Open Access

A new common coupled fixed point theorem in generalized metric space and applications to integral equations

Fixed Point Theory and Applications20132013:266

https://doi.org/10.1186/1687-1812-2013-266

Received: 8 June 2013

Accepted: 19 September 2013

Published: 7 November 2013

Abstract

In the present paper, we prove a common coupled fixed point theorem in the setting of a generalized metric space in the sense of Mustafa and Sims. Our results improve and extend the corresponding results of Shatanawi. We also present an application to integral equations.

Keywords

G-metric spacecommon coupled coincidence fixed pointcommon fixed pointintegral equation

1 Introduction and preliminaries

The study of fixed points of mappings satisfying certain contractive conditions has been in the center of rigorous research activity. For a survey of common fixed point theory in metric and cone metric spaces, we refer the reader to [19]. In 2006, Bhaskar and Lakshmikantham [10] initiated the study of a coupled fixed point in ordered metric spaces and applied their results to prove the existence and uniqueness of solutions for a periodic boundary value problem. For more works in coupled and coincidence point theorems, we refer the reader to [1113].

Some authors generalized the concept of metric spaces in different ways. Mustafa and Sims [14] introduced the notion of G-metric space, in which the real number is assigned to every triplet of an arbitrary set as a generalization of the notion of metric spaces. Based on the notion of G-metric spaces, many authors (for example, [1533]) obtained some fixed point and common fixed point theorems for mappings satisfying various contractive conditions. Fixed point problems have also been considered in partially ordered G-metric spaces [3439].

The purpose of this paper is to obtain some common coupled coincidence point theorems in G-metric spaces satisfying some contractive conditions.

The following definitions and results will be needed in the sequel.

Definition 1.1 [14]

Let X be a nonempty set, and let G : X × X × X R + be a function satisfying the following axioms:
  1. (G1)

    G ( x , y , z ) = 0 if x = y = z ;

     
  2. (G2)

    0 < G ( x , x , y ) for all x , y X with x y ;

     
  3. (G3)

    G ( x , x , y ) G ( x , y , z ) for all x , y , z X with z y ;

     
  4. (G4)

    G ( x , y , z ) = G ( x , z , y ) = G ( y , z , x ) = (symmetry in all three variables);

     
  5. (G5)

    G ( x , y , z ) G ( x , a , a ) + G ( a , y , z ) for all x , y , z , a X (rectangle inequality),

     

then the function G is called a generalized metric, or more specifically, a G-metric on X, and the pair ( X , G ) is called a G-metric space.

Definition 1.2 [14]

Let ( X , G ) be a G-metric space, and let { x n } be a sequence of points in X, a point x in X is said to be the limit of the sequence { x n } if lim m , n G ( x , x n , x m ) = 0 , and one says that the sequence { x n } is G-convergent to x.

Thus, if x n x in a G-metric space ( X , G ) , then for any ϵ > 0 , there exists N N such that G ( x , x n , x m ) < ϵ for all n , m N .

Proposition 1.3 [14]

Let ( X , G ) be a G-metric space, then the following are equivalent:
  1. (1)

    { x n } is G-convergent to x.

     
  2. (2)

    G ( x n , x n , x ) 0 as n .

     
  3. (3)

    G ( x n , x , x ) 0 as n .

     
  4. (4)

    G ( x n , x m , x ) 0 as n , m .

     

Definition 1.4 [14]

Let ( X , G ) be a G-metric space. A sequence { x n } is called G-Cauchy sequence if for each ϵ > 0 , there exists a positive integer N N such that G ( x n , x m , x l ) < ϵ for all n , m , l N ; i.e., if G ( x n , x m , x l ) 0 as n , m , l .

Definition 1.5 [14]

A G-metric space ( X , G ) is said to be G-complete if every G-Cauchy sequence in ( X , G ) is G-convergent in X.

Proposition 1.6 [14]

Let ( X , G ) be a G-metric space, then the following are equivalent:
  1. (1)

    The sequence { x n } is G-Cauchy.

     
  2. (2)

    For every ϵ > 0 , there exists k N such that G ( x n , x m , x m ) < ϵ for all n , m k .

     

Proposition 1.7 [14]

Let ( X , G ) be a G-metric space. Then the function G ( x , y , z ) is jointly continuous in all three of its variables.

Definition 1.8 [14]

Let ( X , G ) and ( X , G ) be G-metric space, and let f : ( X , G ) ( X , G ) be a function. Then f is said to be G-continuous at a point a X if and only if for every ϵ > 0 , there is δ > 0 such that x , y X and G ( a , x , y ) < δ implies that G ( f ( a ) , f ( x ) , f ( y ) ) < ϵ . A function f is G-continuous at X if and only if it is G-continuous at all a X .

Proposition 1.9 [14]

Let ( X , G ) and ( X , G ) be G-metric spaces, then a function f : X X is G-continuous at a point x X if and only if it is G-sequentially continuous at x; that is, whenever ( x n ) is G-convergent to x, ( f ( x n ) ) is G-convergent to f ( x ) .

Proposition 1.10 [14]

Let ( X , G ) be a G-metric space. Then for any x, y, z, a in X, it follows that
  1. (i)

    if G ( x , y , z ) = 0 , then x = y = z ;

     
  2. (ii)

    G ( x , y , z ) G ( x , x , y ) + G ( x , x , z ) ;

     
  3. (iii)

    G ( x , y , y ) 2 G ( y , x , x ) ;

     
  4. (iv)

    G ( x , y , z ) G ( x , a , z ) + G ( a , y , z ) ;

     
  5. (v)

    G ( x , y , z ) 2 3 ( G ( x , y , a ) + G ( x , a , z ) + G ( a , y , z ) ) ;

     
  6. (vi)

    G ( x , y , z ) G ( x , a , a ) + G ( y , a , a ) + G ( z , a , a ) .

     

Definition 1.11 [10]

An element ( x , y ) X × X is called a coupled fixed point of a mapping F : X × X X if F ( x , y ) = x and F ( y , x ) = y .

