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A generalization on weak contractions in partially ordered b-metric spaces and its application to quadratic integral equations

Abstract

We introduce the notion of almost generalized (ψ,φ,L)-contractive mappings, and establish the coincidence and common fixed point results for this class of mappings in partially ordered complete b-metric spaces. Our results extend and improve several known results from the context of ordered metric spaces to the setting of ordered b-metric spaces. As an application, we prove the existence of a unique solution to a class of nonlinear quadratic integral equations.

1 Introduction

Fixed points theorems in partially ordered metric spaces were firstly obtained in 2004 by Ran and Reurings [1], and then by Nieto and Rodríguez-López [2]. In this direction several authors obtained further results under weak contractive conditions (see, e.g., [38]). Berinde initiated in [9] the concept of almost contractions and obtained several interesting fixed point theorems. This has been a subject of intense study since then; see, e.g., [1020]. Some authors used related notions as ‘condition (B)’ (Babu et al. [21]) and ‘almost generalized contractive condition’ for two maps (Ćirić et al. [22]), and for four maps (Aghajani et al. [23]). See also a note by Pacurar [15]. On the other hand, the concept of b-metric space was introduced by Czerwik in [24]. After that, several interesting results of the existence of fixed point for single-valued and multivalued operators in b-metric spaces have been obtained (see [2540]). Pacurar [41] proved some results on sequences of almost contractions and fixed points in b-metric spaces. Recently, Hussain and Shah [42] obtained results on KKM mappings in cone b-metric spaces. Using the concepts of partially ordered metric spaces, almost generalized contractive condition, and b-metric spaces, we define a new concept of almost generalized (ψ,φ,L)-contractive condition. In this paper, some coincidence and common fixed point theorems for mappings satisfying almost generalized (ψ,φ,L)-contractive condition in the setup of partially ordered complete b-metric spaces are proved. Consistent with [43] and [[40], p.264], the following definitions and results will be needed in the sequel.

Definition 1.1 [43]

Let X be a (nonempty) set and s1 be a given real number. A function d:X×X R + is said to be a b-metric space iff for all x,y,zX, 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 does not necessarily need to be a metric (see, also, [40]).

Example 1.1 [44]

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)=|xy| is the usual Euclidean metric, then ρ(x,y)= ( x y ) s is a b-metric on with s=2, but it is not a metric on .

Also, the following example of a b-metric space is given in [45].

Example 1.2 [45]

Let X be the set of Lebesgue measurable functions on [0,1] such that 0 1 | f ( x ) | 2 dx<. Define D:X×X[0,) by D(f,g)= 0 1 | f ( x ) g ( x ) | 2 dx. 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, where the b-metric D is defined with D(x,y)=d(x,y), d is a cone metric (also see [4649]).

Khamsi [50] also showed that each cone metric space over a normal cone has a b-metric structure.

Definition 1.2 [6]

We shall say that the mapping T is g-nondecreasing if

gxgyTxTy.

2 Main results

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

  1. (a)

    ψ is continuous,

  2. (b)

    ψ is nondecreasing,

  3. (c)

    ψ(0)=0<ψ(t) for every t>0.

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

  1. (i)

    φ is right continuous,

  2. (ii)

    φ is nondecreasing,

  3. (iii)

    φ(t)<t for every t>0.

Let (X,d,) be a partially ordered b-metric space and T:XX and g:XX be two mappings. Set

M(x,y)=max { d ( g x , g y ) , d ( g x , T x ) , d ( g y , T y ) , d ( g x , T y ) + d ( g y , T x ) 2 s }

and

N(x,y)=min { d ( g x , T x ) , d ( g y , T y ) , d ( g x , T y ) , d ( g y , T x ) } .

Now, we introduce the following definition.

Definition 2.1 Let (X,d,) be a partially ordered b-metric space. We say that T:XX is an almost generalized (ψ,φ,L)-contractive mapping with respect to g:XX for some ψΨ, φΦ, and L0 if

ψ ( s 3 d ( T x , T y ) ) φ ( ψ ( M ( x , y ) ) ) +Lψ ( N ( x , y ) )
(2.1)

for all x,yX with gxgy.

