# Existence Results for a System of Coupled Hybrid Differential Equations with Fractional Order

Authors: Mohamed Hannabou; Khalid Hilal
Journal: International Journal of Differential Equations
 (2020)

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## Abstract

This paper studies the existence of solutions for a system of coupled hybrid fractional differential equations. We make use of the standard tools of the fixed point theory to establish the main results. The existence and uniqueness result is elaborated with the aid of an example.


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## Body

## 1. Introduction

Fractional calculus is the study of theory and applications of integrals and derivatives of an arbitrary (noninteger) order.This branch of mathematical analysis, extensively investigated in the recent years, has emerged as an effective and powerful tool for the mathematical modeling of several engineering and scientific phenomena. One of the key factors for the popularity of the subject is the nonlocal nature of fractional-order operators.Due to this reason, fractional order operators are used for describing the hereditary properties of many materials and processes. It clearly reflects from the related literature that the focus of investigation has shifted from classical integer-order models to fractional order models. For applications in applied and biomedical sciences and engineering, we refer the reader to the books [1–4].For some recent work on the topic, see [5–12] and the references therein. The study of coupled systems of fractional order differential equations is quite important as such systems appear in a variety of problems of applied nature, especially in biosciences. For details and examples, the reader is referred to the papers [13, 14] and the references cited therein.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 [15, 16].Motivated by some recent studies on hybrid fractional differential equations, we consider the following value problem of coupled hybrid fractional differential equations:(1)DCαxtf1t,xtyt,=h1t,xt,yt,a.e.t∈J=0,T,0<α<1,DCβytf2t,xtyt,=h2t,xt,yt,a.e.t∈J=0,T,0<β<1,ax0f10,x0,y0+bxTf1T,xT,yT=c,ay0f20,x0,y0+byTf2T,xT,yT=c,where DCαandDCβ denote the Caputo fractional derivative of orders αandβ, respectively, a,b,c∈ℝ,fi∈C0,T×ℝ×ℝ,ℝ/0, and fi∈C0,T×ℝ×ℝ,ℝ/0, i=1,2.The aim of this paper is to obtain some existence results for the given problem. Our first theorem describes the uniqueness of solutions for problem (1) by means of Banach’s fixed point theorem. In the second theorem, we apply Leray–Schauder’s alternative criterion to show the existence of solutions for the given problem. The paper is organized as follows. Section 2 contains some basic concepts and an auxiliary lemma, an important result for establishing our main results. In Section 3, we present the main results.


## 2. Coupled System of Hybrid Differential Equations with Fractional Order

In this section, some basic definitions on fractional calculus and an auxiliary lemma are presented [1, 2].Definition 1.
(see [6]). The fractional integral of the function h∈L1a,b,ℝ+ of order α∈ℝ+ is defined by(2)Iaαht=∫att−sα−1Γαhsds,where Γ is the gamma function.Definition 2.
(see [6]). For a function h given on the interval a,b, the Caputo fractional-order derivative of h is defined by(3)Dca+αht=1Γn−α∫att−sn−α−1Γαhnsds,where n=α+1 and α denote the integer part of α.Lemma 1.
(Auxiliary Lemma). Givenh∈C0,T,ℝ and a,b,andc are real constants with a+b≠0, the integral solution of the problem(4)DCαxtft,xt,yt=ht,0<t<T,0<α<1,ax0f0,x0,y0+bxTfT,xT,yT=c,is(5)xt=ft,xt,yt1Γα∫0tt−sα−1hsds−1a+bbΓα∫0TT−sα−1hsds−c,t∈0,T.Proof.
Applying the Caputo integral operator of the orderα, we obtain the first equation in (4).
Again, substituting(6)Dαxtft,xt,yt=ht,we get(7)xtft,xt,yt=x0f0,x0,y0+Iαht.
Then,(8)bxTfT,xT,yT=bx0f0,x0,y0+bΓα∫0TT−sα−1hsds.Thus,(9)ax0f0,x0,y0+bxTfT,xT,yT=a+bx0f0,x0,y0+bΓα∫0TT−sα−1hsds,implies that(10)x0f0,x0,y0=1a+bc−bΓα∫0TT−sα−1hsds.
Consequently,(11)xt=ft,xt,yt1Γα∫0tt−sα−1hsds−1a+bbΓα∫0TT−sα−1hsds−c.
This completes the proof.


