Coincidence of crisp and fuzzy functions
Akbar Azam, Maliha Rashid and Nayyar Mehmood
Department of Mathematics, COMSATS Institute of Information
Technology, Chack Shahzad, Islamabad - 44000, Pakistan.
Email: akbarazam@yahoo.com (A. Azam)
Email: thinkmaliha@gmail.com (M. Rashid)
Email: nayyarmaths@gmail.com(N. Mehmood)
Abstract_ _ We prove a common coincidence point theorem for a non-fuzzy
mapping and a pair of fuzzy mappings under a ' contraction condition on a
metric space in the context of Hausdor¤ metric on the family of fuzzy sets.
Further, we establish a common coincidence point theorem on a metric space
with the d1 metric on fuzzy sets, which extends a number of recent results.
As applications, we obtain a common coincidence point theorem for a single
valued mapping and a pair of set-valued mappings and apply it to achieve some
existence and uniqueness theorems of solution for a class of nonlinear integral
equations.
_____________________________
2010 Mathematics Subject Classi…cation: 46S40; 47H10; 54H25.
Keywords and Phrases: coincidence point; multivalued mapping; fuzzy mapping; integral equation.
* This article was presented at ICRAPAM-2014, Turkey. _ _ _
1
Introduction
In continuation of many investigations in …xed point theory, recently, Shatanawi
[23](see also [21]), obtained some …xed point results for a generalized -weak
contraction mappings in orbitally metric spaces. Weiss [26] and Butnariu [11]
introduced and studied the …xed point theorems in fuzzy mathematics. Heilpern
[16] initiated the concept of fuzzy mappings and established a …xed point theorem for fuzzy approximate quantity-valued contractions in a metric linear space.
Subsequently several other authors generalized this result and studied the existence of …xed points and common …xed points of fuzzy approximate quantityvalued mappings satisfying a contractive type condition in the same structure
of metric linear spaces. Shi-sheng [24] studied an important role of Hausdor¤
metric between fuzzy subsets and studied common …xed points of a pair fuzzy
set valued mappings under a contraction in a metric space (which is not necessarily linear). Frigon and Regan [15] generalized Heilpern theorem under a
contractive condition for 1-level sets (i.e., [T x]1 ) of a fuzzy contraction T on
a complete metric space, where 1-level sets are not assumed to be convex and
compact. Amemiya and Takahashi [2] obtain …xed points of fuzzy mappings by
using the concept of w-distance (see [2] ) in complete metric spaces. Choudhury
and Dutta [12] studied …xed points of fuzzy mappings satisfying contractive inequality involving a generalized altering distance function. Recently Kamran
1
[17] and Azam et al. [6, 7, 8] continue to study …xed points of fuzzy set valued
mappings in connection with d1 metric and Hausdor¤ metric for fuzzy subsets.
In this paper we establish common coincidence point theorems for a crisp and
a pair of fuzzy mappings satisfying ' contractive condition. As applications,
we obtain a common coincidence point theorem for a single valued mapping and
a pair of set-valued mappings. Moreover, we apply it to establish some existence
and uniqueness theorems of solution for a class of nonlinear integral equations.
2
Preliminaries
Let (Z; d) be a metric space, (V; dv ) be a metric linear space; C 2Z be the
family of nonempty compact subset of Z and A; B 2 C 2Z .
