# Investigation of a Mild Solution to Coupled Systems of Impulsive Hybrid Fractional Differential Equations

Authors: Mohamed Hannabou; Hilal Khalid
Journal: International Journal of Differential Equations
 (2019)

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## Abstract

The study of coupled systems of hybrid fractional differential equations requires the attention of scientists for the exploration of their different important aspects. Our aim in this paper is to study the existence and uniqueness of the solution for impulsive hybrid fractional differential equations. The novelty of this work is the study of a coupled system of impulsive hybrid fractional differential equations with initial and boundary hybrid conditions. We used the classical fixed-point theorems such as the Banach fixed-point theorem and Leray–Schauder alternative fixed-point theorem for existence results. We also give an example of the main results.


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## Body

## 1. Introduction

Fractional differential equations appear naturally in a number of fields such as physics, engineering, biophysics, blood flow phenomena, aerodynamics, electron-analytical chemistry, biology, and control theory. An excellent account of fractional differential equations is given in this study. Undergoing abrupt changes at certain moments of time like earthquake, harvesting, and shock, these perturbations can be well approximated as instantaneous change of state or impulses. Furthermore, these processes are modelled by impulsive differential equations.On the other hand, impulsive differential equations appear as a natural description of many evolutionary phenomena in the real world. The majority of processes in applied sciences are represented by differential equations. However, the situation is different in certain physical phenomena undergoing abrupt changes during their evolution as mechanical systems with impact, biological systems (heartbeat, blood flow, and so on), the dynamics of populations, natural disasters, etc. These changes are often of very short duration and are therefore produced instantly in the form of pulses. The modeling of such phenomena requires the use of forms that explicitly and simultaneously involve the continuous evolution of the phenomenon as well as instantaneous changes.Such models are said to be “impulsive;” they are evolutionary of continuous processes governed by differential equations combined with difference equations representing the effect impulsive has undergone.For some recent developments on the topic, see [1–5] and the references therein.Hybrid fractional differential equations have also been studied by several researchers. This class of equations involves the fractional derivative of an unknown hybrid function with the nonlinearity depending on it. Some recent results on hybrid differential equations can be found in a series of papers [6–11].By applying the Banach fixed-point theorem and Kransnoselskii fixed-point theorem, the existence results for the solution are obtained. Shah et al. [12] studied the coupled system of fractional impulsive boundary problems:(1)Dαxt=Φt,xt,yta.e.t∈J=0,1,t≠ti,1<α≤2,Δxt=ti=Iixti,Δ′xt=ti=Ii¯xti,ti∈0,1,i=1,2,…,m,x0=hx,x1=gx,Dβyt=Ψt,xt,yta.e.t∈j′,1<β≤2,Δxt=tj=Jixtj,Δ′xt=ti=Jj¯xtj,tj∈0,1,j=1,2,…,m,x0=κx,x1=fx,where 1<α,β≤2, Φ, Ψ, Ii, Ii¯, Jj, and Jj¯ are continuous functions, g, h, κ, and f are fixed continuous functionals, and Δxt=ti=xti+−xti−, Δ′xt=ti=x′ti+−x′ti− and Δyt=tj=ytj+−ytj−, Δ′yt=tj=y′tj+−y′tj−.Motivated by some recent studies to the boundary value problem of a class of impulsive hybrid fractional differential, we consider the problem of coupled hybrid fractional differential equations:(2)Dαutf1t,ut,vt=g1t,ut,vt,t∈0,1,t≠ti,i=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,Dα and Dβ stand for Caputo fractional derivative of order α and 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 and ψv=∑j=1nδjvηj, where ξi,ηj∈0,1 for i=1,2,…,n, j=1,2,…,m; and Ik:ℝ⟶ℝ and utk+=limε⟶0+utk+ε and utk−=limε⟶0−utk−ε represent the right and left limits of ut at t=tk, k=i,j.We assume that∑i=1nλiuξiα−1<1 and ∑j=1mδjvηjβ−1<1.This paper is arranged as follows. In Section2, we recall some concepts and some fractional calculation laws and establish preparation results. In Section 3, we present the main results. Section 4 is devoted to an example of the main results.


