Open access peer-reviewed chapter

# Contraction Mappings and Applications

Written By

Nawab Hussain and Iram Iqbal

Submitted: 25 June 2018 Reviewed: 18 September 2018 Published: 12 July 2019

DOI: 10.5772/intechopen.81571

From the Edited Volume

## Recent Advances in Integral Equations

Edited by Francisco Bulnes

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## Abstract

The aim of the chapter is to find the existence results for the solution of non-homogeneous Fredholm integral equations of the second kind and non-linear matrix equations by using the fixed point theorems. Here, we derive fixed point theorems for two different type of contractions. Firstly, we utilize the concept of manageable functions to define multivalued α∗−η∗ manageable contractions and prove fixed point theorems for such contractions. After that, we use these fixed point results to find the solution of non-homogeneous Fredholm integral equations of the second kind. Secondly, we introduce weak F contractions named as α-F-weak-contraction to prove fixed point results in the setting of metric spaces and by using these results we find the solution for non-linear matrix equations.

### Keywords

• contraction mapping
• fixed point
• integral equations
• matrix equations
• manageable function

## 1. Introduction

Let Hn denote the set of all n×n Hermitian matrices, Pn the set of all n×n Hermitian positive definite matrices, Sn the set of all n×n positive semidefinite matrices. Instead of XPn we will write X>0. Furthermore, X0 means XSn. Also we will use XY XY instead of XY0YX0. The symbol . denotes the spectral norm, that is,

A=λ+AA,

where λ+AA is the largest eigenvalue of AA. We denote by .1 the Ky Fan norm defined by

A1=i=1nsiA,

where siA, i=1,,n, are the singular values of A. Also,

A1=trAA1/2,

which is trA for (Hermitian) nonnegative matrices. Then the set Hn endowed with this norm is a complete metric space. Moreover, Hn is a partially ordered set with partial order , where XYYX. In this section, denote dXY=YX1=trYX. Now, consider the non-linear matrix equation

X=Q+i=1mAiγXAi,E1

where Q is a positive definite matrix, Ai, i=1,,m, are arbitrary n×n matrices and γ is a mapping from Hn to Hn which maps Pn into Pn. Assume that γ is an order-preserving mapping (γ is order preserving if A,BHn with AB implies that γAγB). There are various kinds of problems in control theory, dynamical programming, ladder networks, etc., where the matrix equations plays a crucial role. Matrix Eq. (1) have been studied by many authors see [1, 2, 3].

At the same time, integral equations have been developed to solve boundary value problems for both ordinary and partial differential equations and play a very important role in nonlinear analysis. Many problems of mathematical physics, theory of elasticity, viscodynamics fluid and mixed problems of mechanics of continuous media reduce to the Fredholm integral Eq. A rich literature on existence of solutions for nonlinear integral equations, which contain particular cases of important integral and functional equations can be found, for example, see [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]. An important technique to solve integral equations is to construct an iterative procedure to generate approximate solutions and find their limit, a host of attractive methods have been proposed for the approximate solutions of Fredholm integral equations of the second kind, see [15, 16, 17, 18, 19]. We consider a non-homogeneous Fredholm integral equation of second kind of the form

zr=bcBrszsds+gr,E2

where tbc, B:bc×bc×IRnIRn and g:IRnIRn.

An advancement in this direction is to find the solution of such mathematical models by using fixed point theorems. In this technique, we generate a sequence by iterative procedure for some self-map T and then look for a fixed point of T, that is actually the solution of given mathematical model. The simplest case is when T is a contraction mapping, that is a self-mapping satisfying

dTxTykdxy,

where k 01. The contraction mapping principle [20] guarantees that a contraction mapping of a complete metric space to itself has a unique fixed point which may be obtained as the limit of an iteration scheme defined by repeated images under the mapping of an arbitrary starting point in the space. The multivalued version of contraction mapping principle can be found in [21]. In general, fixed point theorems allow us to obtain existence theorems concerning investigated functional-operator equations.

In this chapter, we prove the existence of solution for matrix Eq. (1) and integral Eq. (2) by using newly developed fixed point theorems.

