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Coupled coincidence point results for a generalized compatible pair with applications

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Abstract

In this paper, employing a new concept of generalized compatibility of a pair of mappings defined on a product space, certain coupled coincidence point results of mappings involved herein are obtained. We also deduce certain coupled fixed point results without mixed monotone property of F. Our results generalize some recent comparable results in the literature. Moreover, some examples and an application to integral equations are given here to illustrate the usability of the obtained results.

MSC:46S40, 47H10, 54H25.

1 Introduction

The existence of fixed points in ordered metric spaces has been investigated by Ran and Reurings [1]. Recently, many researchers have obtained fixed point and coupled fixed point results in partially ordered metrics spaces (see, e.g., [216]).

The study of coupled fixed points in partially ordered metric spaces was initiated by Guo and Lakshmikantham [17], and then attracted many researchers, see for example [1820] and references therein. Bhaskar and Lakshmikantham [21] introduced the notions of mixed monotone mapping and coupled fixed point. As an application, they studied the existence and uniqueness of a solution for a periodic boundary value problem associated with a first order ordinary differential equation. Lakshmikantham and Ćirić in [22] introduced the concepts of coupled coincidence and coupled common fixed point for mappings satisfying nonlinear contractive conditions in partially ordered complete metric spaces and generalized the concept of the mixed monotone property. For more on coupled fixed point theory we refer to the reviews (see, e.g. [4, 2333]). Recently, Alotaibi and Alsulami [34] presented some coupled coincidence point results involving the (ϕ,ψ)-contractive condition for mappings having the mixed g-monotone property in a partially ordered metric space which are generalizations of the results of Luong and Thuan [35].

In this paper, we introduce the notion of generalized compatibility of a pair {F,G}, of mappings F,G:X×XX. We then employ this notion to obtain coupled coincidence point results for such a pair of mappings involving (ϕ,ψ)-contractive condition without mixed G-monotone property of F. Thus the derived coupled fixed point results do not have the mixed monotone property of F. Our results represent new versions of the results of Alotaibi and Alsulami [34], Luong and Thuan [35] and Bhaskar and Lakshmikantham [21]. We also provide an example and an application to an integral equation to support our results presented here.

2 Preliminaries

We now recall some basic definitions and important results for our use in the sequel.

Definition 1 [21]

Let (X,) be a partially ordered set. The mapping F:X×XX is said to have the mixed monotone property if F is monotone non-decreasing in its first argument and is monotone non-increasing in its second argument, that is, for all x 1 , x 2 X, x 1 x 2 implies F( x 1 ,y)F( x 2 ,y), for any yX and for all y 1 , y 2 X, y 1 y 2 implies F(x, y 1 )F(x, y 2 ), for any xX.

Lakshmikantham and Ćirić [22] generalized the concept of a mixed monotone property as follows.

Definition 2 [22]

Let (X,) be a partially ordered set and g a self mapping on X. A mapping F:X×XX is said to have the mixed g-monotone property if for all x 1 , x 2 X, g x 1 g x 2 implies that F( x 1 ,y)F( x 2 ,y), for any yX and for all y 1 , y 2 X, g y 1 g y 2 implies that F(x, y 1 )F(x, y 2 ), for any xX.

Definition 3 [22]

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

Definition 4 [22]

Let X be a nonempty set, g a self mapping on X and F:X×XX and g:XX. We say that F and g are commutative if g(F(x,y))=F(gx,gy), for all x,yX.

Definition 5 [26]

Let (X,d) be a metric space, F:X×XX a mapping and g a self mapping on X. A hybrid pair F, g is 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 =xand lim n + F( y n , x n )= lim n + g y n =y

with x,yX.

As given in [34, 35], Φ denotes the set of all functions ϕ:[0,)[0,) such that:

  1. 1.

    ϕ is continuous and increasing,

  2. 2.

    ϕ(t)=0 if and only if t=0,

  3. 3.

    ϕ(t+s)ϕ(t)+ϕ(s), for all t,s[0,+).

Let Ψ be the set of all the functions ψ:[0,)[0,) such that lim t r ψ(t)>0 for all r>0 and lim t 0 + ψ(t)=0.

The main result in [34] is given by the next theorem.

Theorem 6 Let (X,) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX be a mapping having the g-mixed monotone property on X such that there exist two elements x 0 , y 0 X with

g x 0 F( x 0 , y 0 )andg y 0 F( y 0 , x 0 ).

Suppose there exist ϕΦ and ψΨ such that

ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ϕ ( d ( g x , g u ) + d ( g y , g v ) ) ψ ( d ( g x , g u ) + d ( g y , g v ) 2 )
(2.1)

for all x,y,u,vX, with gxgu and gygv. Suppose F(X×X)g(X), g is continuous and compatible with F and also suppose either

  1. (a)

    F is continuous or

  2. (b)

    X has the following property:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-decreasing sequence { y n }y, then y n y for all n.

Then there exist x,yX such that g(x)=F(x,y) and g(y)=F(y,x), that is, F and g have a coupled coincidence point in X.

Definition 7 Suppose that F,G:X×XX are two mappings. F is said to be G-increasing with respect to if for all x,y,u,vX, with G(x,y)G(u,v) we have F(x,y)F(u,v).

Example 8 Let X=(0,) be endowed with the natural ordering of real numbers ≤. Define mappings F,G:X×XX by F(x,y)=ln(x+y) and G(x,y)=x+y for all (x,y)X×X. Note that F is G-increasing with respect to ≤.

Example 9 Let X=N endowed with the partial order defined by x,yX×X, xy if and only if y divides x. Define the mappings F,G:X×XX by F(x,y)= x 2 y 2 and G(x,y)=xy for all (x,y)X×X. Then F is G-increasing with respect to .

Definition 10 An element (x,y)X×X is called a coupled coincidence point of mappings F,G:X×XX if F(x,y)=G(x,y) and F(y,x)=G(y,x).

Example 11 Let F,G:R×RR be defined by F(x,y)=xy and G(x,y)= 2 3 (x+y) for all (x,y)X×X. Note that (0,0), (1,2), and (2,1) are a coupled coincidence points of F and G.

