# A Common Fixed Point Theorem for GeneralisedF-Kannan Mapping in Metric Space with Applications

Authors: Lucas Wangwe; Santosh Kumar
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

This paper is aimed at proving a common fixed point theorem forF-Kannan mappings in metric spaces with an application to integral equations. The main result of the paper will extend and generalise the recent existing fixed point results in the literature. We also provided illustrative examples and some applications to integral equation, nonlinear fractional differential equation, and ordinary differential equation for damped forced oscillations to support the results.


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## Body

## 1. Introduction and Preliminaries

In 1922, Banach [1] established a fixed point theorem in a metric space which states that if X is a complete metric space and T:X⟶X is a contraction map, i.e., dTx,Ty≤kdx,y for all x,y∈X and k∈0,1, then T has a unique fixed point or Tx=x has a unique solution. Since then, researchers are finding the ways to determine the fixed points of the maps by changing one or more conditions such as contractive condition, continuity of the maps, and completeness of the space etc. Kannan [2] gave an alternative contractive condition which was different from the Banach contraction condition. In 1968, Kannan [2] used this new contractive condition and proved the following theorem for self-mappings in complete metric spaces as a result of a generalisation of the Banach fixed point theorem.Theorem 1 (see [2]).
LetX,d be a complete metric space and a self-mapping T:X⟶X be a mapping such that
(1)dTx,Ty≤kdx,Tx+dy,Ty,for all x,y∈X and 0≤k≤1/2. Then, T has a unique fixed point z∈X and for any x∈X the sequence of iterate Tnx converges to z.An equivalent form of (1),
(2)dTx,Ty≤k2dx,Tx+dy,Ty,for some k∈0,1.In 1959, Connell [3] gave an example of a metric space X which is not complete and every contraction on X has a fixed point. Also, Subrahmanyam [4] proved the converse of the Banach fixed point theorem using Kannan mapping. Moreover, the assumption of continuity of the mapping and the compactness condition on metric space is required for the existence of a fixed point for a strict type Kannan contraction.In 2000, Branciari [5] introduced a class of generalised metric spaces by replacing triangular inequality to similar ones which involve four or more points instead of three and improved the Banach contraction mapping principle. This motivated several researchers to prove fixed point results in such spaces. For more details on the fixed point theory of a generalised metric space, we refer to the reader [6–11]. In 2008, Azam and Arshad [12] using the concept of Branciari [5] investigated fixed points for the mappings given by Kannan [2] by applying the rectangular property in a generalised metric space.We will require the following definitions and preliminary results to prove our results.Definition 2 (see [12]).
LetX be a nonempty set. Suppose that the mapping d:X×X⟶R satisfies
(i)
dx,y≥0, for all x,y∈X and dx,y=0 if and only if x=y(ii)
dx,y=dy,x, for all x,y∈X(iii)
dx,y≤dx,w+dw,z+dz,y, for all x,y∈X and for all distinct points w,z∈X−x,y [rectangular property]Then,d is called a generalised metric and X,d is a generalised metric space.Definition 3 (see [5]).
LetX,d be a metric space. A mapping T:X⟶X is said to be sequentially convergent if we have, for every sequence yn, if Tyn is convergence then yn also is convergence. T is said to be subsequentially convergent if we have, for every sequence yn, if Tyn is convergence then yn has a convergent subsequence.Definition 4 (see [13]).
LetX be a topological space. If xn is a sequence of points of X, and if n1<n2<⋯<ni<⋯ is an increasing sequence of positive integers, then the sequence yi defined by setting yi=xni is called a subsequence of the sequence xn. The space X is said to be sequentially compact if every sequence of points of X has a convergent subsequence.For more details on the sequentially convergent property, one can see [14, 15].In 2011, Moradi and Alimohammadi [16] generalised Kannan’s results, by using the sequentially convergent mappings and rectangular property in a metric space. Since then, several researchers involved in investigations of Kannan’s contraction mappings using a rectangular property in different spaces. For more details, one can see [12, 17–20] and the references therein. Furthermore, Morandi and Alimohammadi [16] investigated and extended Kannan’s mapping [2] by using the concept due to Branciari [5]. They proved results on two self-mappings as follows.Theorem 5 (see [16]).
LetX,d be a complete metric space and T,S:X⟶X be mappings such that T is continuous, one-to-one, and subsequentially convergent. If λ∈0,1/2 and
(3)dTSx,TSy≤λdTx,TSx+dTy,TSy,x,y∈X,then S has a unique fixed point. Also, if T is sequentially convergent then for every x0∈X, the sequence of iterates Snx0 converges to this fixed point.Wardowski [21] gave an interesting generalisation of the Banach fixed point theorem using a different type of contraction called F-contraction. Since then, many researchers following his approach to construct new fixed point theorems for which one can see [22–27] and the references therein.Wardowski [21] gave the following definitions.LetF be a function defined as F:ℝ+⟶ℝ, which satisfies the following conditions:(F1)F is strictly increasing, i.e., for all α,β∈ℝ+ such that α<β, Fα<Fβ(F2) For each sequenceαnn∈ℕ of positive numbers limn→∞αn=0 if and only if limn→∞Fαn=−∞(F3) There existsk∈0,1 such that
