# Unified Approach to the Existence of Solutions for a Coupled System of Fractional Differential-Integral Equations with Infinite Points and Riemann–Stieltjes Integral Boundary Conditions

Authors: Ying Chen; Lishan Liu; Fang Wang
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this article, by using the Schauder fixed point theorem, we first study the existence of solutions for a new coupled system of Caputo fractional differential equations with multipoint boundary value conditions under the assumption that the nonlinear term satisfies two types of the Carathéodory conditions. Using this result, we provide the existence of solutions of the system with infinite points and Riemann–Stieltjes integral boundary conditions, respectively, instead of doing it directly. Finally, we give three examples to illustrate the feasibility of main results.


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## Body

## 1. Introduction

The fractional differential equation is an important branch of differential equations, which is widely used in mathematics, physics, engineering, and other fields, and it solves robotics, signal processing, and conversion problems [1–4]. In fact, the fractional differential equation is an important tool for mathematical modeling due to its memorability and genetic properties. Recently, nonlocal integer and fractional orders fractional differential equations have attracted attention of a large number of scholars, see [5–24]. Indeed, the nonlocal problems for (fractional) differential equations have received great attention, see[5–14] and the references therein. In this paper, we consider the coupled system of Caputo fractional differential equations(1)Dc0+αxt=g1t,xt,yt,∫0tf1s,xs,ysds,∫0th1s,xs,ysds,a⋅e⋅t∈0,1,Dc0+βyt=g2t,xt,yt,∫0tf2s,xs,ysds,∫0th2s,xs,ysds,a⋅e⋅t∈0,1,with the multipoint conditions(2)x0+∑i=1maixτi=x0,ai>0,τi∈0,1,y0+∑j=1mbjyτj=y0,bj>0,τj∈0,1,where Dc0+α and Dc0+β are the Caputo fractional derivatives, α,β∈0,1, x0, y0∈ℝ. Furthermore, we consider system (1) with the following Riemann–Stieltjes integral conditions:(3)x0+∫01xsdl1s=x0,y0+∫01ysdl2s=y0,where li:0,1⟶ℝi=1,2 is an increasing function, x0, y0∈ℝ. We also discuss system (1) with infinite-point boundary conditions(4)x0+∑i=1∞aixτi=x0,ai>0,τi∈0,1,0≤∑i=1∞ai<+∞,y0+∑j=1∞bjyτj=y0,bj>0,τj∈0,1,0≤∑j=1∞bj<+∞,where x0, y0∈ℝ.There are many methods for searching the existence of solutions of fractional boundary value problems, such as partial order methods [15], topology degree theory [10–14, 16–18], the critical point theorem [19], and so on. Recently, the authors in [10] studied the existence and continuous dependence of solutions by the Schauder fixed point theorem for the following coupled system:(5)dxdt=g1t,yt,∫0tf1s,ysds,a⋅et∈0,T,dxdt=g2t,yt,∫0tf2s,ysds,a⋅et∈0,T,x0+∑i=1maixτi=x0,ai>0,τi∈0,T,y0+∑j=1mbjyτj=y0,bj>0,τj∈0,T.Moreover, the authors studied the following Riemann–Stieltjes integral conditions(6)x0+∫0Txsdh1s=x0,y0+∫0Tysdh2s=y0,and also discussed the following infinite-point conditions:(7)x0+∑i=1∞aixτi=x0,ai>0,τi∈0,T,∑i=1∞ai<+∞,y0+∑j=1∞bjyτj=y0,bj>0,τj∈0,T,∑j=1∞bj<+∞.By using the Schauder fixed point theorem and Banach contraction mapping principle, the existence and uniqueness of solutions of the system are proved.By applying the Guo-Krasnosel’skii fixed point theorem of the cone expansion-compression type, Zhang in [11] considered the several local existence and multiplicity of positive solutions for the following problem:(8)D0+αut+qtft,ut=0,t∈0,1,u0=u′0=⋯=un−20=0,ui1=∑j=1∞αjuξj,where D0+α is Riemann–Liouville fractional derivative, α>2,n−1<α≤n,i∈1,n−2 is a fixed integer, αj≥0,0<ξ1<ξ2<⋯<ξj−1<ξj<⋯<1j=1,2,⋯,△−∑1∞αjξjα−1>0,△=α−1α−2⋯α−i, q:0,1⟶ℝ+ and f may be singular with respect to both the time and space variables.In [12], the existence of solutions for the following problem was obtained by using the Schauder’s fixed point theorem(9)dxdt=ft,Dαxt,a⋅e⋅t∈0,1,∫01xϕsdgs=x0,where Dα is the Caputo fractional derivative and α∈0,1, g:0,1⟶0,1 is an increasing function.In this paper, we discuss the coupled system (1), in which the nonlinear terms gi include two space variables xandy and the integral terms. By using the Schauder fixed point theorem, the existence of solutions for system (1) with the multipoint boundary conditions (2) is proved under the assumption that the nonlinear term satisfies two types of the Carathéodory conditions, which consists sublinear and linear conditions. Furthermore, based on the above results, we give the existence of solutions of system (1) with infinite point and Riemann–Stieltjes integral boundary conditions, respectively, which avoids proving again. Compared with the articles mentioned above, system (1) is more general. For example, the nonlinear terms in [10] contain a single space variable, while we consider two space variables.The outline of this paper is organized as follows. In Section 2, we state some preliminary settings. In Section 3, we show the proof of existence of solutions. In Section 4, three examples are given in order to illustrate our results. In Section 5, we give the conclusions of this paper.


