# Solving a Nonlinear Fredholm Integral Equation via an Orthogonal Metric

Authors: Arul Joseph Gnanaprakasam; Gunaseelan Mani; Vahid Parvaneh; Hassen Aydi
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this paper, we prove fixed point theorems using orthogonal triangularα-admissibility on orthogonal complete metric spaces. Some of the well-known outcomes in the literature are generalized and expanded by the obtained results. An instance to help our outcome is being presented.


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## Body

## 1. Introduction

One of the most important results of mathematical analysis is the famous fixed point result, called the Banach contraction principle (BCP). In several branches of mathematics, it is the most commonly used fixed point result, and it is generalized in many different directions. The substitution of the metric space by other generalized metric spaces is one natural way of reinforcing the BCP. As a generalization of the BCP, Wardowski [1] gave a fixed point result in the setting of complete metric spaces. In other branches of mathematics, the notion of an orthogonal set has many applications and has several kinds of orthogonality. Gordji et al. [2] have imported the current concept of orthogonality on metric spaces and established some fixed point results equipped with the new orthogonality. Furthermore, they used these results to ensure the presence and uniqueness of the solution of a first-ordinary differential equation, while the BCP cannot be applied to this problem. In generalized orthogonal metric spaces, Eshaghi Gordji and Habibi [3] continued in this direction and gave further fixed point theorems. The new definition of orthogonal F-contraction mappings was introduced by Sawangsup et al. [4], and some related fixed point theorems on orthogonal-complete metric spaces have been proved. Many authors have investigated orthogonal contractive form mappings, and significant results have been obtained. For more details, see the works of Eshaghi and Habibi [5], Gungor and Turkoglu [6], Yamaod and Sintunavarat [7], Javed et al. [8], Sawangsup and Sintunavarat [9], Senapati et al. [10], Gunaseelan et al. [11], Beg et al. [12], Uddin et al. [13], Ali et al. [14], etc. In this paper, we prove fixed point theorems using orthogonally triangular α-admissibility on orthogonal metric spaces. At the end, an application is presented.


## 2. Preliminaries

The goal of this section is to immediate some concepts and results used in the article. In this article,℧, ℝ+, and ℕ denote, respectively, the nonempty set, the positive real numbers, and the set of positive integers.In 2013, Hussain et al. [15] introduced the concepts of α-admissible mappings and proved some fixed point theorems.On the other hand, the definition of an orthogonal set (or,O-set), some examples, and some premises of orthogonal sets were introduced by Gordji et al. [2], as follows.Definition 1 (see [2]).
Let℧≠ϕ and ⊥⊆℧×℧ be a binary relation. If ⊥ satisfies the consecutive condition:
(1)∃R0∈℧:∀R∈℧,R⊥R0or∀R∈℧,R0⊥R,then, it is said to be an orthogonal set (briefly O-set). We indicate this O-set by ℧,⊥.Example 1 (see [2]).
Let℧=0,∞ and define R⊥Y if RY∈R,Y. Then, by setting R0=0 or R0=1, ℧,⊥ is an O-set.Definition 2 (see [2]).
The triplet℧,⊥,φ is said to be an O-metric space if ℧,⊥ is an O-set and ℧,φ is a metric space.Definition 3 (see [2]).
Let℧,⊥,φ be an O-metric space. Then, if every Cauchy O-sequence is convergent, ℧ is said to be an orthogonal complete (briefly, O-complete).Definition 4 (see [2]).
Let℧,⊥ be an O-set. A mapping G:℧⟶℧ is said to be ⊥-preserving if GR⊥GY, whenever R⊥Y.Definition 5 (see [16]).
Let℧,⊥ be an O-set and φ be a metric on ℧, G:℧⟶℧ and α:X×X⟶0,∞ be two mappings. We say that G is orthogonally α-admissible whenever R⊥Y and αR,Y≥1 imply that αGR,GY≥1.Definition 6.
Let℧,⊥ be an O-set and φ be a metric on ℧. Given G:℧⟶℧ and α:R×R⟶−∞,∞. We say that G is an orthogonally triangular α-admissible mapping if
(i)
R⊥Y and αR,Y≥1 imply that αGR,GY≥1(ii)
R⊥Z, αR,Z≥1 and Z⊥Y, αZ,Y≥1 imply that R⊥Y(2)αGR,GY≥1.


