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Solving Fractional Differential Equations via Fixed Points of Chatterjea Maps

Nawab Hussain1,*, Saud M. Alsulami1, Hind Alamri1,2,*

1 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia
2 Department of Mathematics, College of Science, Taif University, P.O. Box 11099, Taif, 21944, Saudi Arabia

* Corresponding Authors: Nawab Hussain. Email: email; Hind Alamri. Email: email

(This article belongs to this Special Issue: Computational Aspects of Nonlinear Operator and Fixed Point Theory with Applications)

Computer Modeling in Engineering & Sciences 2023, 135(3), 2617-2648. https://doi.org/10.32604/cmes.2023.023143

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 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


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 [110]).

Let (Λ,d) be a metric space. A mapping S is said to be a contraction if there exists α[0,1) such that

d(Sμ,Sω)αd(μ,ω),(1)

for each μ,ωΛ. A self-mapping S on Λ is nonexpansive if α=1. A point νΛ is said to be a fixed point of S if S(ν)=ν. We denote the set of all fixed points of S as Fix(S).

Kannan [11] established a fixed point theorem for mapping satisfying:

d(Sμ,Sω)α{d(μ,Sμ)+d(ω,Sω)},(2)

for each μ,ωΛ where α[0,12). We know that if Λ is complete, then every contraction and every Kannan mapping has a unique fixed point. A mapping S is called Kannan nonexpansive if α=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:

d(Sμ,Sω)αd(μ,ω)+βd(μ,Sμ)+γd(ω,Sω),μ,ωΛ,(3)

where α,β,γ are non-negative numbers. If α+β+γ<1, then the mapping S is known as a Reich contraction. A mapping satisfying (3) is said to be a Reich type nonexpansive mapping, if α+β+γ=1 (see [14,15]).

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

d(Sμ,Sω)αd(μ,ω)+βd(ω,Sμ)+γd(μ,Sω),(4)

for each μ,ωΛ where α,β,γ are non-negative numbers such that α+β+γ<1. A mapping satisfying (4) is said to be Chatterjea type nonexpansive mapping if α+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 [5,8,9,1719]).

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] If S is a self-mapping on metric space (Λ,d), then the set

O(S,μ,n)={μ,Sμ,,Snμ},n0,

is said to be an orbit of S at μ. A metric space Λ is said to be S-orbital complete if every Cauchy sequence contained in the set

O(S,μ,)={μ,Sμ,S2μ,},

for some μΛ converges in Λ.

In addition, S is said to be orbital continuous at a point νΛ, if for any sequence {μn}O(S,μ,n), then, limnμn=ν implies limnSμn=Sν. Every continuous mapping S is orbital continuity, but the converse is not true, see [20].

Definition 1.2. [21] Let S and T be two self-mappings of a metric space (Λ,d), and {μn} be a sequence in Λ such that μ2n+1=Tμ2n,μ2n+2=Sμ2n+1, n0. Then, the set

O(S,T,μ0,n)={μn,n=1,2,},

is called the (S,T)-orbit at μ0. The mapping T (or S) is called (S,T)-orbital continuous if limnSμn=ν implies limnTSμn=Tν or (limnSμn=ν implies limnSSμn=Sν). The mappings S and T are said to be orbital continuous if S is (S,T)-orbital continuous and T is (S,T)-orbital continuous.

Ćirić in [22] proved that continuity of S implies orbital continuity but the converse is not true.

Definition 1.3. [23] A mapping S:ΛΛ of metric space Λ is said to be κ-continuous, if limnSκ1μn=ν, then limnSκμn=Sν such that κ>1.

Note that, 1-continuity is equivalent to continuity and for any κ=1,2,, κ-continuity implies κ+1-continuity while the converse is not true. Further, continuity of the mapping Sκ and κ-continuity of S are independent conditions when κ>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-mapping S of a metric space (Λ,d) is called asymptotically regular if limnd(Snμ,Sn+1μ)=0 for all μΛ. A mapping S is called asymptotically regular with respect to T at μ0Λ if there exists a sequence μnΛ such that Tμn+1=Sμn, n0 and limnd(Tμn+1,Tμn+2)=0. The mapping S has an approximate fixed point sequence if there exists a sequence μnΛ, such that d(μn,Sμn)0 as n. The self-mappings S and T are called compatible [6] if limnd(STμn,TSμn)=0, whenever {μn} is a sequence in Λ such that limnSμn=limnTμn=ν for some νΛ. C(S,T)={μΛ:Sμ=Tμ} denotes the set of coincidence points of S and T.

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

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

d(Sμ,Sω)αd(μ,ω)+β{d(μ,Sμ)+d(ω,Sω)},(5)

for all μ,ωΛ where α[0,1) and β[0,). Then S has a unique fixed point νΛ and Snμν for any μΛ.

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

Theorem 1.2. Let (Λ,d) be a complete metric space and S:ΛΛ be an asymptotically regular mapping. Assume that there exist α[0,1) and β[0,) satisfying (5) for all μ,ωΛ. Then S has a unique fixed point νΛ, provided that S is either κ-continuous for some κ1 or orbitally continuous. Moreover, Snμν for any μΛ.

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 with Sλ. 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(1ψn)hn+ψnϑn,n0,

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

(i)   ψn[0,1] and n=0ψn=,

(ii)   lim supnϑn0,or

(iii)   n=0|ψnϑn| is convergent.

Then, limnhn=0.

Theorem 2.1. Let (Λ,d) be a complete metric space and S,T:ΛΛ be asymptotically regular. Assume that there exist non-negative numbers α,β,γ where α+β+γ=1, satisfying

d(Sμ,Tω)αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω),(6)

for all μ,ωΛ and α[0,1). Further, S and T are either κ-continuous for some κ1 or orbitally continuous. Then S and T have a unique common fixed point ν. Moreover, for any μΛ, limnSnμ=ν=limnTnμ.

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

Step 1: We shall prove that limnd(Snμ,Tnμ)=0, for any μΛ. The result is trivial if S=T. Suppose that ST and α=0. Then (6) becomes

d(Sμ,Tω)βd(μ,Sμ)+γd(ω,Tω),

for all μ,ωΛ. Defining μ1=Snμ and ω1=Tnμ, for any μΛ, we get

d(Sn+1μ,Tn+1μ)βd(Snμ,Sn+1μ)+γd(Tnμ,Tn+1μ).

As n, the asymptotic regularity of S and T, implies that

limnd(Sn+1μ,Tn+1μ)=0.

Using triangle inequality and asymptotic regularity of S and T, obtain

d(Snμ,Tnμ)d(Snμ,Sn+1μ)+d(Sn+1μ,Tn+1μ)+d(Tn+1μ,Tnμ)0.asn.

Thereafter, suppose that ST and α0. Define μ1=Snμ and ω1=Tnμ. Then, (6) becomes

d(Sn+1μ,Tn+1μ)αd(Snμ,Tnμ)+βd(Snμ,Sn+1μ)+γd(Tnμ,Tn+1μ).

Let hn=d(Snμ,Tnμ), ψn=1α and ϑn=β1αd(Snμ,Sn+1μ)+γ1αd(Tnμ,Tn+1μ).By asymptotically regularity of S and T, we have limnϑn=0. Furthermore, n=1ψn=. Hence, by Lemma 2.1, we get that limnd(Snμ,Tnμ)=0 for any μΛ.

Step 2: Let μn=Snμ for any μΛ. Now, we show that {μn} is a Cauchy sequence converging to νΛ. Moreover, {Tnμ}νΛ. Suppose on contrary that {μn} is not a Cauchy sequence. Then there exists an ε>0 and two subsequences of integers {mκ} and {nκ} such that for every mκ>nκκ, we have

d(Smκ,Snκ)ε,(7)

where κ=1,2,.

