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6DIFFERENTIAL EQUATIONS AND
CONTROL PROCESSES N 4, 2008 Electronic Journal, reg. N P2375 at 07.03.97 ISSN 1817-2172
http://www.neva.ru/journaal http://www.math.spbu.ru/diffjournaal/ e-mail: [email protected]
Weak maximal and minimal solutions for Hammerstein and Urysohn integral equations in reflexive Banach spaces
El-Sayed A.M.A & Hashem H.H.G E-mail : [email protected] & [email protected] Faculty of Science, Alexandria University, Alexandria, Egypt
Abstract: We present existence theorem of weak solutions for Hammerstein and Urysohn integral equations. Also, we shall define the weak maximal and weak minimal solutions for these integral equations and finally prove the existence of the weak maximal and minimal solutions.
Keywords: Weak solution; Urysohn integral equation ; Hammerstein integral equation; Weak maximal and weak minimal solutions.
1 Introduction and Preliminaries
Let L\(I) be the space of Lebesgue integrable functions defined on the interval I = [0,1]. Let E be a reflexive Banach space with the norm ||.|| and dual
E* and denote by C[I, E] the Banach space of strongly continuous functions
x : I ^ E with sup-norm ||.||q. The existence of weak solutions of the integral equation
f t
x(t) = xo + Jo f (s,x(s)) ds, t E I
was proved by O’Regan [9] where f : I x E ^ E, x0 E E. .
The existence of weak solutions to the Hammerstein integral equation
C1
x(t) = h(t) + jo k(t,s) f (s,x(s)) ds, t E [0,1]
was proved by O’Regan ( see[8]) where x takes values in reflexive Banach spaces and f is weakly-weakly continuous.
In this paper, we shall study the existence of a weak solution to the Hammerstein integral equation
C1
x(t) = g(t) + 0 k(t,s) f (s,x(s)) ds, t e [0,1] (1)
and Urysohn integral equation
c 1
x(t) = g(t) + j0 u(t,s,x(s)) ds, t E [0,1] (2)
Where x takes values in reflexive Banach spaces and f : I x E ^ E is weakly measurable in t and weakly sequentially continuous in x .
The function u : I x I x E ^ E is weakly measurable in s and weakly sequentially continuous in x. Also, the existence of the weak maximal and minimal solutions will be proved.
Now, we shall present some auxiliary results that will be need in this work. Let E be a Banach space (need not be reflexive) and let x : I ^ E, then
(1) x(.) is said to be weakly continuous (measurable) at t0 E I if for every 0 E E*, 0(x(.)) is continuous (measurable) at t0.
(2) A function h : E ^ E is said to be sequentially continuous if h maps weakly convergent sequences in E to weakly convergent sequences in E.
If x is weakly continuous on I, then x is strongly measurable and hence weakly measurable (see[3] and [2]). Note that in reflexive Banach space weakly measurable functions are Pettis integrable if and only if 0(x(.)) is Lebesgue integrable on I for every 0 E E* (see[3] pp. 78). Now we state a fixed point theorem and some propositions which will be used in the sequel (see[9]).
Theorem 1.1. Let E be a Banach space and let Q be a nonempty, bounded, closed and convex subset of the space E and let T : Q ^ Q be a weakly sequentially continuous and assume that TQ(t) is relatively weakly compact in E for each t E [0,1] . Then, T has a fixed point in the set Q. proposition 1.1. 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. proposition 1.2. Let E be a normed space with y = 0. Then there exist a
0 E E* with H0H = 1 and HyH = 0(y).
2 Hammerstein integral equation
Let E be a reflexive Banach space and D С E. Consider the following assumptions
(1) g e C[I, E] ;
(2) f : I x D —— E satisfies the following
(i) For each t e I, ft = f (t,.) is weakly sequentially continuous;
(ii) For each x e D, f(. ,x(.)) is weakly measurable on I ;
(iii) The weak closure of the range of f (I x D) is weakly compact in E
(or equivalently: there exists an M such that \\f(t,x)|| < M
(t, x) e I x D; )
(3) k : I x I — R+ is integrable in s and continuous in t , the operator
С1
Ky(t) = J0 k(t,s) y(s) ds
maps Li(I) into Li(I) and /J k(t,s) ds < Mi, t e I.
