Solving Fractional Differential Equations via Fixed Points of Chatterjea Maps

Nawab Hussain,Saud M.Alsulami and Hind Alamri,2,＊

1Department of Mathematics,King Abdulaziz University,P.O.Box 80203,Jeddah,21589,Saudi Arabia

2Department of Mathematics,College of Science,Taif University,P.O.Box 11099,Taif,21944,Saudi Arabia

ABSTRACT In this paper, we present the existence and uniqueness of fixed points and common fixed points for Reich and Chatterjea pairs of self-maps in complete metric spaces.Furthermore,we study fixed point theorems for Reich and Chatterjea nonexpansive mappings in a Banach space using the Krasnoselskii-Ishikawa iteration method associated with Sλ and consider some applications of our results to prove the existence of solutions for nonlinear integral and nonlinear fractional differential equations.We also establish certain interesting examples to illustrate the usability of our results.

KEYWORDS Common fixed points;Reich and Chatterjea mappings;Krasnoselskii-Ishikawa iteration;complete metric space;Banach space;integral equation;nonlinear fractional differential equation

1 Introduction

Fixed point theory plays an important role in various branches of mathematics as well as in nonlinear functional analysis, and is very useful for solving many existence problems in nonlinear differential and integral equations with applications in engineering and behavioural sciences.Recently,many authors have provided the extended fixed point theorems for the different classes of contraction type mappings,such as Kannan,Reich,Chatterjea and Ćirić-Reich-Rus mappings(see[1-10]).

Let(Λ,d)be a metric space.A mappingSis said to be a contraction if there existsα∈[0,1)such that

for eachμ,ω∈Λ.A self-mappingSon Λ is nonexpansive ifα= 1.A pointν∈Λ is said to be a fixed point ofSifS(ν)=ν.We denote the set of all fixed points ofSasFix(S).

Kannan[11]established a fixed point theorem for mapping satisfying:for eachμ,ω∈Λ whereWe know that if Λ is complete, then every contraction and every Kannan mapping has a unique fixed point.A mappingSis called Kannan nonexpansive ifα= 1/2 in(2).Nonexpansive mappings are always continuous but Kannan nonexpansive mappings are discontinuous(see[12]).

In 1980,Gregus[13]combined nonexpansive and Kannan nonexpansive mappings as follows:

whereα,β,γare non-negative numbers.Ifα+β+γ ＜1,then the mappingSis known as a Reich contraction.A mapping satisfying(3)is said to be a Reich type nonexpansive mapping,ifα+β+γ=1(see[14,15]).

In[15],the authors considered the Rhoades mapping satisfying the following condition:

for eachμ,ω∈Λ whereα,β,γare non-negative numbers such thatα+β+γ ＜1.A mapping satisfying(4)is said to be Chatterjea type nonexpansive mapping ifα+β+γ=1.Reich[16]showed the generalized Banach’s theorem and observed that Kannan’s theorem is a particular case of it with a suitable selection of the constant.Reich type mappings and generalized nonexpansive mappings have been important research area on their own for many authors which has been applied in various spaces such as metric space,Banach space,and partially ordered Banach spaces(see[5,8,9,17-19]).

In 1971, Ćirić[20]introduced the notion of orbital continuity.Sastry et al.[21]defined the notion of orbital continuity for a pair of mappings.We now recall some relevant definitions.

Definition 1.1.[20]IfSis a self-mapping on metric space(Λ,d),then the set

O(S,μ,n)={μ,S μ,...,S nμ},n≥0,

is said to be an orbit ofSatμ.A metric space Λ is said to beS-orbital complete if every Cauchy sequence contained in the set

O(S,μ,∞)={μ,S μ,S2μ,...},

for someμ∈Λ converges in Λ.

In addition,Sis said to be orbital continuous at a pointν∈Λ, if for any sequence {μn} ⊂O(S,μ,n),then,limn→∞μn=νimplies limn→∞S μn=S ν.Every continuous mappingSis orbital continuity,but the converse is not true,see[20].

Definition 1.2.[21] LetSandTbe two self-mappings of a metric space(Λ,d), and {μn} be a sequence in Λ such thatμ2n+1=T μ2n,μ2n+2=S μ2n+1,n≥0.Then,the set

O(S,T,μ0,n)={μn,n=1,2,...},

is called the(S,T)-orbit atμ0.The mappingT(orS) is called(S,T)-orbital continuous if limn→∞S μn=νimplies limn→∞T S μn=T νor (limn→∞S μn=νimplies limn→∞S S μn=S ν).The mappingsSandTare said to be orbital continuous ifSis(S,T)-orbital continuous andTis(S,T)-orbital continuous.

Definition 1.3.[23] A mappingS: Λ →Λ of metric space Λ is said to beκ-continuous, if limn→∞S κ-1μn=ν,then limn→∞S κμn=S νsuch thatκ ＞1.

Note that,1-continuity is equivalent to continuity and for anyκ= 1,2,...,κ-continuity impliesκ+1-continuity while the converse is not true.Further,continuity of the mappingS κandκ-continuity ofSare independent conditions whenκ ＞1,for more detail and examples(see[23]).