(4)limn→∞αnkFαn=0Definition 6 (see [21]).
A mappingT:X⟶X is said to be a F-contraction if there exists τ>0, such that
(5)∀x,y∈X,dTx,Ty>0⟹τ+FdTx,Ty≤Fdx,y.In 2012, Wardowski [21] introduced a generalization of Banach contraction principle, which is as follows:Theorem 7 (see [21]).
LetX,d be a complete metric space and T:X⟶X be a F-contraction. Then, T has a unique fixed point x∗∈X and for every x0∈X a sequence Tnx0n∈ℕ is convergent to x∗.In 2019, Goswami et al. [22] defined F-contractive type mappings in b-metric spaces and proved some fixed point results with suitable examples. Recently, Batra et al. [28] noticed that the definition introduced by Goswami et al. [22] is not meaningful in general. They provided suitable examples to support their opinion on this definition. Therefore, Batra et al. [28] gave the notions of F-contraction and Kannan mapping to define a new class of contractions called F-Kannan mappings which is in true sense a generalization of Kannan mappings.Motivated by Batra et al. [28], we use the following notations: Let X be a nonempty set and X,d denotes the metric space with metric d. Define the cardinality of a set A by cardA and FixT denotes the set of all fixed points of a mapping T.Batra et al. [28] gave a new generalization family of contraction called F-Kannan mapping and introduced the following definition.Definition 8 (see [28]).
LetF be a mapping satisfying (F1)–(F3). A mapping T:X⟶X is said to be an F-Kannan mapping if the following holds:
(K1)(6)Tx≠Ty⟹Tx≠xorTy≠y
(K2)∃ϒ>0 such that
(7)ϒ+FdTx,Ty≤Fdx,Tx+dy,Ty2for all x,y∈X, with Tx≠Ty.The remark presented below is due to Batra et al. [28].Remark 9 (see [28]).
By properties ofF, it follows that every F-Kannan mapping T on a metric space X,d satisfies the following condition:
(8)dTx,Ty≤dx,Tx+dy,Ty2,for every x,y∈X.Further, it is concluded thatCardFixT≤1. Let T be a self-map of a metric space X,d. We say that T is a Picard operator (PO) if T has unique fixed point x∗ and limn→∞Tnx=x∗ for all x∈X.Then, the family of all functionsF:ℝ+⟶ℝ satisfying the condition F1−F3 is denoted by ∇F.One can use the following examples in Batra et al. [28] of such functions F:ℝ+⟶ℝ which satisfy F1−F3.Example 10 (see [28]).
LetF1:ℝ+⟶ℝ be defined as F1z=lnz. Then, clearly, (F1)-(F3) are satisfied by F1z. In fact, (F3) holds for every k∈0,1,
(9)dTx,Ty≤e−ϒdx,Tx+dy,Ty2,for all x,y∈X with Tx≠Ty.
Thus, ifT:X⟶X is a Kannan mapping with constant k∈0,1 satisfying
(10)dTx,Ty≤kdx,Tx+dy,Ty2,for every x,y∈X.Example 11 (see [28]).
LetF2:ℝ+⟶ℝ be defined as F2z=lnz+z,z>0. Then, (F1)-(F3) are satisfied by F2z.
(11)dTx,Tydx,Tx+dy,Ty/2edTx,Ty−dx,Tx+dy,Ty/2≤e−ϒ,for all x,y∈X with Tx≠Ty.The following lemma was proved by Batra et al. [28]Lemma 12 (see [28]).
LetX,d be a metric space and T:X⟶X be an F-Kannan mapping. Then, dTnx,Tn+1x⟶0 as n⟼∞ for all x∈X.Batra et al. [28] introduced an F-Kannan mapping using the properties by Subrahmanyam [4] which is an extension of Goswami et al. [22] and Wardowski [21] results. They proved the following result.Theorem 13 (see [28]).
LetX,d be a complete metric space and suppose T:X⟶X is an F-Kannan mapping, then T is a Picard operator (PO).In 2017, Gopal et al. [29] specified the fundamental properties for a fixed point theorem which ensures the existence of a common fixed point for suitable assumptions. Those assumptions are sufficient and include conditions of commutativity, containment of ranges of mappings, continuity of at least one mapping or weaker notion, contractive, and all essential common fixed point theorem attempts to obtain or soften required values of one or more such conditions.Definition 14 (see [30]).
LetS,T be a pair of self-mappings on a metric space X,d. Then, the coincidence point of the pair S,T is a point x∈X such that Sx=Tx=x∗, then x∗ is called coincidence point of the pair S,T. If x∗=x, then x is said to be a common fixed point.Definition 15 (see [31]).
LetS,T be self-mappings of a nonempty set X. A point x∈X is the coincidence point of S and T if t=Sx=Tx. The set of coincidence point of S and T is denoted by CS,T.Definition 16 (see [31, 32]).
LetS,T be a pair of self-mappings on a metric space X,d. Then, the pair (S,T) is said to be
(i)
Commuting if, for allx∈X, STx=TSx,(ii)
Weakly commuting if, for alldSTx,TSx≤dSx,Tx,(iii)
Compatible iflimn⟶∞dSTxn,TSxn=0, whenever xn is a sequence in X such that limn⟶∞Txn=limn⟶∞Sxn=t,(iv)
Weakly compatible if, for allSTx=TSx, for every coincidence point x∈X.This paper is aimed at extending and generalising the results due to Batra et al. [28], and Morandi and Alimohammadi [16] using a pair of two self-mappings in F-Kannan mapping. Doing so, we will be able to extend several other results of the same setting in the literature. We will provide some applications of the theorem to the integral equation, nonlinear fractional differential equation, and ordinary differential equation for damped forced oscillations to support the results.