## 2. Background material from fixed point theory

Let X be a set of points, a distance function on X is a map d:X×X0 that is symmetric, and satisfies dii=0 for all i X. The distance is said to be a metric if the triangle inequality holds, i.e.,

dijdik+dkj,

for all i,j,kX and Xd is called metric space.

Denote by 2X, the family of all nonempty subsets of X, CLX, the family of all nonempty and closed subsets of X, CBX, the family of all nonempty, closed, and bounded subsets of X and KX, the family of all nonempty compact subsets of X. It is clear that, KXCBXCLX2X, let

where DxB=infdxy:yB. Then H is called generalized Pompeiu Hausdorff distance on CLX. It is well known that H is a metric on CBX, which is called Pompeiu Hausdorff metric induced by d.

If T:XX is a single valued self-mapping on X, then T is said to have a fixed point x if Tx=x and if T:X2X is multivalued mapping, then T is said to have a fixed point x if xTx. We denote by FixT, the set of all fixed points of mapping T.

Definition 2.1 [22] Let T:X2X be a multivalued map on a metric space Xd, α,η:X×XIR+ be two functions where η is bounded, then T is an α-admissible mapping with respect to η, if

αyzηyzimpliesthatαTyTzηTyTz,y,zX,

where

αAB=infyA,zBαyz,ηAB=supyA,zBηyz.

Further, Definition 2.1 is generalized in the following way.

Definition 2.2 [23] Let T:X2X be a multivalued map on a metric space Xd, α,η:X×X0 be two functions. We say that T is generalized α-admissible mapping with respect to η, if

αyzηyzimpliesthatαuvηuv,foralluTy,vTz.

If ηyz=1 for all y,zX, then T is said to be generalized α-admissible mapping.

## 3. Some fixed point results

Consistent with Du and Khojasteh [24], we denote by Man(IR)̂, the set of all manageable functions ϑ:IR×IRIR fulfilling the following conditions:

ϑ1 ϑts<st for all s,t>0;

ϑ2 for any bounded sequence tn0+ and any nondecreasing sequence sn0+, it holds that

limnsuptn+ϑtnsnsn<1.E3

Example 3.1 [24] Let r01. Then ϑr:IR×IRIR defined by ϑrts=rst is a manageable function.

Example 3.2 Let ϑ:IR×IRIR defined by

ϑts=ψstifts0+×0+,ftsotherwise,

where ψ:0+0+ satisfying n=1ψnt<+ for all t>0 and f:IR×IRIR is any function. Then ϑtsMan(IR)̂. Indeed, by using Lemma 1 of [25], we have for any s,t>0, ϑts=ψst<st, so, ϑ1 holds. Let tn0+ be a bounded sequence and let sn0+ be a nonincreasing sequence. Then limnsn=infnINsn=a for some a0+, we get

limnsuptn+ϑtnsnsn=limnsupψsnsn<limnsnsn=1,

so, ϑ2 is also satisfied.

Definition 3.3 Let Xd be a metric space and T:X2X be a closed valued mapping. Let α,η:X×XIR+ be two functions and ϑMan(IR)̂. Then T is called a multivalued αη-manageable contraction with respect to ϑ if for all y,zX

αTyTzηTyTzimpliesϑHTyTzd(yz)0.E4

Now we prove first result of this section.

Theorem 3.4 Let Xd be a complete metric space and let T:X2X be a closed valued map satisfying following conditions:

1. T is α-admissible map with respect to η;

2. T is αη manageable contraction with respect to ϑ;

3. there exists z0X and z1Tz0 such that αz0z1ηz0z1;

4. for a sequence znX, limnzn=x and αznzn+1ηznzn+1 for all nIN, implies αznxηznx for all nIN.

Then FixT.

