# On Coupled Systems for Hilfer Fractional Differential Equations with Nonlocal Integral Boundary Conditions

Authors: Athasit Wongcharoen; Sotiris K. Ntouyas; Jessada Tariboon
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, we study a coupled system involving Hilfer fractional derivatives with nonlocal integral boundary conditions. Existence and uniqueness results are obtained by applying Leray-Schauder alternative, Krasnoselskii’s fixed point theorem, and Banach’s contraction mapping principle. Examples illustrating our results are also presented.


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## Body

## 1. Introduction

The theory of fractional differential equations has been widely used in pure mathematics and applications in the fields of physics, biology, and engineering. There are many interesting results for qualitative analysis and applications. We refer the interested reader, in fractional calculus, to the classical reference texts such as [1–7]. In the literature, there exist several different definitions of fractional integrals and derivatives, and the most popular of them are Riemann–Liouville, Caputo, and other less-known such as Hadamard fractional derivative and the Erdeyl–Kober fractional derivative. A generalization of derivatives of both Riemann–Liouville and Caputo was given by Hilfer in [8] as(1)Dα,βHut=Iβn−αDnI1−βn−αut,where n−1<α<n, 0≤β≤1, t>a≥0, and Dn=dn/dtn.He named it as generalized fractional derivative of orderα and a type β. Many authors call it the Hilfer fractional derivative. We notice that when β=0, the Hilfer fractional derivative corresponds to the Riemann–Liouville fractional derivative:(2)Dα,0Hut=DnIn−αut.Whenβ=1, the Hilfer fractional derivative corresponds to the Caputo fractional derivative:(3)Dα,1Hut=In−αDnut.Such derivative interpolates between the Riemann–Liouville and Caputo derivative. Some properties and applications of the Hilfer derivative are given in [9, 10] and the references cited therein.Initial value problems involving Hilfer fractional derivatives were studied by several authors, see, for example [11–16] and references therein. Nonlocal boundary value problems for Hilfer fractional derivative were studied in [17].To the best of our knowledge, there is no work carried out on systems of boundary value problems with Hilfer fractional derivative in the literature. This paper come to fill this gap. Thus, the objective of the present work is to introduce a new class of coupled systems of Hilfer-type fractional differential equations with nonlocal integral boundary conditions and develop the existence and uniqueness of solutions. In precise terms, we consider the following coupled system:(4)Dα,βHxt=ft,xt,yt,t∈a,b,Dα1,β1Hyt=gt,xt,yt,t∈a,b,xa=0,xb=∑i=1mθiIφiyξi,ya=0,yb=∑j=1nζjIψjxzj,where Dα,βH and Dα1,β1H are the Hilfer fractional derivatives of orders α and α1, 1<α,α1<2, and parameters β and β1, respectively, 0≤β,β1≤1, and Iφi, Iψj are the Riemann–Liouville fractional integrals of order φi>0 and ψj>0, respectively, the points ξi, zj∈a,b, a≥0, f,g:a,b×ℝ×ℝ⟶ℝ are continuous functions and θi, ζj∈ℝ, i=1,2,…,m, j=1,2,…,n are given real constants.The paper is organized as follows. We present our main results in Section3, by applying Leray–Schauder alternative, Krasnoselskii’s fixed point theorem, and Banach’s contraction mapping principle, while Section 2 contains some preliminary concepts related to our problem. Examples are constructed to illustrate the main results.


## 2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus and present preliminary results needed in our proofs later [2, 5].Definition 1.
The Riemann–Liouville fractional integral of orderα>0 of a continuous function u:a,∞⟶ℝ is defined by(5)Iαut=1Γα∫att−sα−1usds,provided the right-hand side exists on a,∞.Definition 2.
The Riemann–Liouville fractional derivative of orderα>0 of a continuous function is defined by(6)DαRLut≔DnIn−αut=1Γn−αddtn∫att−sn−α−1usds,where n=α+1, α denotes the integer part of real number α, provided the right-hand side is pointwise defined on a,∞.Definition 3.
The Caputo fractional derivative of orderα>0 of a continuous function is defined by(7)DαCut≔In−αDnut=1Γn−α∫att−sn−α−1ddsnusds,n−1<α<n,provided the right-hand side is pointwise defined on a,∞.Lemma 1 (see [18]).
