Solving Fractional Differential Equations via Fixed Points of Chatterjea Maps

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 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,ω∈Λ. A self-mapping S on Λ is nonexpansive if α=1. A point ν∈Λ 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,ω∈Λ 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,ω∈Λ,(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,ω∈Λ 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 [5,8,9,17–19]).

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

O(S,μ,n)={μ,Sμ,…,Snμ},n≥0,

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 μ∈Λ converges in Λ.

In addition, S is said to be orbital continuous at a point ν∈Λ, if for any sequence {μn}⊂O(S,μ,n), then, limn→∞μn=ν implies limn→∞Sμn=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, n≥0. 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 limn→∞Sμn=ν implies limn→∞TSμn=Tν or (limn→∞Sμn=ν implies limn→∞SSμn=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:Λ→Λ of metric space Λ is said to be κ-continuous, if limn→∞Sκ−1μn=ν, then limn→∞Sκμn=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 limn→∞d(Snμ,Sn+1μ)=0 for all μ∈Λ. A mapping S is called asymptotically regular with respect to T at μ0∈Λ if there exists a sequence μn∈Λ such that Tμn+1=Sμn, n≥0 and limn→∞d(Tμn+1,Tμn+2)=0. The mapping S has an approximate fixed point sequence if there exists a sequence μn⊂Λ, such that d(μn,Sμn)→0 as n→∞. The self-mappings S and T are called compatible [6] if limn→∞d(STμn,TSμn)=0, whenever {μn} is a sequence in Λ such that limn→∞Sμn=limn→∞Tμn=ν for some ν∈Λ. C(S,T)={μ∈Λ: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:Λ→Λ be a continuous asymptotically regular mapping satisfying

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

for all μ,ω∈Λ where α∈[0,1) and β∈[0,∞). Then S has a unique fixed point ν∈Λ and Snμ→ν for any μ∈Λ.

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:Λ→Λ be an asymptotically regular mapping. Assume that there exist α∈[0,1) and β∈[0,∞) satisfying (5) for all μ,ω∈Λ. Then S has a unique fixed point ν∈Λ, provided that S is either κ-continuous for some κ≥1 or orbitally continuous. Moreover, Snμ→ν for any μ∈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 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≤(1−ψn)hn+ψnϑn,n≥0,

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

(i)   ψn⊂[0,1] and ∑n=0∞ψn=∞,

(ii)   lim supn→∞ϑn≤0,or

(iii)   ∑n=0∞|ψnϑn| is convergent.

Then, limn→∞hn=0.

Theorem 2.1. Let (Λ,d) be a complete metric space and S,T:Λ→Λ 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,ω∈Λ and α∈[0,1). Further, S and T are either κ-continuous for some κ≥1 or orbitally continuous. Then S and T have a unique common fixed point ν. Moreover, for any μ∈Λ, limn→∞Snμ=ν=limn→∞Tnμ.

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

Step 1: We shall prove that limn→∞d(Snμ,Tnμ)=0, for any μ∈Λ. The result is trivial if S=T. Suppose that S≠T and α=0. Then (6) becomes

d(Sμ,Tω)≤βd(μ,Sμ)+γd(ω,Tω),

for all μ,ω∈Λ. Defining μ1=Snμ and ω1=Tnμ, for any μ∈Λ, 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

limn→∞d(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 S≠T and α≠0. 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=1−α and ϑn=β1−αd(Snμ,Sn+1μ)+γ1−αd(Tnμ,Tn+1μ).By asymptotically regularity of S and T, we have limn→∞ϑn=0. Furthermore, ∑n=1∞ψn=∞. Hence, by Lemma 2.1, we get that limn→∞d(Snμ,Tnμ)=0 for any μ∈Λ.

Step 2: Let μn=Snμ for any μ∈Λ. Now, we show that {μn} is a Cauchy sequence converging to ν∈Λ. Moreover, {Tnμ}→ν∈Λ. 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>nκ≥κ, 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(κ−1)μ,Snκμ)<ε.

Thus, we have

ε≤d(Smκμ,Snκμ)≤d(Smκμ,Sm(κ−1)μ)+d(Sm(κ−1)μ,Snκμ)<d(Smκμ,Sm(κ−1)μ)+ε.