## 3. Main Results

Inspired by theα-admissibility and fixed point theorems proved by Hussain et al. [15], we prove some fixed point theorems in an orthogonal complete metric space.Theorem 7.
Let℧,⊥,φ be an O-complete metric space. Given an orthogonal element R0. The mappings G:℧⟶℧ and α:℧×℧⟶0,∞ are such that
(3)∀R,Y∈℧withR⊥Y,φGR,GY>0⇒φGR,GY+lαR,GRαY,GY≤ℏφR,YφR,Y+l,
wherel≥1. Suppose there is ℏ:0,∞⟶0,1 so that for each bounded positive sequence Ωn, ℏΩn⟶1 implies Ωn⟶0. Suppose that
(1)
G is ⊥-preserving(2)
G is orthogonally triangular α-admissible(3)
There existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1(4)
EitherG is orthogonally continuous, or if Rn is a sequence in ℧ such that Rn⟶R,αRn,Rn+1≥1 for all n, then Rn⊥R and αR,GR≥1
Then,G has a fixed point.Proof.
By condition (3), there existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1. Let
(4)R1≔GR0,R2≔GR1=G2R0⋯⋯,Rn+1≔GRn=Gn+1R0,for all n≥0. Since G is ⊥-preserving, then, Rn is an O-sequence in ℧. Condition (2) implies that αRn,Rn+1≥1 for all n≥0. If Rn=Rn+1 for some n≥0, then, Rn is a fixed point of G. Assume that Rn≠Rn+1, ∀n≥0. Since G is α-admissible and αR0,GR0≥1, we deduce that αR1,R2=αGR0,G2R0≥1. By continuing this process, we get αRn,GRn≥1 for all n≥0. By the inequality (3), we obtain
(5)φGRn−1,GRn+l≤φGRn−1,GRn+lαRn−1,GRn−1αRn,GRn≤ℏφRn−1,RnφRn−1,Rn+l,then
(6)φRn,Rn+1≤ℏφRn−1,RnφRn−1,Rn.
This implies thatφRn,Rn+1≤φRn−1,Rn. It follows that the sequence defined by φRn,Rn+1 is decreasing, hence, there is d≥0 so that limn⟶∞φRn,Rn+1=d. We claim that d=0. From (6), we have
(7)φRn,Rn+1φRn−1,Rn≤ℏφRn−1,Rn≤1,which implies limn⟶∞ℏφRn−1,Rn=1. We deduce that
(8)limn⟶∞φRn,Rn+1=0.
Now, we claim thatRn is a Cauchy. Assume that there are ε>0 and subsequences mj and nj so that
(9)nj>mj>j,φRnj,Rmj≥εandφRnj,Rmj−1<ε.
Consider(10)ε≤φRnj,Rmj≤φRnj,Rmj−1+φRmj−1,Rmj<ε+φRmj−1,Rmj,j∈ℕ.
Taking the limit asj⟶+∞ and using (8), we get
(11)limj⟶+∞φRnj,Rmj=ε.
Again,(12)φRnj,Rmj≤φRmj,Rmj+1+φRmj+1,Rnj+1+φRnj+1,Rnj,(13)φRnj+1,Rmj+1≤φRmj,Rmj+1+φRmj,Rnj+φRnj+1,Rnj.
Lettingj⟶+∞, together with (8) and (11), we deduce that
(14)limj⟶+∞φRnj+1,Rmj+1=ε.
Since there existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1, using condition (2), we derive that R1⊥R2,αR1,R2=αGR0,G2R0≥1. By continuing this process, we get
(15)Rn⊥Rn+1,αRn,Rn+1≥1,for all n≥0. Suppose that m<n. We have
(16)Rm⊥Rm+1,αRm,Rm+1≥1,Rm+1⊥Rm+2,αRm+1,Rm+2≥1.
Recall thatG is orthogonally triangular α-admissible, so we have
(17)Rm⊥Rm+2,αRm,Rm+2≥1.
Again,(18)Rm⊥Rm+2,αRm,Rm+2≥1,Rm+2⊥Rm+3,αRm+2,Rm+3≥1.
G is orthogonally triangular α-admissible, so we have
(19)Rm⊥Rm+3,αRm,Rm+3≥1.
By continuing this process, we getRm⊥Rn, αRm,Rn≥1 and so Rnj⊥RmjαRnj,Rmj≥1.
From (3), (11), and (14), we have
