# Existence and Uniqueness of Solutions for a Fractional Hybrid System with Nonseparated Integral Boundary Hybrid Conditions

Authors: M. Hannabou; M. Bouaouid; K. Hilal
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we are going to investigate the existence and uniqueness of solutions of a coupled system of nonlinear fractional hybrid equations with nonseparated type integral boundary hybrid conditions. We are going to use Banach’s and Leray-Schauder alternative fixed point theorems to obtain the main results. Lastly, we are giving two examples to show the effectiveness of the main results.


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## Body

## 1. Introduction

Fractional differential equations appear naturally in a number of fields bymany fields of scince such as physics, engineering, biophysics, blood flow phenomena, aerodynamics, electron-analytical chemistry, biology, and economy; for more details, we refer the readers to [1–4] and many other references therein which give an excellent account on the study of fractional differential equations.Hybrid fractional differential equations have also been studied by several researchers. This class of equations involves the fractional derivative of an unknown function hybrid with the nonlinearity depending on it. Some recent results on hybrid differential equations can be found in a series of papers (see [5, 6]).On the other hand, coupled systems of fractional differential equations are very important to study byattract the attention of many researches, because they appear naturally in many problems (see [3, 7–10]).In [11], Sitho et al. discussed the following existence bysome results for the following hybrid fractional integrodifferential equations:
(1)Dαxt−∑i=1mIβihit,xtft,xt=gt,xta.e.t∈J=0,T,0<α≤1,x0=0,where Dα denotes the Riemann-Liouville fractional derivative of order α,0<α≤1,Iϕ is the Riemann-Liouville fractional integral of order ϕ>0,ϕ∈β1,β2,⋯,βm, f∈CJ×ℝ,ℝ\0,g∈CJ×ℝ,ℝ, with hi∈CJ×ℝ,ℝ with hi0,0=0,i=1,2,⋯,m.In [12], Hilal and Kajouni considered the following boundary value problems for hybrid differential equations with fractional order (BVPHDEF for short) involving Caputo differential operators of order 0<α<1:(2)Dαxtft,xt=gt,xta.e.t∈J=0,T,ax0f0,x0+bxTfT,xT=c,where f∈CJ×ℝ,ℝ\0,g∈CJ×ℝ,ℝ and a,b,c are real constants with a+b≠0.Dhage and Lakshmikantham [13] discussed the following first-order hybrid differential equation:
(3)ddtxtft,xt=gt,xta.e.t∈J=0,T,xt0=x0∈ℝ,where f∈CJ×ℝ,ℝ\0 and g∈CJ×ℝ,ℝ. They established the existence, uniqueness results, and some fundamental differential inequalities for hybrid differential equations initiating the study of theory of such systems and proving the utilization of the theory of inequalities, the existence of extremal solutions, and comparison results.Zhao et al. [5] discussed the following fractional hybrid differential equations involving Riemann-Liouville differential operators:
(4)Dqxtft,xt=gt,xta.e.t∈J=0,T,x0=0,where f∈CJ×ℝ,ℝ\0 and g∈CJ×ℝ,ℝ. They established the existence byof solutions and some fundamental differential inequalities are also established and the existence of extremal solutions.Benchohra et al. [14] studied the following boundary value problems for differential equations with fractional order:
(5)cDαyt=ft,yt,foreacht∈J=0,T,0<α<1,ay0+byT=c,where cDα is the Caputo fractional derivative, f:0,T×ℝ⟶ℝ is a continuous function, and a,b,c are real constants with a+b≠0.Hannabou and Hilal [15] considered the boundary value problem of a class of impulsive hybrid fractional coupled differential equations:
(6)Dαutf1t,ut,vt=g1t,ut,vt,t∈0,1,t≠tii=1,2,⋯,n,0<α<1,uti+=uti−+Iiuti−,ti∈0,1,i=1,2,⋯,n,Dβutf2t,ut,vt=g2t,ut,vt,t∈0,1,t≠tj,j=1,2,⋯,m,0<β<1,vtj+=vtj−+Ijvtj−,tj∈0,1,j=1,2,⋯,m,u0f10,u0,v0=ϕu,v0f20,u0,v0=ψv.where Dα,Dβ stand for Caputo fractional derivatives of order α,β, respectively, fi∈C0,1×ℝ×ℝ,ℝ\0,gi∈C0,1×ℝ×ℝ,ℝ,i=1,2 and ϕ,ψ:C0,1,ℝ⟶ℝ are continuous functions defined by ϕu=∑i=1nλiuξi,ψv=∑j=1nδjvηj, where ξi,ηj∈0,1 for i=1,2,⋯,n, j=1,2,⋯,m, and Ik:ℝ⟶ℝ ,utk+=limε⟶0+utk+ε and utk−=limε⟶0−utk+εrepresent the right, left limits of ut at t=tk,k=i,j.The present paper is a continuation of the work in [16] in order to study the existence and uniqueness of solutions for a coupled system of fractional hybrid equation of the following forme:
(7)cDβ1cDα1+λ1x1tf1t,x1t,x2t=ω1t,x1t,x2t,a.e.t∈J=0,1,cDβ2cDα2+λ2x2tf2t,x1t,x2t=ω2t,x1t,x2t,a.e.t∈J=0,1,subject to the fractional nonseparated integral boundary hybrid conditions(8)x10f10,x10,x20+μ1x11f11,x11,x21=σ11∫01g1s,x1sds,cDα1x10f10,x10,x20+μ1Dcα1x11f11,x11,x21=σ21∫01h1s,x1sds,cD2α1x10f10,x10,x20+μ1Dc2α1x11f1,x11,x21=σ31∫01k1s,x1sds,x20f20,x10,x20+μ2x21f21,x11,x21=σ12∫01g2s,x2sds,cDα2x20f20,x10,x20+μ2Dcα2x21f21,x11,x21=σ22∫01h2s,x2sds,cD2α2x20f20,x10,x20+μ2Dc2α2x21f21,x11,x21=σ32∫01k2s,x2sds,where 0<αi<1,1<βi≤2,λi,μi,σ1i,σ2i,σ3i∈ℝ∗ with μi≠−1 for i=1,2,cDβi,cDαi are the Caputo fractional derivatives, f1,f2:0;1×ℝ2⟶ℝ\0, ω1,ω2:0;1×ℝ2⟶ℝ and g1,h1,k1,g2,h2,k2:0;1×ℝ⟶ℝ are given continuous functions.By a solution of the problems (7)–(8), we mean a function x∈CJ,ℝ such that
(i)
the functiont↦x/fit,x1,x2i=1,2 is continuous for each x1,x2∈ℝ2,(ii)
x1,x2 satisfies the equations in (7)–(8).This paper is organized as follows: in the second section, we recall some notations and several known results. In the third section, we show the existence and uniqueness of solutions of problem (7)–(8), these results can be viewed as extension of the result given in [12]. In the fourth section, we give some examples to demonstrate the application of our main results.


