# On System of Nonlinear Sequential Hybrid Fractional Differential Equations

Authors: Muath Awadalla; Kinda Abuasbeh
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

In this study, the existence and uniqueness of the solution for a system consisting of sequential fractional differential equations that contain Caputo–Hadamard (CH) derivative are verified. To study the existence and uniqueness of these solutions, some of the most important results from the fixed point theorems in Banach space were used. A practical example is also given to support the theoretical side that was obtained.


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## Body

## 1. Introduction

Many problems in various fields can be successfully formulated by fractional differential equations, such as theoretical physics, Biology, viscosity, electrochemistry, and other physical processes (see [1–7].) In the last decade, the fractional differential equation has attracted the attention of mathematicians, physicists, and engineers as well [8, 9].A fractional differential equation is an equation that contains fractional derivatives and differentials of some mathematical functions and appears in the form of variables. The goal of solving these equations is to find these mathematical functions whose derivatives achieve these equations. Before starting to search for solutions to these equations, studying the conditions of existence and uniqueness is a major matter. To study these conditions, most researchers use the most important fixed point theorems in Banach space, such as Banach contraction principle and Leray Schauder’s theorem (see [10–24]).In 2014, Zhang et al. [19] published a study investigating the existence results for(1)D1pHxtgt,xt∈Gt,xt,t∈1,e,x1=x′e=0,where D1pH,p∈1,2 is the Hadamard fractional derivative, g∈C1,e×ℝ,ℝ\0.In 2016, Algoudi et al. [25] published a study investigating the existence results for the following boundary value problem (sequential Hadamard type):(2)D1pH+λD1p-1Hxt=f1t,xt,yt,D1rHyt,D1qH+λD1q-1Hyt=f2t,xt,D1vHt,yt,x1=0,xe=Iθ1Hyη,y1=0,ye=Iθ2Hxξ,where D1·H,p,q∈1,2,r,v∈0,1 is the Hadamard fractional derivative and Iθ1H is the Hadamard fractional integral with order θ1,θ2>0,f1,f2∈C1,e×ℝ3,ℝ,η,ξ∈1,e.Some researchers went deeper into their research and verified the stability of the solutions to these equations (see [26, 27]). Furthermore, many specialists in the field have paid attention to hybrid fractional differential equations; the importance of fractional hybrid differential equations is that they have a different dynamic than ordinary differential equations and that the hybrid type describes the nonlinear relationship in the derivative of the hybrid function (see [28–31]).Some focus on having solutions to a system of equations (see [32–36]).Based on what has been studied in the articles mentioned above, existence and uniqueness of the following nonlinear coupled differential equations are investigated. Unlike the previous studies, the main results of this article are different; that is, we generalize the problem mentioned in [36] by converting one single fractional differential equation into a system, using different fractional derivatives; here, we consider the problems in the context of sequential type.(3)Da+qH+λDa+q−1Hxtft,xt,yt=ψt,xt,yt,Da+qH+μDa+q−1Hytgt,xt,yt=φt,xt,yt,xa+=x′a+=0,ya+=y′a+=0,where DγH,γ=p,q is the Hadamard fractional derivative of order 1<p,q<2,a≤t≤T,f,g∈Ca,T×ℝ2,ℝ\0, and λ,μ∈ℝ.In this work, we will follow the steps of researchers and specialists in the field. By organizing our results on existence of a solution to the problem as follows, Section2 contains some fundamental results of fractional calculus and important result for establishing our main results. In Section 3, we introduce our main results. In Section 4, a practical example shows the applicability of our results is given. In Section 5, conclusion and future work are presented.


