# Fractional Hybrid Differential Equations and Coupled Fixed-Point Results forα-Admissible Fψ1,ψ2−Contractions in M−Metric Spaces

Authors: Erdal Karapinar; Shimaa I. Moustafa; Ayman Shehata; Ravi P. Agarwal
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we investigate the existence of a unique coupled fixed point forα−admissible mapping which is of Fψ1,ψ2−contraction in the context of M−metric space. We have also shown that the results presented in this paper would extend many recent results appearing in the literature. Furthermore, we apply our results to develop sufficient conditions for the existence and uniqueness of a solution for a coupled system of fractional hybrid differential equations with linear perturbations of second type and with three-point boundary conditions.


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## Body

## 1. Introduction

Fixed-point theory is an outstanding source which gives responsible techniques for the existence of fixed points for self-mappings under different conditions. One of the newest branches of fixed-point theory concerned with the study of coupled fixed points, brought by Guo and Lakshmikantham [1]. In [2], Bhaskar and Lakshmikantham established some fixed and coupled fixed-point theorems for contractions in two variables defined on partially ordered metric spaces with applications to ordinary differential equations. Thereafter, these results were extended by several authors (see [3–6]).Inspired by the notion of partial metric (or,p−metric) which is one of the vital generalizations of the standard metric, Asadi et al. [7] proposed the concept of M−metric which refines the p−metric and produces useful basic topological concepts. For some fixed-point results and various contractive definitions that have been employed in M−metric space, we refer the reader to [8–12].In [13] (see also, [14–16]), Monfared et al. established some fixed-point results for α−admissible mappings which are Fψ,φ−contractions in complete M−metric spaces. Now, we state one of their main results.Theorem 1.
LetX,μ be a complete M−metric space and T:X⟶X be an α−admissible mapping. Suppose that the following condition is satisfied:(1)ψμTx,Ty+lαx,Txαy,Ty≤Fψμx,y,φμx,y+l,for all x,y∈X and l≥1, where F∈C, ψ is an altering distance function, and φ is an ultra-altering distance function. Suppose that either(a)
T is continuous(b)
Ifxn is a sequence in X such that xn⟶x, αxn,xn+1≥1, ∀n, then αx,Tx≥1
If there existsx0∈X such that αx0,Tx0≥1, then T has a fixed point.Hybrid differential equations have been of great interest as they include several dynamic systems as special cases. The papers [17, 18] discussed the existence and uniqueness results and some fundamental differential inequalities for first-order hybrid differential equations with perturbations of 1st and 2nd type, respectively.Fractional calculus is a field of mathematics that deals with the derivatives and integrals of arbitrary order. Indeed, it is found to be more realistic in describing and modeling several natural phenomena than the classical one. In fact, fractional differential equations (FDEs) play a major role in modeling many real-life problems such as physical phenomena, computer networking, medicine (the modeling of human tissue), mechanics (theory of viscoelasticity), electrical engineering (transmission of ultrasound waves) and many others (see [19–21]).Fractional hybrid differential equations (FHDEs) can be employed in modeling and describing nonhomogenous physical phenomena that take place in their form. FHDEs have been studied using a Riemann–Liouville differential operator of orderα>0 in many literature studies (see [22–26]).In [27], Shaob et al. used Bashiri fixed-point theorem [22] to prove the existence only of a solution to a three-point boundary value problem for a coupled system of FHDEs in Banach spaces.In line with the above studies, our purpose in this paper is to introduce the notion ofα−admissible mapping with two variables and generalize Theorem 1 to coupled fixed-point version. Then, we apply our main results to prove the existence and uniqueness of a solution to the following system of FHDEs involving Riemann–Liouville fractional derivative:(2)Dαxt−ft,xt=gt,yt,Iβyt,xi0=∂ift,xt∂tit=0=0,xτ=δxη,(3)Dαyt−ft,yt=gt,xt,Iβxt,yi0=∂ift,yt∂tit=0=0,yτ=δyη,for all i=0,1,…,n−2, t∈J=o,τ,τ>0, α∈n−1,n, β>0, 0<η<τ, δ≠τ/ηα−1, f∈CJ×ℝ, and g∈CJ×ℝ2.