## 2. Preliminaries

In this section, we recall some basic definitions and properties of the fractional calculus theory and preparation results.Throughout, this paper denotesJ0=0,t1, J1=t1,t2, … , Jp−1=tp−1,tp, Jp=tp,1, and we introduce the spaces: for ti∈0,1 such that t1<t2<⋯<tn and I′=I\t1,t2,…,tn, define the space X=u:0,1⟶ℝ:u∈CI′and leftuti+and right limituti−exist anduti−−uti,1≤i≤n.Then, clearlyX,⋅ is the Banach space under the norm u=maxt∈0,1ut.Similarly, fortj∈0,1 such that t1<t2<⋯<tm and J′=J\t1,t2,…,tm, define the space Y=v:0,1⟶ℝ:v∈CJ′and leftvtj+and right limitvtj−exist andvtj−−vtj,1≤j≤n.Then, clearlyY,⋅ is the Banach space under the norm v=maxt∈0,1vt.Consequently, the productX×Y is a Banach space under the norms u,v=u+v and u,v=maxu,v.Definition 1 (see [13]).
The fractional integral of the functionh∈L1a,b,ℝ+ of order α∈ℝ+ is defined by(3)Iaαht=∫att−sα−1Γαhsds,where Γ is the gamma function.Definition 2 (see [13]).
For a functionh given on the interval a,b, the Riemann–Liouville fractional-order derivative of h is defined by(4)DCa+αht=1Γn−αddtn∫att−sn−α−1Γαsds,where n=α+1 and α denotes the integer part of α.Definition 3 (see [13]).
For a functionh given on the interval a,b, the Caputo fractional-order derivative of h is defined by(5)DCa+αht=1Γn−α∫att−sn−α−1Γαhnsds,where n=α+1 and α denotes the integer part of α.