## 2. Preliminaries and Lemmas

In this section, we will provide some definitions and lemmas to be used in the following proofs.Definition 1.
(see [1]) Let f∈ACn0,1. Then, the Caputo fractional derivative of order α>0 of f is given by the following:(10)Dc0+αft=1Γn−α∫0tfnst−sα−n+1ds,n−1<α<n,fns,α=n,which exists almost everywhere on 0,1.Definition 2.
(see [1]) For a function f:0,+∞⟶ℝ, the Riemann–Liouville fractional integral of order α>0 of f is given by the following:(11)I0+αft=1Γα∫0tt−sα−1fsds,provided that the right-hand side is pointwise defined on 0,+∞.Lemma 1.
(see [1]) For α>0, the fractional differential equation Dc0+αut=0 has a general solution(12)ut=C1+C2t+C3t2+⋯+Cntn−1,Ci∈ℝ,i=1,2,…,n,n−1<α≤n.Lemma 2.
(see [1]) Assume that u∈Cn0,1, then(13)I0+αDc0+αut=ut−C1−C2t−⋯−Cntn−1,Ci∈ℝ,i=1,2,…,n,n−1<α≤n.Lemma 3.
(see [4]) Let f∈Lpa,b, g∈Lqa,b, where p>1, 1/p+1/q=1. Then, fg∈L1a,b, and(14)∫abftgtdt≤fpgq,where fp=∫abftpdt1/p.Lemma 4.
Letu∈C0,1∩Lp10,1,v∈C0,1∩Lp20,1, 1/p1∈0,α,1/p2∈0,β. Then, the system(15)Dc0+αxt=ut,a⋅e⋅t∈0,1,Dc0+βyt=vt,a⋅e⋅t∈0,1,with the boundary conditions(16)x0+∑i=1maixτi=x0,ai>0,τi∈0,1,y0+∑j=1mbjyτj=y0,bj>0,τj∈0,1,has a unique solution x,y, which can be expressed by(17)xt=11+∑i=1maix0−∑i=1mai∫0τiτi−sα−1Γαusds+∫0tt−sα−1Γαusds,yt=11+∑j=1mbiy0−∑j=1mbi∫0τjτi−sβ−1Γβvsds+∫0tt−sβ−1Γβvsds.Proof.
According to the known conditions and Lemma3, we have(18)∫0tt−sα−1Γαusds≤1Γα∫0tt−sp1α−1/p1−1ds1−1/p1∫0tusp1ds1/p1≤1Γαp1−1p1α−11−1/p1uLp1,where u∈C0,1∩Lp10,1 and 1/p1∈0,α, it can be seen p1−1/p1α−1>0.Then,(19)∫0tt−sα−1Γαusds<∞,in the same way(20)∫0tt−sβ−1Γβvsds<∞.
Therefore, it is concluded that the integral exists.
Next, using Lemma2 to integrate both sides at the same time, we have(21)xt=∫0tt−sα−1Γαusds+C1,C1∈ℝ,yt=∫0tt−sβ−1Γβvsds+C2,C2∈ℝ.By the boundary conditions (2), the expression of C1,C2 can be obtained as follows:(22)C1=x0−∑i=1mai∫0τiτi−sα−1/Γαusds1+∑i=1mai,C2=y0−∑j=1mbj∫0τjτj−sβ−1/Γβvsds1+∑j=1mbj.
Consequently, substitutingC1,C2 into formula (21), we obtain the expression of the solution as follows (17).
In order to prove later, we make some assumptions here as follows:
H1gi:0,1×ℝ4⟶ℝi=1,2 satisfies the Carathéodory condition, i.e.,(i)
gi⋅,x1,x2,x3,x4 is measurable for any x1,x2,x3,x4∈ℝ.(ii)
git,⋅,⋅,⋅,⋅ is continuous for almost all t∈0,1.(iii)
There exist functionsa11,b11,b12,b13,b14∈Lp10,1,ℝ+,a21,b21,b22,b23,b24∈Lp20,1,ℝ+ with 1/p1∈0,α,1/p2∈0,β, αi,βi∈0,1i=1,2,3,4, such that(23)g1t,x1,x2,x3,x4≤a11t+b11tx1α1+b12tx2α2+b13tx3α3+b14tx4α4,g2t,x1,x2,x3,x4≤a21t+b21tx1β1+b22tx2β2+b23tx3β3+b24tx4β4.
H1′gi:0,1×ℝ4⟶ℝi=1,2 satisfies the Carathéodory condition, i.e.,(i)
gi⋅,x1,x2,x3,x4 is measurable for any x1,x2,x3,x4∈ℝ.(ii)
git,⋅,⋅,⋅,⋅ is continuous for almost all t∈0,1.(iii)
There exist functionsa11,b11,b12,b13,b14∈Lp10,1,ℝ+,a21,b21,b22,b23,b24∈Lp20,1,ℝ+ with 1/p1∈0,α,1/p2∈0,β, such that(24)g1t,x1,x2,x3,x4≤a11t+b11tx1+b12tx2+b13tx3+b14tx4,g2t,x1,x2,x3,x4≤a21t+b21tx1+b22tx2+b23tx3+b24tx4.