Definition 2.1. By a weak solution of (1) we mean a function x e C[I, E]
such that
(' 1
0(x(t)) = ф(д(t)) + 0 k(t,s) Ф(!'(s,x(s))) ds, t e [0,1] for all ф e E*.
Theorem 2.1. Let the assumptions (1)-(3) be satisfied. Then equation (1) has at least one weak solution x e C[I, E].
Proof: Define the operator T by
С1
Tx(t) = g(t) + jo k(t,s) f (s,x(s)) ds, t e I.
For any x e C[I,E] and since f(.,x(.)) is weakly measurable on I
and \\f(t,x)\\ < M, then ф^(.,x(.))) is Lebesgue integrable on
I Уф e E* and since k(t,.) is Lebesgue integrable on I, then we have ф(k(t,.) f (.,x(.))) = k(t,.) ф^ (.,x(.))) is Lebesgue integrable on I Уф e E *, then k(t,.) f(.,x(.)) is Pettis integrable on I. Thus T is well defined.
We shall prove that T : C[I, E] — C[I, E].
Let t1, t2 e I and t1 < t2 ( without loss of generality assume that
Tx(t2) - Tx(ti) = 0)
c 1 с 1
Tx(t2)-Tx(ti) = g(t2) - g(t1)+ JQ k(t2,s) f (s,x(s)) ds - Jq k(t1,s) f (s,x(s)) ds
f 1
= g(t2) - g(t1) + 0 [k(t2, s) - k(t1,s)]f(s,x(s)) ds Therefore as a consequence of proposition 1.2, we obtain
||Tx(t2) — T x(t1) || = 0(Tx(t2) — Tx(t1)) c 1
= 0(g(t2) — g(t1)) + 0 [k(t2,s) — k(t1,s)]0(f(s,x(s))) ds
11
= ||g(t2) — g(t1)|1 + 0 [k(t2,s) — k(t1,s)] llf(s,x(s))|1 ds
c 1
< ||g(t2) — g(t1)|| + m J0 mt2,s) — k(t1,s) | ds. let the set Q be defined as
Q = { x E C[I, E] : HxH < M2 }, M2 = HgH + M.MX.
Let x E Q, then we have
(' 1
||Tx(t)|| = 0(Tx(t)) = 0(g(t)) + 0 k(t,s) 0(f(s,x(s))) ds
(' 1 c 1
M + J0 k(t,s) Hf (s,x(s))|| ds < ||g|| + M J0 k(t,s) ds < ||g|| + M M1
and hence Tx E Q implies TQ c Q (i.e. T : Q ^ Q )
Then Q is nonempty, uniformly bounded and strongly equi-continuous subset of C[I, E] . also it can be shown that Q is convex and closed.
As a consequence of Proposition 1.1, then TQ is relatively weakly compact. It remains to prove that T is weakly sequentially continuous.
Let {xn} be a sequence in Q converges weakly to x Vt E I and since f (t,x(t)) is weakly sequentially continuous in x, then f (t,xn(t)) converges weakly to f (t,x(t)) thus 0(f (t,xn(t))) converges strongly to 0(f (t,x(t)))
( see assumption (iii)). Applying Lebesgue Dominated Convergence Theorem for Pettis integral, we get
0(Id k(t,s) f (s,xn(s)) ds) = Z,1 k(t,s) 0(f (s,xn(s))) ds (' 1
^ j0 k(t,s) 0(f (s,x(s))) ds V0 E E*, t E I.