Choosing mκ, the smallest number exceeding nk for which (7) holds. In addition, we assume that

d(Sm(κ1)μ,Snκμ)<ε.

Thus, we have

εd(Smκμ,Snκμ)d(Smκμ,Sm(κ1)μ)+d(Sm(κ1)μ,Snκμ)<d(Smκμ,Sm(κ1)μ)+ε.

As κ, it follows by asymptotic regularity of S that

limκd(Smκμ,Snκμ)=ε.(8)

Further, by asymptotic regularity of S and the following inequality

d(Sm(κ1)μ,Sn(κ1)μ)d(Sm(κ1)μ,Smκμ)+d(Smκμ,Snκμ)+d(Snκμ,Sn(κ1)μ),

the implication is that

limkd(Sm(κ1)μ,Sn(κ1)μ)=ε.(9)

Now, using (6) we get

d(Smκμ,Snκμ)d(Smκμ,Tnκμ)+d(Tnκμ,Snκμ)d(Tnκμ,Snκμ)+α[d(Sm(κ1)μ,Sn(κ1)μ)+d(Sn(κ1)μ,Tn(κ1)μ)]+βd(Sm(κ1)μ,Smκμ)+γd(Tn(κ1)μ,Tnκμ).

Taking limit as κ, on the both sides of the above inequality, and using (8), (9) and asymptotic regularity of S and T, we obtain that, εαε, which is a contradiction. Hence, {μn} is a Cauchy sequence in complete space Λ, there exists a point ν in Λ such that μnν. Moreover,

d(Tnμ,ν)d(Tnμ,Snμ)+d(Snμ,ν),

from Step 1 and μnν, we get that Tnμ converges to νΛ.

Step 3: We will prove that μ is the unique common fixed point of S and T. Assume that S is κ-continuous. Since limnSκ1μn=ν, κcontinuity of S implies that

limnSκμn=Sν.

For the uniqueness of the limit, we get Sν=ν.

Similarly, let S be orbitally continuous. Since limnμn=ν, orbital continuity of S implies

limnSμn=Sν,

we obtain Sν=ν.

In addition, since limnTnμ=ν, this gives Tν=ν whenever T is κ-continuous or orbitally continuous. Hence, νFix(S)Fix(T). Now, we prove the uniqueness of the common fixed point, suppose that there is νν in Fix(S)Fix(T). Let μ=ν and ω=ν. Then, (6) implies d(ν,ν)αd(ν,ν), which is a contradiction. We have ν=ν. In the other words, the point ν is the unique common fixed point of S and T.

Example 2.2. Consider Λ=[0,1], equipped with the metric d defined by d(μ,ω)=|μω|. Let S and T be such that

Sμ={μ7ifμ[0,1),0ifμ=1,andTμ={3μ7ifμ[0,1),17ifμ=1.

Clearly, the two mappings S and T are asymptotically regular. For μ=12Λ we have limnd(Snμ,Sn+1μ)=0. Similarly, we can show that T is asymptotically regular. The mappings S and T are orbitally continuous at 0 and discontinuous at 1. Now, we show that S and T satisfy the condition (6) with α=112, β=16 and γ=34. In fact, we have the following four cases:

Case 1: Let μ,ω[0,1), such that ωμ. We get

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=112|μω|+16|μμ7|+34|ω37ω|=112|μω|+17|μ|+37|ω|112|μω|+17|μ+3ω|17|μ3ω|=d(Sμ,Tω).

Case 2: Let μ[0,1) and ω=1. We have

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=112|μ1|+16|μμ7|+34|117|=112(1μ)+μ7+9146184+584μ>17|μ1|=d(Sμ,Tω).

Case 3: Let ω[0,1) and μ=1. We obtain

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=112|1ω|+16+34|ω37ω|=112(1ω)+16+37ω>37ω=d(Sμ,Tω).

Case 4: If μ=ω=1. We get

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=16+34|117|=16+914>214=17=d(Sμ,Tω).

Therefore, in all the cases, S and T satisfy the condition (6) for all μ,ωΛ. Moreover, all the assumptions of Theorem 2.1 hold, hence, the mappings S and T have a unique common fixed point at 0. Further, limnSnμ=0=limnTnμ for any μΛ.

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

Corollary 2.1. Let (Λ,d) be a complete metric space and S and T are self-mappings on Λ which Sp and Tq are asymptotically regular for some positive integers p and q, respectively. Assume that there exist non-negative numbers α,β,γ where α+β+γ=1, such that, the following condition holds

d(Spμ,Tqω)αd(μ,ω)+βd(μ,Spμ)+γd(ω,Tqω),(10)

for all μ,ωΛ. Then, S and T have a unique common fixed point νΛ, provided that both Sp and Tq are either κ-continuous for some κ1 or orbitally continuous.

Proof. Take f=Sp and g=Tq. From (10) obtain

d(fμ,gω)αd(μ,ω)+βd(μ,fμ)+γd(ω,gω),(11)

for all μ,ωΛ. By Theorem 2.1, we obtain f and g have a unique common fixed point ν. Then,

f(Sν)=Sp(Sν)=Sp+1ν=S(Spν)=S(fν)=Sν,

which implies that Sν is a fixed point of f. Similarly, we derive that Tν is a fixed point of g. By (11), we obtain

d(fSμ,gTω)=d(Sμ,Tω)αd(Sμ,Tω)+βd(Sμ,fSμ)+γd(Tω,gTω)=αd(Sμ,Tω)<d(Sμ,Tω),

which implies that Tν=Sν. From the uniqueness of the common fixed point of f and g, it follows that Sν=Tν=ν. Assuming that νν are two common fixed points of S and T such that fν=gν=ν. The uniqueness of the common fixed point of f and g implies that ν=ν. Hence, the common fixed point of S and T is unique.

Example 2.3. Let Λ=c0={ν=(νn)nN:limnνn=0} be the space of all real sequences convergent to zero, endowed with the usual metric d defined by d(μ,ω)=supn|μnωn| for all μ=(μn)n and ω=(ωn)nΛ. Then (Λ,d) is a complete metric space. Let S and T be such that

S(μ)=S(μn){(μ12,μ22,μ32,)ifthereisatleastoneμnwith|μn|1,(μn2n+1,0,0,.)otherwise,

and

T(μ)=T(μn)={(μ14,μ24,μ34,)ifthereisatleastoneμnwith|μn|1,(μn4n+1,0,0,)otherwise.

Choosing μ=(1,1,1,) we have limnd(Snμ,Sn+1μ)=0 and limnd(Tnμ,Tn+1μ)=0. Clearly, S and T are asymptotically regular, and orbitally continuous. There are non-negative numbers α,β,γ such that α+β+γ=1, we get

d(Sp(μ),Tq(ω))αd(μ,ω)+βd(μ,Sp(μ))+γd(ω,Tq(ω)),

for some p,qN.

Obviously, as all the assumptions of Corollary 2.1 hold, S and T have μ=(0,0,0,)Λ, as their unique common fixed point. Also, limnSpμ=0=limnTqμ for any μΛ.

Theorem 2.4. Let (Λ,d) be a complete metric space and S:ΛΛ be an asymptotically regular mapping. Assume that there exist non-negative numbers α,β,γ such that α+β+γ=1, and α<1, satisfying

d(Sμ,Sω)αd(μ,ω)+βd(μ,Sμ)+γd(ω,Sω),(12)

for all μ,ωΛ. The one of the following conditions hold:

(i) The mapping S is continuous. Further, Snμν for each μΛ, as n.

(ii) For κ1, S is κ-continuous or orbitally continuous.

Then, S has a unique fixed point νΛ.

Proof. First, we shall prove condition (i). Let μ0Λ be arbitrary and define a sequence {μn} by μn+1=Sμn for all n0.

