Open access peer-reviewed chapter

# Contraction Mappings and Applications

By Nawab Hussain and Iram Iqbal

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

DOI: 10.5772/intechopen.81571

## 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 Hndenote the set of all n×nHermitian matrices, Pnthe set of all n×nHermitian positive definite matrices, Snthe set of all n×npositive semidefinite matrices. Instead of XPnwe will write X>0. Furthermore, X0means XSn. Also we will use XYXYinstead of XY0YX0. The symbol .denotes the spectral norm, that is,

A=λ+AA,

where λ+AAis the largest eigenvalue of AA. We denote by .1the 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 trAfor (Hermitian) nonnegative matrices. Then the set Hnendowed with this norm is a complete metric space. Moreover, Hnis 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 Qis a positive definite matrix, Ai, i=1,,m, are arbitrary n×nmatrices and γis a mapping from Hnto Hnwhich maps Pninto Pn. Assume that γis an order-preserving mapping (γis order preserving if A,BHnwith ABimplies 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×IRnIRnand 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 Tand then look for a fixed point of T, that is actually the solution of given mathematical model. The simplest case is when Tis 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 Xbe a set of points, a distance function on Xis a map d:X×X0that is symmetric, and satisfies dii=0for all i X. The distance is said to be a metric if the triangle inequality holds, i.e.,

dijdik+dkj,

for all i,j,kXand Xdis 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 Xand KX, the family of all nonempty compact subsets of X. It is clear that, KXCBXCLX2X, let

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

If T:XXis a single valued self-mapping on X, then Tis said to have a fixed point xif Tx=xand if T:X2Xis multivalued mapping, then Tis said to have a fixed point xif xTx. We denote by FixT, the set of all fixed points of mapping T.

Definition 2.1[22] Let T:X2Xbe a multivalued map on a metric space Xd, α,η:X×XIR+be two functions where ηis bounded, then Tis 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:X2Xbe a multivalued map on a metric space Xd, α,η:X×X0be two functions. We say that Tis generalized α-admissible mapping with respect to η, if

αyzηyzimpliesthatαuvηuv,foralluTy,vTz.

If ηyz=1for all y,zX, then Tis 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×IRIRfulfilling the following conditions:

ϑ1ϑts<stfor all s,t>0;

ϑ2for any bounded sequence tn0+and any nondecreasing sequence sn0+, it holds that

limnsuptn+ϑtnsnsn<1.E3

Example 3.1[24] Letr01. Thenϑr:IR×IRIRdefined byϑrts=rstis a manageable function.

Example 3.2Letϑ:IR×IRIRdefined by

ϑts=ψstifts0+×0+,ftsotherwise,

where ψ:0+0+satisfying n=1ψnt<+for all t>0and f:IR×IRIRis any function. Then ϑtsMan(IR)̂. Indeed, by using Lemma 1 of [25], we have for any s,t>0, ϑts=ψst<st,so, ϑ1holds. Let tn0+be a bounded sequence and let sn0+be a nonincreasing sequence. Then limnsn=infnINsn=afor some a0+, we get

limnsuptn+ϑtnsnsn=limnsupψsnsn<limnsnsn=1,

so, ϑ2is also satisfied.

Definition 3.3Let Xdbe a metric space and T:X2Xbe a closed valued mapping. Let α,η:X×XIR+be two functions and ϑMan(IR)̂. Then Tis called a multivalued αη-manageable contraction with respect to ϑif for all y,zX

αTyTzηTyTzimpliesϑHTyTzd(yz)0.E4

Now we prove first result of this section.

Theorem 3.4LetXdbe a complete metric space and letT:X2Xbe a closed valued map satisfying following conditions:

1. Tisα-admissible map with respect toη;

2. Tisαηmanageable contraction with respect toϑ;

3. there existsz0Xandz1Tz0such thatαz0z1ηz0z1;

4. for a sequenceznX, limnzn=xandαznzn+1ηznzn+1for allnIN, impliesαznxηznxfor allnIN.

ThenFixT.