Definition 12 Let F,G:X×XX. We say that the pair {F,G} is generalized compatible if

{ d ( F ( G ( x n , y n ) , G ( y n , x n ) ) , G ( F ( x n , y n ) , F ( y n , x n ) ) ) 0 as  n + , d ( F ( G ( y n , x n ) , G ( x n , y n ) ) , G ( F ( y n , x n ) , F ( x n , y n ) ) ) 0 as  n + ,

whenever ( x n ) and ( y n ) are sequences in X such that

{ lim n F ( x n , y n ) = lim n G ( x n , y n ) = t 1 , lim n F ( y n , x n ) = lim n G ( y n , x n ) = t 2 .

Example 13 Let (R,||) be a usual metric space. Define mappings F,G:X×XX by F(x,y)= x 2 y 2 and G(x,y)= x 2 + y 2 for all x,yX. Let ( x n ) and ( y n ) be two sequences in X such that

{ F ( x n , y n ) t 1 G ( x n , y n ) t 1 as  n + , F ( x n , y n ) t 2 G ( x n , y n ) t 2 as  n + .

We can prove easily that t 1 = t 2 =0 and

{ d ( F ( G ( x n , y n ) , G ( y n , x n ) ) , G ( F ( x n , y n ) , F ( y n , x n ) ) ) 0 as  n + , d ( F ( G ( y n , x n ) , G ( x n , y n ) ) , G ( F ( y n , x n ) , F ( x n , y n ) ) ) 0 as  n + .

Thus the pair {F,G} satisfies the generalized compatibility.

Definition 14 Let F,G:X×XX be two maps. We say that the pair {F,G} is commuting if

F ( G ( x , y ) , G ( y , x ) ) =G ( F ( x , y ) , F ( y , x ) )

for all x,yX.

Obviously, a commuting pair is a generalized compatible but not conversely in general.

3 Main results

Now we prove our main result.

Theorem 15 Let (X,) be a partially ordered set such that there exists a complete metric d on X. Assume that F,G:X×XX are two generalized compatible mappings such that F is G-increasing with respect to , G is continuous and has the mixed monotone property, and there exist two elements x 0 , y 0 X with

G( x 0 , y 0 )F( x 0 , y 0 )andG( y 0 , x 0 )F( y 0 , x 0 ).

Suppose that there exist ϕΦ and ψΨ such that

ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ϕ ( d ( G ( x , y ) , G ( u , v ) ) + d ( G ( y , x ) , G ( v , u ) ) ) ψ ( d ( G ( x , y ) , G ( u , v ) ) + d ( G ( y , x ) , G ( v , u ) ) 2 )
(3.1)

for all x,y,u,vX, with G(x,y)G(u,v) and G(y,x)G(v,u). Suppose that for any x,yX, there exist u,vX such that

{ F ( x , y ) = G ( u , v ) , F ( y , x ) = G ( v , u ) .
(3.2)

Also suppose that either

  1. (a)

    F is continuous or

  2. (b)

    X has the following properties:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-increasing sequence { y n }y, then y y n for all n.

Then F and G have a coupled coincidence point in X.

Proof Let x 0 , y 0 X be such that G( x 0 , y 0 )F( x 0 , y 0 ) and F( y 0 , x 0 )G( y 0 , x 0 ) (such points exist by hypothesis). From (3.2), there exists ( x 1 , y 1 )X×X such that F( x 0 , y 0 )=G( x 1 , y 1 ) and F( y 0 , x 0 )=G( y 1 , x 1 ). Continuing this process, we can construct two sequences { x n } and { y n } in X such that

F( x n , y n )=G( x n + 1 , y n + 1 ),F( y n , x n )=G( y n + 1 , x n + 1 )for all nN.
(3.3)

First we show that for all nN, we have

G( x n , y n )G( x n + 1 , y n + 1 )andG( y n + 1 , x n + 1 )G( y n , x n ).
(3.4)

As G( x 0 , y 0 )F( x 0 , y 0 ) and F( y 0 , x 0 )G( y 0 , x 0 ) and as F( x 0 , y 0 )=G( x 1 , y 1 ) and F( y 0 , x 0 )=G( y 1 , x 1 ), we have G( x 0 , y 0 )G( x 1 , y 1 ) and G( y 1 , x 1 )G( y 0 , x 0 ). Thus (3.4) holds for n=0. Suppose now that (3.4) holds for some fixed nN. Since F is G-increasing with respect to , we have

G( x n + 1 , y n + 1 )=F( x n , y n )F( x n + 1 , y n + 1 )=G( x n + 2 , y n + 2 )

and

G( y n + 2 , x n + 2 )=F( y n + 1 , x n + 1 )F( y n , x n )=G( y n + 1 , x n + 1 ).

Hence (3.4) holds for all nN. For all nN, denote

δ n =d ( G ( x n , y n ) , G ( x n + 1 , y n + 1 ) ) +d ( G ( y n , x n ) , G ( y n + 1 , x n + 1 ) ) .
(3.5)

We can suppose that δ n >0 for all nN. If not, ( x n , y n ) will be a coincidence point and the proof is finished. We claim that for any nN, we have

ϕ( δ n + 1 )ϕ( δ n ).
(3.6)

Since G( x n , y n )G( x n + 1 , y n + 1 ) and G( y n , x n )G( y n + 1 , x n + 1 ), letting x= x n , y= y n , u= x n + 1 and v= y n + 1 in (3.1) and using (3.3), we get

ϕ ( d ( G ( x n + 1 , y n + 1 ) , G ( x n + 2 , y n + 2 ) ) ) = ϕ ( d ( F ( x n , y n ) , F ( x n + 1 , y n + 1 ) ) ) 1 2 ϕ ( d ( G ( x n , y n ) , G ( x n + 1 , y n + 1 ) ) + d ( G ( y n , x n ) , G ( y n + 1 , x n + 1 ) ) ) ψ ( d ( G ( x n , y n ) , G ( x n + 1 , y n + 1 ) ) + d ( G ( y n , x n ) , G ( y n + 1 , x n + 1 ) ) 2 ) = 1 2 ϕ ( δ n ) ψ ( δ n 2 ) .
(3.7)

