Научная статья на тему 'WEAK MAXIMAL AND MINIMAL SOLUTIONS FOR HAMMERSTEIN AND URYSOHN INTEGRAL EQUATIONS IN REFLEXIVE BANACH SPACES'

WEAK MAXIMAL AND MINIMAL SOLUTIONS FOR HAMMERSTEIN AND URYSOHN INTEGRAL EQUATIONS IN REFLEXIVE BANACH SPACES Текст научной статьи по специальности «Математика»

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Текст научной работы на тему «WEAK MAXIMAL AND MINIMAL SOLUTIONS FOR HAMMERSTEIN AND URYSOHN INTEGRAL EQUATIONS IN REFLEXIVE BANACH SPACES»

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DIFFERENTIAL EQUATIONS AND

CONTROL PROCESSES N4, 2008 Electronic Journal, reg. N P2375 at 07.03.97 ISSN 1817-2172

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http://www.neva.ru/journal http://www.math.spbu.ru/diffjournal/ e-mail: [email protected]

Weak solutions of a fractional-order nonlocal boundary value problem in reflexive Banach spaces

El-Sayed A. M. A. & Abd El-Salam Sh. A.

E-mail addresses: [email protected] & [email protected] Faculty of Science, Alexandria University, Alexandria, Egypt

Abstract In this work, we establish an existence result (based on O'Regan fixed point theorem) for a nonlinear fractional-order nonlocal boundary value problem.

Keywords: Fractional calculus; Nonlocal boundary value problem. 1 Preliminaries and Introduction

Let L\(I) be the space of Lebesgue integrable functions on the interval I = [0,1]. Unless otherwise stated, E is a reflexive Banach space with norm ||.|| and dual E*. We will denote by Ew the space E endowed with the weak topology a(E,E*) and denote by C[I, E] the Banach space of strongly continuous functions u : I ^ E with sup-norm ||.||0.

We recall that the fractional integral operator of order /3 > 0 with left-hand point a is defined by (see [4], [9], [10] and [15])

We recall the following definitions. Let E be a Banach space and let u : I ^ E. Then

(1) u(.) is said to be weakly continuous (measurable) at to E I if for every y E E* we have y(u(.)) continuous (measurable) at t0.

(2) A function h : E ^ E is said to be weakly sequentially continuous if h takes weakly convergent sequences in E to weakly convergent sequences in E.

If u is weakly continuous on I, then u is strongly measurable (see [3]), hence weakly measurable.

Note that in reflexive Banach spaces weakly measurable functions are Pettis integrable (see [1], [3] and [8] for the definition) if and only if y(u(.)) is Lebesgue integrable on I for every y E E* (see [3]).

Now, we present some auxiliary results that will be needed in this paper. Firstly, we state O'Regan fixed point theorem ([7]).

Theorem 1.1 Let E be a Banach space with Q a nonempty, bounded, closed, convex, equicontinuous subset of C[I, E]. Suppose T : Q ^ Q is weakly sequentially continuous and assume TQ(t) is weakly relatively compact in E for each t E I, holds. Then the operator T has a fixed point in Q.

The following theorems can be found in [2], [16] and [5] respectively.

Theorem 1.2 (Dominated convergence theorem for Pettis integral) Let u : I ^ E. Suppose there is a sequence (un) of Pettis integrable functions from I into E such that y(un) = y(u) a.e. for y E E*. If there is a

scalar function ^ E L\(I) with ||un(-)|| < a.e. for all n, then u is Pettis integrable and

/ un(s) ds ^ / u(s) ds weakly V t E I. j j j j

Theorem 1.3 A subset of a reflexive Banach space is weakly compact if and only if it is closed in the weak topology and bounded in the norm topology.

Theorem 1.4 Let Q be a weakly compact subset of C[I, E]. Then Q(t) is weakly compact subset of E for each t E I.

Finally, we state some results which is an immediate consequence of the Hahn-Banach theorem.

Theorem 1.5 Let E be a normed space with u0 = 0. then there exists a y E E* with ||y|| = 1 and y(u0) = ||u0||.

Theorem 1.6 If uQ G E is such that p(uQ) = 0 for every p G E*, then uQ = 0.