## 3. Main Result

LetW=wt∈C10,1 denote a Banach space equipped with the norm w=maxwt,t∈0,1, where W=U×V. Note that the product space U×V,X,Y with the norm x,y=x+y, x,y∈U×V, is also a Banach space.We define an operatorℱ:U×V⟶U×V by(12)ℱx,yt=ℱ1x,ytℱ2x,yt,where(13)ℱ1x,yt=f1t,xt,yt×1Γα∫0tt−sα−1hsds−1a+bbΓα∫0TT−sα−1h1t,xt,ytds−c,ℱ2x,yt=f2t,xt,yt×1Γβ∫0tt−sβ−1hsds−1a+bbΓβ∫0TT−sβ−1h2t,xt,ytds−c.In the sequel, we need the following assumptions:A
1 The functions fii=1,2 are continuous and bounded; that is, there exist positive numbers μfi such that fit,x,y≤μfi, ∀t,x,y∈0,1×ℝ×ℝ.A
2 There exist real constants ρ0,σ0>0 and ρi,σi≥0i=1,2 such that(14)h1t,x,y≤ρ0+ρ1x+ρ2y,h2t,x,y≤σ0+σ1x+σ2y,∀x,y∈ℝFor brevity, let us set(15)ν1=μf1TαΓα+1,ν2=μf2TβΓβ+1,ν3=ca+bμf1+μf2,(16)ν0=min1−1+ba+bν1ρ1+ν2ρ2,1−1+ba+bν2σ1+ν2σ2.


### 3.1. First Result

Now, we are in a position to present our first result that deals with the existence and uniqueness of solutions for problem (1). This result is based on Banach’s contraction mapping principle.Theorem 1.
Suppose that conditionA1 holds and that h1,h1:0,T×ℝ2⟶ℝ are continuous functions. In addition, there exist positive constants ηi and ζi, i=1,2, such that(17)h1t,x2,y2−h1t,x1,y1≤η1x1−x2+η2y1−y2,h2t,x1,y1−h2t,x2,y2≤ζ1x1−x2+ζ2y1−y2,∀t∈0.T,x1,x2,y1,y2∈ℝ.
If1+b/a+bν1η1+η2+ν2ζ1+ζ2<1,ν1 and ν2 are given by (15): then, problem (1) has a unique solution.Proof.
Let us setsupt∈0,Th1t,0,0=k1<∞ and supt∈0,Th2t,0,0=k2<∞ and define a closed ball:(18)B¯r=x,y∈U×V:x,y≤r,where(19)r≥1+b/a+bν1k1+ν2k2+ν31−1+b/a+bν1η1+η2+ν2ζ1+ζ2.
Then, we show thatΘB¯r⊂B¯r. For x,y∈B¯r, we obtain(20)ℱ1x,yt≤μf1maxt∈0,T1Γα∫0tt−sα−1h1s,xs,ysds+1a+bbΓα∫0TT−sα−1h1t,xs,ysds−c≤μf1maxt∈0,T1Γα∫0tt−sα−1×h1s,xs,ys−h1s,0,0+h1s,0,0ds+1a+bbΓα∫0TT−sα−1×h1t,xs,ys−h1s,0,0+h1s,0,0ds−c≤ν1η1x+η2y+k1+bν1a+bη1x+η2y+k1+cν1a+b.
Hence,(21)ℱx,y≤ν11+ba+bη1r+η2r+k1+cν1a+b.
Working in a similar manner, one can find that(22)ℱx,y≤ν21+ba+bζ1r+ζ2r+k2+cν1a+b.
From (21) and (22), it follows that ℱx,y≤r.
Next, forx1,y1, x2,y2∈U×V and for any t∈0,T, we have(23)ℱ1x2,y2t−ℱ1x1,y1t≤μf11Γα∫0tt−sα−1h1s,x2s,y2s−h1s,x1s,y1sds+1a+bbΓα∫0TT−sα−1h1s,x2s,ys−h1s,x1s,ysds≤ν1η1x2−x1+η2y2−y1+k1+bν1a+bη1x+η2y≤ν11+ba+bη1x2−x1+η2y2−y1+k1≤ν11+ba+bη1+η2x2−x1+y2−y1,which yields(24)ℱ1x2,y2−F1x1,y1≤ν11+ba+bη1+η2·x2−x1+y2−y1.
Similarly, one can get(25)ℱ2x2,y2−ℱ2x1,y1≤ν21+ba+bζ1+ζ2·x2−x1+y2−y1.
From (24) and (25), we deduce that(26)ℱx2,y2−ℱx1,y1≤1+ba+bν2η1+η2+ν2·ζ1+ζ2×x2−x1+y2−y1.
In view of condition1+b/va+bν1η1+η2+ν2ζ1+ζ2<1, it follows that ℱ is a contraction.
So Banach’s fixed point theorem applies and hence the operatorℱ has a unique fixed point. This, in turn, implies that problem (1) has a unique solution on 0,T. This completes the proof.