Proof.Let z1Tz0be such that αz0z1ηz0z1. Since Tis 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 z0z1and z1Tz1. Thus 0<dz1Tz1HTz0Tz1. Define λ:IR×IRIRby

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>0and

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

This implies that there exists z2Tz1such that

dz1z2<1λHTz0Tz1d(z0z1)dz1Tz1.E12

By induction, we form a sequence znin Xsatisfying 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 dznTznnINis a bounded sequence. By combining (15) and (17), for each nIN, we get

dznzn+1<λHTzn1Tznd(zn1zn)dzn1zn.E18

Which means that dzn1znnINis 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 limnsupin (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 znnINis 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 γ01and n0IN, such that

σnγforallnIN,nn0.E23

For any nn0, since σn01for all nINand γ01, (22, 23) implies that

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

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

dznzmj=nm1dzjzj+1<βn.E25

Since γ01, limnβn=0. Hence limnsupdznzm:m>n=0. This shows that znis a Cauchy sequence in X. Completeness of Xensures the existence of zXsuch that znzas n. Now, since αznzηznzfor 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 nin above inequality, we get dzTz=0. Hence zFixT. □

Let ΔFbe the set of all functions F:IR+IRsatisfying following conditions:

F1Fis strictly increasing;

F2for all sequence αnR+, limnαn=0if and only if limnFαn=;

F3there exist 0<k<1such that limn0+αkFα=0,

ΔF, if Falso satisfies the following:

F4FinfA=infFAfor all A0with inf A>0,

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

dTxTy>0impliesτ+FdTxTyFdxy.

Theorem 3.6[26] LetXdbe a complete metric space and letT:XXbe anF-contraction. ThenThas a unique fixed pointxXand for everyx0Xa sequenceTnx0nINis convergent tox.

Definition 3.7([27]). Let Xdbe a metric space and T:XCBXbe a mapping. Then Tis a multivalued F-contraction, if FΔFand there exists τ>0such that for all x,yX,

HTxTy>0τ+FHTxTyFdxy.

Theorem 3.8([27]). LetXdbe a complete metric space andT:XKXbe a multivaluedF-contraction, thenThas a fixed point inX.

Theorem 3.9([27]). LetXdbe a complete metric space andT:XCXbe a multivaluedF-contraction. SupposeFΔF, thenThas a fixed point inX.

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 3.10Let T:X2Xbe a multivalued mapping on a metric space Xd, then Tis said to be an multivalued α-F-weak-contraction on X, if there exists σ>0, τ:0σ, FΔFand α:X×X0+such that for all zX, yFσzwith DzTz>0satisfying

τdzy+FαzyDyTyFMzy,E28

where,

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

and

Fσz=yTz:FdzyFDzTz+σ.

Note that Fσzin both cases when FΔFand FΔF[32].

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

τdzy+FαzyDyTyFdxy,E30

Theorem 3.12LetXdbe a complete metric space andT:XKXbe an multivaluedα-F-weak-contraction satisfying the following assertions:

2. the mapzDzTzis lower semi-continuous;

3. there existsz0Xandz1Tz0such thatαz0z11;

4. τsatisfieslimts+infτt>σfor alls0.

ThenThas a fixed point inX.

Proof.Let z0X, since TzKXfor every zX, the set Fσzis non-empty for any σ>0, then there exists z1Fσz0and by hypothesis αz0z11. Assume that z1Tz1, otherwise z1is the fixed point of T. Then, since Tz1is 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 Tz0and Tz1are 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 znnINin X, where zn+1Fσzn, zn+1Tzn+1, αznzn+11, Mznzn+1dznzn+1and

τdznzn+1+FDzn+1Tzn+1Fdznzn+1.E38

Since zn+1Fσznand Tznand Tzn+1are 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+1for nIN, then dn>0and from (41) dnis decreasing. Therefore, there exists δ0such that limndn=δ. Now let δ>0. From (41), we get

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

Let τdpn=minτd0τd1τdnfor 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>nsuch that τdpn>τdpm. Then we obtain a subsequence dpnkof dpnwith τdpnk>τdpnk+1for all k. Since dpnkδ+, we deduce that limkinfτdpnk>σ. Hence FdnkFd0+nστdpnkfor all k. Consequently, limkFdnk=and by F2, we obtain limkdpnk)=0, which contradicts that limndn>0.