In this work we study the existence of solutions, in the Banach space C[I, E], of the nonlocal boundary value problem

D13 u(t) + f (t,u(t)) = 0,

17 u(t)|i=Q =

3 G (1,2), t G (0,1), 0,

Y G (0,1], a u(n) = u(1),

0 < n < 1,

0 < a n3-1 < 1.

Now consider the fractional-order integral equation

(1)

t*

i J'Q 1 f (s,u(s)) ds

u(t) = - f (t,u(t)) - 1-a-nf

+ 1-1 VI3-1 fo ( r(/?) f (s,u(s)) ds, ft G (1, 2),t G (0,1).

In [7] the author studied the integral equation

y(t) = xo + / f(s,y(s)) ds, t G [0,T], xo G E o

where E = (E, |.|) is a real Banach space, under the assumptions that f(t,.) is weakly sequentially continuous for each t G [0,T] and f (.,y(.)) is Pettis integrable on [0,T] for each continuous function y : [0,T] ^ E and |f (t,y)| < hr(t) for a.e. t G [0,T] and all y G E with |y| < r,r > 0, hr G L1[0,T]. Also, in [6] the author studied the Volterra-Hammerstein integral equation

y(t) = h(t) + i k(t,s) f(s,y(s)) ds, t G [0,T], T > 0,

o

under the assumptions that f : [0,T] x B ^ B is weakly-weakly continuous and h : [0, T] ^ B is weakly continuous, where B is a reflexive Banach space. Here we study the existence of weak solution of the fractional-order integral equation (2) such that the function f : I x Br ^ E satisfies the following conditions:

(1) For each t G I, ft = f (t,.) is weakly sequentially continuous.

(2) For each u G Er,f (.,u(.)) is weakly measurable on I.

(3) for any r > 0, the weak closure of the range of f (I x Br ) is weakly compact in E (or equivalently; there exists an Mr such that ||f (t,u)|| < Mr for all (t,u) G I x Br).

Definition 1.1 by a weak solution of (2) we mean a function u G C[I, E] such that for all y G E *

<fi(u(t)) = - Ip Hf(t,u(t))) - x at'-'n y(f(s,u(s))) ds

+ x-ar]fJ-1 JQ ( r(/3)) y(f (s,u(s))) ds,f3 G (1, 2), t G (0,1).

2 Fractional-order integrals in reflexive Banach spaces

Here, we define the fractional-order integral operator in reflexive Banach spaces. Definition given below is an extension of such a notion for real-valued functions.

Definition 2.1 Let u : I ^ E be a weakly measurable function, such that y(u(.)) G Li(I), and let a > 0. Then the fractional (arbitrary) order Pettis integral (shortly FPI) Iau(t) is defined by

Ia u(t) = ^ (t — —1 u(s) ds. Jo r(a)

In the above definition the sign "J"" denotes the Pettis integral. Such an integral is well defined (see [11]):

Lemma 2.1 Let u : I ^ E be a weakly measurable function, such that <p(u(.)) G L1(I), and let a > 0. The fractional (arbitrary) order Pettis integral

Ia u(t) = ^ (t — ,s)a—1 u(s) ds Jo r(a)

exists for almost every t G I as a function from I into E and y(Iau(t)) = I ay(u(t)).

The following lemma can be found in [12]

Lemma 2.2 Let u : I — E be weakly continuous function on [0,1]. Then, FPI of u exists for almost every t G [0,1] as a weakly continuous function from [0,1] to E. Moreover,

y(Iau(t)) = Iay(u(t)), for all y G E*.

Definition 2.2 Let u : I — E. We define the fractional-Pseudo derivative (shortly FPD) of u of order a G (n — 1,n),n G N by

da

— u(t) = DnIn-a u(t).

In the above definition the sign "D" denotes the Pseudo differential operator (see [8]).

The following lemma can be found in [13]

Lemma 2.3 Let u : [0,1] — E be weakly continuous function on [0,1] such that the real-valued function In—ayu is n-times differentiate. Then, the FPD of u of order a G (n — 1,n), exists.

Definition 2.3 A function u : I — E is called Pseudo solution of (1) if u G C [I, E ] has FPD of order ft G (1, 2), IY u(t)|t=o = 0,y g (0,1],au(n) = u(1), 0 < n < 1,0 < anf—1 < 1 and satisfies

d2

y(I2-lS u(t)) + y(f (t,u(t))) = 0, a.e. on (0,1), for each y G E*.