Ifg∈Cna,b, n−1<α<n, and 0≤β≤1, then(1)
IαHDα,βgt=gt−∑k=1nt−aγ−k/Γγ−k+1d/dtn−kI1−βn−αga(2)
DHα,βIαgt=gt
The following lemma deals with a linear variant of problem (4).Lemma 2.
Letφi, ψj>0, ξi, zj∈a,b, a≥0, θi, ζj∈ℝ, i=1,2,…,m, j=1,2,…,n, 1<α,α1<2, 0≤β,β1≤1, γ=α+2β−αβ, γ1=α1+2β1−α1β1, h,h1∈Ca,b,ℝ, and(8)Λ=b−aγ+γ1−2ΓγΓγ1−∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjzj−aγ+ψj−1Γγ+ψj≠0.Then, the system(9)Dα,βHxt=ht,t∈a,b,Dα1,β1Hyt=h1t,t∈a,b,xa=0,xb=∑i=1mθiIφiyξi,ya=0,yb=∑j=1nζjIψjxzj,is equivalent to the following integral equations:(10)xt=Iαht+t−aγ−1ΛΓγb−aγ1−1Γγ1∑i=1mθiIα1+φih1ξi−Iαhb+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjhzj−Iα1h1b(11)yt=Iα1h1t+t−aγ1−1ΛΓγ1b−aγ−1Γγ∑j=1nζjIα+ψjhzj−Iα1h1b+∑j=1nζjzj−aγ+ψj−1Γγ+ψj∑i=1mθiIα1+φih1ξi−Iαhb.Proof.
Operating fractional integralIα on both sides of the first equation in (9) and using Lemma 1, we obtain(12)xt−∑k=12t−aγ−kΓγ−k+1ddt2−kI1−β2−αxa=Iαht.
Then, we have, since1−β2−α=2−γ,(13)xt=t−aγ−1ΓγddtI2−γxtt=a+t−aγ−2Γγ−1I2−γxtt=a+Iαht=t−aγ−1ΓγDHγ−1,βxtt=a+t−aγ−2Γγ−1I2−γxtt=a+Iαht.
Then,(14)xt=c1Γγt−aγ−1+c2Γγ−1t−aγ−2+Iαht,where(15)c1=HDγ−1,βxtt=a,c2=I2−γxtt=a.
By a similar way, we obtain(16)yt=d1Γγ1t−aγ1−1+d2Γγ1−1t−aγ1−2+Iα1h1t.
By setting(17)d1=HDγ1−1,βytt=a,d2=I2−γytt=a,from the boundary conditions xa=0 and ya=0, we obtain c2=0 and d2=0. Then, we obtain(18)xt=c1Γγt−aγ−1+Iαht,yt=d1Γγ1t−aγ1−1+Iα1h1t,(19)∑i=1mθiIφiyξi=d1∑i=1mθiξi−aγ1+φi−1Γγ1+φi+∑i=1mθiIα1+φih1ξi,∑j=1nζiIψixzj=c1∑j=1nζjzj−aγ+ψj−1Γγ+ψj+∑j=1nζjIα+ψjhzj.
Fromxb=∑i=1mθiIφiyξi and yb=∑j=1nζjIψjxzj, we have(20)c1=1Λb−aγ1−1Γγ1∑i=1mθiIα1+φih1ξi−Iαhb+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjhzj−Iα1h1b,d1=1Λb−aγ−1Γγ∑j=1nζjIα+ψjhzj−Iα1h1b+∑j=1nζjzj−aγ+ψj−1Γγ+ψj∑i=1mθiIα1+φih1ξi−Iαhb.
Substituting the values ofc1 and d1 in (18), we obtain solutions (10) and (11). The converse follows by direct computation. This completes the proof.