## 2. Preliminaries

In 1994, Matthews [28] introduced the notion of a p−metric space as a part of the study of denotational semantics of dataflow networks. In p−metric spaces, self-distance of an arbitrary point need not be equal to zero.Definition 1.
(see [28]). A p−metric on a nonempty set X is a mapping p:X×X⟶0,∞ such that, for all x,y,z∈X,p1px,x=py,y=px,y⇔x=yp2px,x≤px,yp3px,y=py,xp4px,y≤px,z+pz,y−pz,z
Then,X,p is called a p−metric space.
Notice that, every metric space can be defined to bep−metric space with zero self-distance. After that, Asadi et al. generalized the above definition by relaxing the axiom p2 as follows.Definition 2.
(see [7]). For a nonempty set X, a function μ:X×X⟶0,∞ is called an M−metric if it fulfils the following:m1μx,x=μy,y=μx,y⇔x=ym2mxy≤μx,y, where mxy=minμx,x,μy,ym3μx,y=μy,xm4μx,y−mxy≤μx,z−mxz+μz,y−mzy
Then, the pairX,μ is called an M−metric space.Lemma 1.
(see [7]). Every p−metric is an M−metric.
Here, we give an example to show that the converse might not be held.Example 1.
(see [7]). Let X=1,2,3 and define(4)μ1,2=μ2,1=10,μ1,1=1,μ2,2=9,μ1,3=μ3,1=μ3,2=μ2,3=7,μ3,3=5.
Soμ is M−metric but it is not p−metric for μ2,2≰μ2,3. Also, μ is not metric for self-distances are not zero.
Thus, the class ofM−metric spaces is effectively larger than that of both ordinary metric and p−metric spaces.Notation 1.
LetX,μ be an M−metric space; then define(5)μwx,y=μx,y−2mxy+Mxy,whereMxy=maxμx,x,μy,y.
Hence,μw is an ordinary metric induced by the M−metric μ.
EachM−metric μ on X generates a T0 topology τμ on X formed by the set(6)Bμx,ε:x∈X,ε>0,where(7)Bμx,ε=y∈X:μx,y<mxy+ε.
The notions of convergent sequence, Cauchy sequence, and completeM−metric space X,μ are given as follows:(1)
A sequencexn in X,μ converges to a point x∈X if(8)limn⟶∞μxn,x−mxnx=0.(2)
A sequencexn in X,μ is called μ−Cauchy if(9)limn,m⟶∞μxn,xm−mxnxm,limn,m⟶∞Mxnxm−mxnxmexist and are finite.(3)
X,μ is said to be complete if every μ−Cauchy sequence xn in it converges, with respect to τμ, to a point x∈X, and(10)limn,m⟶∞μxn,xm−mxnxm=limn,m⟶∞Mxnxm−mxnxm=0.Lemma 2 (see [7]).
LetX,μ be an M−metric space; then,(1)
xn is a μ−Cauchy sequence in X,μ if and only if it is Cauchy sequence in the metric space X,μw.(2)
X,μ is complete if and only if X,μw is complete. Furthermore,(11)limn⟶∞μwxn,x=0⟺limn⟶∞μxn,x−mxnx=limn⟶∞Mxnx−mxnx=0.Lemma 3 (see [7]).
Assume thatxn⟶x and yn⟶y in an M−metric space X,μ; then,(12)limn⟶∞μxn,yn−mxnyn=μx,y−mxy.As a consequence of Lemma3, we have(13)xn⟶x,inX,μ⟹limn⟶∞μxn,y−mxny=μx,y−mxy,xn⟶xandxn⟶y,inX,μ⟹limn⟶∞μxn,xn−mxnxn=μx,y−mxy.Definition 3.