d(x; A) = inf d(x; y);
y2A
d(A; B) =
inf
x2A; y2B
Then the Hausdor¤ metric H on C 2Z
d(x; y):
induced by d is de…ned as:
H (A; B) = max sup d (a; B) ; supd (A; b) :
a2A
b2B
A fuzzy set in Z is a function with domain Z and values in [0; 1], I Z is the
collection of all fuzzy sets in Z: If A is a fuzzy set and x 2 Z, then the function
values A(x) is called the grade of membership of x in A. The -level set of a
fuzzy set A, is denoted by [A] , and is de…ned as
[A] = fx : A(x) > g if
2 (0; 1];
[A]0 = fx : A(x) > 0g;
b = fx : A(x) = maxA (y)g:
A
y2Z
For crisp subset A of Z; we denote the characteristic function of A by A : A
fuzzy set A in a metric linear space V is said to be an approximate quantity if
and only if [A] is compact and convex in V for each 2 [0; 1] and supA(x) = 1:
x2V
De…ne some sub-collections of I Z and I V as follows:
W(V ) = A 2 I V : A is an approximate quantity in V ;
n
o
b 2 C 2Z ;
K (Z) = A 2 I Z : A
C (Z) = A 2 I Z : [A] 2 C 2Z ; for each
2
2 [0; 1] :
For A; B 2 I Z ; A B means A(x) 6 B(x) for each x 2 Z : If there exists
an 2 [0; 1] such that [A] ; [B] 2 C 2Z then de…ne
p (A; B)
D (A; B)
=
inf
x2[A] ;y2[B]
d(x; y);
= H([A] ; [B] ):
Z
If [A] ; [B] 2 C 2 for each 2 [0; 1] then de…ne p(A; B); d1 (A; B) : C (Z)
C (Z) ! R; (induced by the Hausdor¤ metric H) as follows:
p(A; B)
=
supp (A; B);
d1 (A; B)
=
supD (A; B):
Let W be an arbitrary set, Z be a metric space. A mapping T is called
fuzzy mapping if T is a mapping from W into I Z . A fuzzy mapping T is
a fuzzy subset on W
Z with membership function T (x)(y). The function
T (x)(y) is the grade of membership of y in T (x): The family of all mappings
W
from W into I Z is denoted by I Z
: A point u 2 W is called coincidence
point of E : W ! Z; T : W ! I Z if T (u) (Eu) > T (u) (x) for all x 2 W:
The set fu 2 W : T (u) (Eu) > T (u) (x) for all x 2 W g of all coincidence points
of E and T is denoted by Coin(E; T ) : A point x 2Coin(I; T ) (where I is identity
map) is a …xed point of T [11, 24].
Lemma 1 Let A and B be nonempty closed and bounded subsets of a metric
space (Z; d) : If a 2 A, then d (a; B) 6 H (A; B) :
Proof. For a 2 A
d (a; B) 6 sup d (a; B) :
a2A
It follows that
d (a; B) 6 max sup d (a; B) ; supd (A; b) :
a2A
b2B
Lemma 2 [3] Let (V; dv ) be a complete metric linear space, T : V
a fuzzy mapping and xo 2 V: Then there exists x1 2 Z such that
! W(V ) be
T (x0 ):
fx1 g
Lemma 3 [1] Let ' : R+ ! R+ be a nondecreasing function satisfying the
following conditions:
P1
' is continuous from right and i=0 'i (t) < 1 for all t > 0 ('i denote the ith
iterative function of '): Then ' (t) < t:
In the following we always suppose that ' is a function satisfying the conditions of Lemma 3 and Tb is the mapping induced by fuzzy mappings T i.e.,
Tbx = fy : T (x) (y) = maxT (x) (t)g:
t2Z
Lemma 4 [7] Let W be a nonempty set, (Z; d) be a metric space, x 2 W and
T : W ! I Z be fuzzy mappings such that Tbx 2 C 2Z and for all x 2 Z: Then
x 2 Tbx i¤ T (x ) (x ) > T (x ) (x) for all x 2 Z.