H2fi:0,1×ℝ2⟶ℝi=1,2 satisfies the Carathéodory condition, i.e.,(i)
fi⋅,x1,x2 is measurable for any x1,x2∈ℝ.(ii)
fit,⋅,⋅ is continuous for almost all t∈0,1.(iii)
There exist functionsa12,c11,c12∈Lp10,1,ℝ+,a22,c21,c22∈Lp20,1,ℝ+ with 1/p1∈0,α,1/p2∈0,β such that(25)f1t,x1,x2≤a12t+c11tx1+c12tx2f2t,x1,x2≤a22t+c21tx1+c22tx2.
H3hi:0,1×ℝ2⟶ℝi=1,2 satisfies the Carathéodory condition, i.e.,(i)
hi⋅,x1,x2 is measurable for any x1,x2∈ℝ.(iii)
hit,⋅,⋅ is continuous for almost all t∈0,1.(iii)
There exist functionsa13,d11,d12∈Lp10,1,ℝ+,a23,d21,d22∈Lp20,1,ℝ+ with 1/p1∈0,α,1/p2∈0,β such that(26)h1t,x1,x2≤a13t+d11tx1+d12tx2h2t,x1,x2≤a23t+d21tx1+d22tx2.
H4(27)0<1+A−1∑i=1maip1−1p1α−11−1/p1b11Lp1+b12Lp1+b13Lp1a12Lp1+c11Lp1+c12Lp1+b14Lp1a13Lp1+d11Lp1+d12Lp1<Γα,0<1+B−1∑j=1mbjp2−1p2β−11−1/p2b21Lp2+b22Lp2+b23Lp2a22Lp2+c21Lp2+c22Lp2+b24Lp2a23Lp2+d21Lp2+d22Lp2<Γβ,where A=1+∑i=1mai,B=1+∑j=1mbj.
H5(28)0<p1−1p1α−11−1/p11+l11−l101+l11−l10b11Lp1+b12Lp1+b13Lp1a12Lp1+c11Lp1+c12Lp1+b14Lp1a13Lp1+d11Lp1+d12Lp1<Γα,0<p2−1p2β−11−1/p21+l21−l201+l21−l20b21Lp2+b22Lp2+b23Lp2a22Lp2+c21Lp2+c22Lp2+b24Lp2a23Lp2+d21Lp2+d22Lp2<Γβ.
H6(29)0<1+∑i=1∞ai1+∑i=1∞aip1−1p1α−11−1/p1b11Lp1+b12Lp1+b13Lp1a12Lp1+c11Lp1+c12Lp1+b14Lp1a13Lp1+d11Lp1+d12Lp1<Γα,0<1+∑j=1∞bj1+∑j=1∞bjp2−1p2β−11−1/p2b21Lp2+b22Lp2+b23Lp2a22Lp2+c21Lp2+c22Lp2+b24Lp2a23Lp2+d21Lp2+d22Lp2<Γβ.


## 3. Main Results

LetX=C0,1 be the Banach space equipped with the norm xX=maxt∈0,1xt. Let Y=X×X be the Banach space of ordered pairs x,y equipped with the norm x,yY=maxxX,yX. In the following part, from Lemma 3, we define the operator T:Y⟶Y as follows:(30)Tx,y=T1x,y,T2x,y,x,y∈Y,where(31)T1x,yt=A−1x0−∑i=1mai∫0τiτi−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds+∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds,T2x,yt=B−1y0−∑j=1mbj∫0τjτi−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds+∫0tt−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds.In the following part, for convenience, we set(32)D1=A−1x0+A−1∑i=1mai+11Γαp1−1p1α−11−1/p1a11Lp1,D2=B−1y0+B−1∑j=1mbj+11Γβp1−1p2β−11−1/p2a21Lp2,G1=b11Lp1+b12Lp1+b13Lp1a12Lp1+c11Lp1+c12Lp1α3+b14Lp1a13Lp1+d11Lp1+d12Lp1α4,G2=b21Lp2+b22Lp2+b23Lp2a22Lp2+c21Lp2+c22Lp2β3+b24Lp2a23Lp2+d21Lp2+d22Lp2β4,W1=b11Lp1+b12Lp1+b13Lp1a12Lp1+c11Lp1+c12Lp1+b14Lp1a13Lp1+d11Lp1+d12Lp1,W2=b21Lp2+b22Lp2+b23Lp2a22Lp2+c21Lp2+c22Lp2+b24Lp2a23Lp2+d21Lp2+d22Lp2,E1=A−1∑i=1mai+11Γαp1−1p1α−11−1/p1G1,E2=A−1∑j=1mbj+11Γβp1−1p1β−11−1/p2G2,Z1=A−1∑i=1mai+11Γαp1−1p1α−11−1/p1W1,Z2=A−1∑j=1mbj+11Γβp1−1p2β−11−1/p2W2.Theorem 1.
Assume that the assumptionsH1-H3 hold. Then, systems (1) and (2) have at least one solution.Proof.