Similarly we have

ϕ ( d ( G ( y n + 2 , x n + 2 ) , G ( y n + 1 , x n + 1 ) ) ) = ϕ ( d ( F ( y n + 1 , x n + 1 ) , F ( y n , x n ) ) ) 1 2 ϕ ( d ( G ( y n + 1 , x n + 1 ) , G ( y n , x n ) ) + d ( G ( x n + 1 , y n + 1 ) , G ( x n , y n ) ) ) ψ ( d ( G ( y n + 1 , x n + 1 ) , G ( y n , x n ) ) + d ( G ( x n + 1 , y n + 1 ) , G ( x n , y n ) ) 2 ) = 1 2 ϕ ( δ n ) ψ ( δ n 2 ) .
(3.8)

Summing (3.7) and (3.8), we obtain

ϕ( δ n + 1 )ϕ( δ n )2ψ ( δ n 2 ) .
(3.9)

Since ϕ is non-decreasing, it follows that the sequence ( δ n ) is monotone decreasing. Therefore, there is some δ0 such that lim n + δ n = δ + . We shall show that δ=0. Assume on contrary that δ>0. Then taking the limit as n+ (equivalently, δ n δ) in (3.9), using the fact that lim n r ψ(t)>0 for all r>0 and ϕ is continuous, we have

ϕ ( δ ) = lim n + ϕ ( δ n ) lim n + [ ϕ ( δ n 1 ) 2 ψ ( δ n 1 2 ) ] = ϕ ( δ ) 2 lim δ n 1 δ ψ ( δ n 1 2 ) < ϕ ( δ )

a contradiction. Thus δ=0, that is

lim n + [ d ( G ( x n , y n ) , G ( x n + 1 , y n + 1 ) ) + d ( G ( y n , x n ) , G ( y n + 1 , x n + 1 ) ) ] = lim n + ϕ ( δ n ) = 0 .
(3.10)

We shall prove that (G( x n , y n ),G( y n , x n )) is a Cauchy sequence in X×X endowed with the metric Λ defined by Λ((x,y),(u,v))=d(x,u)+d(y,v) for all (x,y),(u,v)X×X. If (G( x n , y n ),G( y n , x n )) is not a Cauchy sequence in (X×X,Λ). Then there exists ε>0 for which we can find two sequences of positive integers (m(k)) and (n(k)) such that for all positive integer k with n(k)>m(k)>k, we have

{ Λ ( ( G ( x m ( k ) , y m ( k ) ) , G ( y m ( k ) , x m ( k ) ) ) , ( G ( x n ( k ) , y n ( k ) ) , G ( y n ( k ) , x n ( k ) ) ) ) > ε , Λ ( ( G ( x m ( k ) , y m ( k ) ) , G ( y m ( k ) , x m ( k ) ) ) , ( G ( x n ( k ) 1 , y n ( k ) 1 ) , G ( y n ( k ) 1 , x n ( k ) 1 ) ) ) ε .
(3.11)

By definition of the metric Λ, we have

d k =d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) , y n ( k ) ) ) +d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) , x n ( k ) ) ) >ε
(3.12)

and

d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) 1 , y n ( k ) 1 ) ) +d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) 1 , x n ( k ) 1 ) ) ε.
(3.13)

Further from (3.12) and (3.13), for k0, we have

ε < d k d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) 1 , y n ( k ) 1 ) ) + d ( G ( x n ( k ) 1 , y n ( k ) 1 ) , G ( x n ( k ) , y n ( k ) ) ) + d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) 1 , x n ( k ) 1 ) ) + d ( G ( y n ( k ) 1 , x n ( k ) 1 ) , G ( y n ( k ) , x n ( k ) ) ) ε + δ n ( k ) 1 .

Taking the limit as k+ in the above inequality, we have, by (3.10),

lim k + d k = ε + .
(3.14)

Again, for all k0, we have

d k = d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) , y n ( k ) ) ) + d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) , x n ( k ) ) ) d ( G ( x m ( k ) , y m ( k ) ) , G ( x m ( k ) + 1 , y m ( k ) + 1 ) ) + d ( G ( x m ( k ) + 1 , y m ( k ) + 1 ) , G ( x n ( k ) + 1 , y n ( k ) + 1 ) ) + d ( G ( x n ( k ) + 1 , y n ( k ) + 1 ) , G ( x n ( k ) , y n ( k ) ) ) + d ( G ( y m ( k ) , x m ( k ) ) , G ( y m ( k ) + 1 , x m ( k ) + 1 ) ) + d ( G ( y m ( k ) + 1 , x m ( k ) + 1 ) , G ( y n ( k ) + 1 , x n ( k ) + 1 ) ) + d ( G ( y n ( k ) + 1 , x n ( k ) + 1 ) , G ( y n ( k ) , x n ( k ) ) ) = d ( G ( x m ( k ) , y m ( k ) ) , G ( x m ( k ) + 1 , y m ( k ) + 1 ) ) + d ( G ( y m ( k ) , x m ( k ) ) , G ( y m ( k ) + 1 , x m ( k ) + 1 ) ) + d ( G ( x m ( k ) + 1 , y m ( k ) + 1 ) , G ( x n ( k ) + 1 , y n ( k ) + 1 ) ) + d ( G ( y m ( k ) + 1 , x m ( k ) + 1 ) , G ( y n ( k ) + 1 , x n ( k ) + 1 ) ) + d ( G ( x n ( k ) + 1 , y n ( k ) + 1 ) , G ( x n ( k ) , y n ( k ) ) ) + d ( G ( y n ( k ) + 1 , x n ( k ) + 1 ) , G ( y n ( k ) , x n ( k ) ) ) .