### 3.2. Second Result

In our second result, we discuss the existence of solutions for problem (1) by means of Leray–Schauder alternative.Lemma 2.
(Leray–Schauder alternative [17]). Let F:G⟶G be a completely continuous operator (i.e., a map that is restricted to any bounded set in G is compact). Let ℙF=x∈G:x=λFxfor  some0<λ<1. Then, either the set ℙF is unbounded or F has at least one fixed point.Theorem 2.
Assume that conditionsA1 and A2 hold. Furthermore, it is assumed that1+b/a+bν1ρ1+ν2ρ2<1 and 1+b/a+bν1ρ1+ν2ρ2<1, where ν1 and ν2 are given by (15). Then, boundary value problem (1) has at least one solution.Proof.
We will show that the operatorℱ:U×V⟶U×V satisfies all the assumptions of Lemma 2.
In the first step, we prove that the operatorℱ is completely continuous. Clearly, it follows the continuity of functions f1,f2,h1, and h2 that the operator ℱ is continuous.
Letℳ⊂U×V be bounded. Then, we can find positive constants N1 and N2 such that(27)h1t,xs,ys≤N1,h2t,xs,ys≤N2,∀x,y∈ℳ.
Thus, for anyx,y∈ℳ, we can get(28)ℱ1x,yt≤μf11Γα∫0tt−sα−1h1s,xs,ysds+1a+bbΓα∫0TT−sα−1h1t,xs,ysds−c≤μf1N11+ba+b+cμf1a+b,which yields(29)ℱ1x,y≤μf1N11+ba+b+cμf1a+b.
In a similar manner, one can show that(30)ℱ2x,y≤μf2N21+ba+b+cμf2a+b.
From inequalities (29) and (30), we deduce that the operator ℱ is uniformly bounded.
Now, we show that the operatorℱ is equicontinuous.
For that, we takeτ1, τ2∈0,T with τ1<τ2 and obtain(31)ℱ1x,yτ2−ℱ1x,yτ1≤μf1N1TαΓα+1∫0τ1τ1−sα−1ds−∫0τ2τ2−sα−1ds=μf1N1Tα∫0τ1τ1−sα−1−τ2−sα−1Γα+1ds−∫τ1τ2τ2−sα−1Γα+1ds.
Similarly, one can get(32)ℱ2xτ2,yτ2−ℱ2xτ1,yτ1≤μf2N2TβΓβ+1∫0τ1τ1−sβ−1ds−∫0τ2τ2−sβ−1ds=μf2N2Tβ∫0τ1τ1−sβ−1−τ2−sβ−1Γβ+1ds−∫τ1τ2τ2−sβ−1Γβ+1ds,which tends to 0 independent of x,y.
This implies that the operatorℱx,y is equicontinuous. Thus, by the above findings, the operator ℱx,y is completely continuous.
In the next step, it will be established that the set(33)P=x,y∈U×Vx,y=λℱx,y,0≤λ≤1,is bounded.
Letx,y∈P; then we have x,y=λℱx,y. Thus, for any t∈0,T, we can write(34)xt=λℱ1x,yt,yt=λℱ2x,yt.
Then,(35)xt≤ν11+ba+bρ0+ρ1x+ρ2y+cμf1a+b,yt≤ν21+ba+bσ0+σ1x+σ2y+cμf2a+b,which imply that(36)x≤ν11+ba+bρ0+ρ1x+ρ2y+cμf1a+b,y≤ν21+ba+bσ0+σ1x+σ2y+cμf2a+b.
Consequentlyd, we have(37)x+y=1+ba+bν1ρ0+ν2σ0+ν1ρ1+ν2σ1x+ν1ρ2+ν2σ2y,which, in view of (16), can be expressed as(38)x,y≤1+b/a+bν1ρ0+ν2σ0+ν3ν0.
This shows that the setP is bounded. Hence, all the conditions of Lemma 2 are satisfied and, consequently, the operator ℱ has at least one fixed point, which corresponds to a solution of problem (1). This completes the proof.