Let us set upα˜=maxα1,α2,α3,α4, β˜=maxβ1,β2,β3,β4, then α˜,β˜∈0,1. Accordingly, we infer that, there exists R1>0 large enough such that(33)D1+E1R1α˜≤R1,D2+E2R2β˜≤R1,without loss of generality, we may assume that R1>1.
First, we prove thatT:QR1⟶QR1, where QR1=x,y∈Y:x,yY≤R1. It should be noted that for any x,y∈QR1,(34)1Γα∫0tt−sα−1g1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds≤1Γα∫0tt−sα−1a11s+b11sxsα1+b12sysα2+b13s∫0sa12θdθ+∫0sc11θxθdθ+∫0sc12θyθdθα3+b14s∫0sa13θdθ+∫0sd11θxθdθ+∫0sd12θyθdθα4ds,≤1Γα∫0tt−sα−1a11sds+R1α1∫0tt−sα−1b11sds+R1α2∫0tt−sα−1b12sds+R1α3∫0tt−sα−1b13s∫0sa12θdθ+∑i=12∫0sc1iθdθα3ds+R1α4∫0tt−sα−1b14s∫0sa13θdθ+∑i=12∫0sd1iθdθα4ds,≤1Γα∫0tt−sp1α−1/p1−1ds1−1/p1∫0ta11p1sds1/p1+R1α1∫0tt−sp1α−1/p1−1ds1−1/p1∫0tb11p1sds1/p1+R1α2∫0tt−sp1α−1/p1−1ds1−1/p1∫0tb12p1sds1/p1+R1α3Γα∫0tt−sα−1b13s∫0sa12p1θdθ1/p1∫0s1p1/p1−1dθ1−1/p1+∫0sc11p1θdθ1/p1∫0s1p1/p1−1dθ1−1/p1+∫0sc12p1θdθ1/p1∫0s1p1/p1−1dθ1−1/p1α3ds+R1α4Γα∫0tt−sα−1b14s∫0sa13p1θdθ1/p1∫0s1p1/p1−1dθ1−1/p1+∫0sc11p1θdθ1/p1∫0s1p1/p1−1dθ1−1/p1+∫0sd12p1θdθ1/p1∫0s1p1/p1−1dθ1−1/p1α4ds,≤1Γαp1−1p1α−11−1/p1a11Lp1+R1α1p1−1p1α−11−1p1b11Lp1+R1α2p1−1p1α−11−1p1b12Lp1+R1α3Γαa12Lp1+c11Lp1+c12Lp1α3∫0tt−sα−1b13sds+R1α4Γαa13Lp1+d11Lp1+d12Lp1α4∫0tt−sα−1b14sds,≤1Γαp1−1p1α−11−1/p1a11Lp1+R1α1Γαp1−1p1α−11−1/p1b11Lp1+R1α2Γαp1−1p1α−11−1/p1b12Lp1+R1α3Γαp1−1p1α−11−1/p1b13Lp1a12Lp1+c11Lp1+c12Lp1α3+R1α4Γαp1−1p1α−11−1p1b14Lp1a13Lp1+d11Lp1+d12Lp1α4,≤1Γαp1−1p1α−11−1/p1a11Lp1+1Γαp1−1p1α−11−1/p1G1R1α˜,t∈0,1.
Thus, we have(35)T1x,yt≤A−1x0+∑i=1mai∫0τiτi−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds|∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds|≤A−1x0+A−1∑i=1mai+11Γαp1−1p1α−11−1/p1a11Lp1+A−1∑i=1mai+11Γαp1−1p1α−11−1/p1G1R1α˜=D1+E1R1α˜≤R1,t∈0,1,which means T1x,yX≤R1. In the same way, we get that(36)T2x,yX≤D2+E2R1α˜≤R1.
Based on (35) and (36), we conclude that(37)Tx,yY=maxT1x,yX,T2x,yX≤R1.
This proves thatT:QR1⟶QR1.
Second, we prove that operatorT:QR1⟶QR1 is continuous. Let xn,yn,x,y∈QR1, and xn,yn⟶x,yn⟶∞ in Y. Utilizing the assumptions H1-H3, we have(38)gis,xns,yns,∫0sfiθ,xnθ,ynθdθ,∫0shiθ,xnθ,ynθdθ⟶gis,xs,ys,∫0sfiθ,xθ,yθdθ,∫0shiθ,xθ,yθdθ,a⋅e⋅s∈0,1,i=1,2.
On the other hand, we have(39)|gis,xns,yns,∫0sfiθ,xnθ,ynθdθ,∫0shiθ,xnθ,ynθdθ−gis,xs,ys,∫0sfiθ,xθ,yθdθ,∫0shiθ,xθ,yθdθ≤2ai1s+2R1α˜bi1s+bi2s+bi3s+bi4s,s∈0,1,i=1,2.Based on H1, we know 2ai1+2R1α˜bi1+bi2+bi3+bi4∈L10,1. Thus, by the Lebesgue dominated convergence theorem, we have(40)limn⟶∞Tixn,yn=Tix,yinX,i=1,2.
Accordingly, we conclude thatTxn,yn⟶Tx,y in Y as n⟶∞, that is, the operator T:QR1⟶QR1 is continuous.