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

d(μn+κ,μn)d(μn+κ,μn+κ+1)+d(μn+κ+1,μn+1)+d(μn+1,μn)d(μn+κ,μn+κ+1)+αd(μn+κ,μn)+βd(μn+κ,μn+κ+1)+γd(μn,μn+1)+d(μn+1,μn).

Thus,

(1α)d(μn+κ,μn)(1+β)d(μn+κ,μn+κ+1)+(1+γ)d(μn,μn+1)0,

as n and α<1. This shows that {μn} is a Cauchy sequence in complete metric space Λ. There exists νΛ such that μnν. The continuity of S and μn+1=Sμn, implies that ν=Sν. Let νν be another fixed point of S. Then

0<d(ν,ν)=d(Sν,Sν)αd(ν,ν)+βd(ν,Sν)+γd(ν,Sν)=αd(ν,ν)<d(ν,ν),

which is a contradiction. Hence, S has a unique fixed point νΛ. Now, we show that Snμν. From (12) we have

d(Snμ,ν)=d(Snμ,Sn+1ν)d(Snμ,Sn+1μ)+d(Sn+1μ,Sn+1ν)d(Snμ,Sn+1μ)+αd(Snμ,Snν)+βd(Snμ,Sn+1μ)+γd(Snν,Sn+1ν).

Hence,

(1α)d(Snμ,ν)(1+β)d(Snμ,Sn+1μ)0,asn.

This shows that Snμν for any μΛ as α<1.

Next, we will consider condition (ii). Choose μ0 as an arbitrary point in Λ. We consider a sequence {μn}Λ given by μn+1=Sμn for any n0. Then, from (i) we have proven that {μn} is a Cauchy sequence in a complete metric space. There exists a point νΛ such that μnν as n. and Sμnν. Moreover, for all κ1 we have Sκμnν as n. Suppose that S is κ-continuous. Since Sκ1μnν, we get limnSκμn=Sν. This implies ν=Sν, that is ν is a fixed point of S. Finally, we assume that S is orbitally continuous. Since μnν, orbital continuity implies that limnSμn=Sν. This yields Sν=ν, that is S has a fixed point at ν.

Theorem 2.5. Let (Λ,d) be a complete metric space and S:ΛΛ be a continuous mapping satisfying (12). Assume that S has an approximate fixed point sequence. Then, S has a unique fixed point ν. In particular, μnν as n.

Proof. Suppose that there exist m,nN such that m>n. Then, by triangle inequality and (12), we obtain

d(μn,μm)d(μn,Sμn)+d(Sμn,Sμm)+d(Sμm,μm)d(μn,Sμn)+αd(μn,μm)+βd(μn,Sμn)+(γ+1)d(μm,Sμm),

which implies

(1α)d(μn,μm)(1+β)d(μn,Sμn)+(1+γ)d(μm,Sμm).

As n,m, we have limn,md(μn,μm)=0. Since Λ is a complete metric space, then {μn} is a Cauchy sequence. Hence, the sequence {μn} converges to νΛ. Since limnd(μn,Sμn)=0, from the continuity of S we get that ν is a fixed point of S. The uniqueness of the fixed point follows from (12).

Theorem 2.6. Let (Λ,d) be a complete metric space and S,T:ΛΛ. Suppose that S is asymptotically regular with respect to T. Assume that there exist non-negative numbers α,β,γ such that α+β+γ=1, as α<1, satisfying

d(Sμ,Sω)αd(Tμ,Tω)+βd(Tμ,Sμ)+γd(Tω,Sω),(13)

for each μ,ωΛ. Further, suppose that S and T are (S,T)-orbitally continuous and compatible. Then C(S,T)ϕ and S and T have a unique common fixed point.

Proof. Since S is asymptotically regular with respect to T at μ0Λ, so, there exists a sequence {ωn}Λ such that ωn=Sμn=Tμn+1 for each n0, and limnd(Tμn+1,Tμn+2)=limnd(ωn,ωn+1)=0. We show that {ωn} is a Cauchy sequence. From (13) and triangle inequality, for any n and any κ>0, we have

d(Sμn+κ,Sμn)=d(ωn+κ,ωn)d(ωn+κ,ωn+κ+1)+d(ωn+κ+1,ωn+1)+d(ωn+1,ωn)d(ωn+κ,ωn+κ+1)+αd(ωn+κ,ωn)+βd(ωn+κ,ωn+κ+1)+γd(ωn,ωn+1)+d(ωn+1,ωn)=(1+β)d(ωn+κ,ωn+κ+1)+αd(ωn+κ,ωn)+(γ+1)d(ωn,ωn+1).

Thus,

(1α)d(ωn+κ,ωn)(1+β)d(ωn+κ,ωn+κ+1)+(1+γ)d(ωn+1,ωn).

Since S is asymptotically regular with respect to T, then limnd(ωn+κ,ωn)=0. Therefore, {ωn} is a Cauchy sequence in a complete metric space. There exists a point νΛ such that ωnν as n. Moreover, ωn=Sμn=Tμn+1ν.

Suppose that S and T are compatible mappings. By the orbital continuity of S and T,

limnSSμn=limnSTμn=Sν,

further

limnTSμn=limnTTμn=Tν.

The compatibility of S and T implies limnd(STμn,TSμn)=0. Taking limit as n we have Sν=Tν, which means, C(S,T)ϕ. Since ν is a coincidence point, the compatibility of S and T implies the commutativity of ν. Hence, TSν=SSν=TTν. Using (13), we obtain

d(Sν,SSν)αd(Tν,TSν)+βd(Tν,Sν)+γd(TSν,SSν)=αd(Sν,SSν),

which is a contradiction, thence, Sν=SSν. Hence Sν=SSν=TSν and Sν is a common fixed point of S and T. 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 and S, T be asymptotically regular self-mapping on Λ. Assume that there exist non-negative numbers α,β,γ such that 2α+β+2γ=1, satisfying

d(Sμ,Tω)αd(μ,ω)+βd(ω,Sμ)+γd(μ,Tω),(14)

for each μ,ωΛ. Suppose further that S and T are either κ-continuous for some κ1 or orbitally continuous. Then, S and T have a unique common fixed point ν. Moreover, for any μΛ, limnSnμ=ν=limnTnμ.

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 limnd(Snμ,Tnμ)=0, for any μΛ. The result is trivial if S=T. Suppose ST and α=0, from (14) we obtain

d(Sμ,Tω)βd(ω,Sμ)+γd(μ,Tω),

for each μ,ωΛ. Define μ1=Snμ and ω1=Tnμ, for any μΛ. Then

d(Sn+1μ,Tn+1μ)βd(Tnμ,Sn+1μ)+γd(Snμ,Tn+1μ)β{d(Tnμ,Tn+1μ)+d(Tn+1μ,Sn+1μ)}+γ{d(Snμ,Sn+1μ)+d(Sn+1μ,Tn+1μ)},

which implies

d(Sn+1μ,Tn+1μ)β2α+γd(Tnμ,Tn+1μ)+γ2α+γd(Snμ,Sn+1μ).

As n, since S and T are asymptotic regularity, we have

limnd(Sn+1μ,Tn+1μ)=0.(15)

By the asymptotic regularity of S and T and triangle inequality, we have

d(Snμ,Tnμ)d(Snμ,Sn+1μ)+d(Sn+1μ,Tn+1μ)+d(Tn+1μ,Tnμ)0,asn.