3
3
Main Results
Theorem 5 Let W be a nonempty set, (Z; d) be a metric space and G; J : W !
b Jx
b 2 C 2Z for all x 2 W;
I Z ; E : W ! Z such that Gx;
i
[ h
b [ Jx
b
Gx
EW:
(1)
x2W
Suppose that for all x; y 2 W;
h
i
b + d(Ey; Gx)
b
b Jy)
b 6 ' max d(Ex; Ey); d(Ex; Gx);
b d(Ey; Jy);
b 1 d(Ex; Jy)
H(Gx;
2
(2)
h
i
[
b [ Jx
b is complete. Then
If E is injective and either EW or
Gx
x2W
Coin (E; G)
\
Coin (E; J) 6= :
b Jx
b are singleton subsets of Z for all x then Coin(E; G) TCoin(E; J)
Moreover, if Gx;
is singleton.
b 0 2 C 2Z : Using (1) and the
Proof. Choose x0 2 W; then by hypotheses Gx
b 0 ; we obtain x1 2 W such that Ex1 2 Gx
b 0 and
compactness of Gx
b 0 :
d (Ex0 ; Ex1 ) = d Ex0 ; Gx
b 1 and
Similarly, we can choose x2 2 W such that Ex2 2 Jx
b 1 :
d (Ex1 ; Ex2 ) = d Ex1 ; Jx
By induction we produce sequences fxn g and fExn g of points of W and EW respectively
such that
Ex2a+1
Ex2a+2
such that
b 2a ;
2 Gx
b 2a+1 ;
2 Jx
a = 0; 1; 2;
b 2a );
= d(Ex2a ; Gx
b 2a+1 ); a = 0; 1; 2;
d(Ex2a+1 ; Ex2a+2 ) = d(Ex2a+1 ; Jx
d(Ex2a ; Ex2a+1 )
:
Using Lemma 1 along with the above equations, we have
b 2a
d(Ex2a ; Ex2a+1 ) 6 H Jx
b
1 ; Gx2a
;
b 2a ; Jx
b 2a+1 ); a = 0; 1; 2;
d(Ex2a+1 ; Ex2a+2 ) 6 H(Gx
4
:
:
Assume that Ex2a = Ex2a+1 for some a > 0; then as E is injective,
b 2a = Jx
b 2a+1
x2a = x2a+1 ; Jx
and
b 2a ; Jx
b 2a+1 )
d(Ex2a+1 ; Ex2a+2 ) 6 H(Gx
)!
(
b
b 2a+1 );
d(Ex2a ; Ex
2a+1 ); d(Ex2a ; Gx2a ); d(Ex2a+1 ; Jx
h
i
6 ' max
1
b
b
2 d(Ex2a ; Jx2a+1 ) + d(Ex2a+1 ; Gx2a )
6 ' max d(Ex2a+1 ; Ex2a+2 );
1
[d(Ex2a ; Ex2a+2 ) + d(Ex2a+1 ; Ex2a+1 )]
2
6 ' (d(Ex2a+1 ; Ex2a+2 )) :
Since, ' (t) < t for all t > 0; therefore, the above inequality implies that
b 2a ; Ex2a =
d (Ex2a+1 ; Ex2a+2 ) = 0; which further implies that Ex2a = Ex2a+1 2 Gx
b
b
Ex2a+1 = Ex2a+2 2 Jx2a+1 = Jx2a . It follows that
Then Lemma 4 yields
b 2a \ Jx
b 2a :
Ex2a 2 Gx
G (x2a ) (Ex2a ) > G (x2a ) (x) ; J (x2a ) (Ex2a ) > J (x2a ) (x) ;
for all x 2 W: Thus in this sequel of the proof we can suppose that Exn+1 6=
Exn for n = 0; 1; 2
: Again by using inequality(1), we have
b 2a ; Jx
b 2a+1 )
d(Ex2a+1 ; Ex2a+2 ) 6 H(Gx
d(Ex2a ; Ex2a+1 ); d(Ex2a+1 ; Ex2a+2 );
6 ' max
1
[d(Ex
2a ; Ex2a+2 ) + d(Ex2a+1 ; Ex2a+1 )]
2
6 ' (max fd(Ex2a ; Ex2a+1 ); d(Ex2a+1 ; Ex2a+2 )g) :
If max fd(Ex2a ; Ex2a+1 ); d(Ex2a+1 ; Ex2a+2 )g = d(Ex2a+1 ; Ex2a+2 ); then the