Hence, for all k0,

d k δ m ( k ) + δ n ( k ) + d ( G ( x m ( k ) + 1 , y m ( k ) + 1 ) , G ( x n ( k ) + 1 , y n ( k ) + 1 ) ) + d ( G ( y m ( k ) + 1 , x m ( k ) + 1 ) , G ( y n ( k ) + 1 , x n ( k ) + 1 ) ) .
(3.15)

Using the property of ϕ, we have

ϕ ( d k ) = ϕ ( δ m ( k ) + δ n ( k ) + d ( G ( x m ( k ) + 1 , y m ( k ) + 1 ) , G ( x n ( k ) + 1 , y n ( k ) + 1 ) ) + d ( G ( y m ( k ) + 1 , x m ( k ) + 1 ) , G ( y n ( k ) + 1 , x n ( k ) + 1 ) ) ) ϕ ( δ m ( k ) + δ n ( k ) ) + ϕ ( d ( G ( x m ( k ) + 1 , y m ( k ) + 1 ) , G ( x n ( k ) + 1 , y n ( k ) + 1 ) ) ) + ϕ ( d ( G ( y m ( k ) + 1 , x m ( k ) + 1 ) , G ( y n ( k ) + 1 , x n ( k ) + 1 ) ) ) .

From (3.1), (3.4), and (3.12), for all k0, we have

ϕ ( d ( G ( x m ( k ) + 1 , y m ( k ) + 1 ) , G ( x n ( k ) + 1 , y n ( k ) + 1 ) ) ) = ϕ ( d ( F ( x m ( k ) , y m ( k ) ) , F ( x n ( k ) , y n ( k ) ) ) ) 1 2 ϕ ( d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) , y n ( k ) ) ) + d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) , x n ( k ) ) ) ) ψ ( d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) , y n ( k ) ) ) + d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) , x n ( k ) ) ) 2 ) = 1 2 ϕ ( d k ) ψ ( d k 2 ) 1 2 ϕ ( d k ) .
(3.16)

Also from (3.1), (3.4), and (3.12), for all k0, we have

ϕ ( d ( G ( y m ( k ) + 1 , x m ( k ) + 1 ) , G ( y n ( k ) + 1 , x n ( k ) + 1 ) ) ) = ϕ ( d ( F ( y m ( k ) , x m ( k ) ) , F ( y n ( k ) , x n ( k ) ) ) ) 1 2 ϕ ( d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) , x n ( k ) ) ) + d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) , y n ( k ) ) ) ) ψ ( d ( G ( y m ( k ) , x m ( k ) ) , G ( y n ( k ) , x n ( k ) ) ) + d ( G ( x m ( k ) , y m ( k ) ) , G ( x n ( k ) , y n ( k ) ) ) 2 ) = 1 2 ϕ ( d k ) ψ ( d k 2 ) 1 2 ϕ ( d k ) .
(3.17)

Inserting (3.16) and (3.17) in (3.15), we have

ϕ( d k )ϕ( δ m ( k ) + δ n ( k ) )+ϕ( d k )2ψ ( d k 2 ) .
(3.18)

Letting k+ in the above inequality, we obtain

ϕ(ε)ϕ(0)+ϕ(ε)2 lim k + ψ ( d k 2 ) =ϕ(ε)2 lim d k ε + ψ ( d k 2 ) <ϕ(ε),
(3.19)

which is a contradiction. Hence (G( x n , y n ),G( y n , x n )) is a Cauchy sequence in (X×X,Λ), which implies that (G( x n , y n )) and (G( y n , x n )) are Cauchy sequences in (X,d). Now, since (X,d) is complete, there exist x,yX such that

lim n + G ( x n , y n ) = lim n + F ( x n , y n ) = x and lim n + G ( y n , x n ) = lim n + F ( y n , x n ) = y .
(3.20)

Since the pair {F,G} satisfies the generalized compatibility, from (3.20), we get

lim n + d ( F ( G ( x n , y n ) , G ( y n , x n ) ) , G ( F ( x n , y n ) , F ( y n , x n ) ) ) =0
(3.21)

and

lim n + d ( F ( G ( y n , x n ) , G ( x n , y n ) ) , G ( F ( y n , x n ) , F ( x n , y n ) ) ) =0.
(3.22)

Suppose that F is continuous. For all n0, we have

d ( G ( x , y ) , F ( G ( x n , y n ) , G ( y n , x n ) ) ) d ( G ( x , y ) , G ( F ( x n , y n ) , F ( y n , x n ) ) ) + d ( G ( F ( x n , y n ) , F ( y n , x n ) ) , F ( G ( x n , y n ) , G ( y n , x n ) ) ) .

Taking the limit as n+, using (3.20), (3.21), and the fact that F and G are continuous, we have

G(x,y)=F(x,y).
(3.23)

Similarly, using (3.20), (3.21), and the fact that F and G are continuous, we have

G(y,x)=F(y,x).
(3.24)

Thus (x,y) is a coupled coincidence point of F and G. Now, suppose that (b) holds. By (3.4) and (3.20), we have (G( x n , y n )) is non-decreasing sequence, G( x n , y n )x and (G( y n , x n )) is non-decreasing sequence, G( y n , x n )y as n+. Thus for all nN, we have

G( x n , y n )xandG( y n , x n )y.
(3.25)

Since the pair {F,G} satisfies the generalized compatibility and G is continuous, by (3.21) and (3.22), we have

lim n + G ( G ( x n , y n ) , G ( y n , x n ) ) = G ( x , y ) = lim n + G ( F ( x n , y n ) , F ( y n , x n ) ) = lim n + F ( G ( x n , y n ) , G ( y n , x n ) )
(3.26)

and

lim n + G ( G ( y n , x n ) , G ( x n , y n ) ) = G ( y , x ) = lim n + G ( F ( y n , x n ) , F ( x n , y n ) ) = lim n + F ( G ( y n , x n ) , G ( x n , y n ) ) .
(3.27)

Now, we have

d ( G ( x , y ) , F ( x , y ) ) lim n + d ( G ( F ( x n , y n ) , F ( y n , x n ) ) , F ( x , y ) ) = lim n + d ( F ( G ( x n , y n ) , G ( y n , x n ) ) , F ( x , y ) ) .

Since G has the mixed monotone property, it follows from (3.25) that G(G( x n , y n ),G( y n , x n ))G(x,y) and G(G( y n , x n ),G( x n , y n ))G(y,x). Now using (3.1), (3.26), and (3.27), we get

ϕ ( d ( G ( x , y ) , F ( x , y ) ) ) lim n + 1 2 ϕ ( d ( G ( G ( x n , y n ) , G ( y n , x n ) ) , G ( x , y ) ) + d ( G ( G ( y n , x n ) , G ( x n , y n ) ) , G ( y , x ) ) ) ψ ( d ( G ( G ( x n , y n ) , G ( y n , x n ) ) , G ( x , y ) ) + d ( G ( G ( y n , x n ) , G ( x n , y n ) ) , G ( y , x ) ) 2 ) .