## 3. Main Results

LetC=Ca,b,ℝ, a≥0, denote the Banach space of all continuous functions from a,b to ℝ. The space X=x:xt∈C2a,b,ℝ endowed with the norm x=supxt,t∈a,b is a Banach space. Let Y=y:yt∈C2a,b,ℝ with the norm y=supyt,t∈a,b. It is obvious that the product space X×Y,x,y is Banach space with the norm x,y=x+y.In view of Lemma2, we define two operators K:X×Y⟶X×Y by(21)Kx,yt=K1x,ytK2x,yt,where(22)K1x,yt=Iαfx,yt+t−aγ−1ΛΓγb−aγ1−1Γγ1∑i=1mθiIα1+φigx,yξi−Iαfx,yb+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjfx,yzj−Iα1gx,yb,K2x,yt=Iα1gx,yt+t−aγ−1ΛΓγb−aγ1−1Γγ1∑j=1nζjIα+ψjfx,yzj−Iα1gx,yb+∑j=1nζjzj−aγ+ψj−1Γγ+ψj∑i=1mθiIα1+φigx,yξi−Iαfx,yb,where(23)fx,yt=ft,xt,yt,gx,yt=gt,xt,yt,t∈a,b.For computational convenience, we set(24)M1=b−aαΓα+1+b−aγ−1ΛΓγb−aγ1+α−1Γγ1Γα+1+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjzj−aα+ψjΓα+ψj+1,(25)M2=b−aγ−1ΛΓγ+b−aγ−1Γγ1∑i=1mθiξi−aα1+φiΓα1+φi+1+b−aα1Γα1+1∑i=1mθiξi−aγ1+φi−1Γγ1+φi,(26)M3=b−aγ−1ΛΓγ1b−aγ−1Γγ∑j=1nζjzj−aα+ψjΓα+ψj+1+b−aαΓα1+1∑i=1mζjzj−aγ+ψj−1Γγ+ψj,(27)M4=b−aα1Γα1+1+b−aγ1−1ΛΓγ1b−aγ+α1−1ΓγΓα1+1+∑j=1nζjzj−aγ+ψj−1Γγ+ψj∑i=1mθiξi−aα1+φiΓα1+φi+1.Banach’s contraction mapping principle is applied in the first result to prove existence and uniqueness of solutions of system (4).Theorem 1.
Suppose thatf,g:a,b×ℝ×ℝ⟶ℝ are continuous functions. In addition, we assume that
H1 There exist constants ℓi,ni,i=1,2, such that, for all t∈a,b and xi,yi∈ℝ,i=1,2,(28)ft,x1,y1−ft,x2,y2≤l1x1−x2+l2y1−y2(29)gt,x1,y1−gt,x2,y2≤n1x1−x2+n2y1−y2.Then, system (4) has a unique solution on a,b, if(30)M1+M3ℓ1+ℓ2+M2+M4n1+n2<1.Proof. Definesupt∈a,bft,0,0=N1<∞ and supt∈a,bgt,0,0=N2<∞ such that(31)r>M1+M3N1+M2+M4N21−M1+M3ℓ1+ℓ2+M2+M4n1+n2.Now, we will show that the setKBr⊂Br, where Br=x,y∈X×Y:x,y≤r. For any x,y∈Br, t∈a,b, we find that(32)ft,xt,yt=ft,xt,yt−ft,0,0+ft,0,0≤ft,xt,yt−ft,0,0+ft,0,0≤ℓ1x+ℓ2y+N1,gt,xt,yt≤n1x+n2y+N2.Forx,y∈Br, we have(33)K1x,yt≤Iαfx,yb+b−aγ−1ΛΓγb−aγ1−1Γγ1∑i=1mθiIα1+φigx,yξi+Iαfx,yb+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjfx,yzj+Iα1gx,yb≤Iαfx,y−f0,0+f0,0b+b−aγ−1ΛΓγb−aγ1−1Γγ1×∑i=1mθiIα1+φigx,y−g0,0+g0,0ξi+Iαfx,y−f0,0+f0,0b+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjfx,y−f0,0+f0,0zj+Iα1gx,y−g0,0+g0,0b≤Iαℓ1x+ℓ2y+N1b+b−aγ−1ΛΓγb−aγ1−1Γγ1×∑i=1mθiIα1+φin1x+n2y+N2ξi+Iαℓ1x+ℓ2y+N1b+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjℓ1x+ℓ2y+N1zj+Iα1n1x+n2y+N2b≤M1ℓ1x+ℓ2y+N1+M2n1x+n2y+N2=M1ℓ1+M2n1x+M1ℓ2+M2n2y+M1N1+M2N2≤M1ℓ1+ℓ2+M2n1+n2r+M1N1+M2N2.Hence,(34)K1x,y≤M1ℓ1+ℓ2+M2n1+n2r+M1N1+M2N2.Similarly, we