Proof. Let z1Tz0 be such that αz0z1ηz0z1. Since T is α-admissible map with respect to η, then αTz0Tz1ηTz0Tz1. Therefore, from (4) we have

ϑHTz0Tz1d(z0z1)0.E5

If z1=z0, then z0FixT, also if z1Tz1, then z1FixT. So, we adopt that z0z1 and z1Tz1. Thus 0<dz1Tz1HTz0Tz1. Define λ:IR×IRIR by

λts=t+ϑtssift,s>00otherwise.E6

By ϑ1, we know that

0<λts<1forallt,s>0.E7

Also note that if ϑts0, then

0<tts.E8

So, from (5) and (7), we get

0<λHTz0Tz1d(z0z1)<1.E9

Let

ε1=1λHTz0Tz1d(z0z1)1dz1Tz1.E10

Since dz1Tz1>0. So, by using (9), we get ε1>0 and

dz1Tz1<dz1Tz1+ε1=1λHTz0Tz1d(z0z1)dz1Tz1.E11

This implies that there exists z2Tz1 such that

dz1z2<1λHTz0Tz1d(z0z1)dz1Tz1.E12

By induction, we form a sequence zn in X satisfying for each nIN, znTzn1, znzn1, znTzn, αzn1znηzn1zn,

0<dxnTxnHTzn1Tzn,E13
ϑHTzn1Tznd(zn1zn)0,E14

and

dznzn+1<1λHTzn1Tznd(zn1zn)dznTzn,E15

by taking

εn=1λHTzn1Tznd(zn1zn)1dznTzn.E16

By using (7), (8), (13), and (15), we get for each nIN

dznTzndzn1znλHTzn1Tznd(zn1zn)dzn1zn,E17

this implies that dznTznnIN is a bounded sequence. By combining (15) and (17), for each nIN, we get

dznzn+1<λHTzn1Tznd(zn1zn)dzn1zn.E18

Which means that dzn1znnIN is a monotonically decreasing sequence of non-negative reals and so it must be convergent. So, let limndznzn+1=c0. From ϑ2, we get

limnsupλHTznTznd(zn1zn)<1.E19

Now, if c>0, then by taking the limnsup in (18) and using (19), we have

climnsupλHTzn1Tznd(zn1zn)c<c.E20

This contradiction shows that c=0. Hence, limndznzn+1=0. Next, we prove that znnIN is a Cauchy sequence in X. Let, for each nIN,

σn=λHTzn1Tznd(zn1zn),E21

then from Eq. (9), we have σn01. By (18), we obtain

dznzn+1<σndzn1zn.E22

(19) implies that limnσn<1, so there exists γ01 and n0IN, such that

σnγforallnIN,nn0.E23

For any nn0, since σn01 for all nIN and γ01, (22, 23) implies that

dznzn+1<σndzn1zn<σnσn1dzn2zn1γnn0+1dz0z1.E24

Put βn=γnn0+11γdz0z1, nIN. For m,nIN with m>nn0, we have from (24) that

dznzmj=nm1dzjzj+1<βn.E25

Since γ01, limnβn=0. Hence limnsupdznzm:m>n=0. This shows that zn is a Cauchy sequence in X. Completeness of X ensures the existence of zX such that znz as n. Now, since αznzηznz for all nIN, αTznTzηTznTz, and so from (4), we have ϑHTznTzd(znz)0. Then from (7, 8), we have

HTznTzλHTznTzd(znz)dznz<dznz.E26

Since 0<dzTzHTznTz+dznz, so by using (26), we get

0<dzTz<2dznz.E27

Letting limit n in above inequality, we get dzTz=0. Hence zFixT. □

Let ΔF be the set of all functions F:IR+IR satisfying following conditions:

F1 F is strictly increasing;

F2 for all sequence αnR+, limnαn=0 if and only if limnFαn=;

F3 there exist 0<k<1 such that limn0+αkFα=0,

ΔF, if F also satisfies the following:

F4 FinfA=infFA for all A0 with inf A>0,

Definition 3.5 [27] Let Xd be a metric space. A mapping T:XX is said to be F-contraction of there exists τ>0 such that

dTxTy>0impliesτ+FdTxTyFdxy.

Theorem 3.6 [26] Let Xd be a complete metric space and let T:XX be an F-contraction. Then T has a unique fixed point xX and for every x0X a sequence Tnx0nIN is convergent to x.

