# Existence of Solution for a Fractional Langevin System with Nonseparated Integral Boundary Conditions

Authors: Lahcen Ibnelazyz; Karim Guida; Khalid Hilal; Said Melliani
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we investigate the existence and uniqueness of a coupled system of nonlinear fractional Langevin equations with nonseparated type integral boundary conditions. We use Banach’s and Krasnoselskii’s fixed point theorems to obtain the results. Lastly, we give two examples to show the effectiveness of the main results.


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## Body

## 1. Introduction

In the recent few decades, fractional differential equations have been studied by many researchers, and this is due to the importance of this field and its applications in many problems of physics, chemistry, biology, and economy (for more details, we refer the readers to [1–6] and many other references therein).In particular, fractional Langevin differential equations have been one of the important subjects in the field of fractional differential equations for their rich history (for more information, see [7–9]). Fractional Langevin equations are applied widely in many domains like engineering, physics, and biology (for more details, we give the following references [10–13]).On the other hand, coupled systems of fractional differential equations are very important to study because they appear naturally in many problems (see [14–18]).Recently, in [19], the existence and uniqueness of solutions for a coupled system of Riemann–Liouville and Hadamard fractional derivatives of Langevin equation with fractional integral conditions were proved. The existence and uniqueness of the coupled system of nonlinear fractional Langevin equations with multipoint and nonlocal integral boundary conditions have been studied in [20].So, in this current article, we study the existence and uniqueness of solutions for a coupled system of fractional Langevin equation as follows:(1)Dβ1cDα1c+λ1x1t=f1t,x1t,x2t,Ip2x2t,t∈0,1,Dβ2cDα2c+λ2x2t=f2t,x1t,x2t,Ip1x1t,t∈0,1,subject to the fractional nonseparated integral boundary conditions:(2)x10+μ1x11=σ11∫01g1s,x1sds,Dα1cx10+μ1Dα1cx11=σ21∫01h1s,x1sds,D2α1cx10+μ1D2α1cx11=σ31∫01k1s,x1sds,x20+μ2x21=σ12∫01g2s,x2sds,Dα2cx20+μ2Dα2cx21=σ22∫01h2s,x2sds,D2α2cx20+μ2D2α2cx21=σ32∫01k2s,x2sds,where 0<αi<1,1<βi≤2,pi>0,λi,μi,σ1i,σ2i,σ3i∈ℝ∗ with μi≠−1 for i=1,2, Dβic, Dαic are Caputo’s fractional derivatives, and f1,f2:0;1×ℝ×ℝ⟶ℝ, g1,h1,k1,g2,h2,k2:0;1×ℝ⟶ℝ are given continuous functions.To our knowledge, coupled fractional Langevin equations involving nonseparated type integral boundary conditions have not been extensively investigated yet. The main results shown in this paper can be viewed as the extension of the results in [21].This paper is organized as follows. In Section2, we recall some notations and several known results. In Section 3, we show the existence and uniqueness of solutions to problems (1) and (2). In Section 4, we give two examples to demonstrate the application of our main results.


