# Existence and Stability for a Coupled Hybrid System of Fractional Differential Equations with Atangana–Baleanu–Caputo Derivative

Authors: Liyuan Zhao; Yirong Jiang
Journal: Journal of Mathematics
 (2022)

---

## Abstract

The aim of this article is to investigate a coupled hybrid system of fractional differential equations with the Atangana–Baleanu–Caputo derivative which contains a Mittag–Leffler kernel function in its kernel. We firstly apply the Dhage fixed point principle to obtain the existence of mild solutions. Then, we study the Ulam–Hyers stability of the introduced fractional coupled hybrid system. Finally, an example is presented to exhibit the validity of our results.


---

## Body

## 1. Introduction

Dhage and Lakshmikantham in 2010 [1] introduced initially first order hybrid differential equations (HDEs) and studied some basic results on the existence and uniqueness of solutions. Furthermore, differential inequalities obtained with respect to HDEs were utilized to examine comparison results and some qualitative properties of the solution. Since the work of [1], many researchers in mathematics and other fields committed to the study of all kinds of HDEs. In particular, some scholars showed that fractional-order hybrid differential equations described the hereditary and memory properties of biology, chemistry, physics etc., better than integer order HDEs. Thus, some scholars would be interesting to the study of fractional-order hybrid differential equations. For instance, Baleanu [2] applied Caputo fractional-order hybrid differential equations to analyze a thermostat model. Other contributions on fractional-order hybrid differential equations involving the Hadamard derivative [3], Riemann derivative [4], Hilfer derivative [5], and Atangana derivative [6, 7] (ABC).On the other hand, thanks to the fact that a large number of practical and real world phenomena in the fields of biology, physics, chemistry, and computer network, can be modeled by coupled systems of different types of fractional differential equations. Here, we sketch some references, but not a list of all references is included, such as Hadamard type [8], Caputo [9] and Riemann–Liouville types [10], and Ψ-Hilfer type [11].Motivated by this fact and the work referenced above, the intent of this work is to study the existence and stability for a coupled hybrid system of fractional differential equations with Atangana–Baleanu–Caputo derivative described by the following equation:(1)Dα0ABCxtg1t,xt,yt=f1t,xt,yt,t∈0,T,0<α<1,Dβ0ABCytg2t,xt,yt=f2t,xt,yt,0<β<1,a1x0g10,x0,y0+b1xTg1T,xT,yT=c1,a2x0g20,x0,y0+b2xTg2T,xT,yT=c2,f10,x0,y0=f20,x0,y0≡0,where 0ABCDα,0ABCDβ denote the Atangana–Baleanu–Caputo fractional derivatives of order α and β, respectively, g1,g2∈C0,T×ℝ×ℝ,ℝ,0, f1,f2∈C0,T×ℝ×ℝ,ℝ and ai+bi≠0,i=1,2. The main contributions of this paper are as follows:(1)
The above-given results [2–7] consider single fractional-order hybrid differential equations, while we consider coupled fractional-order hybrid systems.(2)
The work in [8–11] investigated the existence of solutions or initial value boundary value problem. However, it is well known that stability is one of the important dynamic behaviors for fractional-order hybrid differential equations. Moreover, we should mention that boundary value problem (1) are rather general and include some common cases such as nonseparated boundary and antiperiodic boundary conditions, and so on, by choosing different values of ai,bi,cii=1,2. Thus, it is meaningful to study existence and stability for (1).The remainder of the paper is organized as follows: Section2 contains some necessary concepts, assumptions, and facts. In Section 3, we prove an existence result and the Ulam–Hyers stability for the problem (1). Finally, an example is constructed to show the correctness of our results.


