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Fixed points of ( ψ , ϕ , θ ) -contractive mappings in partially ordered b-metric spaces and application to quadratic integral equations

Fixed Point Theory and Applications20132013:245

https://doi.org/10.1186/1687-1812-2013-245

Received: 16 April 2013

Accepted: 20 August 2013

Published: 7 November 2013

Abstract

We prove some coupled coincidence and coupled common fixed point theorems for mappings satisfying ( ψ , ϕ , θ ) -contractive conditions in partially ordered complete b-metric spaces. The obtained results extend and improve many existing results from the literature. As an application, we prove the existence of a unique solution to a class of nonlinear quadratic integral equations.

MSC:47H10, 54H25.

Keywords

coupled common fixed point coupled fixed point coupled coincidence point mixed g-monotone property b-metric space partially ordered set

1 Introduction and preliminaries

In [1, 2], Czerwik introduced the notion of a b-metric space, which is a generalization of the usual metric space, and generalized the Banach contraction principle in the context of complete b-metric spaces. After that, many authors have carried out further studies on b-metric spaces and their topological properties (see, e.g., [114]). In this paper, some coupled coincidence and coupled common fixed point theorems for mappings satisfying ( ψ , ϕ , θ ) -contractive conditions in partially ordered complete b-metric spaces are proved. Also, we apply our results to study the existence of a unique solution to a large class of nonlinear quadratic integral equations. There are many papers in the literature concerning coupled fixed points introduced by Bhaskar and Lakshmikantham [15] and their applications in the existence and uniqueness of solutions for boundary value problems. A number of articles on this topic have been dedicated to the improvement and generalization; see [1620] and references therein. Also, to see some results on common fixed points for generalized contraction mappings, we refer the reader to [2123]. For the sake of convenience, some definitions and notations are recalled from [1, 3, 24] and [25].

Definition 1.1 [1]

Let X be a (nonempty) set and s 1 be a given real number. A function d : X × X R + is said to be a b-metric space iff for all x , y , z X , the following conditions are satisfied:
  1. (i)

    d ( x , y ) = 0 iff x = y ,

     
  2. (ii)

    d ( x , y ) = d ( y , x ) ,

     
  3. (iii)

    d ( x , y ) s [ d ( x , z ) + d ( z , y ) ] .

     

The pair ( X , d ) is called a b-metric space with the parameter s.

It should be noted that the class of b-metric spaces is effectively larger than that of metric spaces since a b-metric is a metric when s = 1 .

The following example shows that, in general, a b-metric need not necessarily be a metric (see also [14]).

Example 1.2 [3]

Let ( X , d ) be a metric space and ρ ( x , y ) = ( d ( x , y ) ) p , where p > 1 is a real number. Then ρ is a b-metric with s = 2 p 1 . However, if ( X , d ) is a metric space, then ( X , ρ ) is not necessarily a metric space. For example, if X = R is the set of real numbers and d ( x , y ) = | x y | is the usual Euclidean metric, then ρ ( x , y ) = ( x y ) s is a b-metric on with s = 2 , but is not a metric on .

Also, the following example of a b-metric space is given in [26].

Example 1.3 [26]

Let X be the set of Lebesgue measurable functions on [ 0 , 1 ] such that 0 1 | f ( x ) | 2 d x < . Define D : X × X [ 0 , ) by D ( f , g ) = 0 1 | f ( x ) g ( x ) | 2 d x . As ( 0 1 | f ( x ) g ( x ) | 2 d x ) 1 2 is a metric on X, then, from the previous example, D is a b-metric on X, with s = 2 .

Khamsi [27] also showed that each cone metric space over a normal cone has a b-metric structure.

Since, in general, a b-metric is not continuous, we need the following simple lemma about the b-convergent sequences in the proof of our main result.

Lemma 1.4 [3]

Let ( X , d ) be a b-metric space with s 1 , and suppose that { x n } and { y n } are b-convergent to x, y, respectively. Then we have
1 s 2 d ( x , y ) lim inf d ( x n , y n ) lim sup d ( x n , y n ) s 2 d ( x , y ) .
In particular, if x = y , then we have lim d ( x n , y n ) = 0 . Moreover, for each z X , we have
1 s d ( x , z ) lim inf d ( x n , z ) lim sup d ( x n , z ) s d ( x , z ) .

In [25], Lakshmikantham and Ćirić introduced the concept of mixed g-monotone property as follows.

Definition 1.5 [25]

Let ( X , ) be a partially ordered set and F : X × X X and g : X X . We say F has the mixed g-monotone property if F is non-decreasing g-monotone in its first argument and is non-increasing g-monotone in its second argument, that is, for any x , y X ,
x 1 , x 2 X , g x 1 g x 2 F ( x 1 , y ) F ( x 2 , y )
and
y 1 , y 2 X , g y 1 g y 2 F ( x , y 1 ) F ( x , y 2 ) .

Note that if g is an identity mapping, then F is said to have the mixed monotone property (see also [15]).

Definition 1.6 [25]

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

Similarly, note that if g is an identity mapping, then ( x , y ) is called a coupled fixed point of the mapping F (see also [15]).

Definition 1.7 [24]

An element x X is called a common fixed point of a mapping F : X × X X and g : X X if
F ( x , x ) = g x = x .
(1.1)

Definition 1.8 [25]

Let X be a nonempty set and F : X × X X and g : X X . One says that F and g are commutative if for all x , y X ,
F ( g x , g y ) = g ( F ( x , y ) ) .

Definition 1.9 [28]

The mappings F and g, where F : X × X X and g : X X , are said to be compatible if
lim n d ( g ( F ( x n , y n ) ) , F ( g x n , g y n ) ) = 0
and
lim n d ( g ( F ( y n , x n ) ) , F ( g y n , g x n ) ) = 0 ,

whenever { x n } and { y n } are sequences in X such that lim n F ( x n , y n ) = lim n g x n = x and lim n F ( y n , x n ) = lim n g y n = y for all x , y X .