Definition 1.12 [11]

An element ( x , y ) X × X is called a coupled coincidence point of the mappings F : X × X X and g : X X if F ( x , y ) = g x and F ( y , x ) = g y .

Definition 1.13 [11]

Let X be a nonempty set. Then we say that the mappings F : X × X X and g : X X are commutative if g F ( x , y ) = F ( g x , g y ) .

2 Main results

We start our work by proving the following crucial lemma.

Lemma 2.1 Let ( X , G ) be a G-metric space. Let F 1 , F 2 , F 3 : X × X X and g : X X be four mappings such that
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) a 1 G ( g x , g u , g w ) + a 2 G ( g y , g v , g z ) + a 3 G ( g x , g u , g u ) + a 4 G ( g y , g v , g v ) + a 5 G ( g u , g w , g w ) + a 6 G ( g v , g z , g z ) + a 7 G ( g w , g x , g x ) + a 8 G ( g z , g y , g y )
(2.1)
for all x , y , u , v , w , z X , where a i 0 , i = 1 , 2 , , 8 and a 1 + a 2 + a 3 + a 4 + a 7 + a 8 < 1 . Suppose that ( x , y ) is a common coupled coincidence point of the mappings pair ( F 1 , g ) , ( F 2 , g ) and ( F 3 , g ) . Then
F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) = g x = g y = F 1 ( y , x ) = F 2 ( y , x ) = F 3 ( y , x ) .
Proof Since ( x , y ) is a common coupled coincidence point of the mappings pair ( F 1 , g ) , ( F 2 , g ) and ( F 3 , g ) , we have g x = F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) and g y = F 1 ( y , x ) = F 2 ( y , x ) = F 3 ( y , x ) . Assume that g x g y . Then by (2.1), we get
G ( g x , g y , g y ) = G ( F 1 ( x , y ) , F 2 ( y , x ) , F 3 ( y , x ) ) a 1 G ( g x , g y , g y ) + a 2 G ( g y , g x , g x ) + a 3 G ( g x , g y , g y ) + a 4 G ( g y , g x , g x ) + a 5 G ( g y , g y , g y ) + a 6 G ( g x , g x , g x ) + a 7 G ( g y , g x , g x ) + a 8 G ( g x , g y , g y ) = ( a 1 + a 3 + a 8 ) G ( g x , g y , g y ) + ( a 2 + a 4 + a 7 ) G ( g y , g x , g x ) .
Also by (2.1), we have
G ( g y , g x , g x ) = G ( F 1 ( y , x ) , F 2 ( x , y ) , F 3 ( x , y ) ) a 1 G ( g y , g x , g x ) + a 2 G ( g x , g y , g y ) + a 3 G ( g y , g x , g x ) + a 4 G ( g x , g y , g y ) + a 5 G ( g x , g x , g x ) + a 6 G ( g y , g y , g y ) + a 7 G ( g x , g y , g y ) + a 8 G ( g y , g x , g x ) = ( a 1 + a 3 + a 8 ) G ( g y , g x , g x ) + ( a 2 + a 4 + a 7 ) G ( g x , g y , g y ) .
Therefore,
G ( g x , g y , g y ) + G ( g y , g x , g x ) ( a 1 + a 2 + a 3 + a 4 + a 7 + a 8 ) [ G ( g x , g y , g y ) + G ( g y , g x , g x ) ] .
Since 0 a 1 + a 2 + a 3 + a 4 + a 7 + a 8 < 1 , we get
G ( g x , g y , g y ) + G ( g y , g x , g x ) < G ( g x , g y , g y ) + G ( g y , g x , g x ) ,
which is a contradiction. So, g x = g y , and hence,
F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) = g x = g y = F 1 ( y , x ) = F 2 ( y , x ) = F 3 ( y , x ) .

 □

Theorem 2.1 Let ( X , G ) be a G-metric space. Let F 1 , F 2 , F 3 : X × X X and g : X X be four mappings such that
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) a 1 G ( g x , g u , g w ) + a 2 G ( g y , g v , g z ) + a 3 G ( g x , g u , g u ) + a 4 G ( g y , g v , g v ) + a 5 G ( g u , g w , g w ) + a 6 G ( g v , g z , g z ) + a 7 G ( g w , g x , g x ) + a 8 G ( g z , g y , g y )
(2.2)
for all x , y , u , v , w , z X , where a i 0 , i = 1 , 2 , , 8 and a 1 + a 2 + a 3 + a 4 + 2 a 5 + 2 a 6 + a 7 + a 8 < 1 . Suppose that F 1 , F 2 , F 3 and g satisfy the following conditions:
  1. (i)

    F 1 ( X × X ) g X , F 2 ( X × X ) g X , F 3 ( X × X ) g X ;

     
  2. (ii)

    gX is G-complete;

     
  3. (iii)

    g is G-continuous and commutes with F 1 , F 2 , F 3 .

     
Then there exist unique x X such that
g x = F 1 ( x , x ) = F 2 ( x , x ) = F 3 ( x , x ) = x .
Proof Let x 0 , y 0 X . Since F 1 ( X × X ) g X , F 2 ( X × X ) g X , F 3 ( X × X ) g X , we can choose x 1 , x 2 , x 3 , y 1 , y 2 , y 3 X such that g x 1 = F 1 ( x 0 , y 0 ) , g y 1 = F 1 ( y 0 , x 0 ) , g x 2 = F 2 ( x 1 , y 1 ) , g y 2 = F 2 ( y 1 , x 1 ) , g x 3 = F 3 ( x 2 , y 2 ) and g y 3 = F 3 ( y 2 , x 2 ) . Combining this process, we can construct two sequences { x n } and { y n } in X such that
g x 3 n = F 3 ( x 3 n 1 , y 3 n 1 ) , g y 3 n = F 3 ( y 3 n 1 , x 3 n 1 ) , n = 1 , 2 , 3 , , g x 3 n + 1 = F 1 ( x 3 n , y 3 n ) , g y 3 n + 1 = F 1 ( y 3 n , x 3 n ) , n = 0 , 1 , 2 , 3 , , g x 3 n + 2 = F 2 ( x 3 n + 1 , y 3 n + 1 ) , g y 3 n + 2 = F 2 ( y 3 n + 1 , x 3 n + 1 ) , n = 0 , 1 , 2 , 3 , .