## 3. Main Results

In this section, we will prove the existence of a mild solution for (2).To obtain the existence of a mild solution, we will need the following assumptions:H1: the function u⟶u/f1t,u,v is increasing in ℝ for every t∈0,t1H2: the function v⟶v/f2t,u,v is increasing in ℝ for every t∈0,t1H3: 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=1,2H4: for all u,u¯,v,v¯∈ℝ for all t∈0,1 there exist positive numbers Mgi>0, such that(6)git,u,v−git,u¯,v¯≤Mgiu−u¯+v−v¯,i=1,2.H5: there exist constants A,B>0 such that for all u,u¯,v,v¯∈ℝ(7)Iiu−Iiu¯≤Au−u¯,i=1,2,…,n,Ijv−Ijv¯≤Bv−v¯,j=1,2,…,m.H6: for any u,v∈C0,1,R, there exist constants Kϕ,Kψ>0, such that(8)ϕu−ϕv≤Kϕu−v,ψu−ψv≤Kψu−v.H7: for any u,v∈C0,1,R, there exist constants Mϕ,Mψ>0 and Nu,Nv>0, such that(9)ϕu≤Mϕu−v,φu≤Mψu−v,Iiu≤Nuu,i=1,2,…,n,Ijv≤Nvv,j=1,2,…,m.H8: there exist constants C,D>0, such that Iiui≤C,i=1,2,…,n and Ijvj≤D,j=1,2,…,mH9: there exist constants ρ,μ>0, such that ϕu≤ρ, ∀u∈X, ψv≤μ, and ∀v∈YH10: there exist constants ρ0,δ0>0 and ρi,δi>0i=1,2, such that(10)g1t,u,v≤ρ0+ρ1u+ρ2v,g2t,u,v≤δ0+δ1u+δ2v,for all u,v∈X×Y.For brevity, let us set(11)Δ1=L1Kϕ+nA+Mg1Γα+1,Δ2=L2Kψ+mB+Mg2Γβ+1.Lemma 1.
Letα∈0,1 and h:0,T0⟶ℝ be continuous. A function u∈C0,T0,ℝ is a solution to the fractional integral equation:(12)ut=u0−∫0at−sα−1Γαhsds+∫0tt−sα−1Γαhsds,if and only if u is a solution to the following fractional Cauchy problems:(13)Dαut=hta.e.t∈0,T0,ua=u0,a>0.Lemma 2.
Let us assume that hypothesesH1 and H3 hold. Let α∈0,1 and h:J⟶ℝ be continuous. A function u is a solution to the fractional integral equation:(14)ut=f1t,ut,vtϕu+θt∑i=1nIiuti−f1t,uti,vti+∫0tt−sα−1Γαhsds,t∈ti,ti+1,where(15)θt=0,t∈t0,t1,1,t∈¯t0,t1,if and only if u is a solution of the following impulsive problem:(16)Dαutf1t,ut,vt=ht,t∈0,1,t≠ti,i=1,2,…,n,0<α<1,uti+=uti−+Iiuti−,ti∈0,1,i=1,2,…,n,u0f10,u0,v0=ϕu.Proof 1.
Let us assume thatu satisfies (16). If t∈t0,t1, then(17)Dαutf1t,ut,vt=ht,t∈t0,t1,(18)u0f10,u0,v0=ϕu.
ApplyingIα on both sides of (17), we can obtain(19)utf1t,ut,vt=u0f10,u0,v0+∫0tt−sα−1Γαhsds,=ϕu+∫0tt−sα−1Γαhsds,then(20)ut=f1t,ut,vtϕu+∫0tt−sα−1Γαhsds.
Ift∈t1,t2, then(21)Dαutf1t,ut,vt=ht,t∈t1,t2,(22)ut1+=ut1−+I1ut1−.
According to Lemma1 and the continuity of f1t,ut,vt, we have(23)utf1t,ut,vt=ut1+f1t1,ut1,vt1−∫0t1t1−sα−1Γαhsds+∫0tt−sα−1Γαhsds,=ut1−+I1ut1−f1t1,ut1,vt1−∫0t1t1−sα−1Γαhsds+∫0tt−sα−1Γαhsds.
Since(24)ut1−=f1t1,ut1,vt1ϕu+∫0t1t1−sα−1Γαhsds,there exists(25)utf1t,ut,vt=ϕu+∫0t1t1−sα−1Γαhsds+I1ut1−f1t1,ut1,vt1−∫0t1t1−sα−1Γαhsds+∫0tt−sα−1Γαhsds,=ϕu+I1ut1−f1t1,ut1,vt1+∫0tt−sα−1Γαhsds,so(26)ut=f1t,ut,vtϕu+I1ut1−f1t1,ut1,vt1+∫0tt−sα−1Γαhsds.
Ift∈t2,t3, then we have(27)utf1t,ut,vt=ut2+f1t2,ut2,vt2−∫0t2t2−sα−1Γαhsds+∫0tt−sα−1Γαhsds,=ut2−+I2ut2−f1t2,ut2,vt2−∫0t2t2−sα−1Γαhsds+∫0tt−sα−1Γαhsds.
For(28)ut2−=f1t2,ut2,vt2ϕu+ut1−+I1ut1−f1t1,ut1,vt1+∫0t2t2−sα−1Γαhsds,we have(29)utf1t,ut,vt=ϕu+ut1−+I1ut1−f1t1,ut1,vt1+∫0t2t2−sα−1Γαhsds+I2ut2−f1t2,ut2,vt2−∫0t2t2−sα−1Γαhsds+∫0tt−sα−1Γαhsds,=ϕu+I1ut1−f1t1,ut1,vt1+I2ut2−f1t2,ut2,vt2+∫0tt−sα−1Γαhsds,so(30)ut=f1t,ut,vtϕu+∑i=12Iiuti−fiti,uti,vti+∫0tt−sα−1Γαhsds.