above inequality implies that d(Ex2a+1 ; Ex2a+2 ) 6 ' (d(Ex2a+1 ; Ex2a+2 )) <
d(Ex2a+1 ; Ex2a+2 ); as Exn+1 6= Exn for n = 0; 1; 2
; which is a contradiction. It follows that
max fd(Ex2a ; Ex2a+1 ); d(Ex2a+1 ; Ex2a+2 )g = d(Ex2a ; Ex2a+1 ):
It further implies that d(Ex2a+1 ; Ex2a+2 ) 6 ' (d(Ex2a ; Ex2a+1 )) : Consequently,
d(Exn+1 ; Exn ) 6 ' (d(Exn ; Exn
1 ))
6 '2 (d(Exn
1 ; Exn 2 ))
6
6 'n (d(Ex1 ; Ex0 )) ;
for each n > 1: Now for each positive integer m; n (n > m); we have
d(Exm ; Exn ) 6 d(Exm ; Exm+1 ) + d(Exm+1 ; Exm+2 ) +
+ d(Exn 1 ; Exn )
m
m+1
6 ' (d(Ex1 ; Ex0 )) + '
(d(Ex1 ; Ex0 )) +
+ 'n 1 (d(Ex1 ; Ex0 ))
6
n
X1
'i (d(Ex1 ; Ex0 )) 6
i=m
1
X
i=m
5
'i (d(Ex1 ; Ex0 )) :
P1
Since, i=1 'i (t) < 1 for each t > 0: It yields that fExn g is a Cauchy sequence.
As EWh is complete,
there exists u 2 W such that Ex
! Eui (this holds also
i
[
[ h n
b
b
b
b
if
Gx [ Jx is complete with lim Exn 2
Gx [ Jx
EW ). Now,
n!1
x2W
x2W
Lemma 1 implies that
b
b
d(Eu; Gu)
6 d(Eu; Ex2n ) + d(Ex2n ; Gu)
b 2n 1 ; Gu)
b
6 d(Eu; Ex2n ) + H(Jx
(
)!
b d(Ex2n 1 ; Jx
b 2n 1 );
d(Eu; Ex
);
d(Eu;
Gu);
2n
1
h
i
6 d(Eu; Ex2n ) + ' max
1
b
b
2 d(Eu; Jx2n 1 ) + d(Ex2n 1 ; Gu)
(
)!
b
d(Eu; Ex
Ex2n );
2n 1 ); d(Eu; Gu); d(Ex2n 1 ; i
h
6 d(Eu; Ex2n ) + ' max
:
1
b
2 d(Eu; Ex2n ) + d(Ex2n 1 ; Gu) :
Letting n ! 1; and using the fact that ' is continuous, we have,
b 6 ' d(Eu; Gu)
b
d(Eu; Gu)
:
Using the above inequality along with Lemma 4, we have
G (u) (Eu) > G (u) (x) ;
for all x 2 W: Similarly, by using
we have
b 6 d(Eu; x2n+1 ) + d(Ex2n+1 ; Ju);
b
d(Eu; Ju)
J (u) (Eu) > J (u) (x) ;
T
for all x 2 W: Hence u 2 (Coin (E; G) Coin (E; J)) : Finally, we show that, if
b Jy
b are singleton subsets of Z for all x; y 2 Z ; then (Coin (E; G) T Coin (E; J)) is
Gx;
T
singleton. For this, assume that there exist two distinct points u; v 2 (Coin (E; G) Coin (E; J)) : It
follows that
b = Ju
b and Ev 2 Gv
b = Jv:
b
Eu 6= Ev; Eu 2 Gu
Using inequality (1) , we obtain
d(Eu; Ev)
b Jv)
b
= H(Gu;
i
1h
b + d(Ev; Gu)
b
d(Eu; Jv)
2
1
6 ' max d(Eu; Ev); d(Eu; Eu); d(Ev; Ev); [d(Eu; Ev) + d(Ev; Eu)]
2
6 ' (d(Eu; Ev)) < d(Eu; Ev);
b d(Ev; Jv);
b
6 ' max d(Eu; Ev); d(Eu; Gu);
which is a contradiction, hence d(Eu; Ev) = 0 and so u = v:
Next we furnish an example to support our result.