(see [29]). A mapping F:0,∞2⟶R is called a C−class function if it is continuous and satisfies the following axioms:(1)
Fs,t≤s(2)
Fs,t=s implies that either s=0 or t=0 for all t∈0,∞
LetC denote the C−class functions.Definition 4.
(see [20, 21]). The fractional integral of order α>0 of a function x:0,∞⟶ℝ is given by(14)Iαxt=1Γα∫0tt−sα−1xsds,provided that the right side is pointwise defined on 0,∞.Definition 5.
(see [20, 21]). The fractional derivative of order α>0 of a continuous function x:0,∞⟶ℝ is given by(15)Dαxt=1Γn−αddtn∫0txst−sα−n+1ds,where n=α+1, provided that the right side is pointwise defined on 0,∞.Lemma 4 (see [30]).
Riemann–Liouville fractional integral and derivative have the following properties:(1)
IαIβxt=Iα+βxt and DαIβxt=Iβ−αxt, for all β≥α>0, x∈L0,1(2)
IαDαxt=xt+c1tα−1+⋯+cntα−n, where n=α+1 and x,Dαx∈C0,1∩L0,1(3)
Iα:C0,1⟶C0,1, α>0


## 3. Fixed-Point Results

First, we introduce the following concepts that generalize the corresponding ones used in [13] and will be beneficial in the sequel.Definition 6.
LetT:X×X⟶X and α:X×X⟶0,∞; then, T is called an α−admissible mapping if(16)αx,u≥1,αy,v≥1⟹αTx,y,Tu,v≥1,∀x,y,u,v∈X2.
Note that, if equation (16) holds, then we have αTy,x,Tv,u≥1 too. Consider the following classes of functions:(17)Ψ1=ψ:0,∞2⟶0,∞,ψis continuous, strictly increasing andψt1,t2=0⟹t1=t2=0,Ψ2=ψ:0,∞×0,∞⟶0,∞,ψis continuous andψt1,t2=0⟹t1=t2=0,Φ=φ:0,∞⟶0,∞,φs+t≤φs+φtandφt2≤φt2∀s,t≥0.Theorem 2.
LetX,μ be a complete M−metric space and T:X×X⟶X be an α−admissible mapping for which there exist F∈C, ϕ∈Φ, ψ1∈Ψ1, and ψ2∈Ψ2 such that ψ1t.t≤ϕt and for all x,y,u,v∈X2 with αx,u≥1,αy,v≥1; we have(18)ϕμTx,y,Tu,v+lmaxαx,u,αy,v≤Fψ1Kx,u+Ky,v2,ψ2Kx,u+Ky,v2+l,where(19)Kx,u=μu,Tu,v1+μx,Tx,y1+μx,u,μx,u,Ky,v=μv,Tv,u1+μy,Ty,x1+μy,v,μy,v.
Suppose that either(a)
T is continuous.(b)
For a convergent sequencexn in X,μ, we have(20)xn⟶x,αxn,xn+1≥1⟹αxn,x≥1,∀n,xn⟶x,xn⟶y⟹αx,y≥1.
If there existx0,y0∈X such that αx0,Tx0,y0≥1 and αy0,Ty0,x0≥1, then T has a coupled fixed point.Proof.
Starting withx0,y0∈X, define the sequences xn,yn⊂X by(21)xn+1=Txn,yn,yn+1=Tyn,xn,∀n∈ℕ0.
By induction methodology forn∈ℕ0, we shall prove that(22)αxn,xn+1≥1,αyn,yn+1≥1,∀n.
Indeed, we haveαx0,x1≥1 and αy0,y1≥1. Suppose that (22) holds for some n and we are going to prove it for n+1. Since T is α−admissible mapping, then by (21), we obtain αxn+1,xn+2≥1 and αyn+1,yn+2≥1. Thus, (22) holds for all n. From (18)–(22), we have(23)ϕμxn,xn+1+l≤ϕμTxn−1,yn−1,Txn,yn+lmaxαxn−1,xn,αyn−1,yn≤Fψ1Kxn−1,xn+Kyn−1,yn2,ψ2Kxn−1,xn+Kyn−1,yn2+l,where(24)Kxn−1,xn=μxn,Txn,yn1+μxn−1,Txn−1,yn−11+μxn−1,xn,μxn−1,xn=μxn,xn+1,μxn−1,xn,Kyn−1,yn=μyn,yn+1,μyn−1,yn.