Definition 3.7 ([27]). Let Xd be a metric space and T:XCBX be a mapping. Then T is a multivalued F-contraction, if FΔF and there exists τ>0 such that for all x,yX,

HTxTy>0τ+FHTxTyFdxy.

Theorem 3.8 ([27]). Let Xd be a complete metric space and T:XKX be a multivalued F-contraction, then T has a fixed point in X.

Theorem 3.9 ([27]). Let Xd be a complete metric space and T:XCX be a multivalued F-contraction. Suppose FΔF, then T has a fixed point in X.

For more in this direction, see, [28, 29, 30, 31]. Here, we give the concept of multivalued α-F-weak-contractions and prove some fixed point results.

Definition 3.10 Let T:X2X be a multivalued mapping on a metric space Xd, then T is said to be an multivalued α-F-weak-contraction on X, if there exists σ>0, τ:0σ, FΔF and α:X×X0+ such that for all zX, yFσz with DzTz>0 satisfying

τdzy+FαzyDyTyFMzy,E28

where,

Mzy=maxdzyD(zTz)D(yTy)DyTz+DzTy2DyTy1+DzTz1+dzyDyTz1+DzTy1+dzy.E29

and

Fσz=yTz:FdzyFDzTz+σ.

Note that Fσz in both cases when FΔF and FΔF [32].

Definition 3.11 Let T:XPX be a multivalued mapping on a metric space Xd, then T is said to be an multivalued α-F-contraction on X, if there exists σ>0, τ:0σ, FΔF and α:X×X0+ such that for all zX, yFσz with DzTz>0 satisfying

τdzy+FαzyDyTyFdxy,E30

Theorem 3.12 Let Xd be a complete metric space and T:XKX be an multivalued α-F-weak-contraction satisfying the following assertions:

1. T is multivalued α-orbital admissible mapping;

2. the map zDzTz is lower semi-continuous;

3. there exists z0X and z1Tz0 such that αz0z11;

4. τ satisfies limts+infτt>σ for all s0.

Then T has a fixed point in X.

Proof. Let z0X, since TzKX for every zX, the set Fσz is non-empty for any σ>0, then there exists z1Fσz0 and by hypothesis αz0z11. Assume that z1Tz1, otherwise z1 is the fixed point of T. Then, since Tz1 is closed, Dz1Tz1>0, so, from (28), we have

τdz0z1+Fαz0z1Dz1Tz1FMz0z1,E31

where

Mz0z1=maxdz0z1D(z0Tz0)D(z1Tz1)Dz1Tz0+Dz0Tz12Dz1Tz11+Dz0Tz01+dz0z1Dz1Tz01+Dz0Tz11+dz0z1.E32

Since Tz0 and Tz1 are compact, so we have

Mz0z1=maxdz0z1d(z0z1)d(z1z2)dz1z1+dz0z22dz1z21+dz0z11+dz0z1dz1z11+dz0z21+dz0z1=maxdz0z1d(z1z2)dz0z22.E33

Since dz0z22dz0z1+dz1z22maxdz0z1dz1z2, it follows that

Mz0z1maxdz0z1dz1z2.E34

Suppose that dz0z1<dz1z2, then (31) implies that

τdz0z1+FDz1Tz1τdz0z1+Fαz0z1Dz1Tz1Fdz1z2,E35

consequently,

τdz0z1+Fdz1z2Fdz1z2,E36

or, Fdz1z2Fdz1z2τdz0z1, which is a contradiction. Hence Mdz0z1dz0z1, therefore by using F1, (31) implies that

τdz0z1+Fαz0z1dz1z2Fdz0z1.E37

On continuing recursively, we get a sequence znnIN in X, where zn+1Fσzn, zn+1Tzn+1, αznzn+11, Mznzn+1dznzn+1 and

τdznzn+1+FDzn+1Tzn+1Fdznzn+1.E38

Since zn+1Fσzn and Tzn and Tzn+1 are compact, we have

τdznzn+1+Fdzn+1zn+2Fdznzn+1E39

and

Fdznzn+1Fdznzn+1+σ.E40

Combining (39) and (40) gives

Fdzn+1zn+2Fdznzn+1+στdznzn+1E41

Let dn=dznzn+1 for nIN, then dn>0 and from (41) dn is decreasing. Therefore, there exists δ0 such that limndn=δ. Now let δ>0. From (41), we get