As κ→∞, it follows by asymptotic regularity of S that

limκ→∞d(Smκμ,Snκμ)=ε.(8)

Further, by asymptotic regularity of S and the following inequality

d(Sm(κ−1)μ,Sn(κ−1)μ)≤d(Sm(κ−1)μ,Smκμ)+d(Smκμ,Snκμ)+d(Snκμ,Sn(κ−1)μ),

the implication is that

limk→∞d(Sm(κ−1)μ,Sn(κ−1)μ)=ε.(9)

Now, using (6) we get

d(Smκμ,Snκμ)≤d(Smκμ,Tnκμ)+d(Tnκμ,Snκμ)≤d(Tnκμ,Snκμ)+α[d(Sm(κ−1)μ,Sn(κ−1)μ)+d(Sn(κ−1)μ,Tn(κ−1)μ)]+βd(Sm(κ−1)μ,Smκμ)+γd(Tn(κ−1)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, ε≤αε, which is a contradiction. Hence, {μn} is a Cauchy sequence in complete space Λ, there exists a point ν in Λ such that μn→ν. Moreover,

d(Tnμ,ν)≤d(Tnμ,Snμ)+d(Snμ,ν),

from Step 1 and μn→ν, we get that Tnμ converges to ν∈Λ.

Step 3: We will prove that μ is the unique common fixed point of S and T. Assume that S is κ-continuous. Since limn→∞Sκ−1μn=ν, κ−continuity of S implies that

limn→∞Sκμn=Sν.

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

Similarly, let S be orbitally continuous. Since limn→∞μn=ν, orbital continuity of S implies

limn→∞Sμn=Sν,

we obtain Sν=ν.

In addition, since limn→∞Tnμ=ν, this gives Tν=ν whenever T is κ-continuous or orbitally continuous. Hence, ν∈Fix(S)∩Fix(T). Now, we prove the uniqueness of the common fixed point, suppose that there is ν≠ν∗ 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)=|μ−ω|. 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=12∈Λ we have limn→∞d(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 ω≤μ. We get

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=112|μ−ω|+16|μ−μ7|+34|ω−37ω|=112|μ−ω|+17|μ|+37|ω|≥112|μ−ω|+17|μ+3ω|≥17|μ−3ω|=d(Sμ,Tω).

Case 2: Let μ∈[0,1) and ω=1. We have

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=112|μ−1|+16|μ−μ7|+34|1−17|=112(1−μ)+μ7+914≥6184+584μ>17|μ−1|=d(Sμ,Tω).

Case 3: Let ω∈[0,1) and μ=1. We obtain

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=112|1−ω|+16+34|ω−37ω|=112(1−ω)+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|1−17|=16+914>214=17=d(Sμ,Tω).

Therefore, in all the cases, S and T satisfy the condition (6) for all μ,ω∈Λ. Moreover, all the assumptions of Theorem 2.1 hold, hence, the mappings S and T have a unique common fixed point at 0. Further, limn→∞Snμ=0=limn→∞Tnμ for any μ∈Λ.

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,ω∈Λ. Then, S and T have a unique common fixed point ν∈Λ, provided that both Sp and Tq are either κ-continuous for some κ≥1 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,ω∈Λ. 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 ν∗≠ν 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)n∈N:limn→∞νn=0} be the space of all real sequences convergent to zero, endowed with the usual metric d∞ defined by d∞(μ,ω)=supn|μn−ωn| for all μ=(μn)n and ω=(ωn)n∈Λ. 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 limn→∞d(Snμ,Sn+1μ)=0 and limn→∞d(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,q∈N.

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, limn→∞Spμ=0=limn→∞Tqμ for any μ∈Λ.

Theorem 2.4. Let (Λ,d) be a complete metric space and S:Λ→Λ 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,ω∈Λ. The one of the following conditions hold:

(i) The mapping S is continuous. Further, Snμ→ν for each μ∈Λ, as n→∞.

(ii) For κ≥1, S is κ-continuous or orbitally continuous.

Then, S has a unique fixed point ν∈Λ.

Proof. First, we shall prove condition (i). Let μ0∈Λ be arbitrary and define a sequence {μn} by μn+1=Sμn for all n≥0.

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,

(1−α)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 ν∈Λ such that μn→ν. The continuity of S and μn+1=Sμn, implies that ν=Sν. Let ν∗≠ν 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 ν∈Λ. Now, we show that Snμ→ν. 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,

(1−α)d(Snμ,ν)≤(1+β)d(Snμ,Sn+1μ)→0,asn→∞.

