# Fixed-Point Theorem for NonlinearF-Contraction via w-Distance

Authors: Abdelkarim Kari; Mohamed Rossafi; El Miloudi Marhrani; Mohamed Aamri
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

The aim of this paper is to introduce a notion ofϕ,F-contraction defined on a metric space with w-distance. Moreover, fixed-point theorems are given in this framework. As an application, we prove the existence and uniqueness of a solution for the nonlinear Fredholm integral equations. Some illustrative examples are provided to advocate the usability of our results.


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## Body

## 1. Introduction

By a contraction on a metric spaceX,d, we understand a mapping T:X⟶X satisfying for all x,y∈X:dTx,Ty≤kdx,y, where k is a real in 0,1.In 1922, Banach proved the following theorem.Theorem 1 (see [1]).
LetX,d be a complete metric space. Let T:X⟶X be a contraction. Then,(i)
T has a unique fixed point x∈X(ii)
For everyx0∈X, the sequence xn, where xn+1=Txn, converges to x(iii)
We have the following estimate: for everyx∈X, dxn,x≤kn/1−kdx0,x1, n∈ℕAs a result of its intelligibility and profitableness, the previous theorem has become a very celebrated and popular tool in solving the existence problems in many branches of mathematical analysis.Many mathematicians extended the Banach contraction principle in two major directions, one by stating the conditions on the mappingT and second by taking the set X as more general structure.Recently, Kari et al. [2] give some fixed-point results for generalized θ−ϕ-contraction in the framework of α,η-compete rectangular b-metric spaces.In 2012, Wardowski [3] introduced the concept of F-contraction, using this concept, he proved the existence and uniqueness of a fixed point in complete metric spaces. This direction has been studied and generalized in different spaces, and various fixed-point theorems are developed [4, 5]. Cosentino and Vetro [6] presented some fixed-point results of Hardy-Rogers type for self-mappings on complete metric spaces or complete ordered metric spaces. In 2016, Piri and Kumam [7] introduced the modified generalized F-contractions, by combining the ideas of Dung and Hang [8], Piri and Kumam [9], Wardowski [3], and Wardowski and Van Dung [10], and gave some fixed-point result for these type mappings on complete metric space.In 1996, Kada et al. initiated the notion ofw-distance on a metric space; then, many authors used this concept to prove some results of fixed-point theory [11, 12].Recently, Wongyat and Sintunavarat [13] introduced a special w-distance called ceiling distance and proved some fixed point for generalized contraction mappings with respect to this distance.Later, Wardowski [14] studied a new type of contractions called nonlinear F-contraction.In this paper, we shall obtain a fixed-point theorem forϕ,F-contraction with respect to w-distance on complete metric spaces. Various examples are constructed to illustrate our results. As an application, we prove the existence and uniqueness of a solution for the nonlinear Fredholm integral equations.The paper is structured as follows:In Section2, we briefly recall some definitions and basic properties used to prove our main results.In Section3, we present our results.Section4 is devoted to the application of the result in nonlinear integral equations.


## 2. Preliminaries

Kada et al. [15] introduced the concept of w-distance on a metric space as follows:Definition 2 (see [15]).
LetX,d be a metric space. A function q:X×X⟶ℝ+ is called a w-distance on X, if it satisfies the following three conditions for all x,y,z∈X:
(W1) qx,y≤qx,z+qz,y
(W2) qx,.:X⟶ℝ+ is lower semicontinuous on for all x∈X
(W3) For each ε>0, there exists δ>0 such that qx,y≤δ and qx,z≤δ imply dy,z≤εRemark 3.
Each metric on a nonempty setX is a w-distance on X.Example 1 (see [13]).
LetX,d be a metric space. The function q:X×X⟶ℝ+ defined by qx,y=c for every x,y∈X is a w-distance on X, where c is a positive real number. But q is not a metric since qx,x=c≠0 for any x∈X.The following lemma is a useful tool for proving our results.Lemma 4 (see [15]).