## 2. Main Results

We present the main result of this paper by assuming a map to be sequentially convergent with a pair of two self-mappings inF-Kannan mappings. We shall start with the extension of Definition 8 using a pair of two self-mappings in the F-Kannan mapping setting.Definition 17.
LetF be a mapping satisfying (F1)–(F3). A pair of two self-mapping T,S:X⟶X is said to be an F-Kannan mapping if the following holds:
(FK1)(12)TSx≠TSy⟹TSx≠xorTSy≠y
(FK2) There existsϒ>0 such that
(13)ϒ+FdTSx,TSy≤FdTx,TSx+dTy,TSy2for all x,y∈X, with TSx≠TSy.By following Batra et al. [28], we presented the remark as below.Remark 18.
By properties ofF, it follows that every F-Kannan mapping T on a metric space X,d satisfies the following condition:
(14)dTSx,TSy≤dTx,TSx+dTy,TSy2,for every x,y∈X.We give the following examples in the context of a pair of two mappings:Example 19.
LetF1:ℝ+⟶ℝ be defined as F1z=lnz. Then, clearly, (F1)–(F3) are satisfied by F1z. In fact, (F3) holds for every k∈0,1. Moreover, condition (13) above takes the form:
(15)dTSx,TSy≤e−ϒdTx,TSx+dTy,TSy2,for all x,y∈X with TSx≠TSy.Thus, ifT,S:X⟶X is a Kannan mapping with constant k∈0,1 satisfying
(16)dTSx,TSy≤kdTx,TSx+dTy,TSy2,for every x,y∈X. Then, it also satisfies (15) and (13) with ϒ=ln1/k.Example 20.
LetF2:ℝ+⟶ℝ be defined as F2z=lnz+z,z>0. Then, (F1)–(F3) are satisfied by F2z. Condition (13) above takes the form:
(17)dTSx,TSydTx,TSx+dTy,TSy/2edTSx,TSy−dTx,TSx+dTy,TSy/2≤e−ϒ,for all x,y∈X with TSx≠TSy.We prove the following lemma which will be useful in proving of the main theorem.Lemma 21.
LetX,d be a metric space and T,S:X⟶X be an F-Kannan mapping. Then,
(18)dTSix0,TSi+1x0⟶0asi⟶∞,for all x∈X.Proof.
Letx0∈X be arbitrary. If TSix0=TSi+1x0 for some i∈ℕ, then sequence xin∈ℕ converges in X and hence the sequence dTSix0,TSi+1x0⟶0asi⟶∞ for all x∈X.
Assume thatTSix0≠TSi+1x0 for any i∈ℕ. Then, by (13) with ϒ>0, we get
(19)ϒ+FdTSix0,TSi+1x0≤FdTSi−1x0,TSix0+dTSix0,TSi+1x02.
From Remark18, we have
(20)dTSix0,TSi+1x0≤dTSi−1x0,TSix0+dTSix0,TSi+1x02.
Using (20) in (19), as results yield to
(21)ϒ+FdTSix0,TSi+1x0≤FdTSix0,TSi+1x0.
Lettingi⟶∞ in (21), we get
(22)ϒ+0≤0,ϒ≤0,which is a contradiction. Hence, dTSix0,TSi+1x0⟶0 as i⟶∞.Motivated by Morandi and Alimohammadi [16], we prove the extended version of Theorem 5 using F-Kannan mappings with a pair of two self-mappings in metric space.Theorem 22.
LetX,d be a complete metric space and T,S:X⟶X be an F-Kannan mapping such that T is continuous, injection, and subsequentially convergent. If λ∈0,1/2, ϒ>0 and
(23)ϒ+FdTSx,TSy≤FdTx,TSx+dTy,TSy2,then S has a unique fixed point. Also, if T is subsequentially convergent then for every x0∈X the sequence of iterates Six0 converges to this fixed point.Proof.
Assumex0∈X be an arbitrary point in X. Let the sequence xii≥1 be defined by xi+1=Sxi and xi=Six0, for i=1,2,⋯.
Using inequality (13), we obtain
(24)dTxi,Txi+1=dTSxi−1,TSxi.ϒ+FdTxi,Txi+1≤FdTxi−1,TSxi−1+dTxi,TSxi2,FdTxi,Txi+1≤FdTxi−1,TSxi−1+dTxi,TSxi2−ϒ.
SinceF is strictly increasing, by using Remark 9, we deduce
(25)dTxi,Txi+1<dTxi−1,TSxi−1+dTxi,TSxi2,dTxi,Txi+1<dTxi−1,Txi+dTxi,Txi+12,and hence
(26)2dTxi,Txi+1−dTxi,Txi+1<dTxi−1,Txi,dTxi,Txi+1<dTxi−1,Txi.
By (F1), this implies that(27)FdTxi,Txi+1<FdTxi−1,Txi.
Consequently, we get(28)ϒ+FdTxi,Txi+1≤FdTxi−1,Txi,so,
(29)FdTxi,Txi+1≤FdTxi−1,Txi−ϒ.
Similarly, we obtain(30)dTxi+1,Txi+2=dTSxi,TSxi+1,FdTxi+1,Txi+2≤FdTxi,TSxi+dTxi+1,TSxi+12−ϒ,which implies that
(31)FdTxi+1,Txi+2≤FdTxi,Txi+1−ϒ.
Applying (29) in (31), we obtain
(32)FdTxi+1,Txi+2≤FdTxi−1,Txi−2ϒ,by (F1).
Using induction and (29), we deduce
(33)FdTxi,Txi+1≤FdTxi−1,Txi−iϒ.
Lettingi⟶∞ in (33) and using property (F2) of F results in
(34)limn→∞dTxi,Txi+1=0.
By Lemma21, we have dTxi,Txi+1⟶0 as i⟶∞. Denote dTxi,Txi+1=αi, for all i=1,2,3,⋯ and i∈ℕ, for F-Kannan mappings.
Using conditionF3 of the function F there exists k∈0,1 such that
(35)limn→∞αikFαi=0.
From (33), for every i∈ℕ, we have
(36)αikFαi≤⋯≤αnkFαi−1−iϒαik,(37)αikFαi−αikFαi−1≤−iϒαik,(38)αikFαi−Fαi−1≤−iϒαik≤0.
On taking limit asi⟶∞ in (36), we get
(39)limi→∞iαik=0.
From (39), there exist i1∈ℕ such that iαik≤1, for all i≥i1, which follows that
(40)αi≤i−1/k,∀i≥i1.
Therefore,∑i=0∞dTxi,Txi+1 converges.
By (40), we prove that Txi is a Cauchy sequence since X,d is complete. Consider i,j∈N such that j≥i,
