# Some Existence Results for a System of Nonlinear Fractional Differential Equations

Authors: Eskandar Ameer; Hassen Aydi; Hüseyin Işık; Muhammad Nazam; Vahid Parvaneh; Muhammad Arshad
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, we show that a sequence satisfying a Suzuki-type JS-rational contraction or a generalized Suzuki-type Ćirić JS-contraction, under some conditions, is a Cauchy sequence. This paper presents some common fixed point theorems and an application to resolve a system of nonlinear fractional differential equations. Some examples and consequences are also given.


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## Body

## 1. Introduction

The application of Banach contraction principle (BCP) [1] is wide spread. Recently, fixed point theory is being applied to show the existence of solutions of different mathematical models expressed in the forms of differential, integral, functional, fractional differential, and matrix equations (both linear and nonlinear). There are several common fixed point theorems, in the literature, which generalize BCP and have been applied to show the existence of solutions of different mathematical models involving two or more functions (see [2–19], for details).Suzuki [20] presented a new generalization of BCP, so called a Suzuki-contraction, and established an existence theorem which characterized the metric completeness. Piri and Kumam [21, 22] investigated results in [20] under the F-contraction structure both in metric and b-metric spaces, respectively. Further generalizations of Suzuki contractions have been studied by Aydi et al. [23] who introduced a Suzuki-type multivalued contraction to obtain fixed point theorems in the setting of weak partial metric spaces. Abbas et al. [2] discussed generalized Suzuki-type multivalued contractions under the effect of a binary relation and obtained some fixed point results.On the contrary, Jleli et al. [24, 25] introduced another generalization of BCP, known as a JS-contraction (also known as a θ-contraction) and Li and Jiang [26] modified the JS-contraction to JS-quasi contractions in order to obtain more general fixed point results, as compared to [24]. Altun et al. [27] obtained some fixed point theorems for JS-contraction type mappings via a binary operation. Several honorable researchers have published their valuable investigations on JS-contractions in well ranked journals (see [28–31]).In this paper, we structure two common fixed point theorems comprising four self-mappings involving generalized Suzuki-typeθ-rational contractions and generalized Suzuki-type Ćirić θ-contractions. The existence of the solution of a mathematical model in terms of fractional differential equations is shown through a common fixed point theorem.


## 2. Advances onθ-Contractions

LetΘ:=θ:0,∞⟶1,∞such thatθsatisfiesΘ1,Θ2,Θ3, whereΘ1θ is nondecreasing.Θ2 For each sequence t⌣n⊂0,∞,(1)limn⟶∞θt⌣n=1,if and only iflimn⟶∞t⌣n=0.Θ3 There exist q∈0,1 and ℓ∈0,∞ such that limtˇ⟶0+θt⌣−1/t⌣q=ℓ.Jleli and Samet [24] introduced the following.Definition 1.
Letχ,ρ be a metric space. The mapping T⌣:χ⟶χ is called a θ-contraction (or a JS-contraction) if there exist a constant k∈0,1 and θ∈Θ such that(2)r,j∈χ,ρT⌣r,T⌣j≠0⇒θρT⌣r,T⌣j≤θρr,jk.
The following examples show that the setΘ is not empty.Example 1.
LetΦ,ψ:0,∞⟶1,∞ be defined by, for all t∈0,∞,(3)Φt=et,ψt=etet.
Then,Φ and ψ are in Θ.
Jleli and Samet [24] established the following fixed point theorem.Theorem 1 (see [24]).
Letχ,ρ be a complete metric space and T⌣:χ⟶χ be a θ-contraction. Then, T⌣ has a unique fixed point.
Consistent with [28], let μ:=θ:0,∞⟶1,∞|θsatisfyΘ1,Θ2,Θ3′, whereΘ′3θ is continuous on 0,∞.
Note thatΘ3 and Θ′3 are independent of each other. For this, see examples in [28]. The set of functions μ is not empty, as shown in the following.Example 2 (see [28]).
Letϕ,φ,ψ,F,G,H:0,∞⟶1,∞ be defined by, for all t∈0,∞,(4)ϕt=et,φt=cosht,ψt=etet,Ft=1+lnt+1,Gt=et,Ht=etet.
Here,ϕ,φ,ψ,F,G,H∈μ.
Hussain et al. [29] modified the class Θ of mappings as follows:(5)LetΨ=θ:0,∞⟶1,∞:θsatisfiesΨ1−Ψ5,whereΨ1θ is non decreasing.Ψ2θt⌣=1if and only ift⌣=0.Ψ3 For each sequence t⌣n⊂0,∞:(6)limn⟶∞θt⌣n=1,if and only iflimn⟶∞t⌣n=0.Ψ4 There exist q∈0,1 and ℓ∈0,∞ such that limtˇ⟶0+θt⌣−1/t⌣q=ℓ.Ψ5θt⌣1+t⌣2≤θt⌣1θt⌣2.
The functionsft=et and gt=5t (for all t∈0,∞) are in Ψ. E∗: for all t1,t2,t3,t4∈ℝ+ with t1⋅t2⋅t3⋅t4=0, there exists k∈0,1 such that Et1,t2,t3,t4=k.
LetΔE denote the set of all functions E:ℝ+4⟶ℝ+ satisfying the condition E∗.Example 3 (see [31]).
IfEt1,t2,t3,t4=λeυmint1,t2,t3,t4, where υ∈ℝ+ and λ∈0,1, then E∈ΔE.Example 4 (see [31]).
IfEt1,t2,t3,t4=m⋅mint1,t2,t3,t4+λ, where m∈ℝ+ and λ∈0,1, then E∈ΔE.