LetX,d be a metric space, q be a w-distance on X, xn and yn be two sequences in X, and x,y,z∈X.(i)
Iflimn⟶+∞qxn,x=limn⟶+∞qxn,y=0 then x=y. In particular, if qz,x=qz,y=0, then x=y(ii)
Ifdxn,yn≤αn and dxn,y≤βn for all n∈ℕ, where αn and βn are sequences in 0,+∞ converging to 0, then yn converges to y(iii)
If for eachε>0, there exists Nε∈ℕ such that m>n>Nε implies qxn,xm<ε, then xn is a Cauchy sequenceDefinition 5 (see [13]).
Aw-distance q on a metric space X,d is said to be a ceiling distance of d if and only if(1)qx,y≥dx,y,for all x,y∈X.Example 2 (see [13]).
LetX=ℝ with the metric d:X×X⟶ℝ+ defined by dx,y=x−y for all x,y∈X, and let a,b≥1. Define the function q:X×X⟶ℝ+ by(2)qx,y=maxay−x,bx−y,for all x,y∈X. Then, q is a ceiling distance of d.The following definition was introduced by Wardowski.Definition 6 (see [3]).
LetF be the family of all functions F:ℝ+⟶ℝ such that(i)
F is strictly increasing(ii)
For each sequencexnn∈ℕ of positive numbers,(3)limn⟶∞xn=0,ifandonlyiflimn⟶∞Fxn=−∞(iii)
There existsk∈0,1 such that limx⟶0xkFx=0Recently, Piri and Kuman [9] extended the result of Wardowski [3] by changing the condition (iii) in Definition 6 as follows.Definition 7 (see [9]).
LetΓ be the family of all functions F:ℝ+⟶ℝ such that(i)
F is strictly increasing(ii)
For each sequencexnn∈ℕ of positive numbers,(4)limn⟶∞xn=0,ifandonlyiflimn⟶∞Fxn=−∞(iii)
F is continuousThe following definition introduced by Wardowski [14] will be used to prove our result.Definition 8 (see [14]).
LetF be the family of all functions F:ℝ+⟶ℝ and Φ be the family of all functions ϕ:0,+∞⟶0,+∞ satisfy the following conditions:(i)
F is strictly increasing(ii)
For each sequencexnn∈ℕ of positive numbers,(5)limn⟶∞xn=0,ifandonlyiflimn⟶∞Fxn=−∞(iii)
liminfα⟶s+ϕs>0 for all s>0(iv)
There existsk∈0,1 such that(6)limx⟶0+xkFx=0By replacing the condition (iii) in Definition8, we introduce a new class of F-contraction.Definition 9.
LetI be the family of all functions F:ℝ+⟶ℝ and Φ be the family of all functions ϕ:0,+∞⟶0,+∞ satisfy the following conditions:(i)
F is strictly increasing(ii)
liminfα⟶s+ϕs>0 for all s>0(iii)
For each sequencexnn∈ℕ of positive numbers,(7)limn⟶∞xn=0,ifandonlyiflimn⟶∞Fxn=−∞(iv)
F is continuousExample 3.
(i)
LetF1t=−1/t+t, F2t=−1/et−1+et. Then, F1,F2∈I(ii)
Letϕ1t=1/t+1, ϕ2t=1/t+1, and ϕ3t=1/t2+1. Then, ϕ1,ϕ2 and ϕ3∈ΦDefinition 10 (see [14]).
LetX,d be a metric space. A mapping T:X⟶X is called a ϕ,F-contraction on X,d, if there exist F∈F and ϕ∈Φ such that(8)FdTx,Ty+ϕdx,y≤Fdx,y,for all x,y∈X for which Tx≠Ty.


## 3. Main Result

In this paper, using the idea introduced by Wongyat and Sintunavarat [13], we presented the concept of ϕ,F-contraction on a complete metric space with w-distance.Definition 11.
Letq be a w-distance on a metric space X,d. A mapping T:X⟶X is said to be a w-generalized ϕ,F-contraction of type F on X,d if there exist F∈F and ϕ∈Φ such that(9)FqTx,Ty+ϕqx,y≤Fqx,y,for all x,y∈X for which Tx≠Ty.Theorem 12.