## 2. Preliminaries

In this section, we introduce some useful definitions, lemmas, and notations of fractional calculus.Definition 1 (see [36]).
The Hadamard fractional integral of orderθ for a continuous function ψ:a,∞⟶ℝ is defined as(4)IθHψt=1Γθ∫atlntxθ−1ψxxdx,θ>0.Definition 2 (see [36]).
The Hadamard fractional derivative of orderθ>0 for a continuous function ψ:a,∞⟶ℝ is defined as(5)DθHψt=δnIθHψt=tddtn1Γn−θ∫atlntrn−θ−1ψrrdr,n−1<θ<n,n=θ+1,whereδ=td/dt,θdenotes the integer part of the realnumberθ.
LetCa,T,ℝ denote the Banach space of all real valued continuous functions defined on a,T and Cδna,T,ℝ denote the Banach space of all real valued functions φ such that δnφ∈Ca,T,ℝ (see [38]).Lemma 1 (see [37]).
Letz∈Cδna,T,ℝ,whereCδna,T=z:a,T⟶ℝ:δnz∈Ca,T.
Then,(6)IθHDθHzt=zt−∑j=1nljlntaθ−j.Lemma 2.
Givenx∈Cδ2a,T,ℝ,andh1∈Ca,T,ℝ, and(7)Da+pH+λDa+p−1Hxtft,xt,yt=h1t,1<p<2,0<a≤t≤T,xa+=x′a+=0,λ∈ℝ.Then, the solution of problem (7) is given by(8)xt=ft,xt,yt×t−μ∫atsμ−1Ia+p−1Hφs,xs,ysds.Proof.
ApplyingIa+pH to (2), we get(9)Ia+pHh1t=xtft,xt,yt+b1lntap−1+b2lntap−2+λIa+1Hxtft,xt,yt+c1lntap−2,where b1,b2,c1∈ℝ. The condition xa+=0 implies that b2=0. The first derivative of (4) is calculated as follows:(10)1tIa+pHh1t=xtft,xt,yt′+b1p−1lntap−2+λtxtft,xt,yt+c1lntap−2.
Note thatx′a+=0 implies xt/ft,xt,yt′|t=a+=0, and b1=0=c1.
The integrating factorηt=e∫λ/tdt; then, multiplying ηt by (10), we get(11)ddttλxtft,xt,yt=tλ−1Ia+p−1Hh1t,then integrating (11) and again using xa+=0, we conclude that(12)xtft,xt,yt=t−λ∫atsλ−1Ia+p−1Hh1sds,consequently,(13)xt=ft,xt,yt×t−λ∫atsλ−1Ia+p−1Hh1sds.
In a like manner of Lemma2, one can easily find the solution yt as(14)yt=gt,xt,yt×t−λ∫atsλ−1Ia+p−1Hh1sds.Remark 1.
Forλ=0, the solution is still valid, as IθHh1t=I1HIθ−1Hh1t=∫atIθ−1h1x/xdx similar logic is applied for the case when μ=0. Here, these cases will not be taken for consideration in this study.