## 2. Preliminaries

LetC0,T,ℝ be the space of all continuous functions on 0,T endowed with the norm x=supt∈0,Txt. We denote by X=AC0,T,ℝ the space of all absolutely continuous functions. Obviously, the product space E=X×X is a Banach space with norm x,yE=x+y. In the remainder of this paper, we recall the necessary basic notions and properties related to fractional calculus.Definition 1.
(see [12]). The fractional Riemann–Liouville integral of order α>0 for a continuous function x on 0,b is given by the following equation:(2)I0αxt=1Γα∫att−sα−1xsds,α>0,b≥t>0.Definition 2 (see [13, 14]).
Letx be differentiable on 0,b , such that x∈L10,b,0<b,α∈0,1. Then, the left Atangana–Baleanu–Caputo derivative (ABC) of x for order α is defined by the following equation:(3)Dα0ABCxt=Nα1−α∫0tx′sEα−αt−sα1−αds,and in the left ABR sense (Riemann–Liouville type) is defined as follows:(4)Dα0ABRxt=Nα1−αddt∫0txsEα−αt−sα1−αds,and the fractional integral associated the above operators is(5)I0αABxt=1−αNαxt+αNαI0αxt,where Eα is a parameter Mittag–Leffler function defined by Eαz∑n=0∞zn/Γnα+1, Nα is a normalization function satisfying N0=N1=1.Remark 1.
In works [13, 14], it has been verified that I0αABDα0ABRxt=xt and 0ABRDαI0αABxt=xt. Also, the authors in the work [13] have established that Dα0ABCxt=Dα0ABRxt−Nα/1−αx0Eα−α/1−αtα.Lemma 1 (see [15]).
Ifxt be well defined on 0,b, then I0αABI0αABCxt=xt−∑k=0Nxk0/k!tk.Definition 3 (see [16]).
An elementx,y∈X×X is said to be a coupled fixed point of a mapping T:X×X⟶X if Tx,y=x and Ty,x=y.Lemma 2 (see [17]).
Suppose thatS is a nonempty, closed, convex, and bounded subset of the Banach algebra X, S^=S×S and P,G:X⟶X and F:S⟶X are three operators such that,(i)
P and G are Lipschitzian with a Lipschitz constants σ and δ, respectively(ii)
F is completely continuous(iii)
y=PyFx+Gy⇒y∈S,∀x∈S(iv)
4σM+δ<1, where M=supFx:x∈S
Then, the operator equationTy,x=PyFx+Gy admits at least one coupled fixed point in S^.Lemma 3 (see [18]).
LetS∗ be a nonempty, closed, convex, and bounded subset of the Banach space X and let ℙ,G:X⟶X and F:S∗⟶X be three operators such that,(i)
ℙ and G are Lipschitzian with a Lipschitz constants Lℙ and LG, respectively(ii)
F is completely continuous(iii)
y=ℙyFx+Gy⇒y∈S,∀x∈S∗(iv)
LℙMF+LG<1, where MF=supFx:x∈S∗
Then, the operator equationℙyFx+Gy=y possesses a solution in S∗.