have(35)K2x,yt≤M3ℓ1x+ℓ2y+N1+M4n1x+n2y+N2,and hence(36)K2x,y≤M3ℓ1+ℓ2+M4n1+n2r+M3N1+M4N2.Consequently, it follows that(37)Kx,y≤M1ℓ1+ℓ2+M2n1+n2r+M1N1+M2N2+M3ℓ1+ℓ2+M4n1+n2r+M3N1+M4N2≤r,which implies KBr⊂Br. Next, we will show that the operator K is a contraction mapping. For any x1,y1,x2,y2∈X×Y, we obtain(38)K1x1,y1t−K1x2,y2t≤Iαfx1,y1−fx2,y2b+b−aγ−1ΛΓγb−aγ1−1Γγ1∑i=1mθiIα1+φigx1,y1−gx2,y2ξi+Iαfx1,y1−fx2,y2b+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjfx1,y1−fx2,y2zj+Iα1gx1,y1−gx2,y2b≤ℓ1x1−x2+ℓ2y1−y2Iα1b+b−aγ−1ΛΓγb−aγ1−1Γγ1n1x1−x2+n2y1−y2∑i=1mθiIα1+φi1ξi+ℓ1x1−x2+ℓ2y1−y2Iα1b+ℓ1x1−x2+ℓ2y1−y2∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψj1zj+n1x1−x2+n2y1−y2Iα11b≤M1ℓ1x1−x2+ℓ2y1−y2+M2n1x1−x2+n2y1−y2=M1ℓ1+M2n1x1−x2+M1ℓ2+M2n2y1−y2.Therefore, we obtain the following inequality:(39)K1x1,y1−K1x2,y2≤M1ℓ1+ℓ2+M2n1+n2x1−x2+y1−y2.In addition, we also obtain(40)K2x1,y1−K2x2,y2≤M3ℓ1+ℓ2+M4n1+n2x1−x2+y1−y2.From (39) and (40), it yields(41)Kx1,y1−Kx2,y2≤M1+M3ℓ1+ℓ2+M2+M4n1+n2×x1−x2+y1−y2.AsM1+M3ℓ1+ℓ2+M2+M4n1+n2<1, therefore, K is a contraction operator. By Banach’s fixed point theorem, the operator K has a unique fixed point, which is the unique solution of (4) on a,b. The proof is completed.Now, we prove our second existence result via Leray–Schauder alternative.Lemma 3 (Leray–Schauder alternative, see [19]).
LetF:E⟶E be a completely continuous operator. Let(42)ξF=x∈E:x=λFxfor some,0<λ<1.
Then, either the setξF is unbounded, or F has at least one fixed point.Theorem 2.
Assume that there exist real constantsui,vi≥0 for i=1,2 and u0,v0>0 such that, for any xi∈ℝi=1,2, we have(43)ft,x1,x2≤u0+u1x1+u2x2,gt,x1,x2≤v0+v1x1+v2x2.IfM1+M3u1+M2+M4v1<1 and M1+M3u2+M2+M4v2<1, where M1,M2,M3,andM4 are given in (24)–(27), then (4) has at least one solution on a,b.Proof.
By continuity of the functionsfandg on a,b×ℝ×ℝ, the operator K is continuous. We will show that the operator K:X×Y⟶X×Y is completely continuous. Let Φ⊂X×Y be bounded. Then, there exist positive constants L1 and L2 such that(44)ft,x,y≤L1,gt,x,y≤L2,∀x,y∈Φ.
Then, for anyx,y∈Φ, we have(45)K1x,yt≤Iαfx,yb+b−aγ−1ΛΓγb−aγ1−1Γγ1∑i=1mθiIα1+φigx,yξi+Iαfx,yb+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjfx,yzj+Iα1gx,yb≤L1Iα1b+b−aγ−1ΛΓγb−aγ1−1Γγ1L2∑i=1mθiIα1+φi1ξi+L1Iα1b+∑i=1mθiξi−aγ1+φi−1Γγ1+φiL1∑j=1nζjIα+ψj1zj+L2Iα11b≤L1M1+L2M2,which yields(46)K1x,y≤L1M1+L2M2.