Then T is weakly sequentially continuous. ( i.e. Txn(t) ^ Tx(t) weakly
Vt E I ) Since all conditions of Theorem 1.1 are satisfied, then the operator T has at least one fixed point x E Q which competes the proof. ■
3 Urysohn integral equation
Let E be a reflexive Banach space and D c E. Consider the following assumptions:
(1) g E C[I, E] ;
(2) u : I x I x D ^ E satisfies the following
(i) For each t,s E I x I, u(t, s,.) is weakly sequentially continuous;
(ii) For each x E D and t E I u(t, .,x(.)) is weakly measurable on I ;
(iii) For each x E D and s E I u(.,s,x(s)) is continuous on I ;
(3) Hu(t, s, x(s))|| < k(t,s), k : I x I ^ R+ is integrable in s and
continuous in t , the operator
C1
Ky(t) = 0 k(t,s) y(s) ds maps L1(I) into L1(I) and /J k(t,s)ds < M1, t E I.
Definition 3.1. By a weak solution of (2) we mean a function x E C[I, E]
such that
C1
0(x(t)) = 0(g(t)) + 0 0(u(t,s,x(s))) ds, t E [0,1]
for all 0 E E*.
Theorem 3.1. Let the assumptions (1)-(3) be satisfied. Then equation (2) has at least one weak solution x E C[I, E] .
Proof: Define the operator T by
i 1
Tx(t) = g(t) + J0 u(t,s,x(s)) ds, t E I.
For any x E C[I, E] since u(t, ., x(.)) is weakly measurable on I, then 0(u(t,., x(.))) is strongly measurable on I V0 E E* and since ||u(t,s,x)|| < k(t,s), then 0(u(t, ,x(.))) is Lebesgue integrable on
I V0 E E* and hence u(t,.,x(.)) Pettis integrable on I. Thus T is well
defined.
We shall prove that T : C[I, E] C[I, E].
Let t1, t2 E I, t1 < t2 ( without loss of generality assume that
Tx(t2) — Tx(t1) = 0)and x E C[I,E]
11 11
Tx(t2) — Tx(t1) = g(t2) — g(t1) + u(t2,s,x(s)) ds — J0 u(t1,s,x(s)) ds
(1
= g(t2) — g(t1) + J0 [u(t2,s,x(s)) — u(t1,s,x(s)) ] ds
Therefore as a consequence of proposition 1.2, we obtain
||Tx(t2) — Tx(t1)|| = 0(Tx(t2) — Tx(t1))
11
= 0(g(t2) — g(t1)) + J0 0[u(t2,s,x(s)) — u(t1, s, x(s)) ] ds
11
< ||g(t2) — g(t1)|| + J0 ||u(t2,s,x(s)) — u(t1,s,x(s))|| ds (3)
let the set Q be defined as
Q = { x E C[I, E] : HxH < M2 }, M2 = HgH + M.
Let x E Q, then we have
c 1
||Tx(t)|| = 0(Tx(t)) = 0(g(t)) + j0 0(u(t,s,x(s))) ds
r 1 r 1
= 0(g(t)) + J0 ||u(t,s,x(s))|| ds < ||g|| + J0 k(t,s) ds < Hg|| + M
and hence Tx E Q implies TQ c Q (i.e. T : Q ^ Q ).
Then Q is nonempty, uniformly bounded and strongly equi-continuous subset of C[I, E] . also it can be shown that Q is convex and closed.
As a consequence of Proposition 1.1, then TQ is relatively weakly compact. It remains to prove that T is weakly sequentially continuous.
Let {xn} be a sequence in Q converges weakly to x Vt E I and since u(t,s,x(s)) is weakly sequentially continuous in x, then u(t,s,xn(s)) converges weakly to u(t,s,x(s)) thus 0(u(t,s,xn(s))) converges strongly to
0(u(t,s,x(s))). using assumption (3) and applying Lebesgue Dominated Convergence Theorem for Pettis integral, we get
0(^ u(t,s,xn(s)) ds) = ^ 0(u(t,s,xn(s))) ds (' 1
^ J0 0(u(t,s,x(s))) ds V0 E E*, t E I.
Then T is weakly sequentially continuous. ( i.e. Txn(t) ^ Tx(t) weakly
Vt E I ) Since all conditions of Theorem 1.1 are satisfied, then the operator
T has at least one fixed point x E Q which competes the proof. ■
4 The weak maximal and weak minimal solutions
Now we give the following definition
Definition 4.1 Let q(t) be a weak solution of (1) Then q(t) is said to be a weak maximal solution of (1) if every weak solution x(t) of (1) satisfies the inequality
0(x(t)) < 0(q(t)), V0 E E*. A weak minimal solution s(t) can be defined by similar way by reversing the above inequality i.e. 0(x(t)) > 0(s(t)), V0 E E *.
