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 X∈Pn we will write X>0. Furthermore, X≥0 means X∈Sn. Also we will use X≥Y X≤Y instead of X−Y≥0Y−X≥0. The symbol ‖.‖ denotes the spectral norm, that is,

‖A‖=λ+A∗A,

where λ+A∗A is the largest eigenvalue of A∗A. 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=trA∗A1/2,

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 X≼Y⇔Y≼X. In this section, denote dXY=Y−X1=trY−X. Now, consider the non-linear matrix equation

X=Q+∑i=1mAi∗γXAi,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,B∈Hn with A≼B implies that γA≼γB). 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 t∈bc, B:bc×bc×IRn→IRn and g:IRn→IRn.

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

dTxTy≤kdxy,

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×X→0∞ 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.,

dij≤dik+dkj,

for all i,j,k∈X 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, KX⊆CBX⊆CLX⊆2X, let

HAB=maxsupx∈ADxBsupy∈BD(yA),

where DxB=infdxy:y∈B. 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:X→X is a single valued self-mapping on X, then T is said to have a fixed point x if Tx=x and if T:X→2X is multivalued mapping, then T is said to have a fixed point x if x∈Tx. We denote by FixT, the set of all fixed points of mapping T.

Definition 2.1 [22] Let T:X→2X be a multivalued map on a metric space Xd, α,η:X×X→IR+ be two functions where η is bounded, then T is an α∗-admissible mapping with respect to η, if

αyz≥ηyzimpliesthatα∗TyTz≥η∗TyTz,y,z∈X,

where

α∗AB=infy∈A,z∈Bαyz,η∗AB=supy∈A,z∈Bηyz.

Further, Definition 2.1 is generalized in the following way.

Definition 2.2 [23] Let T:X→2X be a multivalued map on a metric space Xd, α,η:X×X→0∞ be two functions. We say that T is generalized α∗-admissible mapping with respect to η, if

αyz≥ηyzimpliesthatαuv≥ηuv,forallu∈Ty,v∈Tz.

If ηyz=1 for all y,z∈X, 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×IR→IR fulfilling the following conditions:

ϑ1 ϑts<s−t for all s,t>0;

ϑ2 for any bounded sequence tn⊂0+∞ and any nondecreasing sequence sn⊂0+∞, it holds that

limn→∞suptn+ϑtnsnsn<1.E3

Example 3.1 [24] Let r∈01. Then ϑr:IR×IR→IR defined by ϑrts=rs−t is a manageable function.

Example 3.2 Let ϑ:IR×IR→IR defined by

ϑts=ψs−tifts∈0+∞×0+∞,ftsotherwise,

where ψ:0+∞→0+∞ satisfying ∑n=1∞ψnt<+∞ for all t>0 and f:IR×IR→IR is any function. Then ϑts∈Man(IR)̂. Indeed, by using Lemma 1 of [25], we have for any s,t>0, ϑts=ψs−t<s−t, so, ϑ1 holds. Let tn⊂0+∞ be a bounded sequence and let sn⊂0+∞ be a nonincreasing sequence. Then limn→∞sn=infn∈INsn=a for some a∈0+∞, we get

limn→∞suptn+ϑtnsnsn=limn→∞supψsnsn<limn→∞snsn=1,

so, ϑ2 is also satisfied.

Definition 3.3 Let Xd be a metric space and T:X→2X be a closed valued mapping. Let α,η:X×X→IR+ be two functions and ϑ∈Man(IR)̂. Then T is called a multivalued α∗−η∗-manageable contraction with respect to ϑ if for all y,z∈X

α∗TyTz≥η∗TyTzimpliesϑHTyTzd(yz)≥0.E4

Now we prove first result of this section.

Theorem 3.4 Let Xd be a complete metric space and let T:X→2X be a closed valued map satisfying following conditions:

T is α∗-admissible map with respect to η;
T is α∗−η∗ manageable contraction with respect to ϑ;
there exists z0∈X and z1∈Tz0 such that αz0z1≥ηz0z1;
for a sequence zn⊂X, limn→∞zn=x and αznzn+1≥ηznzn+1 for all n∈IN, implies αznx≥ηznx for all n∈IN.

Then FixT≠∅.