On the other hand,the concept of asymptotic regularity has been introduced by Browder et al.[24]in connection with the study of fixed points of nonexpansive mappings.Asymptotic regularity is a fundamentally important concept in metric fixed point theory.A self-mappingSof a metric space(Λ,d) is called asymptotically regular iffor allμ∈Λ.A mappingSis called asymptotically regular with respect toTatμ0∈Λ if there exists a sequenceμn∈Λ such thatT μn+1=S μn,n≥0 and limn→∞d(T μn+1,T μn+2) = 0.The mappingShas an approximate fixed point sequence if there exists a sequenceμn⊂Λ, such thatd(μn,S μn) →0 asn→∞.The self-mappingsSandTare called compatible[6]if limn→∞d(S T μn,T S μn) = 0,whenever{μn}is a sequence in Λ such that limn→∞S μn= limn→∞T μn=νfor someν∈Λ.C(S,T) = {μ∈Λ :S μ=T μ}denotes the set of coincidence points ofSandT.

In[25],Gòrnicki proved the following fixed point theorem:

Theorem 1.1.Let(Λ,d)be a complete metric space andS:Λ →Λ be a continuous asymptotically regular mapping satisfying

for allμ,ω∈Λ whereα∈[0,1)andβ∈[0,∞).ThenShas a unique fixed pointν∈Λ andS nμ→νfor anyμ∈Λ.

Recently, Bisht [26] showed that the continuity assumption considered in Theorem 1.1 can be weakened by the notion of orbital continuity orκ-continuity.

Theorem 1.2.Let(Λ,d)be a complete metric space andS:Λ →Λ be an asymptotically regular mapping.Assume that there existα∈[0,1)andβ∈[0,∞)satisfying(5)for allμ,ω∈Λ.ThenShas a unique fixed pointν∈Λ,provided thatSis eitherκ-continuous for someκ≥1 or orbitally continuous.Moreover,S nμ→νfor anyμ∈Λ.

This paper is organised as follows: First, we establish some fixed point theorems for Reich and Chatterjea nonexpansive mappings to include asymptotically regular or continuous mappings in complete metric spaces.After that, we prove some fixed point theorems and common fixed points for Reich and Chatterjea type nonexpansive mappings in Banach space using the Krasnoselskii-Ishikawa method associated withSλ.In addition, several examples are provided to illustrate our results.Further, we study the existence of solutions for nonlinear integral equations and nonlinear fractional differential equations.Our work generalizes and complements the comparable results in the current literature.

2 Asymptotic Behaviour of Mappings in Complete Metric Spaces

In this section,we study fixed point and common fixed point theorems for Reich and Chatterjea type nonexpansive mappings in complete metric space.

To start with the following lemma,which is useful to prove the results of this section:

Lemma 2.1.[27]Let{hn}be a sequence of non-negative real numbers satisfying

hn+1≤(1-ψn)hn+ψnϑn,n≥0,

where{ψn},{ϑn}are sequences of real numbers such that:

(i)ψn⊂[0,1]and

(ii) lim supn→∞ϑn≤0,or

Then,limn→∞hn=0.

Theorem 2.1.Let(Λ,d)be a complete metric space andS,T:Λ →Λ be asymptotically regular.Assume that there exist non-negative numbersα,β,γwhereα+β+γ=1,satisfying

for allμ,ω∈Λ andα∈[0,1).Further,SandTare eitherκ-continuous for someκ≥1 or orbitally continuous.ThenSandThave a unique common fixed pointν.Moreover,for anyμ∈Λ,limn→∞S nμ=ν=limn→∞T nμ.

Proof.The proof of the theorem is organized in three steps:

Step 1:We shall prove that limn→∞d(S nμ,T nμ) = 0, for anyμ∈Λ.The result is trivial ifS=T.Suppose thatS≠Tandα=0.Then(6)becomes

d(S μ,T ω)≤βd(μ,S μ)+γ d(ω,T ω),

for allμ,ω∈Λ.Definingμ1=S nμandω1=T nμ,for anyμ∈Λ,we get

d(S n+1μ,T n+1μ)≤βd(S nμ,S n+1μ)+γ d(T nμ,T n+1μ).

Asn→∞,the asymptotic regularity ofSandT,implies that

Using triangle inequality and asymptotic regularity ofSandT,obtain

Thereafter,suppose thatS≠Tandα≠0.Defineμ1=S nμandω1=T nμ.Then,(6)becomes

Lethn=d(S nμ,T nμ),ψn=1-αandBy asymptotically regularity ofSandT,we have limn→∞ϑn=0.Furthermore,Hence,by Lemma 2.1,we get that limn→∞d(S nμ,T nμ)=0 for anyμ∈Λ.

Step 2:Letμn=S nμfor anyμ∈Λ.Now,we show that{μn}is a Cauchy sequence converging toν∈Λ.Moreover,{T nμ}→ν∈Λ.Suppose on contrary that{μn}is not a Cauchy sequence.Then there exists anε ＞0 and two subsequences of integers{mκ}and{nκ}such that for everymκ ＞nκ≥κ,we have

whereκ=1,2,....

Choosingmκ,the smallest number exceedingnkfor which(7)holds.In addition,we assume that

d(S m(κ-1)μ,S nκμ)＜ε.