Next, suppose that ST and α0. Let μ1=Snμ and ω1=Tnμ, for any μΛ. Then, by (14) we obtain

d(Sn+1μ,Tn+1μ)αd(Snμ,Tnμ)+βd(Tnμ,Sn+1μ)+γd(Snμ,Tn+1μ)αd(Snμ,Tnμ)+β{d(Tnμ,Tn+1μ)+d(Tn+1μ,Sn+1μ)}+γ{d(Snμ,Sn+1μ)+d(Sn+1μ,Tn+1μ)},

obtain

(1βγ)d(Sn+1μ,Tn+1μ)αd(Snμ,Tnμ)+βd(Tnμ,Tn+1μ)+γd(Snμ,Sn+1μ)

which implies

d(Sn+1μ,Tn+1μ)α2α+γd(Snμ,Tnμ)+β2α+γd(Tnμ,Tn+1μ)+γ2α+γd(Snμ,Sn+1μ).

Let hn=d(Snμ,Tnμ), ψn=1α2α+γ and ϑn=βαd(Tnμ,Tn+1μ)+γαd(Snμ,Sn+1μ). By the asymptotically regularity of S and T, we have limnϑn=0. Furthermore, n=1ψn=. Hence, by Lemma 2.1, we get that limnd(Snμ,Tnμ)=0 for any μΛ.

Step 2: Let μn=Snμ for any μ in Λ and n0. Then, we show that {μn} is a Cauchy sequence which is convergent to νΛ. Moreover, {Tnμ}νΛ.

Assume that {Snμ} is not a Cauchy sequence. Then, there exists an ε>0 and sequences of integers {mκ} and {nκ} such that mκ>nκκ, for κ=1,2,, we have

d(Smκ,Snκ)ε.(16)

Choosing mκ, the smallest number exceeding nκ for which (16) holds, we also assume that

d(Sm(κ1)μ,Snκμ)<ε.

Now, we have that

εd(Smκμ,Snκμ)d(Smκμ,Sm(κ1)μ)+d(Sm(κ1)μ,Snκμ)<d(Smκμ,Sm(κ1)μ)+ε.

As κ, it follows by asymptotic regularity of S that

limκd(Smκμ,Snκμ)=ε.(17)

Furthermore, by asymptotic regularity of S and the above inequality, we obtain

d(Sm(κ1)μ,Sn(κ1)μ)d(Sm(κ1)μ,Smκμ)+d(Smκμ,Snκμ)+d(Snκμ,Sn(κ1)μ),

implying that

limκd(Sm(κ1)μ,Sn(κ1)μ)=ε.(18)

Then, using (14), we have

d(Smκμ,Snκμ)d(Smκμ,Tnκμ)+d(Tnκμ,Snκμ)d(Tnκμ,Snκμ)+αd(Sm(κ1)μ,Tn(κ1)μ)+βd(Tn(κ1)μ,Smκμ)+γd(Sm(κ1)μ,Tnκμ)d(Tnκμ,Snκμ)+α{d(Sm(κ1)μ,Sn(κ1)μ)+d(Sn(κ1)μ,Tn(κ1)μ)}+β{d(Tn(κ1)μ,Snκ1μ)+d(Sn(κ1)μ,Sm(κ1)μ)+d(Sm(κ1)μ,Smκμ)}+γ{d(Sm(κ1)μ,Sn(κ1)μ)+d(Sn(κ1)μ,Snκμ)+d(Snκμ,Tnκμ)},

as κ, From (17), (18) and the asymptotic regularity of S and T, we have, ε(α+β+γ)ε. This is a contradiction. Hence, {μn} is a Cauchy sequence in complete space Λ. There exists ν in Λ such that μnν. Moreover

d(Tnμ,ν)d(Tnμ,Snμ)+d(Snμ,ν),

from (15) and μnν implies Tnμ converges to νΛ.

Step 3: We prove the uniqueness of the common fixed point of S and T.

Let S be κ-continuous. Since limnSκ1μn=ν, we have that

limnSκμn=Sν.

Since the limit is unique, it implies Sν=ν.

Similarly, suppose that S is orbitally continuous. Since limnμn=ν, we have that

limnSμn=Sμ,

implies that Sν=ν. Furthermore, since limnTnμ=ν, we have that Tν=ν whenever T is κ-continuous or orbitally continuous. Hence, νFix(S)Fix(T). For the uniqueness of fixed point, suppose that νν are two common fixed points of S and T. Let μ=ν and ω=ν. Then, (14) implies d(ν,ν)(α+β+γ)d(ν,ν). Hence, this is a contradiction. We must have, ν as the unique common fixed point of S and T.

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

d((f1,f2),(g1,g2))=f1g1+f2F1g2G1.

where F1(t)=1+0tf1(u)du and G1(t)=1+0tg1(u)du for each t[0,1]. Let S and T be such that

S(f1,f2)={(15,15)if(f1,f2)(15,0),(15,215)if(f1,f2)=(15,0),

and

T(f1,f2)={(15,15)if(f1,f2)(15,0),(15,315)if(f1,f2)=(15,0).

If we choose f=(f1,f2)=(12,12) we have limnd(Snf,Sn+1f)=limnd(Tnf,Tn+1f)=0. Then, S and T are asymptotically regular, orbitally continuous at (15,15) and discontinuous at (15,0). Now, we show that S and T satisfy the condition (14). We choose α=115, β=15 and γ=13.

In fact, we have the following four cases:

Case 1: Let (f1(t),f2(t))(15,0), and (g1(t),g2(t))(15,0), we have

αd(f,g)+βd(g,S(f))+γd(f,T(g))=115d(f,g)+15d((g1,g2),(15,15))+13d((f1,f2),(15,15))0=d(S(f),T(g)).

Case 2: If (f1(t),f2(t))(15,0), and (g1(t),g2(t))=(15,0), implies

d(S(f),T(g))=d((15,15),(15,315))=||1515||+||15+01125du3153150115du||=||125375||=0.

Therefore,

αd(f,g)+βd(g,S(f))+γd(f,T(g))=115d(f,g)+15d(g,S(f))+13d(f,T(g))0=d(S(f),T(g)).

Case 3: If (f1(t),f2(t))=(15,0), and (g1(t),g2(t))(15,0), we obtain

d(S(f),T(g))=d((15,215),(15,15))=||1515||+||215+2150115du15150115du||=||675||=675.

Thus,

αd(f,g)+βd(g,S(f))+γd(f,T(g))=115d(f,g)+15d(g,S(f))+13d(f,T(g))=115d(f,g)+15d(g,S(f))+13d((15,0),(15,15))=115d(f,g)+15d(g,S(f))+13||625||=115d(f,g)+15d(g,S(f))+675d(S(f),T(g)).

Case 4: If (f1(t),f2(t))=(15,0), and (g1(t),g2(t))=(15,0), we obtain

d(S(f),T(g))=d((15,215),(15,315))=||1515||+||215+2150115du3153150115du||=||475||=475.

However,

αd(f,g)+βd(g,S(f))+γd(f,T(g))=115d((15,0),(15,0))+15d((15,0),(15,215))+13d((15,0),(15,315))=15||1275||+13||1875||=4125+675=675>475=d(S(f),T(g)).

Therefore, in all cases, S and T satisfy the condition (14) for all μ,ωΛ. Moreover, all the assumptions of Theorem 2.7 hold. Point (15,15) is the unique common fixed point of S and T. Moreover, limnSnμ=0=limnTnμ for any μΛ.

Corollary 2.2. Let (Λ,d) be a complete metric space and S and T be self-mappings on Λ where Sp and Tq are asymptotically regular for some positive integers p and q, respectively. Assume that there exist non-negative numbers α,β,γ with 2α+β+2γ=1 such that the following condition holds

d(Spμ,Tqω)αd(μ,ω)+βd(ω,Spμ)+γd(μ,Tqω),(19)

for all μ,ωΛ. Then, S and T have a unique common fixed point νΛ, provided that both Sp and Tq are either κ-continuous for some κ1 or orbitally continuous.