Hence,(25)ϕμxn,xn+1≤Fψ1wn2,wn−12,ψ2wn2,wn−12,where wn=μxn,xn+1+μyn,yn+1. Similarly, we have(26)ϕμyn,yn+1≤Fψ1wn2,wn−12,ψ2wn2,wn−12.
Adding (25) and (26) and using properties on F and ϕ, we obtain(27)ψ1wn2,wn2≤ϕwn2≤Fψ1wn2,wn−12,ψ2wn2,wn−12≤ψ1wn2,wn−12.
Sinceψ1 is strictly increasing, then wn≤wn−1,∀n. Hence, the sequence wn is monotone decreasing and bounded as follows. Therefore, there exist some w≥0 such that(28)limn⟶∞wn=w.
Now, we shall prove thatw=0. Assume that w>0. Using the properties of ψ1,ψ2,andF and letting n⟶∞ in (27) yield that(29)ψ1w2,w2≤Fψ1w2,w2,ψ2w2,w2<ψ1w2,w2,which is contradiction. Thus, w=0 and(30)limn⟶∞μxn,xn+1=limn⟶∞μyn,yn+1=0.
In what follows, we prove thatxn and yn are μ−Cauchy sequences in X,μ. Since we have(31)0≤mxn,xn+1≤μxn,xn+1⟶0,asn⟶∞⟹limn⟶∞mxn,xn+1=minlimn⟶∞μxn,xn,limn⟶∞μxn+1,xn+1=0⟹limn⟶∞μxn,xn=0,then(32)limn,m⟶∞mxn,xm=minlimn⟶∞μxn,xn,limm⟶∞μxm,xm=0,limn,m⟶∞Mxn,xm=maxlimn⟶∞μxn,xn,limm⟶∞μxm,xm=0.
That is,(33)limn⟶∞Mxn,xm−mxn,xm=0.
On the other hand, we have(34)μxn,xm−mxn,xm≤μxn,xn+1−mxn,xn+1+μxn+1,xn+1−mxn+1,xn+2+⋯+μxm−1,xm−mxm−1,xm⟶0,asn,m⟶∞.
Therefore, (33) and (34) imply that xn is an μ−Cauchy sequence. In a similar way, we can show that yn is also a μ−Cauchy sequence. By the completeness of the space X,μ, there exist x,y∈X such that(35)limn⟶∞μxn,x−mxn,x=limn⟶∞μyn,y−myn,y=0,limn⟶∞Mxn,x−mxn,x=limn⟶∞Myn,y−myn,y=0.
With respect to the sequencexn, we obtain(36)μxn,xn⟶0⟹mxn,x⟶0⟹μxn,x,Mxn,x⟶0,asn⟶∞,but(37)Mxn,x=maxμxn,xn,μx,x⟶μx,x.
Thus, the uniqueness of the limit implies that(38)μx,x=0.
Now, suppose thata holds. According to Lemma 2, since xn and yn are Cauchy sequences in a complete M−metric space X,μ, then they converge to some x,y in the metric space X,μw. Also, as F is continuous, Fxn,yn converges to Fx,y in X,μw, that is, limn⟶∞μwFxn,yn,Fx,y=0 which is equivalent to(39)μFxn,yn,Fx,y−mFxn,yn,Fx,y⟶0,MFxn,yn,Fx,y−mFxn,yn,Fx,y⟶0,asn⟶∞.
Also, we have(40)μxn+1,xn+1⟶0⟹mFxn,yn,Fx,y⟶0⟹MFxn,yn,Fx,y⟶0,asn⟶∞,but(41)MFxn,yn,Fx,y=maxμxn+1,xn+1,μFx,y,Fx,y⟶μFx,y,Fx,y.