Thus,we have

Asκ→∞,it follows by asymptotic regularity ofSthat

Further,by asymptotic regularity ofSand the following inequality

the implication is that

Now,using(6)we get

Taking limit asκ→∞,on the both sides of the above inequality,and using(8),(9)and asymptotic regularity ofSandT, we obtain that,ε≤αε, which is a contradiction.Hence, {μn} is a Cauchy sequence in complete space Λ,there exists a pointνin Λ such thatμn→ν.Moreover,

from Step 1 andμn→ν,we get thatT nμconverges toν∈Λ.

Step 3:We will prove thatμis the unique common fixed point ofSandT.Assume thatSisκ-continuous.Since limn→∞S κ-1μn=ν,κ-continuity ofSimplies that

For the uniqueness of the limit,we getS ν=ν.

Similarly,letSbe orbitally continuous.Since limn→∞μn=ν,orbital continuity ofSimplies

we obtainS ν=ν.

In addition,since limn→∞T nμ=ν,this givesT ν=νwheneverTisκ-continuous or orbitally continuous.Hence,ν∈Fix(S)∩Fix(T).Now,we prove the uniqueness of the common fixed point,suppose that there isν≠ν*inFix(S)∩Fix(T).Letμ=νandω=ν*.Then,(6)impliesd(ν,ν*)≤αd(ν,ν*), which is a contradiction.We haveν=ν*.In the other words, the pointνis the unique common fixed point ofSandT.

Example 2.2.Consider Λ = [0,1],equipped with the metricddefined byd(μ,ω) = |μ-ω|.LetSandTbe such that

Clearly, the two mappingsSandTare asymptotically regular.Forwe haveSimilarly,we can show thatTis asymptotically regular.The mappingsSandTare orbitally continuous at 0 and discontinuous at 1.Now,we show thatSandTsatisfy the condition(6)withIn fact,we have the following four cases:

Case 1:Letμ,ω∈[0,1),such thatω≤μ.We get

Case 2:Letμ∈[0,1)andω=1.We have

Case 3:Letω∈[0,1)andμ=1.We obtain

Case 4:Ifμ=ω=1.We get

Therefore,in all the cases,SandTsatisfy the condition(6)for allμ,ω∈Λ.Moreover,all the assumptions of Theorem 2.1 hold,hence,the mappingsSandThave a unique common fixed point at 0.Further,limn→∞S nμ=0=limn→∞T nμfor anyμ∈Λ.

In the special case of our result,we can generate the Theorem 1.4 of Gnicki[28].

Corollary 2.1.Let(Λ,d)be a complete metric space andSandTare self-mappings on Λ whichS pandT qare asymptotically regular for some positive integerspandq, respectively.Assume that there exist non-negative numbersα,β,γwhereα+β+γ=1,such that,the following condition holds

for allμ,ω∈Λ.Then,SandThave a unique common fixed pointν∈Λ,provided that bothS pandT qare eitherκ-continuous for someκ≥1 or orbitally continuous.

Proof.Takef=S pandg=T q.From(10)obtain

for allμ,ω∈Λ.By Theorem 2.1,we obtainfandghave a unique common fixed pointν.Then,

which implies thatS νis a fixed point off.Similarly,we derive thatT νis a fixed point ofg.By(11),we obtain

which implies thatT ν=S ν.From the uniqueness of the common fixed point offandg,it follows thatS ν=T ν=ν.Assuming thatν*≠νare two common fixed points ofSandTsuch thatf ν*=gν*=ν*.The uniqueness of the common fixed point offandgimplies thatν=ν*.Hence,the common fixed point ofSandTis unique.

Example 2.3.Let Λ =c0= {ν=(νn)n∈N:limn→∞νn=0} be the space of all real sequences convergent to zero, endowed with the usual metricd∞defined byfor allμ=(μn)nandω=(ωn)n∈Λ.Then(Λ,d∞)is a complete metric space.LetSandTbe such that

and

Choosingμ=(1,1,1,...)we haveClearly,SandTare asymptotically regular, and orbitally continuous.There are non-negative numbersα,β,γsuch thatα+β+γ=1,we get

for somep,q∈N.

Obviously,as all the assumptions of Corollary 2.1 hold,SandThaveμ=(0,0,0,...) ∈Λ,as their unique common fixed point.Also,limn→∞S pμ=0=limn→∞T qμfor anyμ∈Λ.

Theorem 2.4.Let(Λ,d)be a complete metric space andS:Λ →Λ be an asymptotically regular mapping.Assume that there exist non-negative numbersα,β,γsuch thatα+β+γ=1,andα ＜1,satisfying

for allμ,ω∈Λ.The one of the following conditions hold:

(i)The mappingSis continuous.Further,S nμ→νfor eachμ∈Λ,asn→∞.

(ii)Forκ≥1,Sisκ-continuous or orbitally continuous.

Then,Shas a unique fixed pointν∈Λ.

Proof.First,we shall prove condition(i).Letμ0∈Λ be arbitrary and define a sequence{μn}byμn+1=S μnfor alln≥0.

Using the triangle inequality and asymptotic regularity in(12)we get for anynandκ ＞0,

Thus,

asn→∞andα ＜1.This shows that{μn}is a Cauchy sequence in complete metric space Λ.There existsν∈Λ such thatμn→ν.The continuity ofSandμn+1=S μn,implies thatν=S ν.Letν*≠νbe another fixed point ofS.Then

which is a contradiction.Hence,Shas a unique fixed pointν∈Λ.Now,we show thatS nμ→ν.From(12)we have

Hence,

This shows thatS nμ→νfor anyμ∈Λ asα ＜1.