Similarly, we obtain that(47)K2x,y≤L1M3+L2M4.
Hence, from the above inequalities, we obtain that the setKΦ is uniformly bounded. Next, we are going to prove that the set KΦ is equicontinuous. For any x,y∈Φ and τ1,τ2∈a,b such that τ1<τ2, we have(48)K1x,yτ2−K1x,yτ1≤Iαfx,yτ2−Iαfx,yτ1+τ2−aγ−1−τ1−aγ−1ΛΓγb−aγ1−1Γγ1∑i=1mθiIα1+φigx,yξi+Iαfx,yb+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjfx,yzj+Iα1gx,yb≤L1∫aτ1τ2−sα−1−τ1−aα−1Γαds+∫τ1τ2τ2−sα−1Γαds+τ2−aγ−1−τ1−aγ−1ΛΓγb−aγ1−1Γγ1L2∑i=1mθiIα1+φi1ξi+L1Iα1b+∑i=1mθiξi−aγ1+φi−1Γγ1+φiL1∑j=1nζjIα+ψj1zj+L2Iα11b≤L1Γα+12τ2−τ1α+τ2−aα−τ1−aα+τ2−aγ−1−τ1−aγ−1ΛΓγb−aγ1−1Γγ1L2∑i=1mθiξi−aα1+φiΓα1+φi+1+L1b−aαΓα+1+∑i=1mθiξi−aγ1+φi−1Γγ1+φiL1∑j=1nζjzj−aα+ψjα+ψj+1+L2b−aα1Γα+1.
Therefore, we obtain(49)K1x,yτ2−K1x,yτ1⟶0,asτ1⟶τ2.
Analogously, we can obtain the following inequality:(50)K2x,yτ2−K2x,yτ1⟶0,asτ1⟶τ2.
Hence, the setKΦ is equicontinuous. By applying the Arzelá–Ascoli theorem, the set KΦ is relative compact which implies that the operator K is completely continuous. Lastly, we shall show that the set ξ=x,y∈X×Y:x,y=λKx,y,0≤λ≤1 is bounded. Let any x,y∈ξ, then x,y=λKx,y. For any t∈a,b, we have(51)xt=λK1x,yt,yt=λK2x,yt.
Then, we obtain(52)x≤u0+u1x+u2yM1+v0+v1x+v2yM2,y≤u0+u1x+u2yM3+v0+v1x+v2yM4,which imply that(53)x+y≤M1+M3u0+M2+M4v0+M1+M3u1+M2+M4v1x+M1+M3u2+M2+M4v2y.
Thus, we obtain(54)x,y≤M1+M3u0+M2+M4v0M∗,where M∗=min1−M1+M3u1−M2+M4v1,1−M1+M3u2−M2+M4v2, which shows that the set ξ is bounded. Therefore, by applying Lemma 3, the operator K has at least one fixed point. Therefore, we deduce that problem (4) has at least one solution on a,b. The proof is complete.
The last existence theorem is based on Krasnoselskii’s fixed point theorem.Lemma 4 (Krasnoselskii’s fixed point theorem, see [20]).
LetM be a closed, bounded, convex, and nonempty subset of a Banach space X. Let AandB be operators such that (i) Ax+By∈M, where x,y∈M, (ii) A is compact and continuous, and (iii) B is a contraction mapping. Then, there exists z∈M such that z=Az+Bz.Theorem 3.
Assume thatf,g:a,b×ℝ×ℝ⟶ℝ are continuous functions satisfying assumption H1 in Theorem 1. In addition, we suppose and there exist two positive constants P,Q such that, for all t∈a,b and xi,yi∈ℝ,i=1,2,(55)ft,x1,x2≤P,gt,x1,x2≤Q.
If(56)b−aαΓα+1ℓ1+ℓ2+b−aα1Γα+1n1+n2<1,then problem (4) has at least one solution on a,b.Proof.