(41)dTxi,Txj≤dTxi,Txi+1+dTxi+1,Txi+2+dTxi+2,Txi+3+⋯+dTxj−1,Txj,≤αi+αi+1+αi+2+⋯+αj−1,=∑ij−1dTxi,Txi+1≤∑j=i∞αi≤∑i=1∞i−1/k.
This shows that the series∑i=1∞i−1/k converges, which implies that
(42)limi→∞dTxi,Txj=0.
So,Txi=Txj for every j≥i in X. Hence, Txi is a Cauchy sequence in X. The completeness of X ensures the existence of x∗∈X such that
(43)dTx∗,x∗=limi,j→∞dTxi,Txj=0,=limi→∞dTxi,x∗=0.
By (43), it follows that xi+1⟶x∗ as i⟶∞. By continuity of S and T, we have
(44)x∗=limi→∞xi=limi→∞xi+1=limi→∞Sxi=Sx∗,x∗=limi→∞xi=limi→∞xi+1=limi→∞Txi=Tx∗.
SinceX is a complete metric space, there exists x∗∈X such that
(45)limi→∞Txi=x∗.
Now, we prove thatx∗ is a fixed point of T. Thus, by (iii) of Definition 2, we have
(46)dx∗,Tx∗≤dx∗,Txi+dTxi,Txi+1+dTxi+1,Tx∗.
By Remark18, it implies that
(47)dTxi+1,Tx∗≤dTxi,Txi+1+dTxi+1,Tx∗2.
Applying (47) in (46), we obtain
(48)dx∗,Tx∗≤dx∗,x∗+dx∗,x∗+dx∗,Tx∗+dx∗,Tx∗2.
Lettingi⟶∞ and using Lemma 21 in above inequality, we get
(49)dx∗,Tx∗≤dx∗,Txi+dTxi,Txi+1+dTxi,Txi+1+dTxi+1,Tx∗2,≤dx∗,x∗+dx∗,x∗+dx∗,x∗+dx∗,Tx∗2,dx∗,Tx∗≤dx∗,Tx∗2,2dx∗,Tx∗≤dx∗,Tx∗,2dx∗,Tx∗≤dx∗,Tx∗,dx∗,Tx∗≤0.
That is,Tx∗=x∗.
Next, we prove thatx∗ is a unique fixed point of T. Assume the contrary, i.e, there exists w∗∈CardFixT such that x∗≠w∗. Let Txi⟶w∗ and w∗ is a fixed point of T. Using Remark 18 and Lemma 21, it follows that x∗=w∗ which is a contradiction. Thus, T is a PO on X.
Moreover,T is a subsequentially convergent, xi has a convergent subsequence, and there exists w∗∈X and xikk=1∞ so that limk→∞xik=w∗. Since T is continuous and
(50)limk→∞xik=w∗.
Due to the continuity ofT, it implies that
(51)limk→∞Txik=Tw∗.
By (45), we conclude that
(52)Tw∗=x∗.
Using Remark18 and (ii) of Definition 2, we get
(53)dTxi,Txi+1=dTSxi−1,TSxi,≤λdTxi−1,TSxi−1+dTxi,TSxi,=λdTxi−1,TSxi−1+dTxi,Txi+1,≤λ1−λdTxi−1,TSxi−1.
Thus, using equation (13) and (iii) of Definition 2, we have
(54)ϒ+FdTxi+1,Txi+2=ϒ+FdTSxi,TSxi+1.FdTSw∗,Tw∗≤FdTSw∗,Txi+dTxi,Txi+1+dTxi+1,Tw∗.
AsF is sequentially increasing, this implies that
(55)dTSw∗,Tw∗<dTSw∗,Txi+dTxi,Txi+1+dTxi+1,Tw∗,dTSw∗,Tw∗≤dTSw∗,TSikx0+dTSikx0,TSik+1x0+dTSik+1x0,Tw∗.
By Lemma21 when TSix0≠TSi+1x0 for any i∈ℕ and (13), we obtain
(56)dTSw∗,TSikx0≤λdTw∗,TSw∗+dTSik−1x0,TSikx0≤λdTw∗,TSw∗+λλ1−λik−1dTx0,Tx1,(57)dTSik+1x0,TSikx0≤λ1−λikdTx1,Tx0.
Using (56) and (57) in (55), we obtain
(58)dTSw∗,Tw∗≤λdTw∗,TSw∗+λλ1−λik−1dTx0,Tx1+λ1−λikdTx1,Tx0+dTxik+1,Tw∗,
which follows(59)dTSw∗,Tw∗≤λ1−λikdTx0,Tx1+λ1−λik+1dTx1,Tx0+11−λdTxik+1,Tw∗.
Lettingk⟶∞ in (59), we obtain
(60)dTSw∗,Tw∗=0.
SinceT is injection, Sw∗=w∗. So, S has a fixed point. As T is sequentially convergent, we conclude that xi converges to the fixed point of S. Implying that Sx⊂X and Tx⊂X, then, there exists a point w∗⊂X such that w∗∈Sw∗∩Tw∗, that is, w∗ is a common fixed point of S and T. which satisfies all fundamental property of Definition 16.Dasgupta et al. [33] and Moradi and Alimohammadi [16], in their work, considered an example in which Kannan Theorem is not applicable. At the same time, generalised Kannan mappings imply the existence of a fixed point for the considered mapping. In this work, we use one more example of this type which satisfies F-Kannan mapping.Example 23.
Consider the sequenceX=0∪1,1/2,1/3,⋯ and d be a Euclidean metric on X. Then, X,d is a complete metric space. Let the mapping S:X⟶X be determined as follows:
(61)S0=0,S1n=1n+1.for n≥1. Let there exists λ∈0,1/2, so that for all x,y∈X condition (1) is satisfied although is not true for every λ>0, which is a contradiction, hence, Kannan’s theorem cannot be applicable.The mappingT:X⟶X be determined as
(62)T0=0,T1n=1nn.For alln>1, T is continuous, injection, and subsequentially convergent.Now, letm,n∈ℕ, m>n. Then, we prove that T,S is an F-Kannan mapping with respect to F2z=lnz+z and ϒ=1.By using (KF2) withF2z, we note that (23) becomes
(63)dTSx,TSydTx,TSx+dTy,TSy/2edTSx,TSy−dTx,TSx+dTy,TSy/2≤e−ϒ.To see this, we now calculatedSx,Ty for x=1/m,y=1/n,n≥1.
(64)dTSx,TSy=dTS1m,TS1n,≤1n+1n+1−1m+1m+1,(65)dTx,TSx=dT1m,TS1m,≤131m−1m+1m+1,(66)dTy,TSy=dT1n,TS1n,≤131n−1n+1n+1.Applying (64),(65), and (66) in (63) becomes
(67)TS1/m,TS1/nT1/m,TS1/m+T1/n,TS1/n/6eTS1/m,TS1/n−T1/m,TS1/m+T1/n,TS1/n/6≤e−ϒ.By Theorem22, S has unique fixed point that is x∗=0.