2 Main results

Throughout the paper, let Ψ be a family of all functions ψ : [ 0 , ) [ 0 , ) satisfying the following conditions:
  1. (a)

    ψ is continuous,

     
  2. (b)

    ψ non-decreasing,

     
  3. (c)

    ψ ( t ) = 0 if and only if t = 0 .

     
We denote by Φ the set of all functions ϕ : [ 0 , ) [ 0 , ) satisfying the following conditions:
  1. (a)

    ϕ is lower semi-continuous,

     
  2. (b)

    ϕ ( t ) = 0 if and only if t = 0 ,

     

and Θ the set of all continuous functions θ : [ 0 , ) [ 0 , ) with θ ( t ) = 0 if and only if t = 0 .

Let ( X , d , ) be a partially ordered b-metric space, and let T : X × X X and g : X X be two mappings. Set
M s , T , g ( x , y , u , v ) = max { d ( g x , g u ) , d ( g y , g v ) , d ( g x , T ( x , y ) ) , 1 2 s d ( g u , T ( u , v ) ) , d ( g y , T ( y , x ) ) , 1 2 s d ( g v , T ( v , u ) ) , d ( g x , T ( u , v ) ) + d ( g u , T ( x , y ) ) 2 s , d ( g y , T ( v , u ) ) + d ( g v , T ( y , x ) ) 2 s }
and
N T , g ( x , y , u , v ) = min { d ( g x , T ( x , y ) ) , d ( g u , T ( u , v ) ) , d ( g u , T ( x , y ) ) , d ( g x , T ( u , v ) ) } .

Now, we introduce the following definition.

Definition 2.1 Let ( X , d , ) be a partially ordered b-metric space and ψ Ψ , ϕ Φ and θ Θ . We say that T : X × X X is an almost generalized ( ψ , ϕ , θ ) -contractive mapping with respect to g : X X if there exists L 0 such that
ψ ( s 3 d ( T ( x , y ) , T ( u , v ) ) ) ψ ( M s , T , g ( x , y , u , v ) ) ϕ ( M s , T , g ( x , y , u , v ) ) + L θ ( N T , g ( x , y , u , v ) )
(2.1)

for all x , y , u , v X with g x g u and g y g v .

Now, we establish some results for the existence of a coupled coincidence point and a coupled common fixed point of mappings satisfying almost generalized ( ψ , ϕ , θ ) -contractive condition in the setup of partially ordered b-metric spaces. The first result in this paper is the following coupled coincidence theorem.

Theorem 2.2 Suppose that ( X , d , ) is a partially ordered complete b-metric space. Let T : X × X X be an almost generalized ( ψ , ϕ , θ ) -contractive mapping with respect to g : X X , and T and g are continuous such that T has the mixed g-monotone property and commutes with g. Also, suppose T ( X × X ) g ( X ) . If there exists ( x 0 , y 0 ) X × X such that g x 0 T ( x 0 , y 0 ) and g y 0 T ( y 0 , x 0 ) , then T and g have coupled coincidence point in X.

