Z

¯t+h

tq+h

g(s)ds

≤ h

²

α

30 + h

²

α

30 + h

²

α 30 + h

²

α

30 + h

²

α

30 + h

²

α 30 + h

²

α

30 < h

²

α

4 .

Since φ

q

is an arbitrary measurable selection of F (t, x

q

, y

q

) on [0, ¯ t + h], it follows that

e x

q

+ hy

q

+ h 2

Z

tq+h tq

F (t, x

q

, y

q

)ds, K

!

≤ h

²

α 4 .

On the other hand, by (3.6), we have

t

_q+1

≤ t < t

q

+ h ≤ T, and hence h > t

_q+1

− t

q

= h

q

. Finally h > h

q

, 0 < h ≤ α and

e x

_q

+ hy

q

+ h 2

Z

tq+h tq

F (t, x

q

, y

_q

)ds, K

!

≤ h

²

α 4 .

This contradicts the maximality of h

q

. Therefore, there exists an integer m ≥ 1 such that t

_m−1

< T ≤ t

m

and for which assertions (a)–(e) are fulfilled.

Now, we take t

_m

= T and we define the function θ : [0, T ] → [0, T ], z : [0, T ] → E and f ∈ L

¹

([0, T ], E) by setting for all t ∈ [t

q

, t

_q+1

[

θ(t) = t

q

, f (t) = f

q

(t), z(t) = x

q

+ (t − t

q

)y

q

+ (t − t

q

)

²

2 u

_q

.

Claim 3.2. For all q ∈ {0, . . . , m} we have

y

_q

− y

₀

− Z

tq

0

f (s)ds

≤ αt

_q

.

P roof. Obviously, for q = 0 the above assertion is fulfilled. By induction, assume that

y

_j

− y

₀

− Z

tj

0

f (s)ds

≤ αt

_j

.

For any j = 1, . . . , q − 1. By (d) we have

y

q

− y

₀

− Z

tq

0

f (s)ds

=

=

y

_q−1

− y

₀

− Z

t_q−1

0

f (s)ds + h

_q−1

u

_q−1

− Z

tq

t_q−1

f (s)ds

≤

y

_q−1

− y

₀

− Z

t_q−1

0

f (s)ds

+

h

_q−1

u

_q−1

− Z

tq

t_q−1

f (s)ds

≤ αt

_q−1

+ αh

_q−1

= αt

_q−1

+ αt

q

− αt

_q−1

= αt

q

.

Now let t ∈ [t

_q

, t

_q+1

], then by Claim 3.2, relations (3.1), (3.4) and (d), we have

˙z(t) − y

0

− Z

t

0

f(s)ds

=

y

_q

− y

0

− Z

tq

0

f (s)ds + (t − t

q

)u

q

− Z

t

tq

f (s)ds

≤

y

q

− y

₀

− Z

tq

0

f (s)ds

+ kh

q

u

q

k + Z

tq+1

tq

g(s)ds

≤ αt

_q

+ 2 Z

tq+1

tq

g(s)ds + αh

_q

≤ ε 4 + ε

2 + ε 4 = ε.

The proof of Lemma 3.1 is complete.

Proof of Theorem 2.3. Let (ε

_n

)

_n≥1

be a strictly decreasing sequence of positive scalars such that P

∞

n=1

ε

_n

< ∞. In view of Lemma 3.1, we can define inductively sequences (f

n

(·))

_n≥1

⊂ L

¹

([0, T ], E), (z

n

(·))

_n≥1

⊂ C

¹

([0, T ], E) and (θ

_n

(·))

_n≥1

⊂ S([0, T ], [0, T ]), where S([0, T ], [0, T ]) denotes the space of step functions from [0, T ] into [0, T ] such that

(1) f

_n

(t) ∈ F t, z

_n

(θ

_n

(t)), ˙z

_n

(θ

_n

(t))
for all t ∈ [0, T ];

(2)

f

_n+1

(t) − f

n

(t) ≤ d 

f

n

(t), F t, z

_n+1

(θ

_n+1

(t)), ˙z

_n+1

(θ

_n+1

(t))
 + ε

_n+1

for all t ∈ [0, T ];

(3)

˙ z

n

(t) − y

₀

− R

t

0

f

n

(τ )dτ

≤ ε

n

for all t ∈ [0, T ].

By (1) and (2) we have f

_n+1

(t) − f

n

(t)

≤ H 

F t, z

_n

(θ

_n

(t)), ˙z

_n

(θ

_n

(t))
, F t, z

n+1

(θ

_n+1

(t)), ˙z

_n+1

(θ

_n+1

(t))


+ ε

_n+1

≤ m(t)

˙z

n

(θ

n

(t)) − ˙z

_n+1

(θ

_n+1

(t))

+ ε

_n+1

≤ m(t) 

˙z

n

(θ

n

(t))− ˙z

n

(t) +

˙z

n

(t)− ˙z

_n+1

(t) +

˙z

_n+1

(t) − ˙z

_n+1

(θ

_n+1

(t))

 + ε

_n+1

.