The following class of mappings was considered in [32] to develop some fixed point theorems. We also use this class of mappings in our investigations:(7)Φ=ϒ:ℝ+×ℝ+⟶ℝandϒs,t=12s−t.Definition 2 (see [5]).
Letχ,ρ be a metric space. The pair f,g is said to be compatible if and only if(8)limn⟶∞ρfgrn,gfrn=0,whenever rn is a sequence in χ so that(9)limn⟶∞frn=limn⟶∞grn=t,for somet∈χ.
The following lemma is important in the sequel.Lemma 1 (see [5]).
Letχ,ρ be a metric space. If there exist two sequences rn and un such that(10)limn⟶∞ρrn,un=0,limn⟶∞rn=t,for somet∈χ,then(11)limn⟶∞un=t.


## 3. Generalized Suzuki-Type Contractions and Fixed Point Results

Jleli and Samet [24, 25] have employed θ-contractions to obtain some fixed point theorems. Suzuki [20] extended Edelstein contraction to develop a new generalization of Banach contraction called Suzuki contraction. The contractions developed in [20, 24] were then followed by Liu et al. [33] to introduce a Suzuki-type θ-contraction. In this section, we introduce more general forms of Suzuki-type θ-contractions involving four self-mappings. We construct some conditions under which a sequence, whose terms satisfy generalized Suzuki-type θ-rational contractions or generalized Suzuki-type Ćirić θ-contractions, is a Cauchy sequence. We obtain two common fixed point theorems, which then are applied to show the existence of the solution of a system of fractional differential equations. We start with the following definition.Definition 3.
Letχ,ρ be a metric space, and f,g,S^,T⌣:χ⟶χ be four single-valued maps. These maps form a new generalized Suzuki-type θ,E-rational contraction if, for all r,j∈χ, with ρfr,gj>0, for some θ∈μ, E∈ΔE, and ϒ∈Φ,(12)ϒρfr,S^r,ρS^r,T⌣j<0⇒θρfr,gj≤θAr,jEGr,j,where(13)Ar,j=maxρS^r,T⌣j,ρfr,S^r,ρgj,T⌣j,ρfr,T⌣j+ρgj,S^r2,1+ρfr,S^rρgj,T⌣j1+ρS^r,T⌣j,Gr,j=ρfr,S^r,ρgj,T⌣j,ρfr,T⌣j,ρgj,S^r.
Our first result is as follows.Theorem 2.
Letχ,ρ be a complete metric space, and f,g,S^,T⌣:χ⟶χ be four single-valued maps. Suppose that the mappings f,g,S^,T⌣ form a new generalized Suzuki-type θ,E-rational contraction with fχ⊆T⌣χ and gχ⊆S^χ. If f,S^ and g,T⌣ are compatible pair and T⌣ and S^ are continuous on χ,ρ, then f,g,S^, and T⌣ have a unique common fixed point in χ.Proof.
Letr0∈χ be an arbitrary point. As fχ⊆T⌣χ, there exists r0∈χ such that(14)fr0=T⌣r1.
Sincegr1∈S^χ, we can choose r2∈χ such that gr1=S^r2. In general r2n+1 and r2n+2 are chosen in χ such that(15)fr2n=T⌣r2n+1,gr2n+1=S^r2n+2.
We construct a sequenceℜn in χ such that(16)ℜ2n=fr2n=T⌣r2n+1,ℜ2n+1=gr2n+1=S^r2n+2,for n=0,1,2,…. Assume that ρfr2n,gr2n+1>0 and since(17)12ρfr2n,S^r2n=12ρℜ2n,ℜ2n−1<ρℜ2n−1,ℜ2n=ρS^r2n,T⌣r2n+1.
We have(18)ϒρfr2n,S^r2n,ρS^r2n,T⌣r2n+1<0.
Hence, from the contractive condition (12),(19)1<θρℜ2n,ℜ2n+1=θρfr2n,gr2n+1≤θAr2n,r2n+1EGr2n,r2n+1.
For everyn≥1, where(20)Gr2n,r2n+1=ρℜ2n,ℜ2n−1,ρℜ2n+1,ℜ2n,ρℜ2n,ℜ2n,ρℜ2n+1,ℜ2n−1=ρℜ2n,ℜ2n−1,ρℜ2n+1,ℜ2n,0,ρℜ2n+1,ℜ2n−1.
Asρℜ2n,ℜ2n−1⋅ρℜ2n+1,ℜ2n⋅ρℜ2n+1,ℜ2n−1⋅0=0, so by E∗, there exists z∈0,1 such that(21)EGr2n,r2n+1=Eρℜ2n,ℜ2n−1,ρℜ2n+1,ℜ2n,0,ρℜ2n+1,ℜ2n−1=z,(22)Ar2n,r2n+1=maxρS^r2n,T⌣r2n+1,ρfr2n,S^r2n,ρgr2n+1,T⌣r2n+1,ρfr2n,T⌣r2n+1+ρgr2n+1,S^r2n2,1+ρfr2n,S^r2nρgr2n+1,T⌣r2n+11+ρS^r2n,T⌣r2n+1,=maxρℜ2n−1,ℜ2n,ρℜ2n,ℜ2n−1,ρℜ2n+1,ℜ2n,ρℜ2n,ℜ2n+ρℜ2n+1,ℜ2n−12,1+ρℜ2n,ℜ2n−1ρℜ2n+1,ℜ2n1+ρℜ2n−1,ℜ2n,=maxρℜ2n−1,ℜ2n,ρℜ2n+1,ℜ2n,ρℜ2n+1,ℜ2n−12.
Since(23)ρℜ2n+1,ℜ2n−12≤ρℜ2n,ℜ2n−12+ρℜ2n+1,ℜ2n2≤maxρℜ2n−1,ℜ2n,ρℜ2n+1,ℜ2n.one writes(24)Ar2n,r2n+1=maxρℜ2n−1,ℜ2n,ρℜ2n+1,ℜ2n.