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

Theorem 2.1 Suppose that (X,d,) is a partially ordered complete b-metric space. Let T:XX be an almost generalized (ψ,φ,L)-contractive mapping with respect to g:XX, and T and g are continuous such that T is a monotone g-nondecreasing mapping, commutative with g and T(X)g(X). If there exists x 0 X such that g x 0 T x 0 , then T and g have a coincidence point in X.

Proof By the given assumptions, there exists x 0 X such that g x 0 T x 0 . Since T(X)g(X), we can define x 1 X such that g x 1 =T x 0 , then g x 0 T x 0 =g x 1 . Also there exists x 2 X such that g x 2 =T x 1 . Since T is a monotone g-nondecreasing mapping, we have

g x 1 =T x 0 T x 1 =g x 2 .

Continuing in this way, we construct a sequence { x n } in X such that for all n=0,1,2, ,

g x n + 1 =T x n
(2.2)

for which

g x 0 g x 1 g x 2 g x n g x n + 1 .
(2.3)

If there exists k 0 N such that g x k 0 + 1 =g x k 0 , then g x k 0 =T x k 0 . This means that x k 0 is a coincidence point of T, g, and the proof is finished. Thus, g x n + 1 g x n for all nN. From (2.2) and (2.3) and the inequality (2.1) with (x,y)=( x n , x n + 1 ), we have

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 , T x n + 1 ) ) φ ( ψ ( M ( x n , x n + 1 ) ) ) + L ψ ( N ( x n , x n + 1 ) ) ,
(2.4)

where

M ( x n , x n + 1 ) = max { d ( g x n , g x n + 1 ) , d ( g x n , T x n ) , d ( g x n + 1 , T x n + 1 ) , d ( g x n , T x n + 1 ) + d ( g x n + 1 , T x n ) 2 s } = max { d ( g x n , g x n + 1 ) , d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) , d ( g x n , g x n + 2 ) 2 s }

and

N( x n , x n + 1 )=min { d ( g x n , T x n ) , d ( g x n + 1 , T x n + 1 ) , d ( g x n , T x n + 1 ) , d ( g x n + 1 , T x n ) } =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 ) } ,

then we get

M ( x n , x n + 1 ) = max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) } , N ( x n , x n + 1 ) = 0 .
(2.5)

By (2.4) and (2.5), we have

ψ ( d ( g x n + 1 , g x n + 2 ) ) φ ( ψ ( max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) } ) ) .
(2.6)

Suppose that max{d(g x n ,g x n + 1 ),d(g x n + 1 ,g x n + 2 )}=d(g x n + 1 ,g x n + 2 )>0 for some nN, then by (2.6)

ψ ( d ( g x n + 1 , g x n + 2 ) ) φ ( ψ ( d ( g x n + 1 , g x n + 2 ) ) ) <ψ ( d ( g x n + 1 , g x n + 2 ) ) ;

a contradiction. Hence,

max { d ( g x n , g x n + 1 ) , d ( g x n + 1 , g x n + 2 ) } =d(g x n ,g x n + 1 )

and thus

ψ ( d ( g x n + 1 , g x n + 2 ) ) φ ( ψ ( d ( g x n , g x n + 1 ) ) ) <ψ ( d ( g x n , g x n + 1 ) ) .

Thus, we get

ψ ( d ( g x n + 1 , g x n + 2 ) ) <ψ ( d ( g x n , g x n + 1 ) )

for all nN. Now, from

ψ ( d ( g x n , g x n + 1 ) ) φ ( ψ ( d ( g x n 1 , g x n ) ) ) φ 2 ( ψ ( d ( g x n 2 , g x n 1 ) ) ) φ n ( ψ ( d ( g x 0 , g x 1 ) ) )

and the property of φ, we obtain lim n ψ(d(g x n ,g x n + 1 ))=0, and consequently

lim n d(g x n ,g x n + 1 )=0.
(2.7)

Now, we shall prove that {g x n } is a Cauchy sequence in (X,d). Suppose, on the contrary, that {g x n } is not a Cauchy sequence. Then there exist ϵ>0 and subsequences {g x m ( k ) }, {g x n ( k ) } of {g x n } with m(k)>n(k)k such that

d(g x n ( k ) ,g x m ( k ) )ϵ.
(2.8)