This shows that Snμ→ν for any μ∈Λ 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 n≥0. Then, from (i) we have proven that {μn} is a Cauchy sequence in a complete metric space. There exists a point ν∈Λ such that μn→ν as n→∞. and Sμn→ν. Moreover, for all κ≥1 we have Sκμn→ν as n→∞. Suppose that S is κ-continuous. Since Sκ−1μn→ν, we get limn→∞Sκμn=Sν. This implies ν=Sν, that is ν is a fixed point of S. Finally, we assume that S is orbitally continuous. Since μn→ν, orbital continuity implies that limn→∞Sμn=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:Λ→Λ 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, μn→ν as n→∞.

Proof. Suppose that there exist m,n∈N 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

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

As n,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 to ν∈Λ. Since limn→∞d(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:Λ→Λ. 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,ω∈Λ. 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 μ0∈Λ, so, there exists a sequence {ωn}∈Λ such that ωn=Sμn=Tμn+1 for each n≥0, and limn→∞d(Tμn+1,Tμn+2)=limn→∞d(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,

(1−α)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 limn→∞d(ωn+κ,ωn)=0. Therefore, {ωn} is a Cauchy sequence in a complete metric space. There exists a point ν∈Λ such that ωn→ν as n→∞. Moreover, ωn=Sμn=Tμn+1→ν.

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

limn→∞SSμn=limn→∞STμn=Sν,

further

limn→∞TSμn=limn→∞TTμn=Tν.

The compatibility of S and T implies limn→∞d(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,ω∈Λ. Suppose further that S and T are either κ-continuous for some κ≥1 or orbitally continuous. Then, S and T have a unique common fixed point ν. Moreover, for any μ∈Λ, limn→∞Snμ=ν=limn→∞Tnμ.

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(Snμ,Tnμ)=0, for any μ∈Λ. The result is trivial if S=T. Suppose S≠T and α=0, from (14) we obtain

d(Sμ,Tω)≤βd(ω,Sμ)+γd(μ,Tω),

for each μ,ω∈Λ. Define μ1=Snμ and ω1=Tnμ, for any μ∈Λ. 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

limn→∞d(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 S≠T and α≠0. Let μ1=Snμ and ω1=Tnμ, for any μ∈Λ. 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

(1−β−γ)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=1−α2α+γ and ϑn=βαd(Tnμ,Tn+1μ)+γαd(Snμ,Sn+1μ). By the asymptotically regularity of S and T, we have limn→∞ϑn=0. Furthermore, ∑n=1∞ψn=∞. Hence, by Lemma 2.1, we get that limn→∞d(Snμ,Tnμ)=0 for any μ∈Λ.

Step 2: Let μn=Snμ for any μ in Λ and n≥0. Then, we show that {μn} is a Cauchy sequence which is convergent to ν∈Λ. Moreover, {Tnμ}→ν∈Λ.

Assume that {Snμ} is not a Cauchy sequence. Then, there exists an ε>0 and sequences of integers {mκ} and {nκ} such that mκ>nκ≥κ, 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(κ−1)μ,Snκμ)<ε.

Now, we have that

ε≤d(Smκμ,Snκμ)≤d(Smκμ,Sm(κ−1)μ)+d(Sm(κ−1)μ,Snκμ)<d(Smκμ,Sm(κ−1)μ)+ε.

As κ→∞, it follows by asymptotic regularity of S that

limκ→∞d(Smκμ,Snκμ)=ε.(17)

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

d(Sm(κ−1)μ,Sn(κ−1)μ)≤d(Sm(κ−1)μ,Smκμ)+d(Smκμ,Snκμ)+d(Snκμ,Sn(κ−1)μ),

implying that

limκ→∞d(Sm(κ−1)μ,Sn(κ−1)μ)=ε.(18)

Then, using (14), we have

d(Smκμ,Snκμ)≤d(Smκμ,Tnκμ)+d(Tnκμ,Snκμ)≤d(Tnκμ,Snκμ)+αd(Sm(κ−1)μ,Tn(κ−1)μ)+βd(Tn(κ−1)μ,Smκμ)+γd(Sm(κ−1)μ,Tnκμ)≤d(Tnκμ,Snκμ)+α{d(Sm(κ−1)μ,Sn(κ−1)μ)+d(Sn(κ−1)μ,Tn(κ−1)μ)}+β{d(Tn(κ−1)μ,Snκ−1μ)+d(Sn(κ−1)μ,Sm(κ−1)μ)+d(Sm(κ−1)μ,Smκμ)}+γ{d(Sm(κ−1)μ,Sn(κ−1)μ)+d(Sn(κ−1)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 μn→ν. Moreover

d(Tnμ,ν)≤d(Tnμ,Snμ)+d(Snμ,ν),

from (15) and μn→ν implies Tnμ converges to ν∈Λ.