Proof By the given assumptions, there exists ( x 0 , y 0 ) X × X such that g x 0 T ( x 0 , y 0 ) and g y 0 T ( y 0 , x 0 ) . Since T ( X × X ) g ( X ) , we can define ( x 1 , y 1 ) X × X such that g x 1 = T ( x 0 , y 0 ) and g y 1 = T ( y 0 , x 0 ) , then g x 0 T ( x 0 , y 0 ) = g x 1 and g y 0 T ( y 0 , x 0 ) = g y 1 . Also, there exists ( x 2 , y 2 ) X × X such that g x 2 = T ( x 1 , y 1 ) and g y 2 = T ( y 1 , x 1 ) . Since T has the mixed g-monotone property, we have
g x 1 = T ( x 0 , y 0 ) T ( x 0 , y 1 ) T ( x 1 , y 1 ) = g x 2
and
g y 2 = T ( y 1 , x 1 ) T ( y 0 , x 1 ) T ( y 0 , x 0 ) = g y 1 .
Continuing in this way, we construct two sequences { x n } and { y n } in X such that
g x n + 1 = T ( x n , y n ) and g y n + 1 = T ( y n , x n ) for all  n = 0 , 1 , 2 ,
(2.2)
for which
g x 0 g x 1 g x 2 g x n g x n + 1 , g y 0 g y 1 g y 2 g y n g y n + 1 .
(2.3)
From (2.2) and (2.3) and inequality (2.1) with ( x , y ) = ( x n , y n ) and ( u , v ) = ( x n + 1 , y n + 1 ) , we obtain
ψ ( d ( g x n + 1 , g x n + 2 ) ) ψ ( s 3 d ( g x n + 1 , g x n + 2 ) ) = ψ ( s 3 d ( T ( x n , y n ) , T ( x n + 1 , y n + 1 ) ) ) ψ ( M s , T , g ( x n , y n , x n + 1 , y n + 1 ) ) ϕ ( M s , T , g ( x n , y n , x n + 1 , y n + 1 ) ) + L θ ( N T , g ( x n , y n , x n + 1 , y n + 1 ) ) ,
(2.4)
where
M s , T , g ( x n , y n , x n + 1 , y n + 1 ) = max { d ( g x n , g x n + 1 ) , d ( g y n , g y n + 1 ) , d ( g x n , T ( x n , y n ) ) , 1 2 s d ( g x n + 1 , T ( x n + 1 , y n + 1 ) ) , d ( g y n , T ( y n , x n ) ) , 1 2 s d ( g y n + 1 , T ( y n + 1 , x n + 1 ) ) , d ( g x n , T ( x n + 1 , y n + 1 ) ) + d ( g x n + 1 , T ( x n , y n ) ) 2 s , d ( g y n , T ( y n + 1 , x n + 1 ) ) + d ( g y n + 1 , T ( y n , x n ) ) 2 s } = max { d ( g x n , g x n + 1 ) , d ( g y n , g y n + 1 ) , 1 2 s d ( g x n + 1 , g x n + 2 ) , 1 2 s d ( g y n + 1 , g y n + 2 ) , d ( g x n , g x n + 2 ) 2 s , d ( g y n , g y n + 2 ) 2 s }
and
N T , g ( x n , y n , x n + 1 , y n + 1 ) = min { d ( g x n , T ( x n , y n ) ) , d ( g x n + 1 , T ( x n + 1 , y n + 1 ) ) , d ( g x n + 1 , T ( x n , y n ) ) , d ( g x n + 1 , T ( x n + 1 , y n + 1 ) ) } = 0 .
Since
d ( g x n , g x n + 2 ) 2 s d ( g x n , g x n + 1 ) + d ( g x n + 1 , g x n + 2 ) 2 max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) }
and
d ( g y n , g y n + 2 ) 2 s d ( g y n , g y n + 1 ) + d ( g y n + 1 , g y n + 2 ) 2 max { d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } ,
then we get
M s , T , g ( x n , y n , x n + 1 , y n + 1 ) max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , M s , T , g ( x n , y n , x n + 1 , y n + 1 ) d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } , N T , g ( x n , y n , x n + 1 , y n + 1 ) = 0 .
(2.5)
By (2.4) and (2.5), we have
ψ ( d ( g x n + 1 , g x n + 2 ) ) ψ ( max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } ) ϕ ( max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } ) .
(2.6)
Similarly, we can show that
ψ ( d ( g y n + 1 , g y n + 2 ) ) ψ ( max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } ) ϕ ( max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } ) .
(2.7)
Now, denote
δ n = max { d ( g x n , g x n + 1 ) , d ( g y n , g y n + 1 ) } .
(2.8)
Combining (2.6), (2.7) and the fact that max { ψ ( a ) , ψ ( b ) } = ψ ( max { a , b } ) for a , b [ 0 , + ) , we have
ψ ( δ n + 1 ) = max { ψ ( d ( g x n + 1 , g x n + 2 ) ) , ψ ( d ( g y n + 1 , g y n + 2 ) ) } .
(2.9)
So, using (2.6), (2.7), (2.8) together with (2.9), we obtain
ψ ( δ n + 1 ) ψ ( max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } ) ϕ ( max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } ) .
(2.10)
Now we prove that for all n N ,
max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } = δ n and δ n + 1 δ n .
(2.11)

For this purpose, consider the following three cases.

Case 1. If max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } = δ n , then by (2.10) we have
ψ ( δ n + 1 ) ψ ( δ n ) ϕ ( δ n ) < ψ ( δ n ) ,
(2.12)

so (2.11) obviously holds.

Case 2. If max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } = d ( g x n + 1 , g x n + 2 ) > 0 , then by (2.6) we have
ψ ( d ( g x n + 1 , g x n + 2 ) ) ψ ( d ( g x n + 1 , g x n + 2 ) ) ϕ ( d ( g x n + 1 , g x n + 2 ) ) < ψ ( d ( g x n + 1 , g x n + 2 ) ) ,

which is a contradiction.

Case 3. If max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g y n , g y n + 1 ) , d ( g y n + 1 , g y n + 2 ) } = d ( g y n + 1 , g y n + 2 ) > 0 , then from (2.7) we have
ψ ( d ( g y n + 1 , g y n + 2 ) ) ψ ( d ( g y n + 1 , g y n + 2 ) ) ϕ ( d ( g y n + 1 , g y n + 2 ) ) < ψ ( d ( g y n + 1 , g y n + 2 ) ) ,

which is again a contradiction.