Thus, the uniqueness of the limit implies that(42)μFx,y,Fx,y=0.
By Lemma3, we obtain(43)μxn+1,Fx,y−mxn+1,Fx,y⟶μx,Fx,y−mx,Fx,y=μx,Fx,y.
Compared with (39), we obtain(44)μx,Fx,y=0.
From (38), (42), and (44), we obtain(45)x=Fx,y.
Proceeding as above, one can obtain(46)y=Fy,x.
Suppose thatb holds, then αxn,x≥1 and αyn,y≥1. Setting x,y=xn,yn and u,v=x,y in (18), we obtain(47)ϕμTxn,yn,Tx,y+lmaxαxn,x,αyn,y≤Fψ1Kxn,x+Kyn,y2,ψ2Kxn,x+Kyn,y2+l,where(48)Kxn,x=μx,Tx,y1+μxn,xn+11+μxn,x,μxn,x⟶μx,Tx,y,0,Kyn,y=μy,Ty,x1+μyn,yn+11+μyn,y,μyn,y⟶μy,Ty,x,0,asn⟶∞.
That is,(49)ϕμxn+1,Tx,y≤Fψ1Kxn,x+Kyn,y2,ψ2Kxn,x+Kyn,y2.
In a similar way, one can obtain(50)ϕμyn+1,Ty,x≤Fψ1Kxn,x+Kyn,y2,ψ2Kxn,x+Kyn,y2.
Adding (49) and (50) and using properties on F and ϕ, we obtain(51)ψ1μxn+1,Tx,y+μyn+1,Ty,x2,μxn+1,Tx,y+μyn+1,Ty,x2≤ψ1Kxn,x+Kyn,y2.
Taking limits atn⟶∞ yields(52)ψ1μx,Tx,y+μy,Ty,x2,μx,Tx,y+μy,Ty,x2≤ψ1μx,Tx,y+μy,Ty,x2,0.
Therefore, we haveμx,Tx,y+μy,Ty,x=0. Again from (18) and taking into account that αx,x,αy,y≥1, we obtain that μTx,y,Tx,y=μTy,x,Ty,x=0. Consequently, x=Tx,y and y=Ty,x.
For the uniqueness of the coupled fixed point in Theorem2, we consider the following condition:(53)ifx,yandu,vare two coupled fixed points ofT,thenαx,u≥1,orαy,v≥1.Theorem 3.
Adding condition (53) to the hypotheses of Theorem 2, we obtain that T has a unique coupled fixed point.Proof.
Theorem2 asserts that T has at least one coupled fixed point. Assume that x,y and u,v are two coupled fixed points of T, then αx,u≥1 or αy,v≥1. Now, we apply (18) and use the properties of ϕ,ψ1,ψ2,andF to obtain(54)ψ1μx,x+μy,y2,μx,x+μy,y2≤ϕμx,x+μy,y2≤Fψ1μx,x+μy,y2,μx,x+μy,y2,ψ2μx,x+μy,y2,μx,x+μy,y2≤ψ1μx,x+μy,y2,μx,x+μy,y2.
Hence, we have(55)Fψ1μx,x+μy,y2,μx,x+μy,y2,ψ2μx,x+μy,y2,μx,x+μy,y2=ψ1μx,x+μy,y2,μx,x+μy,y2⟹μx,x=μy,y=0.
Similarly, we have(56)μu,u=μv,v=0,μx,u=μy,v=0.
Hence, by (m1) x=uandy=v, i.e., x,y is the unique coupled fixed point of T.
If we defineFs,t=s−t and(57)αx,y=1,x≺¯y;0,otherwise,then we get the following corollary which is a generalization of the main results in [31].Corollary 1.