If g x 3 n = g x 3 n + 1 , then g x = F 1 ( x , y ) , where x = x 3 n , y = y 3 n . If g x 3 n + 1 = g x 3 n + 2 , then g x = F 2 ( x , y ) , where x = x 3 n + 1 , y = y 3 n + 1 . If g x 3 n + 2 = g x 3 n + 3 , then g x = F 3 ( x , y ) , where x = x 3 n + 2 , y = y 3 n + 2 . On the other hand, if g y 3 n = g y 3 n + 1 , then g y = F 1 ( y , x ) , where y = y 3 n , x = x 3 n . If g y 3 n + 1 = g y 3 n + 2 , then g y = F 2 ( y , x ) , where y = y 3 n + 1 , x = x 3 n + 1 . If g y 3 n + 2 = g y 3 n + 3 , then g y = F 3 ( y , x ) , where y = y 3 n + 2 , x = x 3 n + 2 . Without loss of generality, we can assume that g x n g x n + 1 and g y n g y n + 1 , for all n = 0 , 1 , 2 ,  .

By (2.2) and (G3), we have
G ( g x 3 n , g x 3 n + 1 , g x 3 n + 2 ) = G ( F 3 ( x 3 n 1 , y 3 n 1 ) , F 1 ( x 3 n , y 3 n ) , F 2 ( x 3 n + 1 , y 3 n + 1 ) ) = G ( F 1 ( x 3 n , y 3 n ) , F 2 ( x 3 n + 1 , y 3 n + 1 ) , F 3 ( x 3 n 1 , y 3 n 1 ) ) a 1 G ( g x 3 n , g x 3 n + 1 , g x 3 n 1 ) + a 2 G ( g y 3 n , g y 3 n + 1 , g y 3 n 1 ) + a 3 G ( g x 3 n , g x 3 n + 1 , g x 3 n + 1 ) + a 4 G ( g y 3 n , g y 3 n + 1 , g y 3 n + 1 ) + a 5 G ( g x 3 n + 1 , g x 3 n 1 , g x 3 n 1 ) + a 6 G ( g y 3 n + 1 , g y 3 n 1 , g y 3 n 1 ) + a 7 G ( g x 3 n 1 , g x 3 n , g x 3 n ) + a 8 G ( g y 3 n 1 , g y 3 n , g y 3 n ) ( a 1 + a 3 + a 5 + a 7 ) G ( g x 3 n 1 , g x 3 n , g x 3 n + 1 ) + ( a 2 + a 4 + a 6 + a 8 ) G ( g y 3 n 1 , g y 3 n , g y 3 n + 1 ) .
(2.3)
Similarly, we have
G ( g y 3 n , g y 3 n + 1 , g y 3 n + 2 ) ( a 1 + a 3 + a 5 + a 7 ) G ( g y 3 n 1 , g y 3 n , g y 3 n + 1 ) + ( a 2 + a 4 + a 6 + a 8 ) G ( g x 3 n 1 , g x 3 n , g x 3 n + 1 ) .
(2.4)
By combining (2.3) and (2.4), we get
G ( g x 3 n , g x 3 n + 1 , g x 3 n + 2 ) + G ( g y 3 n , g y 3 n + 1 , g y 3 n + 2 ) ( i = 1 8 a i ) [ G ( g x 3 n 1 , g x 3 n , g x 3 n + 1 ) + G ( g y 3 n 1 , g y 3 n , g y 3 n + 1 ) ] .
(2.5)
In the same way, we can show that
G ( g x 3 n 1 , g x 3 n , g x 3 n + 1 ) + G ( g y 3 n 1 , g y 3 n , g y 3 n + 1 ) ( i = 1 8 a i ) [ G ( g x 3 n 2 , g x 3 n 1 , g x 3 n ) + G ( g y 3 n 2 , g y 3 n 1 , g y 3 n ) ]
(2.6)
and
G ( g x 3 n 2 , g x 3 n 1 , g x 3 n ) + G ( g y 3 n 2 , g y 3 n 1 , g y 3 n ) ( i = 1 8 a i ) [ G ( g x 3 n 3 , g x 3 n 2 , g x 3 n 1 ) + G ( g y 3 n 3 , g y 3 n 2 , g y 3 n 1 ) ] .
(2.7)
It follows from (2.5), (2.6) and (2.7) that for all n N , we have
G ( g x n , g x n + 1 , g x n + 2 ) + G ( g y n , g y n + 1 , g y n + 2 ) ( i = 1 8 a i ) [ G ( g x n 1 , g x n , g x n + 1 ) + G ( g y n 1 , g y n , g y n + 1 ) ] = k [ G ( g x n 1 , g x n , g x n + 1 ) + G ( g y n 1 , g y n , g y n + 1 ) ] k 2 [ G ( g x n 2 , g x n 1 , g x n ) + G ( g y n 2 , g y n 1 , g y n ) ] k n [ G ( g x 0 , g x 1 , g x 2 ) + G ( g y 0 , g y 1 , g y 2 ) ] .
(2.8)
Where k = i = 1 8 a i [ 0 , 1 ) . From (G3), we have G ( g x n , g x n + 1 , g x n + 1 ) G ( g x n , g x n + 1 , g x n + 2 ) and G ( g y n , g y n + 1 , g y n + 1 ) G ( g y n , g y n + 1 , g y n + 2 ) . Hence, by the (G3) and (2.8), we get
G ( g x n , g x n + 1 , g x n + 1 ) + G ( g y n , g y n + 1 , g y n + 1 ) G ( g x n , g x n + 1 , g x n + 2 ) + G ( g y n , g y n + 1 , g y n + 2 ) k n [ G ( g x 0 , g x 1 , g x 2 ) + G ( g y 0 , g y 1 , g y 2 ) ] .
(2.9)
Therefore, for all n , m N , n < m , by (G5) and (2.9), we have
G ( g x n , g x m , g x m ) + G ( g y n , g y m , g y m ) [ G ( g x n , g x n + 1 , g x n + 1 ) + G ( g y n , g y n + 1 , g y n + 1 ) ] + [ G ( g x n + 1 , g x n + 2 , g x n + 2 ) + G ( g y n + 1 , g y n + 2 , g y n + 2 ) ] + + [ G ( g x m 1 , g x m , g x m ) + G ( g y m 1 , g y m , g y m ) ] ( k n + k n + 1 + + k m 1 ) [ G ( g x 0 , g x 1 , g x 2 ) + G ( g y 0 , g y 1 , g y 2 ) ] k n 1 k [ G ( g x 0 , g x 1 , g x 2 ) + G ( g y 0 , g y 1 , g y 2 ) ] 0 as  n , m .
(2.10)
Which implies that
G ( g x n , g x m , g x m ) 0 and G ( g y n , g y m , g y m ) 0 as  n , m .