Thus, in all the cases, (2.11) holds for each n N . It follows that the sequence { δ n } is a monotone decreasing sequence of nonnegative real numbers and, consequently, there exists δ 0 such that
lim n δ n = δ .
(2.13)
We show that δ = 0 . Suppose, on the contrary, that δ > 0 . Taking the limit as n in (2.12) and using the properties of the function ϕ, we get
ψ ( δ ) ψ ( δ ) ϕ ( δ ) < ψ ( δ ) ,
which is a contradiction. Therefore δ = 0 , that is,
lim n δ n = lim n max { d ( g x n , g x n + 1 ) , d ( g y n , g y n + 1 ) } = 0 ,
which implies that
lim n d ( g x n , g x n + 1 ) = 0 and lim n d ( g y n , g y n + 1 ) = 0 .
(2.14)
Now, we claim that
lim n , m max { d ( g x n , g x m ) , d ( g y n , g y m ) } = 0 .
(2.15)
Assume, on the contrary, that there exist ϵ > 0 and subsequences { g x m ( k ) } , { g x n ( k ) } of { g x n } and { g y m ( k ) } , { g y n ( k ) } of { g y n } with m ( k ) > n ( k ) k such that
max { d ( g x n ( k ) , g x m ( k ) ) , d ( g y n ( k ) , g y m ( k ) ) } ϵ .
(2.16)
Additionally, corresponding to n ( k ) , we may choose m ( k ) such that it is the smallest integer satisfying (2.16) and m ( k ) > n ( k ) k . Thus,
max { d ( g x n ( k ) , g x m ( k ) 1 ) , d ( g y n ( k ) , g y m ( k ) 1 ) } < ϵ .
(2.17)
Using the triangle inequality in a b-metric space and (2.16) and (2.17), we obtain that
ϵ d ( g x m ( k ) , g x n ( k ) ) s d ( g x m ( k ) , g x m ( k ) 1 ) + s d ( g x m ( k ) 1 , g x n ( k ) ) < s d ( g x m ( k ) , g x m ( k ) 1 ) + s ϵ .
Taking the upper limit as k and using (2.14), we obtain
ϵ lim sup k d ( g x n ( k ) , g x m ( k ) ) s ϵ .
(2.18)
Similarly, we obtain
ϵ lim sup k d ( g y n ( k ) , g y m ( k ) ) s ϵ .
(2.19)
Also,
ϵ d ( g x n ( k ) , g x m ( k ) ) s d ( g x n ( k ) , g x m ( k ) + 1 ) + s d ( g x m ( k ) + 1 , g x m ( k ) ) s 2 d ( g x n ( k ) , g x m ( k ) ) + s 2 d ( g x m ( k ) , g x m ( k ) + 1 ) + s d ( g x m ( k ) + 1 , g x m ( k ) ) s 2 d ( g x n ( k ) , g x m ( k ) ) + ( s 2 + s ) d ( g x m ( k ) , g x m ( k ) + 1 ) .
So, from (2.14) and (2.18), we have
ϵ s lim sup k d ( g x n ( k ) , g x m ( k ) + 1 ) s 2 ϵ .
(2.20)
Similarly, we obtain
ϵ s lim sup k d ( g y n ( k ) , g y m ( k ) + 1 ) s 2 ϵ .
(2.21)
Also,
ϵ d ( g x m ( k ) , g x n ( k ) ) s d ( g x m ( k ) , g x n ( k ) + 1 ) + s d ( g x n ( k ) + 1 , g x n ( k ) ) s 2 d ( g x m ( k ) , g x n ( k ) ) + s 2 d ( g x n ( k ) , g x n ( k ) + 1 ) + s d ( g x n ( k ) + 1 , g x n ( k ) ) s 2 d ( g x m ( k ) , g x n ( k ) ) + ( s 2 + s ) d ( g x n ( k ) , g x n ( k ) + 1 ) .
So, from (2.14) and (2.18), we have
ϵ s lim sup k d ( g x m ( k ) , g x n ( k ) + 1 ) s 2 ϵ .
(2.22)
In a similar way, we obtain
ϵ s lim sup k d ( g y m ( k ) , g y n ( k ) + 1 ) s 2 ϵ .
(2.23)
Also,
d ( g x n ( k ) + 1 , g x m ( k ) ) s d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + s d ( g x m ( k ) + 1 , g x m ( k ) ) .
So, from (2.14) and (2.22), we have
ϵ s 2 lim sup k d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) .
(2.24)
Similarly, we obtain
ϵ s 2 lim sup k d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) .
(2.25)
Linking (2.14), (2.18), (2.19), (2.20), (2.21), (2.22) together with (2.23), we get
ϵ s 2 = min { ϵ , ϵ , ϵ s + ϵ s 2 s , ϵ s + ϵ s 2 s } max { lim sup k d ( g x n ( k ) , g x m ( k ) ) , lim sup k d ( g y n ( k ) , g y m ( k ) ) , lim sup k d ( g x n ( k ) , g x m ( k ) + 1 ) + lim sup k d ( g x m ( k ) , g x n ( k ) + 1 ) 2 s , lim sup k d ( g y n ( k ) , g y m ( k ) + 1 ) + lim sup k d ( g y m ( k ) , g y n ( k ) + 1 ) 2 s } max { s ϵ , s ϵ , s 2 ϵ + s 2 ϵ 2 s , s 2 ϵ + s 2 ϵ 2 s } = s ϵ .
So,
ϵ s 2 lim sup k M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ϵ s .
(2.26)
Similarly, we have
ϵ s 2 lim inf k M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ϵ s
(2.27)
and
lim k N T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) = 0 .
(2.28)
Since m ( k ) > n ( k ) , from (2.2) we have
g x n ( k ) g x m ( k ) , g y n ( k ) g y m ( k ) .
Thus,
ψ ( s 3 d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) = ψ ( s 3 d ( T ( x n ( k ) , y n ( k ) ) , T ( x m ( k ) , y m ( k ) ) ) ) ψ ( s 3 d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) ψ ( M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) ψ ( s 3 d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) = ϕ ( M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) ψ ( s 3 d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) = + L θ ( N T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) , ψ ( s 3 d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) ) = ψ ( s 3 d ( T ( y n ( k ) , x n ( k ) ) , T ( y m ( k ) , x m ( k ) ) ) ) ψ ( s 3 d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) ) ψ ( M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) ψ ( s 3 d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) ) = ϕ ( M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) ψ ( s 3 d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) ) = + L θ ( N T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) .
Since ψ is a non-decreasing function, we have
max { ψ ( s 3 d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) , ψ ( s 3 d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) ) } = ψ ( s 3 max { d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) , d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) } ) .
Taking the upper limit as k and using (2.25) and (2.26), we get
ψ ( s ϵ ) ψ ( s 3 max { lim sup k d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) , lim sup k d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) } ) ψ ( lim sup k M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) ϕ ( lim inf k M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) + L θ ( lim sup k N T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) ψ ( s ϵ ) ϕ ( lim inf k M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) ,
which implies that
ϕ ( lim inf k M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) ) = 0 ,
so
lim inf k M s , T , g ( x n ( k ) , y n ( k ) , x m ( k ) , y m ( k ) ) = 0 ,
a contradiction to (2.27). Therefore, (2.15) holds and we have
lim n , m d ( g x n , g x m ) = 0 and lim n , m d ( g y n , g y m ) = 0 .
Since X is a complete b-metric space, there exist x , y X such that
lim n g x n + 1 = x and lim n g y n + 1 = y .
(2.29)
From the commutativity of T and g, we have
g ( g x n + 1 ) = g ( T ( x n , y n ) ) = T ( g x n , g y n ) , g ( g y n + 1 ) = g ( T ( y n , x n ) ) = T ( g y n , g x n ) .
(2.30)
Now, we shall show that
g x = T ( x , y ) and g y = T ( y , x ) .
Letting n in (2.30), from the continuity of T and g, we get
g x = lim n g ( g x n + 1 ) = lim n T ( g x n , g y n ) = T ( lim n g x n , lim n g y n ) = T ( x , y ) , g y = lim n g ( g y n + 1 ) = lim n T ( g y n , g x n ) = T ( lim n g y n , lim n g x n ) = T ( y , x ) .