To apply Lemma4, we decompose the operator K into four operators K1,1,K1,2,K2,1, and K2,2 as(57)K1,1x,yt=t−aγ−1ΛΓγb−aγ1−1Γγ1∑i=1mθiIα1+φigx,yξi−Iαfx,yb+∑i=1mθiξi−aγ1+φi−1Γγ1+φi∑j=1nζjIα+ψjfx,yzj−Iα1gx,yb,K1,2x,yt=Iαfx,yt,K2,1x,yt=t−aγ1−1ΛΓγ1b−aγ−1Γγ∑j=1nζjIα+ψjfx,yzj−Iα1gx,yb+∑j=1nζjzj−aγ+ψj−1Γγ+ψj∑i=1mθiIα1+φigx,yξi−Iαfx,yb,K2,2x,yt=Iα1gx,yt.
Note thatK1x,yt=K1,1x,yt+K1,2x,yt and K2x,yt=K2,1x,yt+K2,2x,yt. Also, observe that the ball Bδ is a closed, bounded, and convex subset of the Banach space C. Let Bδ=x,y∈X×Y:x,y≤δ be a ball, where a constant δ≥maxM1P+M2Q,M3P+M4Q. Now, we will show that KBδ⊂Bδ for satisfying condition (i) of Lemma 4. Setting x=x1,x2andy=y1,y2∈Bδ and using condition (55), then we have, as in Theorem 2 that(58)K1,1x1,x2t+K1,2y1,y2t≤M1P+M2Q≤δ.
Similarly, we can find that(59)K2,1x1,x2t+K2,2y1,y2t≤M3P+M4Q≤δ.
That yieldsK1x+K2y∈Bδ. To show that the operator K1,2,K2,2 is a contraction mapping satisfying condition (iii) of Lemma 4, for x1,y1,x2,y2∈Bδ, we have(60)K1,2x1,y1t−K1,2x2,y2t≤Iαfx1,y1−fx2,y2t≤ℓ1x1−x2+ℓ2y1−y2Iα1b≤b−aαΓα+1ℓ1+ℓ2x1−x2+y1−y2(61)K2,2x1,y1t−K2,2x2,y2t≤Iα1gx1,y1−gx2,y2t≤b−aα1Γα1+1n1+n2x1−x2+y1−y2.
It follows from (60) and (61) that(62)K1,2,K2,2x1,y1−K1,2,K2,2x2,y2≤b−aαΓα+1ℓ1+ℓ2+b−aα1Γα1+1n1+n2x1−x2+y1−y2,which is a contraction by inequality in (56). Therefore, condition (iii) of Lemma 4 is satisfied. Next, we will show that the operator K1,1,K2,1 satisfies condition (ii) of Lemma 4. By applying the continuity of the functions f,g on a,b×ℝ×ℝ, we can conclude that the operator K1,1,K2,1 is continuous. For each x,y∈Bδ, one has(63)K1,1x,yt≤b−aγ−1ΛΓγb−aγ1−1Γγ1Q∑i=1mθiIα1+φi1ξi+PIα1b+∑i=1mθiξi−aγ1+φi−1Γγ1+φiP∑j=1nζjIα+ψj1zj+QIα11b=b−aγ−1ΛΓγb−aγ1−1Γγ1Q∑i=1mθiξi−aα1+φiΓα1+φi+1+Pb−aαΓα+1+∑i=1mθiξi−aγ1+φi−1Γγ1+φiP∑j=1nζjzj−aα+ψjΓα+ψj+1+Qb−aα1Γα1+1≔P∗,and similarly(64)K2,1x,yt≤Q∗.
Then, we obtain the following fact:(65)K1,1,K2,1x,y≤P∗+Q∗,which implies that the set K1,1,K2,1Bδ is uniformly bounded. In the next step, we will show that the set K1,1,K2,1Bδ is equicontinuous. For τ1,τ2∈a,b such that τ1<τ2 and for any x,y∈Bδ, we can prove that(66)K1,1x,yτ2−K1,1x,yτ1≤τ2−aγ−1−τ1−aγ−1ΛΓγb−aγ1−1Γτ1Q∑i=1mθiIα1+φi1ξi+PI1b+∑i=1mθiξi−aγ1+φi−1Γγ1+φiP∑j=1nζjIα+ψj1zj+QIα11b≤τ2−aγ−1−τ1−aγ−1ΛΓγb−aγ1−1Γγ1Q∑i=1mθiξi−aα1+φiΓα1+φi+1+Pb−aαΓα+1+∑i=1mθiξi−aγ1+φi−1Γγ1+φiP∑j=1nζjzj−aα+ψjΓα+ψj+1+Qb−aα1Γα1+1.