In the following part, we prove thatTQR1 is equicontinuous. Based on H1, we know ai1, bi1, bi2, bi3, bi4∈L10,1. Thus, for any x,y∈QR1, t1,t2∈0,1t1<t2, we have(41)Tix,yt2−Tix,yt1≤∫0t2t2−sα−1Γα×gis,xs,ys,∫0sfiθ,xθ,yθdθ,∫0shiθ,xθ,yθdθds−∫0t1t1−sα−1Γα×gis,xs,ys,∫0sfiθ,xθ,yθdθ,∫0shiθ,xθ,yθdθds|,≤∫0t1t2−sα−1−t1−sα−1Γα×gis,xs,ys,∫0sfiθ,xθ,yθdθ,∫0shiθ,xθ,yθdθds+∫t1t2t2−sα−1Γα×gis,xs,ys,∫0sfiθ,xθ,yθdθ,∫0shiθ,xθ,yθdθds,≤1Γαai1L1+R1α˜bi1L1+bi2L1+bi3L1ai2Lpi+ci1Lpi+ci2Lpiα˜+bi4L1ai3Lpi+di1Lpi+di2Lpiα˜×∫0t1t2−sα−1−t1−sα−1ds+∫t1t2t2−sα−1ds,≤1αΓαai1L1+R1α˜bi1L1+bi2L1+bi3L1ai2Lpi+ci1Lpi+ci2Lpiα˜+bi4L1ai3Lpi+di1Lpi+di2Lpiα˜t2α−t1α,i=1,2.Since the function tα is continuous on 0,1, we get that, for any ε>0, there exists δ>0 such that, for any t1,t2∈0,1, t1−t2<δ,(42)t2α−t1α<ε.
Consequently, for anyt1,t2∈0,1, t1−t2<δ, we have(43)Tix,yt2−Tix,yt1≤1αΓαai1L1+R1α˜bi1L1+bi2L1+bi3L1ai2Lpi+ci1Lpi+ci2Lpiα˜+bi4L1ai3Lpi+di1Lpi+di2Lpiα˜ε,which means that TiQR1i=1,2 is equicontinuous on 0,1, and also, TQR1 is equicontinuous on 0,1. Based on (35) and (36), we know that the set TQR1 is uniformly bounded. According to the Arzal a′-Ascoli theorem, we prove that operator T is a completely continuous operator. Thus, by the Schauder fixed point theorem, we deduce that systems (1) and (2) have at least one solution x,y∈QR1, which can be expressed as follows:(44)xt=11+∑i=1maix0−∑i=1mai∫0τiτi−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds+∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds,yt=11+∑j=1mbjy0−∑j=1mbj∫0τjτj−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds+∫0tt−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds.Theorem 2.
Assume that the assumptionsH1-H3 hold. Then, systems (1) and (3) have at least one solution.Proof.
First, we make an interval partition:0=t0<t1<⋯<tm=1 such that τi∈ti−1,ti,τj∈tj−1,tji,j=1,2,…,m. Let ai=l1ti−l1ti−1,bj=l2tj−l2tj−1. Then, we have(45)x0+∑i=1maixτi=x0+∑i=1ml1ti−l1ti−1xτi=x0,y0+∑j=1mbjyτj=y0+∑j=1ml2tj−l2tj−1yτj=y0.By Theorem 1, we know, systems (1) and (2) have a solution x,y∈QR1, which can be expressed in (44) . Let m⟶∞ in (45), we then have(46)x0+limm⟶∞∑i=1ml1ti−l1ti−1xτi=x0+∫01xsdl1s=x0,y0+limm⟶∞∑j=1ml2tj−l2tj−1yτj=y0+∫01ysdl2s=y0,that is, the boundary conditions (3). Therefore, we take the limit to the solution (44), and we then get the solution of systems (1) and (3), which can be expressed as follows:(47)xt=limm⟶∞11+∑i=1maix0−∑i=1mai∫0τiτi−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds+∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds,=11+l11−l10x0−limm⟶∞∑i=1ml1ti−l1ti−1∫0τiτi−sα−1Γα×g1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds+∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds,=11+l11−l10x0−∫01∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθdsdl1t+∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds,yt=limm⟶∞11+∑j=1mbjy0−∑j=1mbj∫0τjτj−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθdsdl2t+∫0tt−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds,=11+l21−l20y0−∫01∫0tt−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθdsdl2t+∫0tt−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds.Theorem 3.
Assume that the assumptionsH1-H3 hold. Then, systems (1) and (4) have at least one solution.Proof.