This implies that ( x , y ) is a coupled coincidence point of T and g. This completes the proof. □

Corollary 2.3 Let ( X , d , ) be a partially ordered complete b-metric space, and let T : X × X X be a continuous mapping such that T has the mixed monotone property. Suppose that there exist ψ Ψ , ϕ Φ , θ Θ and L 0 such that
ψ ( s 3 d ( T ( x , y ) , T ( u , v ) ) ) ψ ( M s ( x , y , u , v ) ) ϕ ( M s ( x , y , u , v ) ) + L θ ( N ( x , y , u , v ) ) ,
where
M s ( x , y , u , v ) = max { d ( x , u ) , d ( y , v ) , d ( x , T ( x , y ) ) , 1 2 s d ( u , T ( u , v ) ) , d ( y , T ( y , x ) ) , 1 2 s d ( v , T ( v , u ) ) , d ( x , T ( u , v ) ) + d ( u , T ( x , y ) ) 2 s , d ( y , T ( v , u ) ) + d ( v , T ( y , x ) ) 2 s }
and
N ( x , y , u , v ) = min { d ( x , T ( x , y ) ) , d ( u , T ( u , v ) ) , d ( u , T ( x , y ) ) , d ( x , T ( u , v ) ) }

for all x , y , u , v X with x u and y v . If there exists ( x 0 , y 0 ) X × X such that x 0 T ( x 0 , y 0 ) and y 0 T ( y 0 , x 0 ) , then T has a coupled fixed point in X.

Proof Take g = I X and apply Theorem 2.2. □

The following result is the immediate consequence of Corollary 2.3.

Corollary 2.4 Let ( X , d , ) be a partially ordered complete b-metric space. Let T : X × X X be a continuous mapping such that T has the mixed monotone property. Suppose that there exists ϕ Φ such that
d ( T ( x , y ) , T ( u , v ) ) 1 s 3 M s ( x , y , u , v ) 1 s 3 ϕ ( M s ( x , y , u , v ) ) ,
(2.31)
where
M s ( x , y , u , v ) = max { d ( x , u ) , d ( y , v ) , d ( x , T ( x , y ) ) , 1 2 s d ( u , T ( u , v ) ) , d ( y , T ( y , x ) ) , 1 2 s d ( v , T ( v , u ) ) , d ( x , T ( u , v ) ) + d ( u , T ( x , y ) ) 2 s , d ( y , T ( v , u ) ) + d ( v , T ( y , x ) ) 2 s }

for all x , y , u , v X with x u and y v . If there exists ( x 0 , y 0 ) X × X such that x 0 T ( x 0 , y 0 ) and y 0 T ( y 0 , x 0 ) , then T has a coupled fixed point in X.

3 Uniqueness of a common fixed point

In this section we shall provide some sufficient conditions under which T and g have a unique common fixed point. Note that if ( X , ) is a partially ordered set, then we endow the product X × X with the following partial order relation, for all ( x , y ) , ( z , t ) X × X ,
( x , y ) ( z , t ) x z , y t .

From Theorem 2.2, it follows that the set C ( T , g ) of coupled coincidences is nonempty.