Then we obtain G(x,y)=F(x,y). Similarly, we can show that G(y,x)=F(y,x). □

The commuting maps {F,G} are obviously generalized compatible, thus we obtain the following.

Corollary 16 Let (X,) be a partially ordered set such that there exists a complete metric d on X. Assume that F,G:X×XX are two commuting mappings such that F is G-increasing with respect to , G is continuous and has the mixed monotone property, and there exist two elements x 0 , y 0 X with

G( x 0 , y 0 )F( x 0 , y 0 )andG( y 0 , x 0 )F( y 0 , x 0 ).

Suppose that the inequalities (3.1) and (3.2) hold and either

  1. (a)

    F is continuous or

  2. (b)

    X has the following properties:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-increasing sequence { y n }y, then y y n for all n.

Then F and G have a coupled coincidence point in X.

Definition 17 Let (X,) be a partially ordered set, F:X×XX and g:XX. We say that F is g-increasing with respect to if for any x,yX,

g x 1 g x 2 impliesF( x 1 ,y)F( x 2 ,y)

and

g y 1 g y 2 impliesF(x, y 1 )F(x, y 2 ).

Now, we deduce an analogous result to Theorem 3.1 of Alotaibi and Alsulami [34] (Theorem 6) without g-mixed monotone property of F.

Corollary 18 Let (X,) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that F is g-increasing with respect to , and there exist ϕΦ and ψΨ such that

ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ϕ ( d ( g x , g u ) + d ( g y , g v ) ) ψ ( d ( g x , g u ) + d ( g y , g v ) 2 )

for all x,y,u,vX, with gxgu and gygv. Suppose that F(X×X)g(X), g is continuous and monotone increasing with respect to , and the pair {F,g} is compatible. Also suppose that either

  1. (a)

    F is continuous or

  2. (b)

    X has the following properties:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-increasing sequence { y n }y, then y y n for all n.

If there exist two elements x 0 , y 0 X with

g x 0 F( x 0 , y 0 )andg y 0 F( y 0 , x 0 ).

Then F and g have a coupled coincidence point.

Corollary 19 Let (X,) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×XX and g:XX be two mappings such that F is g-increasing with respect to , and there exist ϕΦ and ψΨ such that

ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ϕ ( d ( g x , g u ) + d ( g y , g v ) ) ψ ( d ( g x , g u ) + d ( g y , g v ) 2 )

for all x,y,u,vX, with gxgu and gygv. Suppose that F(X×X)g(X), g is continuous and monotone increasing with respect to , and the pair {F,g} is commuting. Also suppose that either

  1. (a)

    F is continuous or

  2. (b)

    X has the following properties:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-increasing sequence { y n }y, then y y n for all n.

If there exist two elements x 0 , y 0 X with

g x 0 F( x 0 , y 0 )andg y 0 F( y 0 , x 0 ).

Then F and g have a coupled coincidence point.

Definition 20 Let (X,) be a partially ordered set, F:X×XX. We say that F is increasing with respect to if for any x,yX,

x 1 x 2 impliesF( x 1 ,y)F( x 2 ,y)

and

y 1 y 2 impliesF(x, y 1 )F(x, y 2 ).

The following result provides the conclusion of the main results of Luong and Thuan [35] without the mixed monotone property of the involved mapping F.

Corollary 21 [35]

Let (X,) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Assume that F:X×XX is an increasing map with respect to and there exist two elements x 0 , y 0 X with

x 0 F( x 0 , y 0 )and y 0 F( y 0 , x 0 ).

Suppose there exist ϕΦ and ψΨ such that

ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) 1 2 ϕ ( d ( x , u ) + d ( y , v ) ) ψ ( d ( x , u ) + d ( y , v ) 2 )

for all x,y,u,vX, with xu and yv. Also suppose that either

  1. (a)

    F is continuous or

  2. (b)

    X has the following properties:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-increasing sequence { y n }y, then y y n for all n.

Then F has a coupled fixed point.

Corollary 22 Let (X,) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Assume that F:X×XX be an increasing map with respect to and there exist two elements x 0 , y 0 X with

x 0 F( x 0 , y 0 )and y 0 F( y 0 , x 0 ).

Suppose there exists ψΨ such that

d ( F ( x , y ) , F ( u , v ) ) d ( x , u ) + d ( y , v ) 2 ψ ( d ( x , u ) + d ( y , v ) 2 )

for all x,y,u,vX, with xu and yv. Also suppose that either

  1. (a)

    F is continuous or

  2. (b)

    X has the following properties:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-increasing sequence { y n }y, then y y n for all n.

Then F has a coupled fixed point.

The conclusion of the main results of Bhaskar and Lakshmikantham [21] without the mixed monotone property of the involved mapping F is obtained in the following corollary.

Corollary 23 Let (X,) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Assume that F:X×XX be an increasing map with respect to and there exist two elements x 0 , y 0 X with

x 0 F( x 0 , y 0 )and y 0 F( y 0 , x 0 ).

Suppose there exists a real number k[0,1) such that

d ( F ( x , y ) , F ( u , v ) ) k 2 ( d ( x , u ) + d ( y , v ) )

for all x,y,u,vX, with xu and yv. Also suppose that either

  1. (a)

    F is continuous or

  2. (b)

    X has the following properties:

  3. (i)

    if a non-decreasing sequence { x n }x, then x n x for all n,

  4. (ii)

    if a non-increasing sequence { y n }y, then y y n for all n.

Then F has a coupled fixed point.

Now we prove the uniqueness of the coupled coincidence 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),(u,v)X×X:

(x,y)(u,v)if and only ifG(x,y)G(u,v)andG(y,x)G(v,u),

where G:X×XX×X is one-one.

Theorem 24 In addition to the hypotheses of Theorem  15, suppose that for every (x,y), (z,t) in X×X, there exists another (u,v) in X×X which is comparable to (x,y) and (z,t), then F and G have a unique coupled coincidence point.