## 2. Preliminaries and Notations

In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper. LetX be a Banach space of all continuous functions defined frome endowed with norm. Then, the product space is also a Banach space equipped with the norm.We denote byL1J,ℝ the space of Lebesgue integrable real-valued functions on J equipped with the norm .L1 defined by
(9)xL1=∫0Txsds.Definition 1 (see [17]).
The fractional integral of the functionh∈L1a,b,ℝ+ of order α∈ℝ+ is defined by
(10)Iaαht=∫att−sα−1Γαhsds,where Γ is the gamma function.Definition 2 (see [17]).
Leth be a fonction difened on a,b, the Riemann-Liouville fractional derivative of order α is defined by
(11)Dca+αht=1Γn−αddtn∫att−sn−α−1Γαsds,where n=α+1 and α denotes the integer part of α.Definition 3 (see [17]).
Leth be a function defined on a,b, the Caputo fractional derivative of order α is defined by
(12)Dca+αht=1Γn−α∫att−sn−α−1Γαhnsds,where n=α+1 and α denotes the integer part of α.Lemma 4 (see [2]).
Letα,β≥0, then the following relations hold:
(13)Iαtβ=Γβ+1Γα+β+1tα+β,cDαtβ=Γβ+1Γβ−α+1tβ−α.Lemma 5 (see [2]).
Letn∈ℕ and n−1<α<n. If f is a continuous function, then we have
(14)IαcDαft=ft+a0+a1t+a2t2+⋯+an−1tn−1.Lemma 6 (Leray-Schauder alternative, see [18]).
LetF:G⟶G be a completely continuous operator (i.e., a map that is restricted to any bounded set in G is compact). Let PF=u∈G:u=λFuforsome0<λ<1. Then, either the set PF is unbounded or F has at least one fxed point.
We make the following assumption:
H0 The function x↦x/fit,x1,x2i=1,2 is increasing in ℝ almost everywhere for t∈J.Lemma 7.
Assume that hypothesisH0 holds. Then, for any y1,y2∈CJ,ℝ2. The function x∈CJ,ℝ is a solution of the coupled system,
(15)cDβ1cDα1+λ1x1tf1t,x1t,x2t=y1t,t∈0,1,cDβ2cDα2+λ2x2tf2t,x1t,x2t=y2t,t∈0,1,
subject to the boundary condition (8), has a solution given by
(16)x2t=f2t,x1t,x2t1Γα2+β2∫0tt−sα2+β2−1y2sds+A12t∫01h2s,x2sds+A22t∫01g2s,x2sds+A32t∫01k2s,x2sds+A42tΓβ2−1−α2−1∫011−sβ2−α2−1y2sds+A51tΓβ2∫011−sβ2−1y2sds+μ2λ21+μ2Γα2∫011−sα2−1x2sds−μ21+μ2Γα2+β2∫011−sα2+β2y2sds,−λ2Γα2∫0tt−sα2−1x2sds,where
(17)A1it=tαiσ2i1−λiΓ2−αiΓαi+11+μi+tαi+1λiσ2iΓ2−αiΓ2+αiμi+μi1+μi2Γαi+1−11+μiΓαi+2Γ2−αiλiσ2i−μiσ2i1+μi2Γαi+1,A2it=tαiλiσ1iΓαi+11+μi−σ1i1+μi+μiλiσ1i1+μi2Γαi+1,A3it=−tαiΓ2−αiσ3iΓαi+11+μi+tαi+1Γ2−αiσ3iΓαi+2μi+μiΓ2−αiσ3i1+μi2Γαi+1−Γ2−αiσ3i1+μiΓαi+2,A4it=tαiΓ2−αiμiΓαi+11+μi−tαi+1Γ2−αiΓαi+2+Γ2−αiμi1+μiΓαi+2−μi21+μi2Γαi+1,A5it=−tαiμiΓαi+11+μi−μi21+μi2Γαi+1,fori=1,2.Proof.
Using Lemma5, we obtain
(18)cDα1+λ1x1tf1t,x1t,x2t=Iβ1y1t+a0+a1t,cDα1x1tf1t,x1t,x2t=Iβ1y1t+a0+a1t−λ1x1tf1t,x1t,x2t,x1tf1t,x1t,x2t=Iα1+β1y1t+Iα1a0+Iα1a1t−Iα1λ1x1tf1t,x1t,x2t+a2,where a0,a1,a2∈ℝ.According to the conditioncD2α1x10/f10,x10,x20+μ1cD2α1x11/f11,x1,x21=σ31∫01k1s,x1sds, we find that
(19)a1=Γ2−α1σ31μ1∫01k1s,x1sds+λ1σ21μ1∫01h1s,x1sds−1Γβ1−α1∫011−sβ1−α1y1sds.Using the facts thatcDα1x10/f10,x10,x20+μ1cDα1x11/f11,x11,x21=σ21∫01h1s,x1sds and x10/f10,x10,x20+μ1x11/f11,x11,x21=σ11∫01g1s,x1sds, we have
(20)a0=−Γ2−α1σ311+μ1∫01k1s,x1sds+1−λ1Γ2−α1σ211+μ1∫01h1s,x1sds+λ1σ111+μ1∫01g1s,x1sds+Γ2−α1μ11+μ1Γβ1−α1∫011−sβ1−α1−1y1sds−μ11+μ1Γβ1∫011−sβ1−1y1sds,a2=μ1λ11+μ1Γα1∫011−sα1−1x1sds+σ111+μ1−μ1λ1σ11Γα1+11+μ12∫01g1s,x1sds−μ11+μ1Γα1+β1∫011−sα1+β1−1y1sds+μ121+μ12Γβ1Γα1+1∫011−sβ1−1y1sds+Γ2−α1Γβ1−α1μ11+μ1Γα1+2−μ121+μ12Γα1+1∫011−sβ1−α1−1y1sds+μ121+μ12Γα1+1−μ11+μ1Γα1+2Γ2−α1λ1σ21μ1−μ1σ211+μ12Γα1+1×∫01h1s,x1sds.Substituting the values ofa0,a1, and a2, we obtain
(21)x1t=f1t,x1t,x2t1Γα1+β1∫0tt−sα1+β1−1y1sds+A11t∫01h1s,x1sds+A21t∫01g1s,x1sds+A32t∫01k1s,x1sds+A41tΓβ1−α1∫011−sβ−α−1y1sds+A51tΓβ1∫011−sβ1−1y1sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds−μ11+μ1Γα1+β1∫011−sα1+β1−1y1sds−λ1Γα1∫0tt−sα−1x1sds.Analogously, we can deduce that(22)x2t=f2t,x1t,x2t1Γα2+β2∫0tt−sα2+β2−1y2sds+A12t∫01h2s,x2sds+A22t∫01g2s,x2sds+A32t∫01k2s,x2sds+A42tΓβ2−1−α2−1∫011−sβ2−α2−1y2sds+A51tΓβ2∫011−sβ2−1y2sds+μ2λ21+μ2Γα2∫011−sα2−1x2sds−μ21+μ2Γα2+β2∫011−sα2+β2y2sds−λ2Γα2∫0tt−sα2−1x2sds.By a direct computation, the converse of the lemma can be easily verified.