## 3. Main Results

To start for verifying the main results, the following assumptions are needed for us in the sequel:(H1) Two functionsg1,g2∈C0,T×ℝ×ℝ,ℝ,0 are such that,(i)
Two functionsx↦x/g1t,x,y and y↦y/g2t,x,y are increasing in ℝ a.e. For each t∈0,T(ii)
functionsx/g1t,x,y,y/g2t,x,y∈AC0,T,ℝ(H2) There exist constantsLfi>0,Lgi>0 such that i=1,2(6)fit,x1,y2−fit,x2,y2≤Lfix1−x2+y1−y2,git,x1,y2−git,x2,y2≤Lgix1−x2+y1−y2,∀t,xi,yi∈0,T×ℝ×ℝ.(H3) There exist constantsMfi,Mgi∈ℝ+ such that i=1,2(7)fit,x,y≤Mfi,git,x,y≤Mgi∀t,x,y∈0,T×ℝ×ℝ.(H4)fi:0,T×ℝ×ℝ⟶ℝ are absolutely continuous and there exist constants Kfit∈L10,T,ℝ+ such that i=1,2(8)fit,x,y≤Kfit,∀t∈0,T.In order to study problem (1), we firstly consider the following problem:(9)Dα0ABCxtg1t,xt,yt=ψ1t,t∈0,T,0<α<1,Dβ0ABCytg2t,xt,yt=ψ2t,0<β<1,a1x0g10,x0,y0+b1xTg1T,xT,yT=c1,a2x0g20,x0,y0+b2xTg2T,xT,yT=c2,ψ10=ψ20≡0,Similar to Definition1 in [6], we give a definition of (9)Definition 4.
A functionx,y∈AC0,T,ℝ×AC0,T,ℝ is called a solution of (9), if functions x/g1t,x,y,y/g2t,x,y∈AC0,T,ℝ,∀x,y∈ℝ, ψ1,ψ2∈L10,T,ℝ×L10,T,ℝ, and x,y satisfies (9).An application of Lemma 4 in [7], Lemma 4 in [9], and Theorem 3.1 in [6], we derive an equivalent fractional integral equation to (9). In order to make this paper readable, we present the proof.Let x^,y^∈E be a solution of (42), x,y∈E be a solution of problem (1) satisfyingLemma 4.
Let0<α,β<1, ψ1,ψ2∈AC0,T,ℝ and assume that (H1) holds. Then, the solution of problem (9) if and only if it is a solution of the following system:(10)xt=g1t,xt,yt1−αNαψ1t+αNαI0αψ1t−1a1+b1b11−αNαψ1T+αNαI0αψ1T−c1,yt=g2t,xt,yt1−βNβψ1t+βNαI0βψ2t−1a2+b2b21−βNβψ2T+βNαI0βψ2T−c2.Proof.
The main idea of the proof comes from Lemma 4 in [7] and Theorem 3.1 in [6]. In the view of Lemma 4 in [7], if x,y is a solution of (9), then x,y satisfies fractional integral system (10).
Conversely, since(11)Dα0ABCxtg1t,xt,yt=ψ1t,t∈0,T,0<α<1,Dβ0ABCytg2t,xt,yt=ψ2t,0<β<1.
Byx,y∈AC0,T,ℝ×AC0,T,ℝ, applying I0αAB,I0βAB on both sides, and thanks to Remark 1, we obtain the following equation:(12)I0αAB0ABCDαxtg1t,xt,yt=xtg1t,xt,yt−x0g10,x0,y0,t∈0,T,0<α<1,I0βAB0ABCDβytg2t,xt,yt=ytg2t,xt,yt−y0g20,x0,y0,0<β<1.
Hence(13)I0αABψ1t=xtg1t,xt,yt−x0g10,x0,y0,t∈0,T,0<α<1,I0βABψ2t=ytg2t,xt,yt−y0g20,x0,y0,0<β<1.
Sinceψ1,ψ2∈AC0,T,ℝ, we know that I0αABψ1t,I0βABψ2t∈AC0,T,ℝ. Because x/g1t,x,y,y/g2t,x,y∈AC0,T,ℝ (see condition (H1)), the two functions in both sides of (13) are continuous and thus (13) hold for every t∈0,T. It follows from Lemma 1 that(14)xtg1t,xt,yt=I0αABψ1t+x0,t∈0,T,ytg2t,xt,yt=I0βABψ2t+y0,t∈0,T.