If, for somen, Ar2n,r2n+1=ρℜ2n,ℜ2n+1, then from (19), (21), and (24), we have(25)θρℜ2n,ℜ2n+1≤θρℜ2n,ℜ2n+1z<θρℜ2n,ℜ2n+1,a contradiction. Therefore,(26)θρℜ2n,ℜ2n+1≤θρℜ2n−1,ℜ2nz,forn≥1.
Thus, for alln∈ℕ, we have(27)θρℜn,ℜn+1≤θρℜn−1,ℜnz,forn≥1.
It implies that(28)1<θρℜn,ℜn+1≤θρℜn−1,ℜnz2⋮≤θρℜ0,ℜ1zn,for all n∈ℕ. Taking the limit as n⟶∞ in (28) and using the fact that θ∈μ, we have(29)limn⟶∞θρℜn,ℜn+1=1.
ByΘ2, we obtain(30)limn⟶∞ρℜn,ℜn+1=0.
Now, we will show thatℜn is a Cauchy sequence. We suppose on the contrary that ℜn is not a Cauchy sequence; then, there exist ε>0 and subsequences nkk=1∞ and mkk=1∞ of natural numbers such that, for mk>nk>k, we have ρℜnk,ℜmk≥ε. Then, ρℜnk,ℜmk−1<ε for all k∈ℕ. Therefore,(31)ε≤ρℜnk,ℜmk≤ρℜnk,ℜmk−1+ρℜmk−1,ℜmk<ε+ρℜmk−1,ℜmk.
By taking the upper limit ask⟶∞ in (31) and using (30), we obtain(32)limk⟶∞ρℜnk,ℜmk=ε.
Using the triangular inequality, we have(33)ρℜnk,ℜmk≤ρℜnk,ℜnk+1+ρℜnk+1,ℜmk,(34)ρℜnk+1,ℜmk≤ρℜnk+1,ℜnk+ρℜnk,ℜmk.
By taking upper limit ask⟶∞ in (33) and (34) and applying (30) and (32),(35)ε≤limk⟶∞ρℜnk+1,ℜmk≤ε.
Thus,(36)limk⟶∞ρℜnk+1,ℜmk=ε.
Similarly, we can obtain(37)limk⟶∞ρℜnk+1,ℜmk+1=limk⟶∞ρℜnk,ℜmk+1=ε.
Assume thatρfrnk+1,grmk+1>0, and since(38)12ρfrnk,S^rmk=12ρℜnk,ℜmk−1<ρℜnk,ℜmk−1=ρS^rmk,T⌣rnk+1,we obatin(39)ϒρfrnk,S^rmk,ρS^rnk,T⌣rmk+1<0.
Hence, from (12),(40)θρℜnk+1,ℜmk+1=θρfrnk+1,grmk+1≤θArnk,rmkEGrnk,rmk,where(41)Grnk,rmk=ρℜnk+1,ℜnk,ρℜmk+1,ℜmk,ρℜnk+1,ℜmk,ρℜmk+1,ℜnk.
Taking limit atk⟶∞ in (41), we obtain(42)limk⟶∞Grnk,rmk=limk⟶∞ρℜnk+1,ℜnk,limk⟶∞ρℜmk+1,ℜmk,limk⟶∞ρℜnk+1,ℜmk,limk⟶∞ρℜmk+1,ℜnk=0,0,ε,ε.
ByE∗, there exists z∈0,1 such that E0,0,ε,ε=z. Using the continuity of E and (41),(43)Elimk⟶∞Grnk,rmk≤z.
Also,(44)Arnk,rmk=maxρS^rnk+1,T⌣rmk+1,ρfrnk+1,S^rnk+1,ρgrmk+1,T⌣rmk+1,ρfrnk+1,T⌣rmk+1+ρgrmk+1,S^rnk+12,1+ρfrnk+1,S^rnk+1ρgrmk+1,T⌣rmk+11+ρS^rnk+1,T⌣rmk+1,=maxρℜnk,ℜmk,ρℜnk+1,ℜnk,ρℜmk+1,ℜmk,ρℜnk+1,ℜmk+ρℜmk+1,ℜnk2,1+ρℜnk+1,ℜnkρℜmk+1,ℜmk1+ρℜnk,ℜmk.
Takingk⟶∞ and using (30), (32), (36), and (37), we obtain(45)limk⟶∞Arnk,rmk≤ε.
Thus, fromΘ′3, (37), (40), (43), and (45), we have(46)θε=θlimk⟶∞ρℜnk+1,ℜmk+1≤θlimk⟶∞Arnk,rmkElimk⟶∞Grnk,rmk≤θεz<θε,which is a contradiction. Thus, ℜn is a Cauchy sequence. Since χ is a complete metric space, there exists r∗∈χ such that limn⟶∞ρℜn,r∗=0 and(47)limn⟶∞fr2n=limn⟶∞T⌣r2n+1=limn⟶∞gr2n+1=limn⟶∞S^r2n+2=r∗.
Now, we shall prove thatℜ∗ is a common fixed point of f,g, S^, and T⌣. As S^ is continuous, so(48)limn⟶∞S^fr2n=S^r∗=limn⟶∞S^S^r2n+2.
Sincef,S^ is a compatible pair,(49)limn⟶∞ρgT⌣r2n+1,T⌣gr2n+1=0.
From Lemma1, we have(50)limn⟶∞fS^r2n=S^r∗.
Putr1=S^r2n and r2=r2n+1 in (12) and if ρS^r∗,r∗>0, we obtain(51)12ρfS^r2n,S^S^r2n<ρS^S^r2n,T⌣r2n+1.