6
Example 6 Let Z = [0; 1); d(x; y) = jx yj, whenever x; y 2 Z,
(0; 1] and de…ne a pair of mappings P; Q : (0 ; 1) ! I Z as follows:
8
2
if 0 6 t 6 3x 9x2
>
>
<
2
if 3x 9x2 6 t 6 3x
2
P(x)(t) =
>
if 3x < t < 6x
>
: 3
0
if 6x 6 t < 1;
8
2
if 0 6 t 6 3x 9x2
>
>
<
2
if 3x 9x2 6 t 6 6x
4
Q (x) (t) =
>
if 6x < t < 9x
>
: 10
0
if 9x 6 t < 1:
;
2
Now de…ne G; J : Z ! I Z ; E : Z ! Z as follows:
8
if x = 0
< f0g
P
(x)
if
x 2 0; 13
G(x)(t) =
:
if x > 13
f 12 g
8
if x = 0
< f0g
Q (x) if x 2 0; 31
J (x) (t) =
:
if x > 31 ;
f1g
2
De…ne
Ex = 3x for all x 2 Z:
: [0; 1) ! [0; 1) as follows:
t
(t) =
1
2
t2
2
if t 2 [0; 1]
if t > 1:
b Jx
b are compact for x 2 Z: If
It is obvious that (t) < t for t > 0; and Gx;
1
b = Jy
b and
x = y = 0 or x; y > 3 ; then Gx
If x = 0; y 2 0; 13
b Jy
b = 0:
H Gx;
h
b (0) = f0g, Jy
b = 0; 3y
; then G
b Jy
b
H Gx;
9y 2
= 3 (x
2
3 (x + y)
= 3 (x y) 1
2
3jx yj
6 3 jx yj 1
2
=
3x
6 3 jx
9x2
2
yj
i
and
9y 2
= (3y) = (jEy
2
b (0) ; Jy
b = 3y
H G
If x; y 2 0; 31 ; then
9y 2
2
3y +
(3 jx
y)
E0j):
9
x2
y2
2
2
yj)
2
7
= (j3 (x
y) j) = (jEx
Eyj):
If x 2 0; 13 ; y > 13 ; then
b Jy
b
H Gx;
= H
0; 3x
(1
3x)
6 j(1
3x)j
=
=
1
3
(3
1
9x2
1
9x2
;
= 1
3x
2
2
2
2
1
1
1 9x2 = (1 3x) 1
(1 + 3x)
2
2
1
1
2
1
(1 3x) = j(1 3x)j
j1 3xj
2
2
x
) 6 (3 jy
xj) = (jEy
Exj):
b Jx
b are not singleton subsets of Z for all x, however, all other
Note that Gx;
b 0 = 1 and
assumptions of Theorem 5 are satis…ed. Let x0 = 1; then Ex1 2 Gx
2
b 0 = d 3;
d (Ex0 ; Ex1 ) = d (3; Ex1 ) = d Ex0 ; Gx
1
2
b 1 = 0; 3 and
imply that Ex1 = 12 : Furthermore, Ex2 2 Jx
8
1
; Ex2
2
d (Ex1 ; Ex2 ) = d
b 1 =d
= d Ex1 ; Jx
3
1
; 0;
2
8
;
;
yield Ex2 = 38 : This process achieves lim Exn = 0:
n!1
The following theorem improves/generalizes the results of [3, 10, 14, 16, 18,
19, 22].
Theorem 7 Let W be a nonempty set, (Z; d) be a metric space and G; J : W !