## 3. Main Results

In view of Lemma7, we define the operator U:X×X⟶X×X by Ux1,x2=U1x1,x2,U2x1,x2.Where,(23)U1x1,x2t=f1t,x1t,x2t1Γα1+β1∫0tt−sα1+β1−1w1s,x1s,x2sds+A11t∫01h1s,x1sds+A21t∫01g1s,x1sds+A32t∫01k1s,x1sds+A41tΓβ1−α1∫011−sβ1−α1−1w1s,x1s,x2sds+A51tΓβ1∫011−sβ1−1w1s,x1s,x2sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds−μ11+μ1Γα1+β1∫011−sα1+β1−1w1s,x1s,x2sds−λ1Γα1∫0tt−sα−1x1sds,(24)U2x1,x2t=f2t,x1t,x2t1Γα2+β2∫0tt−sα2+β2−1w2s,x1s,x2sds+A12t∫01h2s,x2sds+A22t∫01g2s,x2sds+A32t∫01k2s,x2sds+A42tΓβ2−1−α2−1∫011−sβ2−α2−1w2s,x1s,x2sds+A51tΓβ2∫011−sβ2−1w2s,x1s,x2sds+μ2λ21+μ2Γα2∫011−sα2−1x2sds−μ21+μ2Γα2+β2∫011−sα2+β2w2s,x1s,x2sds−λ2Γα2∫0tt−sα2−1x2sds.For computational convenience, we set(25)r11=maxL1q111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61+λ1Γα1+1,L1q21×1Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1,r12=maxL2q121Γα2+β2+1+A42Γβ2−α2+1+A52Γβ2+1+μ21+μ2Γα2+β2+1+μ2λ21+μ2Γα2+1+A12q52+A22q42+A32q62+λ2Γα2+1,q22L2×1Γα2+β2+1+A42Γβ2−α2+1+A52Γβ2+1+μ21+μ2Γα2+β2+1,Aij=maxt∈0,1Aijtfori=1,2,..,5,j=1,2.Before giving the main results, we impose the following assumptions:H1 The functions fi are continuous and bounded; that is, there exist positive numbers Li>0 such that fit,u,v≤Li for all t,u,v∈0,1×ℝ×ℝi=1,2.H2w1,w2:0,1×ℝ2⟶ℝ and h1,g1,k1,h2,g2,k2:0,1×ℝ⟶ℝ are continuous functions.H3 There exist positive constants q11,q21,q31,q12,q22,q32 such that for all t∈0,1 and x1,x2,y1,y2∈ℝ, we have
(26)w1t,x1,y1−w1t,x2,y2≤q11x1−x2+q21y1−y2,w2t,x1,y1−w2t,x2,y2≤q12x1−x2+q22y1−y2.H4 There exist positive constants q41,q51,q61,q42,q52,q62 such that
(27)g1t,x1−g1t,x2≤q41x1−x2,g2t,x1−g2t,x2≤q42x1−x2,h1t,x1−h1t,x2≤q51x1−x2,h2t,x1−h2t,x2≤q52x1−x2,k1t,x1−k1t,x2≤q61x1−x2,k2t,x1−k2t,x2≤q62x1−x2,∀x1,x2∈ℝ,t∈0,1.H5fit,x,y≤mit; hit,x≤ρit; kit,x≤ψit; git,x≤ϕit, ∀t,x,y∈0,1×ℝ×ℝ with mi,ρi,ϕi,ψi∈C0,1;ℝ+, for i=1,2.