## 3. Some Applications

In this section, we will provide three applications of the theorem proved in the previous section.


### 3.1. Existence of a Solution for an Integral Equation

In this section, as inspired by Nashine et al. [34], we establish the existence of a solution for the following Volterra type integral equation: This problem is equivalent to the integral equation
(68)ut=ht+∫abft,s,usds,∀t,s∈a,b,where f:a,b×a,b×ℝ⟶ℝ and h:a,b⟶ℝ are continuous functions. Let X=Ca,b,ℝ be the space of all continuous functions defined on Ca,b. Notice that Ca,b endowed with metric
(69)dx,y=x−y∞=A^maxt∈a,bxt−A^yt,is a complete metric space and X can be equipped with the partial order ≼ given by x,y∈X,x≼y⟹xt≼ytandx∞,y∞≤1, or xt=yt for all t∈a,b. It was shown by Nieto and Rodrigurz-Lopez [35] that X,≼ is regular. For more applications to nonlinear integral equations, one can see [36–38] and the references therein.Now, we define a mappingS:X⟶X by
(70)Sxt=ht+∫atft,s,xsds,t∈a,b.Ifx∈a,b is a fixed point of S, then x∈a,b is a solution of (68).We prove our results, by establishing the existence of a common fixed point for a pair of self-mappings.Theorem 24.
LetS,T:Ca,b⟶Ca,b be self-maps of a metric space X,d such that the following condition holds
(1)
For allt,s∈a,b and u∈Ca,b, we have
(71)ft,s,ut≤ft,s∫abfs,τ,uτdτ+ht(2)
There exists two functionsf1,f2:a,b×X×X⟶X with constants α such that, for all t∈a,b, we have
(72)f1t,s,a−f2t,s,b≤αt,sqb−a,q≤1(3)
For(73)supt∈a,b∫abαt,sds≤1Then, integral equation (68) has a solution x∗∈Ca,b,ℝ.Proof.
From (a), for allt∈a,b, we have
(74)Sxt=ht+∫abft,s,xsds,∀t,s∈a,b.≤ft,s∫abfs,τ,xτdτ+hsds+ht=ht+∫abft,s,Sxsds,∀t,s∈a,b=SSxt.
Therefore,Sx≤SSx for all x∈a,b.
By using condition (2) of Theorem24, we obtain
(75)Sxt−Syt=∫abft,s,xs−ft,s,ysds,≤∫abαt,sqdsαxt−yt,=qdx,y∫abαt,sds,≤qdx,y.
Thus,dSx,Sy≤qdx,y for all x,y∈X. Hence, x=x∗ is a common fixed point of S and T, also a solution to integral equation. Then, the integral equation (68) has a solution x∗∈Ca,b,ℝ.