Proof. Take f=Sp and g=Tq. Then, for each μ,ωΛ, the (19) becomes

d(fμ,gω)αd(μ,ω)+βd(ω,fμ)+γd(μ,gω)αd(μ,ω)+β{d(ω,μ)+d(μ,fμ)}+γ{d(μ,ω)+(ω,gω)}(α+β+γ)d(ω,μ)+βd(μ,fμ)+γd(ω,gω)d(ω,μ)+βd(μ,fμ)+γd(ω,gω).(20)

According to Theorem 2.7, f and g have a unique common fixed point ν. Now, we show that Sν is a fixed point of f, obtain

f(Sν)=Sp(Sν)=Sp+1ν=S(fν)=Sν,

implies that Sν is a fixed point of f. Similarly, we can prove that Tν is also a fixed point of g. Using (20), given that Sν=Tν, since f and g have a unique common fixed point, it follows that Sν=Tν=ν. Suppose that ν is another common fixed point of S and T such that νν. we have, fν=gν=ν. The uniqueness of the common fixed point of f and g implies ν=ν. Then ν is a unique common fixed point of S and T.

Theorem 2.9. Let (Λ,d) be a complete metric space. The mapping S:ΛΛ is asymptotically regular. Assume that there exist non-negative numbers α,β and γ with 2α+β+2γ=1 such that

d(Sμ,Sω)αd(μ,ω)+βd(ω,Sμ)+γd(μ,Sω),(21)

for all μ,ωΛ, provided that one of the following conditions hold:

(i)   The mapping S is a continuous. Further Snμν for each μΛ, as n0.

(ii)   For κ1, S is κ-continuous or orbitally continuous.

Then S has a unique fixed point μΛ

Proof. First, we will proof the condition (i) holds. Let μ0Λ be arbitrary. Then define a sequence {μn} by μn+1=Sμn for all n0.

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

d(μn+κ,μn)d(μn+κ,μn+κ+1)+d(μn+κ+1,μn+1)+d(μn+1,μn)d(μn+κ,μn+κ+1)+αd(μn+κ,μn)+βd(μn,μn+κ+1)+γd(μn+κ,μn+1)+d(μn+1,μn)d(μn+κ,μn+κ+1)+αd(μn+κ,μn)+β{d(μn,μn+κ)+d(μn+κ,μn+κ+1)}+γ{d(μn+κ,μn)+d(μn,μn+1)}+d(μn+1,μn)=(1+β)d(μn+κ,μn+κ+1)+(α+β+γ)d(μn+κ,μn)+(1+γ)d(μn+1,μn).

Then,

d(μn+κ,μn)1+βα+γd(μn+κ,μn+κ+1)+1+γα+γd(μn,μn+1)0,

as n. This shows that {μn} is a Cauchy sequence in a complete metric space Λ, there exists νΛ such that μnν. As S is continuous and μn+1=Sμn, we obtain that ν=Sν. Suppose that νν is another fixed point of S. Then, we have

0<d(ν,ν)=d(Sν,Sν)αd(ν,ν)+βd(ν,Sν)+γd(ν,Sν)=(α+β+γ)d(ν,ν)<d(ν,ν),

which is a contradiction. Hence, S has a unique fixed point νΛ. Now, we show that Snν. From (21), we have

d(Snμ,ν)=d(Snμ,Snν)d(Snμ,Sn+1μ)+d(Sn+1μ,Sn+1ν)d(Snμ,Sn+1μ)+αd(Snμ,Snν)+βd(Snν,Sn+1μ)+γd(Snμ,Sn+1ν)d(Snμ,Sn+1μ)+αd(Snμ,ν)+β{d(ν,Sn+1μ)+d(Snμ,Sn+1μ)}+γd(Snμ,ν)=(1+β)d(Snμ,Sn+1μ)+(α+β+γ)d(Snμ,ν).

Hence,

d(Snμ,ν)1+βα+γd(Snμ,Sn+1μ)0,asn.

This shows that Snν for any μΛ.

We next prove the condition (ii) holds. Suppose that μ0 is any point in Λ. The sequence {μn}Λ is given by μn+1=Sμn=Snμ, for each n0. Then, in the proof condition (i) we have shown that {μn} is a Cauchy sequence in complete space Λ. There exists a point νΛ such that μnν as n, in addition, Sμnν. Moreover, for each κ1 we have Sκμnν as n. Suppose that S is κ-continuous, and Sκ1μnν, implies limnSκμn=Sν. Hence, ν is a fixed point of S.

Finally, we show that ν as a fixed point of S. Assume that S is orbitally continuous. Since μnν, orbital continuity implies that limnSμn=Sν. Then ν is a fixed point of S.

Theorem 2.10. Let (Λ,d) be a complete metric space and S:ΛΛ be a continuous mapping satisfying (21). Suppose that S has an approximate fixed point sequence. Then, S has a unique fixed point ν. In particular, μnν as n.

Proof. Let m>n for all n,mN. Then, using triangle inequality and (21), we obtain

d(μn,μm)d(μn,Sμn)+d(Sμn,Sμm)+d(Sμm,μm)d(μn,Sμn)+αd(μn,μm)+βd(μm,Sμn)+γd(μn,Sμm)+d(Sμm,μm)d(μn,Sμn)+αd(μn,μm)+β{d(μm,μn)+d(μn,Sμn)}+γ{d(μn,μm)+d(μm,Sμm)}+d(Sμm,μm)(1+β)d(μn,Sμn)+(α+β+γ)d(μn,μm)+(1+γ)d(μm,Sμm).

This implies

(α+γ)d(μn,μm)(1+β)d(μn,Sμn)+(1+γ)d(μm,Sμm).

As n,m, we have limn,md(μn,μm)=0. Hence, {μn} is a Cauchy sequence in Λ. By completeness of Λ, this sequence {μn} converges to νΛ. Since limnd(μn,Sμn)=0, continuity of S implies that ν is a fixed point of S. The uniqueness of the fixed point follows from the inequality (21).

Theorem 2.11. Let (Λ,d) be a complete metric space and S,T:ΛΛ. Suppose that S is asymptotically regular with respect to T. Assume that there exist non-negative numbers α,β,γ such that 2α+β+2γ=1, satisfying

d(Sμ,Sω)αd(Tμ,Tω)+βd(Tω,Sμ)+γd(Tμ,Sω),(22)

for each μ,ωΛ. Furthermore, suppose that S and T are (S,T)-orbitally continuous and compatible. Then C(S,T)ϕ and S and T have a unique common fixed point.

Proof. Since S is asymptotically regularity with respect to T at μ0Λ. So, there exists a sequence {ωn}Λ such that ωn=Sμn=Tμn+1 for each n0, and limnd(Tμn+1,Tμn+2)=limnd(ωn,ωn+1)=0. We show that {ωn} is a Cauchy sequence. By triangle inequality and (22), for each n and κ>0, we obtain

d(Sμn+κ,Sμn)=d(ωn+κ,ωn)d(ωn+κ,ωn+κ+1)+d(ωn+κ+1,ωn+1)+d(ωn+1,ωn)d(ωn+κ,ωn+κ+1)+αd(ωn+κ,ωn)+βd(ωn,ωn+κ+1)+γd(ωn+κ,ωn+1)+d(ωn+1,ωn)d(ωn+κ,ωn+κ+1)+αd(ωn+κ,ωn)+β{d(ωn,ωn+κ)+d(ωn+κ,ωn+κ+1)}+γ{d(ωn+κ,ωn)+d(ωn,ωn+1)}+d(ωn+1,ωn)(1+β)d(ωn+κ,ωn+κ+1)+(α+β+γ)d(ωn+κ,ωn)+(1+γ)d(ωn,ωn+1).

Thus,

d(ωn+κ,ωn)1+βα+βd(ωn+κ,ωn+κ+1)+1+γα+βd(ωn+1,ωn).