(20)φRnj+1,Rmj+1+l≤φRnj+1,Rmj+1+lαRnj,GRnjαRmj,GRmj=φGRnj,GRmj+lαRnj,GRnjαRmj,GRmj≤ℏφRnj,RmjφRnj,Rmj+l.
Hence,(21)φRnj+1,Rmj+1φRnj,Rmj≤ℏφRnj,Rmj≤1.
Lettingj⟶+∞ in the above inequality, we get
(22)limj⟶+∞ℏφRnj,Rmj=1.
That is, at the limitj⟶+∞, φRnj,Rmj=0<ε, which is a contradiction. Hence, Rn is a Cauchy sequence in the complete metric space ℧, hence, there is Z∈℧ so that Rn⟶Z. First, we suppose that G is orthogonally continuous, then, we have
(23)GZ=limn⟶∞GRn=limn⟶∞Rn+1=Z.
So,Z is a fixed point of G. Assume the condition (4) holds. Then, Rn⊥Z and αZ,GZ≥1. Now, by (3), we have
(24)φGZ,Rn+1+l≤φGZ,GRn+lαZ,GZαRn,GRn≤ℏφZ,RnφZ,Rn+l.
That is,φGZ,Rn+1≤ℏφZ,RnφZ,Rn, and so we get
(25)φGZ,Z≤φGZ,Rn+1+φZ,Rn+1≤ℏφZ,RnφZ,Rn+φZ,Rn+1.
Takingn⟶∞, we find φGZ,Z=0, that is, Z=GZ.☐Theorem 8.
Let℧,⊥,φ be an O-complete metric space. Given an orthogonal element R0. The mappings G:℧⟶℧ and α:℧×℧⟶0,∞ are such that
(26)∀R,Y∈℧withR⊥YφGR,GY>0⇒αR,GRαY,GY+1φGR,GY≤2ℏφR,YφR,Y,where l≥1. Assume that there exists a function ℏ:0,∞⟶0,1 such that, for any bounded sequence Ωn of positive reals, ℏΩn⟶1 implies Ωn⟶0, satisfying the conditions:
(1)
G is ⊥-preserving(2)
G is orthogonally triangular α-admissible(3)
There existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1(4)
EitherG is orthogonally continuous, or if Rn is a sequence in ℧ such that Rn⟶R,αRn,Rn+1≥1 for all n, then Rn⊥R and αR,GR≥1
Then,G has a fixed point.Proof.
By condition (3), there existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1. Let
(27)R1≔GR0,R2≔GR1=G2R0⋯⋯,Rn+1≔GRn=Gn+1R0,for all n≥0. Since G is ⊥-preserving, Rn is an O-sequence in ℧. Condition (2) implies that αRn,Rn+1≥1 for all n≥0. If Rn=Rn+1 for some n≥0, then Rn is a fixed point of G. Assume that Rn≠Rn+1, ∀n≥0. By Theorem 7, we conclude that αRn,GRn≥1 for all n≥0. From (26), we obtain
(28)2φGRn−1,GRn≤αRn−1,GRn−1αRn,GRn+1φGRn−1,GRn≤2ℏφRn−1,RnφRn−1,Rn,which yields that
(29)φRn,Rn+1≤ℏφRn−1,RnφRn−1,Rn.
So, we conclude thatφRn,Rn+1≤φRn−1,Rn. It follows that the sequence dn≔φRn,Rn+1 is decreasing, so there is d≥0 so that dn⟶d as n⟶∞. We claim that d=0. Assume that d>0. Considering (29), we have
(30)φRn,Rn+1φRn−1,Rn≤ℏφRn−1,Rn≤1,which implies limn⟶∞ℏφRn−1,Rn=1.
Therefore,d=limn⟶∞dn=limn⟶∞φRn−1,Rn=0. It is a contradiction. Thus,
(31)limn⟶∞φRn,Rn+1=0.
Now, we claim thatRn is a Cauchy sequence. Assume there are ε>0 and sequences mj and nj so that
(32)nj>mj>j,φRnj,Rmj≥εandφRnj,Rmj−1<ε.
As in the proof of Theorem7, one writes
(33)limj⟶+∞φRnj,Rmj=ε,(34)limj⟶+∞φRnj+1,Rmj+1=ε.
Since there existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1, using condition (2), we derive that
(35)R1⊥R2,αR1,R2=αGR0,G2R0≥1.
By continuing this process, we get(36)Rn⊥Rn+1,αRn,Rn+1≥1,for all n≥0. Suppose that m<n. Recall that
(37)Rm⊥Rm+1,αRm,Rm+1≥1,Rm+1⊥Rm+2,αRm+1,Rm+2≥1.