LetX,d be a complete metric space and q:X×X⟶0,+∞ be a w-distance on X and a ceiling distance of d, supposing that T:X⟶X is a w-generalized ϕ,F-contraction of type F. Then, T has a unique fixed point on X.Proof.
Letx0∈X be an arbitrary point in X; define a sequence xnn∈ℕ by(10)xn+1=Txn=Tn+1x0,for all n∈ℕ. If there exists n0∈ℕ such that dxn0,xn0+1=0, then the proof is finished.
We can suppose thatdxn,xn+1>0 for all n∈ℕ.
Sinceq is a ceiling distance of d, we obtain qxn,xn+1>0 for all n∈ℕ.
Substitutingx=xn−1 and y=xn, from (9), for all n∈ℕ, we have(11)Fqxn,xn+1+ϕqxn−1,xn≤Fqxn−1,xn,∀n∈ℕ.
Imply that(12)Fqxn,xn+1≤Fqxn−1,xn−ϕqxn−1,xn<Fqxn−1,xn.
SinceF is increasing, then qxn,xn+1<qxn−1,xn. Therefore, qxn,xn+1n∈ℕ is monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists α≥0 such that(13)limn⟶∞qxn,xn+1=α.
The inequality (11) implies(14)Fqxn,xn+1≤Fqxn−1,xn−ϕqxn−1,xn≤Fqxn−2,xn−1−ϕqxn−1,xn−ϕqxn−2,xn−1≤⋯≤Fqx0,x1−∑i=0nϕqxi,xi+1.
Sinceliminfα⟶s+ϕα>0, we have liminfn⟶∞ϕqxn−1,xn>0; then from the definition of the limit, there exists n0∈ℕ and A>0 such that for all n≥n0, ϕqxn−1,xn>A, hence(15)Fqxn,xn+1≤Fqx0,x1−∑i=0n0−1ϕqxi,xi+1−∑i=n0−1nϕqxi,xi+1≤Fqx0,x1−∑i=n0−1nA=Fqx0,x1−n−n0A,for all n≥n0. Taking the limit as n⟶∞ in the above inequality, we get(16)limn⟶∞Fqxn,xn+1≤limn⟶∞Fqx0,x1−n−n0A,that is, limn⟶∞Fqxn,xn+1=−∞; then, from the condition (ii) of Definition 8, we conclude that(17)limn⟶∞qxn,xn+1=0.
Next, we shall prove thatxnn∈ℕ is a Cauchy sequence, i.e., limn,m⟶∞dxn,xm=0, for all n,m∈ℕ.
The condition (iv) of Definition8 implies that there exists k∈0,1 such that(18)limn⟶∞qxn,xn+1kqxn,xn+1=0.
From the inequality (12), we get(19)qxn,xn+1kFqxn,xn+1≤qxn,xn+1kFqx0,x1−n−n0A.
Hence,(20)qxn,xn+1kFqxn,xn+1−qxn,xn+1kFqx0,x1≤−qxn,xn+1kn−n0A≤0.
Taking limitn⟶∞ in the above inequality, we conclude that(21)limn⟶∞qxn,xn+1kn−n0A=0.
Then, there existsn1∈ℕ, such that for all n≥n1,(22)qxn,xn+1≤1n−n0Ak.
Therefore, form>n≥maxn0,n1, we have(23)qxn,xm≤qxn,xn+1+qxm+1,xm+2+⋯+qxm−1,xm=∑i=1m−1qxi,xi+1≤∑i=1∞qxi,xi+1≤∑i=1∞1i−n0Ak.
Since0<k<1, then(24)limn,m⟶∞qxn,xm=0.
By Lemma4, we can conclude that xn is a Cauchy sequence in X. By the completeness of X,d, there exists z∈X such that(25)limn⟶∞dxn,z=0.
Now, we show thatdTz,z=0; arguing by contradiction, we assume that(26)dTz,z>0⇒qTz,z>0.
From (24), for each r>0, there is nr∈ℕ such that(27)qxnr,xn<1r,for all nr>r. Since qxnr,. is lower semicontinuous and xn⟶x as n⟶∞, we get(28)qxnr,z≤liminfx⟶∞qxnr,xn≤1r,implying that(29)liminfn⟶∞qxnr,z=0.