Case 2.There is n0INsuch that τdpn0>τdpmfor all m>n0. Then FdmFd0+mστdpn0for all m>n0. Hence limmFdm=, so limmdm=0, which contradicts that limmdm>0. Thus,

limndn=0.

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

dnrFdndnrFd0dnrnστdpn0.E45

Letting nin (45), we obtain limnndnr=0. This implies that there exists n1INsuch that ndnr1, or, dn1n1/r, for all n>n1. Next, for m>nn1we have

dznzmi=nm1dzizi+1i=nm11i1/k,

since 0<k<1, i=nm11i1/kconverges. Therefore, dznzm0as m,n. Thus, znis a Cauchy sequence. Since Xis complete, there exists zXsuch that znzas n. From Eqs. (44) and F2, we have limnDznTzn=0. Since zDzTzis lower semi-continuous, then

0DzTzlimninf DznTzn=0.

Thus, Thas a fixed point.□

In the following theorem we take CXinstead of KX, then we need to take FΔFin Definition 3.10.

Theorem 3.13LetXdbe a complete metric space andT:XCXbe an multivaluedα-F-weak-contraction withFΔFsatisfying all the assertions of Theorem 3.12. ThenThas a fixed point inX.

Proof.Let z0X, since TzCXfor every zXand FΔF, the set Fσzis non-empty for any σ>0, then there exists z1Fσz0and by hypothesis αz0z11. Assume that z1Tz1, otherwise z1is the fixed point of T. Then, since Tz1is 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 3.14LetXdbe a complete metric space,T:XKXbe a continuous mapping andFΔF. Assume that the following assertions hold:

2. there existsz0Xandz1Tz0such thatαz0z11;

3. there existsτ:00such that

limts+inf τt>0foralls0

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

τdzy+αzyFHTzTyFMzy,E48

where Mzyis defined in (29).

ThenThas a fixed point inX.

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, σcannot be equal to zero.

Theorem 3.15LetXdbe a complete metric space,T:XCXbe a continuous mapping andFΔFsatisfying all assertions of Theorem 3.14. ThenThas a fixed point inX.

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

Theorem 3.16LetXdbe a complete metric space,T:XXbe a continuous mapping andFΔF. Assume that the following assertions hold:

2. there existsz0,z1Xsuch thatαz0z11;

3. there existsτ:00such that

limts+inf τt>0foralls0

and for all zXwith 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

ThenThas a fixed point inX.

Now, let Xdbe a partially ordered metric space. Recall that T:X2Xis monotone increasing if TyTzfor 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 3.17LetXdbe a complete partially ordered metric space and letT:X2Xbe a closed valued mapping satisfying the following assertions for ally,zXwithyz:

1. Tis monotone increasing;

2. ϑHTyTzd(yz)0;

3. there existsz0Xandz1Tz0such thatz0z1;

4. for a sequenceznX, limnzn=zandznzn+1for allnIN, we haveznzfor allnIN.

ThenFixT.

Proof.Define α,η:X×X0by

αyz=1yz0otherwiseηyz=12yz0otherwise,

then for y,zXwith yz, αyzηyzimplies αTyTz=1>12=ηTyTzand αTyTz=ηTyTz=0otherwise. Thus, all the conditions of Theorem 3.4 are satisfied and hence Thas a fixed point.□

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

Theorem 3.18LetXdbe a complete partially ordered metric space and letT:XXbe a self-map fulfilling the following assertions:

1. Tis monotone increasing;

2. ϑdTyTzd(yz)0;

3. there existsz0Xandz1=Tz0such thatz0z1;

4. for a sequenceznX, limnzn=zandznzn+1for allnIN, we haveznzfor allnIN.

for all y,zXwith yzand ϑMan(IR)̂. Then FixT.