Thus, { g x n } and { g y n } are all G-Cauchy in gX. Since gX is G-complete, we get that { g x n } and { g y n } are G-convergent to some x g X and y g X , respectively. Since g is G-continuous, we have { g g x n } is G-convergent to gx and { g g y n } is G-convergent to gy. That is,
g g x n g x and g g y n g y as  n .
(2.11)
Also, since g commutes with F 1 , F 2 and F 3 , respectively, we have
g g x 3 n = g F 3 ( x 3 n 1 , y 3 n 1 ) = F 3 ( g x 3 n 1 , g y 3 n 1 ) , g g y 3 n = g F 3 ( y 3 n 1 , x 3 n 1 ) = F 3 ( g y 3 n 1 , g x 3 n 1 ) , g g x 3 n + 1 = g F 1 ( x 3 n , y 3 n ) = F 1 ( g x 3 n , g y 3 n ) , g g y 3 n + 1 = g F 1 ( y 3 n , x 3 n ) = F 1 ( g y 3 n , g x 3 n ) , g g x 3 n + 2 = g F 2 ( x 3 n + 1 , y 3 n + 1 ) = F 2 ( g x 3 n + 1 , g y 3 n + 1 ) , g g y 3 n + 2 = g F 2 ( y 3 n + 1 , x 3 n + 1 ) = F 2 ( g y 3 n + 1 , g x 3 n + 1 ) .
Thus, from condition (2.2), we have
G ( g g x 3 n , g g x 3 n + 1 , F 2 ( x , y ) ) = G ( F 1 ( g x 3 n , g y 3 n ) , F 2 ( x , y ) , F 3 ( g x 3 n 1 , g y 3 n 1 ) ) a 1 G ( g g x 3 n , g x , g g x 3 n 1 ) + a 2 G ( g g y 3 n , g y , g g y 3 n 1 ) + a 3 G ( g g x 3 n , g x , g x ) + a 4 G ( g g y 3 n , g y , g y ) + a 5 G ( g x , g g x 3 n 1 , g g x 3 n 1 ) + a 6 G ( g y , g g y 3 n 1 , g g y 3 n 1 ) + a 7 G ( g g x 3 n 1 , g g x 3 n , g g x 3 n ) + a 8 G ( g g y 3 n 1 , g g y 3 n , g g y 3 n ) .
Letting n , using (2.11) and the fact that G is continuous on its variables, we get that
G ( g x , g x , F 2 ( x , y ) ) = 0 .
Hence, g x = F 2 ( x , y ) . Similarly, we may show that g y = F 2 ( y , x ) . Also for the same reason, we may show that g x = F 1 ( x , y ) , g y = F 1 ( y , x ) , g x = F 3 ( x , y ) and g y = F 3 ( y , x ) . Therefore, ( x , y ) is a common coupled coincidence point of the pair ( F 1 , g ) , ( F 2 , g ) and ( F 3 , g ) . By Lemma 2.1, we obtain
g x = F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) = F 1 ( y , x ) = F 2 ( y , x ) = F 3 ( y , x ) = g y .
(2.12)
Since the sequences { g x 3 n 1 } , { g x 3 n } and { g x 3 n + 1 } are all a subsequence of { g x n } , then they are all G-convergent to x. Similarly, we may show that { g y 3 n 1 } , { g y 3 n } and { g y 3 n + 1 } are all G-convergent to y. From (2.2), we have
G ( g x 3 n , g x , g x ) = G ( F 1 ( x , y ) , F 2 ( x , y ) , F 3 ( x 3 n 1 , y 3 n 1 ) ) a 1 G ( g x , g x , g x 3 n 1 ) + a 2 G ( g y , g y , g y 3 n 1 ) + a 3 G ( g x , g x , g x ) + a 4 G ( g y , g y , g y ) + a 5 G ( g x , g x 3 n 1 , g x 3 n 1 ) + a 6 G ( g y , g y 3 n 1 , g y 3 n 1 ) + a 7 G ( g x 3 n 1 , g x , g x ) + a 8 G ( g y 3 n 1 , g y , g y ) .
Letting n , and using the fact that G is continuous on its variables, we get that
G ( x , g x , g x ) ( a 1 + a 7 ) G ( g x , g x , x ) + ( a 2 + a 8 ) G ( g y , g y , y ) + a 5 G ( g x , x , x ) + a 6 G ( g y , y , y ) .
Similarly, we may show that
G ( y , g y , g y ) ( a 1 + a 7 ) G ( g y , g y , y ) + ( a 2 + a 8 ) G ( g x , g x , x ) + a 5 G ( g y , y , y ) + a 6 G ( g x , x , x ) .
Thus, using the Proposition 1.10(iii), we have
G ( x , g x , g x ) + G ( y , g y , g y ) ( a 1 + a 2 + a 7 + a 8 ) [ G ( g x , g x , x ) + G ( g y , g y , y ) ] + ( a 5 + a 6 ) [ G ( g x , x , x ) + G ( g y , y , y ) ] ( a 1 + a 2 + 2 a 5 + 2 a 6 + a 7 + a 8 ) [ G ( g x , g x , x ) + G ( g y , g y , y ) ] .
Since 0 a 1 + a 2 + a 3 + a 4 + 2 a 5 + 2 a 6 + a 7 + a 8 < 1 , so the last inequality happens only if G ( x , g x , g x ) = 0 and G ( y , g y , g y ) = 0 . Hence, x = g x and y = g y . From (2.12), we have x = g x = g y = y , thus, we get
g x = F 1 ( x , x ) = F 2 ( x , x ) = F 3 ( x , x ) = x .
To prove the uniqueness, let z X with z x such that
z = g z = F 1 ( z , z ) = F 2 ( z , z ) = F 3 ( z , z ) .
Again using condition (2.2) and Proposition 1.10(iii), we have
G ( z , z , x ) = G ( F 1 ( z , z ) , F 2 ( z , z ) , F 3 ( x , x ) ) a 1 G ( g z , g z , g x ) + a 2 G ( g z , g z , g x ) + a 3 G ( g z , g z , g z ) + a 4 G ( g z , g z , g z ) + a 5 G ( g z , g x , g x ) + a 6 G ( g z , g x , g x ) + a 7 G ( g x , g z , g z ) + a 8 G ( g x , g z , g z ) ( a 1 + a 2 + 2 a 5 + 2 a 6 + a 7 + a 8 ) G ( z , z , x ) .