### 3.2. Existence of a Solution for Nonlinear Fractional Differential Equation

The purpose of this section is to provide an application of Theorem22 to get a common solution of a nonlinear fractional differential equation, where we can apply F-Kannan mappings in metric spaces.Here, we investigates the Caputo derivative with the fractional order of the nonlinear fractional differential equation. This form of fractional derivative for a continuous functionf:0,∞⟶ℝ is given as
(76)CDtαft=1Γn−α∫att−sn−α−1fnsds,n−1<α,n=α+1,where α denotes the integer part of the real number α (see [37, 39]). Also, the Riemann-Liouville fractional integral of order α is given by
(77)Isαft=1Γα−1∫0tt−sαfsdsα>0.The Caputo fractional differential equation has several applications in mathematics, i.e., in image processing, digital data processing, electrical signal, acoustics, physics, and probability theory (one can see in [40]). The following nonlinear fractional differential equation is inspired by Kilbas et al. [41], Baleanu et al. [39], Budhia et al. [42], and Kanwal et al. [37]:
(78)CDsαxt=ft,xt,t∈0,1,1<α≤2,x0=0,x1=∫0vxsds0<v<1,where CDsα denotes the Caputo fractional derivative of order α and f:0,1⟶X is a continuous function.Consider the spaceX=CII=0,1 of the continuous function defined on I. Suppose that X,∥ is a Banach space, and I≔0,T,T>0. Let CI,X be the Banach space of all continuous functions from I into X with the norm ∥x∥≔supxt=L,t∈I for x∈CI,X (one can see in [43]).This space defines the metric by(79)dx,y=supt∈0,Txt−yt,∀x,y∈X. This is a complete metric space.Nonlinear fractional equation (78) can be written as
(80)xt=1Γα∫0tt−sα−1fs,xsds−2t2−ν2Γα∫011−sα−1fs,xsds+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xzdzds.A functionx∈CI,X is a solution of the fractional differential integral equation (80) if and only if x is a solution of the nonlinear fractional differential equation (78).Now, we prove the following theorem.Theorem 25.
Suppose the following condition hold:(i)
f∈CI×X,X is sequentially continuous(ii)
There exists a continuous functionf:0,1×ℝ⟶ℝ+, such that
(81)∣ft,xs−ft,ys≤e−ϒLxs−ysfor all t∈0,1 and for all x,y∈X such that dxt,yt≥0 and a constant L with a constant κ∈L, such that
(82)Lκ≤1,κ=tα2−ν2α+1+2tα+να+1+12−ν2Γααα+1.Then, fractional differential equation (78) has a common solution as a fixed point x∗∈CI,X.Proof.
Let us defineT,S:CI⟶CI by
(83)TSxt=1Γα∫0tt−sα−1fs,xsds−2t2−ν2Γα∫011−sα−1fs,xsds+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xzdzds,for t∈0,1, then TS is sequentially continuous. Suppose that
(84)Sxt=∫0ss−zα−1fz,xzdz,this implies that S∈TS and S posses a fixed point x∗∈TS. To prove the existence of fixed point of TS, we prove that TS is sequentially and contraction. To show TS is sequentially continuous, let TSx≠TSy, for all x,y∈0,T. By condition (ii), we have
(85)TSx−TSy=1Γα∫0tt−sα−1fs,xsds−2t2−ν2Γα∫011−sα−1fs,xsds+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xzdzds−1Γα∫0tt−sα−1fs,ysds+2t2−ν2Γα∫011−sα−1fs,ysds−2t2−ν2Γα∫0ν∫0ss−zα−1fz,yzdzds,≤1Γα∫0tt−sα−1fs,xs−fs,ysds+2t2−ν2Γα∫011−sα−1fs,xs−fs,ysds,+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xz−fz,yzdzds,≤1Γα∫0tt−sα−1xs−ysds+2t2−ν2Γα∫011−sα−1xs−ysds+2t2−ν2Γα∫0ν∫0ss−zα−1xz−yzdzds,=e−ϒLΓαx−y∞∫0tt−sα−1ds+2te−ϒL2−ν2Γαx−y∞∫011−sα−1ds+2te−ϒL2−ν2Γαx−y∞∫0ν∫0ss−zα−1dzds,≤e−ϒLtααΓα+2te−ϒL2−ν2αΓα+2te−ϒLνα+12−ν2αα+1Γαx−y∞,≤e−ϒLtααΓα+2t2−ν2αΓα+2tνα+12−ν2αα+1Γαx−y∞,=e−ϒLtα2−ν2α+1+2tα+να+1+12−ν2Γααα+1x−y∞,≤e−ϒLκx−y∞.
This implies that(86)TSx,TSy∞≤e−ϒLκx−y∞.
SinceLκ<1, we have
(87)TSx,TSy∞≤e−ϒx−y∞.
Thus, for eachx,y∈X, we have
(88)dTSx,TSy≤e−ϒMx,y.
Taking logarithms on both sides of (88) using F1z=lnz and the property of F, we get
(89)lndTSx,TSy≤lne−ϒMx,y.
Equivalently to(90)ϒ+FdTSx,TSy≤FMx,y.
ForLκ∈0,1, ϒ>0, and Mx,y is an F-Kannan mapping. Therefore, TS is a contraction mapping on X. Since all the conditions of Theorem 25 are satisfied, therefore, there exists x∗∈CI a common fixed point of T and S, that is, x∗ is a solution to fractional nonlinear differential equation (78).


### 3.3. Existence of Common Solution of Ordinary Differential Equations for Damped Forced Oscillations

This section investigates the solution of the forced damped oscillation differential equations problem in the setting of metric spaces. Nieto and Rodríguez-López [35] initiated the proof of the existing solution of an ordinary differential equation. Since then, several authors are interested in this line of research. In details, one can see the literature in [41, 44–48] and the references therein.In 2020, Shoib et al. [49], considered the forced damped oscillation differential problem of an object of mass m moves to and fro on the x-axis around an equilibrium position x=0. The object has position xt at time t, it undergoes the applied force ft such that
(91)md2xdt2+bdxdt+kx=ft.where m, b, and k are constant positive numbers. If the intial conditions are assumed to be
(92)x0=0,x′0=0.Problem (91) can be written as
(93)xt=∫0TGt,sKt,xsds,∀t,s∈0,T,where T>0. Let X=CI,I=0,T.The green function for forced damped oscillation is defined by(94)Gt,s=−seτT+s−t,0≤s≤t≤T,−teτt−s,0≤t≤s≤T,where τ can be written in terms of b, k, and m.Inspired by Shoaib et al. [49] and Gupta et al. [45], we find the common solution of a forced damped oscillation differential equation using the fixed point method.Theorem 26.
Suppose the following assumption hold:(i)
There exists a continuous functionK:0,T2⟶0,∞, such that
(95)Kt,s,xt−Kt,s,yt≤e−ϒlnx−y2+1x−y,for t,s∈0,T, ϒ>0 and x,y∈ℝ, where,
(96)lnx−y2+1x−y=dTx,TSx+dTy,TSy2(ii)
There exist a functonG:I×I⟶ℝ+, such that(97)supt∈0,T∫0TG2t,sds≤1TThen, problem (93) has a fixed point x∗∈X, which is a solution of (91).Proof.
LetTS:C10,1⟶C10,1 be an operator defined by
(98)TSxt=∫0TGt,sKt,s,xsds.
Considerx>y, for all x,y∈CI and (93), we have
(99)TSxt−TSyt=supt∈0,1∫0TGt,sKt,s,xs−Kt,s,ysds,≤supt∈0,T∫0te−ϒGt,slnxs−ys2+1xs−ysds.
Recall Cauchy-Schwartz inequality defined by(100)∑i=1∞xiyi2≤∑i=1∞xi2∑i=1∞yi2.
Using (99) in (100), we obtain
(101)∫0Te−ϒGt,slnxs−ys2+1xs−ys≤∫0TG2t,s1/2∫0Te−2ϒlnxs−ys2+1xs−ys21/2.
Then,(102)∫0TG2t,s=1T.
The second integral gives(103)∫0Te−2ϒlnxs−ys2+1xs−ysds≤e−2ϒlnxs−ys2+1xs−ys×T.
Using (102) and (103) in the R.H.S of (101), we get
(104)≤1T1/2e−2ϒlndx,y2+1dx,y×T1/2.
Hence, from (99),
(105)dTSx,TSy≤e−ϒlndx,y2+1dx,y.
Taking logarithms on both sides of (105) using F1z=lnz and the property of F, we get
(106)lndTSx,TSy≤lne−ϒlndx,y2+1dx,y.
This implies that(107)ϒ+FdTSx,TSy≤Flndx,y2+1dx,y.
Let us choose a mapSx=lndx,y2+1, it is obvious, S∈TS.
Thus, we conclude that forx>y all conditions of Theorem 26 are satisfied. Hence, TS has a unique common fixed point x∗ which is the solution of integral equation (93).