By boundary condition (1) and solving the system, we can obtain xt and yt which have given in (10).Theorem 1.
If hypotheses (H1)–(H4) hold andLg1+Lg2Δ1+Δ2<1, then the coupled system (1) has a solution x,y∈E, where(15)Δ1=1+b1a1+b11−αNα+TαNαΓαKf1L1+c1a1+b1,Δ2=1+b2a2+b21−βNβ+TβNβΓβKf2L1+c2a2+b2.Proof.
We chooseR∗ so that(16)Mg1+Mg2Δ1+Δ21−Lg1+Lg2Δ1+Δ2≤R∗,and specify a subset S∗ of the Banach space X×X by(17)S∗=x,y∈X×X:x,yE≤R∗.
Evidently,S∗ is a convex, bounded and closed subset of the Banach space E. According to Lemma 4, if x,y∈S∗⊆X×X is a solution of (1), then it is a solution of the next system(18)xt=g1t,xt,yt1−αNαf1t,xt,yt+αNαI0αf1t,xt,yt−1a1+b1b11−αNαf1T,xT,yT+αNαI0αf1T,xT,yT−c1,yt=g2t,xt,yt1−βNβf2t,xt,yt+βNβI0βf2t,xt,yt−1a2+b2b21−βNβf2T,xT,yT+βNβI0βf2T,xT,yT−c2.
Fori=1,2, we define operators ℙ=ℙ1,ℙ1:E⟶E,F=F1,F2:S∗⟶E by(19)ℙix,yt=git,xt,yt,F1x,yt=1−αNαf1t,xt,yt+αNαI0αf1t,xt,yt−1a1+b1b11−αNαf1T,xT,yT+αNαI0αf1T,xT,yT−c1,F2x,yt=1−βNβf2t,xt,yt+βNβI0βf2t,xt,yt−1a2+b2b21−βNβf2T,xT,yT+βNβI0βf2T,xT,yT−c2.
Then, the coupled hybrid system of (18) transformed into the coupled system of operator equations as follows:(20)ℙx,ytFx,yt=x,yt,t∈0,T.
This implies that(21)ℙ1x,ytF1x,yt=xt,t∈0,T,ℙ2x,ytF2x,yt=yt,t∈0,T.
By Lemma3, we divide our proof into four steps:Step 1.
ℙ=ℙ1,ℙ1 are Lipschitz operators with constants Lℙ=Lg1+Lg2 on E. Let xi,yi,∈E,i=1,2. Then it follow from (H2) that(22)ℙ1x1,y1t−ℙ1x2,y2t=g1t,x1t,y1t−g1t,x2t,y2t≤Lg1x1t−x2t+y1t−y2t≤Lg1x1−x2+y1−y2.
Thus, we operate the supremum norm overt and obtain(23)ℙ1x1,y1−ℙ1x2,y2≤Lg1x1−x2+y1−y2.
Along the same lines, one has(24)ℙ2x1,y1−ℙ2x2,y2≤Lg2x1−x2+y1−y2.
Thus, we use the definition of operatorℙ to get(25)ℙx1,y1−ℙx2,y2=ℙ1x1,y1,ℙ2x1,y1−ℙ1x2,y2,ℙ2x2,y2=ℙ1x1,y1−ℙ1x2,y2,ℙ2x1,y1−ℙ2x2,y2≤ℙ1x1,y1−ℙ1x2,y2+ℙ2x1,y1−ℙ2x2,y2≤Lg1x1−x2+y1−y2+Lg2x1−x2+y1−y2=Lg1+Lg2x1−x2+y1−y2=Lℙx1,y1−x2,y2.
Therefore, we can confirm thatℙ is also Lipschitzian with Lipschitz constant Lℙ=Lg1+Lg2.Step 2.
Now, we show thatF=F1,F2 is a continuous and compact operator from S∗ into E . We firstly prove the continuity of F , let xn,ynn∈ℕ be a sequence in S∗ converging to a point x,y∈S∗. Then, Lebesgue dominated convergence theorem yields(26)limn⟶∞F1xn,ynt=1−αNαlimn⟶∞f1t,xnt,ynt+αNαlimn⟶∞I0αf1t,xnt,ynt−1a1+b1b11−αNαlimn⟶∞f1T,xnT,ynT+αNαlimn⟶∞I0αf1T,xnT,ynT−c1=1−αNαlimn⟶∞f1t,xnt,ynt+αNα1Γα∫0tt−sα−1limn⟶∞f1s,xns,ynsds−1a1+b1b11−αNαlimn⟶∞f1T,xnT,ynT+αNα∫0TT−sα−1limn⟶∞f1s,xns,ynsds−c1=F1x,yt.