Hence,(52)ϒρfS^r2n,S^S^r2n,ρS^S^r2n,T⌣r2n+1<0,and by (12), we obtain(53)θρfS^r2n,gr2n+1≤θAr2n,r2n+1EGr2n,r2n+1,where(54)Ar2n,r2n+1=maxρS^S^r2n,T⌣r2n+1,ρfS^r2n,S^S^r2n,ρgr2n+1,T⌣r2n+1,ρfS^r2n,T⌣r2n+1+ρgr2n+1,S^S^r2n2,1+ρfS^r2n,S^S^r2nρgr2n+1,T⌣r2n+11+ρS^S^r2n,T⌣r2n+1,Gr2n,r2n+1=ρfS^r2n,S^S^r2n,ρgr2n+1,T⌣r2n+1,ρfS^r2n,T⌣r2n+1,ρgr2n+1,S^S^r2n.
Setting the upper limit in (53), we obtain(55)θρS^r∗,r∗≤θρS^r∗,r∗z<θρS^r∗,r∗,a contradiction. Thus, ρS^r∗,r∗=0 and S^r∗=r∗. Again since T⌣ is continuous,(56)limn⟶∞T⌣gr2n+1=T⌣r∗=limn⟶∞T⌣T⌣r2n+1.
Sinceg,T⌣ is a compatible pair,(57)limn⟶∞ρgT⌣r2n+1,T⌣gr2n+1=0.
From Lemma1, we have(58)limn⟶∞gT⌣r2n+1=T⌣r∗.
Putr1=r2n and r2=T⌣r2n+1 in (12) and if ρr∗,T⌣r∗>0, we obtain(59)12ρfr2n,S^r2n<ρS^r2n,T⌣T⌣r2n+1.
Therefore,(60)ϒρfr2n,S^r2n,ρS^r2n,T⌣T⌣r2n+1<0.
By (12), one obtains(61)θρfr2n,gT⌣r2n+1≤θAr2n,r2n+1EGr2n,r2n+1,where(62)Ar2n,r2n+1=maxρS^r2n,T⌣T⌣r2n+1,ρfr2n,S^r2n,ρgT⌣r2n+1,T⌣T⌣r2n+1,ρfr2n,T⌣T⌣r2n+1+ρgT⌣r2n+1,S^r2n2,1+ρfr2n,S^r2nρgT⌣r2n+1,T⌣T⌣r2n+11+ρS^r2n,T⌣T⌣r2n+1,Gr2n,r2n+1=ρfr2n,S^r2n,ρgT⌣r2n+1,T⌣T⌣r2n+1,ρfr2n,T⌣T⌣r2n+1,ρgT⌣r2n+1,S^r2n.
Setting the upper limit, we obtain(63)θρr∗,T⌣r∗≤θρr∗,T⌣r∗z<θρr∗,T⌣r∗,a contradiction. Thus, ρr∗,T⌣r∗=0 and so r∗=T⌣r∗. Suppose ρfr∗,r∗>0, we obtain(64)12ρS^r∗,T⌣r2n+1<ρfr∗,S^r∗.
Hence,(65)ϒρfr∗,S^r∗,ρS^r∗,T⌣r2n+1<0,and by (12), we obtain(66)θρfr∗,gr2n+1≤θAr∗,r2n+1EGr∗,r2n+1,where(67)Ar∗,r2n+1=maxρS^r∗,T⌣r2n+1,ρfr∗,S^r∗,ρgr2n+1,T⌣r2n+1,ρfr∗,T⌣r2n+1+ρgr2n+1,S^r∗2,1+ρfr∗,S^r∗ρgr2n+1,T⌣r2n+11+ρS^r∗,T⌣r2n+1,Gr∗,r2n+1=ρfr∗,S^r∗,ρgr2n+1,T⌣r2n+1,ρfr∗,T⌣r2n+1,ρgr2n+1,S^r∗.
Taking the upper limit in (66), and as r∗=T⌣r∗=S^r∗, so we obtain(68)θρfr∗,r∗≤θρfr∗,r∗z<θρfr∗,r∗,a contradiction. Thus, ρfr∗,r∗=0 and r∗=fr∗. Finally, suppose ρr∗,gr∗>0, and as r∗=T⌣r∗=S^r∗=fr∗, so, we obtain(69)ϒ0,ρr∗,gr∗<0.
Thus, from (12)(70)θρr∗,gr∗=θρfr∗,gr∗≤θρr∗,gr∗z<θρr∗,gr∗,a contradiction. Thus, ρr∗,gr∗=0 and r∗=gr∗. Therefore, r∗ is a common of the four mappings S^,g,f and T⌣. Next, assume that υ∗is another common of the four mappings S^,g,f, and T⌣ such that r∗≠υ∗, one obtains(71)ϒ0,ρS^r∗,T⌣υ∗<0.
Then, by (12), we obtain(72)θρr∗,υ∗=θρfr∗,gυ∗≤θAr∗,υ∗EGr∗,υ∗,where(73)Ar∗,υ∗=maxρS^r∗,T⌣υ∗,ρfr∗,S^r∗,ρgυ∗,T⌣υ∗,ρfr∗,T⌣υ∗+ρgυ∗,S^r∗2,1+ρfr∗,S^r∗ρgυ∗,T⌣υ∗1+ρS^r∗,T⌣υ∗,Gr∗,υ∗=ρfr∗,S^r∗,ρgυ∗,T⌣υ∗,ρfr∗,Tˇυ∗,ρgυ∗,S^r∗.