Proof. Let z1∈Tz0 be such that αz0z1≥ηz0z1. Since T is α∗-admissible map with respect to η, then α∗Tz0Tz1≥η∗Tz0Tz1. Therefore, from (4) we have

ϑHTz0Tz1d(z0z1)≥0.E5

If z1=z0, then z0∈FixT, also if z1∈Tz1, then z1∈FixT. So, we adopt that z0≠z1 and z1∉Tz1. Thus 0<dz1Tz1≤HTz0Tz1. Define λ:IR×IR→IR by

λts=t+ϑtssift,s>00otherwise.E6

By ϑ1, we know that

0<λts<1forallt,s>0.E7

Also note that if ϑts≥0, then

0<t≤sλts.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 z2∈Tz1 such that

dz1z2<1λHTz0Tz1d(z0z1)dz1Tz1.E12

By induction, we form a sequence zn in X satisfying for each n∈IN, zn∈Tzn−1, zn≠zn−1, zn∉Tzn, α∗zn−1zn≥η∗zn−1zn,

0<dxnTxn≤HTzn−1Tzn,E13

ϑHTzn−1Tznd(zn−1zn)≥0,E14

and

dznzn+1<1λHTzn−1Tznd(zn−1zn)dznTzn,E15

by taking

εn=1λHTzn−1Tznd(zn−1zn)−1dznTzn.E16

By using (7), (8), (13), and (15), we get for each n∈IN

dznTzn≤dzn−1znλHTzn−1Tznd(zn−1zn)≤dzn−1zn,E17

this implies that dznTznn∈IN is a bounded sequence. By combining (15) and (17), for each n∈IN, we get

dznzn+1<λHTzn−1Tznd(zn−1zn)dzn−1zn.E18

Which means that dzn−1znn∈IN is a monotonically decreasing sequence of non-negative reals and so it must be convergent. So, let limn→∞dznzn+1=c≥0. From ϑ2, we get

limn→∞supλHTznTznd(zn−1zn)<1.E19

Now, if c>0, then by taking the limn→∞sup in (18) and using (19), we have

c≤limn→∞supλHTzn−1Tznd(zn−1zn)c<c.E20

This contradiction shows that c=0. Hence, limn→∞dznzn+1=0. Next, we prove that znn∈IN is a Cauchy sequence in X. Let, for each n∈IN,

σn=λHTzn−1Tznd(zn−1zn),E21

then from Eq. (9), we have σn∈01. By (18), we obtain

dznzn+1<σndzn−1zn.E22

(19) implies that limn→∞σn<1, so there exists γ∈01 and n0∈IN, such that

σn≤γforalln∈IN,n≥n0.E23

For any n≥n0, since σn∈01 for all n∈IN and γ∈01, (22, 23) implies that

dznzn+1<σndzn−1zn<σnσn−1dzn−2zn−1⋯≤γn−n0+1dz0z1.E24

Put βn=γn−n0+11−γdz0z1, n∈IN. For m,n∈IN with m>n≥n0, we have from (24) that

dznzm≤∑j=nm−1dzjzj+1<βn.E25

Since γ∈01, limn→∞βn=0. Hence limn→∞supdznzm:m>n=0. This shows that zn is a Cauchy sequence in X. Completeness of X ensures the existence of z∈X such that zn→z as n→∞. Now, since αznz≥ηznz for all n∈IN, α∗TznTz≥η∗TznTz, and so from (4), we have ϑHTznTzd(znz)≥0. Then from (7, 8), we have

HTznTz≤λHTznTzd(znz)dznz<dznz.E26

Since 0<dzTz≤HTznTz+dznz, so by using (26), we get

0<dzTz<2dznz.E27

Letting limit n→∞ in above inequality, we get dzTz=0. Hence z∈FixT. □

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

F1 F is strictly increasing;

F2 for all sequence αn⊆R+, limn→∞αn=0 if and only if limn→∞Fαn=−∞;

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

ΔF∗, if F also satisfies the following:

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

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

dTxTy>0impliesτ+FdTxTy≤Fdxy.

Theorem 3.6 [26] Let Xd be a complete metric space and let T:X→X be an F-contraction. Then T has a unique fixed point x∗∈X and for every x0∈X a sequence Tnx0n∈IN is convergent to x∗.