## 3. Main Results

In this section, we will present the main results to be obtained from this study.The spaceH=xt,yt:x,y∈Cδ2×Cδ2 is a Banach space with the norm defined as x,yH=x+y∀x,y∈H. Based on Lemma 1, we define an operator ℵ:H⟶H as(15)ℵx,yt=ℵ1x,ytℵ2x,yt,where(16)ℵ1x,yt=ft,xt,yt×t−μ∫atsμ−1Ia+p−1Hφs,xs,ysds,ℵ2x,yt=gt,xt,yt×t−μ∫atsμ−1Ia+q−1Hφs,xs,ysds.To obtain our results for problem (3), we assume that the following conditions hold:C1 Assume that both f,g are continuous and ∃λf,λg>0 such that(17)ft,x,y≤λf,gt,x,y≤λg∀t,x,y∈a,T×ℝ2.C2 Suppose that ψ,φ are continuous and ∃υi,τi>0,i=1,2 such that(18)ψt,x1,y1−ψt,x2,y2≤υ1x1−x2+υ2y1−y2,φt,x1,y1−φt,x2,y2≤τ1x1−x2+τ2y1−y2,∀t∈a,T,xi,yi∈ℝ,i=1,2.C3∃ω0,θ0>0,andωi,θi≥0i=1,2 such that(19)ψt,x,y≤ω0+ω1x+ω2y,φt,x,y≤θ0+θ1x+θ2y,∀t∈a,T,x,y∈ℝ,i=1,2.C4 Define a bounded subset S⊂H, that is, ∃σi>0,i=1,2 such that ψt,xt,yt≤σ1, and φt,xt,yt≤σ2, ∀x,y∈S.UsingC4, note that(20)t−λ∫atsλ−1Ia+p−1Hψs,xs,ysds=t−λ∫atsλ−1Ia+p−1Hψs,xs,ysds,=t−λ∫atsλ−1Ia+p−1Hψs,xs,ysds≤σ1lnt/ap−11−a/tλλΓp.To ease our computations, we set(21)Λ1=supa≤t≤T−λ∫atsμ−1∫aslnsrq−2drrds≤lnT/ap−11−a/TλλΓp,Λ2=supa≤t≤Tt−μ∫atsμ−1∫aslnsrq−2drrds≤lnT/aq−11−a/TμμΓq.Theorem 1.
Assume thatC1 and C2 hold if λfΛ1υ1+υ2+λgΛ2τ1+τ2<1. Then, problem (1) has a unique solution.Proof.
Considerℵ defined by (9) and let Bγ¯=x,y∈H:x,y≤γ be a closed ball in H with γ≥λfΛ1Nψ+λgΛ2Nφ/1−λfΛ1υ1+υ2+λgΛ2τ1+τ2, where Nψ=supa≤t≤Tψt,0,0,Nφ=supa≤t≤Tφt,0,0.
Observe thatψt,x,y=ψt,x,y−ψt,0,0+ψt,0,0≤υ1x+υ2y+Nψ≤υ1+υ2γ+Nψ.
First, we show thatℵB¯γ⊂Bγ¯. For any x,y∈Bγ¯,t∈a,T, we have(22)ℵ1x,yt=ft,xt,yt×t−λ∫atsλ−1Ia+p−1Hψs,xs,ysds,=λf×t−λ∫atsλ−1IHa+p−1ψs,xs,ysds,≤λfυ1+υ2γ+Nψ×supa≤t≤T−λ∫atsλ−1∫aslnsrp−2drrds,≤λfΛ1υ1+υ2γ+Nψ.
In a same manner, we find that(23)ℵ2x,y≤λgΛ2τ1+τ2γ+Nφ.
From (14) and (15), we deduce that ℵx,y≤γ
Next forx1,y1,x2,y2∈H,∀t∈a,T, we have (24)ℵ1x1,y1t−ℵ1x2,y2t≤λf×supa≤t≤Tt−λ∫atsλ−1Ia+p−1Hψs,x1s,y1s−ψs,x2s,y2sds≤λfυ1x1−x2+υ2y1−y2supa≤t≤T−λ∫atsλ−1∫aslnsrp−2drrds,≤λfΛ1υ1x1−x2+υ2y1−y2,≤λfΛ1υ1+υ2x1−x2+y1−y2.
Similarly, we can find(25)ℵ2x1,y1−ℵ2x2,y2≤λgΛ2τ1+τ2x1−x2+y1−y2.
Combining (24) and (25) yields(26)ℵ2x1,y1−ℵ2x2,y2≤λfΛ1υ1+υ2+λgΛ2τ1+τ2×x1−x2+y1−y2.Theorem 2.
. AssumeC1 and C3 and C4 hold if λfΛ1ω1+λgΛ2θ1<1 and λfΛ1ω2+λgΛ2θ2<1. Then, problem (1) has at least one solution.Proof.
We first prove that the operatorℵ:H⟶H is completely continuous; obviously, the operator is continuous as a result that f,g,ψ, and φ are all assumed to be continuous.
ByC4, ∀x,y∈S, we have(27)ℵ1x,yt≤λf×supa≤t≤Tt−λ∫atsλ−1Ia+p−1Hψs,xs,ysds≤λfΛ1σ1,(28)ℵ2x,y<λgΛ2σ2.
Combining inequalities (27) and (28) yields ℵ2x,y≤λfΛ1σ1+λgΛ2σ2; that is, the operator ℵ is uniformly bounded.
Next, we prove equicontinuity for the operatorℵ; for this, we let t1,t2∈a,T,t1<t2.
Then,(29)ℵ1x,yt2−ℵ1x,yt1≤λf×supa≤t≤Tt2−λ∫at2sλ−1Ia+p−1Hψs,xs,ysds−supa≤t≤Tt1−λ∫at2sλ−1Ia+p−1Hψs,xs,ysds≤λfσ1λΓpt1λ−aλlnt1ap−1+t2λlnt2ap−1⋅t2λ−t1λ,(30)ℵ2x,yt2−ℵ2x,yt1≤λg×supa≤t≤Tt2−μ∫at2sμ−1IHa+q−1ψs,xs,ysds−supa≤t≤Tt1−μ∫at1sμ−1IHa+q−1ψs,xs,ysds≤λgσ2μΓqt1μ−aμlnt1aq−1+t2μlnt2aq−1⋅t2μ−t1μ.
R.H.Ss are both independent onx,y; in addition, R.H.Ss of both (29) and (30) approach to zero when t1⟶t2 and they imply that the operator ℵx,y is equicontinuous; consequently, the operator ℵx,y is completely continuous.
To finish, we establish the bounded set given byΩ=x,y∈H:x,y=βℵx,y,β∈0,1, and then ∀t∈0,1, with x,y=βℵx,y, we obtain(31)xt=βℵ1x,yt,yt=βℵ2x,yt.
ByC3, we get(32)x≤λfΛ1ω0+ω1x+ω2y,(33)y≤λgΛ2θ0+θ1x+θ2y.
Consequently, we have(34)x+y≤λfΛ1ω0+λgΛ2θ0+λfΛ1ω1+λgΛ2θ1x+λfΛ1ω2+λgΛ2θ2y.
Inequality (34) can be written as follows:(35)x,y≤λfΛ1ω0+λgΛ2θ0/Λ0,where Λ0=min1−λfΛ1ω1+λgΛ2θ1,1−λfΛ1ω2+λgΛ2θ2. By (35), we conclude that Ω is bounded. Hence, Leray–Schauder alternative applies; that is, problem (1) has at least one solution. This completes the proof.