Since 0 a 1 + a 2 + a 3 + a 4 + 2 a 5 + 2 a 6 + a 7 + a 8 < 1 , we get G ( z , z , x ) < G ( z , z , x ) , which is a contradiction. Thus, F 1 , F 2 , F 3 and g have a unique common fixed point. □

Remark 2.1 Theorem 2.1 extends and improves Theorem 3.2 of Shatanawi [26].

The following corollary can be obtained from Theorem 2.1 immediately.

Corollary 2.1 Let ( X , G ) be a G-metric space. Let F 1 , F 2 , F 3 : X × X X and g : X X be mappings such that
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) a 1 G ( g x , g u , g w ) + a 2 G ( g y , g v , g z )
(2.13)
for all x , y , u , v , w , z X , where a i 0 , i = 1 , 2 and a 1 + a 2 < 1 . Suppose that F 1 , F 2 , F 3 and g satisfy the following conditions:
  1. (1)

    F 1 ( X × X ) g X , F 2 ( X × X ) g X , F 2 ( X × X ) g X ;

     
  2. (2)

    gX is G-complete;

     
  3. (3)

    g is G-continuous and commutes with F 1 , F 2 , F 3 .

     
Then there exist unique x X such that
g x = F 1 ( x , x ) = F 2 ( x , x ) = F 3 ( x , x ) = x .

Remark 2.2 If F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) and a 1 = a 2 = k , then Corollary 2.1 is reduced to Theorem 3.2 of Shatanawi [26].

Now, we give an example to support Corollary 2.1.

Example 2.1 Let X = [ 0 , 1 ] . Define G : X × X × X R + by
G ( x , y , z ) = | x y | + | y z | + | z x |
for all x , y , z X . Then ( X , G ) is a complete G-metric space. Define a map
F 1 , F 2 , F 3 : X × X X
by
F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) = x + y 8
for all x , y X . Also, define g : X X by g x = x 2 for x X . Then F ( X × X ) g X . Through calculation, we have
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) G ( x + y 8 , u + v 8 , w + z 8 ) = 1 8 ( | x u + y v | + | u w + v z | + | w x + z y | ) 1 8 ( | x u | + | y v | + | u w | + | v z | + | w x | + | z y | ) = 1 4 ( G ( g x , g u , g w ) + G ( g y , g v , g z ) ) .

Then the mappings F 1 , F 2 , F 3 and g are satisfying condition (2.13) of Corollary 2.1 with a 1 = a 2 = 1 4 . So that all the conditions of Corollary 2.1 are satisfied. By Corollary 2.4, F 1 , F 2 , F 3 and g have a unique common fixed point. Moreover, 0 is the unique common fixed point for all of the mappings F 1 , F 2 , F 3 and g.

If a 1 = a 2 = 0 , then Theorem 2.1 is reduced to the following.

Corollary 2.2 Let ( X , G ) be a G-metric space. Let F 1 , F 2 , F 3 : X × X X and g : X X be four mappings such that
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) c 1 G ( g x , g u , g u ) + c 2 G ( g y , g v , g v ) + c 3 G ( g u , g w , g w ) + c 4 G ( g v , g z , g z ) + c 5 G ( g w , g x , g x ) + c 6 G ( g z , g y , g y )
(2.14)
for all x , y , u , v , w , z X , where c i 0 , i = 1 , 2 , , 6 and c 1 + c 2 + 2 c 3 + 2 c 4 + c 5 + c 6 < 1 . Suppose that F 1 , F 2 , F 3 and g satisfy the following conditions:
  1. (i)

    F 1 ( X × X ) g X , F 2 ( X × X ) g X , F 3 ( X × X ) g X ;

     
  2. (ii)

    gX is G-complete;

     
  3. (iii)

    g is G-continuous and commutes with F 1 , F 2 , F 3 .

     
Then there exist unique x X such that
g x = F 1 ( x , x ) = F 2 ( x , x ) = F 3 ( x , x ) = x .

If we take F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) in Corollary 2.2, then the following corollary is obtained.