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

HTxTy>0⇒τ+FHTxTy≤Fdxy.

Theorem 3.8 ([27]). Let Xd be a complete metric space and T:X→KX 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:X→CX be a multivalued F-contraction. Suppose F∈ΔF∗, 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:X→2X 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:0∞→σ∞, F∈ΔF and α:X×X→0+∞ such that for all z∈X, y∈Fσz with DzTz>0 satisfying

τdzy+FαzyDyTy≤FMzy,E28

where,

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

and

Fσz=y∈Tz:Fdzy≤FDzTz+σ.

Note that Fσz≠∅ in both cases when F∈ΔF and F∈ΔF∗ [32].

Definition 3.11 Let T:X→PX 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:0∞→σ∞, F∈ΔF and α:X×X→0+∞ such that for all z∈X, y∈Fσz with DzTz>0 satisfying

τdzy+FαzyDyTy≤Fdxy,E30

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

T is multivalued α-orbital admissible mapping;
the map z→DzTz is lower semi-continuous;
there exists z0∈X and z1∈Tz0 such that αz0z1≥1;
τ satisfies limt→s+infτt>σ for all s≥0.

Then T has a fixed point in X.

Proof. Let z0∈X, since Tz∈KX for every z∈X, the set Fσz is non-empty for any σ>0, then there exists z1∈Fσz0 and by hypothesis αz0z1≥1. Assume that z1∉Tz1, otherwise z1 is the fixed point of T. Then, since Tz1 is closed, Dz1Tz1>0, so, from (28), we have

τdz0z1+Fαz0z1Dz1Tz1≤FMz0z1,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 dz0z22≤dz0z1+dz1z22≤maxdz0z1dz1z2, it follows that

Mz0z1≤maxdz0z1dz1z2.E34

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

τdz0z1+FDz1Tz1≤τdz0z1+Fαz0z1Dz1Tz1≤Fdz1z2,E35

consequently,

τdz0z1+Fdz1z2≤Fdz1z2,E36

or, Fdz1z2≤Fdz1z2−τdz0z1, which is a contradiction. Hence Mdz0z1≤dz0z1, therefore by using F1, (31) implies that

τdz0z1+Fαz0z1dz1z2≤Fdz0z1.E37

On continuing recursively, we get a sequence znn∈IN in X, where zn+1∈Fσzn, zn+1∉Tzn+1, αznzn+1≥1, Mznzn+1≤dznzn+1 and

τdznzn+1+FDzn+1Tzn+1≤Fdznzn+1.E38

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

τdznzn+1+Fdzn+1zn+2≤Fdznzn+1E39

and

Fdznzn+1≤Fdznzn+1+σ.E40

Combining (39) and (40) gives

Fdzn+1zn+2≤Fdznzn+1+σ−τdznzn+1E41

Let dn=dznzn+1 for n∈IN, then dn>0 and from (41) dn is decreasing. Therefore, there exists δ≥0 such that limn→∞dn=δ. Now let δ>0. From (41), we get

Fdn+1≤Fdn+σ−τdn≤Fdn−1+2σ−τdn−τdn−1⋯≤Fd0+nσ−τdn−τdn−1−⋯−τd0.E42

Let τdpn=minτd0τd1⋯τdn for all n∈IN. From (42), we get

Fdn+1≤Fd0+nσ−τdpn.E43

From (38), we also get

FDzn+1Tzn+1≤FDz0Tz0+nσ−τdpn.E44

Now consider the sequence τdpn. We distinguish two cases.

Case 1. For each n∈IN, there is m>n such that τdpn>τdpm. Then we obtain a subsequence dpnk of dpn with τdpnk>τdpnk+1 for all k. Since dpnk→δ+, we deduce that limk→∞infτdpnk>σ. Hence Fdnk≤Fd0+nσ−τdpnk for all k. Consequently, limk→∞Fdnk=−∞ and by F2, we obtain limk→∞dpnk)=0, which contradicts that limn→∞dn>0.

Case 2. There is n0∈IN such that τdpn0>τdpm for all m>n0. Then Fdm≤Fd0+mσ−τdpn0 for all m>n0. Hence limm→∞Fdm=−∞, so limm→∞dm=0, which contradicts that limm→∞dm>0. Thus,

limn→∞dn=0.