Similarly, we prove(27)limn⟶∞F2xn,ynt=F2x,yt.
Therefore,Fxn,yn=F1xn,yn,F2xn,yn converges to Fx,y.
In what follows, the compactness ofF is explored on S∗. Firstly, we prove the uniform boundedness. Let x,y∈S∗. Then using (H4), one has for t∈0,T,(28)F1x,yt≤1−αNαf1t,xt,yt+αNα1Γα∫0tt−sα−1f1s,xs,ysds−1a1+b1b11−αNαf1T,xT,yT+αNα∫0TT−sα−1f1s,xs,ysds+c1≤1−αNαKf1L1+TαNαΓαKf1L1+1a1+b1b11−αNαKf1L1+b1TαNαΓαKf1L1+c1.
Consequently, one has(29)F1x,yt≤1+b1a1+b11−αNα+TαNαΓαKf1L1+c1a1+b1=Δ1.
Similarly, we prove(30)F2x,yt≤1+b2a2+b21−βNβ+TβNβΓβKf2L1+c2a2+b2=Δ2.
Consequently, we get(31)Fx,yt=F1x,yt+F2x,yt≤Δ1+Δ2.
Hence, it yields thatF is uniformly bounded. Next we prove that F is equicontinious, Let t1,t2∈0,T,t1<t2, any x,y∈S∗, one has(32)F1x,yt2−F1x,yt1≤1−αNαf1t2,xt2,yt2−f1t1,xt1,yt1+|αNα1Γα∫0t1t2−sα−1−t1−sα−1f1s,xs,ysds+αNα1Γα∫t1t2t2−sα−1f1s,xs,ysds|.
It is easy to see thatf1t2,xt2,yt2−f1t1,xt1,yt1⟶0 as t1−t2⟶0, in view of (H4), we obtain(33)F1x,yt2−F1x,yt1≤αNαKf1L1Γα+1∫0t1t2−sα−1−t1−sα−1ds+αNαKf1L1Γα+1∫0t2t2−sα−1ds≤Kf1L1Γα+1t2−t1α+t2α−t1α.
Hence, forε>0, there exists δ>0 such that t1−t2<δ⇒F1x,yt2−F1x,yt1<ε. Similarly, we prove(34)F2x,yt2−F2x,yt1⟶0,t2⟶t1.
Consequently, we get(35)Fx,yt2−Fx,yt1⟶0.
This implies thatF is equicontinious on S∗, and so F is relatively compact. In consequence, we apply the Arzelá–Ascoli theorem to show that F is a complete continuous.Step 3.
we show that the item (iii) of Lemma3 is satisfied. Let x,y∈S∗ satisfy (20).(36)x,y=ℙ1x,yF1x,y,ℙ2x,yF2x,y.
Then, one has(37)xt=ℙ1x,yF1x,y≤g1t,xt,yt1−αNαf1t,xt,yt+αNα1Γα∫0tt−sα−1f1s,xs,ysds+1a1+b1b11−αNαf1T,xT,yT+αNα∫0TT−sα−1f1s,xs,ysds+c1≤g1t,xt,yt−g1t,0,0+g1t,0,0×1−αNα+TαNαΓα+b1a1+b11−αNα+TαNαΓαKf1L1+c1a1+b1≤Lg1xt+yt+Mg1×1−αNα+TαNαΓα+b1a1+b11−αNα+TαNαΓαKf1L1+c1a1+b1.
Hence, we have(38)x≤Lg1x+y+Mg1×1−αNα+TαNαΓα+b1a1+b11−αNα+TαNαΓαKf1L1+c1a1+b1=Lg1x+y+Mg1×Δ1.
Similarly, we prove(39)y≤Lg2x+y+Mg2×Δ2.
It follows from (38) and (39) that(40)x+y≤Mg1+Mg2Δ1+Δ21−Lg1+Lg2Δ1+Δ2≤R∗.
Formula (40) implies that item (iii) of Lemma 3 holds.Step 4.
we show that the condition (iv) of Lemma3 is fulfilled. Since(41)MF=supFx,y:x,y∈S∗=supF1x,y+F1x,y:x,y∈S∗≤Δ1+Δ2.Thus, we have LℙMF=Lg1+Lg2Δ1+Δ2<1, that is, condition (iv) of Lemma 3 holds.
From steps1 to step 4, it follows that all the conditions of Lemma 3, coupled system of hybrid (1) has a solution.
In what follows, we show the Ulam-Hyers stability of (1).Definition 5.
