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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].
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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