SinceG is orthogonally triangular α-admissible, we have
(38)Rm⊥Rm+2,αRm,Rm+2≥1.
Again,(39)Rm⊥Rm+2,αRm,Rm+2≥1,Rm+2⊥Rm+3,αRm+2,Rm+3≥1.
G is orthogonally triangular α-admissible, so
(40)Rm⊥Rm+3,αRm,Rm+3≥1.
By continuing this process, we getRm⊥Rn, αRm,Rn≥1 and so Rnj⊥RmjαRnj,Rmj≥1.
Now, from (26), (33), and (34), we have
(41)2φRnj+1,Rmj+1≤αRnj,GRnjαRmj,GRmj+1φRnj+1,Rmj+1=αRnj,GRnjαRmj,GRmj+1φGRnj,GRmj≤2ℏφRnj,RmjφRnj,Rmj.
Hence,(42)φRnj+1,Rmj+1φRnj,Rmj≤ℏφRnj,Rmj≤1.
Lettingj⟶+∞, we get
(43)limj⟶+∞ℏφRnj,Rmj=1.
That is,limj⟶+∞φRnj,Rmj=0<ε. It is a contradiction, so Rn is a Cauchy sequence, hence, there is Z∈℧ so that Rn⟶Z. First, assume that G is orthogonally continuous, then
(44)GZ=limn⟶∞GRn=limn⟶∞GRn+1=Z.
So,Z is a fixed point of G. If (4) is verified, so Rn⊥Z and αZ,GZ≥1. Now, by (26), we have
(45)2φGZ,Rn+1≤αZ,GZαRn,GRn+1φGZ,GRn≤2ℏφZ,RnφZ,Rn.
That is,φGZ,Rn+1≤ℏφZ,RnφZ,Rn, and so we get
(46)φGZ,Z≤φGZ,Rn+1+φZ,Rn+1≤ℏφZ,RnφZ,Rn+φZ,Rn+1.
At the limitn⟶∞, one gets φGZ,Z=0, that is, Z=GZ.☐Theorem 9.
Let℧,⊥,φ be an O-complete metric space. Given an orthogonal element R0. The mappings G:℧⟶℧ and α:℧×℧⟶0,∞ are such that
(47)∀R,Y∈℧withR⊥YφGR,GY>0⇒αR,GRαY,GYφGR,GY≤ℏφR,YφR,Y,where l≥1. Assume that there exists a function ℏ:0,∞⟶0,1 such that, for any bounded sequence Ωn of positive reals, ℏΩn⟶1 implies Ωn⟶0, satisfying the conditions:
(1)
G is ⊥-preserving(2)
G is orthogonally triangular α-admissible(3)
There existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1(4)
EitherG is orthogonally continuous, or if Rn is a sequence in ℧ such that Rn⟶R,αRn,Rn+1≥1 for all n, then Rn⊥R and αR,GR≥1
Then,G has a fixed point.Proof.
By condition (3), there existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1. Let
(48)R1≔GR0,R2≔GR1=G2R0,⋯,Rn+1≔GRn=Gn+1R0,for all n≥0. Since G is ⊥-preserving, Rn is an O-sequence in ℧. Condition (2) yields that αRn,Rn+1≥1 for all n≥0. If Rn=Rn+1 for any n≥0, then Rn is a fixed point of G. Assume that Rn≠Rn+1, ∀n≥0. By Theorem 7, we conclude that αRn,GRn≥1 for all n≥0. From (47), we obtain
(49)αRn−1,GRn−1αRn,GRnφGRn−1,GRn≤ℏφRn−1,RnφRn−1,Rn,then
(50)φRn,Rn+1≤ℏφRn−1,RnφRn−1,Rn.
So, we conclude thatφRn,Rn+1≤φRn−1,Rn. Thus, φRn,Rn+1 is decreasing; hence, there is d≥0 so that φRn,Rn+1⟶d as n⟶∞. Regarding (50), we obtain
(51)φRn,Rn+1φRn−1,Rn≤ℏφRn−1,Rn≤1,
which implies thatlimn⟶∞ℏφRn−1,Rn=1. Hence, d=limn⟶∞dn=limn⟶∞φRn−1,Rn=0, which is a contradiction. Hence, we derive that
(52)limn⟶∞φRn,Rn+1=0.
We claim thatRn is Cauchy sequence. Assume that there are ε>0 and subsequences mj and nj so that
(53)nj>mj>j,φRnj,Rmj≥εandφRnj,Rmj−1<ε.
As in Theorem7, one has
(54)limj⟶+∞φRnj,Rmj=ε,(55)limj⟶+∞φRnj+1,Rmj+1=ε.
Since there existsR0∈℧ such that R0⊥GR0 and αR0,GR0≥1, using condition (2), we derive that