Additionally, corresponding to n(k), we may choose m(k) such that it is the smallest integer satisfying (2.8) and m(k)>n(k)k. Thus,

d(g x n ( k ) ,g x m ( k ) 1 )<ϵ.
(2.9)

Using the triangle inequality in b-metric space and (2.8) and (2.9) we obtain

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 (2.7) we obtain

ϵ lim sup k d(g x n ( k ) ,g x m ( k ) )sϵ.
(2.10)

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 (2.7) and (2.10), we have

ϵ s lim sup k d(g x n ( k ) ,g x m ( k ) + 1 ) s 2 ϵ.
(2.11)

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 (2.7) and (2.10), we have

ϵ s lim sup k d(g x m ( k ) ,g x n ( k ) + 1 ) s 2 ϵ.
(2.12)

Also

d(g x n ( k ) + 1 ,g x m ( k ) )sd(g x n ( k ) + 1 ,g x m ( k ) + 1 )+sd(g x m ( k ) + 1 ,g x m ( k ) ),

so from (2.7) and (2.12), we have

ϵ s 2 lim sup k d(g x n ( k ) + 1 ,g x m ( k ) + 1 ).
(2.13)

Linking (2.7), (2.10), (2.11) together with (2.12) we get

lim sup k M ( x n ( k ) , x m ( k ) ) = max { lim sup k d ( g x n ( k ) , g x m ( k ) ) , lim sup k d ( g x n ( k ) , g x n ( k ) + 1 ) , lim sup k d ( g x m ( k ) , g x m ( k ) + 1 ) , 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 } max { s ϵ , 0 , 0 , s 2 ϵ + s 2 ϵ 2 s } = s ϵ .

So,

lim sup k M( x n ( k ) , x m ( k ) )ϵs.
(2.14)

Similarly, we have

lim sup k N( x n ( k ) , x m ( k ) )=0.
(2.15)

Since m(k)>n(k) from (2.2), we have

g x n ( k ) g x 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 ) , T x m ( k ) ) ) φ ( ψ ( M ( x n ( k ) , x m ( k ) ) ) ) + L ψ ( N ( x n ( k ) , x m ( k ) ) ) .

Passing to the upper limit as k, and using (2.13), (2.14), and (2.15), we get

ψ ( s ϵ ) ψ ( s 3 lim sup k d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) = lim sup k ψ ( s 3 d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) ) = lim sup k ψ ( s 3 d ( T x n ( k ) , T x m ( k ) ) ) lim sup k φ ( ψ ( M ( x n ( k ) , x m ( k ) ) ) ) + lim sup k L ψ ( N ( x n ( k ) , x m ( k ) ) ) = φ ( ψ ( lim sup k M ( x n ( k ) , x m ( k ) ) ) ) + L ψ ( lim sup k N ( x n ( k ) , x m ( k ) ) ) φ ( ψ ( ϵ s ) ) < ψ ( s ϵ ) ,

which is a contradiction. Thus, we proved that {g x n } is a Cauchy sequence in (X,d). Since X is a complete b-metric space, there exists xX such that

lim n g x n + 1 =x.
(2.16)

From the commutativity of T and g, we have

g(g x n + 1 )=g ( T ( x n ) ) =T(g x n ).
(2.17)

Letting n in (2.17) and from the continuity of T and g, we get

gx= lim n g(g x n + 1 )= lim n T(g x n )=T ( lim n g x n ) =T(x).

This implies that x is a coincidence point of T and g. This completes the proof. □

Now, we will prove the following result.

Theorem 2.2 Suppose that (X,d,) is a partially ordered complete b-metric space. Let T:XX be an almost generalized (ψ,φ,L)-contractive mapping with respect to g:XX, T is a g-nondecreasing mapping and T(X)g(X). Also suppose

if  { g x n } X  is a nondecreasing sequence with  g x n g z  in  g X , then  g x n g z , g z g ( g z ) n  hold .
(2.18)

Also suppose gX is closed. If there exists x 0 X such that g x 0 T x 0 , then T and g have a coincidence. Further, if T and g commute at their coincidence points, then T and g have a common fixed point.