It implies that(74)θρr∗,υ∗≤θρr∗,υ∗z<θρr∗,υ∗,a contradiction. Thus, r∗=υ∗. Therefore, r∗ is the unique common fixed point of S^,g,f, and T⌣.Corollary 1.
Letχ,ρ be a complete metric space, and f,g,S^,T⌣:χ⟶χ be four single-valued maps. Suppose that if, for all r,j∈χ with ρfr,gj>0 for some θ∈μ, z∈0,1 and ϒ∈Φ,(75)ϒρfr,S^r,ρS^r,T⌣j<0⟹θρfr,gj≤θAr,jz,where(76)Ar,j=maxρS^r,T⌣j,ρfr,S^r,ρgj,T⌣j,ρfr,T⌣j+ρgj,S^r2,1+ρfr,S^rρgj,T⌣j1+ρS^r,T⌣j,with fχ⊆T⌣χ and gχ⊆S^χ. If f,S^ and g,T⌣ are compatible pairs and T⌣ and S^ are continuous on χ,ρ, then f,g, S^, and T⌣ have a unique common fixed point in χ.Example 5.
Letχ=0,1 and define the function ρ:χ×χ⟶0,+∞ by ρr,j=r−j. Clearly, χ,ρ is a complete metric space. Let θt=et, t>0, and ϒs,t=s/2−t, then θ∈μ and ϒ∈Φ. Define the mappings S^,g,f,T⌣:χ⟶χ by(77)S^r=r38,gr=r38,fr=r316,T⌣r=r34.
Clearly,fχ⊆T⌣χ and gχ⊆S^χ, if rn is a sequence in χ such that(78)limn⟶∞frn=limn⟶∞S^rn=t,for somet∈χ,then(79)limn⟶∞frn−t=limn⟶∞S^kn−t=0,and equivalently,(80)limn⟶∞rn316−t=limn⟶∞rn38−t=0.
Thus,(81)limn⟶∞rn16−316t=limn⟶∞rn8−38t=0.
We concludet1/16=t1/8 (by uniqueness of limit); hence, t∈0,1. Using continuity of f and S^, one obtains(82)limn⟶∞ρfS^rn,S^frn=limn⟶∞fS^rn−S^frn=ft−S^t=0−0=0,for t=0∈χ. Hence, the pair f,S^ is compatible. Similarly, the pair g,T⌣ is compatible. Define E:ℝ4⟶ℝ+ as Es1,s2,s3,s4=9/10. Next, for all r,j∈χ with ρfr,gj>0,(83)ϒρfr,S^r,ρS^r,T⌣j<0,and so(84)θρfr,gj≤θAr,jEGr,j.
Hence, all hypotheses of Theorem2 are satisfied. Thus, f,g, S^, and T⌣ have a unique common fixed point, which is r∗=0.
We can easily obtain the following results by chasing the proof of Theorem2.Corollary 2.
Letχ,ρ be a complete metric space, and f,g,S^,T⌣:χ⟶χ be four single-valued maps. Suppose that if, for all r,j∈χ, with ρfr,gj>0, for some θ∈μ, ϒ∈Φ and E:ℝ+4⟶ℝ+,(85)Υρfr,S^r,ρS^r,T⌣j<0⟹θρfr,gj≤θmaxρfr,S^r,ρgj,T⌣jEGr,j,where(86)Gr,j=ρfr,S^r,ρgj,T⌣j,ρfr,T⌣j,ρgj,S^r,with fχ⊆T⌣χ and gχ⊆S^χ. If f,S^ and g,T⌣ are compatible pairs, and T⌣ and S^ are continuous on χ,ρ, then f,g, S^, and T⌣ have a unique common fixed point in χ.The following corollary generalizes the result introduced by Sehgal [34].Corollary 3.
Letχ,ρ be a complete metric space, and f,g,S^,T⌣:χ⟶χ be four single-valued maps. Suppose that if, for all r,j∈χ, with ρfr,gj>0, for some θ∈μ, ϒ∈Φ and E:ℝ+4⟶ℝ+, ϒρfr,S^r,ρS^r,T⌣j<0 implies that(87)θρfr,gj≤θmaxρS^r,T⌣j,ρfr,S^r,ρgj,T⌣jEGr,j,where(88)Gr,j=ρfr,S^r,ρgj,T⌣j,ρfr,T⌣j,ρgj,S^r,with fχ⊆T⌣χ and gχ⊆S^χ. If f,S^ and g,T⌣ are compatible pairs, and T⌣ and S^ are continuous on χ,ρ, then f,g, S^, and T⌣ have a unique common fixed point in χ.


## 4. On Suzuki-Type Ćirić JS-Contractions

We start this section with the following definition.Definition 4.
Letχ,ρ be a metric space and f,g,S^,T⌣:χ⟶χ be four single-valued maps. These maps form a new generalized Suzuki-type Ćirić JS-contraction if, for all r,j∈χ, for some θ∈Ψ, ϒ∈Φ, ϒρfr,S^r,ρS^r,T⌣j<0 implies that(89)θρfr,gj≤θρS^r,T⌣ja⋅θρfr,S^rb⋅θρgj,T⌣jc⋅θρfr,T⌣j+ρgj,S^rd,where a,b,c,d≥0 with a+b+c+2d<1.
Our second result is as follows.Theorem 3.
Letχ,ρ be a complete metric space and f,g,S^,T⌣:χ⟶χ be four single-valued maps. Suppose that the mappings f,g,S^,T⌣ form a generalized Suzuki-type Ćirić JS-contraction with fχ⊆T⌣χ and gχ⊆S^χ. If f,S^ and g,T⌣ are compatible pairs and T⌣ and S^ are continuous on χ,ρ, then f,g, S^, and T⌣ have a unique common fixed point in χ.Proof.