From F3, there exists 0<r<1 such that limn→∞dnrFdn=0. By (43), we get for all n∈IN

dnrFdn−dnrFd0≤dnrnσ−τd−pn≤0.E45

Letting n→∞ in (45), we obtain limn→∞ndnr=0. This implies that there exists n1∈IN such that ndnr≤1, or, dn≤1n1/r, for all n>n1. Next, for m>n≥n1 we have

dznzm≤∑i=nm−1dzizi+1≤∑i=nm−11i1/k,

since 0<k<1, ∑i=nm−11i1/k converges. Therefore, dznzm→0 as m,n→∞. Thus, zn is a Cauchy sequence. Since X is complete, there exists z∗∈X such that zn→z∗ as n→∞. From Eqs. (44) and F2, we have limn→∞DznTzn=0. Since z→DzTz is lower semi-continuous, then

0≤DzTz≤limn→∞inf DznTzn=0.

Thus, T has a fixed point.□

In the following theorem we take CX instead of KX, then we need to take F∈ΔF∗ in Definition 3.10.

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

Proof. Let z0∈X, since Tz∈CX for every z∈X and F∈ΔF∗, the set Fσz is non-empty for any σ>0, then there exists z1∈Fσz0 and by hypothesis αz0z1≥1. Assume that z1∉Tz1, otherwise z1 is the fixed point of T. Then, since Tz1 is closed, Dz1Tz1>0, so, from (28), we have

τdz0z1+αz0z1FDz1Tz1≤FMz0z1,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 z∈X.□

Theorem 3.14 Let Xd be a complete metric space, T:X→KX be a continuous mapping and F∈ΔF. Assume that the following assertions hold:

T is generalized α∗-admissible mapping;
there exists z0∈X and z1∈Tz0 such that αz0z1≥1;
there exists τ:0∞→0∞ such that

limt→s+inf τt>0foralls≥0

and for all z∈X with HTzTy>0, there exist a function α:X×X→−∞∪0+∞ satisfying

τdzy+αzyFHTzTy≤FMzy,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:X→CX be a continuous mapping and F∈ΔF∗ 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:X→X be a continuous mapping and F∈ΔF. Assume that the following assertions hold:

T is α-admissible mapping;
there exists z0,z1∈X such that αz0z1≥1;
there exists τ:0∞→0∞ such that

limt→s+inf τt>0foralls≥0

and for all z∈X with dTzTy>0, there exist a function α:X×X→−∞∪0+∞ satisfying

τdzy+αzyFdTzTy≤Fmzy,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:X→2X is monotone increasing if Ty≼Tz for all y,z∈X, for which y≼z (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:X→2X be a closed valued mapping satisfying the following assertions for all y,z∈X with y≼z:

T is monotone increasing;
ϑHTyTzd(yz)≥0;
there exists z0∈X and z1∈Tz0 such that z0≼z1;
for a sequence zn⊂X, limn→∞zn=z and zn≼zn+1 for all n∈IN, we have zn≼z for all n∈IN.

Then FixT≠∅.

Proof. Define α,η:X×X→0∞ by

αyz=1y≼z0otherwiseηyz=12y≼z0otherwise,

then for y,z∈X with y≼z, αyz≥ηyz implies α∗TyTz=1>12=η∗TyTz and α∗TyTz=η∗TyTz=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:X→X be a self-map fulfilling the following assertions:

T is monotone increasing;
ϑdTyTzd(yz)≥0;
there exists z0∈X and z1=Tz0 such that z0≼z1;
for a sequence zn⊂X, limn→∞zn=z and zn≼zn+1 for all n∈IN, we have zn≼z for all n∈IN.

for all y,z∈X with y≼z and ϑ∈Man(IR)̂. Then FixT≠∅.