Fdn+1Fdn+στdnFdn1+2στdnτdn1Fd0+τdnτdn1τd0.E42

Let τdpn=minτd0τd1τdn for all nIN. From (42), we get

Fdn+1Fd0+nστdpn.E43

From (38), we also get

FDzn+1Tzn+1FDz0Tz0+nστdpn.E44

Now consider the sequence τdpn. We distinguish two cases.

Case 1. For each nIN, there is m>n such that τdpn>τdpm. Then we obtain a subsequence dpnk of dpn with τdpnk>τdpnk+1 for all k. Since dpnkδ+, we deduce that limkinfτdpnk>σ. Hence FdnkFd0+nστdpnk for all k. Consequently, limkFdnk= and by F2, we obtain limkdpnk)=0, which contradicts that limndn>0.

Case 2. There is n0IN such that τdpn0>τdpm for all m>n0. Then FdmFd0+mστdpn0 for all m>n0. Hence limmFdm=, so limmdm=0, which contradicts that limmdm>0. Thus,

limndn=0.

From F3, there exists 0<r<1 such that limndnrFdn=0. By (43), we get for all nIN

dnrFdndnrFd0dnrnστdpn0.E45

Letting n in (45), we obtain limnndnr=0. This implies that there exists n1IN such that ndnr1, or, dn1n1/r, for all n>n1. Next, for m>nn1 we have

dznzmi=nm1dzizi+1i=nm11i1/k,

since 0<k<1, i=nm11i1/k converges. Therefore, dznzm0 as m,n. Thus, zn is a Cauchy sequence. Since X is complete, there exists zX such that znz as n. From Eqs. (44) and F2, we have limnDznTzn=0. Since zDzTz is lower semi-continuous, then

0DzTzlimninf DznTzn=0.

Thus, T has a fixed point.□

In the following theorem we take CX instead of KX, then we need to take FΔF in Definition 3.10.

Theorem 3.13 Let Xd be a complete metric space and T:XCX be an multivalued α-F-weak-contraction with FΔF satisfying all the assertions of Theorem 3.12. Then T has a fixed point in X.

Proof. Let z0X, since TzCX for every zX and FΔF, the set Fσz is non-empty for any σ>0, then there exists z1Fσz0 and by hypothesis αz0z11. Assume that z1Tz1, otherwise z1 is the fixed point of T. Then, since Tz1 is closed, Dz1Tz1>0, so, from (28), we have

τdz0z1+αz0z1FDz1Tz1FMz0z1,E46

where

Mz0z1=maxdz0z1D(z0Tz0)D(z1Tz1)Dz1Tz0+Dz0Tz12Dz1Tz11+Dz0Tz01+dz0z1Dz1Tz01+Dz0Tz11+dz0z1.E47

The rest of the proof can be completed as in the proof of Theorem 3.12 by considering the closedness of Tz, for all zX.□

Theorem 3.14 Let Xd be a complete metric space, T:XKX be a continuous mapping and FΔF. Assume that the following assertions hold:

1. T is generalized α-admissible mapping;

2. there exists z0X and z1Tz0 such that αz0z11;

3. there exists τ:00 such that

limts+inf τt>0foralls0

and for all zX with HTzTy>0, there exist a function α:X×X0+ satisfying

τdzy+αzyFHTzTyFMzy,E48

where Mzy is defined in (29).

Then T has a fixed point in X.

Proof. By following the steps in the proof of Theorem 3.12, we get the required result.□

Note that Theorem 3.14 cannot be obtained from Theorem 3.12, because in Theorem 3.12, σ cannot be equal to zero.

Theorem 3.15 Let Xd be a complete metric space, T:XCX be a continuous mapping and FΔF satisfying all assertions of Theorem 3.14. Then T has a fixed point in X.