Letr0∈χ be an arbitrary point. As fχ⊆T⌣χ, there exists r0∈χ such that(90)fr0=T⌣r1.
Sincegr1∈S^χ, we can choose r2∈χ such that gr1=S^r2. In general, r2n+1 and r2n+2 are chosen in χ such that(91)fr2n=T⌣r2n+1,gr2n+1=S^r2n+2.
We construct a sequenceℜn in χ such that(92)ℜ2n=fr2n=T⌣r2n+1,ℜ2n+1=gr2n+1=S^r2n+2,for n=0,1,2,…. Since(93)12ρfr2n,S^r2n=12ρℜ2n,ℜ2n−1<ρℜ2n−1,ℜ2n=ρS^r2n,T⌣r2n+1,we have(94)ϒρfr2n,S^r2n,ρS^r2n,T⌣r2n+1<0.
Then, from (89),(95)θρℜ2n,ℜ2n+1=θρfr2n,gr2n+1≤θρS^r2n,Tˇr2n+1a⋅θρfr2n,S^r2nb⋅θρgr2n+1,Tˇr2n+1c⋅θρfr2n,Tˇr2n+1+ρgr2n+1,S^r2nd=θρℜ2n−1,ℜ2na⋅θρℜ2n,ℜ2n−1b⋅θρℜ2n+1,ℜ2nc⋅θρℜ2n,ℜ2n+ρℜ2n+1,ℜ2n−1d=θρℜ2n−1,ℜ2na+b⋅θρℜ2n+1,ℜ2nc⋅θρℜ2n−1,ℜ2n+ρℜ2n,ℜ2n+1d.
ByΨ5, we have, for every n∈ℕ,(96)θρℜ2n−1,ℜ2n+ρℜ2n,ℜ2n+1≤θρℜ2n−1,ℜ2n⋅θρℜ2n,ℜ2n+1.
Hence, (95) becomes(97)1<θρℜ2n,ℜ2n+1≤θρℜ2n−1,ℜ2na+b+d⋅θρℜ2n+1,ℜ2nc+d.
It implies that(98)θρℜ2n,ℜ2n+11−c−d≤θρℜ2n−1,ℜ2na+b+d.
Therefore,(99)θρℜ2n,ℜ2n+1≤θρℜ2n−1,ℜ2na+b+d/1−c−d,and so(100)θρℜn,ℜn+1≤θρℜn−1,ℜna+b+d/1−c−d.
It implies that(101)1<θρℜn,ℜn+1≤θρℜn−1,ℜna+b+d/1−c−d2⋮≤θρℜ0,ℜ1a+b+d/1−c−dn,for all n∈ℕ. Taking the limit as n⟶∞ in (101) and knowing that θ∈Ψ, we have(102)limn⟶∞θρℜn,ℜn+1=1.
ByΨ3, we obtain(103)limn⟶∞ρℜn,ℜn+1=0.
From conditionΨ4, there exist q∈0,1 and ℓ∈0,∞ such that(104)limn⟶∞θρℜn,ℜn+1−1ρℜn,ℜn+1q=ℓ.
Suppose thatℓ<∞. Let V=ℓ/2>0. From the definition of the limit, there exists n0∈ℕ such that(105)θρℜn,ℜn+1−1ρℜn,ℜn+1q−ℓ≤V,for alln≥n0.
This implies(106)θρℜn,ℜn+1−1ρℜn,ℜn+1q≥ℓ−V=V,for alln≥n0.
Then,(107)nρℜn,ℜn+1q≤Pnθρℜn,ℜn+1−1,for alln≥n0,where P=1/V. Suppose now that ℓ=∞. Let V>0 be an arbitrary positive number. From the definition of the limit, there exists n0≥1 such that(108)θρℜn,ℜn+1−1ρℜn,ℜn+1q≥V,for alln≥n0,which implies(109)nρℜn,ℜn+1q≤Pnθρℜn,ℜn+1−1,n≥n0,where P=1/V. Thus, in all cases, there exist P>0 and n0≥1 such that(110)nρℜn,ℜn+1q≤Pnθρℜn,ℜn+1−1,n≥n0.
By using (101), we obtain(111)nρℜn,ℜn+1q≤Pnθρℜ0,ℜ1a+b+d/1−c−dn−1,for alln≥n0.
Settingn⟶∞ in inequality (111), we obtain(112)limn⟶∞nρℜn,ℜn+1q=0.
Thus, there existsn1∈ℕ such that(113)ρℜn,ℜn+1≤1n1/q,for alln≥n1.
To proveℜn is a Cauchy sequence, we use (113) and for m>n≥n1,(114)ρℜn,ℜm≤∑i=nm−1ρℜi,ℜi+1≤∑i=n∞ρℜi,ℜi+1≤∑i=n∞1i1/q.
The convergence of the series∑i=n∞1/i1/q entails limn,m⟶∞ρℜn,ℜm=0. Thus, ℜn is a Cauchy sequence. Since χ is a complete metric space, there exists r∗∈χ such that limn⟶∞dℜn,r∗=0 and(115)limn⟶∞fr2n=limn⟶∞T⌣r2n+1=limn⟶∞gr2n+1=limn⟶∞S^r2n+2=r∗.
Now, we shall prove thatℜ∗ is a common fixed point of f,g, S^, and T⌣. As S^ is continuous, so(116)limn⟶∞S^fr2n=S^r∗=limn⟶∞S^S^r2n+2.
Sincef,S^ is a compatible pair,(117)limn⟶∞ρfS^r2n,S^fr2n=0.
From Lemma1, we have(118)limn⟶∞fS^r2n=S^r∗.
Putr=S^r2n and j=r2n+1 in (89) and if ρS^r∗,r∗>0, we obtain(119)12ρfS^r2n,S^S^r2n<ρS^S^r2n,T⌣r2n+1.