C (Z) ; E : W ! Z . Let
[
([Gx]1 [ [Jx]1 ) EW
x2W
and for all x; y 2 W
d(Ex; Ey); p(Ex; G (x)); p(Ey; J (y));
:
1
2 [p(Ex; J (y)) + p(Ey; G (x))]
[
If E is injective and either EW or
([Gx]1 [ [Jx]1 ) is complete. Then there
d1 (G (x) ; J (y)) 6 ' max
x2W
exists a point u 2 Z such that fEug Gu; fEug Ju:
Proof. Pick x 2 Z; then by assumptions [Gx]1 ; [Jx]1 are nonempty compact
b = [Gx] and Jx
b = [Jx] for all x; y 2 Z; it follows
subsets of Z: Now, as Gx
1
1
b Jy)
b = D1 (G (x) ; J (y)) 6 d1 (G (x) ; J (y)): Thus
that H(Gx;
b Jy)
b 6 ' max
H(Gx;
d(Ex; Ey); p(Ex; G (x)); p(Ey; J (y));
1
2 [p(Ex; J (y)) + p(Ey; G (x))]
8
:
b
[Gx] :It follows that Gx
b
[Gx] for each
2 [0; 1] ; therefore, d(Ex; [Gx] ) 6 d(Ex; Gx) for each
2
b
[0; 1] and it implies that p(Ex; G (x)) 6 d(Ex; Gx):
This further implies that
(
)!
b d(Ey; Jy);
b
d(Ex;h Ey); d(Ex; Gx);
i
b Jy)
b 6 ' max
H(Gx;
:
1
b
b
2 d(Ex; Jy) + d(Ey; Gx)
By de…nition of
cuts; for
;we have[Gx]
b \ Ju:
b Hence, Eu 2
Now, by Theorem 5 there exists u 2 Z such that Eu 2 Gu
[Gu]1 \ [Ju]1 ; which implies that fEug Gu ; fEug Ju:
In the following we present some known results as corollaries of the above
theorem.
Corollary 8 [3] Let (V; dv ) be a complete metric linear space, A; B : V !
W(V ) be fuzzy mappings and there exists q 2 0; 12 such that for all x; y 2 V
d(x; y); p(x; A (x)); p(y; B (y));
p(x; B (y)); p(y; A (x))
d1 (A (x) ; B (y)) 6 q max
Then there exists a point u 2 Z such that fug Au ;
Proof. q 2 0; 21 implies that q = a2 where a < 1:Then
fug
Bu:
a
max
2
a
max
2
d(x; y); p(x; A (x)); p(y; B (y));
p(x; B (y)); p(y; A (x))
6 a max
d(x; y); p(x; A (x)); p(y; B (y));
d1 (A (x) ; B (y)) 6
6
:
d(x; y); p(x; A (x)); p(y; B (y));
p(x; B (y)) + p(y; A (x))
p(x;B(y))+ p(y;A(x))
2
It follows that
6
where
fug
max
d(x; y); p(x; A (x)); p(y; B (y));
p(x;B(y))+ p(y;A(x))
2
;
(r) = ar: For E = I; W = Z; Theorem 7 …nds u 2 V such that
Au ; fug Bu:
Corollary 9 [22] Let (V; dv ) be a complete metric linear space, G; J : V
W(V ) and there exists a 2 [0; 1) such that for all x; y 2 V .
d1 (G (x) ; J (y)) 6 a max
d(x; y); p(x; G (x)); p(y; J (y));
1
2 [p(x; J (y)) + p(y; G (x))]
Then there exists a point u 2 Z such that fug Gu ;
Proof. In Theorem 7, choose x = ax; E = I; W = Z.
9
fug
Ju:
:
!
Corollary 10 [16] Let (V; dv ) be a complete metric linear space, T : V
W(V ) be fuzzy mapping such that for all x; y 2 Z;
!
d1 (T (x) ; T (y)) 6 d(x; y)
where 0 6
4
< 1. Then there exists u 2 Z such that
fug
T u:
Applications
As applications of Theorem 5 we establish a common coincidence point result
for a single valued mapping and a pair of set-valued mappings. Moreover, we
apply it to establish some hypothesis which guarantee the existence of unique
solutions of a class of nonlinear integral equations.