## 4. Example

Consider the following initial value problem:(36)Da+7/4H+Da+3/4Hxt1/3sinxt+2=3lnt+111t3+15x1+x+144tan−1y,0<1≤t≤e,Da+5/4H+μDa+1/4Hyt1/3sinyt+1=e−tcost+120tan−1y+tan−1x,x1=x′1=0,y1=y′1=0.Here,1=λ=μ, p=7/4,q=5/4,ft,x,y=1/3sinx+2,gt,x,y=1/3cosy+1, (37)ψt,xt,yt=3lnt+111t3+15x1+x+144tan−1y,φt,xt,yt=e−tcost+120tan−1y+tan−1x.Observe that(38)ψt,x1,y1−ψt,x2,y2≤144x2−x1+144y2−y1,φt,x1,y1−φt,x2,y2≤120x2−x1+120y2−y1,λfΛ1υ1+υ2+λgΛ2τ1+τ2≤0.1363483<1.Thus, problem (36) satisfies all the conditions of Theorem 1; accordingly, we conclude that the B.V.P has a unique solution on 1,e.


## 5. Conclusion and Future Work

In this article, the existence and uniqueness theory of solutions for sequential fractional differential system involving Hadamard fractional derivatives of order 1 <p, q < 2 with initial conditions were investigated. For the future work, the researcher may generalize our system by taking an n×1 system of sequential fractional differential equations and may apply another type of fractional derivatives such as Psi-Hilfer and Psi-Caputo fractional derivatives.



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