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

Let S be κ-continuous. Since limn→∞Sκ−1μn=ν, we have that

limn→∞Sκμn=Sν.

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

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

limn→∞Sμn=Sμ,

implies that Sν=ν. Furthermore, since limn→∞Tnμ=ν, we have that Tν=ν whenever T is κ-continuous or orbitally continuous. Hence, ν∈Fix(S)∩Fix(T). For the uniqueness of fixed point, suppose that ν≠ν∗ 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×Λ→R+ by

d((f1,f2),(g1,g2))=‖f1−g1‖∞+‖f2F1−g2G1‖∞.

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 limn→∞d(Snf,Sn+1f)=limn→∞d(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))=||15−15||∞+||15+∫01125du−315−315∫0115du||∞=||125−375||∞=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))=||15−15||∞+||215+215∫0115du−15−15∫0115du||∞=||−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))+675≥d(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))=||15−15||∞+||215+215∫0115du−315−315∫0115du||∞=||−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,ω∈Λ. Moreover, all the assumptions of Theorem 2.7 hold. Point (15,15) is the unique common fixed point of S and T. Moreover, limn→∞Snμ=0=limn→∞Tnμ for any μ∈Λ.

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,ω∈Λ. Then, S and T have a unique common fixed point ν∈Λ, provided that both Sp and Tq are either κ-continuous for some κ≥1 or orbitally continuous.

Proof. Take f=Sp and g=Tq. Then, for each μ,ω∈Λ, 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 ν∗≠ν. 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:Λ→Λ 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,ω∈Λ, provided that one of the following conditions hold:

(i)   The mapping S is a continuous. Further Snμ→ν for each μ∈Λ, as n→0.

(ii)   For κ≥1, S is κ-continuous or orbitally continuous.

Then S has a unique fixed point μ∈Λ

Proof. First, we will proof the condition (i) holds. Let μ0∈Λ be arbitrary. Then define a sequence {μn} by μn+1=Sμn for all n≥0.

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 ν∈Λ such that μn→ν. As S is continuous and μn+1=Sμn, we obtain that ν=Sν. Suppose that ν∗≠ν 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 ν∈Λ. Now, we show that Sn→ν. 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 Sn→ν for any μ∈Λ.

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 n≥0. Then, in the proof condition (i) we have shown that {μn} is a Cauchy sequence in complete space Λ. There exists a point ν∈Λ such that μn→ν as n→∞, in addition, Sμn→ν. Moreover, for each κ≥1 we have Sκμn→ν as n→∞. Suppose that S is κ-continuous, and Sκ−1μn→ν, implies limn→∞Sκμn=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 μn→ν, orbital continuity implies that limn→∞Sμn=Sν. Then ν is a fixed point of S.

Theorem 2.10. Let (Λ,d) be a complete metric space and S:Λ→Λ 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, μn→ν as n→∞.

Proof. Let m>n for all n,m∈N. 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,m→∞d(μn,μm)=0. Hence, {μn} is a Cauchy sequence in Λ. By completeness of Λ, this sequence {μn} converges to ν∈Λ. Since limn→∞d(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:Λ→Λ. 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,ω∈Λ. 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 μ0∈Λ. So, there exists a sequence {ωn}∈Λ such that ωn=Sμn=Tμn+1 for each n≥0, and limn→∞d(Tμn+1,Tμn+2)=limn→∞d(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 ν∈Λ such that ωn→ν as n→∞, and ωn=Sμn=Tμn+1→ν.

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

limn→∞SSμn=limn→∞STμn=Sν,

further,

limn→∞STμn=limn→∞SSμn=Sν,

Then, compatibility of S and T yields limn→∞d(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:Λ→Λ by

Sμ={2ifμ=2,3if2<μ≤5,2ifμ>5,andTμ={2ifμ=2,11if2<μ≤5,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|2−ω+13|+15|ω+13−2|+13|2−2|=175|ω−5|+115|ω−5|≥0=d(Sμ,Sω).

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

αd(Tμ,Tω)+βd(Tω,Sμ)+γd(Tμ,Sω)=115|11−ω+13|+15|ω+13−3|+13|11−2|=175|32−ω|+115|ω−8|+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+13−ω+13|+15|ω+13−2|+13|μ+13−2|=175|μ−ω|+115|ω−5|+19|μ−5|>0=d(Sμ,Sω).

Otherwise, if μ=1 and ω=2 or 2<ω≤5 (or 2<μ,ω≤5), 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,n∈N.(23)

with an arbitrary μ0∈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 of S.