## 3.1. Existence of a Solution for an Integral Equation

In this section, as inspired by Nashine et al. [34], we establish the existence of a solution for the following Volterra type integral equation: This problem is equivalent to the integral equation
(68)ut=ht+∫abft,s,usds,∀t,s∈a,b,where f:a,b×a,b×ℝ⟶ℝ and h:a,b⟶ℝ are continuous functions. Let X=Ca,b,ℝ be the space of all continuous functions defined on Ca,b. Notice that Ca,b endowed with metric
(69)dx,y=x−y∞=A^maxt∈a,bxt−A^yt,is a complete metric space and X can be equipped with the partial order ≼ given by x,y∈X,x≼y⟹xt≼ytandx∞,y∞≤1, or xt=yt for all t∈a,b. It was shown by Nieto and Rodrigurz-Lopez [35] that X,≼ is regular. For more applications to nonlinear integral equations, one can see [36–38] and the references therein.Now, we define a mappingS:X⟶X by
(70)Sxt=ht+∫atft,s,xsds,t∈a,b.Ifx∈a,b is a fixed point of S, then x∈a,b is a solution of (68).We prove our results, by establishing the existence of a common fixed point for a pair of self-mappings.Theorem 24.
LetS,T:Ca,b⟶Ca,b be self-maps of a metric space X,d such that the following condition holds
(1)
For allt,s∈a,b and u∈Ca,b, we have
(71)ft,s,ut≤ft,s∫abfs,τ,uτdτ+ht(2)
There exists two functionsf1,f2:a,b×X×X⟶X with constants α such that, for all t∈a,b, we have
(72)f1t,s,a−f2t,s,b≤αt,sqb−a,q≤1(3)
For(73)supt∈a,b∫abαt,sds≤1Then, integral equation (68) has a solution x∗∈Ca,b,ℝ.Proof.
From (a), for allt∈a,b, we have
(74)Sxt=ht+∫abft,s,xsds,∀t,s∈a,b.≤ft,s∫abfs,τ,xτdτ+hsds+ht=ht+∫abft,s,Sxsds,∀t,s∈a,b=SSxt.
Therefore,Sx≤SSx for all x∈a,b.
By using condition (2) of Theorem24, we obtain
(75)Sxt−Syt=∫abft,s,xs−ft,s,ysds,≤∫abαt,sqdsαxt−yt,=qdx,y∫abαt,sds,≤qdx,y.
Thus,dSx,Sy≤qdx,y for all x,y∈X. Hence, x=x∗ is a common fixed point of S and T, also a solution to integral equation. Then, the integral equation (68) has a solution x∗∈Ca,b,ℝ.


## 3.2. Existence of a Solution for Nonlinear Fractional Differential Equation

The purpose of this section is to provide an application of Theorem22 to get a common solution of a nonlinear fractional differential equation, where we can apply F-Kannan mappings in metric spaces.Here, we investigates the Caputo derivative with the fractional order of the nonlinear fractional differential equation. This form of fractional derivative for a continuous functionf:0,∞⟶ℝ is given as
(76)CDtαft=1Γn−α∫att−sn−α−1fnsds,n−1<α,n=α+1,where α denotes the integer part of the real number α (see [37, 39]). Also, the Riemann-Liouville fractional integral of order α is given by
(77)Isαft=1Γα−1∫0tt−sαfsdsα>0.The Caputo fractional differential equation has several applications in mathematics, i.e., in image processing, digital data processing, electrical signal, acoustics, physics, and probability theory (one can see in [40]). The following nonlinear fractional differential equation is inspired by Kilbas et al. [41], Baleanu et al. [39], Budhia et al. [42], and Kanwal et al. [37]:
(78)CDsαxt=ft,xt,t∈0,1,1<α≤2,x0=0,x1=∫0vxsds0<v<1,where CDsα denotes the Caputo fractional derivative of order α and f:0,1⟶X is a continuous function.Consider the spaceX=CII=0,1 of the continuous function defined on I. Suppose that X,∥ is a Banach space, and I≔0,T,T>0. Let CI,X be the Banach space of all continuous functions from I into X with the norm ∥x∥≔supxt=L,t∈I for x∈CI,X (one can see in [43]).This space defines the metric by(79)dx,y=supt∈0,Txt−yt,∀x,y∈X. This is a complete metric space.Nonlinear fractional equation (78) can be written as
(80)xt=1Γα∫0tt−sα−1fs,xsds−2t2−ν2Γα∫011−sα−1fs,xsds+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xzdzds.A functionx∈CI,X is a solution of the fractional differential integral equation (80) if and only if x is a solution of the nonlinear fractional differential equation (78).Now, we prove the following theorem.Theorem 25.
Suppose the following condition hold:(i)
f∈CI×X,X is sequentially continuous(ii)
There exists a continuous functionf:0,1×ℝ⟶ℝ+, such that
(81)∣ft,xs−ft,ys≤e−ϒLxs−ysfor all t∈0,1 and for all x,y∈X such that dxt,yt≥0 and a constant L with a constant κ∈L, such that
(82)Lκ≤1,κ=tα2−ν2α+1+2tα+να+1+12−ν2Γααα+1.Then, fractional differential equation (78) has a common solution as a fixed point x∗∈CI,X.Proof.
Let us defineT,S:CI⟶CI by
(83)TSxt=1Γα∫0tt−sα−1fs,xsds−2t2−ν2Γα∫011−sα−1fs,xsds+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xzdzds,for t∈0,1, then TS is sequentially continuous. Suppose that
(84)Sxt=∫0ss−zα−1fz,xzdz,this implies that S∈TS and S posses a fixed point x∗∈TS. To prove the existence of fixed point of TS, we prove that TS is sequentially and contraction. To show TS is sequentially continuous, let TSx≠TSy, for all x,y∈0,T. By condition (ii), we have
(85)TSx−TSy=1Γα∫0tt−sα−1fs,xsds−2t2−ν2Γα∫011−sα−1fs,xsds+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xzdzds−1Γα∫0tt−sα−1fs,ysds+2t2−ν2Γα∫011−sα−1fs,ysds−2t2−ν2Γα∫0ν∫0ss−zα−1fz,yzdzds,≤1Γα∫0tt−sα−1fs,xs−fs,ysds+2t2−ν2Γα∫011−sα−1fs,xs−fs,ysds,+2t2−ν2Γα∫0ν∫0ss−zα−1fz,xz−fz,yzdzds,≤1Γα∫0tt−sα−1xs−ysds+2t2−ν2Γα∫011−sα−1xs−ysds+2t2−ν2Γα∫0ν∫0ss−zα−1xz−yzdzds,=e−ϒLΓαx−y∞∫0tt−sα−1ds+2te−ϒL2−ν2Γαx−y∞∫011−sα−1ds+2te−ϒL2−ν2Γαx−y∞∫0ν∫0ss−zα−1dzds,≤e−ϒLtααΓα+2te−ϒL2−ν2αΓα+2te−ϒLνα+12−ν2αα+1Γαx−y∞,≤e−ϒLtααΓα+2t2−ν2αΓα+2tνα+12−ν2αα+1Γαx−y∞,=e−ϒLtα2−ν2α+1+2tα+να+1+12−ν2Γααα+1x−y∞,≤e−ϒLκx−y∞.
This implies that(86)TSx,TSy∞≤e−ϒLκx−y∞.
SinceLκ<1, we have
(87)TSx,TSy∞≤e−ϒx−y∞.
Thus, for eachx,y∈X, we have
(88)dTSx,TSy≤e−ϒMx,y.
Taking logarithms on both sides of (88) using F1z=lnz and the property of F, we get
(89)lndTSx,TSy≤lne−ϒMx,y.
Equivalently to(90)ϒ+FdTSx,TSy≤FMx,y.
ForLκ∈0,1, ϒ>0, and Mx,y is an F-Kannan mapping. Therefore, TS is a contraction mapping on X. Since all the conditions of Theorem 25 are satisfied, therefore, there exists x∗∈CI a common fixed point of T and S, that is, x∗ is a solution to fractional nonlinear differential equation (78).