From Theorems 3.14 and 3.15, we get the following fixed point result for single valued mappings:

Theorem 3.16 Let Xd be a complete metric space, T:XX be a continuous mapping and FΔF. Assume that the following assertions hold:

2. there exists z0,z1X such that αz0z11;

3. there exists τ:00 such that

limts+inf τt>0foralls0

and for all zX with dTzTy>0, there exist a function α:X×X0+ satisfying

τdzy+αzyFdTzTyFmzy,E49

where

mzy=maxdzyd(zTz)d(yTy)dyTz+dzTy2dyTy1+dzTz1+dzydyTz1+dzTy1+dzy.E50

Then T has a fixed point in X.

Now, let Xd be a partially ordered metric space. Recall that T:X2X is monotone increasing if TyTz for all y,zX, for which yz (see [33]). There are many applications in differential and integral equations of monotone mappings in ordered metric spaces (see [34, 35, 36] and references therein).

Theorem 3.17 Let Xd be a complete partially ordered metric space and let T:X2X be a closed valued mapping satisfying the following assertions for all y,zX with yz:

1. T is monotone increasing;

2. ϑHTyTzd(yz)0;

3. there exists z0X and z1Tz0 such that z0z1;

4. for a sequence znX, limnzn=z and znzn+1 for all nIN, we have znz for all nIN.

Then FixT.

Proof. Define α,η:X×X0 by

αyz=1yz0otherwiseηyz=12yz0otherwise,

then for y,zX with yz, αyzηyz implies αTyTz=1>12=ηTyTz and αTyTz=ηTyTz=0 otherwise. Thus, all the conditions of Theorem 3.4 are satisfied and hence T has a fixed point.□

In case of single valued mapping Theorem 3.17 reduced to the following:

Theorem 3.18 Let Xd be a complete partially ordered metric space and let T:XX be a self-map fulfilling the following assertions:

1. T is monotone increasing;

2. ϑdTyTzd(yz)0;

3. there exists z0X and z1=Tz0 such that z0z1;

4. for a sequence znX, limnzn=z and znzn+1 for all nIN, we have znz for all nIN.

for all y,zX with yz and ϑMan(IR)̂. Then FixT.

Definition 3.19 Let T:X2X be a multivalued mapping on a partially ordered metric space Xd, then T is said to be an ordered F-τ-contraction on X, if there exists σ>0 and τ:0σ, FΔF such that for all zX, yFσz with zy and DzTz>0 satisfying

τdzy+FDyTyFMzy,E51

where,

Mzy=maxdzyD(zTz)D(yTy)DyTz+DzTy2DyTy1+DzTz1+dzyDyTz1+DzTy1+dzy.E52

Theorem 3.20 Let Xd be a complete partially ordered metric space and T:XKX be an ordered F-τ-contraction satisfying the following assertions:

1. T is monotone increasing;

2. the map zDzTz is lower semi-continuous;

3. there exists z0X and z1Tz0 such that z0z1;

4. τ satisfies

limts+inf τt>σforalls0

Then T has a fixed point in X.

Proof. By using the similar arguments as in the proof of Theorem 3.17 and using Theorem 3.12, we get the result.□

Theorem 3.21 Let Xd be a complete partially ordered metric space and T:XCX be an ordered F-τ-contraction with FΔF satisfying all the assertions of Theorem 3.20. Then T has a fixed point in X.

Theorem 3.22 Let Xd be a complete partially ordered metric space, T:XKX be a continuous mapping and FΔF. Assume that the following assertions hold:

1. T is monotone increasing;

2. there exists z0X and z1Tz0 such that z0z1;

3. there exists τ:00 such that

limts+inf τt>0foralls0

and for all z,yX with zy and HTzTy>0 satisfying

τdzy+FHTzTyFMzy,E53

where Mzy is defined in (52).

Then T has a fixed point in X.

Proof. By defining α:X×X0 as in the proof of Theorem 3.17 and by using Theorem (3.14), we get the required result.□

Theorem 3.23 Let Xd be a complete partially ordered metric space, T:XCX be a continuous mapping and FΔF satisfying all assertions of Theorem 3.22. Then T has a fixed point in X.