Theorem 3.1 By adding to the hypotheses of Theorem  2.2, the condition: for every ( x , y ) and ( z , t ) in X × X , there exists ( u , v ) X × X such that ( T ( u , v ) , T ( v , u ) ) is comparable to ( T ( x , y ) , T ( y , x ) ) and to ( T ( z , t ) , T ( t , z ) ) , then T and g have a unique coupled common fixed point; that is, there exists a unique ( x , y ) X × X such that
x = g x = T ( x , y ) , y = g y = T ( y , x ) .
Proof We know, from Theorem 2.2, that there exists at least a coupled coincidence point. Suppose that ( x , y ) and ( z , t ) are coupled coincidence points of T and g, that is, T ( x , y ) = g x , T ( y , x ) = g y , T ( z , t ) = g z and T ( t , z ) = g t . We shall show that g x = g z and g y = g t . By the assumptions, there exists ( u , v ) X × X such that ( T ( u , v ) , T ( v , u ) ) is comparable to ( T ( x , y ) , T ( y , x ) ) and to ( T ( z , t ) , T ( t , z ) ) . Without any restriction of the generality, we can assume that
( T ( x , y ) , T ( y , x ) ) ( T ( u , v ) , T ( v , u ) ) and ( T ( z , t ) , T ( t , z ) ) ( T ( u , v ) , T ( v , u ) ) .
Put u 0 = u , v 0 = v and choose ( u 1 , v 1 ) X × X such that
g u 1 = T ( u 0 , v 0 ) , g v 1 = T ( v 0 , u 0 ) .
For n 1 , continuing this process, we can construct sequences { g u n } and { g v n } such that
g u n + 1 = T ( u n , v n ) , g v n + 1 = T ( v n , u n ) for all  n .
Further, set x 0 = x , y 0 = y and z 0 = z , t 0 = t and in the same way define sequences { g x n } , { g y n } and { g z n } , { g t n } . Then it is easy to see that
g x n T ( x , y ) , g y n T ( y , x ) and g z n T ( z , t ) , g t n T ( t , z )
(3.1)
for all n 1 . Since ( T ( x , y ) , T ( y , x ) ) = ( g x , g y ) = ( g x 1 , g y 1 ) is comparable to ( T ( u , v ) , T ( v , u ) ) = ( g u , g v ) = ( g u 1 , g v 1 ) , then it is easy to show ( g x , g y ) ( g u , g v ) . Recursively, we get that
( g x n , g y n ) ( g u n , g v n ) for all  n .
(3.2)
Thus from (2.1) we have
ψ ( d ( g x , g u n + 1 ) ) ψ ( s 3 d ( g x , g u n + 1 ) ) = ψ ( s 3 d ( T ( x , y ) , T ( u n , v n ) ) ) ψ ( M s , T , g ( x , y , u n , v n ) ) ϕ ( M s , T , g ( x , y , u n , v n ) ) + L θ ( N T , g ( x , y , u n , v n ) ) ,
where
M s , T , g ( x , y , u n , v n ) = max { d ( g x , g u n ) , d ( g y , g v n ) , d ( g x , T ( x , y ) ) , 1 2 s d ( g u n , T ( u n , v n ) ) , d ( g y , T ( y , x ) ) , 1 2 s d ( g v n , T ( v n , u n ) ) , d ( g x , T ( u n , v n ) ) + d ( g u n , T ( x , y ) ) 2 s , d ( g y , T ( v n , u n ) ) + d ( g v n , T ( y , x ) ) 2 s } max { d ( g x , g u n ) , d ( g y , g v n ) , d ( g y , g v n + 1 ) , d ( g x , g u n + 1 ) } .
It is easy to show that
M s , T , g ( x , y , u n , v n ) max { d ( g x , g u n ) , d ( g y , g v n ) }
and
N T , g ( x , y , u n , v n ) = 0 .
Hence,
ψ ( d ( g x , g u n + 1 ) ) ψ ( max { d ( g x , g u n ) , d ( g y , g v n ) } ) ϕ ( max { d ( g x , g u n ) , d ( g y , g v n ) } ) .
(3.3)
Similarly, one can prove that
ψ ( d ( g y , g v n + 1 ) ) ψ ( max { d ( g x , g u n ) , d ( g y , g v n ) } ) ϕ ( max { d ( g x , g u n ) , d ( g y , g v n ) } ) .
(3.4)
Combining (3.3), (3.4) and the fact that max { ψ ( a ) , ψ ( b ) } = ψ ( max { a , b } ) for a , b [ 0 , + ) , we have
ψ ( max { d ( g x , g u n + 1 ) , d ( g y , g v n + 1 ) } ) = max { ψ ( d ( g x , g u n + 1 ) ) , ψ ( d ( g y , g v n + 1 ) ) } ψ ( max { d ( g x , g u n ) , d ( g y , g v n ) } ) ϕ ( max { d ( g x , g u n ) , d ( g y , g v n ) } ) ψ ( max { d ( g x , g u n ) , d ( g y , g v n ) } ) .
(3.5)
Using the non-decreasing property of ψ, we get that
max { d ( g x , g u n + 1 ) , d ( g y , g v n + 1 ) } max { d ( g x , g u n ) , d ( g y , g v n ) }
implies that max { d ( g x , g u n ) , d ( g y , g v n ) } is a non-increasing sequence. Hence, there exists r 0 such that
lim n max { d ( g x , g u n ) , d ( g y , g v n ) } = r .
Passing the upper limit in (3.5) as n , we obtain
ψ ( r ) ψ ( r ) ϕ ( r ) ,
which implies that ϕ ( r ) = 0 , and then r = 0 . We deduce that
lim n max { d ( g x , g u n ) , d ( g y , g v n ) } = 0 ,
which concludes
lim n d ( g x , g u n ) = lim n d ( g y , g v n ) = 0 .
(3.6)
Similarly, one can prove that
lim n d ( g z , g u n ) = lim n d ( g t , g v n ) = 0 .
(3.7)
From (3.6) and (3.7), we have g x = g z and g y = g t . Since g x = T ( x , y ) and g y = T ( y , x ) , by the commutativity of T and g, we have
g ( g x ) = g ( T ( x , y ) ) = T ( g x , g y ) , g ( g y ) = g ( T ( y , x ) ) = T ( g y , g x ) .
(3.8)
Denote g x = a and g y = b . Then from (3.8) we have
g ( a ) = T ( a , b ) , g ( b ) = T ( b , a ) .
(3.9)
Thus, ( a , b ) is a coupled coincidence point. It follows that g a = g z and g b = g y , that is,
g ( a ) = a , g ( b ) = b .
(3.10)
From (3.9) and (3.10), we obtain
a = g ( a ) = T ( a , b ) , b = g ( b ) = T ( b , a ) .
(3.11)
Therefore, ( a , b ) is a coupled common fixed point of T and g. To prove the uniqueness of the point ( a , b ) , assume that ( c , d ) is another coupled common fixed point of T and g. Then we have
c = g c = T ( c , d ) , d = g d = T ( d , c ) .

Since ( c , d ) is a coupled coincidence point of T and g, we have g c = g x = a and g d = g y = b . Thus c = g c = g a = a and d = g d = g b = b , which is the desired result. □

Theorem 3.2 In addition to the hypotheses of Theorem  3.1, if g x 0 and g y 0 are comparable, then T and g have a unique common fixed point, that is, there exists x X such that x = g x = T ( x , x ) .