Define S0=I (the identity map on Λ) and Sn=Sn−1∘S, called the nth iterate of S for n∈N. The Krasnoselskii-Ishikawa iteration method associated with S is the sequence {μn}n=0∞ defined by

μn+1=(1−λ)μn+λSμn,(24)

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

Sλ=(1−λ)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:Λ→Λ satisfying the following inequality

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

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

‖Sλpμ−Sλqμ‖≤ηδ[O(Sλ,μ,n)],(27)

where η=max{2α,β,γ} and δ[A]=sup{‖μ−ω‖:μ,ω∈A}.

Proof. Let us choose λ=1−2α1−α, where 2α<1, clearly, 0<λ<1. Considering the operator given by (25), we have

λ‖μ−Sμ‖=‖μ−Sλμ‖,(28)

and

λ‖ω−Sω‖=‖ω−Sλω‖,(29)

for all μ,ω∈Λ. Moreover, we obtain

‖Sλμ−Sλω‖=‖(1−λ)(μ−ω)+λ(Sμ−Sω)≤(1−λ)‖μ−ω‖+λ‖Sμ−Sω‖,(30)

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

λ‖Sμ−Sω‖≤αλ‖μ−ω‖+βλ‖μ−Sμ‖+λγ‖ω−Sω‖.(31)

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

λ‖Sμ−Sω‖≤αλ‖μ−ω‖+β‖μ−Sλμ‖+γ‖ω−Sλω‖.(32)

Now, (30) and (32) imply that

‖Sλμ−Sλω‖≤(1−λ)‖μ−ω‖+αλ‖μ−ω‖+β‖μ−Sλμ‖+γ‖ω−Sλω‖≤(1−λ+αλ)‖μ−ω‖+β‖μ−Sλμ‖+γ‖ω−Sλω‖.(33)

Since λ=1−2α1−α and 2α<1, we assume that 1−λ+αλ=2α<1. Then (33) becomes

‖Sλμ−Sλω‖≤2α‖μ−ω‖+β‖μ−Sλμ‖+γ‖ω−Sλω‖.(34)

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

‖Sλpμ−Sλqμ‖=‖SλSλp−1μ−SλSλq−1μ‖≤2α‖Sλp−1μ−Sλq−1μ‖+β‖Sλp−1μ−Sλpμ‖+γ‖Sλq−1μ−Sλqμ‖,

which implies that

‖Sλpμ−Sλqμ‖≤ηδ[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 μ∈Λ, then for any n∈N, there exists a positive integer κ≤n, such that

‖μ−Sλκμ‖=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,∞)]≤11−η‖μ−Sλκμ‖,(35)

holds for each μ∈S, and η=max{2α,β,γ}.

Proof. Let μ∈Λ 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)]:n∈N}.

Then, (35) implies that

δ[O(Sλ,μ,n)]≤11−η‖μ−Sλκμ‖.

Let n be any positive integer. From Remark 3.1, there exists Sλκ∈O(Sλ,μ,n) where 1≤κ≤n, such that

‖μ−Sλκμ‖=δ[O(Sλ,μ,n)].

Applying a triangle inequality and Lemma 3.1, we obtain

‖μ−Sλκμ‖≤‖μ−Sλμ‖+‖Sλμ−Sλκμ‖≤‖μ−Sλμ‖+ηδ[O(Sλ,μ,n)]=‖μ−Sλμ‖+η‖μ−Sλκμ‖.

Therefore,

δ[O(Sλ,μ,n)]=‖μ−Sλκμ‖≤11−η‖μ−Sλμ‖.

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=(1−λ)μn−1+λSμn−1,n≥1,

converges to a unique fixed point ν for any μ0∈Λ, provided that Λ is Sλ-orbitally complete.

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

‖Sλμ−Sλω‖≤2α‖μ−ω‖+β‖μ−Sλμ‖+γ‖ω−Sλω‖,(36)

where λ=1−2α1−α 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=Sλμn−1=Sλnμ0,n≥0.(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

‖μn−μm‖=‖Sλnμ0−Sλmμ0‖=‖SλSλn−1μ0−Sλm−n+1Sλn−1μ0‖=‖Sλμn−1−Sλm−n+1μn−1‖≤ηδ[O(Sλ,μn−1,m−n+1)],(38)

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

‖μn−1−μn+κ−1‖=‖Sλμn−2−Sλκ+1μn−2‖≤ηδ[O(Sλ,μn−2,κ+1)],

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

‖μn−μm‖≤ηn1−η‖μ0−Sλμ0‖.