Corollary 2.3 Let ( X , G ) be a G-metric space. Let F : X × X X and g : X X be four mappings such that
G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) c 1 G ( g x , g u , g u ) + c 2 G ( g y , g v , g v ) + c 3 G ( g u , g w , g w ) + c 4 G ( g v , g z , g z ) + c 5 G ( g w , g x , g x ) + c 6 G ( g z , g y , g y )
(2.15)
for all x , y , u , v , w , z X , where c i 0 , i = 1 , 2 , , 6 and c 1 + c 2 + 2 c 3 + 2 c 4 + c 5 + c 6 < 1 . Suppose that F and g satisfy the following conditions:
  1. (i)

    F ( X × X ) g X ;

     
  2. (ii)

    gX is G-complete;

     
  3. (iii)

    g is G-continuous and commutes with F.

     
Then there exist unique x X such that
g x = F ( x , x ) = x .

Now, we give an example to support Corollary 2.3.

Example 2.2 Let X = [ 0 , 1 ] . Define G : X × X × X R + by
G ( x , y , z ) = | x y | + | y z | + | z x |
for all x , y , z X . Then ( X , G ) is a complete G-metric space. Define a map F : X × X X by
F ( x , y ) = x y 8
for all x , y X . Also, define g : X X by g x = x for x X . Then F ( X × X ) g X . Through calculation, we have
G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) = 1 8 ( | x y u v | + | u v w z | + | w z x y | ) 1 8 ( | y | | x u | + | u | | y v | + | v | | u w | + | w | | v z | + | z | | w x | + | x | | z y | ) 1 8 ( | x u | + | y v | + | u w | + | v z | + | w x | + | z y | ) = 1 16 ( G ( g x , g u , g u ) + G ( g y , g v , g v ) + G ( g u , g w , g w ) + G ( g v , g z , g z ) + G ( g w , g x , g x ) + c 6 G ( g z , g y , g y ) ) .

Then the mappings F 1 , F 2 , F 3 and g are satisfying condition (2.15) of Corollary 2.3 with c 1 = c 2 = c 3 = c 4 = c 5 = c 6 = 1 16 . So that all the conditions of Corollary 2.3 are satisfied. By Corollary 2.3, F and g have a unique common fixed point. Moreover, 0 is the unique common fixed point for all of the mappings F and g.

If we take F 1 ( x , y ) = F 2 ( x , y ) = F 3 ( x , y ) in Theorem 2.1, then the following corollary is obtained.

Corollary 2.4 Let ( X , G ) be a G-metric space. Let F : X × X X and g : X X be mappings such that
G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) a 1 G ( g x , g u , g w ) + a 2 G ( g y , g v , g z ) + a 3 G ( g x , g u , g u ) + a 4 G ( g y , g v , g v ) + a 5 G ( g u , g w , g w ) + a 6 G ( g v , g z , g z ) + a 7 G ( g w , g x , g x ) + a 8 G ( g z , g y , g y )
(2.16)
for all x , y , u , v , w , z X , where a i 0 , i = 1 , 2 , , 8 and a 1 + a 2 + a 3 + a 4 + 2 a 5 + 2 a 6 + a 7 + a 8 < 1 . Suppose that F and g satisfy the following conditions:
  1. (1)

    F ( X × X ) g X ;

     
  2. (2)

    gX is G-complete;

     
  3. (3)

    g is G-continuous and commutes with F.

     

Then there exist unique x X such that g x = F ( x , x ) = x .

Now, we introduce an example to support Corollary 2.4.

Example 2.3 Let X = [ 1 , 1 ] . Define G : X × X × X R + by
G ( x , y , z ) = | x y | + | y z | + | z x |
for all x , y , z X . Then ( X , G ) is a complete G-metric space. Define a map
F : X × X X
by
F ( x , y ) = 1 16 x 2 + 1 16 y 2 1

for all x , y X . Also, define g : X X by g x = x for x X .

Clearly, we can get F ( X × X ) = [ 1 , 7 8 ] g X , and g is G-continuous and commutes with F.

By the definition of the mappings of F and g, for all x , y , z , u , v , w [ 1 , 1 ] , we have
G ( F ( x , y ) , F ( u , v ) , F ( w , z ) ) G ( 1 16 x 2 + 1 16 y 2 1 , 1 16 u 2 + 1 16 v 2 1 , 1 16 w 2 + 1 16 z 2 1 ) = 1 16 ( | x 2 u 2 + y 2 v 2 | + | u 2 w 2 + v 2 z 2 | + | w 2 x 2 + z 2 y 2 | ) 1 16 ( | x 2 u 2 | + | y 2 v 2 | + | u 2 w 2 | + | v 2 z 2 | + | w 2 x 2 | + | z 2 y 2 | ) 1 16 ( 2 | x u | + 2 | y v | + 2 | u w | + 2 | v z | + 2 | w x | + 2 | z y | ) = 1 16 G ( g x , g u , g u ) + 1 16 G ( g y , g v , g v ) + 1 16 G ( g u , g w , g w ) + 1 16 G ( g v , g z , g z ) + 1 16 G ( g w , g x , g x ) + 1 16 G ( g z , g y , g y ) .

Then the mappings F and g are satisfying condition (2.16) of Corollary 2.4 with a 1 = a 2 = 0 , a 3 = a 4 = a 5 = a 6 = a 7 = a 8 = 1 16 . So that all the conditions of Corollary 2.4 are satisfied. By Corollary 2.4, F and g have a unique common fixed point. Here x = 4 2 6 is the unique common fixed point of mappings F and g; that is, F ( x , x ) = g x = x .

3 Application to integral equations

Throughout this section, we assume that X = C [ 0 , 1 ] is the set of all continuous functions defined on [ 0 , 1 ] . Define G : X × X × X R + by
G ( x , y , z ) = sup t [ 0 , 1 ] | x ( t ) y ( t ) | + sup t [ 0 , 1 ] | y ( t ) z ( t ) | + sup t [ 0 , 1 ] | z ( t ) x ( t ) |

for all x , y , z X . Then ( X , G ) is a G-complete metric space.