Proof Following the proof of Theorem 3.1, T and g have a unique coupled common fixed point ( x , y ) . We only have to show that x = y . Since g x 0 and g y 0 are comparable, we may assume that g x 0 g y 0 . By using the mathematical induction, one can show that
g x n g y n for all  n 0 ,
(3.12)
where { g x n } and { g y n } are defined by (2.2). From (2.29) and Lemma 1.4, we have
ψ ( s d ( x , y ) ) = ψ ( s 3 1 s 2 d ( x , y ) ) lim sup n ψ ( s 3 d ( g x n + 1 , g y n + 1 ) ) = lim sup n ψ ( s 3 d ( T ( x n , y n ) , T ( y n , x n ) ) ) lim sup n ψ ( M s , T , g ( x n , y n , y n , x n ) ) lim inf n ϕ ( M s , T , g ( x n , y n , y n , x n ) ) + lim sup n L θ ( N T , g ( x n , y n , y n , x n ) ) ψ ( d ( x , y ) ) lim inf n ϕ ( M s ( x n , y n , y n , x n ) ) < ψ ( d ( x , y ) ) ,

a contradiction. Therefore, x = y , that is, T and g have a common fixed point. □

Remark 3.3 Since a b-metric is a metric when s = 1 , from the results of Jachymski [29], the condition
ψ ( d ( F ( x , y ) , F ( u , v ) ) ) ψ ( max { d ( g x , g u ) , d ( g y , g v ) } ) ϕ ( max { d ( g x , g u ) , d ( g y , g v ) } )
is equivalent to
d ( F ( x , y ) , F ( u , v ) ) φ ( max { d ( g x , g u ) , d ( g y , g v ) } ) ,

where ψ Ψ , ϕ Φ and φ : [ 0 , ) [ 0 , ) is continuous, φ ( t ) < t for all t > 0 and φ ( t ) = 0 if and only if t = 0 . So, our results can be viewed as a generalization and extension of the corresponding results in [15, 25, 3032] and several other comparable results.

4 Application to integral equations

Here, in this section, we wish to study the existence of a unique solution to a nonlinear quadratic integral equation, as an application to our coupled fixed point theorem. Consider the nonlinear quadratic integral equation
x ( t ) = h ( t ) + λ 0 1 k 1 ( t , s ) f 1 ( s , x ( s ) ) d s 0 1 k 2 ( t , s ) f 2 ( s , x ( s ) ) d s , t I = [ 0 , 1 ] , λ 0 .
(4.1)
Let Γ denote the class of those functions γ : [ 0 , + ) [ 0 , + ) which satisfy the following conditions:
  1. (i)

    γ is non-decreasing and ( γ ( t ) ) p γ ( t p ) for all p 1 .

     
  2. (ii)

    There exists ϕ Φ such that γ ( t ) = t ϕ ( t ) for all t [ 0 , + ) .

     

For example, γ 1 ( t ) = k t , where 0 k < 1 and γ 2 ( t ) = t t + 1 are in Γ.

We will analyze Eq. (4.1) under the following assumptions:

(a1) f i : I × R R ( i = 1 , 2 ) are continuous functions, f i ( t , x ) 0 and there exist two functions m i L 1 ( I ) such that f i ( t , x ) m i ( t ) ( i = 1 , 2 ).

(a2) f 1 ( t , x ) is monotone non-decreasing in x and f 2 ( t , y ) is monotone non-increasing in y for all x , y R and t I .

(a3) h : I R is a continuous function.

(a4) k i : I × I R ( i = 1 , 2 ) are continuous in t I for every s I and measurable in s I for all t I such that
0 1 k i ( t , s ) m i ( s ) d s K , i = 1 , 2 ,

and k i ( t , x ) 0 .

(a5) There exist constants 0 L i < 1 ( i = 1 , 2 ) and γ Γ such that for all x , y R and x y ,
| f i ( t , x ) f i ( t , y ) | L i γ ( x y ) ( i = 1 , 2 ) .
(a6) There exist α , β C ( I ) such that
α ( t ) h ( t ) + λ 0 1 k 1 ( t , s ) f 1 ( s , α ( s ) ) d s 0 1 k 2 ( t , s ) f 2 ( s , β ( s ) ) d s h ( t ) + λ 0 1 k 1 ( t , s ) f 1 ( s , β ( s ) ) d s 0 1 k 2 ( t , s ) f 2 ( s , α ( s ) ) d s β ( t ) .

(a7) max { L 1 p , L 2 p } λ p K 2 p 1 2 4 p 3 .

Consider the space X = C ( I ) of continuous functions defined on I = [ 0 , 1 ] with the standard metric given by
ρ ( x , y ) = sup t I | x ( t ) y ( t ) | for  x , y C ( I ) .
This space can also be equipped with a partial order given by
x , y C ( I ) , x y x ( t ) y ( t ) for any  t I .
Now, for p 1 , we define
d ( x , y ) = ( ρ ( x , y ) ) p = ( sup t I | x ( t ) y ( t ) | ) p = sup t I | x ( t ) y ( t ) | p for  x , y C ( I ) .

It is easy to see that ( X , d ) is a complete b-metric space with s = 2 p 1 [3].

Also, X × X = C ( I ) × C ( I ) is a partially ordered set if we define the following order relation:
( x , y ) , ( u , v ) X × X , ( x , y ) ( u , v ) x u and y v .

For any x , y X and each t I , max { x ( t ) , y ( t ) } and min { x ( t ) , y ( t ) } belong to X and are upper and lower bounds of x, y, respectively. Therefore, for every ( x , y ) , ( u , v ) X × X , one can take ( max { x , u } , min { y , v } ) X × X which is comparable to ( x , y ) and ( u , v ) . Now, we formulate the main result of this section.