Proof From Theorem 15, the set of coupled coincidence points of F and G is nonempty. Suppose (x,y) and (z,t) are coupled coincidence points of F and G, that is,

{ F ( x , y ) = G ( x , y ) , F ( y , x ) = G ( y , x )

and

{ F ( z , t ) = G ( z , t ) , F ( t , z ) = G ( t , z ) .

Now we prove that G(x,y)=G(z,t) and G(y,x)=G(t,z). By assumption, there exists (u,v) in X×X that is comparable to (x,y) and (z,t). We define sequences {G( u n , v n )} and {G( v n , u n )} as follows, with u 0 =u, v 0 =v:

F( u n , v n )=G( u n + 1 , v n + 1 ),F( v n , u n )=G( v n + 1 , u n + 1 )for all nN.

Since (u,v) is comparable to (x,y), we assume that (x,y)(u,v)=( u 0 , v 0 ). Which implies G(x,y)G( u 0 , v 0 ) and G(y,x)G( v 0 , u 0 ). We suppose that (x,y)( u n , v n ) for some n. We prove that

(x,y)( u n + 1 , v n + 1 ).

Since F is G increasing, we have G(x,y)G( u n , v n ) implies F(x,y)F( u n , v n ) and G(y,x)G( v n , u n ) implies F(y,x)F( v n , u n ). Now

G(x,y)=F(x,y)F( u n , v n )=G( u n + 1 , v n + 1 )

and

G(y,x)=F(y,x)F( v n , u n )=G( v n + 1 , u n + 1 ).

Thus we have

(x,y)( u n , v n )for all n.
(3.28)

Using (3.1) and (3.28), we have

ϕ ( d ( G ( x , y ) , G ( u n + 1 , v n + 1 ) ) ) = ϕ ( d ( F ( x , y ) , F ( u n , v n ) ) ) 1 2 ϕ ( d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) ) ψ ( d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) 2 ) .
(3.29)

Similarly

ϕ ( d ( G ( v n + 1 , u n + 1 ) , G ( y , x ) ) ) = ϕ ( d ( F ( v n , u n ) , F ( y , x ) ) ) 1 2 ϕ ( d ( G ( v n , u n ) , G ( y , x ) ) + d ( G ( u n , v n ) , G ( x , y ) ) ) ψ ( d ( G ( v n , u n ) , G ( y , x ) ) + d ( G ( u n , v n ) , G ( x , y ) ) 2 ) .
(3.30)

Using (3.29), (3.30), and the property of ϕ, we have

ϕ ( d ( G ( x , y ) , G ( u n + 1 , v n + 1 ) ) + d ( G ( y , x ) , G ( v n + 1 , u n + 1 ) ) ) ϕ ( d ( F ( x , y ) , F ( u n , v n ) ) ) + ϕ ( d ( G ( v n + 1 , u n + 1 ) , G ( y , x ) ) ) ϕ ( d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) ) 2 ψ ( d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) 2 ) ,
(3.31)

which implies that

ϕ ( d ( G ( x , y ) , G ( u n + 1 , v n + 1 ) ) + d ( G ( y , x ) , G ( v n + 1 , u n + 1 ) ) ) ϕ ( d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) ) .

By using the property of ϕ, we get

d ( G ( x , y ) , G ( u n + 1 , v n + 1 ) ) + d ( G ( y , x ) , G ( v n + 1 , u n + 1 ) ) d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) .

This implies that the sequence {d(G(x,y),G( u n , v n ))+d(G(y,x),G( v n , u n ))} is decreasing. Therefore, there exists l0 such that

lim n + [ d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) ] =l.
(3.32)

Now we show that l=0. We suppose on the contrary that l>0. Taking the limit as n in (3.31) and using the property of ψ, we have

ϕ(l)ϕ(l)2 lim n + ψ ( d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) 2 ) <ϕ(l),

a contradiction. Thus l=0, that is,

lim n + [ d ( G ( x , y ) , G ( u n , v n ) ) + d ( G ( y , x ) , G ( v n , u n ) ) ] =0.
(3.33)

This implies that

lim n + d ( G ( x , y ) , G ( u n , v n ) ) = lim n + d ( G ( y , x ) , G ( v n , u n ) ) =0.

Similarly, we show that

lim n + d ( G ( z , t ) , G ( u n , v n ) ) = lim n + d ( G ( t , z ) , G ( v n , u n ) ) =0.
(3.34)

Using (3.33) and (3.34), we have G(x,y)=G(z,t) and G(y,x)=G(t,z). □

Example 25 Let X=[0,1] endowed with the natural ordering of real numbers. We endow X with the standard metric X

d(x,y)=|xy|

for all x,yX. Then (X,d) is a complete metric space. Define the mappings F,G:X×XX as follows:

F(x,y)= { x 2 y 2 3 if  x y , 0 if  x < y

and

G(x,y)= { x 2 y 2 if  x y , 0 if  x < y .

First we prove that F is G-increasing.

Let (x,y),(u,v)X×X with G(x,y)G(u,v). We consider the following cases.

Case 1: If x<y, then F(x,y)=0F(u,v).

Case 2: If xy, and if uv, then

G ( x , y ) G ( u , v ) x 2 y 2 u 2 v 2 x 2 y 2 3 u 2 v 2 3 F ( x , y ) F ( u , v ) .

But if u<v, then

G(x,y)G(u,v)0 x 2 y 2 0 x 2 = y 2 F(x,y)=0F(u,v).

Thus we see that F is G-increasing.

Now we prove that for any x,yX, there exist u,vX such that

{ F ( x , y ) = G ( u , v ) , F ( y , x ) = G ( v , u ) .

Let (x,y)X×X be fixed. We consider the following cases.

Case 1: If x=y, then we have F(x,y)=0=G(x,y) and F(y,x)=0=G(y,x).

Case 2: If x>y, then we have F(x,y)= x 2 y 2 3 =G( x 3 , y 3 ) and F(y,x)=0=G( y 3 , x 3 ).

Case 3: If x<y, then we have F(x,y)=0=G( x 3 , y 3 ) and F(y,x)= y 2 x 2 3 =G( y 3 , x 3 ).