LetX,μ be an ordered complete M−metric space and T:X×X⟶X be an increasing mapping for which there exist ϕ∈Φ, ψ1∈Ψ1, and ψ2∈Ψ2 such that ψ1t,t≤ϕt and for all x,y,u,v∈X2 with x≺¯u and y≺¯v; we have(58)ϕμTx,y,Tu,v≤ψ1Kx,u+Ky,v2−ψ2Kx,u+Ky,v2,where(59)Kx,u=μu,Tu,v1+μx,Tx,y1+μx,u,μx,u,Ky,v=μv,Tv,u1+μy,Ty,x1+μy,v,μy,v.
Suppose that either(a)
T is continuous.(b)
For a convergent sequencexn in X,μ, we have(60)xn⟶x,xn≺¯xn+1⟹xn≺¯x,∀n,xn⟶x,xn⟶y⟹x≺¯y.
If there existx0,y0∈X such that x0≺¯Tx0,y0 and y0≺¯Ty0,x0, then T has a coupled fixed point.
Now, we introduce the following classes of functionsΨ and Φ by(61)Ψ=ψ:0,∞⟶0,∞,ψis continuous, strictly increasing andψt>0fort>0,Φ=φ:0,∞⟶0,∞,φis continuous andφt>0fort>0.
If we considerϕt=ψt, ψ1s,t=ψt for some ψ∈Ψ and ψ2s,t=φt for some φ∈Φ, then we obtain an extension of the main result in [13].Corollary 2.
LetX,μ be a complete M−metric space and T:X×X⟶X be an α−admissible mapping such that(62)ψμTx,y,Tu,v+lmaxαx,u,αy,v≤Fψμx,u+μy,v2,φμx,u+μy,v2+l,for all x,y,u,v∈X2 with αx,u≥1,αy,v≥1, where F∈C, ψ∈Ψ, and φ∈Φ. Suppose that either(a)
T is continuous.(b)
For a convergent sequencexn in X,μ, we have(63)xn⟶x,αxn,xn+1≥1⟹αxn,x≥1,∀n,xn⟶x,xn⟶y⟹αx,y≥1.
If there existx0,y0∈X such that αx0,Tx0,y0≥1 and αy0,Ty0,x0≥1, then T has a coupled fixed point.Remark 1.
Notice that in [32, 33], it was shown that each coupled fixed-point theorem can be observed from the analogue of single/standard fixed-point theorems. On the other hand, for the usage of it in application, the coupled fixed-point theorem can be used to handle the problem. Therefore, in this paper, we consider the coupled fixed-point results, Theorem 2 and Theorem 3.


## 4. Fractional Differential Equations

In this section, we present sufficient conditions for the existence and uniqueness of the solution of coupled systems (2) and (3). Before starting and proving the main results, we need to fix the analytical framework of our considered problem.Consider the completeM−metric space X,μ, where X=CJ,ℝ and μ is defined by(64)μx,y=supt∈Jxt−yt,∀x,y∈X.In addition, define the operatorT:X×X⟶X as(65)Tx,y=Axt+Byt,where(66)Axt=ft,xt+tα−1δfη,xη−fτ,xττα−1−δηα−1,Byt=Iαgt,yt,Iβyt+tα−1δIαgη,yη,Iβyη−Iαgτ,yτ,Iβyττα−1−δηα−1.Now, we claim that wheneverx,y∈X2 is a coupled fixed point of the operator T, it follows that xt and yt solve (2) and (3).Lemma 5.
Letn−1<α≤n, 0<η<τ, δ≠τ/ηα−1, and h∈L0,τ; then, the boundary value problem(67)Dαxt−ft,xt=ht,∀t∈J,xi0=∂ift,xt∂tit=0=0,xτ=δxη,∀i=0,1,…,n−2,has the integral representation of solution(68)xt=ft,xt+tα−1δfη,xη−fτ,xττα−1−δηα−1+Iαht+tα−1δIαhη−Iαhττα−1−δηα−1.Proof.
Applying the operatorIα on both sides of (67) and using Lemma 4, we obtain(69)xt−ft,xt+c1tα−1+c2tα−2+⋯+cntα−n=Iαht,att=0⟹x0=0,f0,0=0⟹cn=0.