Theorem 4.1 Under assumptions (a1)-(a7), Eq. (4.1) has a unique solution in C ( I ) .

Proof We consider the operator T : X × X X defined by
T ( x , y ) ( t ) = h ( t ) + λ 0 1 k 1 ( t , s ) f 1 ( s , x ( s ) ) d s 0 1 k 2 ( t , s ) f 2 ( s , y ( s ) ) d s for  t I .
By virtue of our assumptions, T is well defined (this means that if x , y X , then T ( x , y ) X ). Firstly, we prove that T has the mixed monotone property. In fact, for x 1 x 2 and t I , we have
T ( x 1 , y ) ( t ) T ( x 2 , y ) ( t ) = h ( t ) + λ 0 1 k 1 ( t , s ) f 1 ( s , x 1 ( s ) ) d s 0 1 k 2 ( t , s ) f 2 ( s , y ( s ) ) d s h ( t ) λ 0 1 k 1 ( t , s ) f 1 ( s , x 2 ( s ) ) d s 0 1 k 2 ( t , s ) f 2 ( s , y ( s ) ) d s = λ 0 1 k 1 ( t , s ) [ f 1 ( s , x 1 ( s ) ) f 1 ( s , x 2 ( s ) ) ] d s 0 1 k 2 ( t , s ) f 2 ( s , y ( s ) ) d s 0 .
Similarly, if y 1 y 2 and t I , then T ( x , y 1 ) ( t ) T ( x , y 2 ) ( t ) . Therefore, T has the mixed monotone property. Also, for ( x , y ) ( u , v ) , that is, x u and y v , we have
| T ( x , y ) ( t ) T ( u , v ) ( t ) | | λ 0 1 k 1 ( t , s ) f 1 ( s , x ( s ) ) d s 0 1 k 2 ( t , s ) [ f 2 ( s , y ( s ) ) f 2 ( s , v ( s ) ) ] d s + λ 0 1 k 2 ( t , s ) f 2 ( s , v ( s ) ) d s 0 1 k 1 ( t , s ) [ f 1 ( s , x ( s ) ) f 1 ( s , u ( s ) ) ] d s | λ 0 1 k 1 ( t , s ) f 1 ( s , x ( s ) ) d s 0 1 k 2 ( t , s ) | f 2 ( s , y ( s ) ) f 2 ( s , v ( s ) ) | d s + λ 0 1 k 2 ( t , s ) f 2 ( s , v ( s ) ) d s 0 1 k 1 ( t , s ) | f 1 ( s , x ( s ) ) f 1 ( s , u ( s ) ) | d s λ 0 1 k 1 ( t , s ) m 1 ( s ) d s 0 1 k 2 ( t , s ) L 2 γ ( y ( s ) v ( s ) ) d s + λ 0 1 k 2 ( t , s ) m 2 ( s ) d s 0 1 k 1 ( t , s ) L 1 γ ( u ( s ) x ( s ) ) d s .
Since the function γ is non-decreasing and x u and y v , we have
γ ( u ( s ) x ( s ) ) γ ( sup t I | x ( s ) u ( s ) | ) = γ ( ρ ( x , u ) )
and
γ ( y ( s ) v ( s ) ) γ ( sup t I | y ( s ) v ( s ) | ) = γ ( ρ ( y , v ) ) ,
hence
| T ( x , y ) ( t ) T ( u , v ) ( t ) | λ K 0 1 k 2 ( t , s ) L 2 γ ( ρ ( y , v ) ) d s + λ K 0 1 k 1 ( t , s ) L 1 γ ( ρ ( u , x ) ) d s λ K 2 max { L 1 , L 2 } [ γ ( ρ ( u , x ) ) + γ ( ρ ( y , v ) ) ] .
Then we can obtain
d ( T ( x , y ) , T ( u , v ) ) = sup t I | T ( x , y ) ( t ) T ( u , v ) ( t ) | p { λ K 2 max { L 1 , L 2 } [ γ ( ρ ( u , x ) ) + γ ( ρ ( y , v ) ) ] } p = λ p K 2 p max { L 1 p , L 2 p } [ γ ( ρ ( u , x ) ) + γ ( ρ ( y , v ) ) ] p ,
and using the fact that ( a + b ) p 2 p 1 ( a p + b p ) for a , b ( 0 , + ) and p > 1 , we have
d ( T ( x , y ) , T ( u , v ) ) 2 p 1 λ p K 2 p max { L 1 p , L 2 p } [ ( γ ( ρ ( u , x ) ) ) p + ( γ ( ρ ( y , v ) ) ) p ] 2 p 1 λ p K 2 p max { L 1 p , L 2 p } [ γ ( d ( u , x ) ) + γ ( d ( y , v ) ) ] 2 p λ p K 2 p max { L 1 p , L 2 p } [ γ ( M s ( x , y , u , v ) ) ] 2 p λ p K 2 p max { L 1 p , L 2 p } [ M s ( x , y , u , v ) ϕ ( M s ( x , y , u , v ) ) ] 1 2 3 p 3 M s ( x , y , u , v ) 1 2 3 p 3 ϕ ( M s ( x , y , u , v ) ) .

This proves that the operator T satisfies the contractive condition (2.31) appearing in Corollary 2.4.

Finally, let α, β be the functions appearing in assumption (a6); then, by (a6), we get
α T ( α , β ) T ( β , α ) β .

Theorem 3.1 gives us that T has a unique coupled fixed point ( x , y ) X × X . Since α β , Theorem 3.2 says that x = y and this implies x = T ( x , x ) . So, x C ( I ) is the unique solution of Eq. (4.1) and the proof is complete. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Karaj Branch, Islamic Azad University

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© Aghajani and Arab; licensee Springer. 2013

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