Next,we will consider condition(ii).Chooseμ0as an arbitrary point in Λ.We consider a sequence{μn} ∈Λ given byμn+1=S μnfor anyn≥0.Then, from (i) we have proven that {μn} is a Cauchy sequence in a complete metric space.There exists a pointν∈Λ such thatμn→νasn→∞.andS μn→ν.Moreover,for allκ≥1 we haveS κμn→νasn→∞.Suppose thatSisκ-continuous.SinceS κ-1μn→ν, we get limn→∞S κμn=S ν.This impliesν=S ν, that isνis a fixed point ofS.Finally,we assume thatSis orbitally continuous.Sinceμn→ν,orbital continuity implies that limn→∞S μn=S ν.This yieldsS ν=ν,that isShas a fixed point atν.

Theorem 2.5.Let(Λ,d)be a complete metric space andS: Λ →Λ be a continuous mapping satisfying(12).Assume thatShas an approximate fixed point sequence.Then,Shas a unique fixed pointν.In particular,μn→νasn→∞.

Proof.Suppose that there existm,n∈N such thatm ＞n.Then,by triangle inequality and(12),we obtain

which implies

Asn,m→∞,we have limn,m→∞d(μn,μm) = 0.Since Λ is a complete metric space,then{μn}is a Cauchy sequence.Hence,the sequence{μn}converges toν∈Λ.Since limn→∞d(μn,S μn) = 0,from the continuity ofSwe get thatνis a fixed point ofS.The uniqueness of the fixed point follows from(12).

Theorem 2.6.Let(Λ,d) be a complete metric space andS,T: Λ →Λ.Suppose thatSis asymptotically regular with respect toT.Assume that there exist non-negative numbersα,β,γsuch thatα+β+γ=1,asα ＜1,satisfying

for eachμ,ω∈Λ.Further,suppose thatSandTare(S,T)-orbitally continuous and compatible.ThenC(S,T)≠φandSandThave a unique common fixed point.

Proof.SinceSis asymptotically regular with respect toTatμ0∈Λ,so,there exists a sequence{ωn} ∈ Λ such thatωn=S μn=T μn+1for eachn≥ 0, and limn→∞d(T μn+1,T μn+2) =limn→∞d(ωn,ωn+1)=0.We show that{ωn}is a Cauchy sequence.From(13)and triangle inequality,for anynand anyκ ＞0,we have

Thus,

SinceSis asymptotically regular with respect toT,then limn→∞d(ωn+κ,ωn)=0.Therefore,{ωn}is a Cauchy sequence in a complete metric space.There exists a pointν∈Λ such thatωn→νasn→∞.Moreover,ωn=S μn=T μn+1→ν.

Suppose thatSandTare compatible mappings.By the orbital continuity ofSandT,

The compatibility ofSandTimplies limn→∞d(S T μn,T S μn) = 0.Taking limit asn→∞we haveS ν=T ν,which means,C(S,T) ≠φ.Sinceνis a coincidence point,the compatibility ofSandTimplies the commutativity ofν.Hence,T S ν=S S ν=T T ν.Using(13),we obtain

which is a contradiction,thence,S ν=S S ν.HenceS ν=S S ν=T S νandS νis a common fixed point ofSandT.The uniqueness of the common fixed point follows from(13).

In the next theorem, we establish a common fixed point result on Chatterjea nonexpansive mapping.

Theorem 2.7.Let(Λ,d) be a complete metric space andS,Tbe asymptotically regular selfmapping on Λ.Assume that there exist non-negative numbersα,β,γsuch that 2α+β+2γ= 1,satisfying

for eachμ,ω∈Λ.Suppose further thatSandTare eitherκ-continuous for someκ≥ 1 or orbitally continuous.Then,SandThave a unique common fixed pointν.Moreover,for anyμ∈Λ,limn→∞S nμ=ν=limn→∞T nμ.

Proof.We follow the lines of Theorem 2.1 to prove this theorem.The proof is divided into three steps as follow:

Step 1:We shall prove that limn→∞d(S nμ,T nμ) = 0, for anyμ∈Λ.The result is trivial ifS=T.SupposeS≠Tandα=0,from(14)we obtain

for eachμ,ω∈Λ.Defineμ1=S nμandω1=T nμ,for anyμ∈Λ.Then

which implies

Asn→∞,sinceSandTare asymptotic regularity,we have

By the asymptotic regularity ofSandTand triangle inequality,we have

Next,suppose thatS≠Tandα≠ 0.Letμ1=S nμandω1=T nμ,for anyμ∈Λ.Then,by(14)we obtain

obtain

which implies

Step 2:Letμn=S nμfor anyμin Λ andn≥0.Then,we show that{μn}is a Cauchy sequence which is convergent toν∈Λ.Moreover,{T nμ}→ν∈Λ.

Assume that{S nμ}is not a Cauchy sequence.Then,there exists anε ＞0 and sequences of integers{mκ}and{nκ}such thatmκ ＞nκ≥κ,forκ=1,2,...,we have

Choosingmκ,the smallest number exceedingnκfor which(16)holds,we also assume that

Now,we have that

Asκ→∞,it follows by asymptotic regularity ofSthat

Furthermore,by asymptotic regularity ofSand the above inequality,we obtain

implying that

Then,using(14),we have

asκ→∞,From(17),(18)and the asymptotic regularity ofSandT,we have,ε≤(α+β+γ)ε.This is a contradiction.Hence, {μn} is a Cauchy sequence in complete space Λ.There existsνin Λ such thatμn→ν.Moreover

from(15)andμn→νimpliesT nμconverges toν∈Λ.