Hence,(120)ϒρfS^r2n,S^S^r2n,ρS^S^r2n,T⌣r2n+1<0.
Then, from (89), we obtain(121)θρfS^r2n,gr2n+1≤θρS^S^r2n,T⌣r2n+1a⋅θρfS^r2n,S^S^r2nb⋅θρgr2n+1,T⌣r2n+1c⋅θρfS^r2n,T⌣r2n+1+ρgr2n+1,S^S^r2nd.
Setting the upper limit, we obtain(122)θρS^r∗,r∗≤θρS^r∗,r∗a⋅θρS^r∗,r∗+ρr∗,S^r∗d≤θρS^r∗,r∗a+2d<θρS^r∗,r∗,a contradiction. Thus, ρS^r∗,r∗=0, and so S^r∗=r∗. Again, since T⌣ is continuous,(123)limn⟶∞T⌣gr2n+1=T⌣r∗=limn⟶∞T⌣T⌣r2n+1.
Sinceg,T⌣ is a compatible pair,(124)limn⟶∞ρgT⌣r2n+1,T⌣gr2n+1=0.
From Lemma1, we have(125)limn⟶∞gT⌣r2n+1=T⌣r∗.
Putr1=r2n and r2=T⌣r2n+1 in (89) and if ρr∗,T⌣r∗>0, we obtain(126)12ρfr2n,S^r2n<ρS^r2n,T⌣T⌣r2n+1.
Therefore,(127)ϒρfr2n,S^r2n,ρS^r2n,T⌣T⌣r2n+1<0,and from (89), one obtains(128)θρfr2n,gT⌣r2n+1≤θρS^r2n,T⌣T⌣r2n+1a⋅θρfr2n,S^r2nb⋅θρgT⌣r2n+1,T⌣T⌣r2n+1c⋅θρfr2n,T⌣T⌣r2n+1+ρgT⌣r2n+1,S^r2nd.
Passing to the upper limit in (128), we obtain(129)θρr∗,T⌣r∗≤θρr∗,T⌣r∗a+2d<θρr∗,T⌣r∗,a contradiction. Thus, ρr∗,T⌣r∗=0 and hence r∗=T⌣r∗. Suppose that ρfr∗,r∗>0, we obtain(130)12ρS^r∗,T⌣r2n+1<ρfr∗,S^r∗,for alln∈ℕ,and so(131)ϒρfr∗,S^r∗,ρS^r∗,T⌣r2n+1<0.
By (89), we obtain(132)θρfr∗,gr2n+1≤θρS^r∗,T⌣r2n+1a⋅θρfr∗,S^r∗b⋅θρgr2n+1,T⌣r2n+1c⋅θρfr∗,T⌣r2n+1+ρgr2n+1,S^r∗d.
Taking the upper limit in (132) and as r∗=T⌣r∗=S^r∗, we obtain(133)θρfr∗,r∗≤θρfr∗,r∗b+d<θρfr∗,r∗,a contradiction. Thus, ρfr∗,r∗=0 and hence r∗=fr∗. Finally, suppose ρr∗,gr∗>0 and as r∗=T⌣r∗=S^r∗=fr∗, we have(134)ϒ0,ρr∗,gr∗<0.
From (89), we deduce(135)θρr∗,gr∗=θρfr∗,gr∗<θρr∗,gr∗,a contradiction. Thus, ρr∗,gr∗=0 and r∗=gr∗. Therefore, r∗ is a common of the four mappings S^,g,f, and T⌣. Next, assume that υ∗ is another common of the four mappings S^,g,f, and T⌣ such that r∗≠υ∗, one obtains(136)ϒ0,ρS^r∗,T⌣υ∗<0.
Then, by (89),(137)θρr∗,υ∗=θρS^r∗,T⌣υ∗≤θρS^r∗,T⌣υ∗a⋅θρfr∗,S^r∗b⋅θρgυ∗,T⌣υ∗c⋅θρfr∗,T⌣υ∗+ρgυ∗,S^r∗d.
It implies that(138)θρr∗,υ∗≤θρr∗,υ∗a+2d<θρr∗,υ∗,a contradiction. Thus, r∗=υ∗, that is, r∗ is the unique common fixed point of S^,g,f, and T⌣.Example 6.
Letχ=0,∞ and define the function ρ:χ×χ⟶0,+∞ by ρr,j=r−j. Clearly, χ,ρ is a complete metric space. Let θt=et, and ϒs,t=s/2−t, then θ∈Ψ and ϒ∈Φ. Define the mappings S^,g,f,T⌣:χ⟶χ by(139)S^r=e7r−1,gr=ln1+r6,fr=ln1+r6,T⌣r=e6r−1.
Clearly,fχ=T⌣χ=gχ=S^χ if rn is a sequence in χ such that(140)limn⟶∞frn=limn⟶∞S^rn=t,for somet∈χ.
Then,(141)limn⟶∞frn−t=limn⟶∞S^rn−t=0,and equivalently,(142)limn⟶∞ln1+rn6−t=limn⟶∞e7rn−1−t=0.
Thus,(143)limn⟶∞rn−6et−6=limn⟶∞rn−ln1+t7=0.
It gives that6et−6=ln1+t/7 (by uniqueness of limit); hence, t=0. Using continuity of f and S^, one obtains(144)limn⟶∞ρfS^rn,S^frn=limn⟶∞fS^rn−S^frn=ft−S^t=0−0=0,for t=0∈χ. Hence, the pair f,S^ is compatible. Similarly, the pair g,T⌣ is compatible. Next, for all r,j∈χ with(145)ϒρfr,S^r,ρS^r,T⌣j<0.