Since S is asymptotically regularity with respect to T, it implies that d(ωn+κ,ωn)0 as n. Therefore, {ωn} is a Cauchy sequence. Since Λ is complete, there exists a point νΛ such that ωnν as n, and ωn=Sμn=Tμn+1ν.

Assuming that S and T are compatible mappings, orbital continuity of S and T implies that

limnSSμn=limnSTμn=Sν,

further,

limnSTμn=limnSSμn=Sν,

Then, compatibility of S and T yields limnd(STμn,TSμn)=0. Taking limit as n we have Sν=Tν, which implies, C(S,T)ϕ. Again the compatibility of S and T implies commutativity at a coincidence point ν. Hence TSν=SSν=TTν. Using (22), we have

d(Sν,SSν)αd(Tν,TSν)+βd(TSν,Sν)+γd(Tν,SSν)<d(Sν,SSν),

that is, Sν=SSν. Then Sν=SSν=TSν and Sν are common fixed points of S and T. The uniqueness of the common fixed point follows from (22).

Example 2.12. Let Λ=[2,20] and d be the usual metric. Define S,T:ΛΛ by

Sμ={2ifμ=2,3if2<μ5,2ifμ>5,andTμ={2ifμ=2,11if2<μ5,μ+13ifμ>5.

Then S and T satisfy all the conditions of Theorem 2.11 for α=115, β=15 and γ=13. In fact, if μ=2, ω>5, we have

αd(Tμ,Tω)+βd(Tω,Sμ)+γd(Tμ,Sω)=115|2ω+13|+15|ω+132|+13|22|=175|ω5|+115|ω5|0=d(Sμ,Sω).

Similarly, if take 2<μ5 and ω>5 we obtain

αd(Tμ,Tω)+βd(Tω,Sμ)+γd(Tμ,Sω)=115|11ω+13|+15|ω+133|+13|112|=175|32ω|+115|ω8|+83>83>1=d(Sμ,Sω).

If μ>5 and ω>5, we get

αd(Tμ,Tω)+βd(Tω,Sμ)+γd(Tμ,Sω)=115|μ+13ω+13|+15|ω+132|+13|μ+132|=175|μω|+115|ω5|+19|μ5|>0=d(Sμ,Sω).

Otherwise, if μ=1 and ω=2 or 2<ω5 (or 2<μ,ω5), it is easy to verify that the mappings S and T satisfy the conditions of Theorem 2.11. The mappings S and T have a unique common fixed point μ=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

μn+1=Sμn=Snμ0,nN.(23)

with an arbitrary μ0Λ. 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 of S.

Define S0=I (the identity map on Λ) and Sn=Sn1S, called the nth iterate of S for nN. The Krasnoselskii-Ishikawa iteration method associated with S is the sequence {μn}n=0 defined by

μn+1=(1λ)μn+λSμn,(24)

for each n0, and λ[0,1]. The Krasnoselskii-Ishikawa sequence {μn}n=0 is exactly the Picard iteration corresponding to an averaged operator:

Sλ=(1λ)I+λS,(25)

However, if λ=1, the Krasnoselskii-Ishikawa iteration given by (24) is reduced to the Picard iteration. Moreover, Fix(S)=Fix(Sλ), for each λ(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 numbers α,β and γ with 2α+2β+γ=1, where 2α<1 and the Reich nonexpansive mapping S:ΛΛ satisfying the following inequality

d(Sμ,Sω)αd(μ,ω)+βd(μ,Sμ)+γd(ω,Sω),(26)

for all μ,ωΛ. Then for any positive integer n, there exists λ(0,1) such that for all μΛ and for each p, q in N, we have

SλpμSλqμηδ[O(Sλ,μ,n)],(27)

where η=max{2α,β,γ} and δ[A]=sup{μω:μ,ωA}.

Proof. Let us choose λ=12α1α, where 2α<1, clearly, 0<λ<1. Considering the operator given by (25), we have

λμSμ=μSλμ,(28)

and

λωSω=ωSλω,(29)

for all μ,ωΛ. Moreover, we obtain

SλμSλω=(1λ)(μω)+λ(SμSω)(1λ)μω+λSμSω,(30)

Since S is a Reich nonexpansive mapping, from (26), we have

λSμSωαλμω+βλμSμ+λγωSω.(31)

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

λSμSωαλμω+βμSλμ+γωSλω.(32)

Now, (30) and (32) imply that

SλμSλω(1λ)μω+αλμω+βμSλμ+γωSλω(1λ+αλ)μω+βμSλμ+γωSλω.(33)

Since λ=12α1α and 2α<1, we assume that 1λ+αλ=2α<1. Then (33) becomes

SλμSλω2αμω+βμSλμ+γωSλω.(34)

Let μΛ be arbitrary and fixed positive integer n. Therefore, using (34), we have

SλpμSλqμ=SλSλp1μSλSλq1μ2αSλp1μSλq1μ+βSλp1μSλpμ+γSλq1μSλqμ,

which implies that

SλpμSλqμηδ[O(Sλ,μ,n)],

such that η=max{2α,β,γ}.

Remark 3.1. It follows from Lemma 3.1 that if S is a Reich nonexpansive mapping given by (26) and μΛ, then for any nN, there exists a positive integer κn, such that

μSλκμ=δ[O(Sλ,μ,n)].

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

δ[O(Sλ,μ,)]11ημSλκμ,(35)

holds for each μS, and η=max{2α,β,γ}.

Proof. Let μΛ be arbitrary and λ(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)]:nN}.

Then, (35) implies that

δ[O(Sλ,μ,n)]11ημSλκμ.

Let n be any positive integer. From Remark 3.1, there exists SλκO(Sλ,μ,n) where 1κn, such that

μSλκμ=δ[O(Sλ,μ,n)].

Applying a triangle inequality and Lemma 3.1, we obtain

μSλκμμSλμ+SλμSλκμμSλμ+ηδ[O(Sλ,μ,n)]=μSλμ+ημSλκμ.

Therefore,

δ[O(Sλ,μ,n)]=μSλκμ11ημSλμ.

Since n was 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-mapping S be a Reich nonexpansive given by (26), with 2α+2β+γ=1 and 2α<1. Then, the Krasnoselskii-Ishikawa iteration {μn} defined by

μn+1=(1λ)μn1+λSμn1,n1,

converges to a unique fixed point ν for any μ0Λ, provided that Λ is Sλ-orbitally complete.

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

SλμSλω2αμω+βμSλμ+γωSλω,(36)

where λ=12α1α and 2α<1.

Let μ0 be an arbitrary point of Λ. Given the iteration (23), the Krasnoselskii-Ishikawa sequence {μn} is exactly the Picard iteration associated with Sλ, that is

μn=Sλμn1=Sλnμ0,n0.(37)

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

μnμm=Sλnμ0Sλmμ0=SλSλn1μ0Sλmn+1Sλn1μ0=Sλμn1Sλmn+1μn1ηδ[O(Sλ,μn1,mn+1)],(38)

where η=max{2α,β,γ}. According to Remark 3.1, there exists an integer κ, 1κmn+1, such that

δ[O(Sλ,μn1,mn+1)]=μn1μn+κ1.

Again, by Lemma 3.1, we have

μn1μn+κ1=Sλμn2Sλκ+1μn2ηδ[O(Sλ,μn2,κ+1)],

which implies that

μn1μn+κ1ηδ[O(Sλ,μn2,mn+2)].

Moreover, by (38) we get

μnμmηδ[O(Sλ,μn1,mn+1)]η2δ[O(Sλ,μn2,mn+2)].

Continuing this process, we obtain

μnμmηδ[O(Sλ,μn1,mn+1)]ηnδ[O(Sλ,μ0,m)],

and it follows from Lemma 3.2 that

μnμmηn1ημ0Sλμ0.