## 3.3. Existence of Common Solution of Ordinary Differential Equations for Damped Forced Oscillations

This section investigates the solution of the forced damped oscillation differential equations problem in the setting of metric spaces. Nieto and Rodríguez-López [35] initiated the proof of the existing solution of an ordinary differential equation. Since then, several authors are interested in this line of research. In details, one can see the literature in [41, 44–48] and the references therein.In 2020, Shoib et al. [49], considered the forced damped oscillation differential problem of an object of mass m moves to and fro on the x-axis around an equilibrium position x=0. The object has position xt at time t, it undergoes the applied force ft such that
(91)md2xdt2+bdxdt+kx=ft.where m, b, and k are constant positive numbers. If the intial conditions are assumed to be
(92)x0=0,x′0=0.Problem (91) can be written as
(93)xt=∫0TGt,sKt,xsds,∀t,s∈0,T,where T>0. Let X=CI,I=0,T.The green function for forced damped oscillation is defined by(94)Gt,s=−seτT+s−t,0≤s≤t≤T,−teτt−s,0≤t≤s≤T,where τ can be written in terms of b, k, and m.Inspired by Shoaib et al. [49] and Gupta et al. [45], we find the common solution of a forced damped oscillation differential equation using the fixed point method.Theorem 26.
Suppose the following assumption hold:(i)
There exists a continuous functionK:0,T2⟶0,∞, such that
(95)Kt,s,xt−Kt,s,yt≤e−ϒlnx−y2+1x−y,for t,s∈0,T, ϒ>0 and x,y∈ℝ, where,
(96)lnx−y2+1x−y=dTx,TSx+dTy,TSy2(ii)
There exist a functonG:I×I⟶ℝ+, such that(97)supt∈0,T∫0TG2t,sds≤1TThen, problem (93) has a fixed point x∗∈X, which is a solution of (91).Proof.
LetTS:C10,1⟶C10,1 be an operator defined by
(98)TSxt=∫0TGt,sKt,s,xsds.
Considerx>y, for all x,y∈CI and (93), we have
(99)TSxt−TSyt=supt∈0,1∫0TGt,sKt,s,xs−Kt,s,ysds,≤supt∈0,T∫0te−ϒGt,slnxs−ys2+1xs−ysds.
Recall Cauchy-Schwartz inequality defined by(100)∑i=1∞xiyi2≤∑i=1∞xi2∑i=1∞yi2.
Using (99) in (100), we obtain
(101)∫0Te−ϒGt,slnxs−ys2+1xs−ys≤∫0TG2t,s1/2∫0Te−2ϒlnxs−ys2+1xs−ys21/2.
Then,(102)∫0TG2t,s=1T.
The second integral gives(103)∫0Te−2ϒlnxs−ys2+1xs−ysds≤e−2ϒlnxs−ys2+1xs−ys×T.
Using (102) and (103) in the R.H.S of (101), we get
(104)≤1T1/2e−2ϒlndx,y2+1dx,y×T1/2.
Hence, from (99),
(105)dTSx,TSy≤e−ϒlndx,y2+1dx,y.
Taking logarithms on both sides of (105) using F1z=lnz and the property of F, we get
(106)lndTSx,TSy≤lne−ϒlndx,y2+1dx,y.
This implies that(107)ϒ+FdTSx,TSy≤Flndx,y2+1dx,y.
Let us choose a mapSx=lndx,y2+1, it is obvious, S∈TS.
Thus, we conclude that forx>y all conditions of Theorem 26 are satisfied. Hence, TS has a unique common fixed point x∗ which is the solution of integral equation (93).



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*Source: 6619877-2021-04-17.xml*