Now, for the properties of the integrals of fractional-orders in reflexive spaces we have the following lemma [11]:

Lemma 2.4 Let u : I ^ E be weakly measurable and y(u(.)) G L1(I). If a, ft G (0,1), we have:

(1) IaIfu(t) = Ia+l3u(t) for a.e. t G I.

(2) lima^11au(t) = 11u(t) weakly uniformly on I if only these integrals exist on I.

(3) lima^0 Iau(t) = u(t) weakly in E for a.e. t G I.

(4) If, for a fixed t G I, y(u(t)) is bounded for each y G E*, then limt->0Iau(t) = 0.

3 Main result

In this section we present our main result by proving the existence of solutions

of the equation (2) in C[I,E].

Let E be a reflexive Banach space. And let

Er = {u E C[I, E] : ||u||0 < r(1M+ p) + r} (r > 0),

where ||.||0 is the sup-norm. We will consider the set

Br = {u(t) E E : u E Er, t E I}. Now, we are in a position to formulate and prove our main result.

Theorem 3.1 Let the assumptions (1) - (3) are satisfied.

(a + 1) Mr

If ----- < r

f (1 - a n^-1) r(1 + P) < '

Then equation (2) has at least one weak solution u E C[I,E]. Proof: Let us define the operator T as

Tu(t) = - I'9 f (t,u(t)) - 1 -a-1 I' (n f (s,u(s)) ds

+ f(s'u(s)) ds,p E (1,2),¿ E L

We will solve equation (2) by finding a fixed point of the operator T. We claim

T : C[I, E] ^ C[I, E].

To prove our claim, first note that assumption (2) implies that for each u E C[I,E], f (.,u(.)) is weakly measurable on I. The fact that f has weakly compact range means that y(f (.,u(.))) is Lebesgue integrable on I for every y E E* and thus the operator T is well defined. Now, we show that if u E C[I, E], then Tu E C[I,E]. Note that there exists r > 0 with

||u||0 = suptel ||u(t) 11 < r(1M+ + r.

Now assumption (3) implies that

||f(t,u(t))|| < Mr for t E [0,1].

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Let t,T G [0,1] with t > t. Without loss of generality, assume Tu(t) — Tu(t) = 0. Then there exists (a consequence of Theorem 1.5) y G E* with ||y|| = 1 and

Thus

|Tu(t) - Tu(t)|| = y(Tu(t) - Tu(t)).

lTu(t) - Tu(t)|| <

/0 y(f (s,u(s))) ds

<

+ + < + + + < + + < +

o

(t-s)!

-1

o r(3) y(f (s,u(s))) ds \

a rn (n-s)1-1

1- a r/1 1 1

1-a n1-1

-1 1

Tl -1

in y(f (s,u(s))) ds \t3 Jo^ y(f (s,u(s))) ds \t3-1 - T3-1

o

(t-s)1-1 - (t-s)1-

o r(3)

t (t-s)!-1

y(f (s,u(s))) ds\

\ir ~irn- y(f (s,u(s))) ds\

a Mr Jn (n-s)I-1 dQ \t3-1 _ T3-1 1-a 'q!-1 Jo r(| db \l '

Mr

o

1 (1-s)

1-a n1-1 Jo r(3) Mr ( (' t

< r(3)

£\(t - s)3

ds \t3-1 - t3-1\ -1 - (t - s)3-1\ ds + JTt(t - s)

)3-1 ds

a Mr n (1-a n1-1) r(1+3)

_Mr_

(1-a n1-1) r(1+3)

M,

t3-1 - T3-1\ t3-1 - T3-11

rMf) (2 (t - T)3 + \t3 - T3\)

\t3-1 - t3-1\.