Now, by triangular inequality, we get,(30)qTxnr,Tz≤qTxnr,z+qz,Tz,(31)qz,Tz≤qz,xnr+qxnr,Tz.
By lettingn⟶∞ in inequality (30) and (31), we obtain(32)qz,Tz≤limn⟶∞qTxn,Tz≤qz,Tz.
Therefore,(33)limn⟶∞qTxnr,Tz=qz,Tz.
LetA=dz,Tz>0, from the definition of the limit, there exists n2∈ℕ such that(34)qTxnr,Tz−qz,Tz≤A,∀n≥n2,which implies that(35)qTxnr,Tz>0,∀n≥n2.
Applying (9) with x=z and y=xnr, we have(36)FqTz,Txnr+ϕqz,xnr≤Fqz,xnr,which implies that(37)FqTz,Txnk≤Fqz,xnr.
SinceF is increasing, we get(38)qTz,Txnr≤qz,xnr.
By lettingn⟶∞ in the above inequality, we obtain(39)limn⟶∞qTxnr,Tz=qz,Tz=0.
Which is a contradiction, thendz,Tz=qz,Tz=0, so Tz=z.
For uniqueness, now, suppose thatz,u∈X are two fixed points of T such that u≠z. Therefore, we have(40)qTz,Tu=qz,u>0.
Applying (9) with x=z and y=u, we have(41)FqTu,Tz=Fqz,u+ϕqz,u≤Fqz,u,implying(42)qu,z<qu,z,which is a contradiction. Therefore, u=z.Example 4.
LetX=0,+∞ with the metric d:X×X⟶0,+∞ defined by(43)dx,y=x−y,for all x,y∈X. Define a mapping T:X⟶X by(44)Tx=x3.
Suppose thatFt=lnt and ϕt=1/t+1, clearly F∈F and ϕ∈Φ. Also, we define a w-distance q:X×X⟶0,+∞ by(45)qx,y=maxx,y,for all x,y∈X. It is easy to see that q is a ceiling distance of d. Now, we will show that T satisfies the condition (9).Case 1.
Ifx≥y, then qx,y=x, qTx,Ty=x/3, and ϕqx,y=1/x+1. Thus,(46)Fqx,y=lnxFqTx,Ty=lnx3.
We prove thatT is a ϕ,F contraction mapping of type F. Indeed,(47)FqTx,Ty+ϕqx,y−Fqx,y=lnx3+1x+1−lnx=1x+1−ln3≤0.
Therefore,(48)FqTx,Ty+ϕqx,y≤Fqx,y.Case 2.
Ifx<y, then qx,y=y, qTx,Ty=y/3, and ϕqx,y=1/y+1. Thus,(49)Fqx,y=lnyFqTx,Ty=lny3.
Therefore,(50)FqTx,Ty+ϕqx,y≤Fqx,y.
Hence, 0 is the unique fixed point ofT.Example 5.
Consider the sequenceSnn∈ℕ∗ defined as follows:(51)S1=1×2,S2=1×2+2×3,⋯,Sn=1×2+2×3+⋯+nn+1=nn+1n+23.
Let the metricd:X×X⟶0,+∞ defined by(52)dx,y=x−y,for all x,y∈X.
Define a mappingT:X⟶X by(53)TSn=S1ifn=1,Sn−1ifn≥2.
Clearly, the Banach contraction is not satisfied. In fact, we can check easily that(54)limn⟶∞dTSn,TS1dSn,S1=limn⟶∞nn−1n+1−6nn+1n+2−6=1.
Suppose thatFt=lnt and ϕt=1/t+1, clearly F∈F and ϕ∈Φ. Also, we define a w-distance q:X×X⟶0,+∞ by(55)qx,y=maxx,y,for all x,y∈X. It is easy to see that q is a ceiling distance of d.
Now, we will show thatT satisfies the condition (9).Case 1.
n=1 and m≥2. In this case, we have(56)qSn,S1=nn+1n+23,qTSn,TS1=nn−1n+23,ϕqSn,S1=3nn+1n+2+3.