On the other hand, for t ∈ [t

q

, t

_q+1

[ we have

˙z

n

(t) − ˙z

n

(θ

n

(t)) =

˙z

n

(t) − ˙z

n

(t

q

)

≤

˙z

n

(t) − y

₀

− Z

t

0

f

_n

(s)ds

+

y

_q

− y

₀

− Z

tq

0

f

_n

(s)ds

+ Z

t

tq

kf

n

(s)kds

≤ ε

n

+ αt

q

+ Z

t

tq

g(s)ds ≤ ε

n

+ ε

n

4 + ε

n

4 ,

hence

˙z

n

(t) − ˙z

n

(θ

n

(t))

≤ 2ε

n

. (3.8)

Thus, we get

f

_n+1

(t) − f

n

(t)

≤ m(t) 

2ε

n

+ 2ε

n+1

+ k ˙z

n

− ˙z

n+1

k

∞



+ ε

n+1

≤ m(t) 

4ε

n

+ k ˙z

n

− ˙z

_n+1

k

∞



+ ε

_n+1

. (3.9)

Relations (3.2) and (3.9) yield

k ˙z

_n+1

(t) − ˙z

_n

(t) k

≤

˙z

_n+1

(t) − y

₀

− Z

t

0

f

_n+1

(s)ds

+

˙z

_n

(t) − y

₀

− Z

t

0

f

_n

(s)ds +

Z

t 0

k f

n+1

(s) − f

n

(s) k ds

≤ ε

_n+1

+ ε

n

+ Z

t

0

m(s) k ˙z

n

(·) − ˙z

_n+1

(·) k

∞

+4ε

n

ds + tε

n

≤ 7ε

n

+ k ˙z

n

(·) − ˙z

_n+1

(·) k

∞

Z

T 0

m(s)ds.

Then

k ˙z

n

(·) − ˙z

n+1

(·) k

∞

≤ 7ε

n

1 − L , (3.10)

where L = R

T

0

m(s)ds. Hence for, n < m, it follows that k ˙z

_m

(·) − ˙z

_n

(·) k

∞

≤ 7

1 − L

m−1

X

i=n

ε

_i

.

Thus the sequence ( ˙z

n

(·))

n≥1

converges uniformly on [0, T ] to a function y(·).

On the other hand, observe that ˙z

n

(θ

n

(t)) converges uniformly to y(t) on [0, T ]. Indeed, by (3.8) since

k ˙z

n

(θ

n

(t)) − y(t) k ≤ k ˙z

n

(t) − ˙z

n

(θ

n

(t)) k + k ˙z

n

(t) − y(t) k then ( ˙z

n

(·)) converges uniformly to y(·).

Thus

(3.11) ˙z

n

(θ

n

(·)) converges uniformly to y(·) on [0, T ].

By construction, we have ˙z

n

(θ

n

(t)) ∈ B(y

₀

, r) for every t ∈ [0, T ], then y(t) ∈ Ω for all t ∈ [0, T ].

Now we return to relation (3.9). By relation (3.10) we have

f

_n+1

(t) − f

n

(t)

≤ m(t) 

4ε

n

+ k ˙z

n

(·) − ˙z

_n+1

(·) k

∞

 + ε

n

≤

 m(t) 

4 + 7 1 − L

 + 1

 ε

_n

.

This implies (as above) that (f

_n

(t))

_n≥1

is a Cauchy sequence and (f

_n

(·))

_n≥1

converges point-wisely to f (·). Further, since kf

n

(t)k ≤ g(t), by (3) and by Lebesgue dominated convergence theorem, we have

y(t) = lim

n→∞

˙z

n

(t) = lim

n→∞

y

₀

+ Z

t

0

f

n

(s)ds

!

= y

₀

+ Z

t

0

f (s)ds.

Hence ˙y(t) = f (t). Since for n < m

k ˙z

m

(·) − ˙z

n

(·)k

∞

≤ 7 1 − L

m−1

X

i=n

ε

_i

,

then we have for all t ∈ [0, T ]

kz

m

(t) − z

n

(t)k ≤ Z

t

0

k ˙z

m

(s) − ˙z

n

(s)kds ≤ 7 1 − L

m−1

X

i=n

ε

_i

.

Thus the sequence (z

_n

(·))

_n≥1

converges uniformly on [0, T ] to a function x(·). Also the relation

z

_n

(t) = x

₀

+ Z

t

0

˙z

n

(s)ds yields

x(t) = x

₀

+ Z

t

0

y(s)ds.

Therefore, ˙x(t) = y(t) for all t ∈ [0, T ].

On the other hand, observe that z

n

(θ

n

(t)) converges uniformly to x(t) on [0, T ]. Indeed, for t ∈ [t

q

, t

_q+1

[ we have

kz

n

(t) − z

n

(θ

n

(t))k ≤ Z

t

θn(t)

k ˙z

n

(s)kds.

Since θ

n

(t) converges to t and ˙z

n

(·) is bounded, it follows that kz

_n

(t) − z

_n

(θ

_n

(t))k converges to 0 as n → ∞. Since

kz

_n

(θ

_n

(t)) − x(t)k ≤ kz

_n

(t) − z

_n

(θ

_n

(t))k + kz

_n

(t) − x(t)k

and (z

_n

(·)) converges uniformly to x(·), then z

_n

(θ

_n

(t)) converges uniformly to x(t) on [0, T ]. By construction, we have z

n

(θ

n

(t)) ∈ K for every t ∈ [0, T ] and K is closed, then x(t) ∈ K for all t ∈ [0, T ].

Finally, observe that by (1), d 

f (t), F (t, x(t), ˙x(t)) 

≤ H 

F t, z

_n

(θ

n

(t)), ˙z

n

(θ

n

(t))
, F t, x(t), ˙x(t)
 + kf (t) − f

_n

(t)k ≤ kf (t) − f

_n

(t)k + m(t)k ˙z

_n

(θ

_n

(t)) − ˙x(t)k.

Since f

n

(t) converges to f (t) and by (3.11), the last term converges to 0. So that ¨ x(t) = f (t) ∈ F (t, x(t), ˙x(t)) a.e. in [0, T ]. Hence the proof is complete.

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Received 25 March 2007