Ift∈ti,ti+1i=3,4,…,n, using the same method, we get(31)ut=f1t,ut,vtϕu+∑i=1kIiuti−fiti,uti,vti+∫0tt−sα−1Γαhsds.
Conversely, assume thatu satisfies (14). If t∈t0,t1, then we have(32)ut=f1t,ut,vtϕu+∫0tt−sα−1Γαhsds.
Then, divided byf1t,ut,vt and applying Dα on both sides of (32), (17) is satisfied.
Again, substitutingt=0 in (32), we have u0/f10,u0v0=ϕu. Since u⟶u/ft,u,v is increasing in ℝ for t∈t0,t1, the map u⟶u/ft,u,v is injective in ℝ. Then, we get (18).
Ift∈t1,t2, then we have(33)ut=f1t,ut,vtϕu+I1ut1−f1t1,ut1,vt1+∫0tt−sα−1Γαhsds.
Then, divided byf1t,ut,vt and applying Dα on both sides of (33), (19) is satisfied. Again by H3, substituting t=t1 in (32) and taking the limit of (33), (33) minus (32) gives (22).
Ift∈ti,ti+1i=2,3,…,n, similarly we get(34)Dαutf1t,ut,vt=ht,t∈tk,tk+1,uti+=uti−+Iiuti−.
This completes the proof.Lemma 3.
Letg1,g2 are continuous, then u,v∈X×Y is a solution of (2) if and only if u,v is the solution of the integral equations:(35)ut=f1t,ut,vtϕu+θt∑i=1nIiuti−f1t,uti,vti+∫0tt−sα−1Γαg1t,ut,vtds,t∈ti,ti+1,vt=f2t,ut,vtψv+ωt∑j=1mIjvtj−f2t,utj,vtj+∫0tt−sβ−1Γβg2t,ut,vtds,t∈tj,tj+1,where(36)θt=0,t∈t0,t1,1,t∈¯t0,t1,ωt=0,t∈t0,t1,1,t∈¯t0,t1.
We define an operatorΦ:X×Y⟶X×Y by(37)Φu,vt=Φ1u,vt,Φ2u,vt,where(38)Φ1u,vt=f1t,ut,vtϕu+θt∑i=1nIiuti−f1t,uti,vti+∫0tt−sα−1Γαg1s,us,vsds,Φ2u,vt=f2t,ut,vtψv+ωt∑j=1mIjvtj−f2t,utj,vtj+∫0tt−sβ−1Γβg2s,us,vsds.
Now, we are in a position to present our first result that deals with the existence and uniqueness of solutions for the problem (2). This result is based on Banach’s contraction mapping principle.Theorem 1.
Suppose that the conditionH1−H7 holds and that g1,g2:0,1×ℝ2⟶ℝ are continuous functions. In addition, there exist positive constants λi, ζi, i=1,2, such that(39)g1t,u,v−g1t,u¯,v¯≤λ1u−u¯+ζ1v−v¯,g2t,u,v−g2t,u¯,v¯≤λ2u−u¯+ζ2v−v¯.
IfmaxΔ1,Δ2<1, Δ1 and Δ2 are given by (11), then the impulsive coupled system (2) has a unique mild solution.Proof 2.
Let us setsupt∈Jg1t,0,0=κ1<∞ and supt∈Jg2t,0,0=κ1<∞ and define a closed ball: B¯=u,v∈X×Y:u,v≤r, where(40)r≥maxL1κ11−L1Mϕ+nNu+1/Γα+1λ1+λ2,L2κ21−L2Mψ+nNv+1/Γβ+1ζ1+ζ2.
Then, we show thatΦB¯⊂B¯. For u,v∈B¯, we obtain(41)Φ1u,vt≤L1ϕu+θt∑i=1nIiuti−f1t,uti,vti+∫0tt−sα−1Γαg1s,us,vsds,≤L1Mϕu+nNuu+∫0tt−sα−1Γαg1s,us,vs−g1s,0,0+g1s,0,0ds,≤L1Mϕu+nNuu+1Γα+1λ1+λ2u+κ1,≤L1Mϕ+nNur+1Γα+1λ1+λ2r+κ1.
Hence,(42)Φ1u,vt≤L1Mϕ+nNur+1Γα+1λ1+λ2r+κ1.
Working in a similar manner, one can find that(43)Φ2u,vt≤L2Mψ+nNvr+1Γβ+1ζ1+ζ2r+κ2.
From (42) and (43), it follows that Φu,v≤r.
Next, foru,v,u¯,v¯∈X×Y, and for any t∈0,1, we have(44)Φ1u,vt−Φ1u¯,v¯t=f1t,ut,vtϕu+θt∑i=1nIiuti−f1t,uti,vti+∫0tt−sα−1Γαg1s,us,vsds−f1t,u¯t,v¯tϕu¯+θt∑i=1nIiu¯ti−f1t,u¯ti,v¯ti+∫0tt−sα−1Γαg1s,u¯s,v¯sds,≤L1Kϕu−u¯+nAu−u¯+Mg1Γα+1u−u¯+v−v¯,≤L1Kϕu−u¯+nAu−u¯+Mg1Γα+1u−u¯+v−v¯,which implies that(45)Φ1u,v−Φ1u¯,v¯≤L1Kϕ+nA+Mg1Γα+1u−u¯+v−v¯,=Δ1u−u¯+v−v¯.