Also, we have(70)x´t−dft,xtdt+c1α−1tα−2+c2α−2tα−3+⋯+cn−1α−n+1tα−n=Iα−1ht,att=0⟹x´0=0,dft,xtdtt=0=∂ft,xt∂tt=0=0⟹cn−1=0,⋮xn−2t−dn−2ft,xtdtn−2+c1α−1…α−n+2tα−n+1+c2α−2…α−n+1tα−n=Iα−n+2ht,att=0⟹xn−20=0,dn−2ft,xtdtn−2t=0=∂n−2ft,xt∂tn−2t=0=0⟹c2=0.
Hence, we obtain(71)xt−ft,xt+c1tα−1=Iαht.
Att=τ and η, we have(72)xτ−fτ,xτ+c1τα−1=Iαhτ,(73)δxη−δfη,xη+c1δηα−1=δIαhη.
By subtracting (73) from (72), we obtain(74)c1=fτ,xτ+Iαhτ−δfη,xη+Iαhητα−1−δηα−1.
Consequently, the general solution of (67) is(75)xt=ft,xt+tα−1δfη,xη−fτ,xττα−1−δηα−1+Iαht+tα−1δIαhη−Iαhττα−1−δηα−1.
Consider the following coupled system of fractional hybrid integral equations (in short, FHIE):(76)xt=ft,xt+tα−1δfη,xη−fτ,xττα−1−δηα−1+Iαgt,yt,Iβyt+tα−1δIαgη,yη,Iβyη−Iαgτ,yτ,Iβyττα−1−δηα−1,(77)yt=ft,yt+tα−1δfη,yη−fτ,yττα−1−δηα−1+Iαgt,xt,Iβxt+tα−1δIαgη,xη,Iβxη−Iαgτ,xτ,Iβxττα−1−δηα−1.Lemma 6.
Assume that the functionρ:J×ℝ⟶ℝ defined by ρt,x=xt−ft,xt satisfies the following:(78)Diρt,xt=0=0⟹x=0,∀i=0,1,…,n−2.
Then,x,y∈C2J,ℝ is a solution of FHDE systems (2) and (3) if and only if x,y is a solution of FHIE systems (76) and (77).Proof.
Letxandy be a solution of (2) and (3). Then, by Lemma 5, we gain that the general solution of (2) has the integral form presented in (76) and the solution of (3) has the form presented in (77). Thus, xandy satisfy (76) and (77).
Conversely, letxandy fulfill (76) and (77). Then, applying Dα on both sides of (76) and using the relation Dαtλ=Γλ+1/Γλ−α+1tλ−α if λ>−1,λ≥α>0, and Dαtλ=0 if λ<α (Remark 2.1 in [34]) yield(79)Dαxt−ft,xt=Dαtα−1δfη,xη−fτ,xττα−1−δηα−1+DαIαgt,yt,Iβyt+Dαtα−1δIαgη,yη,Iβyη−Iαgτ,yτ,Iβyττα−1−δηα−1⟹Dαxt−ft,xt=gt,yt,Iαyt.
Soxt satisfies the differential equation in (2). To see that it also satisfies the boundary conditions in the same equation, fix i=0,1,…,n−2 and apply Di in (76):(80)Dixt=Dift,xt+Iα−igt,yt,Iβyt+ΓαΓα−itα−1−iδfη,xη−fτ,xττα−1−δηα−1+δIαgη,yη,Iβyη−Iαgτ,yτ,Iβyττα−1−δηα−1.
Substitutingt=0 in (80) and taking into account that α−1−i>0 yield(81)xitt=0−dift,xtdtit=0=0⟹xi0=∂ift,xt∂tit=0=0.
Again, puttingt=τ and t=η in (76) implies(82)xτ−δxη=0.
Thus,xt satisfies (2). A completely dual calculation reveals that yt also satisfies (3).
As a consequence of Lemma6, the coupled fixed point of the operator T coincides with the solution of (76) and (77) and then with the solution of (2) and (3).Theorem 4.