Based on Theorem1, we know, for any m∈ℕ+, the solution of systems (1) and (2) can be expressed as follows:(48)xmt=11+∑i=1maix0−∑i=1mai∫0τiτi−sα−1Γαg1s,xms,yms,∫0sf1θ,xmθ,ymθdθ,∫0sh1θ,xmθ,ymθdθds+∫0tt−sα−1Γαg1s,xms,yms,∫0sf1θ,xmθ,ymθdθ,∫0sh1θ,xmθ,ymθdθds,ymt=11+∑j=1mbiy0−∑j=1mbi∫0τjτi−sβ−1Γβg2s,xms,yms,∫0sf2θ,xmθ,ymθdθ,∫0sh2θ,xmθ,ymθdθds+∫0tt−sβ−1Γβg2s,xms,yms,∫0sf2θ,xmθ,ymθdθ,∫0sh2θ,xmθ,ymθdθds.Let m⟶∞ in (48), we then have(49)limm⟶∞xmt=limm⟶∞11+∑i=1maix0−∑i=1mai∫0τiτi−sα−1Γαg1s,xms,yms,∫0sf1θ,xmθ,ymθdθ,∫0sh1θ,xmθ,ymθdθds+limm⟶∞∫0tt−sα−1Γαg1s,xms,yms,∫0sf1θ,xmθ,ymθdθ,∫0sh1θ,xmθ,ymθdθds,limm⟶∞ymt=limm⟶∞11+∑j=1mbiy0−∑i=1mbj∫0τjτi−sβ−1Γβg2s,xms,yms,∫0sf2θ,xmθ,ymθdθ,∫0sh2θ,xmθ,ymθdθds+limm⟶∞∫0tt−sβ−1Γβg2s,xms,yms,∫0sf2θ,xmθ,ymθdθ,∫0sh2θ,xmθ,ymθdθds.
Based on the convergence of∑i=1∞ai and ∑j=1∞bj, it is easy to check that(50)∑i=1∞ai∫0τiτi−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds∑j=1∞bi∫0τjτi−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds,are convergent. Thus, by using the Lebesgue-dominated convergence theorem, we know that the existence of the limits for the right-hand side in (44). On the other hand, we notice that the boundary conditions (2) can be transformed to the boundary conditions (4) as m⟶∞, we obtain that the limits of (44) is the solution of systems (1) and (4), which can be expressed as follows:(51)xt=11+∑i=1∞aix0−∑i=1∞ai∫0τiτi−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds+∫0tt−sα−1Γαg1s,xs,ys,∫0sf1θ,xθ,yθdθ,∫0sh1θ,xθ,yθdθds,,yt=11+∑j=1∞biy0−∑j=1∞bi∫0τjτj−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds+∫0tt−sβ−1Γβg2s,xs,ys,∫0sf2θ,xθ,yθdθ,∫0sh2θ,xθ,yθdθds,.Theorem 4.
Assume that the assumptionsH1′-H3 and H4 hold. Then, systems (1) and (2) have at least one solution.Proof.
It follows fromH4 that Z1,Z2<1. Thus, there exists R1′>0 large enough such that(52)D1+Z1R1′≤R1′,D2+Z2R1′≤R1′.In the following part, we prove that T:QR1′⟶QR1′, where QR1′=x,y∈Y:x,yY≤R1′. Similarly as (35), we obtain that, for any x,y∈QR1′,(53)T1x,yt≤1+A−1∑i=1maip1−1/p1α−11−1/p1Γαb11Lp1+b12Lp1+b13Lp1a12Lp1+c11Lp1+c12Lp1+b14Lp1a13Lp1+d11Lp1+d12Lp1⋅R1′+A−1x0+1+A−1∑i=1maia11Lp1Γαp1−1p1α−11−1/p1=Z1R1′+D1≤R1′,t∈0,1,T2x,yt≤1+B−1∑j=1mbip2−1/p2β−11−1/p2Γβb21Lp2+b22Lp2+b23Lp2a22Lp2+c21Lp2+c22Lp2+b24Lp2a23Lp2+d21Lp2+d22Lp1⋅R1′+B−1y0+1+B−1∑j=1mbia21Lp2Γβp1−1p1β−11−1/p2=Z2R1′+D2≤R1′,t∈0,1.
Thus, we obtain thatTx,yY=maxT1x,yX,T2x,yX≤R1′. By the same way as Theorem 1, we get that T is a completely continuous operator. Then, by the Schauder fixed point theorem, we deduce that there exists at least one solution of systems (1) and (2).Theorem 5.
Assume that the assumptionsH1′-H3, H5 hold. Then, systems (1) and (3) have at least one solution.Proof.
First, we make an interval partition:0=t0<t1<⋯<tm=1 such that τi∈ti−1,ti,τj∈tj−1,tji,j=1,2,…,m. Let ai=l1ti−l1ti−1,bj=l2tj−l2tj−1. In this case, we obtain as follows:(54)∑i=1mai=l11−l10,∑j=1mbj=l21−l20.So, the assumption H5 implies that H4 is true. Thus, based on Theorem 4, we know that systems (1) and (2) have a solution in QR1′. Accordingly, by Theorem 2, we take the limit to the solution of systems (1) and (2), the solution of systems (1) and (3) is then obtained.Theorem 6.
Assume that the assumptionsH1′-H3, H6 hold. Then, systems (1) and (4) have at least one solution.Proof.
The proof is similar to that of Theorem3. So, we omit details.


## 4. Examples

Example 1.