coupled system of hybrid (1) is Ulam–Hyers stable, if for all ϑ=maxϑ1,ϑ2>0 and for all x^,y^∈E of (42)(42)Dα0ABCxtg1t,xt,yt=f1t,xt,yt≤ϑ1,t∈0,T,0<α<1,Dβ0ABCytg2t,xt,yt=f2t,xt,yt≤ϑ2,0<β<1,there is λ=maxλ1,λ2>0 and a solution x,y∈E of (1) such that,(43)x^,y^−x,y≤λϑ.Remark 2.
x^,y^∈E is called a solution of (42) iff there is a function x,y∈E≤ (depending on x^,y^) satisfying
u1t≤ϑ1 and u2t≤ϑ2,t∈0,T;(44)Dα0ABCxtg1t,xt,yt=f1t,xt,yt+u1t,∀t∈0,T,0<α<1,Dβ0ABCytg2t,xt,yt=f2t,xt,yt+u2t,∀t∈0,T,0<β<1.Lemma 5.
Forx^,y^∈E solution of (42), the following inequality holds:(45)x^tg1t,x^t,y^t−φ10−I0αABf1t,x^t,y^t=I0αABu1t≤ϑ11−αNα+TαNαΓαφ10=x^0g10,x^0,y^0;y^tg2t,x^t,y^t−φ20−I0αABf2t,x^t,y^t=I0αABu2t≤ϑ21−αNβ+TβNβΓβ,φ20=y^0g20,x^0,y^0.Theorem 2.
If (H1)–(H4), (42) Θ=1−Λ11−Λ2−Λ2Λ1≠0 are fulfilled, then (1) is Ulam-Hyers stable, where Λ1=Lf1Mg11−α/Nα+Tα/NαΓα+1, Λ2=Lf2Mg21−β/Nβ+Tβ/NβΓβ+1.Proof.
Letx^,y^∈E be a solution of (42), x,y∈E be a solution of problem (1) satisfying(46)x0g1t,x0,y0=φ10=1a1+b1−b1I0αABf1T,xT,yT+c1,y0g2t,x0,y0=φ20=1a2+b2−b2I0βABf2T,xT,yT+c2.By Definition2 and Lemma 2, conditions (H3), (H4), Lf1≤Nα/1−α, we have(47)x^t−xt≤Mg1x^tg1t,x^t,y^t−xtg1t,xt,yt≤Mg1|x^tf1−ABI0αf1t,x^t,y^t+IAB0αf1t,x^t,y^t−φ10−ABI0αf1t,xt,yt|≤Mg1x^tf1t,x^t,y^t−ABI0αf1t,x^t,y^t−φ10+Mg1IAB0αf1t,x^t,y^t−ABI0αf1t,xt,yt≤ϑ11−αNα+TαNαΓα+Lf1Mg1ABI0αx^t−xt+y^t−yt≤ϑ11−αNα+TαNαΓα+Lf1Mg11−αNα+TαNαΓα+1x^−x+y^−y≤ϑ11−αNα+TαNαΓα+Λ1x^−x+y^−y.Using (47), we get(48)1−Λ1x^−x+Λ1y^−y≤ϑ11−αNα+TαNαΓα.Similar the proof of (48), we can obtain(49)1−Λ2y^−y+Λ1x^−x≤ϑ21−βNβ+TβNβΓβ.It follows from (48) and (49) that(50)x^−x≤1−Λ1Θϑ11−αNα+TαNαΓα+Λ1Θϑ21−βNβ+TβNβΓβ,(51)y^−y≤Λ2Θϑ11−αNα+TαNαΓα+1−Λ2Θϑ21−βNβ+TβNβΓβ,Using (50) and (51), we get(52)x^−x+y^−y≤1−Λ1Θ+Λ2Θϑ11−αNα+TαNαΓα+Λ1+1−Λ2Θϑ21−βNβ+TβNβΓβ=1−Λ1Θ+Λ2Θϑ1Ω1+Λ1+1−Λ2Θϑ2Ω2,where Ω1=1−α/Nα+Tα/NαΓα,Ω2=1−β/Nβ+Tβ/NβΓβ. For ϑ=maxϑ1,ϑ2 and λ=1−Λ1+Λ2Ω1+Λ1+1−Λ2/Θ, we can get(53)x^,y^−x,y=x^−x+y^−y≤λϑ.Therefore, by means of Definition2, the solution of problem (1) is Ulam–Hyers stable.


## 4. An Example

We consider the next problem(54)D1/20ABCxt1+sinxt/32+sinyt/32=t9xt1+xt+yt1+ytt∈0,T,12x0+12xπ=0,D1/20ABCyt1+sinxt/32+sinyt/32=t9xt1+xt+yt2+ytt∈0,T,12y0+12yπ=0.Obviously, fori=1,2, Lgi=1/32,Kfit=t, Mgi=Mfi=1, and taking Nα=Nβ=1 as a normalization function, all hypotheses of Theorem 2 are satisfied.In fact, sincet/9L1=π2/18, we can find that LℙMF=Lg1+Lg2Δ1+Δ2=π2/18×7/4×1/16<1.



---

*Source: 4741224-2022-09-26.xml*