Proof As in the proof of Theorem 2.1, we can show that {g x n } is a Cauchy sequence. Since gX is a closed, there exists xX such that

lim n g x n + 1 =gx.
(2.19)

Now we show that x is a coincidence point of T and g. Since from (2.18) and (2.19) we have g x n gx for all n, then by the triangle inequality in a b-metric space and (2.1), we get

d ( g x , T x ) s d ( g x , g x n + 1 ) + s d ( g x n + 1 , T x ) = s d ( g x , g x n + 1 ) + s d ( T x n , T x ) , ψ ( d ( g x , T x ) ) lim n ψ ( s d ( T x n , T x ) ) lim n ψ ( s 3 d ( T x n , T x ) ) ψ ( d ( g x , T x ) ) lim n [ φ ( ψ ( M ( x n , x ) ) ) + L ψ ( N ( x n , x ) ) ] ψ ( d ( g x , T x ) ) φ ( ψ ( d ( g x , T x ) ) ) < ψ ( d ( g x , T x ) ) .

Indeed,

lim n M ( x n , x ) = lim n max { d ( g x n , g x ) , d ( g x n , T x n ) , d ( g x , T x ) , d ( g x n , T x ) + d ( g x , T x n ) 2 s } = d ( g x , T x )

and

lim n N( x n ,x)= lim n min { d ( g x n , T x n ) , d ( g x , T x ) , d ( g x n , T x ) , d ( g x , T x n ) } =0.

Hence d(gx,Tx)=0, that is, Tx=gx. Thus we proved that T and g have a coincidence. Suppose now that T and g commute at x. Set y=Tx=gx. Then

Ty=T(gx)=g(Tx)=gy.

Since from (2.18) we have gxg(gx)=gy and as gx=Tx and gy=Ty, from (2.1) we obtain

ψ ( d ( T x , T y ) ) ψ ( s 3 d ( T x , T y ) ) φ ( ψ ( M ( x , y ) ) ) + L ψ ( N ( x , y ) ) = φ ( ψ ( max { d ( g x , g y ) , d ( g x , T x ) , d ( g y , T y ) , d ( g x , T y ) + d ( g y , T x ) 2 s } ) ) + L ψ ( min { d ( g x , T x ) , d ( g y , T y ) , d ( g x , T y ) , d ( g y , T x ) } ) = φ ( ψ ( d ( T x , T y ) ) ) < ψ ( d ( T x , T y ) ) .

Hence d(Tx,Ty)=0, that is, y=Tx=Ty. Therefore, Ty=gy=y. Thus we proved that T and g have a common fixed point. □

In the following, we deduce some fixed point theorems from our main results given by Theorems 2.1 and 2.2.

Corollary 2.3 Let (X,d,) be a partially ordered complete b-metric space and T:XX is a nondecreasing mapping. Suppose there exist ψΨ, φΦ, and L0 such that

ψ ( s 3 d ( T x , T y ) ) φ ( ψ ( M ( x , y ) ) ) +Lψ ( N ( x , y ) ) ,

where

M(x,y)=max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 s }

and

N(x,y)=min { d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) }

for all x,yX with xy. Also suppose either

  1. (a)

    if { x n }X is a nondecreasing sequence with x n z in X, then x n z, for all n, holds, or

  2. (b)

    T is continuous.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point in X.

Example 2.1 Let X be the set of Lebesgue measurable functions on [0,1] such that 0 1 |x(t)|dt<. Define D:X×X[0,) by

D(x,y)= ( 0 1 | x ( t ) y ( t ) | d t ) 2 .

Then D is a b-metric on X, with s=2. Also, this space can also be equipped with a partial order given by

x,yX,xyx(t)y(t)for any t[a,b].

The operator T:XX defined by

Tx(t)= t n + e t + 2 4 ln ( | x ( t ) | + 1 ) .
(2.20)

Now, we prove that T has a fixed point. For all x,yX with xy, we have

2 3 D ( T x , T y ) = 2 3 ( 0 1 | T x ( t ) T y ( t ) | d t ) 2 2 2 0 1 | 2 4 ln ( | x ( t ) | + 1 ) 2 4 ln ( | y ( t ) | + 1 ) | d t 0 1 | ( ln ( | x ( t ) | + 1 ) ln ( | y ( t ) | + 1 ) ) | d t 0 1 ln ( | x ( t ) | + 1 | y ( t ) | + 1 ) d t 0 1 ln ( 1 + | x ( t ) y ( t ) | | y ( t ) | + 1 ) d t ln ( 1 + 0 1 | x ( t ) y ( t ) | d t ) ln ( 1 + ( 0 1 | x ( t ) y ( t ) | d t ) 2 ) ln ( 1 + D ( x , y ) ) .