Mr (a if + 1) (1-a n1-1) r(1+3)

which proves that Tu G C[I, E]. Now, let

Q = {u G Er : (V t,T G I)

(\\u(t) - u(t)\\ <

Mr

+

r(1 + ft)

Mr (a n3

(2 (t - t

+ 1)

(1 - an3-1) r(1 + ft)

+ \t3 - t3\) \t3-1 - t3-1 \},

Note that Q is nonempty, closed, bounded, convex and equicontinuous subset of C[I, E]. Now, we claim that T : Q — Q and is weakly sequentially continuous. If this is true then according to Theorem 1.3, TQ is bounded in C[I, E] (hence, Theorem 1.4, implies TQ(t) is weakly relatively compact in E for each t G I) and the result follows immediately from Theorem 1.1. It remains to prove our claim. First we show that T maps Q into Q. To see this, note that the

1

inequality (2) shows that TQ is norm continuous. Now, take u G Q; without loss of generality, we may assume that Iaf (t,u(t)) = 0, then, by Theorem 1.5, there exists y G E* with ||y|| = 1 and ||Iaf (t,u(t))|| = y(Iaf (t,u(t))). Thus

||Tu(t)|| <

< l|I^ f(t,u(t))|| ^lix-Oo^ J? f(s,u(s)) dsll

+ ||xzo-^ /g1 f(s,u(s)) ds||

= y(IP f(t,u(t))) + y(x-^ /0 (^f1 f (s,u(s)) ds)

+ y (i-OT^^-i Jg1 f(s,u(s)) ds)

= IP y(f(t,u(t))) + i-^ J? y(f (s,u(s))) ds

+ i-0f"/»-i Jo y(f(s,u(s)/ ds i

< Mr fo r()) ds + C1-arn3-1 /G ^-O) ds + Mr t3-1 r1 (l-s)3-1 ,

+ i-a nT-1 JG r(p) ds

< Mr + _Mr tT-i__(a + 1)

< r(1+P) + (l-a nT-1) r(1+P) (a 'I + 1)

< Mr + Mr (a + l)

< r(1+P) ^ (l-a n3-1) r(1+P)

Mr

r(i+P)

< Mr + r

therefore

Tu g < r ^ + r.

r(i + P)

Thus T : Q ^ Q. Finally, we will show that T is weakly sequentially continuous. To see this, let {unbe a sequence in Q and let un(t) ^ u(t) in Ew for each t G [0,1]. Recall [5] that a sequence {un}^=1 is weakly convergent in C [I, E] if and only if it is weakly pointwise convergent in E. Fix t G I. From the weak sequential continuity of f (t,.), the Lebsegue dominated convergence theorem (see assumption (3)) for the Pettis integral [2] implies for each y G E* that y(Tun(t)) ^ y(Tu(t)) a.e. on I, Tun(t) ^ Tu(t) in Ew. So T : Q ^ Q is weakly sequentially continuous. The proof is complete. ■

Now, we are looking for sufficient conditions to ensure the existence of Pseudo solution to the boundary value problem (1).

Theorem 3.2 If f : I x Br ^ E satisfies the assumptions of Theorem 3.1, then the boundary value problem (1) has at least one solution u G C[I, E].

Proof: Let us remark, that by assumptions (2), (3) the FPI of f of order ft > 1 exists and

y(I3f(t,u(t))) = I3 y(f(t,u(t))), for all y G E*. Let u be a solution of equation (2), then

u(t) = - I3 f (t,u(t)) - 1 a^-3-1 J" (n-ft!-1 f (s,u(s)) ds

t3-1 r1 (1 _ s)3-1 + 1 - a ^ I r(ft)) f(S,U(s)) ds, ft G (1,2), t G I0,1).

It is clear that

IYu(t)\t=o = 0, y G (0,1], a u(n) = u(1). Furthermore, we have

u(t) = - I3 f(t,u(t)) + Kt3-1, (2)

where J

K = ^ f (s,u(s)) ds

+ 1-arnF- /o1íiт^-- f (s,u(s)) ds.

since u G C[I, E], then y(I2-3u(t)) = I2-3y(u(t)), for all y G E* (see Lemma 2.2). From equation (2), we deduce that

yu(t) = - y(I3 f(t,u(t))) + yKt3-1

3

= - I33 y(f (t,u(t))) + yKt33-1. (3)

Operating by 12 3 on both sides of the equation (3) and using the properties of fractional calculus in the space L1[0,1] (see [14] and [15]) result in

I2-3 yu(t) = -12 y(f(t,u(t))) + yK r(ft) t.

Therefore,

y(12-3 u(t)) = -12 y(f (t,u(t))) + yK r(ft) t.

Thus

d2

y(I2-3 u(t)) = - y(f (t,u(t))) a.e. on (0,1).

That is u has the FPD of order ft G (1, 2) and u is a solution of the differential equation (1) which complete the proof. ■

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