Therefore, we haveK1,1x,yτ2−K1,1x,yτ1⟶0 as τ1⟶τ2. Similarly, we can show that K2,1x,yτ2−K2,1x,yτ1⟶0 as τ1⟶τ2.
Thus,K1,1,K2,1x,yτ2−K1,1,K2,1x,yτ1 tends to zero as τ1⟶τ2. Therefore, the set K1,1,K2,1Bδ is equicontinuous. By applying the Arzelá–Ascoli theorem, the operator K1,1,K2,1 is compact on Bδ. By application of Lemma 4, we have that problem (4) has at least one solution on a,b. This completes the proof.Example 1.
Consider the coupled system of Hilfer fractional differential equations with nonlocal integral boundary conditions of the form(67)DH3/2,1/2xt=ft,xt,yt,t∈13,103,DH5/4,2/3yt=gt,xt,yt,t∈13,103,x13=0,x103=12I1/3y23+23I1/2y1+34I3/5y43+45I2/3y53,y13=0,y103=32I3/2x2+43I5/3x73+54I7/4y83+65I9/5x3.
Here,α=3/2, α1=5/4, β=1/2, β1=2/3, a=1/3, b=10/3, m=n=4, θi=i/i+1, ζi=2+i/1+i, φi=i/i+2, and ψi=2i+1/i+1, i=1,2,3,4, ξj=j/3, j=2,3,4,5, and zr=r/3, r=6,7,8,9. Then, we can compute constants as γ=7/4, γ1=41/21, Λ=6.371398411, M1=12.56809051, M2=3.253588460, M3=48.72536839, and M4=22.05071608.(i)
Let the nonlinear functionsf and g be defined by(68)ft,x,y=9e−t9t2+3239x2+2x1+x+33t+539siny+23,(69)gt,x,y=9103t+112tan−1x+3cos2t479+3ty1+y+12.It is easy to check thatf and g satisfy H1 in Theorem 1 as ft,x1,y1−ft,x2,y2≤1/180x1−x2+1/180y1−y2 and gt,x1,y1−gt,x2,y2≤1/160x1−x2+1/160y1−y2. Setting ℓ1=ℓ2=1/180 and n1=n2=1/160, we have(70)M1+M3ℓ1+ℓ2+M2+M4n1+n2=0.9973422390<1.Therefore, by applying Theorem1, we deduce that problem (67) with (68)–(69) has a unique solution x,y on 1/3,10/3.(ii)
Given the functionsf and g by(71)ft,x,y=3e−y53t+50x91+x8+93t+292y41+y3+34,(72)gt,x,y=3e−y2sinx19+x214+3t+3ycos2xy450+3t+47,we can check that(73)ft,x,y≤34+185x+1100y,gt,x,y≤47+195x+168y.which satisfy conditions in Theorem2 by u0=3/4, v0=4/7, u1=1/85, v1=1/95, u2=1/100, and v2=1/68. Hence, we find that M1+M3u1+M2+M4v1=0.9874606169<1 and M1+M3u2+M2+M4v2=0.9850567146<1. The conclusion of Theorem 2 implies that problem (67) with (71)-(72) has at least one solution x,y on 1/3,10/3.(iii)
Define the functionsf and g by(74)ft,x,y=3e−t23t+29x1+x+99t2+134siny+23,(75)gt,x,y=953t+52tan−1x+93t+142y1+y+12.Observe that conditionH1 in Theorem 1 is satisfied for nonlinear functions f and g with Lipschitz constants ℓ1=1/10, ℓ2=1/15, n1=1/20, and n2=1/25. Next, we can find that M1+M3ℓ1+ℓ2+M2+M4n1+n2=12.49296389>1. Then, Theorem 1 cannot be used to obtain the existence criteria for the investigated problem. However, we calculate that ft,x,y≤5/6, gt,x,y≤108+5π/200, and(76)b−aαΓα+1ℓ1+ℓ2+b−aα1Γα1+1n1+n2=0.9650966816<1.Hence, all assumptions of Theorem3 hold. Therefore, by Theorem 3, problem (67) with (74)-(75) has at least one solution x,y on 1/3,10/3.



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*Source: 2875152-2020-07-13.xml*