Consider the following integral equations:
F i ( x , y ) ( t ) = 0 1 k ( t , s ) ( f i ( s , x ( s ) ) + g i ( s , y ( s ) ) ) d s , t [ 0 , 1 ] ( i = 1 , 2 , 3 ) .
(3.1)
Next, we will analyze (3.1) under the following conditions:
  1. (i)

    k : [ 0 , 1 ] × [ 0 , 1 ] R + is continuous.

     
  2. (ii)

    f i , g i : [ 0 , 1 ] × R R ( i = 1 , 2 , 3 ) are continuous functions.

     
  3. (iii)
    There exist constants λ i , μ i > 0 ( i = 1 , 2 , 3 ) such that
    { | f 1 ( t , x ) f 2 ( t , y ) | λ 1 | x y | , | f 2 ( t , x ) f 3 ( t , y ) | λ 2 | x y | , | f 3 ( t , x ) f 1 ( t , y ) | λ 3 | x y | and { | g 1 ( t , x ) g 2 ( t , y ) | μ 1 | x y | , | g 2 ( t , x ) g 3 ( t , y ) | μ 2 | x y | , | g 3 ( t , x ) g 1 ( t , y ) | μ 3 | x y |
     
for all t [ 0 , 1 ] and x , y R .
  1. (iv)
    k ( max { λ 1 , μ 1 } + 2 max { λ 2 , μ 2 } + max { λ 3 , μ 3 } ) < 1 , where
    k = sup { k ( t , s ) : t , s [ 0 , 1 ] } .
     

The aim of this section is to give an existence theorem for a solution of the above integral equations by using the obtained result given by Theorem 2.1.

Theorem 3.1 Under conditions (i)-(iv), integral equation (3.1) has a unique common solution in C [ 0 , 1 ] .