### 3.1. First Result

In our first result, we discuss the existence of solutions for system (7)–(8) by means of the Banach fixed point theorem.Theorem 8.
Suppose thatH1–H3 are satisfied.
Then, there exist a unique solution for systems (7)–(8) provided that r11+r12<1.Proof.
We putsup0≤t≤1fit,0,0=M0i, sup0≤t≤1git,0=M1i, sup0≤t≤1hit,0=M2i, sup0≤t≤1kit,0=M3i, for i=1,2.
LetBr=x1,x2∈X×X:x1,x2≤r with r≥r21+r22/1−r11+r12, where
(28)r21=M01Γα1+β1+1+M01A41Γβ1−α1+1+A11M21+A21M11+A31M31+A51M01Γβ1+1+μ1M011+μ1Γα1+β1+1,r22=M02Γα2+β2+1+M02A42Γβ2−α2+1+A12M22+A22M12+A32M32+A52M02Γβ2+1+μ2M021+μ2Γα2+β2+1.We prove thatUBr⊆Br.Forx1,x2∈Br, t∈0,1, we have
(29)U1x1,x2t≤f1s,x1s,x2s1Γα1+β1∫0tt−sα1+β1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+A11t∫01h1s;x1s−h1s;0+h1s;0ds+A21t∫01g1s;x1s−g1s;0+g1s;0ds+A31t∫01k1s;x1s−k1s;0+k1s;0ds+A41tΓβ1−α1×∫011−sβ1−α1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+A51tΓβ1∫011−sβ1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+μ1λ11+μ1Γα1∫011−sα1−1x1sds+μ11+μ1Γα1+β1×∫011−sα1+β1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+λ1Γα1∫0tt−sα1−1x1sds.Consequently,(30)U1x1,x2≤L11Γα1+β1+1q11x1+q21x2+x2+M01+A11q51x1+M21+A21q41x1+M11+A31q61x1+M31+A41Γβ1−α1+1q11x1+q21x2+x2+M01+A51Γβ1+1q11x1+q21x2+x2+M01+μ1λ11+μ1Γα1+1x1+μ11+μ1Γα1+β1+1q11x1+q21x2+x2+M01+λ1Γα1+1x1≤L1q111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61+λ1Γα1+1x1L1q211Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1x2+M01Γα1+β1+1+M01A41Γβ1−α1+1+A11M21+A21M11+A31M31+A51M01Γβ1+1+μ1M011+μ1Γα1+β1+1≤r11r+r21.In the same way, we obtain that(31)U2x1,x2≤r12r+r22.Therefore, we have(32)Ux1,x2=U1x1,x2+U2x1,x2≤r11+r12r+r21+r22≤r.Now, forx1,x2,x1′,x2′∈X×X and t∈0;1, we get
(33)U1x1,x2t−U1x1′,x2′t≤f1s,x1s,x2s−f1s,x1′s,x2′s1Γα1+β1∫0tt−sα1+β1−1w1s,x1s,x2s−w1s,x1′s,x2′sds+A11t∫01h1s;x1s−h1s;x1′sds+A21t∫01g1s;x1s−g1s;x1′sds+A31t×∫01k1s;x1s−k1s;x1′sds+A41tΓβ1−α1∫011−sβ1−α1−1×w1s,x1s,x2s−w1s,x1′s,x2′sds+A51tΓβ1∫011−sβ1−1w1s,x1s,x2s−w1s,x1′s,x2′sds+μ1λ11+μ1Γα1×∫011−sα1−1x1s−x1′sds+μ11+μ1Γα1+β1∫011−sα1+β1−1w1s,x1s,x2s−w1s,x1′s,x2′sds+λ1Γα1∫0tt−sα1−1x1s−x1′sds≤L11Γα1+β1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+λ1Γα1+1x1−x1′+A11q51x1−x1′+A21q41x1−x1′+A31q61x1−x1′+A41Γβ1−α1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+A51Γβ1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+μ1λ11+μ1Γα1+1x1−x1′+μ11+μ1Γα1+β1+1q11x1−x1′+q21x2−x2′+λ1Γα1+1x1−x1′≤L1q111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61x1−x1′+q21×1Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1x2−x2′+λ1Γα1+1x1−x1′≤r11x1−x1′+x2−x2′.Analogously, one has(34)U2x1,x2t−U2x1′,x2′t≤r12x1−x1′+x2−x2′,and thus
(35)Ux1,x2−Ux′1,x′2≤r11+r12x1−x1′+x2−x2′.Sincer11+r12<1,by thene the operator U is a contraction mapping. Hence, we deduce that systems (7)–(8) have a unique solution.


### 3.2. Second Result

In our second result, we discuss the existence of solutions for system (7)–(8) by means of the so-called Leray-Schauder alternative.Theorem 9.
Assume that conditionsH1–H3 and H5 hold. Furthermore, it is assumed that
(36)π1+π2η0+η1x1+η2x2+π3x1+π4x2<1.Then, system (7)–(8) have at least one solution.Proof. We will show that the operatorY:X×Y⟶X×Y satisfes all the assumptions of Lemma 6.In the first step, we prove that the operatorY is completely continuous.Clearly, it follows by the continuity of functionsf1,f2,g1, and g2 that the operator Y is continuous.LetS⊂X×Y be bounded. Then, we can find positive constants H1 and H2 such that
(37)w1t,x1,x2≤H1,w2t,x1,x2≤H2,∀x1,x2∈S.Thus, for anyx1,x2∈S, we get
(38)Y1x1,x2t≤L11Γα1+β1∫0tt−sα1+β1−1w1s,x1s,x2sds+A11t×∫01h1s;x1sds+A21t∫01g1s;x1sds+A31t∫01k1s;x1sds+A41tΓβ1−α1∫011−sβ1−α1−1w1s,x1s,x2sds+A51tΓβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2sds+μ11+μ1Γα1+β1∫011−sα1+β1−1w1s,x1s,x2sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds+λ1Γα1∫0tt−sα1−1x1sds≤L11Γα1+β1∫0tt−sα1+β1−1H1ds+A11∫01ρ1ds+A21∫01ϕ1ds+A31∫01ψ1ds+A41Γβ1−α1∫011−sβ1−α1−1H1ds+A51Γβ1∫011−sβ1−1H1ds+μ11+μ1Γα1+β1∫011−sα1+β1−1H1ds+μ1λ11+μ1Γα1∫011−sα1−1x1sds+λ1Γα1∫0tt−sα1−1x1sds≤L1H1Γα1+β1+1+A11ρ1+A21ϕ1+A31ψ1+A41H1Γβ1−α1+1+A51H1Γβ1+1+μ1H11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1x1+λ1Γα1+1x1.In a similar manner, we have(39)Y2x1,x2t≤L2H2Γα2+β2+1+A12ρ2+A22ϕ2+A32ψ2+A42H2Γβ2−α2+1+A52H2Γβ2+1+μ2H21+μ2Γα2+β2+1+μ2λ21+μ2Γα2+1x2+λ2Γα2+1x2.From the inequalities above, we deduce that the operatorY is uniformly bounded.Next, we prove thatY is equicontinuous.The continuity off1,f2,h1,h2,g1,g2,k1,k2 implies that the operator Y1 is continuous. Moreover, Y1 is uniformly bounded on Br′.Suppose that0≤t1<t2≤1. Then we have
(40)Y1x1,x2t2−Y1x1,x2t1≤L11Γα1+β1∫0t2t2−sα1+β1−1w1s,x1s,x2sds−∫0t1t1−sα1+β1−1w1s,x1s,x2sds+A11t2−A11t1×∫01h1s;x1sds+A21t2−A21t1∫01g1s;x1sds+A31t2−A31t1∫01k1s;x1sds+A41t2−A41t1Γβ1−α1∫011−sβ1−α1−1w1s,x1s,x2sds+A51t2−A51t1Γβ1∫011−sβ1−1w1s,x1s,x2sds+λ1Γα1∫0t2t2−sα1−1x1sds−∫0t1t1−sα1−1x1sds≤L1m1Γα1+β1+1t2α1+β1−t1α1+β1+ρ1A11t2−A11t1+ϕ1A21t2−A21t1+ψ1A31t2−A31t1m1Γβ1−α1+1A41t2−A41t1+m1Γβ1+1A51t2−A51t1+λ1Γα1∫0t1t1−sα1−1−t2−sα1−1x1sds−∫t1t2t1−sα1−1x1sds.Similarly, one has(41)Y2x1,x2t2−Y2x1,x2t1≤L2m2Γα2+β2+1t2α2+β2−t1α2+β2+ρ2A12t2−A12t1+ϕ2A22t2−A22t1+ψ2A32t2−A32t1m2Γβ2−α2+1A42t2−A42t1+m2Γβ2+1A52t2−A52t1+λ1Γα1∫0t1t1−sα2−1−t2−sα1−1x1sds−∫t1t2t1−sα2−1x1sds,which tend to0 independently of x1,x2. This implies that the operator Yx1,x2 is equicontinuous. Thus, by the above fndings, the operator Yx1,x2 is completely continuous.In the next step, we will prove that the setP=x1,x2∈X×Y/x1,x2=λYx1,x2,0≤λ≤1 is bounded.Letx1,x2∈P. Then, we have x1,x2=λYx1,x2. Thus, for any t∈0,1, we can write
(42)x1t=λY1x1,x2t,x2t=λY2x1,x2t.Then,(43)x1≤L11Γα1+β1+1η0+η1x1+η2x2+A11ρ1+A21ϕ1+A31ψ1+A41Γβ1−α1+1η0+η1x1+η2x2+A51Γβ1+1η0+η1x1+η2x2+μ1λ11+μ1Γα1x1+μ11+μ1Γα1+β1+1η0+η1x1+η2x2+λ1Γα1+1x1≤L1A11ρ1+A21ϕ1+A32ψ1+1Γα1+β1+1+A41Γβ1−α1+1+μ11+μ1Γα1+β1+1+A51Γβ1+1η0+η1x1+η2x2+L1μ1λ11+μ1Γα1+λ1Γα1+1x1,x2≤L21Γα2+β2+1η0+η1x1+η2x2+A12ρ2+A22ϕ2+A32ψ2+A42Γβ2−α2+1η0+η1x1+η2x2+A52Γβ2+1η0+η1x1+η2x2+μ2λ21+μ2Γα2x2+μ21+μ2Γα2+β2+1η0+η1x1+η2x2+λ2Γα2+1x2≤L2A12ρ2+A22ϕ2+A32ψ2+1Γα2+β2+1+A42Γβ2−α2+1+μ21+μ2Γα2+β2+1+A52Γβ2+1η0+η1x1+η2x2+L2μ2λ21+μ2Γα2+λ2Γα2+1x2.In consequence, we have(44)x2+x1≤L1A11ρ1+L1A21ϕ1+L1A32ψ1+L1A12ρ2+L1A22ϕ2+L1A32ψ2+L1Γα1+β1+1+L1A41Γβ1−α1+1+L1μ11+μ1Γα1+β1+1+L1A51Γβ1+1+L2Γα2+β2+1+L2A42Γβ2−α2+1+L2μ21+μ2Γα2+β2+1+L2A52Γβ2+1η0+η1x1+η2x2+L2μ1λ11+μ1Γα1+λ1Γα1+1x1+L2μ2λ21+μ2Γα2+λ2Γα2+1x2,thenx2,x1≤π1+π2η0+η1x1+η2x2+π3x1+π4x2 with
(45)π1=L1A11ρ1+L1A21ϕ1+L1A32ψ1+L1A12ρ2+L1A22ϕ2+L1A32ψ2,π2=L1Γα1+β1+1+L1A41Γβ1−α1+1+L1μ11+μ1Γα1+β1+1+L1A51Γβ1+1+L2Γα2+β2+1+L2A42Γβ2−α2+1+L2μ21+μ2Γα2+β2+1+L2A52Γβ2+1,π3=L1μ1λ11+μ1Γα1+λ1Γα1+1,π4=L2μ2λ21+μ2Γα2+λ2Γα2+1.This shows that the setP is bounded. Hence, all the conditions of Lemma 6 are satisfied, and consequently, the operator Y has at least one fxed point, which corresponds to a solution of system (7)–(8). This completes the proof.