Step 3:We prove the uniqueness of the common fixed point ofSandT.

LetSbeκ-continuous.Since limn→∞S κ-1μn=ν,we have that

Since the limit is unique,it impliesS ν=ν.

Similarly,suppose thatSis orbitally continuous.Since limn→∞μn=ν,we have that

implies thatS ν=ν.Furthermore,since limn→∞T nμ=ν,we have thatT ν=νwheneverTisκcontinuous or orbitally continuous.Hence,ν∈Fix(S)∩Fix(T).For the uniqueness of fixed point,suppose thatν≠ν*are two common fixed points ofSandT.Letμ=νandω=ν*.Then,(14)impliesd(ν,ν*) ≤(α+β+γ)d(ν,ν*).Hence,this is a contradiction.We must have,νas the unique common fixed point ofSandT.

Example 2.8.Let Λ=C0([0,1]×[0,1])be the space of all continuous functions on[0,1].Definedd:Λ×Λ →R+by

In fact,we have the following four cases:

Case 1:Let(f1(t),andwe have

Case 2:If(f1(t)andimplies

Therefore,

Case 3:If(f1(t),,and(g1(t),,we obtain

Thus,

Case 4:If(f1(t),and(g1(t),,we obtain

However,

Therefore, in all cases,SandTsatisfy the condition (14) for allμ,ω∈Λ.Moreover, all the assumptions of Theorem 2.7 hold.Pointis the unique common fixed point ofSandT.Moreover,limn→∞S nμ=0=limn→∞T nμfor anyμ∈Λ.

Corollary 2.2.Let(Λ,d)be a complete metric space andSandTbe self-mappings on Λ whereS pandT qare asymptotically regular for some positive integerspandq, respectively.Assume that there exist non-negative numbersα,β,γwith 2α+β+2γ=1 such that the following condition holds

for allμ,ω∈Λ.Then,SandThave a unique common fixed pointν∈Λ,provided that bothS pandT qare eitherκ-continuous for someκ≥1 or orbitally continuous.

Proof.Takef=S pandg=T q.Then,for eachμ,ω∈Λ,the(19)becomes

According to Theorem 2.7,fandghave a unique common fixed pointν.Now,we show thatS νis a fixed point off,obtain

implies thatS νis a fixed point off.Similarly,we can prove thatT νis also a fixed point ofg.Using(20), given thatS ν=T ν, sincefandghave a unique common fixed point, it follows thatS ν=T ν=ν.Suppose thatν*is another common fixed point ofSandTsuch thatν*≠ν.we have,f ν*=gν*=ν*.The uniqueness of the common fixed point offandgimpliesν*=ν.Thenνis a unique common fixed point ofSandT.

Theorem 2.9.Let(Λ,d)be a complete metric space.The mappingS: Λ →Λ is asymptotically regular.Assume that there exist non-negative numbersα,βandγwith 2α+β+2γ=1 such that

for allμ,ω∈Λ,provided that one of the following conditions hold:

(i) The mappingSis a continuous.FurtherS nμ→νfor eachμ∈Λ,asn→0.

(ii) Forκ≥1,Sisκ-continuous or orbitally continuous.

ThenShas a unique fixed pointμ∈Λ

Proof.First,we will proof the condition(i)holds.Letμ0∈Λ be arbitrary.Then define a sequence{μn}byμn+1=S μnfor alln≥0.

Using the triangle inequality and asymptotic regularity in(21)we obtain for anynandκ ＞0,

Then,

asn→∞.This shows that{μn}is a Cauchy sequence in a complete metric space Λ,there existsν∈Λ such thatμn→ν.AsSis continuous andμn+1=S μn,we obtain thatν=S ν.Suppose thatν*≠νis another fixed point ofS.Then,we have

which is a contradiction.Hence,Shas a unique fixed pointν∈Λ.Now,we show thatS n→ν.From(21),we have

Hence,

This shows thatS n→νfor anyμ∈Λ.

We next prove the condition(ii)holds.Suppose thatμ0is any point in Λ.The sequence{μn}∈Λ is given byμn+1=S μn=S nμ,for eachn≥0.Then,in the proof condition(i)we have shown that{μn}is a Cauchy sequence in complete space Λ.There exists a pointν∈Λ such thatμn→νasn→∞,in addition,S μn→ν.Moreover,for eachκ≥1 we haveS κμn→νasn→∞.Suppose thatSisκ-continuous,andS κ-1μn→ν,implies limn→∞S κμn=S ν.Hence,νis a fixed point ofS.

Finally,we show thatνas a fixed point ofS.Assume thatSis orbitally continuous.Sinceμn→ν,orbital continuity implies that limn→∞S μn=S ν.Thenνis a fixed point ofS.

Theorem 2.10.Let(Λ,d) be a complete metric space andS:Λ →Λ be a continuous mapping satisfying(21).Suppose thatShas an approximate fixed point sequence.Then,Shas a unique fixed pointν.In particular,μn→νasn→∞.