## 3.1. First Result

Now, we are in a position to present our first result that deals with the existence and uniqueness of solutions for problem (1). This result is based on Banach’s contraction mapping principle.Theorem 1.
Suppose that conditionA1 holds and that h1,h1:0,T×ℝ2⟶ℝ are continuous functions. In addition, there exist positive constants ηi and ζi, i=1,2, such that(17)h1t,x2,y2−h1t,x1,y1≤η1x1−x2+η2y1−y2,h2t,x1,y1−h2t,x2,y2≤ζ1x1−x2+ζ2y1−y2,∀t∈0.T,x1,x2,y1,y2∈ℝ.
If1+b/a+bν1η1+η2+ν2ζ1+ζ2<1,ν1 and ν2 are given by (15): then, problem (1) has a unique solution.Proof.
Let us setsupt∈0,Th1t,0,0=k1<∞ and supt∈0,Th2t,0,0=k2<∞ and define a closed ball:(18)B¯r=x,y∈U×V:x,y≤r,where(19)r≥1+b/a+bν1k1+ν2k2+ν31−1+b/a+bν1η1+η2+ν2ζ1+ζ2.
Then, we show thatΘB¯r⊂B¯r. For x,y∈B¯r, we obtain(20)ℱ1x,yt≤μf1maxt∈0,T1Γα∫0tt−sα−1h1s,xs,ysds+1a+bbΓα∫0TT−sα−1h1t,xs,ysds−c≤μf1maxt∈0,T1Γα∫0tt−sα−1×h1s,xs,ys−h1s,0,0+h1s,0,0ds+1a+bbΓα∫0TT−sα−1×h1t,xs,ys−h1s,0,0+h1s,0,0ds−c≤ν1η1x+η2y+k1+bν1a+bη1x+η2y+k1+cν1a+b.
Hence,(21)ℱx,y≤ν11+ba+bη1r+η2r+k1+cν1a+b.
Working in a similar manner, one can find that(22)ℱx,y≤ν21+ba+bζ1r+ζ2r+k2+cν1a+b.
From (21) and (22), it follows that ℱx,y≤r.
Next, forx1,y1, x2,y2∈U×V and for any t∈0,T, we have(23)ℱ1x2,y2t−ℱ1x1,y1t≤μf11Γα∫0tt−sα−1h1s,x2s,y2s−h1s,x1s,y1sds+1a+bbΓα∫0TT−sα−1h1s,x2s,ys−h1s,x1s,ysds≤ν1η1x2−x1+η2y2−y1+k1+bν1a+bη1x+η2y≤ν11+ba+bη1x2−x1+η2y2−y1+k1≤ν11+ba+bη1+η2x2−x1+y2−y1,which yields(24)ℱ1x2,y2−F1x1,y1≤ν11+ba+bη1+η2·x2−x1+y2−y1.
Similarly, one can get(25)ℱ2x2,y2−ℱ2x1,y1≤ν21+ba+bζ1+ζ2·x2−x1+y2−y1.
From (24) and (25), we deduce that(26)ℱx2,y2−ℱx1,y1≤1+ba+bν2η1+η2+ν2·ζ1+ζ2×x2−x1+y2−y1.
In view of condition1+b/va+bν1η1+η2+ν2ζ1+ζ2<1, it follows that ℱ is a contraction.
So Banach’s fixed point theorem applies and hence the operatorℱ has a unique fixed point. This, in turn, implies that problem (1) has a unique solution on 0,T. This completes the proof.