In this section f assumed to satisfy the following assumption:
(4) for any x, y E E satisfying 0(x(t)) < 0(y(t)), V0 E E* implies that
0(f (s,x(s))) < 0(f(s,y(s)))
Lemma 4.1 Let f (t,x), k(t,s) satisfy assumptions of Theorem 2.1
and let x(t), y(t) E C[I, E] on I satisfying
(' 1
0(x(t)) < 0(g(t)) + 0 k(t,s) 0(f (s,x(s))) ds
0(y(t)) > 0(g(t)) + f0 k(t,s) 0(f (s,y(s))) ds, E E*
where one of them is strict.
If (f (t,x)) satisfies assumption (4). Then
0(x(t)) < 0(y(t)). (4)
proof: Let the conclusion (4) be false, then there exists t1 such that
0(x(t1)) = 0(y(t1)) t1 > 0
and
0(x(t)) < 0(y(t)) 0 < t < t1.
Since (f (t,x)) satisfies assumption (4), we get
C1
0(x(t1)) < 0(g(t1)) + 0 k(t1,s) 0(f (s,x(s))) ds
C1
< 0(g(t1)) + 0 k(t1,s) 0(f (s,y(s))) ds
< 0(y(t1)).
Which contradicts the fact that 0(x(t1)) = 0(y(t1)), then
0(x(t)) < 0(y(t)).
Theorem 4.1 Let the assumptions of Theorem 2.1 be satisfied. If f (t,x) satisfies assumption (4), then there exist a weak maximal and weak minimal solutions of (1).
Proof Firstly we shall prove the existence of the weak maximal solution of
(1). Let e > 0 be given. Now consider the integral equation
11
xe(t) = g(t) + J0 k(t,s) fe(s,xe(s))ds, (5)
where
fe(t,xe(t)) = f (t,xe(t)) + e-
Clearly the function fe(t,xe) satisfies the conditions (1)-(3) of Theorem 2.1 and
II fe(t,xe) II < M + e = M'.
Therefore equation (5) has a weak solution xe e C[I, E] according to Theorem
2.1. Let ei and e2 be such that 0 < e2 < e1 < e. Then
11
xei(t) = g(t) + J0 k(t,s) fc!(s,xa(s)) ds,
11
xei(t) = g(t) + J0 k(t,s) ( f (sx(s)) + ei) ds,
implies that
c 1
0(xei(t)) > 0(g(t)) + J0 k(t,s)( 0(f(s,x£i(s))) + e2) ds, (6)
C1
0(xe2(t)) = 0(g(t)) + J0 k(t,s)( 0(f (sx(s))) + e2) ds (7)
Using Lemma 4.1, then (6) and (7) implies
0(xe2 (t)) < 0(xei (t)) fOr t e [0, 1].
As shown before in the proof of Theorem 2.1 the family of functions xe(t) defined by (5) is uniformly bounded and of strongly equi-continuous functions. Hence by Arzela-Ascoli Theorem , there exists a decreasing sequence en such that e ^ 0 as n ^ m, and limn^m xCn(t) exists uniformly in [0,1] and denote this limit by q(t). from the weakly sequentially continuity of the function fCn in the second argument and applying Lebesgue Dominated Convergence Theorem for Pettis integral, we get
11
q(t) = nx^(t) = g(t) + J0 k(t,s) f (s,q(s)) ds
which proves that q(t) as a solution of (1).
Finally, we shall show that q (t) is the weak maximal solution of (1). To do this, let x(t) be any weak solution of (1). Then
C1
0(xe(t)) = 0(g(t)) + J0 k(t,s)(0(f (s,xe(s))) + e) ds
C1
> 0(g(t)) + J0 k(t,s) 0(f (s,xe(s))) ds
and
C1
0(x(t)) = 0(g(t)) + J0 k(t,s) 0(f (s,x(s))) ds
applying Lemma 4.1, we get
0(xe(t)) > 0(x(t)) fort e [0,1].
from the uniqueness of the maximal solution (see [2]), it is clear that xe(t) tends to q(t) uniformly in t e [0,1] as e ^ 0.