Proof.Letm ＞nfor alln,m∈N.Then,using triangle inequality and(21),we obtain

This implies

Asn,m→∞, we have limn,m→∞d(μn,μm) = 0.Hence, {μn} is a Cauchy sequence in Λ.By completeness of Λ,this sequence{μn}converges toν∈Λ.Since limn→∞d(μn,S μn)=0,continuity ofSimplies thatνis a fixed point ofS.The uniqueness of the fixed point follows from the inequality(21).

Theorem 2.11.Let(Λ,d) be a complete metric space andS,T:Λ →Λ.Suppose thatSis asymptotically regular with respect toT.Assume that there exist non-negative numbersα,β,γsuch that 2α+β+2γ=1,satisfying

for eachμ,ω∈ Λ.Furthermore, suppose thatSandTare(S,T)-orbitally continuous and compatible.ThenC(S,T)≠φandSandThave a unique common fixed point.

Proof.SinceSis asymptotically regularity with respect toTatμ0∈ Λ.So, there exists a sequence {ωn} ∈Λ such thatωn=S μn=T μn+1for eachn≥0, and limn→∞d(T μn+1,T μn+2) =limn→∞d(ωn,ωn+1) = 0.We show that{ωn}is a Cauchy sequence.By triangle inequality and(22),for eachnandκ ＞0,we obtain

Thus,

SinceSis asymptotically regularity with respect toT,it implies thatd(ωn+κ,ωn)→0 asn→∞.Therefore,{ωn}is a Cauchy sequence.Since Λ is complete,there exists a pointν∈Λ such thatωn→νasn→∞,andωn=S μn=T μn+1→ν.

Assuming thatSandTare compatible mappings,orbital continuity ofSandTimplies that

Then,compatibility ofSandTyields limn→∞d(S T μn,T S μn) = 0.Taking limit asn→∞we haveS ν=T ν, which implies,C(S,T) ≠φ.Again the compatibility ofSandTimplies commutativity at a coincidence pointν.HenceT S ν=S S ν=T T ν.Using(22),we have

that is,S ν=S S ν.ThenS ν=S S ν=T S νandS νare common fixed points ofSandT.The uniqueness of the common fixed point follows from(22).

Example 2.12.Let Λ=[2,20]anddbe the usual metric.DefineS,T:Λ →Λ by

ThenSandTsatisfy all the conditions of Theorem 2.11 forIn fact,ifμ=2,ω ＞5,we have

Similarly,if take 2＜μ≤5 andω ＞5 we obtain

Ifμ ＞5 andω ＞5,we get

Otherwise, ifμ= 1 andω= 2 or 2＜ω≤5 (or 2＜μ,ω≤5), it is easy to verify that the mappingsSandTsatisfy the conditions of Theorem 2.11.The mappingsSandThave a unique common fixed pointμ=2.

3 Fixed Point and Common Fixed Point Results in Banach Spaces

In this section, we present some fixed point and common fixed point theorems for Reich and Chatterjea nonexpansive mappings in a Banach space.

Consider a fixed point iteration,which is given by

with an arbitraryμ0∈Λ.The iterative method(23)is also known as Picard iteration.For the Banach contraction mapping theorem[1],the Picard iteration converges to the unique fixed point ofS.

DefineS0=I(the identity map on Λ)andS n=S n-1°S,called thenthiterate ofSforn∈N.The Krasnoselskii-Ishikawa iteration method associated withSis the sequencedefined by

for eachn≥0, andλ∈[0,1].The Krasnoselskii-Ishikawa sequenceis exactly the Picard iteration corresponding to an averaged operator:

However, ifλ= 1, the Krasnoselskii-Ishikawa iteration given by (24) is reduced to the Picard iteration.Moreover,Fix(S)=Fix(Sλ),for eachλ∈(0,1].

In the following,we prove basic lemmas for the Reich nonexpansive mapping which in turn are useful to proving the results of this section.

Lemma 3.1.Let(Λ,‖.‖)be a normed space.Assume that there exist non-negative numbersα,βandγwith 2α+2β+γ=1,where 2α ＜1 and the Reich nonexpansive mappingS:Λ →Λ satisfying the following inequality

for allμ,ω∈Λ.Then for any positive integern,there existsλ∈(0,1)such that for allμ∈Λ and for eachp,qin N,we have

whereη=max{2α,β,γ}andδ[A]=sup{‖μ-ω‖:μ,ω∈A}.

Proof.Let us choosewhere 2α ＜1, clearly, 0＜λ ＜1.Considering the operator given by(25),we have

for allμ,ω∈Λ.Moreover,we obtain

SinceSis a Reich nonexpansive mapping,from(26),we have

Using(28),(29)and(31),we obtain

Now,(30)and(32)imply that

Letμ∈Λ be arbitrary and fixed positive integern.Therefore,using(34),we have

which implies that

such thatη=max{2α,β,γ}.

Remark 3.1.It follows from Lemma 3.1 that ifSis a Reich nonexpansive mapping given by(26)andμ∈Λ,then for anyn∈N,there exists a positive integerκ≤n,such that

Lemma 3.2.Let(Λ,‖.‖) be a normed space.Suppose thatSis a Reich nonexpansive mapping give by(26),such that 2α+2β+γ=1,and 2α ＜1.Then,there existsλ∈(0,1)such that

holds for eachμ∈S,andη=max{2α,β,γ}.