## 3.2. Second Result

In our second result, we discuss the existence of solutions for problem (1) by means of Leray–Schauder alternative.Lemma 2.
(Leray–Schauder alternative [17]). Let F:G⟶G be a completely continuous operator (i.e., a map that is restricted to any bounded set in G is compact). Let ℙF=x∈G:x=λFxfor  some0<λ<1. Then, either the set ℙF is unbounded or F has at least one fixed point.Theorem 2.
Assume that conditionsA1 and A2 hold. Furthermore, it is assumed that1+b/a+bν1ρ1+ν2ρ2<1 and 1+b/a+bν1ρ1+ν2ρ2<1, where ν1 and ν2 are given by (15). Then, boundary value problem (1) has at least one solution.Proof.
We will show that the operatorℱ:U×V⟶U×V satisfies all the assumptions of Lemma 2.
In the first step, we prove that the operatorℱ is completely continuous. Clearly, it follows the continuity of functions f1,f2,h1, and h2 that the operator ℱ is continuous.
Letℳ⊂U×V be bounded. Then, we can find positive constants N1 and N2 such that(27)h1t,xs,ys≤N1,h2t,xs,ys≤N2,∀x,y∈ℳ.
Thus, for anyx,y∈ℳ, we can get(28)ℱ1x,yt≤μf11Γα∫0tt−sα−1h1s,xs,ysds+1a+bbΓα∫0TT−sα−1h1t,xs,ysds−c≤μf1N11+ba+b+cμf1a+b,which yields(29)ℱ1x,y≤μf1N11+ba+b+cμf1a+b.
In a similar manner, one can show that(30)ℱ2x,y≤μf2N21+ba+b+cμf2a+b.
From inequalities (29) and (30), we deduce that the operator ℱ is uniformly bounded.
Now, we show that the operatorℱ is equicontinuous.
For that, we takeτ1, τ2∈0,T with τ1<τ2 and obtain(31)ℱ1x,yτ2−ℱ1x,yτ1≤μf1N1TαΓα+1∫0τ1τ1−sα−1ds−∫0τ2τ2−sα−1ds=μf1N1Tα∫0τ1τ1−sα−1−τ2−sα−1Γα+1ds−∫τ1τ2τ2−sα−1Γα+1ds.
Similarly, one can get(32)ℱ2xτ2,yτ2−ℱ2xτ1,yτ1≤μf2N2TβΓβ+1∫0τ1τ1−sβ−1ds−∫0τ2τ2−sβ−1ds=μf2N2Tβ∫0τ1τ1−sβ−1−τ2−sβ−1Γβ+1ds−∫τ1τ2τ2−sβ−1Γβ+1ds,which tends to 0 independent of x,y.
This implies that the operatorℱx,y is equicontinuous. Thus, by the above findings, the operator ℱx,y is completely continuous.
In the next step, it will be established that the set(33)P=x,y∈U×Vx,y=λℱx,y,0≤λ≤1,is bounded.
Letx,y∈P; then we have x,y=λℱx,y. Thus, for any t∈0,T, we can write(34)xt=λℱ1x,yt,yt=λℱ2x,yt.
Then,(35)xt≤ν11+ba+bρ0+ρ1x+ρ2y+cμf1a+b,yt≤ν21+ba+bσ0+σ1x+σ2y+cμf2a+b,which imply that(36)x≤ν11+ba+bρ0+ρ1x+ρ2y+cμf1a+b,y≤ν21+ba+bσ0+σ1x+σ2y+cμf2a+b.
Consequentlyd, we have(37)x+y=1+ba+bν1ρ0+ν2σ0+ν1ρ1+ν2σ1x+ν1ρ2+ν2σ2y,which, in view of (16), can be expressed as(38)x,y≤1+b/a+bν1ρ0+ν2σ0+ν3ν0.
This shows that the setP is bounded. Hence, all the conditions of Lemma 2 are satisfied and, consequently, the operator ℱ has at least one fixed point, which corresponds to a solution of problem (1). This completes the proof.


## 4. Example

An example is provided as follows:(39)DC1/2xt1/2sinxt+1=14t+22xt1+xt+1+132sin2yt,t∈0,1,DC1/2yt1/2cosxt+1=132πsin2πxt+yt161+yt+12,t∈0,1,x0f10,x0,y0+x1f11,x1,y1=1,y0f20,x0,y0+y1f21,x1,y1=1.Here,α=1/2, β=1/2, a=b=c=1,(40)f1t,x,y=12sinxt+1,f2t,x,y=12cosxt+1,h1t,x,y=14t+22xt1+xt+1+132sin2yt,h2t,x,y=132πsin2πxt+yt161+yt+12.Note that(41)h1t,x1,y1−h1t,x2,y2≤116x2−x1+116y2−y1,h2t,x1,y1−h1t,x2,y2≤116x2−x1+116y2−y1,∀t∈0.1,x1,x2,y1,y2∈ℝ,1+ba+bν1η1+η2+ν2ζ1+ζ2=0.12394093<1.Thus, all the conditions of Theorem1 are satisfied and, consequently, there exists a unique solution for problem (39) on 0,1.



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