In this section, we consider the following system:(55)Dc2/3xt=t−1/5+tx2/3t+y1/3tt1/2x2t+1+2t2∫0tsx1/3s+s−1/2ysds2/5∫0t2s−1/4x1/3sy1/5sds2/5,Dc3/4yt=t−3/4+4t−1/2x1/5ty1/7t∫0ts2+2s−1/2xsys1/2ds1/7+t∫0tsxs+s−1/2ys+1ds1/3y2t+1,with the following boundary conditions:(56)x0+12x14+16x19+112x116=1,y0+y14+12y19+14y116=4.Let α=2/3, β=3/4,(57)g1t,x1,x2,x3,x4=t−1/5+tx12/3+x21/3t1/2x12+1+2t2x3x42/5,g2t,x1,x2,x3,x4=t−3/4+4t−1/2x11/5x21/7x31/7+tx41/3x22+1,f1t,x1,x2=tx11/3+t−1/2x2,f2t,x1,x2=t2+2t−1/2x1x21/2,h1t,x1,x2=2t−1/4x11/3x21/5,h2t,x1,x2=tx1+t−1/2x2+1.
Then, systems (55) and (56) can be transformed to systems (1) and (2). In addition, we obtain that(58)g1t,x1,x2,x3,x4≤t−1/5+tx12/3+t−1/2x21/3+2t2x34/5+2t2x44/5,g2t,x1,x2,x3,x4≤t−3/4+2t−1/2x12/5+t−1/2x24/7+t4/5x34/7+tx41/3,f1t,x1,x2≤t+tx1+t−1/2x2,f2t,x1,x2≤t2+t−1/2x1+t−1/2x2,h1t,x1,x2≤2t−1/4+t−1/4x1+t−1/4x2,h2t,x1,x2≤t−1/2+tx1+t−1/2x2.Let 1/p1=1/2, 1/p2=1/2, a11t=t−1/5, b11t=t, b12t=t−1/2, b13t=t2, b14t=t2, a21t=t−3/4, b21t=2t−1/2, b22t=t−1/2, b23t=t−1/2, b24t=t, a12t=t, c11t=t, c12t=t−1/2, a22t=t2, c21t=t−1/2, c22t=t−1/2, a13t=2t−1/4, d11t=t−1/4d12t=t−1/4, a23t=t−1/2, d21t=t, d22t=t−1/2. It then holds that 1/p1∈0,αα=2/3, 1/p2∈0,ββ=3/4, a11, b11, b12, b13, b14, a12, c11, c12, a13, d11d12∈Lp10,1,ℝ+, a21,b21,b22,b23,b24,a22,c21,c22,a23,d21,d22∈Lp20,1,ℝ+, and(59)g1t,x1,x2,x3,x4≤a11t+b11tx1α1+b12tx2α2+b13tx3α3+b14tx4α4,g2t,x1,x2,x3,x4≤a21t+b21tx1β1+b22tx2β2+b23tx3β3+b24tx4β4,f1t,x1,x2≤a12t+c11tx1+c12tx2,f2t,x1,x2≤a22t+c21tx1+c22tx2,h1t,x1,x2≤a13t+d11tx1+d12tx2,h2t,x1,x2≤a23t+d21tx1+d22tx2.
Thus, we get that the conditionsH1-H3 are satisfied. Therefore, by Theorem 1, we infer that systems (55)-(56) have at least one solution.Example 2.
We discuss the following system:(60)Dc2/3xt=t2+2txt1/3+t1/2yt1/5+t−1/3∫0ts−1/2xsys1/3ds1/3+t−1/2∫0txsys4/54+s+xs+ysds1/5,Dc2/3yt=t−3/4+x4ty2t1+xt17/4yt7/4+t1/3∫0ts+xs1/3+ys2+s+y2sds1/4+t−1/3∫0tln1+xs1/5ys1/3ds1/2,with the following boundary condition:(61)x0+∫01xsdl1s=1,y0+∫01ysdl2s=4,where(62)l1t=1,t∈0,12,2,t∈12,1,l2t=12,t∈0,12,3,t∈12,1.Let α=2/3, β=2/3,(63)g1t,x1,x2,x3,x4=t2+2tx11/3+t1/2x21/5+t−1/3x31/3+t−1/2x41/5,g2t,x1,x2,x3,x4=t−3/4+x14x221+x117/4x27/4+t1/3x31/4+t−1/3x41/2,f1t,x1,x2=t−1/2x1x21/3,f2t,x1,x2=t+x11/3+x22+t+x22,h1t,x1,x2=x1x24/54+t+x1+x2,h2t,x1,x2=ln1+x11/5x2|1/3.