Definition 3.19Let T:X2Xbe a multivalued mapping on a partially ordered metric space Xd, then Tis said to be an ordered F-τ-contraction on X, if there exists σ>0and τ:0σ, FΔFsuch that for all zX, yFσzwith zyand DzTz>0satisfying

τdzy+FDyTyFMzy,E51

where,

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

Theorem 3.20LetXdbe a complete partially ordered metric space andT:XKXbe an orderedF-τ-contraction satisfying the following assertions:

1. Tis monotone increasing;

2. the mapzDzTzis lower semi-continuous;

3. there existsz0Xandz1Tz0such thatz0z1;

4. τsatisfies

limts+inf τt>σforalls0

ThenThas a fixed point inX.

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

Theorem 3.21LetXdbe a complete partially ordered metric space andT:XCXbe an orderedF-τ-contraction withFΔFsatisfying all the assertions of Theorem 3.20. ThenThas a fixed point inX.

Theorem 3.22LetXdbe a complete partially ordered metric space,T:XKXbe a continuous mapping andFΔF. Assume that the following assertions hold:

1. Tis monotone increasing;

2. there existsz0Xandz1Tz0such thatz0z1;

3. there existsτ:00such that

limts+inf τt>0foralls0

and for all z,yXwith zyand HTzTy>0satisfying

τdzy+FHTzTyFMzy,E53

where Mzyis defined in (52).

ThenThas a fixed point inX.

Proof.By defining α:X×X0as in the proof of Theorem 3.17 and by using Theorem (3.14), we get the required result.□

Theorem 3.23LetXdbe a complete partially ordered metric space,T:XCXbe a continuous mapping andFΔFsatisfying all assertions of Theorem 3.22. ThenThas a fixed point inX.

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

Theorem 3.24LetXdbe a complete partially ordered metric space,T:XXbe a continuous mapping andFΔF. Assume that the following assertions hold:

1. Tis monotone increasing;

2. there existsz0,z1Xsuch thatz0z1;

3. there existsτ:00such that

limts+inf τt>0foralls0

and for all z,yXwith zyand dTzTy>0satisfying

τdzy+FdTzTyFmzy,E54

where

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

ThenThas a fixed point inX.

## 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:XXby

Tzr=bcBrszsds+gr,rab.E56

Theorem 4.1LetX=CbcIRnwith the usual spermium norm. Suppose that

1. B:bc×bc×IRnIRnand g:IRnIRnare continuous;

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

BrsuB(rsv)prsuv,E57

for each r,sbcand u,vIRnwith uv.

• suprbcbcprsds=q14;

• there exists z0Xand z1Tz0such that z0z1;

• for a sequence znX, limnzn=zand znzn+1for all nIN, we have znzfor all nIN.

• Then the integralEq. (2) has a solution inX.

Proof.Let X=CbcIRnand z=maxrbczr, for zCab. Consider a partial order defined on Xby

y,zCbcIRn,yzifandonlyifyrzr,forrbc.E58

Then X.is a complete partial ordered metric space and for any increasing sequence znin Xconverging to zX, we have znrzrfor any rbc(see [36]). By using Eq. (56), conditions (2, 3) and taking ϑrs=12srfor all y,zXwith yz, we obtain

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

This implies that

12yzTyTz12yz14yz=14yz.

So ϑdTyTzdyz0for all y,zXwith yz. Hence all the conditions of Theorem 3.18 are satisfied. Therefore Thas a fixed point, consequently, integral Eq. (2) has a solution in X.□

### 4.2 Solution of non-linear matrix equation

Theorem 4.2Letγ:HnHnbe an order-preserving mapping which mapsPnintoPnandQPn. Assume that there exists a positive numberNfor whichi=1mAiAiNInandi=1mAiγQAi0such that for allXYwe have

dγXγY1NmYXe2+dXY2dXY,E59

where

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

Then (1) has a solution inPn.

Proof.Define T:HnHnand F:IR+IRby

TX=Q+i=1mAiγXAiE60

and Fr=lnrrespectively. Then a fixed point of Tis a solution of (1). Let X,YHnwith XY, then γXγy. So, for dXY>0and 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 Thas a fixed point and hence (1) has a solution in Pn.□

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Nawab Hussain and Iram Iqbal (July 12th 2019). Contraction Mappings and Applications, Recent Advances in Integral Equations, Francisco Bulnes, IntechOpen, DOI: 10.5772/intechopen.81571. Available from:

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