From Theorems 3.22 and 3.23, we get the following fixed point result for single valued mapping.

Theorem 3.24 Let Xd be a complete partially ordered metric space, T:XX be a continuous mapping and FΔF. Assume that the following assertions hold:

1. T is monotone increasing;

2. there exists z0,z1X such that z0z1;

3. there exists τ:00 such that

limts+inf τt>0foralls0

and for all z,yX with zy and dTzTy>0 satisfying

τdzy+FdTzTyFmzy,E54

where

mzy=maxdzyd(zTz)d(yTy)dyTz+dzTy2dyTy1+dzTz1+dzydyTz1+dzTy1+dzy.E55

Then T has a fixed point in X.

## 4. Existence of solution

In this section, by using the fixed point results proved in the previous section, we obtain the existence of the solution of integral Eq. (2) and matrix Eq. (1).

### 4.1 Solution of Fredholm integral equation of second kind

Let be a partial order relation on IRn. Define T:XX by

Tzr=bcBrszsds+gr,rab.E56

Theorem 4.1 Let X=CbcIRn with the usual spermium norm. Suppose that

1. B:bc×bc×IRnIRn and g:IRnIRn are continuous;

2. there exists a continuous function p:bc×bcbc such that

BrsuB(rsv)prsuv,E57

for each r,sbc and u,vIRn with uv.

3. suprbcbcprsds=q14;

4. there exists z0X and z1Tz0 such that z0z1;

5. for a sequence znX, limnzn=z and znzn+1 for all nIN, we have znz for all nIN.

Then the integral Eq. (2) has a solution in X.

Proof. Let X=CbcIRn and z=maxrbczr, for zCab. Consider a partial order defined on X by

y,zCbcIRn,yzifandonlyifyrzr,forrbc.E58

Then X. is a complete partial ordered metric space and for any increasing sequence zn in X converging to zX, we have znrzr for any rbc (see [36]). By using Eq. (56), conditions (2, 3) and taking ϑrs=12sr for all y,zX with yz, we obtain

TyrTzr=bcB(rsys)dsbcB(rszs)dsbcBrsysBrszsdsbcprsyszsds14yz.

This implies that

12yzTyTz12yz14yz=14yz.

So ϑdTyTzdyz0 for all y,zX with yz. Hence all the conditions of Theorem 3.18 are satisfied. Therefore T has a fixed point, consequently, integral Eq. (2) has a solution in X.□

### 4.2 Solution of non-linear matrix equation

Theorem 4.2 Let γ:HnHn be an order-preserving mapping which maps Pn into Pn and QPn. Assume that there exists a positive number N for which i=1mAiAiNIn and i=1mAiγQAi0 such that for all XY we have

dγXγY1NmYXe2+dXY2dXY,E59

where

mXY=maxdXYd(XTX)d(YTY)dYTY+dXTX2dYTY1+dXTX1+dXYdYTX1+dXTY1+dXY.

Then (1) has a solution in Pn.

Proof. Define T:HnHn and F:IR+IR by

TX=Q+i=1mAiγXAiE60

and Fr=lnr respectively. Then a fixed point of T is a solution of (1). Let X,YHn with XY, then γXγy. So, for dXY>0 and τt=1t+12, we have

dTXTY=TYTX1=trTYTX=i=1mtrAiAiγYγX=tri=1mAiAiγYγXi=1mAiAiγYγX1i=1mAiAiNmYXe2+YX12YX1<mYXe2+YX12YX1,

and so,

lnTYTX1<lnmYXe2+YX12YX1=lnmXY2+YX12YX1.

This implies that

1YX1+12+lnTYTX1<lnmXY.

Consequently,

τdXY+FdTXTY<FmXY.

Also, from i=1mAiγQAi0, we have QTQ. Thus, by using Theorem 3.24, we conclude that T has a fixed point and hence (1) has a solution in Pn.□

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Written By

Nawab Hussain and Iram Iqbal

Submitted: 25 June 2018 Reviewed: 18 September 2018 Published: 12 July 2019