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

‖ν−Sλν‖≤‖ν−μn+1‖+‖μn+1−Sλν‖=‖ν−μn+1‖+‖Sλμn−Sλν‖≤‖ν−μn+1‖+2α‖μn−ν‖+β‖μn−μn+1‖+γ‖ν−Sλν‖.

Hence,

‖ν−Sλν‖≤11−γ(‖ν−μn+1‖+2α‖μn−ν‖+β‖μn−μn+1‖).

Since limn→∞μn=ν, we have ‖ν−Sλν‖=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 ν≠ν∗. By supposition, we obtain

‖ν−ν∗‖=‖Sλν−Sλν∗‖≤2α‖ν−ν∗‖+β‖ν∗−Sλν‖+γ‖ν∗−Sλν∗‖≤2α‖ν−ν∗‖<‖ν−ν∗‖,

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

‖Sμ−Tω‖≤α‖μ−ω‖+β‖μ−Sμ‖+γ‖ω−Tω‖,(39)

for all μ,ω∈Λ. Then, the iteration sequence {μn}n=0∞ defined by (24) converges to a unique common fixed point ν for any μ0∈Λ, 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=(1−λ)μ2n+1+λSμ2n+1=μ2n+2,Tλμ2x=(1−λ)μ2n+λTμ2n=μ2n+1.(41)

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

‖Sλμ2n+1−Tλμ2n‖=‖μ2n+2−μ2n+1‖≤(1−λ)‖μ2n+1−μ2n‖+αλ‖μ2n+1−μ2n‖+β‖μ2n+1−Sλμ2n+1‖+γ‖μ2n−Tλμ2n‖=(1−λ+αλ)‖μ2n+1−μ2n‖+β‖μ2n+1−μ2n+2‖+γ‖μ2n−μ2n+1‖,(42)

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

(1−β)‖μ2n+2−μ2n+1‖≤(2α+γ)|μ2n+1−μ2n‖⇒‖μ2n+2−μ2n+1‖≤2α+γ1−β‖μ2n+1−μ2n‖≤1−2β1−βδ[O(Sλ,Tλ,μ2n,2n+1)]<ηδ[O(SλTλ,μ2n,2n+1)],

where η=1−2β1−β<1 and β<1.

Similarly,

‖μ2n+3−μ2n+2‖≤η‖μ2n+2−μ2n+1‖=ηδ[O(Sλ,Tλ,μ2n+1,2n+2)]≤η2δ[O(Sλ,Tλ,μ2n,2n+1)].

Continuing the process, we get the following:

‖μm+1−μm‖<ηδ[O(Sλ,Tλ,μm−1,m)]<η2δ[O(Sλ,Tλ,μm−2,m−1)]<…<ηmδ[O(Sλ,Tλ,μ0,m+1)],m∈N.(43)

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

‖μm−μn‖<ηδ[O(Sλ,Tλ,μn,n+1)]<….<ηnδ[O(Sλ,Tλ,μ0,m)]≤ηn1−η‖μ0−Tλμ0‖,n<m.

Since limn→∞ηn1−η=0, so, {μn} is a Cauchy sequence on Λ. Since Λ is (Sλ,Tλ)-orbital complete, there exists ν∈Λ such that limn→∞μn=ν. Now, we prove that ν is a fixed point of Sλ. From (41) and (42), we consider the following inequalities:

‖Sλν−ν‖≤‖Sλν−μ2n+1‖+‖μ2n+1−ν‖=‖Sλν−Tλμ2n‖+‖μ2n+1−ν‖≤2α‖μ2n−ν‖+β‖ν−Sλν‖+γ‖μ2n−Tλμ2n‖≤2α1−β‖μ2n−ν‖+γ1−β‖μ2n−μ2n+1‖,

where we have ‖ν−Sλν‖=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 ν≠ν∗. By supposition, we can replace μ2n+2 by ν and μ2n+1 by ν∗ in (42) to obtain

‖ν−ν∗‖=‖Sλν−Tλν∗‖≤2α‖ν−ν∗‖+β‖ν−Sλν‖+γ‖ν∗−Tλν∗‖≤2α‖ν−ν∗‖<‖ν−ν∗‖,

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:Λ→Λ. Assume that there exist non-negative numbers α,β,γ such that 2α+3β+2γ=1, satisfying