Definition 3.19 Let T:X→2X 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:0∞→σ∞, F∈ΔF such that for all z∈X, y∈Fσz with z≼y and DzTz>0 satisfying

τdzy+FDyTy≤FMzy,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:X→KX be an ordered F-τ-contraction satisfying the following assertions:

T is monotone increasing;
the map z→DzTz is lower semi-continuous;
there exists z0∈X and z1∈Tz0 such that z0≼z1;
τ satisfies

limt→s+inf τt>σforalls≥0

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:X→CX be an ordered F-τ-contraction with F∈ΔF∗ 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:X→KX be a continuous mapping and F∈ΔF. Assume that the following assertions hold:

T is monotone increasing;
there exists z0∈X and z1∈Tz0 such that z0≼z1;
there exists τ:0∞→0∞ such that

limt→s+inf τt>0foralls≥0

and for all z,y∈X with z≼y and HTzTy>0 satisfying

τdzy+FHTzTy≤FMzy,E53

where Mzy is defined in (52).

Then T has a fixed point in X.

Proof. By defining α:X×X→0∞ 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:X→CX be a continuous mapping and F∈ΔF∗ 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:X→X be a continuous mapping and F∈ΔF. Assume that the following assertions hold:

T is monotone increasing;
there exists z0,z1∈X such that z0≼z1;
there exists τ:0∞→0∞ such that

limt→s+inf τt>0foralls≥0

and for all z,y∈X with z≼y and dTzTy>0 satisfying

τdzy+FdTzTy≤Fmzy,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:X→X by

Tzr=∫bcBrszsds+gr,r∈ab.E56

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

B:bc×bc×IRn→IRn and g:IRn→IRn are continuous;
there exists a continuous function p:bc×bc→bc such that
Brsu−B(rsv)≤prs∣u−v∣,E57
for each r,s∈bc and u,v∈IRn with u≪v.
supr∈bc∫bcprsds=q≤14;
there exists z0∈X and z1∈Tz0 such that z0≼z1;
for a sequence zn⊂X, limn→∞zn=z and zn≼zn+1 for all n∈IN, we have zn≼z for all n∈IN.

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

Proof. Let X=CbcIRn and ‖z‖=maxr∈bc∣zr∣, for z∈Cab. Consider a partial order defined on X by

y,z∈CbcIRn,y≼zifandonlyifyr≪zr,forr∈bc.E58

Then X‖.‖≼ is a complete partial ordered metric space and for any increasing sequence zn in X converging to z∈X, we have znr≪zr for any r∈bc (see [36]). By using Eq. (56), conditions (2, 3) and taking ϑrs=12s−r for all y,z∈X with y≼z, we obtain

∣Tyr−Tzr∣=∫bcB(rsys)ds−∫bcB(rszs)ds≤∫bc∣Brsys−Brszs∣ds≤∫bcprs∣ys−zs∣ds≤14‖y−z‖.

This implies that

12‖y−z‖−‖Ty−Tz‖≥12‖y−z‖−14‖y−z‖=14‖y−z‖.

So ϑdTyTzdyz≥0 for all y,z∈X with y≼z. 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:Hn→Hn be an order-preserving mapping which maps Pn into Pn and Q∈Pn. Assume that there exists a positive number N for which ∑i=1mAiAi∗≺NIn and ∑i=1mAi∗γQAi≻0 such that for all X≼Y we have

dγXγY≤1NmYXe−2+dXY2dXY,E59

where

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

Then (1) has a solution in Pn.

Proof. Define T:Hn→Hn and F:IR+→IR by

TX=Q+∑i=1mAi∗γXAiE60

and Fr=lnr respectively. Then a fixed point of T is a solution of (1). Let X,Y∈Hn with X≼Y, then γX≼γy. So, for dXY>0 and τt=1t+12, we have

dTXTY=TY−TX1=trTY−TX=∑i=1mtrAiAi∗γY−γX=tr∑i=1mAiAi∗γY−γX≤∑i=1mAiAi∗γY−γX1≤∑i=1mAiAi∗NmYXe−2+Y−X12Y−X1<mYXe−2+Y−X12Y−X1,

and so,

lnTY−TX1<lnmYXe−2+Y−X12Y−X1=lnmXY−2+Y−X12Y−X1.

This implies that

1Y−X1+12+lnTY−TX1<lnmXY.

Consequently,

τdXY+FdTXTY<FmXY.

Also, from ∑i=1mAi∗γQAi≻0, we have Q≼TQ. Thus, by using Theorem 3.24, we conclude that T has a fixed point and hence (1) has a solution in Pn.□

Mammie Motiang

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