Then, systems (60) and (61) can be transformed to systems (1) and (3). In addition, we obtain that(64)g1t,x1,x2,x3,x4≤t2+2tx11/3+t1/2x21/5+t−1/3x31/3+t−1/2x41/5,g2t,x1,x2,x3,x4≤t−3/4+tx11/2+tx21/2+t1/3x31/4+t−1/3x41/2,f1t,x1,x2≤t+t−1/2x12/3+t−1/2x22/3,f2t,x1,x2≤t+t−1/3x11/3+tx21/3,h1t,x1,x2≤2t+1+tx11/2+1+tx21/2,h2t,x1,x2≤t+t−1/2x12/5+t−1/2x22/3.Let 1/p1=1/4, 1/p2=1/2, a11t=t2, b11t=2t, b12t=t1/2, b13t=t−1/3, b14t=t−1/2, a21t=t−3/4, b21t=t, b22t=t, b23t=t1/3, b24t=t−1/3, a12t=t, c11t=t−1/2, c12t=t−1/2, a22t=t, c21t=t−1/3, c22t=t, a13t=2t, d11t=1+td12t=1+t, a23t=t, d21t=t−1/2, d22t=t−1/2. It then holds that 1/p1∈0,αα=2/3,1/p2∈0,ββ=2/3, a11, b11, b12, b13, b14, a12, c11, c12, a13, d11d12∈Lp10,1,ℝ+, a21,b21,b22,b23,b24,a22,c21,c22,a23,d21,d22∈Lp20,1,ℝ+, and(65)g1t,x1,x2,x3,x4≤a11t+b11tx1α1+b12tx2α2+b13tx3α3+b14tx4α4,g2t,x1,x2,x3,x4≤a21t+b21tx1β1+b22tx2β2+b23tx3β3+b24tx4β4,f1t,x1,x2≤a12t+c11tx1+c12tx2,f2t,x1,x2≤a22t+c21tx1+c22tx2,h1t,x1,x2≤a13t+d11tx1+d12tx2,h2t,x1,x2≤a23t+d21tx1+d22tx2.Thus, we get that the conditions H1-H3 are satisfied. Therefore, by Theorem 2, we infer that systems (60)-(61) have at least one solution.Example 3.
We discuss the following system:(66)Dc3/4xt=t−2/3+2t−1/2xt1/4yt1/4+t∫0t2s−1/4xs1/3+sys22+ys4/3ds1/4+t1/2∫0ts−1/2xs1/5+sy1/3sds1/3,Dc2/5yt=t−3/4+t−1/4xt1/21+yt2+tyt1/4+t2∫0ts−1/3xs1/3+sys1/5ds1/5+t−1/2∫0tsxs1/4ys1/4ds1/7,with the following boundary condition:(67)x0+∑i=1∞1i+1x1i3=1,y0+∑i=1∞13i−1x1i3=5.Let α=3/4, β=2/5,(68)g1t,x1,x2,x3,x4=t−2/3+2t−1/2x11/4x21/4+tx31/4+t1/2x41/3,g2t,x1,x2,x3,x4=t−3/4+t−1/3x11/21+x22+tx21/4+t2x31/5+t−1/2x41/7,f1t,x1,x2=2t−1/4x11/3+tx222+x24/3,f2t,x1,x2=t−1/3x11/3+tx21/5,h1t,x1,x2=t−1/2x11/5+tx21/3,h2t,x1,x2=tx11/4x21/4.
Then, systems (66) and (67) can be transformed to systems (1) and (4). In addition, we obtain that(69)g1t,x1,x2,x3,x4≤t−2/3+2t−1/2x11/2+2t−1/2x21/2+tx31/4+t1/2x41/3,g2t,x1,x2,x3,x4≤t−3/4+t−1/3x11/2+tx21/4+t2x31/5+t−1/2x41/7,f1t,x1,x2≤t+2t−1/4x11/3+t|x21/3,f2t,x1,x2≤t+t−1/3x11/3+tx21/5,h1t,x1,x2≤t−1/2+t−1/2x11/5+tx21/3,h2t,x1,x2≤t+tx11/2+tx21/2.Let 1/p1=1/3, 1/p2=1/4, a11t=t−2/3, b11t=2t−1/2, b12t=2t−1/2, b13t=t, b14t=t1/2, a21t=t−3/4, b21t=t−1/3, b22t=t, b23t=t2, b24t=t−1/2, a12t=t, c11t=2t−1/4, c12t=t, a22t=t, c21t=t−1/3, c22t=t, a13t=t−1/2, d11t=t−1/2d12t=t, a23t=t, d21t=t, d22t=t. It then holds that 1/p1∈0,αα=3/4,1/p2∈0,ββ=2/5, a11, b11, b12, b13, b14, a12, c11, c12, a13, d11, d12∈Lp10,1,ℝ+, a21,b21,b22,b23,b24,a22,c21,c22,a23,d21,d22∈Lp20,1,ℝ+, and(70)g1t,x1,x2,x3,x4≤a11t+b11tx1α1+b12tx2α2+b13tx3α3+b14tx4α4,g2t,x1,x2,x3,x4≤a21t+b21tx1β1+b22tx2β2+b23tx3β3+b24tx4β4,f1t,x1,x2≤a12t+c11tx1+c12tx2,f2t,x1,x2≤a22t+c21tx1+c22tx2,h1t,x1,x2≤a13t+d11tx1+d12tx2,h2t,x1,x2≤a23t+d21tx1+d22tx2.
Thus, we infer that the conditionsH1-H3 are satisfied. Therefore, by Theorem 3, we infer that systems (66)-(67) have at least one solution.


## 5. Conclusions

In this paper, by using the Schauder fixed point theorem, the existence of solutions for the coupled system (1) with multipoint boundary conditions (2) is proved, in which the nonlinear term satisfies two types of the Carathéodory condition. Under these two conditions, the power exponent of the control function of the nonlinear term includes sublinear and linear cases, which makes the problem we study more general. Furthermore, we consider the Riemann–Stieltjes integral and infinite-point boundary conditions around it. Based on the results of systems (1)-(2), the existence of solutions for the last two systems is obtained directly. Finally, three examples are given to illustrate our theoretical results.



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