(56)R1⊥R2,αR1,R2=αGR0,G2R0≥1.
By continuing this process, we get(57)Rn⊥Rn+1,αRn,Rn+1≥1,for all n≥0. Suppose that m<n. Recall that
(58)Rm⊥Rm+1,αRm,Rm+1≥1,Rm+1⊥Rm+2,αRm+1,Rm+2≥1.
SinceG is orthogonally triangular α-admissible, we have
(59)Rm⊥Rm+2,αRm,Rm+2≥1.
Again, since(60)Rm⊥Rm+2,αRm,Rm+2≥1,Rm+2⊥Rm+3,αRm+2,Rm+3≥1,using the fact that G is orthogonally triangular α-admissible, we have
(61)Rm⊥Rm+3,αRm,Rm+3≥1.
By continuing this process, we getRm⊥Rn, αRm,Rn≥1 and so Rnj⊥RmjαRnj,Rmj≥1.
Now, from (47), (54), and (55), we have
(62)φRnj+1,Rmj+1≤αRnj,GRnjαRmj,GRmjφRnj+1,Rmj+1=αRnj,GRnjαRmj,GRmjφGRnj,GRmj≤ℏφRnj,RmjφRnj,Rmj.
Hence,(63)φRnj+1,Rmj+1φRnj,Rmj≤ℏφRnj,Rmj≤1.
Lettingj⟶+∞, we get
(64)limj⟶+∞ℏφRnj,Rmj=1.
At the limitj⟶+∞, φRnj,Rmj=0. Hence, Rn is a Cauchy sequence. The completness of ℧ ensures that there is Z∈℧ so that Rn⟶Z. If G is orthogonally continuous, then
(65)GZ=limn⟶∞GRn=limn⟶∞GRn+1=Z.
So,Z is a fixed point of G. If (4) is verified, so Rn⊥Z and αZ,GZ≥1. Now, by (47), we have
(66)φGZ,Rn+1≤αZ,GZαRn,GRnφGZ,GRn≤ℏφZ,RnφZ,Rn.
That is,φGZ,Rn+1≤ℏφZ,RnφZ,Rn, and so we get
(67)φGZ,Z≤φGZ,Rn+1+φZ,Rn+1≤ℏφZ,RnφZ,Rn+φZ,Rn+1.
Takingn⟶∞, one has φGZ,Z=0, that is, Z=GZ.☐Theorem 10.
Assume that all the hypotheses of Theorems7, 8, and 9 hold. Adding the following condition:
(c) If R=GR and Z=GZ then R⊥Z, αR,GR≥1 and αZ,GZ≥1, we obtain the uniqueness of the fixed point of G.Proof.
Suppose thatZ and Z∗ are two fixed points of G such that Z≠Z∗. Then, Z⊥Z∗αZ,GZ≥1 and αZ∗,GZ∗≥1. For Theorem 7, we have
(68)φGZ,GZ∗+l≤φGZ,GZ∗+lαZ,GZαZ∗,GZ∗≤ℏφZ,Z∗φZ,Z∗+l.
For Theorem8, we have
(69)2φGZ,GZ∗≤αZ,GZαZ∗,GZ∗+1φGZ,GZ∗≤2ℏφZ,Z∗φZ,Z∗.
For Theorem9, we have
(70)φGZ,GZ∗≤αZ,GZαZ∗,GZ∗φGZ,GZ∗≤ℏφZ,Z∗φZ,Z∗.
We deduce thatℏφZ,Z∗=1, and so φZ,Z∗=0. That is, Z=Z∗.☐Example 2.
Let℧=0,∞ be equipped with the metric φR,Y=R−Y for all R,Y∈℧. Suppose R⊥Y if RY=R. Let G:℧×℧⟶℧ be defined by
(71)GR=RR+1ifR∈0,1,2RifR∈1,∞.Defineα:R×R⟶0,∞ and ℏ:0,∞⟶0,1 by
(72)αR,Y=1ifR,Y∈0,1,0otherwise,(73)ℏϑ=11+ϑ.Clearly,℧,φ is an O-complete metric space and ⊥-preserving. We claim that G is an orthogonal triangular α-admissible mapping. Let R,Y∈℧. If R⊥Y, αR,Y≥1, then R,Y∈0,1. Also, for R∈0,1, we have GR≤1. It yields that αGR,GY≥1.LetR,Y,Z∈℧. If R⊥Z, αR,Z≥1 and Z⊥Y, αZ,Y≥1, then, R,Y,Z∈0,1. Also, for R∈0,1, we get GR≤1. Thus, R⊥Y, αGR,GY≥1. Hence, the statement is satisfied. Due to the above, α0,G0≥1.Now, letRn be a sequence in ℧ so that αRn,Rn+1≥1 for all n≥0 and Rn⟶R as n⟶∞, then, Rn⊂0,1, hence, R∈0,1. This implies that αR,GR≥1. Let R,Y∈0,1 and Y≥R. We get
(74)φGR,GY+lαR,GRαY,GY=GY−GR=YY+1−RR+1+l=Y−R1+R1+Y+l≤Y−R1+Y−R+l=ℏφR,YφR,Y+l.Otherwise,αR,GRαY,GY=0 and so
(75)φGR,GY+lαR,GRαY,GY=1≤ℏφR,YφR,Y+l.Hence, all the hypotheses of Theorem7 are satisfied, and G has a unique fixed point, R=0.