Now, if we define φ(t)=ln(1+t), ψ(t)= t , and x 0 =0. Thus, by Corollary 2.3 we see that T has a fixed point.

Remark 2.1 Corollary 2.3 extends and generalizes many existing fixed point theorems in the literature [2, 3, 51, 52].

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 and T:XX is a nondecreasing mapping. Suppose there exists φΦ such that

s 3 d(Tx,Ty)φ ( max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 s } )
(2.21)

for all x,yX with xy. Also suppose either

  1. (a)

    if { x n }X is a nondecreasing sequence with x n z in X, then x n z, for all n, holds, or

  2. (b)

    T is continuous.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point in X.

Remark 2.2 Corollary 2.4 is a generalization to [[3], Theorem 1.3].

Taking φ(t)=λt, 0<λ<1, in Corollary 2.4 we obtain the following generalization of the results in [1, 53].

Corollary 2.5 Let (X,d,) be a partially ordered complete b-metric space and T:XX is a nondecreasing mapping. Suppose there exists φΦ such that

s 3 d(Tx,Ty)λmax { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 s }

for all x,yX with xy. Also suppose either

  1. (a)

    if { x n }X is a nondecreasing sequence with x n z in X, then x n z, for all n, holds, or

  2. (b)

    T is continuous.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point in X.

Corollary 2.6 Let (X,d,) be a partially ordered complete b-metric space and T:XX is a nondecreasing mapping. Suppose there exist ψΨ and 0λ<1 such that

ψ ( s 3 d ( T x , T y ) ) λψ ( d ( x , y ) )

for all x,yX with xy. Also suppose either

  1. (a)

    if { x n }X is a nondecreasing sequence with x n z in X, then x n z, for all n, holds, or

  2. (b)

    T is continuous.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point in X.

3 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 the our fixed point theorem. Consider the integral equation

x(t)=h(t)+λ 0 1 k(t,s)f ( s , x ( s ) ) ds,tI=[0,1],λ0.
(3.1)

Let Γ denote the class of those functions γ:[0,+)[0,+) for which γΦ and ( γ ( t ) ) p γ( t p ), for all p1.

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

We will analyze (3.1) under the following assumptions:

(a1) f:I×RR is continuous monotone nondecreasing in x, f(t,x)0 and there exist constant 0L<1 and γΓ such that for all x,yR and xy

|f(t,x)f(t,y)|Lγ(xy).

(a2) h:IR is a continuous function.

(a3) k:I×IR is continuous in tI for every sI and measurable in sI for all tI such that

0 1 k(t,s)dsK

and k(t,s)0.

(a4) There exists αC(I) such that

α(t)h(t)+λ 0 1 k(t,s)f ( s , α ( s ) ) ds.

(a5) L p λ p K p 1 2 3 p 3 .

We 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,yC(I).

This space can also be equipped with a partial order given by

x,yC(I),xyx(t)y(t)for any tI.

Now for p1, 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,yC(I).

It is easy to see that (X,d) is a complete b-metric space with s= 2 p 1 [44].

For any x,yX and each tI, 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,yX, one can take max{x,y},min{x,y}X which are comparable to x, y. Now, we formulate the main result of this section.

Theorem 3.1 Under assumptions (a1)-(a5), (3.1) has a unique solution in C(I).

Proof We consider the operator T:XX defined by

T(x)(t)=h(t)+λ 0 1 k(t,s)f ( s , x ( s ) ) dsfor tI.

By virtue of our assumptions, T is well defined (this means that if xX then T(x)X). For xy, and tI we have

T ( x ) ( t ) T ( y ) ( t ) = h ( t ) + λ 0 1 k ( t , s ) f ( s , x ( s ) ) d s h ( t ) λ 0 1 k ( t , s ) f ( s , y ( s ) ) d s = λ 0 1 k ( t , s ) [ f ( s , x ( s ) ) f ( s , y ( s ) ) ] d s 0 .