So, fora=b=c=0.1 and d=0.3, we have(146)θρfr,gj≤θρS^r,T⌣ja⋅θρfr,S^rb⋅θρgj,T⌣jc⋅θρfr,T⌣j+ρgj,S^rd.
Hence, all hypotheses of Theorem3 are satisfied. Thus, f,g, S^, and T⌣ have a unique common fixed point.


## 5. An Application

We apply the result given by Theorem2 to study the existence of a solution for a system of nonlinear fractional differential equations. Let χ=C0,1,ℝ be the space of all continuous functions on 0,1. The metric on χ is given by(147)ρr,j=r−j∞=maxt∈0,1rt−jt,r,j∈χ.Then,χ=C0,1,ℝ is complete metric space.Consider the following system of fractional differential equations:(148)DαCrt=K1t,S^rt,DαCjt=K2t,T⌣jt,with boundary conditions(149)r0=0,Ir1=r′0,j0=0,Qj1=j′0.Note thatDαC denotes the Caputo fractional derivative of order α, defined by(150)DαCK1t=1Γn−α∫0tt−sn−α−1K1nsds,DαCK2t=1Γn−β∫0tt−sn−α−1K2nsds,where we consider(151)n−1<α,α<1,n=α+1,and IαK1 and IαK2 denote the Riemann–Liouville fractional integral of order α of continuous functions K1 and K2, given by(152)IαK1t=1Γα∫0tt−sα−1K1sds,withα>0,QαK2t=1Γα∫0tt−sα−1K2sds,withα>0.System (148) can be written in the following integral form:(153)rt=1Γα∫0tt−sα−1K1s,S^rsds+2tΓα∫01∫0ss−uα−1K1u,S^rududs,jt=1Γα∫0tt−sα−1K2s,T⌣jsds+2tΓα∫01∫0ss−uα−1K2u,T⌣jududs.Define the mappingsf,g:χ⟶χ by(154)frt=1Γα∫0tt−sα−1K1s,S^rsds2tΓα∫01∫0ss−uα−1K1u,S^rududs,gjt=1Γα∫0tt−sα−1K2s,T⌣jsds+2tΓα∫01∫0ss−uα−1K2u,T⌣jududs.Theorem 4.
Assume that the following conditions hold:(i)
K1,K20,1×ℝ⟶ℝ are continuous functions.(ii)
K1s,.,K2s,.:⟶ℝ are increasing functions.(iii)
For allr,j∈χ with fr≤T⌣j and frs−gjs>0, we have(155)K1s,S^rs−K2s,T⌣js≤e−τΓα+14Ar,j,where(156)Ar,j=maxS^rs−T⌣js,frs−S^rs,gjs−T⌣js,frs−T⌣js+gjs−S^rs2,1+|frs−S^rs|gjs−T⌣js1+S^rs−T⌣js.(iv)
There existsr0,j0∈C0,1,ℝ such that, for all t∈0,1, we have(157)r0t≤1Γα∫0tt−sα−1K1s,S^r0sds+2tΓα∫01∫0ss−uα−1K1u,S^r0ududs,j0t≤1Γα∫0tt−sα−1K2s,T⌣j0sds+2tΓα∫01∫0ss−uα−1K2u,T⌣j0ududs.(v)
If there exists a sequencern in χ such that(158)limn⟶∞ρfS^rn,S^frn=0,limn⟶∞ρgT⌣rn,T⌣grn=0,whenever(159)limn⟶∞frn=limn⟶∞S^rn=t,limn⟶∞grn=limn⟶∞T⌣rn=t,for some t∈χ.
Then, system (148) has a solution.Proof.
Following assumptions (iii) and (iv), we have(160)frt−gjt=1Γα∫0tt−sα−1K1s,S^rsds−1Γα∫0tt−sα−1K2s,Tˇjsds+2tΓα∫01∫0ss−uα−1K1u,S^rududs−2tΓα∫01∫0ss−uα−1K2u,Tˇjududs≤1Γα∫0tt−sα−1K1s,S^rs−K2s,Tˇjsds+2Γα∫01∫0ss−uα−1K1s,S^rs−K2u,Tˇjududs≤1Γαe−τΓα+14⋅∫0tt−sα−1Ar,jds+2Γαe−τΓα+14⋅∫01∫0ss−uα−1Ar,jduds≤1Γαe−τΓα+14⋅Ar,j⋅∫0tt−sα−1ds+2Γαe−τΓα+14⋅Ar,j⋅∫01∫0ss−uα−1duds≤e−τΓα⋅Γα+14Γα⋅Γα+1⋅Ar,j+2e−τBα+1,1Γα⋅Γα+14Γα⋅Γα+1⋅Ar,j≤e−τ4Ar,j+e−τ2Ar,j,where B is the beta function. From the above inequality, we obtain that(161)ρfr,gj≤34e−τAr,j.
It implies that(162)eρfr,gj≤e3/4e−τAr,j=e3/4e−τAr,j=eAr,j3/4e−τ=eAr,jEGr,j,where Es1,s2,s3,s4=3/4eτ<1,Gr,j=ρfr,S^r,ρgj,T⌣j,ρfr,T⌣j,ρgj,S^r and θt=et. Since this inequality holds for all r,j∈χ with fr≤T⌣j, so it is true for any ϒ∈Φ,(163)ϒρfr,S^r,ρS^r,T⌣j<0.
Hence, we have(164)θρfr,gj≤θAr,jEGr,j.
Thus,f,g,S^, and T⌣ are generalized Suzuki-type θ,E-rational contraction mappings. Therefore, all hypotheses of Theorem 2 are satisfied. Hence, f,g,S^, and T⌣ have a common fixed point, that is, system (148) has at least one solution.



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