Thus,(57)FqSn,S1=lnnn+1n+23,FqTSn,TS1=lnnn−1n+13.
On the other hand,(58)FqTSn,TS1+ϕdSn,S1−FqSn,S1=lnnn−1n+13−lnnn+1n+23+3n.n+1n+2+3=lnn−1n+2+3nn+1n+2+3≤0,foralln≥2.
Therefore,(59)FqTSn,TS1+ϕqSn,S1≤FqSn,S1.Case 2.
m>n>1. In this case, we have(60)qSn,Sm=mm+13,qTSn,TSm=mm−13,ϕqSn,Sm=3m2+m+3.
Thus,(61)FqSm,S1=lnmm+1m+23,FqTSn,TS1=lnmm−1m+13.
On the other hand,(62)FqTSn,TSm+ϕqSn,Sn−FqSn,Sn=lnmm−1m+13−lnmm+1m+23+3mm+1m+2+3=lnm−1m+2+3nm+1m+2+3≤0,foralln≥2.
Therefore,(63)FqTSn,TSm+ϕqSn,Sm≤FqSn,Sm.
Thus, the inequality (9) that is satisfied implies that T has a unique fixed point. In this example, S1 is the unique fixed point of T.Definition 13.
Letq be a w-distance on a metric space X,d. A mapping T:X⟶X is said to be a w-generalized ϕ,F-contraction of type I on X,d if there exist F∈I and ϕ∈Φ such that(64)FqTx,Ty+ϕqx,y≤Fqx,y,for all x,y∈X for which Tx≠Ty.Theorem 14.
LetX,d be a complete metric space and q:X×X⟶0,+∞ be a w-distance on X and a ceiling distance of d. Suppose that T:X⟶X is a ϕ,F-contraction of type I. Then, T has a unique fixed point on X.Proof.
As in the proof of Theorem12, we can conclude that(65)limn⟶∞dxn,xn+1=0.
Next, we show thatxn is a Cauchy sequence, i.e.,(66)limn,m⟶∞dxm,xn=0.
Now, we claim thatlimn,m⟶∞qxm,xn=0. Arguing by contradiction, we assume that there exists ε>0 we can find and sequences mkk and nkk of positive integers such that for all positive integers, nk>mk>k,(67)qxmk,xnk≥ε,(68)qxmk,xnk−1<ε.
Again by triangular inequality and using (65), (67), and (68), we get(69)ε≤qxmk,xnk≤qxmk,xnk−1+qxnk−1,xmk<ε+qxnk−1,xmk.
So,(70)limk⟶∞qxmk,xnk=ε.
Again by the triangular inequality, for alln∈ℕ, we have the following two inequalities(71)qxmk+1,xnk+1≤qxmk+1,xmk+qxmk,xnk+qxnk,xnk+1,qxmk,xnk≤qxmk,xmk+1+qxmk+1,xnk+1+qxnk+1,xnk.
Lettingk⟶∞ in the above inequalities, using (65) and (70), we obtain(72)limk⟶∞dxmk+1,xnk+1=ε.
Hence, from the definition of the limit, there existsn0∈ℕ such that(73)dxmk+1,xnk+1≥ε,foralln≥n0.
Applying (64) with x=xmk and y=xnk, we obtain(74)Fqxmk+1,xnk+1+ϕqxmk,xnk≤Fqxmk,xnk.
Lettingk⟶∞ the above inequality, we obtain(75)limk⟶∞Fqxmk+1,xnk+1≤limk⟶∞Fqxmk,xnk−limk⟶∞ϕqxmk,xnk=limk⟶∞Fqxmk,xnk−liminfk⟶∞ϕqxmk,xnk.
SinceF is a continuous and liminfk⟶∞ϕqxmk,xnk>0, we conclude that(76)ε<ε,which is a contradiction. Then,(77)limn,m⟶∞qxm,xn=0.
By the condition (iii) of Lemma4, we can conclude that xn is a Cauchy sequence. Since X,d is a complete metric space, there exists u∈X such that xn⟶x as n⟶∞.