## 3.1. First Result

In our first result, we discuss the existence of solutions for system (7)–(8) by means of the Banach fixed point theorem.Theorem 8.
Suppose thatH1–H3 are satisfied.
Then, there exist a unique solution for systems (7)–(8) provided that r11+r12<1.Proof.
We putsup0≤t≤1fit,0,0=M0i, sup0≤t≤1git,0=M1i, sup0≤t≤1hit,0=M2i, sup0≤t≤1kit,0=M3i, for i=1,2.
LetBr=x1,x2∈X×X:x1,x2≤r with r≥r21+r22/1−r11+r12, where
(28)r21=M01Γα1+β1+1+M01A41Γβ1−α1+1+A11M21+A21M11+A31M31+A51M01Γβ1+1+μ1M011+μ1Γα1+β1+1,r22=M02Γα2+β2+1+M02A42Γβ2−α2+1+A12M22+A22M12+A32M32+A52M02Γβ2+1+μ2M021+μ2Γα2+β2+1.We prove thatUBr⊆Br.Forx1,x2∈Br, t∈0,1, we have
(29)U1x1,x2t≤f1s,x1s,x2s1Γα1+β1∫0tt−sα1+β1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+A11t∫01h1s;x1s−h1s;0+h1s;0ds+A21t∫01g1s;x1s−g1s;0+g1s;0ds+A31t∫01k1s;x1s−k1s;0+k1s;0ds+A41tΓβ1−α1×∫011−sβ1−α1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+A51tΓβ1∫011−sβ1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+μ1λ11+μ1Γα1∫011−sα1−1x1sds+μ11+μ1Γα1+β1×∫011−sα1+β1−1w1s,x1s,x2s−w1s,0,0+w1s,0,0ds+λ1Γα1∫0tt−sα1−1x1sds.Consequently,(30)U1x1,x2≤L11Γα1+β1+1q11x1+q21x2+x2+M01+A11q51x1+M21+A21q41x1+M11+A31q61x1+M31+A41Γβ1−α1+1q11x1+q21x2+x2+M01+A51Γβ1+1q11x1+q21x2+x2+M01+μ1λ11+μ1Γα1+1x1+μ11+μ1Γα1+β1+1q11x1+q21x2+x2+M01+λ1Γα1+1x1≤L1q111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61+λ1Γα1+1x1L1q211Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1x2+M01Γα1+β1+1+M01A41Γβ1−α1+1+A11M21+A21M11+A31M31+A51M01Γβ1+1+μ1M011+μ1Γα1+β1+1≤r11r+r21.In the same way, we obtain that(31)U2x1,x2≤r12r+r22.Therefore, we have(32)Ux1,x2=U1x1,x2+U2x1,x2≤r11+r12r+r21+r22≤r.Now, forx1,x2,x1′,x2′∈X×X and t∈0;1, we get
(33)U1x1,x2t−U1x1′,x2′t≤f1s,x1s,x2s−f1s,x1′s,x2′s1Γα1+β1∫0tt−sα1+β1−1w1s,x1s,x2s−w1s,x1′s,x2′sds+A11t∫01h1s;x1s−h1s;x1′sds+A21t∫01g1s;x1s−g1s;x1′sds+A31t×∫01k1s;x1s−k1s;x1′sds+A41tΓβ1−α1∫011−sβ1−α1−1×w1s,x1s,x2s−w1s,x1′s,x2′sds+A51tΓβ1∫011−sβ1−1w1s,x1s,x2s−w1s,x1′s,x2′sds+μ1λ11+μ1Γα1×∫011−sα1−1x1s−x1′sds+μ11+μ1Γα1+β1∫011−sα1+β1−1w1s,x1s,x2s−w1s,x1′s,x2′sds+λ1Γα1∫0tt−sα1−1x1s−x1′sds≤L11Γα1+β1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+λ1Γα1+1x1−x1′+A11q51x1−x1′+A21q41x1−x1′+A31q61x1−x1′+A41Γβ1−α1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+A51Γβ1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+μ1λ11+μ1Γα1+1x1−x1′+μ11+μ1Γα1+β1+1q11x1−x1′+q21x2−x2′+λ1Γα1+1x1−x1′≤L1q111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61x1−x1′+q21×1Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1x2−x2′+λ1Γα1+1x1−x1′≤r11x1−x1′+x2−x2′.Analogously, one has(34)U2x1,x2t−U2x1′,x2′t≤r12x1−x1′+x2−x2′,and thus
(35)Ux1,x2−Ux′1,x′2≤r11+r12x1−x1′+x2−x2′.Sincer11+r12<1,by thene the operator U is a contraction mapping. Hence, we deduce that systems (7)–(8) have a unique solution.