Therefore, T has the monotone nondecreasing property. Also, for xy, we have

| T ( x ) ( t ) T ( y ) ( t ) | = | h ( t ) + λ 0 1 k ( t , s ) f ( s , x ( s ) ) d s h ( t ) λ 0 1 k ( t , s ) f ( s , y ( s ) ) d s | λ 0 1 k ( t , s ) | f ( s , x ( s ) ) f ( s , y ( s ) ) | d s λ 0 1 k ( t , s ) L γ ( y ( s ) x ( s ) ) d s .

Since the function γ is nondecreasing and xy, we have

γ ( y ( s ) x ( s ) ) γ ( sup t I | x ( s ) y ( s ) | ) =γ ( ρ ( x , y ) ) ,

hence

|T(x)(t)T(y)(t)|λ 0 1 k(t,s)Lγ ( ρ ( x , y ) ) dsλKLγ ( ρ ( x , y ) ) .

Then we obtain

d ( T ( x ) , T ( y ) ) = sup t I | T ( x ) ( t ) T ( y ) ( t ) | p { λ K L γ ( ρ ( x , y ) ) } p = λ p K p L p γ ( ρ ( x , y ) ) p λ p K p L p γ ( ρ ( x , y ) p ) = λ p K p L p γ ( d ( x , y ) ) λ p K p L p φ ( max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 s } ) 1 2 3 p 3 φ ( max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 s } ) .

This proves that the operator T satisfies the contractive condition (2.21) appearing in Corollary 2.4. Also, let α, β be the functions appearing in assumption (a4); then, by (a4), we get αT(α). So, (3.1) has a solution and the proof is complete. □

Example 3.1 Consider the following functional integral equation:

x(t)= t 2 1 + t 4 + 1 27 0 1 e s sin t 2 ( 1 + t ) | x ( s ) | 1 + | x ( s ) | ds
(3.2)

for t[0,1]. Observe that this equation is a special case of (3.1) with

h ( t ) = t 2 1 + t 4 , k ( t , s ) = e s 1 + t , f ( t , x ) = sin t 2 | x | 1 + | x | .

Indeed, by using γ(t)= 1 3 t we see that γΦ and ( γ ( t ) ) p = ( 1 3 t ) p = 1 3 p t p 1 3 t p =γ( t p ), for all p1. Further, for arbitrarily fixed x,yR such that xy and for t[0,1] we obtain

| f ( t , x ) f ( t , y ) | = | sin t 2 | x | 1 + | x | sin t 2 | y | 1 + | y | | 1 2 | x y | = 1 6 γ ( x y ) .

Thus, the function f satisfies assumption (a1) with L= 1 6 . It is also easily seen that h is a continuous function. Further, notice that the function k is continuous in tI for every sI and measurable in sI for all tI and k(t,s)0. Moreover, we have

0 1 k ( t , s ) d s = 0 1 e s 1 + t d s = 1 e 1 1 + t 1 e 1 2 3 = K .

If we put α(t)= 3 t 2 4 ( 1 + t 4 ) , we have

α ( t ) = 3 t 2 4 ( 1 + t 4 ) t 2 1 + t 4 t 2 1 + t 4 + 1 27 0 1 e s sin t 2 ( 1 + t ) | α ( s ) | 1 + | α ( s ) | d s = h ( t ) + λ 0 1 k ( t , s ) f ( s , α ( s ) ) d s .

This shows that assumption (a4) holds. Taking L= 1 6 , K= 2 3 and λ= 1 27 , then inequality L p λ p K p 1 2 3 p 3 appearing in assumption (a5) has the following form:

1 6 p × 1 27 p × 2 p 3 p 1 2 3 p 3 .

It is easily seen that each number p1 satisfies the above inequality. Consequently, all the conditions of Theorem 3.1 are satisfied. Hence the integral equation (3.2) has a unique solution in C(I).

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Allahyari, R., Arab, R. & Shole Haghighi, A. A generalization on weak contractions in partially ordered b-metric spaces and its application to quadratic integral equations. J Inequal Appl 2014, 355 (2014). https://doi.org/10.1186/1029-242X-2014-355

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Keywords

  • fixed point
  • common fixed point
  • coincidence point
  • integral equations
  • b-metric space
  • partially ordered set