Similarly, we can show that(46)Φ2u,v−Φ2u¯,v¯≤Δ2u−u¯+v−v¯.
From (45) and (46), we deduce that(47)Φu,v−Φu¯,v¯≤maxΔ1,Δ2u−u¯+v−v¯.In view of this conditionmaxΔ1,Δ2<1, it follows that Φ is a contraction. So, Banach’s fixed point theorem applies, and hence the operator Φ has a unique fixed point. This, in turn, implies that the problem (2) has a unique solution on J. This completes the proof.In our second result, we discuss the existence of solutions for the problem (2) by means of Leray–Schauder alternative.For brevity, let us set(48)μ1=L1Γα+1,μ2=L2Γβ+1,(49)μ0=min1−μ1ρ1+μ2δ1,1−μ1ρ2+μ2δ2.Lemma 4 (Leray–Schauder alternative, see [14]).
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=λℱufor some0<λ<1. Then either the set Pℱ is unbounded or ℱ has at least one fixed point.Theorem 2.
Let us assume that conditionsH1−H3 and H8−H10 hold. Furthermore, it is assumed that μ1ρ1+μ2δ1<1 and μ1ρ2+μ2δ2<1 , where μ1 and μ2 are given by (48). Then, the boundary value problem (2) has at least one solution.Proof 3.
We will show that the operatorΦ:X×Y⟶X×Y satisfies all the assumptions of Lemma 4.
In the first step, we will prove that the operatorΦ is completely continuous.
Clearly, it follows by the continuity of functionsf1,f2,g1, and g2 that the operator Φ is continuous.
LetS⊂X×Y bounded. Then, we can find positive constants H1 and H2 such that g1t,u,v≤H1abdg2t,u,v≤H2,∀u,v∈S.
Thus, for anyu,v∈S, we can get(50)Φ1u,vt≤L1ρ+∑i=1nC+∫0tt−sα−1ΓαH1ds,≤L1ρ+nC+H1Γα+1,which yields(51)Φ1u,v≤L1ρ+nC+H1Γα+1.
In a similar manner, one can show that(52)Φ2u,v≤L2σ+mD+H2Γβ+1.
From the inequalities (51) and (52), we deduce that the operator Φ is uniformly bounded.
Now, we show that the operatorΦ is equicontinuous.
We takeτ1,τ2∈J with τ1<τ2 and obtain(53)Φ1uτ2,vτ2−Φ1uτ1,vτ1≤L1ϕu+θτ2∑i=1nIiuti−f1t,uti,vti+H1∫0τ2τ2−sα−1Γαds−ϕu+θτ1∑i=1nIiuti−f1t,uti,vti+H1∫0τ1τ1−sα−1Γαds≤L1θτ2−θτ1∑i=1nIiuti−f1t,uti,vti+H1∫​0τ2τ2−sα−1Γαds−∫​0τ1τ1−sα−1Γαds≤L1θτ2−θτ1∑i=1nIiuti−f1t,uti,vti+H1∫0τ1τ1−sα−1−τ2−sα−1Γαds−∫τ2τ2τ2−sα−1Γαds,Φ2uτ2,vτ2−Φ2uτ1,vτ1≤L2ωτ2−ωτ1∑j=1mIjutj−f2t,utj,vtj+H2∫0τ1τ1−sα−1−τ2−sα−1Γαds−∫τ2τ2τ2−sα−1Γαds,which tend to 0 independently of u,v. This implies that the operator Φu,v is equicontinuous. Thus, by the abovementioned findings, the operator Φu,v is completely continuous.
In the next step, it will be established that the setP=u,v∈X×Y/u,v=λΦu,v,0≤λ≤1 is bounded.
Letu,v∈P. Then, we have u,v=λΦu,v. Thus, for any t∈0,1, we can write(54)ut=λΦ1u,vt,vt=λΦ2u,vt,then(55)u≤L1ρ+nC+1Γα+1ρ0+ρ1u+ρ2v,≤L1ρ+nC+μ1ρ0+ρ1u+ρ2v,which implies that(56)v≤L2σ+mD+1Γβ+1δ0+δ1u+δ2v,≤L2σ+mD+μ2δ0+δ1u+δ2v.
In consequence, we have(57)u+v≤L1ρ+nC+L2σ+mD+μ1ρ0+μ2δ0+μ1ρ1+μ2δ1u+μ1ρ2+μ2δ2v,which in view of (49), can be expressed as(58)u,v≤L1ρ+nC+L2σ+mD+μ1ρ0+μ2δ0μ0.
This shows that the set is bounded. Hence, all the conditions of Lemma4 are satisfied and consequently the operator Φ has at least one fixed point, which corresponds to a solution of the system (2). This completes the proof.