## 3.2. Second Result

In our second result, we discuss the existence of solutions for system (7)–(8) by means of the so-called Leray-Schauder alternative.Theorem 9.
Assume that conditionsH1–H3 and H5 hold. Furthermore, it is assumed that
(36)π1+π2η0+η1x1+η2x2+π3x1+π4x2<1.Then, system (7)–(8) have at least one solution.Proof. We will show that the operatorY:X×Y⟶X×Y satisfes all the assumptions of Lemma 6.In the first step, we prove that the operatorY is completely continuous.Clearly, it follows by the continuity of functionsf1,f2,g1, and g2 that the operator Y is continuous.LetS⊂X×Y be bounded. Then, we can find positive constants H1 and H2 such that
(37)w1t,x1,x2≤H1,w2t,x1,x2≤H2,∀x1,x2∈S.Thus, for anyx1,x2∈S, we get
(38)Y1x1,x2t≤L11Γα1+β1∫0tt−sα1+β1−1w1s,x1s,x2sds+A11t×∫01h1s;x1sds+A21t∫01g1s;x1sds+A31t∫01k1s;x1sds+A41tΓβ1−α1∫011−sβ1−α1−1w1s,x1s,x2sds+A51tΓβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2sds+μ11+μ1Γα1+β1∫011−sα1+β1−1w1s,x1s,x2sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds+λ1Γα1∫0tt−sα1−1x1sds≤L11Γα1+β1∫0tt−sα1+β1−1H1ds+A11∫01ρ1ds+A21∫01ϕ1ds+A31∫01ψ1ds+A41Γβ1−α1∫011−sβ1−α1−1H1ds+A51Γβ1∫011−sβ1−1H1ds+μ11+μ1Γα1+β1∫011−sα1+β1−1H1ds+μ1λ11+μ1Γα1∫011−sα1−1x1sds+λ1Γα1∫0tt−sα1−1x1sds≤L1H1Γα1+β1+1+A11ρ1+A21ϕ1+A31ψ1+A41H1Γβ1−α1+1+A51H1Γβ1+1+μ1H11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1x1+λ1Γα1+1x1.In a similar manner, we have(39)Y2x1,x2t≤L2H2Γα2+β2+1+A12ρ2+A22ϕ2+A32ψ2+A42H2Γβ2−α2+1+A52H2Γβ2+1+μ2H21+μ2Γα2+β2+1+μ2λ21+μ2Γα2+1x2+λ2Γα2+1x2.From the inequalities above, we deduce that the operatorY is uniformly bounded.Next, we prove thatY is equicontinuous.The continuity off1,f2,h1,h2,g1,g2,k1,k2 implies that the operator Y1 is continuous. Moreover, Y1 is uniformly bounded on Br′.Suppose that0≤t1<t2≤1. Then we have
(40)Y1x1,x2t2−Y1x1,x2t1≤L11Γα1+β1∫0t2t2−sα1+β1−1w1s,x1s,x2sds−∫0t1t1−sα1+β1−1w1s,x1s,x2sds+A11t2−A11t1×∫01h1s;x1sds+A21t2−A21t1∫01g1s;x1sds+A31t2−A31t1∫01k1s;x1sds+A41t2−A41t1Γβ1−α1∫011−sβ1−α1−1w1s,x1s,x2sds+A51t2−A51t1Γβ1∫011−sβ1−1w1s,x1s,x2sds+λ1Γα1∫0t2t2−sα1−1x1sds−∫0t1t1−sα1−1x1sds≤L1m1Γα1+β1+1t2α1+β1−t1α1+β1+ρ1A11t2−A11t1+ϕ1A21t2−A21t1+ψ1A31t2−A31t1m1Γβ1−α1+1A41t2−A41t1+m1Γβ1+1A51t2−A51t1+λ1Γα1∫0t1t1−sα1−1−t2−sα1−1x1sds−∫t1t2t1−sα1−1x1sds.Similarly, one has(41)Y2x1,x2t2−Y2x1,x2t1≤L2m2Γα2+β2+1t2α2+β2−t1α2+β2+ρ2A12t2−A12t1+ϕ2A22t2−A22t1+ψ2A32t2−A32t1m2Γβ2−α2+1A42t2−A42t1+m2Γβ2+1A52t2−A52t1+λ1Γα1∫0t1t1−sα2−1−t2−sα1−1x1sds−∫t1t2t1−sα2−1x1sds,which tend to0 independently of x1,x2. This implies that the operator Yx1,x2 is equicontinuous. Thus, by the above fndings, the operator Yx1,x2 is completely continuous.In the next step, we will prove that the setP=x1,x2∈X×Y/x1,x2=λYx1,x2,0≤λ≤1 is bounded.Letx1,x2∈P. Then, we have x1,x2=λYx1,x2. Thus, for any t∈0,1, we can write
(42)x1t=λY1x1,x2t,x2t=λY2x1,x2t.Then,(43)x1≤L11Γα1+β1+1η0+η1x1+η2x2+A11ρ1+A21ϕ1+A31ψ1+A41Γβ1−α1+1η0+η1x1+η2x2+A51Γβ1+1η0+η1x1+η2x2+μ1λ11+μ1Γα1x1+μ11+μ1Γα1+β1+1η0+η1x1+η2x2+λ1Γα1+1x1≤L1A11ρ1+A21ϕ1+A32ψ1+1Γα1+β1+1+A41Γβ1−α1+1+μ11+μ1Γα1+β1+1+A51Γβ1+1η0+η1x1+η2x2+L1μ1λ11+μ1Γα1+λ1Γα1+1x1,x2≤L21Γα2+β2+1η0+η1x1+η2x2+A12ρ2+A22ϕ2+A32ψ2+A42Γβ2−α2+1η0+η1x1+η2x2+A52Γβ2+1η0+η1x1+η2x2+μ2λ21+μ2Γα2x2+μ21+μ2Γα2+β2+1η0+η1x1+η2x2+λ2Γα2+1x2≤L2A12ρ2+A22ϕ2+A32ψ2+1Γα2+β2+1+A42Γβ2−α2+1+μ21+μ2Γα2+β2+1+A52Γβ2+1η0+η1x1+η2x2+L2μ2λ21+μ2Γα2+λ2Γα2+1x2.In consequence, we have(44)x2+x1≤L1A11ρ1+L1A21ϕ1+L1A32ψ1+L1A12ρ2+L1A22ϕ2+L1A32ψ2+L1Γα1+β1+1+L1A41Γβ1−α1+1+L1μ11+μ1Γα1+β1+1+L1A51Γβ1+1+L2Γα2+β2+1+L2A42Γβ2−α2+1+L2μ21+μ2Γα2+β2+1+L2A52Γβ2+1η0+η1x1+η2x2+L2μ1λ11+μ1Γα1+λ1Γα1+1x1+L2μ2λ21+μ2Γα2+λ2Γα2+1x2,thenx2,x1≤π1+π2η0+η1x1+η2x2+π3x1+π4x2 with
(45)π1=L1A11ρ1+L1A21ϕ1+L1A32ψ1+L1A12ρ2+L1A22ϕ2+L1A32ψ2,π2=L1Γα1+β1+1+L1A41Γβ1−α1+1+L1μ11+μ1Γα1+β1+1+L1A51Γβ1+1+L2Γα2+β2+1+L2A42Γβ2−α2+1+L2μ21+μ2Γα2+β2+1+L2A52Γβ2+1,π3=L1μ1λ11+μ1Γα1+λ1Γα1+1,π4=L2μ2λ21+μ2Γα2+λ2Γα2+1.This shows that the setP is bounded. Hence, all the conditions of Lemma 6 are satisfied, and consequently, the operator Y has at least one fxed point, which corresponds to a solution of system (7)–(8). This completes the proof.