Theorem 11 Let W be a nonempty set, (Z; d) be a metric space and A; B :
W ! C 2Z ; E : W ! Z. Let
[
x2W
[Ax [ Bx]
EW:
and for all x; y 2 W
H(Ax; By) 6 ' max d(Ex; Ey); d(Ex; Ax); d(Ey; By);
If E is injective and either EW or
[
x2W
point u 2 W such that
1
[d(Ex; By) + d(Ey; Ax)]
2
[Ax [ Bx] is complete. Then there exists
Eu 2 Au \ Bu:
(3)
Moreover, if Ax; Bx are single valued mappings then there exists a unique
solution of the system of equations Ex = Ax; Ex = Bx:
Proof. Pick ; 2 (0; 1] and de…ne two fuzzy mappings G; J : W ! I Z as
follows:
G(x) (t)
Then
99
100
1
100
t 2 Ax
;
t2
= Ax
J(x) (t) =
b = Ax and Jx
b = Bx:
Gx
99
100
1
100
t 2 Bx
t2
= Bx:
Therefore, Theorem 5 can be applied to obtain u 2 Z such that
\
u 2 Coin (E; G) Coin (E; J) :
10
:
It follows that
G (u) (Eu) > G (u) (x) ; J (u) (Eu) > J (u) (x)
for all x 2 W: Hence Eu 2 Au \ Bu: Moreover, if Ax; Bx are single-valued
mapb Jx
b are singleton subsets of Z for all x and so Coin(E; G) TCoin(E; J)
pings, then Gx;
is singleton. It follows that the system of equations Ex = Ax; Ex = Bx have a
unique solution.
Next, we consider the above theorem as a source of existence and uniqueness
theorem for an integral equation of form:
f (x (t))
Z
b
K (t; s; x (s)) ds = g (t) :
(4)
a
Here, x : [a; b] ! Rn is unknown; g : [a; b] ! Rn and f : Rn ! Rn are
given, is a parameter. The kernel K of the integral equation is de…ned on
[a; b] [a; b] Rn : This equation is reduced to Fredholm integral equation of 2nd
kind (see also [13] and the references cited therein) for n = 1 and f = I(identity
mapping).
Theorem 12 Let = (b 1 a) and f; g; K be continuous, where f is injective
and
kK (t; s; x1 ) K (t; s; x2 )k1 ' (kf x1 f x2 k1 )
for all t; s 2 [a; b] and x1 ; x2 2 Rn : If for each x 2 (C [a; b] ; Rn ) there exRb
ists y 2 (C [a; b] ; Rn ) such that (f y) (t) = g (t) +
K (t; s; x (s)) ds and
a
ff x : x 2 (C [a; b] ; Rn )g is complete. Then the integral equation (4) has a unique
solution in (C [a; b] ; Rn ) for any 2 [ ; ] :
Proof. Let Z = Y = (C [a; b] ; Rn ) and d (x; y) = max kx (t)
t2[a;b]
y (t)k1 for all
x; y 2 Z: Let ; : Z ! Z be de…ned as follows:
( x) (t) = g (t) +
Z
b
K (t; s; x (s)) ds;
x = f x:
a
Then by assumptions Z = f x : x 2 Zg is complete. Let x 2 JZ; then
x = x for x 2 Z: By assumptions there exists y 2 Z such that ( x) (t) =
11
(f y) (t) ; that is x = f y and hence Z
k( x) (t)
( y) (t)k1
Z
= j j
j j
j j
j j
Z: Moreover,
b
[K (t; s; x (s))] ds
a
Z
[K (t; s; y (s))] ds
a
kK (t; s; x (s))
Z
1
K (t; s; y (s))k1 ds
b
' (k(f x) (s)
(f y) (s)k1 ) ds
' (k( x) (s)
( y) (s)k1 ) ds
!
a
Z
b
a
sup k( x) (t)
t2[a;b]
j j
b
b
a
'
Z
' (d( x; y))
( y) (t)k1
j j (b
a)
' (d( x; y))
Therefore, for A = B = ; all conditions of Theorem 11.are satis…ed. Hence
there exists a unique w 2 Z such that (w) (t) = (w) (t) for all t; which is the
unique solution of the integral equation (4).