Taking limit as n, we find that {μn} is a Cauchy sequence. Since Λ is Sλ-orbital complete, there exists νΛ such that limnμn=ν. Next, we prove that ν is a fixed point of Sλ. In (36), we consider the following inequalities:

νSλννμn+1+μn+1Sλν=νμn+1+SλμnSλννμn+1+2αμnν+βμnμn+1+γνSλν.

Hence,

νSλν11γ(νμn+1+2αμnν+βμnμn+1).

Since limnμn=ν, we have νSλν=0, that is Sλ has a fixed point. We claim that there is a unique common fixed point of Sλ. Assume on the contrary that, Sλν=ν and Sλν=ν but νν. By supposition, we obtain

νν=SλνSλν2ανν+βνSλν+γνSλν2ανν<νν,

which is a contradiction, hence, ν=ν. Since Fix(Sλ)=Fix(S), we get that S has 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 and S and T be self-mappings on Λ. Assume that there exist non-negative numbers α,β,γ such that 2α+2β+γ=1 and 2α<1, satisfying

SμTωαμω+βμSμ+γωTω,(39)

for all μ,ωΛ. Then, the iteration sequence {μn}n=0 defined by (24) converges to a unique common fixed point ν for any μ0Λ, provided that Λ is (Sλ,Tλ)-orbitally complete.

Proof. Suppose that μ0 is an arbitrary point in Λ. Consider the iterative process {μn}n=0 defined by (24), which is, in fact, the Picard iteration associated with Sλ, that is

μn+1=Sλμn.(40)

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

Sλμ2n+1=(1λ)μ2n+1+λSμ2n+1=μ2n+2,Tλμ2x=(1λ)μ2n+λTμ2n=μ2n+1.(41)

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

Sλμ2n+1Tλμ2n=μ2n+2μ2n+1(1λ)μ2n+1μ2n+αλμ2n+1μ2n+βμ2n+1Sλμ2n+1+γμ2nTλμ2n=(1λ+αλ)μ2n+1μ2n+βμ2n+1μ2n+2+γμ2nμ2n+1,(42)

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

(1β)μ2n+2μ2n+1(2α+γ)|μ2n+1μ2nμ2n+2μ2n+12α+γ1βμ2n+1μ2n12β1βδ[O(Sλ,Tλ,μ2n,2n+1)]<ηδ[O(SλTλ,μ2n,2n+1)],

where η=12β1β<1 and β<1.

Similarly,

μ2n+3μ2n+2ημ2n+2μ2n+1=ηδ[O(Sλ,Tλ,μ2n+1,2n+2)]η2δ[O(Sλ,Tλ,μ2n,2n+1)].

Continuing the process, we get the following:

μm+1μm<ηδ[O(Sλ,Tλ,μm1,m)]<η2δ[O(Sλ,Tλ,μm2,m1)]<<ηmδ[O(Sλ,Tλ,μ0,m+1)],mN.(43)

Now, we show that {μn} is a Cauchy sequence converging to νΛ. Then there exists mN such that n<m, from (43), and Lemma 3.2, we obtain

μmμn<ηδ[O(Sλ,Tλ,μn,n+1)]<.<ηnδ[O(Sλ,Tλ,μ0,m)]ηn1ημ0Tλμ0,n<m.

Since limnηn1η=0, so, {μn} is a Cauchy sequence on Λ. Since Λ is (Sλ,Tλ)-orbital complete, there exists νΛ such that limnμn=ν. Now, we prove that ν is a fixed point of Sλ. From (41) and (42), we consider the following inequalities:

SλννSλνμ2n+1+μ2n+1ν=SλνTλμ2n+μ2n+1ν2αμ2nν+βνSλν+γμ2nTλμ2n2α1βμ2nν+γ1βμ2nμ2n+1,

where we have νSλν=0, as n, that is Sλ has a fixed point. Similarly, the mapping Tλ has a fixed point. We claim that there is a unique common fixed point of Sλ and Tλ. We assume on the contrary, such that Tλν=Sλν=ν and Tλν=Sλν=ν but νν. By supposition, we can replace μ2n+2 by ν and μ2n+1 by ν in (42) to obtain

νν=SλνTλν2ανν+βνSλν+γνTλν2ανν<νν,

this is a contradiction, hence ν=ν. Since Fix(Sλ)=Fix(S), we get that S and T have a unique common fixed point.

Theorem 3.4. Let (Λ,.) be a Banach space and S,T:ΛΛ. Assume that there exist non-negative numbers α,β,γ such that 2α+3β+2γ=1, satisfying

SμTωαμω+βωSμ+γμTω,(44)

for all μ,ωΛ. Then, the iteration sequence {μn}n=1 defined by (24) converges to a unique common fixed point ν for any μ0Λ, provided that Λ is (Sλ,Tλ)-orbitally complete.

Proof. Let μ0 be arbitrary. Since (Sλ,Tλ)-orbitally complete in Λ, we define the operators Sλ and Tλ such that

Sλμ2n=μ2n+1,Tλμ2n+1=μ2n+2.(45)

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

Sλμ2n+1Tλμ2n=μ2n+2μ2n+1(1λ)μ2n+1ν2n+λ(α+β+γ)μ2n+1μ2n+βμ2n+1Sλμ2n+1+γμ2nTλμ2n(2α+β+γ)μ2n+1μ2n+βμ2n+1μ2n+2)+γμ2nμ2n+1.(46)

Let us choose 1λ+λ(α+β+γ)2α+β+γ. Since 2α+3β+2γ=1, we have 2(α+γ)=13β. This implies that

(1β)μ2n+2μ2n+1(2α+2γ+β)|μ2n+1μ2nμ2n+2μ2n+112β1βμ2n+1μ2nηδ[O(Sλ,Tλ,μ2n,2n+1)],

where η=12β1β and β<1.

Similarly, we obtain

μ2n+3μ2n+2ημ2n+2μ2n+1=ηδ[O(Sλ,Tλ,μ2n+1,2n+2)]η2δ[O(Sλ,Tλ,μ2n,2n+1)].

Continuing the process, we get the following:

μm+1μm<ηδ[O(Tλ,Sλ,μm1,m)]<η2δ[O(Tλ,Sλ,μm2,m1)]<<ηmδ[O(Tλ,Sλ,μ0,m+1)],mN.(47)

Next, we show that {μn} is a Cauchy sequence converging to νΛ. There exists mN such that n<m, by Lemma 3.2, and (47), we have

μmμn<ηδ[O(Sλ,Tλ,μn,n+1)]<.<ηnδ[O(Sλ,Tλ,μ0,m)]ηn1ηδμ0Tλμ0,n<m.

Since limnηn1η=0, so, {μn} is a Cauchy sequence in Λ. Since Λ is (Sλ,Tλ)-orbital complete, there exists νΛ such that limnμn=ν. We prove now that ν is a fixed point of Sλ. From (45) and (46), we consider the following inequalities:

SλννSλνμ2n+1+μ2n+1ν=SλνTλμ2n+μ2n+1ν(α+β+γ)μ2nν+βμ2nSλν+γνTλμ2n+μ2n+1ν(α+β+γ)μ2nν+β(μ2nν+νSλν)+(γ+1)νμ2n+1α+2β+γ1βμ2nν+γ+11βνμ2n+1.

Since μ2nν as n we have νSλν=0, as n, that is Sλ has a fixed point. Similarly, the mapping Tλ has a fixed point. We claim that there is a unique common fixed point of Sλ and Tλ. We assume the contrary that Tλν=Sλν=ν and Tλν=Sλν=ν but νν with supposition, we can replace μ2n+2 by ν and μ2n+1 by ν in (41) to obtain

νν=SλνTλν(2α+β+γ)νν+βνSλν+γνTλν(2α+β+γ)νν<νν,

which is a contradiction, hence ν=ν. Since Fix(Sλ)=Fix(S), we obtain that S and T have a unique common fixed point.