As in the proof of Theorem12, we conclude that for each l>0, there exists n2 such that limn⟶∞qTxnl,Tu=qu,Tu, limn⟶∞qxnl,u=0, and qTxnl,Tu>0 for all nl≥n2.
Now applying (64) with x=xnl and y=u, we get(78)FqTxnl,Tu+ϕqxnl,u≤Fqxnl,u,
which implies that(79)limn⟶∞FqTxnl,Tu≤limn⟶∞Fqxnl,u.
Therefore,limn⟶+∞FqTxnl,Tu=0. Hence, qu,Tu=0, so Tu=u.
Following the proof of Theorem12, we know that u is a unique fixed point of T. This complete the proof.Example 6.
LetX=0,+∞ with the metric d:X×X⟶0,+∞ defined by(80)dx,y=x−y,for all x,y∈X. Define a mapping T:X⟶X by(81)Tx=x2.
Suppose thatFt=t−1/t and ϕt=1/t+1, clearly F∈I and ϕ∈Φ. Also, we define a w-distance q:X×X⟶0,+∞ by(82)qx,y=maxx,y,for all x,y∈X. It is easy to see that q is a ceiling distance of d. Now, we will show that T satisfies the condition (64).
We prove thatT is a ϕ,F-contraction mapping of type I.Case 1.
Ifx≥y, then qx,y=x, qTx,Ty=x/2−2/x, and ϕqx,y=1/x+1. Thus,(83)Fqx,y=x−1x,FqTx,Ty=x2−2x.
On the other hand,(84)FqTx,Ty+ϕqx,y−Fqx,y=x2−2x+1x+1−x+1x≤0.
Therefore,(85)FqTx,Ty+ϕqx,y≤Fqx,y.Case 2.
Ifx<y, then qx,y=y, qTx,Ty=y/2−2/y, and ϕqx,y=1/y+1. Thus,(86)Fqx,y=y−1y,FqTx,Ty=y2−2y.
On the other hand,(87)FqTx,Ty+ϕqx,y−Fqx,y=y2−2y+1y+1−y+1y=−y2+22y+1y+1≤0≤0.
Therefore,(88)FqTx,Ty+ϕqx,y≤Fqx,y.
Hence, 0 is the unique fixed point ofT.Example 7.
LetX be the set defined by(89)X=λn:n∈ℕ∗,where(90)λn=nn+12.
Let the metricd:X×X⟶0,+∞ defined by(91)dx,y=x−y,for all x,y∈X. Define a mapping T:X⟶X by(92)Tλn=1ifn=1,nn−12ifn≥2.
Clearly, the Banach contraction is not satisfied. In fact, we can check easily that(93)limn⟶∞dTλn,Tλ1dλn,λ1=limn⟶∞n2−n−2n2+n−2=1.
Suppose thatFt=−1/t+t and ϕt=1/t+1, clearly F∈I and ϕ∈Φ. Also, we define a w-distance q:X×X⟶0,+∞ by(94)qx,y=maxx,y,for all x,y∈X. It is easy to see that q is a ceiling distance of d. Now, we will show that T satisfies the condition (64).Case 1.
n=1 and m≥2. In this case, we have(95)qλn,λ1=nn+12,qTλn,Tλ1=nn−12,ϕqλn,λ1=1nn+1/2+1.
Thus,(96)Fqλn,λ1=2−n2−n+n2+n2,FqTλn,Tλ1=2−n2+n+n2−n2,
On the other hand,(97)FqTλn,Tλ1+ϕqλn,λ1−Fqλn,λ1=−4nn−1n2+n−n+2n2+n+2≤0.
Therefore,(98)FqTλn,Tλ1+ϕλn,λ1≤Fqλn,λ1.Case 2.
m>n>1. In this case, we have(99)qλn,λm=mm+12,qTλn,Tλm=mm−12,ϕqλn,λm=2m2+m+2.
Thus,(100)Fqλn,λm=2−m2−m+m2+m2,FqTλn,Tλm=2−m2+m+m2−m2,
On the other hand,(101)FqTλn,Tλm+ϕλn,λm−Fqλn,λm=−4mm2+mm2−m−m+2m2+m+2≤0.