‖Sμ−Tω‖≤α‖μ−ω‖+β‖ω−Sμ‖+γ‖μ−Tω‖,(44)

for all μ,ω∈Λ. Then, the iteration sequence {μn}n=1∞ defined by (24) converges to a unique common fixed point ν for any μ0∈Λ, 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+1−Tλμ2n‖=‖μ2n+2−μ2n+1‖≤(1−λ)‖μ2n+1−ν2n‖+λ(α+β+γ)‖μ2n+1−μ2n‖+β‖μ2n+1−Sλμ2n+1‖+γ‖μ2n−Tλμ2n‖≤(2α+β+γ)‖μ2n+1−μ2n‖+β‖μ2n+1−μ2n+2‖)+γ‖μ2n−μ2n+1‖.(46)

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

(1−β)‖μ2n+2−μ2n+1‖≤(2α+2γ+β)|μ2n+1−μ2n‖⇒‖μ2n+2−μ2n+1‖≤1−2β1−β‖μ2n+1−μ2n‖≤ηδ[O(Sλ,Tλ,μ2n,2n+1)],

where η=1−2β1−β and β<1.

Similarly, we obtain

‖μ2n+3−μ2n+2‖≤η‖μ2n+2−μ2n+1‖=ηδ[O(Sλ,Tλ,μ2n+1,2n+2)]≤η2δ[O(Sλ,Tλ,μ2n,2n+1)].

Continuing the process, we get the following:

‖μm+1−μm‖<ηδ[O(Tλ,Sλ,μm−1,m)]<η2δ[O(Tλ,Sλ,μm−2,m−1)]<…<ηmδ[O(Tλ,Sλ,μ0,m+1)],m∈N.(47)

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

‖μm−μn‖<ηδ[O(Sλ,Tλ,μn,n+1)]<….<ηnδ[O(Sλ,Tλ,μ0,m)]≤ηn1−ηδ‖μ0−Tλμ0‖,n<m.

Since limn→∞ηn1−η=0, so, {μn} is a Cauchy sequence in Λ. Since Λ is (Sλ,Tλ)-orbital complete, there exists ν∈Λ such that limn→∞μn=ν. We prove now that ν is a fixed point of Sλ. From (45) and (46), we consider the following inequalities:

‖Sλν−ν‖≤‖Sλν−μ2n+1‖+‖μ2n+1−ν‖=‖Sλν−Tλμ2n‖+‖μ2n+1−ν‖≤(α+β+γ)‖μ2n−ν‖+β‖μ2n−Sλν‖+γ‖ν−Tλμ2n‖+‖μ2n+1−ν‖≤(α+β+γ)‖μ2n−ν‖+β(‖μ2n−ν‖+‖ν−Sλν‖)+(γ+1)‖ν−μ2n+1‖≤α+2β+γ1−β‖μ2n−ν‖+γ+11−β‖ν−μ2n+1‖.

Since μ2n→ν as n→∞ we have ‖ν−Sλν‖=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 ν≠ν∗ with supposition, we can replace μ2n+2 by ν and μ2n+1 by ν∗ in (41) to obtain

‖ν−ν∗‖=‖Sλν−Tλν∗‖≤(2α+β+γ)‖ν−ν∗‖+β‖ν−Sλν‖+γ‖ν∗−Tλν∗‖≤(2α+β+γ)‖ν−ν∗‖<‖ν−ν∗‖,

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)=|μ−ω|. Consider the following self-mappings defined by

Sμ={12if0≤μ≤12,12μ+14if12≤μ≤1.Tμ={12if0≤μ≤12,1+μ3if12≤μ≤1.

Take μ0=23, then O(Sλ,Tλ,μn,n)⊂{2/(2p3q−1),p,q∈N}.

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 μ≤ω. We have

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=19|μ−ω|+127|μ−12|+13|ω−12|≥19(ω−μ)+127(12−μ)+13(12−ω)=527−29ω−427μ≥d(Sμ,Tω).

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

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=19|μ−ω|+127|μ−12|+13|ω−13−13ω|≥19(ω−μ)+127(12−μ)+19(2ω−1)=154[18ω−8μ−5]≥|16−13ω|=d(Sμ,Tω).

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

αd(μ,ω)+βd(μ,Sμ)+γd(ω,Tω)=19|μ−ω|+127|μ−μ2−14|+13|ω−13−13ω|≥19(ω−μ)+1108(2μ−1)+19(2ω−1)=13ω−554μ−13108=1108[36ω−10μ−13]≥112|6μ−4ω−1|=d(Sμ,Tω).

Therefore, in all the cases, S and T satisfy the condition of (39) for all μ,ω∈Λ. 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,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:

{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 0≤α<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={ν∈Λ:0≤ν(t)≤1}.