By similar way as done above we can prove that s(t) is the weak minimal
solution of (1). ■
The weak maximal and minimal solutions of (2) can be defined in the same fashion as done above.
Now, the function u assumed to satisfy the following assumption:
(4*) for any x, y e E satisfying 0(x(t)) < 0(y(t)), e E* implies that 0(u(t,s,x(s))) < 0(u(t,s,y(s)))
Now the following lemma can be proved.
Lemma 4.2 Let u(t,s,x) satisfies assumptions of Theorem 3.1 and let x(t), y(t) e C[I, E] on I satisfying
(' 1
0(x(t)) < 0(g(t)) + J0 0(u(t,s,x(s))) ds
(' 1
0(y(t)) > 0(g(t)) + J0 0(u(t,s,y(s))) ds, e E*
where one of them is strict.
If (u(t,s,x)) satisfies assumption (4*). Then
0(x(t)) < 0(y(t)).
Theorem 4.2 Let the assumptions of Theorem 3.1 be satisfied. If u(t,s,x) satisfies assumption (4*), then there exists a weak maximal and weak minimal solutions of (2).
Proof Firstly we shall prove the existence of the weak maximal solution of
(2). Let e > 0 be given. Now consider the integral equation
11
xe(t) = g(t) + Jo ue(t,s,xe(s))ds, (8)
where
ue(t,s,xe(t)) = u(t,s,xe(t)) + e.
Clearly the function ue(t,s,xe) satisfies the conditions (1)-(3) of Theorem
3.1 and
|| ue(t,s,xe) || < k(t,s) + e = k1(t,s) e L1.
Therefore equation (8) has a weak solution xe e C[I, E] according to Theorem
3.1. Let e1 and e2 be such that 0 < e2 < e1 < e. Then
11
xCi(t) = g(t) + Jo u€i(t,s,xei(s)) ds,
11
xCi(t) = g(t) + Jo (u(t,s,x€i(s)) + e1) ds,
implies that
f 1
0(xei(t)) > 0(g(t)) + J0 (0(u(t,s,xd(s))) + e2) ds, (9)
(' 1
0(xe2(t)) = 0(g(t)) + J0 (0(u(t,s,xe2(s))) + e2) ds (10)
Using Lemma 4.2, then (9) and (10) implies
0(xe2 (t)) < 0(xei (t)) for t e [0, 1].
As shown before in the proof of Theorem 3.1 the family of functions xe(t) defined by (8) is uniformly bounded and of strongly equi-continuous functions. Hence by Arzela-Ascoli Theorem , there exists a decreasing sequence en such that e ^ 0 as n ^ m, and limn^m xCn(t) exists uniformly in [0,1] and denote this limit by q(t). from the weakly sequentially continuity of the function u€n in x and applying Lebesgue Dominated Convergence Theorem for Pettis integral, we get
11
q(t) = n 1^m x^n(t) = g(t) + J0 u(t,s,q(s)) ds
which proves that q(t) as a solution of (2).
Finally, we shall show that q (t) is the weak maximal solution of (2). To do
this, let x(t) be any weak solution of (2). Then
11
0(xe(t)) = 0(g(t)) + J0 (0(u(t,s,xe(s))) + e) ds
11
> 0(g(t)) + J0 0(u(t,s,xe(s))) ds
and
f 1
0(x(t)) = 0(g(t)) + J0 0(u(t,s,x(s))) ds
applying Lemma 4.2, we get
0(xe(t)) > 0(x(t)) fort e [0,1].
from the uniqueness of the maximal solution (see [2]), it is clear that xe(t) tends to q(t) uniformly in t e [0,1] as e ^ 0.
By similar way as done above we can prove that s(t) is the weak minimal
solution of (2). ■
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