Proof First, we consider F i : X × X X ( i = 1 , 2 , 3 ). By virtue of our assumptions, F i is well defined (this means that for x , y X then F i ( x , y ) X ( i = 1 , 2 , 3 )). Then we can get
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) = sup t [ 0 , 1 ] | F 1 ( x , y ) F 2 ( u , v ) | + sup t [ 0 , 1 ] | F 2 ( u , v ) F 3 ( w , z ) | + sup t [ 0 , 1 ] | F 3 ( w , z ) F 1 ( x , y ) | = sup t [ 0 , 1 ] | 0 1 k ( t , s ) ( f 1 ( s , x ( s ) ) + g 1 ( s , y ( s ) ) ) d s 0 1 k ( t , s ) ( f 2 ( s , u ( s ) ) + g 2 ( s , v ( s ) ) ) d s | + sup t [ 0 , 1 ] | 0 1 k ( t , s ) ( f 2 ( s , u ( s ) ) + g 2 ( s , v ( s ) ) ) d s 0 1 k ( t , s ) ( f 3 ( s , w ( s ) ) + g 3 ( s , z ( s ) ) ) d s | + sup t [ 0 , 1 ] | 0 1 k ( t , s ) ( f 3 ( s , w ( s ) ) + g 3 ( s , z ( s ) ) ) d s 0 1 k ( t , s ) ( f 1 ( s , x ( s ) ) + g 1 ( s , y ( s ) ) ) d s | = sup t [ 0 , 1 ] | 0 1 k ( t , s ) ( ( f 1 ( s , x ( s ) ) f 2 ( s , u ( s ) ) ) + ( g 1 ( s , y ( s ) ) g 2 ( s , v ( s ) ) ) ) d s | + sup t [ 0 , 1 ] | 0 1 k ( t , s ) ( ( f 2 ( s , u ( s ) ) f 3 ( s , w ( s ) ) ) + ( g 2 ( s , v ( s ) ) g 3 ( s , z ( s ) ) ) ) d s | + sup t [ 0 , 1 ] | 0 1 k ( t , s ) ( ( f 3 ( s , w ( s ) ) f 1 ( s , x ( s ) ) ) + ( g 3 ( s , z ( s ) ) g 1 ( s , y ( s ) ) ) ) d s | sup t [ 0 , 1 ] 0 1 k ( t , s ) ( | f 1 ( s , x ( s ) ) f 2 ( s , u ( s ) ) | + | g 1 ( s , y ( s ) ) g 2 ( s , v ( s ) ) | ) d s + sup t [ 0 , 1 ] 0 1 k ( t , s ) ( | f 2 ( s , u ( s ) ) f 3 ( s , w ( s ) ) | + | g 2 ( s , v ( s ) ) g 3 ( s , z ( s ) ) | ) d s + sup t [ 0 , 1 ] 0 1 k ( t , s ) ( | f 3 ( s , w ( s ) ) f 1 ( s , x ( s ) ) | + | g 3 ( s , z ( s ) ) g 1 ( s , y ( s ) ) | ) d s .
(3.2)
By conditions (iii),
{ | f 1 ( s , x ( s ) ) f 2 ( s , u ( s ) ) | λ 1 | x ( s ) u ( s ) | , | f 2 ( s , u ( s ) ) f 3 ( s , w ( s ) ) | λ 2 | u ( s ) w ( s ) | , | f 3 ( s , w ( s ) ) f 1 ( s , x ( s ) ) | λ 3 | w ( s ) x ( s ) |
and
{ | g 1 ( s , y ( s ) ) g 2 ( s , v ( s ) ) | μ 1 | y ( s ) v ( s ) | , | g 2 ( s , v ( s ) ) g 3 ( s , z ( s ) ) | μ 2 | v ( s ) z ( s ) | , | g 3 ( s , z ( s ) ) g 1 ( s , y ( s ) ) | μ 3 | z ( s ) y ( s ) | .
Taking these inequalities into (3.2), we obtain
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) sup t [ 0 , 1 ] 0 1 k ( t , s ) ( λ 1 | x ( s ) u ( s ) | + μ 1 | y ( s ) v ( s ) | ) d s + sup t [ 0 , 1 ] 0 1 k ( t , s ) ( λ 2 | u ( s ) w ( s ) | + μ 2 | v ( s ) z ( s ) | ) + sup t [ 0 , 1 ] 0 1 k ( t , s ) ( λ 3 | w ( s ) x ( s ) | + μ 3 | z ( s ) y ( s ) | ) max { λ 1 , μ 1 } sup t [ 0 , 1 ] 0 1 k ( t , s ) ( | x ( s ) u ( s ) | + | y ( s ) v ( s ) | ) d s + max { λ 2 , μ 2 } sup t [ 0 , 1 ] 0 1 k ( t , s ) ( | u ( s ) w ( s ) | + | v ( s ) z ( s ) | ) d s + max { λ 3 , μ 3 } sup t [ 0 , 1 ] 0 1 k ( t , s ) ( | w ( s ) x ( s ) | + | z ( s ) y ( s ) | ) d s .
(3.3)
Using the Cauchy-Schwartz inequality in (3.3), we get
0 1 k ( t , s ) ( | x ( s ) u ( s ) | + | y ( s ) v ( s ) | ) d s ( 0 1 k 2 ( t , s ) d s ) 1 2 ( 0 1 ( | x ( s ) u ( s ) | + | y ( s ) v ( s ) | ) 2 d s ) 1 2 k ( sup t [ 0 , 1 ] | x ( t ) u ( t ) | + sup t [ 0 , 1 ] | y ( t ) v ( t ) | ) .
(3.4)
Similarly, we can obtain the following estimate
0 1 k ( t , s ) ( | u ( s ) w ( s ) | + | v ( s ) z ( s ) | ) d s k ( sup t [ 0 , 1 ] | u ( t ) w ( t ) | + sup t [ 0 , 1 ] | v ( t ) z ( t ) | ) ,
(3.5)
0 1 k ( t , s ) ( | w ( s ) x ( s ) | + | z ( s ) y ( s ) | ) d s k ( sup t [ 0 , 1 ] | w ( t ) x ( t ) | + sup t [ 0 , 1 ] | z ( t ) y ( t ) | ) .
(3.6)
Substituting (3.4), (3.5) and (3.6) into (3.3), we obtain that
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) max { λ 1 , μ 1 } k ( sup t [ 0 , 1 ] | x ( t ) u ( t ) | + sup t [ 0 , 1 ] | y ( t ) v ( t ) | ) + max { λ 2 , μ 2 } k ( sup t [ 0 , 1 ] | u ( t ) w ( t ) | + sup t [ 0 , 1 ] | v ( t ) z ( t ) | ) + max { λ 3 , μ 3 } k ( sup t [ 0 , 1 ] | w ( t ) x ( t ) | + sup t [ 0 , 1 ] | z ( t ) y ( t ) | ) = 1 2 max { λ 1 , μ 1 } k 2 sup t [ 0 , 1 ] | x ( t ) u ( t ) | + 1 2 max { λ 1 , μ 1 } k 2 sup t [ 0 , 1 ] | y ( t ) v ( t ) | + 1 2 max { λ 2 , μ 2 } k 2 sup t [ 0 , 1 ] | u ( t ) w ( t ) | + 1 2 max { λ 2 , μ 2 } k 2 sup t [ 0 , 1 ] | v ( t ) z ( t ) | + 1 2 max { λ 3 , μ 3 } k 2 sup t [ 0 , 1 ] | w ( t ) x ( t ) | + 1 2 max { λ 3 , μ 3 } k 2 sup t [ 0 , 1 ] | z ( t ) y ( t ) | = 1 2 max { λ 1 , μ 1 } k G ( x , u , u ) + 1 2 max { λ 1 , μ 1 } k G ( y , v , v ) + 1 2 max { λ 2 , μ 2 } k G ( u , w , w ) + 1 2 max { λ 2 , μ 2 } k G ( v , z , z ) + 1 2 max { λ 3 , μ 3 } k G ( w , x , x ) + 1 2 max { λ 3 , μ 3 } k G ( z , y , y ) .
(3.7)
Taking g x = x for all x X , and
a 1 = a 2 = 0 , a 3 = a 4 = 1 2 max { λ 1 , μ 1 } k , a 5 = a 6 = 1 2 max { λ 2 , μ 2 } k , a 7 = a 8 = 1 2 max { λ 3 , μ 3 } k ,
then inequality (3.7) becomes
G ( F 1 ( x , y ) , F 2 ( u , v ) , F 3 ( w , z ) ) a 1 G ( g x , g u , g w ) + a 2 G ( g y , g v , g z ) + a 3 G ( g x , g u , g u ) + a 4 G ( g y , g v , g v ) + a 5 G ( g u , g w , g w ) + a 6 G ( g v , g z , g z ) + a 7 G ( g w , g x , g x ) + a 8 G ( g z , g y , g y ) .
(3.8)
By condition (iv), we know that
a 1 + a 2 + a 3 + a 4 + 2 ( a 5 + a 6 ) + a 7 + a 8 = k ( max { λ 1 , μ 1 } + 2 max { λ 2 , μ 2 } + max { λ 3 , μ 3 } ) < 1 .

This proves that the operator F i ( i = 1 , 2 , 3 ) and g = I satisfy contractive condition (2.2) appearing in Theorem 2.1 with g = I . Therefore, F 1 , F 2 , F 3 have a unique common coupled fixed point, that is, F 1 ( x , x ) = F 2 ( x , x ) = F 3 ( x , x ) = x , and so, ( x , x ) is the unique solution of equation (3.1). □

Declarations

Acknowledgements

The authors are grateful to the editor and the reviewer for suggestions which improved the contents of the article. This work is supported by the National Natural Science Foundation of China (11271105), the Natural Science Foundation of Zhejiang Province (Y6110287, LY12A01030) and the Physical Experiment Center of Hangzhou Normal University.

Authors’ Affiliations

(1)
Department of Mathematics, Institute of Applied Mathematics, Hangzhou Normal University

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© Gu and Yin; licensee Springer. 2013

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