## 4. Example

Consider the following coupled system of hybrid fractional differential equations:(59)DC1/2utt+ut+vt/40+t2=e−t+sinut+cosvt20,t∈0,1\t1,0<12<1,ut1+=ut1−+−2ut1−,t1≠0,1,CD1/2vtt2+ut−vt/32+t=e−2t+sin2ut+cos2vt20,t∈0,1\t1,0<12<1,vt1+=vt1−+−2vt1−,t1≠0,1,u0f10,u0,v0=∑i=1nciuti,v0f20,u0,v0=∑j=1mdjutj,where f1t,u,v=t+ut+vt/40+t2, f2t,u,v=t2+ut−vt/32+t, g1t,u,v=e−t+sinut+cosvt/40, and g2t,u,v=e−2t+sin2ut+cos2vt/20.Note that(60)g1t,u1,v1−g1t,u2,v2≤140u2−u1+140v2−v1,g2t,u1,v1−g2t,u2,v2≤120u2−u1+120v2−v1,∀t∈0,1,u1,u2,v1,v2∈ℝ,Δ1=L1Kϕ+nA+Mg1Γα+1≃0.3354687<1,Δ2=L2Kψ+mB+Mg2Γβ+1≃0.2548789<1.Thus, all the assumptions in Theorem2 are satisfied, and our results can be applied to the problem (59).



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