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

Gμ(t)=∫0tK1(t,r,μ(r),Sμ(r))dr,

and

G∗ω(t)=∫0tK2(t,r,ω(r),Tω(r))dr,

satisfying Gμ,G∗ω in E, for all μ(t),ω(t)∈E and

‖Gμ(t)−G∗ω(t)‖<α(1−α)‖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,ω∈E. 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)−G∗ω(t)|≤α‖μ(t)−Sμ(t)‖+‖ω(t)−Tω(t)‖+‖Sμ(t)−Tω(t)‖+α(1−α)‖μ(t)−ω(t)‖,

which implies that

(1−α)‖Sμ(t)−Tω(t)‖≤α‖μ(t)−Sμ(t)‖+α‖ω(t)−Tω(t)‖+α(1−α)‖μ(t)−ω(t)‖.

Therefore,

‖Sμ(t)−Tω(t)‖≤α1−α‖μ(t)−Sμ(t)‖+‖ω(t)−Tω(t)‖+α(1−α)1−α‖μ(t)−ω(t)‖.

Now, since 0≤α<1 and β=α1−α≤γ, then

‖Sμ(t)−Tω(t)‖≤α‖μ(t)−ω(t)‖+β‖μ(t)−Tμ(t)‖+γ‖ω(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 [29–35] and the references therein.

The classical Caputo fractional derivative is defined by

Dξμ(t):=1Γ(m−ξ)∫0t(t−r)m−ξ−1dmdrmμ(r)dr

where m−1<Re(ξ)<m,m∈N.

Now, we consider the following fractional differential equation:

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

where Dξ is the Caputo fractional derivative of order ξ and h:[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

G(μ,ω)={1Γ(ξ)(μ(1−ω)ξ−1−(μ−ω)ξ−1),0≤ω≤μ≤1,1Γ(ξ)μ(1−ω)ξ−1,0≤μ≤ω≤1.(50)

Assume that

|h(t,μ)−h(t,ω)|≤q2(1−q)|μ−ω|,

for all t∈[0,1], q∈(0,23) and μ,ω∈R.

Theorem 5.1. Let Λ=C[0,1] and the operators S,T:Λ→Λ 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 μ∈Λ 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,ω∈Λ and t∈[0,1]. Then,

|Sμ(t)−Tω(t)|≤∫01G(t,r)(h(r,μ(r))−h(r,ω(r)))dr≤q2(1−q)∫01G(t,r)|μ(r)−ω(r)|dr⇒‖Sμ(t)−Tω(t)‖≤q2(1−q)‖μ−ω‖≤q2(1−q)‖μ−ω‖+1−3q(1−q)‖μ−Sω‖+q2(1−q)‖ω−Tμ‖,

where q<13. Take 2α=2q2(1−q),2γ=2q2(1−q) and β=1−3q1−q 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,(0≤t≤1,1<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(1−μ∗)ξ−1−(t−μ∗)ξ−1)μ∗2dμ∗+tΓ(ξ)∫t1μ∗2(1−μ∗)ξ−1dμ∗.

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

Sμ(t)=16Γ(ξ)∫0t(t(1−μ∗)ξ−1−(t−μ∗)ξ−1)μ∗2dμ∗+t6Γ(ξ)∫t1μ∗2(1−μ∗)ξ−1dμ∗,

and

Tω(t)=13Γ(ξ)∫0t(t(1−ω∗)ξ−1−(t−ω∗)ξ−1)ω∗2dω∗+t3Γ(ξ)∫t1ω∗2(1−ω∗)ξ−1dω∗.

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

‖Sμ(t)−Tω(t)‖≤‖16μ(t)−13ω(t)‖≤13‖μ(t)−ω(t)‖+16‖μ(t)‖.

On the other hand, we obtain that

α‖μ(t)−ω(t)‖+β‖ω(t)−Sμ(t)‖+γ‖μ(t)−Tω(t)‖=133‖μ(t)−ω(t)‖+4550‖ω(t)−Sμ(t)‖+13660‖μ(t)−Tω(t)‖≤133‖μ(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)‖+4550‖μ(t)−16μ(t)‖+13660‖ω(t)−13ω(t)‖=1920‖μ(t)−ω(t)‖+34‖μ(t)‖+13990‖ω(t)‖

which implies that

‖Sμ(t)−Tω(t)‖≤α‖μ(t)−ω(t)‖+β‖ω(t)−Sμ(t)‖+γ‖μ(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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