Now we prove that G is continuous and has the mixed monotone property.

Clearly G is continuous. Let (x,y)X×X be fixed. Suppose that x 1 , x 2 X are such that x 1 < x 2 . We discuss the following cases.

Case 1: If x 1 <y, then we have G( x 1 ,y)=0G( x 2 ,y).

Case 2: If x 2 > x 1 >y, then we have G( x 1 ,y)= x 1 2 y 2 x 2 2 y 2 =G( x 2 ,y).

Similarly, we can show that if y 1 , y 2 X are such that y 1 < y 2 , then G(x, y 1 )G(x, y 2 ).

Now, we prove that the pair {F,G} satisfies the generalized compatibility hypothesis.

Let ( x n ) and ( y n ) be two sequences in X such that

t 1 = lim n + F( x n , y n )= lim n + G( x n , y n )

and

t 2 = lim n + F( y n , x n )= lim n + G( y n , x n ).

Then we must have t 1 = t 2 =0 and one can easily prove that

{ lim n + d ( F ( G ( x n , y n ) , G ( y n , x n ) ) , G ( F ( x n , y n ) , F ( y n , x n ) ) ) = 0 , lim n + d ( F ( G ( y n , x n ) , G ( x n , y n ) ) , G ( F ( y n , x n ) , F ( x n , y n ) ) ) = 0 .

Now we prove that there exist two elements x 0 , y 0 X with

G( x 0 , y 0 )F( x 0 , y 0 )andG( y 0 , x 0 )F( y 0 , x 0 ).

Since we have G(0, 1 2 )=0=F(0, 1 2 ) and G( 1 2 ,0)= 1 4 1 12 =F( 1 2 ,0). Now, let ϕ:[0,)[0,) be defined as ϕ(t)= 3 4 t, for all t[0,) and let ψ:[0,)[0,) be defined as ψ(t)= 3 4 t, for all t[0,). Clearly ϕΦ and ψΨ. We next verify the contraction (3.1) for all x,y,u,vX, with G(x,y)G(u,v) and G(v,u)G(y,x). We have

ϕ ( d ( F ( x , y ) , F ( u , v ) ) ) = 3 4 [ d ( F ( x , y ) , F ( u , v ) ) ] = 3 4 | F ( x , y ) F ( u , v ) | = 1 4 | G ( x , y ) G ( u , v ) | = 1 2 | G ( x , y ) G ( u , v ) | 2 1 2 | G ( x , y ) G ( u , v ) | + | G ( y , x ) G ( v , u ) | 2 = 3 4 | G ( x , y ) G ( u , v ) | + | G ( y , x ) G ( v , u ) | 2 1 4 | G ( x , y ) G ( u , v ) | + | G ( y , x ) G ( v , u ) | 2 = 3 8 | G ( x , y ) G ( u , v ) | + | G ( y , x ) G ( v , u ) | 1 4 | G ( x , y ) G ( u , v ) | + | G ( y , x ) G ( v , u ) | 2 = 1 2 ϕ ( d ( G ( x , y ) , G ( u , v ) ) + d ( G ( y , x ) , G ( v , u ) ) ) ψ ( d ( G ( x , y ) , G ( u , v ) ) + d ( G ( y , x ) , G ( v , u ) ) 2 ) .

Hence condition (3.1) is satisfied. Thus all the requirements of Theorem 15 are satisfied and (0,0) is a coupled coincidence point of F and G.

4 Applications to the integral equations

Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various applications in differential and integral equations (see [1, 9, 10, 14] and references therein). Motivated by the work in [2, 3538], we study the existence of solutions for a system of nonlinear integral equations using the results proved in the previous section.

Let Θ denote the class of those functions θ:[0,)[0,) which satisfies the following conditions:

  1. (i)

    θ is increasing.

  2. (ii)

    There exists ψΨ such that θ(t)= t 2 ψ( t 2 ) for all t[0,).

Consider the integral equation

x(t)= a b ( K 1 ( t , s ) + K 2 ( t , s ) ) ( f ( s , x ( s ) ) + g ( s , x ( s ) ) ) ds+h(t)
(4.1)

for all t[a,b]. We assume that K 1 , K 2 , f, and g satisfy the following conditions:

  1. (i)

    0 K 1 (t,s), 0 K 2 (t,s) for all t,s[a,b].

  2. (ii)

    There exist λ,μ>0 and θΘ such that for all x,yR, xy,

    0f(t,x)f(t,y)λθ(xy)and0g(t,x)g(t,y)μθ(xy).
  3. (iii)

    We have

    max{λ,μ} sup t [ a , b ] a b ( K 1 ( t , s ) + K 2 ( t , s ) ) ds 1 2 .
  4. (iv)

    There exist continuous functions z,w:[a,b]R such that

    z ( t ) a b K 1 ( t , s ) ( f ( s , z ( s ) ) + g ( s , w ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , w ( s ) ) + g ( s , z ( s ) ) ) d s + h ( t )

and

w ( t ) a b K 1 ( t , s ) ( f ( s , w ( s ) ) + g ( s , z ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , z ( s ) ) + g ( s , w ( s ) ) ) d s + h ( t )

for all t[a,b].

Theorem 26 Consider the integral equation (4.1) with K 1 , K 2 C([a,b]×[a,b],R), f,gC([a,b]×R,R) and hC([a,b],R) and suppose that the conditions (i)-(iv) are satisfied. Then the integral equation (4.1) has a solution in C([a,b],R).

Proof Let X=C([a,b],R) denote the space of continuous functions defined on the interval [a,b]. We endowed X with the metric d:X×XR defined by

d(x,y)= sup t [ a , b ] |x(t)y(t)|for all x,yX.

It is clear that (X,d) is a complete metric space and (X,d,) is a complete ordered metric space if xy whenever x(t)y(t) for all t[a,b]. Suppose { u n } is a monotone non-decreasing in X that converges to uX. Then for every t[a,b] the sequence of real numbers

u 1 (t) u 2 (t) u n (t)

converges to u(t). Therefore for all t[a,b], nN, u n (t)u(t). Hence u n u for all n. Similarly, we can verify that lim n v(t) of a monotone non-increasing sequence v n (t) in X is a lower bound for all the elements in the sequence. That is, v v n for all n. Therefore, condition (b) of Corollary 22 holds. Also, X×X=C([a,b],R)×C([a,b],R) is a partially ordered set if we define the following order relation on X×X, for all x,y,u,vX, with xu and yv.