Assume thatf:J×ℝ⟶ℝ and g:J×ℝ2⟶ℝ are continuous functions and there exist two functions φ0,φ1:J⟶ℝ with bounds φ0 and φ1, respectively, such that(83)ft,x−ft,y≤φ0tx−y,gt,x,u−gt,y,v≤φ1tx−y+u−v.Moreover, define these functions ϕ∈Φ, ψ1∈Ψ1, ψ2∈Ψ2, F∈C, and α:X2⟶0,∞ as(84)ϕs=s,ψ1s,t=ρ+1t,ψ2s,t=t,Fs,t=s−t,αx,y=1,∀s,t∈0,∞,x,y∈X,where(85)ρ=2maxφ01+δ+11−δη/τα−1,φ1ταΓα+11+τβΓβ+11+δ+11−δη/τα−1>0.
Then, problems (2) and (3) have a unique solution.Proof.
We check that the hypothesis of Theorems2 and Theorem 3 is satisfied. For x,y,u,v∈X2, we have(86)μTx,y,Tu,v=supt∈JTx,yt−Tu,vt≤supt∈Jφ0txt−ut+tα−1τα−1−δηα−1δxη−uη+uτ−xτ+Iαφ1tyt−vt+Iβyt−Iβvt+tα−1τα−1−δηα−1δIαφ1tyη−vη+|Iβyη−Iβvη+Iαφ1tyτ−vτ+Iβyτ−Iβvτ≤φ0μx,u+tα−1τα−1−δηα−1δμx,u+μx,u+φ1tαΓα+1μy,v+tβΓβ+1μy,v+tα−1τα−1−δηα−1δηαΓα+1μy,v+ηβΓβ+1μy,v+ταΓα+1μy,v+τβΓβ+1μy,v≤φ01+δ+11−δη/τα−1μx,u+φ1ταΓα+11+τβΓβ+11+δ+11−δη/τα−1μy,v≤ρ+1μx,u+μy,v2−μx,u+μy,v2.
Thus, for anys≥0, we obtain(87)ϕμTx,y,Tu,v≤Fψ1s,μx,u+μy,v2,ψ2s,μx,u+μy,v2.
Therefore, the operatorT satisfies condition (18) of Theorem 2. With simple calculations, we can derive that the other hypothesis of Theorems 2 and Theorem 3 holds. So, the operator T has a unique fixed point, or equivalently, systems (2) and (3) have a unique solution in X2.
Now, we present an illustrated example to justify our results.Example 2.
Consider the following system of two FHDEs with three-point boundary conditions:(88)Dαxt−ft,xt=gt,yy,Iβyt,xi0=∂ift,xt∂tit=0=0,xτ=δxη,(89)Dαyt−ft,yt=gt,xt,Iβxt,yi0=∂ift,yt∂tit=0=0,yτ=δyη,where(90)α=52,β=92,η=12,τ=δ=1,ft,x=x+x2+1−et|x,gt,x,u=x+sinu.
By computation, we can show that(91)ft,x−ft,y=x+x2+1−et|x−x+x2+1−et|x≤x−y+x2+1−y2+1−1≤2x−y,gt,x,u−gt,y,v=x+sinu−y+sinv≤x−y+u−v.
Applying Theorem4, we conclude that problem (89) has one solution.


## 5. Concluding Remarks

In this work, we proved some coupled fixed-point results forα−admissible mappings which are Fψ1,ψ2-contractions in a larger structure such as M−metric spaces. Furthermore, we applied aforesaid fixed-point results to investigate the existence of a unique solution for a coupled system of higher-order fractional hybrid differential equations which are equipped with three-point boundary conditions. The respective results have been verified by providing a suitable example.In fact, the results dealing with solutions of the general systems of fractional differential equations are useful in applications to various problems which are simply modelled by means of these systems.It is believed that several recent studies (see, for example, [35–42]) on fractional calculus and its widespread applications will possibly motivate further research studies on mathematical modeling and analysis of applied problems along the lines which we have developed in this article.



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