## 4. Application

Let℧=Cλ1,λ2 be a set of all real continuous functions on λ1,λ2 equipped with metric φR,Y=R−Y for all R,Y∈Cλ1,λ2. Then, ℧,φ is a complete metric space. Consider the orthogonality relation ⊥ on ℧ given as
(76)R⊥Y⇐RΩYΩ≥RΩorRΩYΩ≥YΩ,∀Ω∈λ1,λ2.Now, we consider the nonlinear Fredholm integral equation(77)RΩ=vΩ+1λ2−λ1∫λ1λ2jΩ,s,Rsds,where Ω,s∈λ1,λ2. Assume that j:λ1,λ2×λ1,λ2×R⟶ℝ and v:λ1,λ2⟶ℝ continuous, where vΩ is a given function in ℧.Theorem 11.
Suppose that℧,d is an O-complete metric space equipped with the metric φR,Y=R−Y for all R,Y∈℧ and G:℧⟶℧ is an orthogonal continuous operator on ℧ defined by
(78)GRΩ=vΩ+1λ2−λ1∫λ1λ2jΩ,s,Rsds,for all R,Y∈℧ with R≠Y and s,Ω∈λ1,λ2 satisfying the following inequality
(79)jΩ,s,GRs−jΩ,s,GYs≤R−Y1+R−Y,then, the integral operator defined by (78) has a unique solution.Proof.
We defineα:R×R⟶0,∞ such that αR,Y=1 for all R,Y∈℧ and ℏ:0,∞⟶0,1 defined by
(80)ℏt=11+t.
Therefore,G is orthogonally triangular α-admissible. Now, we show that G is ⊥-preserving. For each, R,Y∈℧ with R⊥Y and Ω∈a,b, we have
(81)GRΩ=vΩ+1λ2−λ1∫λ1λλ2jΩ,s,Rsds≥1.
Accordingly,GRΩGYΩ≥GYΩ and so GYΩ⊥GYΩ. Then, G is ⊥-preserving. Clearly, G is orthogonally continuous. Let R,Y∈℧, l≥1 with R⊥Y. Suppose that GR≠GY. Using (78), we derive
(82)φGR,GY+lαR,GRαY,GY=φGR,GY+l=GR−GY+l=1λ2−λ1∫λ1λ2jΩ,s,GRsds−∫λ1λ2jΩ,s,GYsds+l≤1λ2−λλ1∫λ1λ2jΩ,s,GRs−jΩ,s,GYsds+l≤1λ2−λ1∫λ1λ2R−Y1+R−Yds+l,=R−Y1+R−Y+l=ℏφR,YφR,Y+l.
Hence, all the conditions of Theorem7 are satisfied, and so the integral operator G defined by (78) has a unique solution.☐


## 5. Conclusion

The idea ofα-admissibility on O-complete metric spaces was introduced in this article, and some fixed point theorems were demonstrated. An illustrative example is provided that shows the validity of the hypotheses and the degree of usefulness of our findings.



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*Source: 1202527-2021-09-11.xml*