Define the mapping F:X×XX by

F ( x , y ) ( t ) = a b K 1 ( t , s ) ( f ( s , x ( s ) ) + g ( s , y ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , y ( s ) ) + g ( s , x ( s ) ) ) d s + h ( t )

for all t[a,b]. Now we shall show that F is increasing. For x 1 x 2 , that is, x 1 (t) x 2 (t) for all t[a,b], we have

F ( x 1 , y ) ( t ) F ( x 2 , y ) ( t ) = a b K 1 ( t , s ) ( f ( s , x 1 ( s ) ) + g ( s , y ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , y ( s ) ) + g ( s , x 1 ( s ) ) ) d s + h ( t ) a b K 1 ( t , s ) ( f ( s , x 2 ( s ) ) + g ( s , y ( s ) ) ) d s a b K 2 ( t , s ) ( f ( s , y ( s ) ) + g ( s , x 2 ( s ) ) ) d s h ( t ) = a b K 1 ( t , s ) ( f ( s , x 1 ( s ) ) f ( s , x 2 ( s ) ) ) d s + a b K 2 ( t , s ) ( g ( s , x 1 ( s ) ) g ( s , x 2 ( s ) ) ) d s 0 .

Hence F( x 1 ,y)(t)F( x 2 ,y)(t) for all t[a,b], that is, F( x 1 ,y)F( x 2 ,y). Similarly, if y 1 y 2 , that is y 1 (t) y 2 (t) for all t[a,b], we have

F ( x , y 1 ) ( t ) F ( x , y 2 ) ( t ) = a b K 1 ( t , s ) ( f ( s , x ( s ) ) + g ( s , y 1 ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , y 1 ( s ) ) + g ( s , x ( s ) ) ) d s + h ( t ) a b K 1 ( t , s ) ( f ( s , x ( s ) ) + g ( s , y 2 ( s ) ) ) d s a b K 2 ( t , s ) ( f ( s , y 2 ( s ) ) + g ( s , x ( s ) ) ) d s h ( t ) = a b K 1 ( t , s ) ( g ( s , y 1 ( s ) ) g ( s , y 2 ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , y 1 ( s ) ) f ( s , y 2 ( s ) ) ) d s 0 .

Hence F(x, y 1 )(t)F(x, y 2 )(t) for all t[a,b], that is, F(x, y 1 )F(x, y 2 ). Thus F(x,y) is increasing. Now, for x,y,u,vX such that xu and vy, we have

d ( F ( x , y ) ( t ) F ( u , v ) ( t ) ) = sup t [ a , b ] | F ( x , y ) ( t ) F ( u , v ) ( t ) | = sup t [ a , b ] | a b K 1 ( t , s ) ( f ( s , x ( s ) ) + g ( s , y ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , y ( s ) ) + g ( s , x ( s ) ) ) d s + h ( t ) ( a b K 1 ( t , s ) ( f ( s , u ( s ) ) + g ( s , v ( s ) ) ) d s + a b K 2 ( t , s ) ( f ( s , v ( s ) ) + g ( s , u ( s ) ) ) d s + h ( t ) ) | = sup t [ a , b ] | a b K 1 ( t , s ) [ ( f ( s , x ( s ) ) f ( s , u ( s ) ) ) + ( g ( s , y ( s ) ) g ( s , v ( s ) ) ) ] d s + a b K 2 ( t , s ) [ ( f ( s , y ( s ) ) f ( s , v ( s ) ) ) + ( g ( s , x ( s ) ) g ( s , u ( s ) ) ) ] d s | sup t [ a , b ] | a b K 1 ( t , s ) [ λ θ ( x ( s ) u ( s ) ) + μ θ ( y ( s ) v ( s ) ) ] d s + a b K 2 ( t , s ) [ λ θ ( y ( s ) v ( s ) ) + μ θ ( x ( s ) u ( s ) ) ] d s | max { λ , μ } sup t [ a , b ] a b ( K 1 ( t , s ) + K 2 ( t , s ) ) × [ θ ( x ( s ) u ( s ) ) + θ ( y ( s ) v ( s ) ) ] d s .

As the function θ is increasing and u(t)x(t) and y(t)v(t) for all t[a,b], then θ(x(s)u(s))θ(d(x,u)), θ(y(s)v(s))θ(d(y,v)), for all s[a,b], we obtain

d ( F ( x , y ) , F ( u , v ) ) max { λ , μ } [ θ ( d ( x , u ) ) + θ ( d ( y , v ) ) ] sup t [ a , b ] a b ( K 1 ( t , s ) + K 2 ( t , s ) ) 1 2 [ θ ( d ( x , u ) ) + θ ( d ( y , v ) ) ] θ ( d ( x , u ) + d ( y , v ) ) d ( x , u ) + d ( y , v ) 2 ψ ( d ( x , u ) + d ( y , v ) 2 ) .

Therefore, for xu and vy, we have

d ( F ( x , y ) , F ( u , v ) ) d ( x , u ) + d ( y , v ) 2 ψ ( d ( x , u ) + d ( y , v ) 2 ) .

Also from condition (iv) we have z(t)F(z,w)(t) and F(w,z)(t)w(t) for all t[a,b], that is, zF(z,w) and F(w,z)w. Thus all of the hypotheses of Corollary 22 are satisfied and the mapping F has a coupled fixed point that is a solution in X=C([a,b],R) of the integral equation (4.1). □

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Acknowledgements

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the first author acknowledges with thanks DSR, KAU for financial support.

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Correspondence to Mujahid Abbas.

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The authors declare that they have no competing interests.

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All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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Hussain, N., Abbas, M., Azam, A. et al. Coupled coincidence point results for a generalized compatible pair with applications. Fixed Point Theory Appl 2014, 62 (2014) doi:10.1186/1687-1812-2014-62

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Keywords

  • coupled coincidence point
  • generalized compatibility
  • mixed monotone property
  • ordered set
  • complete metric space