Therefore,(102)FqTλn,Tλm+ϕλn,λm≤Fqλn,λm.
Thus, the inequality (64) that is satisfied implies that T has a unique fixed point. In this example, λ1 is the unique fixed point of T.Takingq=d in Theorems 12 and 14, we obtain the following result.Corollary 15.
LetX,d be a complete metric space and T:X⟶X be a ϕ,F-contraction of type F. Then, T has a unique fixed point.Corollary 16.
LetX,d be a complete metric space and T:X⟶Xbe a ϕ,F-contraction of type I. Then, T has a unique fixed point.


## 4. Application to Nonlinear Integral Equations

In this section, we endeavor to apply Theorems12 and 14 to prove the existence and uniqueness of the integral equation of Fredholm type:(103)xt=λ∫abKt,s,xsds,where a,b∈ℝ, x∈Ca,b,ℝ, and K:a,b2×ℝ⟶ℝ is a given continuous function.Theorem 17.
Consider the Fredholm integral equation (103) and assume that the kernel function K satisfies the condition Kt,s,xs+Kt,s,ys≤e−1/xt+yt+1xt+yt for all t,s∈a,b and x,y∈ℝ. Then, the equation (4.1) has a unique solution x∈Ca,b for some constant λ depending on the constants a,b.Proof.
LetX=Ca,b and T:X⟶X defined by(104)Txt=λ∫abKt,s,xsds,for all x∈X. Clearly, X with the metric d:X×X⟶0,+∞ given by(105)dx,y=supt∈a,bxt−yt,for all x,y∈X, is a complete metric space. Next, define the function q:X×X⟶0,+∞ by(106)qx,y=supt∈a,bxt∣+∣yt,for all x,y∈X.
Clearly,q is a w-distance on X and a ceiling distance of d. We will find the condition on λ under which the operator has a unique fixed point which will be the solution of the integral equation (103). Assume that x,y∈X and t,s∈a,b. Then, we get(107)Txt+Tyt=λ∫abKt,s,xsds+∫abKt,s,ysds≤λ∫abKt,s,xsds+λ∫abKt,s,ysds≤λ∫abe−1/xs+ys+1xs+ysds,
which implies that(108)supt∈a,bTxt+Tyt=supt∈a,bλ∫abKt,xsds+∫abKt,ysds≤supt∈a,b≤λ∫abKt,s,xsds+λ∫abKt,s,ysds≤λ∫abe−sups∈a,b1/xs+ys+1sups∈a,bxs+ysds.
Since by the definition of thew-distance on X and a ceiling distance of d, we have qTx,Ty>0 and qx,y>0 for any x≠y, then we can take natural logarithm sides and get(109)lnqTx,Ty≤lnλb−ae−1/qx,yqx,y=−1qx,y+1+lnλb−aqx,y,provided that λb−a≤1, which implies that(110)lnqTx,Ty≤−1qx,y+1+lnqx,y.
Hence,(111)FqTx,Ty+ϕqx,y≤Fqx,y,for all x,y∈X. It follows that T satisfies the condition (9). Therefore, there exists a unique solution of the nonlinear Fredholm inequality (103).Example 8.
Leta,b=0,e2. Consider the equation(112)xt=∫0e2xs1+xs2ds.
Here,Ks,t,xs=xs/11+xs2 and we have(113)Ks,t,xs+Ks,t,ys=xs1+xs2+ys1+ys2≤xs+xs≤e−1/xs+ys+1xs+xs.
Then, the condition (9) holds. From Theorem 17, the nonlinear integral equation (103) has a unique solution for λ≤1/e2.
By using direct computation, let(114)β=∫0e2xs1+xs2ds.
Then, we havext=λ and hence,(115)β=∫0e2βλ1+βλ2ds=e2.βλ1+βλ2,which implies that(116)λ2β3−e2λ−1β=0⇔β1+λ2β2−e2λ=0.
Therefore,(117)β=0,λ2β2+1−e2λ=0,we obtain that this equation has a unique solution when λ<1/b−a=1/e2.



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