From the proof of Theorem 12, it follows that the solution w of (4) is the
limit of the sequence yn = f xn (t) obtained by the following iterative scheme:
f xn (t) = g (t) +
Z
b
K (t; s; xn
1
a
(s)) ds; x0 2 (C [a; b] ; Rn ) ; n = 1; 2; 3
(5)
Theorem 13 Let f; g; K be continuous, f is injective and there exists l 2 R
such that
kK (t; s; x1 ) K (t; s; x2 )k1 l kf x1 f x2 k1 ;
for all t; s 2 [a; b] and x1 ; x2 2 Rn : If for each x 2 (C [a; b] ; Rn ) there exRb
ists y 2 (C [a; b] ; Rn ) such that (f y) (t) = g (t) +
K (t; s; x (s)) ds and
a
ff x : x 2 (C [a; b] ; Rn )g is complete. Then the integral equation (4) has a unique
1
1
solution in (C [a; b] ; Rn ) for any 2
l(b a) ; l(b a) :
Proof. Let Z = Y = (C [a; b] ; Rn ) and d (x; y) = max kx (t)
t2[a;b]
x; y 2 Z: Let ; : Z ! Z; ' : R+ ! R+ be de…ned as follows:
( x) (t) = g (t) +
Z
b
K (t; s; x (s)) ds;
a
' (t) = j j l (b
12
a) t;
x = f x:
y (t)k1 for all
Then by assumptions
k( x) (t)
( y) (t)k1
Z = f x : x 2 Zg is complete and Z
Z
= j j
j j
b
[K (t; s; x (s))] ds
a
Z
j jl
b
[K (t; s; y (s))] ds
a
1
b
a
j jl
Z
Z: Since,
Z
kK (t; s; x (s))
b
kf x (s)
a
Z
f y (s)k1 ds
b
a
k( x) (s)
( y) (s)k1 ds
!
sup k( x) (t)
( y) (t)k1
t2[a;b]
j j l (b
K (t; s; y (s))k1 ds
j j l (b
a)
a) d ( x; y) = ' (d ( x; y))
Therefore, for A = B = ; all conditions of Theorem 11 are satis…ed. Hence,
there exists a unique w 2 Z such that (w) (t) = (w) (t) for all t: This is the
unique solution of (4), which can be approximated by using iterative scheme
(5).
In the following we furnish a simple example to verify (5).
Example 14 Consider the integral equation:
Z 1
3
8x3 (t) = t3 +
[tx (s)] ds
(6)
0
Let Z = Y = (C [0; 1] ; R) ; d (x; y) = max jx (t)
t2C[0;1]
jK (t; s; x1 )
K (t; s; x2 )j =
y (t)j for all x; y 2 Z: Since,
3
3
[tx1 ]
[tx2 ]
t3 x31 x32
1 3
t 8x31 8x32
8
l jf x1 f x2 j
3
where, K (t; s; x) = [tx (s)] ; f x = 8x3 l = 81 : Therefore, if 2 ( 8; 8) ; all
conditions of Theorem 13 are satis…ed. Now, we approximate the solution; by
constructing the iterative sequences:
Z 1
3
y n = t3 +
[txn 1 (s)] ds = 8x3n (t) x0 (t) = 0 n = 1; 2; 3;
:
0
It follows that
yn =
n
X
i=1
i
t3
1
(8i 1 ) (4i
"
n
1X
;
x
(t)
=
n
1)
8 i=1
13
i
t3
1
(8i 1 ) (4i
1)
# 31
n = 1; 2; 3
:
As
2 ( 8; 8) ; the series
Hence
1
X
t3
i 1
(8i
1 )(4i
i=1
w (t) =
4
32
1)
is convergent and yn !
32t3
32
:
1
3
t
is the required solution.
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