## 4. Examples

### 4.1. Example 1

Consider the following system of fractional hybrid differentiel equation:(46)cD4/3Dc1/3+1700x1tf1t,x1t,x2t=11+t4t4x1t4x2t+10+x2t4x1t+6+1Γ4/3∫0tt−s1/3ds1+x12s,Dc4/3Dc1/3+1300x2tf2x1t/f1t,x1t,x2t=1500+t2sinx2t+x1t1+x2t+1Γ15/2∫0tt−s13/2x2sds,x10f10,x10,x20+x11f11,x11,x21=1300∫011s+100x1s300+x1sds,Dc1/3x10f10,x10,x20+Dc1/3x11f11,x11,x21=1300∫011s+8000x1s30+x1sds,Dc2/3x10f10,x10,x20+Dc2/3x11f11,x11,x21=1300∫011s+1600x1s30+x1sds.x20f20,x10,x20+x21f21,x11,x21=1200∫01x2s300+x2sds,cD1/3x20f20,x10,x20+Dc1/3x21f21,x11,x21=1200∫011s+23x2s30+x2sds,Dc2/3x20f20,x10,x20+Dc2/3x21f21,x11,x21=1200∫011s+42x2s30+x2sds, where t∈0,1,β1=β2=4/3,α1=α2=1/3,λ1=1/700,λ2=1/300,μ1=μ2=1,σi1=1/300,σi2=1/200,i=1,2,3 and
(47)w1t,x1t,x2t=11+t4t4x1t4x2t+10+x2t4x1t+6+1Γ4/3∫0tt−s1/3ds1+x12s,f1t,x1t,x2t=t+12100sinx1t+x2t1+x1t+3,g1t,x1t=x1t300+x1t,h1t,x1t=1t+23x1t30+x1t,k1t,x1t=1t+42x1t30+x1t,w2t,x1t,x2t=1500+t2sinx2t+x1t1+x2t+1Γ15/2∫0tt−s13/2x2sds,f2t,x1t,x2t=1235+tx22t+x1t1++x2t+t1/3+1,g2t,x2t=1t+100x2t300+x2t,h2t,x2t=1t+8000x2t30+x2t,k2t,x2t=1t+1600x2t30+x2t.Clearly,q11=q21=1/400,q12=q22=1/500, q41=q51=q61=1/300, q42=q52=q62=1/200, and q61=q62=1/400; furthermore, we have
(48)r11+r12≈max0.007421,0.0128+max0.012255,0.007583≈0.026<1.Thus, by Theorem8, system (46) has an unique solution.