Proof.Letμ∈Λ be arbitrary andλ∈(0,1).Since

δ[O(Sλ,μ,1)]≤δ[O(Sλ,μ,2)]≤...,

we see that

δ[O(Sλ,μ,∞)]=sup{δ[O(Sλ,μ,n)]:n∈N}.

Then,(35)implies that

Letnbe any positive integer.From Remark 3.1,there existsS λκ∈O(Sλ,μ,n)where 1 ≤κ≤n,such that

‖μ-S λκ μ‖=δ[O(Sλ,μ,n)].

Applying a triangle inequality and Lemma 3.1,we obtain

Therefore,

Sincenwas arbitrary,the proof is completed.

Now,we state and prove our main results of this section:

Theorem 3.2.Let(Λ,‖.‖)be a Banach space and a self-mappingSbe a Reich nonexpansive given by(26),with 2α+2β+γ=1 and 2α ＜1.Then,the Krasnoselskii-Ishikawa iteration{μn}defined by

μn+1=(1-λ)μn-1+λS μn-1,n≥1,

converges to a unique fixed pointνfor anyμ0∈Λ,provided that Λ isSλ-orbitally complete.

Proof.Following a similar lines of the proof of Lemma 3.1,we have

Letμ0be an arbitrary point of Λ.Given the iteration(23),the Krasnoselskii-Ishikawa sequence{μn}is exactly the Picard iteration associated withSλ,that is

We shall show that the sequence of iterates{μn}given by(37)is a Cauchy sequence.Letnandm(n ＜m)be any positive integers.From Lemma 3.1,we obtain

whereη=max{2α,β,γ}.According to Remark 3.1,there exists an integerκ,1 ≤κ≤m-n+1,such that

δ[O(Sλ,μn-1,m-n+1)]=‖μn-1-μn+κ-1‖.

Again,by Lemma 3.1,we have

which implies that

‖μn-1-μn+κ-1‖≤ηδ[O(Sλ,μn-2,m-n+2)].

Moreover,by(38)we get

‖μn-μm‖≤ηδ[O(Sλ,μn-1,m-n+1)]≤η2δ[O(Sλ,μn-2,m-n+2)].

Continuing this process,we obtain

‖μn-μm‖≤ηδ[O(Sλ,μn-1,m-n+1)]≤...≤ηnδ[O(Sλ,μ0,m)],

and it follows from Lemma 3.2 that

Taking limit asn→∞,we find that{μn}is a Cauchy sequence.Since Λ isSλ-orbital complete,there existsν∈Λ such that limn→∞μn=ν.Next,we prove thatνis a fixed point ofSλ.In(36),we consider the following inequalities:

Hence,

Since limn→∞μn=ν,we have‖ν-Sλν‖=0,that isSλhas a fixed point.We claim that there is a unique common fixed point ofSλ.Assume on the contrary that,Sλν=νandSλν*=ν*butν≠ν*.By supposition,we obtain

which is a contradiction,hence,ν=ν*.SinceFix(Sλ) =Fix(S),we get thatShas a unique fixed point.

In the next theorem,we present a common fixed point result for Reich mappings.

Theorem 3.3.Let(Λ,‖.‖)be a Banach space andSandTbe self-mappings on Λ.Assume that there exist non-negative numbersα,β,γsuch that 2α+2β+γ=1 and 2α ＜1,satisfying

for allμ,ω∈Λ.Then,the iteration sequencedefined by(24)converges to a unique common fixed pointνfor anyμ0∈Λ,provided that Λ is(Sλ,Tλ)-orbitally complete.

Proof.Suppose thatμ0is an arbitrary point in Λ.Consider the iterative processdefined by(24),which is,in fact,the Picard iteration associated withSλ,that is

Now,using the operator defined by(25),we obtain

From(39)and(41),we get the following:

such that 1-λ+αλ=2α ＜1,λ∈(0,1).This implies

Similarly,

Continuing the process,we get the following:

Now,we show that{μn}is a Cauchy sequence converging toν∈Λ.Then there existsm∈N such thatn ＜m,from(43),and Lemma 3.2,we obtain

where we have‖ν-Sλν‖=0,asn→∞,that isSλhas a fixed point.Similarly,the mappingTλhas a fixed point.We claim that there is a unique common fixed point ofSλandTλ.We assume on the contrary,such thatTλν=Sλν=νandTλν*=Sλν*=ν*butν≠ν*.By supposition,we can replaceμ2n+2byνandμ2n+1byν*in(42)to obtain

this is a contradiction,henceν=ν*.SinceFix(Sλ) =Fix(S),we get thatSandThave a unique common fixed point.

Theorem 3.4.Let(Λ,‖.‖)be a Banach space andS,T: Λ →Λ.Assume that there exist nonnegative numbersα,β,γsuch that 2α+3β+2γ=1,satisfying

for allμ,ω∈Λ.Then,the iteration sequencedefined by(24)converges to a unique common fixed pointνfor anyμ0∈Λ,provided that Λ is(Sλ,Tλ)-orbitally complete.