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

Sμ={12if0μ12,12μ+14if12μ1.Tμ={12if0μ12,1+μ3if12μ1.

Take μ0=23, then O(Sλ,Tλ,μn,n){2/(2p3q1),p,qN}.

Let α=19, β=127, and γ=13. Now, we show that S and T satisfy the condition (39). We consider the following cases:

Case 1: Let μ[0,12], and ω[0,12], such that μω. We have

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=19|μω|+127|μ12|+13|ω12|19(ωμ)+127(12μ)+13(12ω)=52729ω427μd(Sμ,Tω).

Case 2: Let μ[0,12], ω[12,1], such that μω. We get

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=19|μω|+127|μ12|+13|ω1313ω|19(ωμ)+127(12μ)+19(2ω1)=154[18ω8μ5]|1613ω|=d(Sμ,Tω).

Case 3: Let μ[12,1], ω[12,1] and μω. We obtain

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=19|μω|+127|μμ214|+13|ω1313ω|19(ωμ)+1108(2μ1)+19(2ω1)=13ω554μ13108=1108[36ω10μ13]112|6μ4ω1|=d(Sμ,Tω).

Therefore, in all the cases, S and T satisfy the condition of (39) for all μ,ωΛ. Moreover, as all the assumptions of Theorem 3.3 hold, so S and T have μ=12 as 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 metric d:Λ×Λ[0,) by d(μ,ω)=supt[0,1]|μ(t)ω(t)|, for each μ,ωΛ. 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:

{Sμ(t)=αμ(t)01K(t,r,μ(r),Sμ(r))dr,Tω(t)=αω(t)01K(t,r,ω(r),Tω(r))dr,(48)

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

(i)   K1(t,r,μ(r),Sμ(r))0 and K2(t,r,ω(r),Tω(r))0 for t,r[0,1] such that Ki(.,.,0,.)0, for i=1,2. In addition, S(E),T(E)E where E={νΛ:0ν(t)1}.

(ii)   The two mappings G and G are defined by

Gμ(t)=0tK1(t,r,μ(r),Sμ(r))dr,

and

Gω(t)=0tK2(t,r,ω(r),Tω(r))dr,

satisfying Gμ,Gω in E, for all μ(t),ω(t)E and

Gμ(t)Gω(t)<α(1α)μ(t)ω(t),

for all μ(t),ω(t)E, such that μ(t)ω(t) and t[0,1].

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

Proof. We have

Sμ(t)=αμ(t)0tK1(t,r,μ(r),Sμ(r))dr.

We also have,

Tω(t)=αω(t)0tK2(t,r,ω(r),Tω(r))dr.

Suppose that μ,ωE. Then, by using our assumptions, we obtain

|Sμ(t)Tω(t)|=|α(μ(t)ω(t))01K1(t,r,μ(r),Sμ(r))dr+01K2(t,r,ω(r),Tω(r))dr|α|μ(t)ω(t)|+|01[K1(t,r,μ(r),Sμ(r))K2(t,r,ω(r),Tω(r))]dr|α|μ(t)ω(t)Sμ(t)+Sμ(t)Tω(t)+Tω(t)|+|Gμ(t)Gω(t)|αμ(t)Sμ(t)+ω(t)Tω(t)+Sμ(t)Tω(t)+α(1α)μ(t)ω(t),

which implies that

(1α)Sμ(t)Tω(t)αμ(t)Sμ(t)+αω(t)Tω(t)+α(1α)μ(t)ω(t).

Therefore,

Sμ(t)Tω(t)α1αμ(t)Sμ(t)+ω(t)Tω(t)+α(1α)1αμ(t)ω(t).

Now, since 0α<1 and β=α1αγ, then

Sμ(t)Tω(t)αμ(t)ω(t)+βμ(t)Tμ(t)+γω(t)Sω(t).

By Theorem 2.1 there exists a common fixed point of S and T which 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 [2935] and the references therein.

The classical Caputo fractional derivative is defined by

Dξμ(t):=1Γ(mξ)0t(tr)mξ1dmdrmμ(r)dr

where m1<Re(ξ)<m,mN.

Now, we consider the following fractional differential equation:

{Dξμ(t)+h(t,μ(t))=0,(0μ1,1<ξ<2),μ(0)=μ(1)=0,(49)

where Dξ is the Caputo fractional derivative of order ξ and h:[0,1]×RR 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

G(μ,ω)={1Γ(ξ)(μ(1ω)ξ1(μω)ξ1),0ωμ1,1Γ(ξ)μ(1ω)ξ1,0μω1.(50)

Assume that

|h(t,μ)h(t,ω)|q2(1q)|μω|,

for all t[0,1], q(0,23) and μ,ωR.

Theorem 5.1. Let Λ=C[0,1] and the operators S,T:ΛΛ be defined as

S(μ(t))=01G(μ,ω)h(μ(t),ω(t))dt,

and

T(μ(t))=01G(μ,ω)h(μ(t),ω(t))dt,

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

Proof. It is easy to see that μΛ is a solution of (49) if and only if μ is a solution of the following integral equation:

μ(t)=01G(t,r)h(s,μ(r))dr.

Let μ,ωΛ and t[0,1]. Then,

|Sμ(t)Tω(t)|01G(t,r)(h(r,μ(r))h(r,ω(r)))drq2(1q)01G(t,r)|μ(r)ω(r)|drSμ(t)Tω(t)q2(1q)μωq2(1q)μω+13q(1q)μSω+q2(1q)ωTμ,

where q<13. Take 2α=2q2(1q),2γ=2q2(1q) and β=13q1q such that 2α+β+2γ=1. By Theorem 2.7 there exists a common fixed point of S and T which is a common solution for the system (49).

Example 5.2. Consider the following fractional differential equation:

{Dξμ(t)+t2=0,(0t1,1<ξ<2),μ(0)=μ(1)=0.(51)

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

μ(t)=1Γ(ξ)0t(t(1μ)ξ1(tμ)ξ1)μ2dμ+tΓ(ξ)t1μ2(1μ)ξ1dμ.

The operator S,T:C[0,1]C[0,1] are defined by

Sμ(t)=16Γ(ξ)0t(t(1μ)ξ1(tμ)ξ1)μ2dμ+t6Γ(ξ)t1μ2(1μ)ξ1dμ,

and

Tω(t)=13Γ(ξ)0t(t(1ω)ξ1(tω)ξ1)ω2dω+t3Γ(ξ)t1ω2(1ω)ξ1dω.

By taking ξ=12, α=133, β=4550 and γ=13660 with 2α+β+2γ=1 and the initial value μ0(t)=t(1t), t[0,1]. We have

Sμ(t)Tω(t)16μ(t)13ω(t)13μ(t)ω(t)+16μ(t).

On the other hand, we obtain that

αμ(t)ω(t)+βω(t)Sμ(t)+γμ(t)Tω(t)=133μ(t)ω(t)+4550ω(t)Sμ(t)+13660μ(t)Tω(t)133μ(t)ω(t)+4550{ω(t)μ(t)+μ(t)Sμ(t)}+13660{μ(t)ω(t)+ω(t)Tω(t)}=(133+4550+13660)μ(t)ω(t)+4550μ(t)16μ(t)+13660ω(t)13ω(t)=1920μ(t)ω(t)+34μ(t)+13990ω(t)

which implies that

Sμ(t)Tω(t)αμ(t)ω(t)+βω(t)Sμ(t)+γμ(t)Tω(t).

By Theorem 2.7 there exists a common fixed point of S and T which 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 with Sλ. 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.

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Cite This Article

Hussain, N., Alsulami, S. M., Alamri, H. (2023). Solving Fractional Differential Equations via Fixed Points of Chatterjea Maps. CMES-Computer Modeling in Engineering & Sciences, 135(3), 2617–2648.


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