### 4.2. Example 2

Consider the following system:(49)Dc2/3Dc1/2+1600x1tf1t,x1t,x2t=1t2+4t2x1t3x1t+1+x2t6x2t+10+1Γ4/3∫0tt−s1/3ds1+x22s,cD4/3Dc1/3+1700x2tf2t,x1t,x2t=11+t4t4x1t4x1t+10+x2t4x2t+6+1Γ4/3∫0tt−s1/3ds1+x12s,x10f10,x10,x20+x11f11,x11,x21=1300∫011s+100x1s300+x1sds,Dc1/2x10f10,x10,x20+Dc1/2x11f11,x11,x21=1300∫011s+203x1s30+x1sds,Dc1x10f10,x10,x20+Dc1x11f11,x11,x21=1300∫011s+402x1s30+x1sds,x20f20,x10,x20+x21f21,x11,x211300∫011s+200x2s300+x2sds,Dc1/3x20f20,x10,x20+Dc1/3x21f21,x11,x21=1300∫011s+8000x2s30+x2sds,Dc2/3x20f20,x10,x20+Dc2/3x21f21,x11,x21=1300∫011s+1600x2s30+x2sds,where t∈0,1,β1=3/2, α1=1/2, β2=4/3, α2=1/3,λ1=1/600, λ2=1/700, μ1=1, μ2=1, σ11=σ21=σ31=σ12=σ22=σ32=1/300, and
(50)w1t,x1t,x2t=1t2+4t2x1t3x1t+1+x2t6x2t+10,f1t,x1t,x2t=t2200cosx1t+x2t1+x1t+3,g1t,x1t=1t+100x1t300+x1t,h1t,x1t=1t+203x1t30+x1t,k1t,x1t=1t+402x1t30+x1t,w2t,x1t,x2t=11+t4t4x1t4x1t+10+x2t4x2t+6,f2t,x1t,x2t=1235+tx22t+x1t1++x2t+t1/3+1g2t,x2t=1t+200x2t300+x2t,h2t,x2t=1t+8000x2t30+x2t,k2t,x2t=1t+1600x2t30+x2t.In this conrete application, we have(51)π1+π2η0+η1x1+η2x2+π3x1+π4x2=0.0029<1.The riview of Theorem9, problem (49) has a least one solution.


## 4.1. Example 1

Consider the following system of fractional hybrid differentiel equation:(46)cD4/3Dc1/3+1700x1tf1t,x1t,x2t=11+t4t4x1t4x2t+10+x2t4x1t+6+1Γ4/3∫0tt−s1/3ds1+x12s,Dc4/3Dc1/3+1300x2tf2x1t/f1t,x1t,x2t=1500+t2sinx2t+x1t1+x2t+1Γ15/2∫0tt−s13/2x2sds,x10f10,x10,x20+x11f11,x11,x21=1300∫011s+100x1s300+x1sds,Dc1/3x10f10,x10,x20+Dc1/3x11f11,x11,x21=1300∫011s+8000x1s30+x1sds,Dc2/3x10f10,x10,x20+Dc2/3x11f11,x11,x21=1300∫011s+1600x1s30+x1sds.x20f20,x10,x20+x21f21,x11,x21=1200∫01x2s300+x2sds,cD1/3x20f20,x10,x20+Dc1/3x21f21,x11,x21=1200∫011s+23x2s30+x2sds,Dc2/3x20f20,x10,x20+Dc2/3x21f21,x11,x21=1200∫011s+42x2s30+x2sds, where t∈0,1,β1=β2=4/3,α1=α2=1/3,λ1=1/700,λ2=1/300,μ1=μ2=1,σi1=1/300,σi2=1/200,i=1,2,3 and
(47)w1t,x1t,x2t=11+t4t4x1t4x2t+10+x2t4x1t+6+1Γ4/3∫0tt−s1/3ds1+x12s,f1t,x1t,x2t=t+12100sinx1t+x2t1+x1t+3,g1t,x1t=x1t300+x1t,h1t,x1t=1t+23x1t30+x1t,k1t,x1t=1t+42x1t30+x1t,w2t,x1t,x2t=1500+t2sinx2t+x1t1+x2t+1Γ15/2∫0tt−s13/2x2sds,f2t,x1t,x2t=1235+tx22t+x1t1++x2t+t1/3+1,g2t,x2t=1t+100x2t300+x2t,h2t,x2t=1t+8000x2t30+x2t,k2t,x2t=1t+1600x2t30+x2t.Clearly,q11=q21=1/400,q12=q22=1/500, q41=q51=q61=1/300, q42=q52=q62=1/200, and q61=q62=1/400; furthermore, we have
(48)r11+r12≈max0.007421,0.0128+max0.012255,0.007583≈0.026<1.Thus, by Theorem8, system (46) has an unique solution.


## 4.2. Example 2

Consider the following system:(49)Dc2/3Dc1/2+1600x1tf1t,x1t,x2t=1t2+4t2x1t3x1t+1+x2t6x2t+10+1Γ4/3∫0tt−s1/3ds1+x22s,cD4/3Dc1/3+1700x2tf2t,x1t,x2t=11+t4t4x1t4x1t+10+x2t4x2t+6+1Γ4/3∫0tt−s1/3ds1+x12s,x10f10,x10,x20+x11f11,x11,x21=1300∫011s+100x1s300+x1sds,Dc1/2x10f10,x10,x20+Dc1/2x11f11,x11,x21=1300∫011s+203x1s30+x1sds,Dc1x10f10,x10,x20+Dc1x11f11,x11,x21=1300∫011s+402x1s30+x1sds,x20f20,x10,x20+x21f21,x11,x211300∫011s+200x2s300+x2sds,Dc1/3x20f20,x10,x20+Dc1/3x21f21,x11,x21=1300∫011s+8000x2s30+x2sds,Dc2/3x20f20,x10,x20+Dc2/3x21f21,x11,x21=1300∫011s+1600x2s30+x2sds,where t∈0,1,β1=3/2, α1=1/2, β2=4/3, α2=1/3,λ1=1/600, λ2=1/700, μ1=1, μ2=1, σ11=σ21=σ31=σ12=σ22=σ32=1/300, and
(50)w1t,x1t,x2t=1t2+4t2x1t3x1t+1+x2t6x2t+10,f1t,x1t,x2t=t2200cosx1t+x2t1+x1t+3,g1t,x1t=1t+100x1t300+x1t,h1t,x1t=1t+203x1t30+x1t,k1t,x1t=1t+402x1t30+x1t,w2t,x1t,x2t=11+t4t4x1t4x1t+10+x2t4x2t+6,f2t,x1t,x2t=1235+tx22t+x1t1++x2t+t1/3+1g2t,x2t=1t+200x2t300+x2t,h2t,x2t=1t+8000x2t30+x2t,k2t,x2t=1t+1600x2t30+x2t.In this conrete application, we have(51)π1+π2η0+η1x1+η2x2+π3x1+π4x2=0.0029<1.The riview of Theorem9, problem (49) has a least one solution.


## 5. Conclusion

It is known that most natural phenomena are modeled by different types of fractional differential equations. This diversity in investigating complicated fractional differential equations increases our ability for exact modeling of different phenomena. This is useful in making modern software which helps us to allow for more cost-free testing and less material consumption. For our contribution in this present work, we investigate a fractional hybrid differential system with mixed integral hybrid and boundary hybrid conditions. We investigatedtwo numerical examples to illustrate our main results



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