Proof.Letμ0be arbitrary.Since(Sλ,Tλ)-orbitally complete in Λ,we define the operatorsSλandTλsuch that

Following similar lines of the proof of Theorem 3.2,we obtain

Let us choose 1-λ+λ(α+β+γ)≤2α+β+γ.Since 2α+3β+2γ=1,we have 2(α+γ)=1-3β.This implies that

Similarly,we obtain

Continuing the process,we get the following:

Next,we show that{μn}is a Cauchy sequence converging toν∈Λ.There existsm∈N such thatn ＜m,by Lemma 3.2,and(47),we have

Sinceμ2n→νasn→∞we have‖ν-Sλν‖=0,asn→∞,that isSλhas a fixed point.Similarly,the mappingTλhas a fixed point.We claim that there is a unique common fixed point ofSλandTλ.We assume the contrary thatTλν=Sλν=νandTλν*=Sλν*=ν*butν≠ν*with supposition,we can replaceμ2n+2byνandμ2n+1byν*in(41)to obtain

which is a contradiction, henceν=ν*.SinceFix(Sλ) =Fix(S), we obtain thatSandThave a unique common fixed point.

Example 3.5.Let Λ = [0,1]be equipped with the usual metricddefined byd(μ,ω) = |μ-ω|.Consider the following self-mappings defined by

Case 1:Letandsuch thatμ≤ω.We have

Case 2:Letsuch thatμ≤ω.We get

Case 3:Letandμ≤ω.We obtain

Therefore,in all the cases,SandTsatisfy the condition of(39)for allμ,ω∈Λ.Moreover,as all the assumptions of Theorem 3.3 hold,soSandThaveas their unique common fixed point.

4 Application to Nonlinear Integral Equations

Let Λ =C([0,1],R)be the set of real continuous functions defined on[0,1].Define the metricd: Λ×Λ →[0,∞)byfor eachμ,ω∈Λ.Then,(Λ,d)is a complete metric space.

We consider the following integral equations formulated as a common fixed point problem of the following nonlinear mappings:

such that 0 ≤α ＜1, where the investigation is essentially based on the properties of the kernelK(.,.,.,.).The following assumptions apply:

(i)K1(t,r,μ(r),S μ(r)) ≥0 andK2(t,r,ω(r),T ω(r)) ≥0 fort,r∈[0,1]such thatKi(.,.,0,.) ≠0,fori=1,2.In addition,S(E),T(E)⊆EwhereE={ν∈Λ:0 ≤ν(t)≤1}.

(ii) The two mappingsGandG*are defined by

and

satisfyingGμ,G*ωinE,for allμ(t),ω(t)∈Eand

‖Gμ(t)-G*ω(t)‖＜α(1-α)‖μ(t)-ω(t)‖,

for allμ(t),ω(t)∈E,such thatμ(t)≠ω(t)andt∈[0,1].

Theorem 4.1.Under the assumption(i) and(ii), then the system of (48) has a common fixed point inE.

Proof.We have

We also have,

Suppose thatμ,ω∈E.Then,by using our assumptions,we obtain

which implies that

Therefore,

Now,since 0 ≤α ＜1 andthen

By Theorem 2.1 there exists a common fixed point ofSandTwhich is a common solution for the system(48).

5 Application to Nonlinear Fractional Differential Equation

Fractional differential equations have applications in various fields of engineering and science including diffusive transport, electrical networks, fluid flow and electricity.Many researchers have studied this topic because it has many applications.Related to this matter, we suggest the recent literature[29-35]and the references therein.

The classical Caputo fractional derivative is defined by

wherem-1＜Re(ξ)＜m,m∈N.

Now,we consider the following fractional differential equation:

whereDξis the Caputo fractional derivative of orderξandh:[0,1]×R →R is a continuous function.Suppose that Λ =C[0,1] be a Banach space of a continuous function endowed with the maximum norm and Green’s function associated with(49)is defined as

Assume that

Theorem 5.1.Let Λ=C[0,1]and the operatorsS,T:Λ →Λ be defined as

for allt∈[0,1].If condition(50)is satisfied then the system(49)has a common solution in Λ.

Proof.It is easy to see thatμ∈Λ is a solution of(49)if and only ifμis a solution of the following integral equation:

Letμ,ω∈Λ andt∈[0,1].Then,

Example 5.2.Consider the following fractional differential equation:

The exact solution of the above problem(51)is given by

The operatorS,T:C[0,1]→C[0,1]are defined by

and

On the other hand,we obtain that

which implies that

By Theorem 2.7 there exists a common fixed point ofSandTwhich is a common solution for the system(51).

6 Conclusion

This paper contains the study of Reich and Chatterjea nonexpansive mappings on complete metric and Banach spaces.The existence of fixed points of these mappings which are asymptotically regular or continuous mappings in complete metric space is discussed.Furthermore,we provide some fixed point and common fixed point theorems for Reich and Chatterjea nonexpansive mappings by employing the Krasnoselskii-Ishikawa iteration method associated withSλ.We have established application of our results to nonlinear integral equations and nonlinear fractional differential equations.

Author Contribution:N.H.and H.A.introduced contractive inequalities,obtained the solution and wrote the paper.S.A.read and analyzed the paper;all authors read and approved the final manuscript.

Acknowledgement:The authors wish to express their appreciation to the reviewers for their helpful suggestions which greatly improved the presentation of this paper.

Funding Statement:The authors received no specific funding for this study.

Conflicts of Interest:The authors declare that they have no conflicts of interest to report regarding the present study.