## 2. Preliminaries and Notations

In this section, we introduce some notations, definitions, and lemmas that we need in our proofs later.Definition 1.
(see [3]). The fractional integral of order α>0 with the lower limit zero for a function f can be defined as(3)Iαft=1Γα∫0tt−sα−1fsds.Definition 2.
(see [3]). The Caputo derivative of order α>0 with the lower limit zero for a function f can be defined as(4)Dαcft=1Γn−α∫0tt−sn−α−1fnsds,where n∈ℕ, 0≤n−1<α<n, t>0.Theorem 1.
(see [22]). Let M be a bounded, closed, convex, and nonempty subset of a Banach space X. Let A and B be operators such that(i)
Ax+By∈M whenever x,y∈M.(ii)
A is compact and continuous.(iii)
B is a contraction mapping.
Then, there existsz∈M such that z=Az+Bz.Lemma 1.
(see [3]). Let α,β≥0; then, the following relations hold:(1)
Iαtβ=Γβ+1/Γα+β+1tα+β(2)
Dαctβ=Γβ+1/Γβ−α+1tβ−αLemma 2.
(see [3]). Let n∈ℕ and n−1<α<n. If f is a continuous function, then we have(5)IαDαcft=ft+a0+a1t+a2t2+⋯+an−1tn−1.Lemma 3.
Lety1,y2∈C0,1,ℝ; the coupled system(6)Dβ1cDα1c+λ1x1t=y1t,t∈0,1,Dβ2cDα2c+λ2x2t=y2t,t∈0,1,subject to the boundary conditions (2) has a solution given by(7)x1t=1Γα1+β1∫0tt−sα1+β1−1y1sds−λ1Γα1∫0tt−sα1−1x1sds+A11t∫01h1s,x1sds+A21t∫01g1s,x1sds+A31t∫01k1s,x1sds+A41tΓβ1−α1∫011−sβ1−α1−1y1sds+A51tΓβ1∫011−sβ1−1y1sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds−μ11+μ1Γα1+β1∫011−sα1+β1−1y1sds,x2t=1Γα2+β2∫0tt−sα2+β2−1y2sds−λ2Γα2∫0tt−sα2−1x2sds+A12t∫01h2s,x2sds+A22t∫01g2s,x2sds+A32t∫01k2s,x2sds+A42tΓβ2−α2∫011−sβ2−α2−1y2sds+A52tΓβ2∫011−sβ2−1y2sds+μ2λ21+μ2Γα2∫011−sα2−1x2sds−μ21+μ2Γα2+β2∫011−sα2+β2−1y2sds,where(8)A1it=tαiσ2i1−λiΓ2−αiΓαi+11+μi+tαi+1λiσ2iΓ2−αiΓ2+αiμi+μi1+μi2Γαi+1−11+μiΓαi+2Γ2−αiλiσ2i−μiσ2i1+μi2Γαi+1,A2it=tαiλiσ1iΓαi+11+μi−σ1i1+μi+μiλiσ1i1+μi2Γαi+1,A3it=−tαiΓ2−αiσ3iΓαi+11+μi+tαi+1Γ2−αiσ3iΓαi+2μi+μiΓ2−αiσ3i1+μi2Γαi+1−Γ2−αiσ3i1+μiΓαi+2,A4it=tαiΓ2−αiμiΓαi+11+μi−tαi+1Γ2−αiΓαi+2+Γ2−αiμi1+μiΓαi+2−μi21+μi2Γαi+1,A5it=−tαiμiΓαi+11+μi−μi21+μi2Γαi+1,fori=1,2.Proof.
Using Lemma2, we obtain(9)Dα1c+λ1x1t=Iβ1y1t+a01+a11t,Dα1cx1t=Iβ1y1t+a01+a11t−λ1x1t,x1t=Iα1+β1y1t+Iα1a01+Iα1a11t−Iα1λ1x1t+a21,where a01,a11,a21∈ℝ.
According to the conditionD2α1cx10+μ1D2α1cx11=σ31∫01k1s,x1sds, we find that(10)a11=Γ2−α1σ31μ1∫01k1s,x1sds+λ1σ21μ1∫01h1s,x1sds−1Γβ1−α1∫011−sβ1−α1y1sds.
Using the facts thatDα1cx10+μ1Dα1cx11=σ21∫01h1s,x1sds and x10+μ1x11=σ11∫01g1s,x1sds, we have(11)a01=−Γ2−α1σ311+μ1∫01k1s,x1sds+1−λ1Γ2−α1σ211+μ1∫01h1s,x1sds+λ1σ111+μ1∫01g1s,x1sds+Γ2−α1μ11+μ1Γβ1−α1∫011−sβ1−α1−1y1sds−μ11+μ1Γβ1∫011−sβ1−1y1sds,a21=μ1λ11+μ1Γα1∫011−sα1−1x1sds+σ111+μ1−μ1λ1σ11Γα1+11+μ12∫01g1s,x1sds−μ11+μ1Γα1+β1∫011−sα1+β1−1y1sds+μ121+μ12Γβ1Γα1+1∫011−sβ1−1y1sds+Γ2−α1Γβ1−α1μ11+μ1Γα1+2−μ121+μ12Γα1+1∫011−sβ1−α1−1y1sds+μ121+μ12Γα1+1−μ11+μ1Γα1+2Γ2−α1λ1σ21μ1−μ1σ211+μ12Γα1+1×∫01h1s,x1sds.
Substituting the value ofa01, a11, and a21, we obtain(12)x1t=1Γα1+β1∫0tt−sα1+β1−1y1sds−λ1Γα1∫0tt−sα1−1x1sds+A11t∫01h1s,x1sds+A21t∫01g1s,x1sds+A31t∫01k1s,x1sds+A41tΓβ1−α1∫011−sβ1−α1−1y1sds+A51tΓβ1∫011−sβ1−1y1sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds−μ11+μ1Γα1+β1∫011−sα1+β1−1y1sds.
Analogously, we can deduce that(13)x2t=1Γα2+β2∫0tt−sα2+β2−1y2sds−λ2Γα2∫0tt−sα2−1x2sds+A12t∫01h2s,x2sds+A22t∫01g2s,x2sds+A32t∫01k2s,x2sds+A42tΓβ2−α2∫011−sβ2−α2−1y2sds+A52tΓβ2∫011−sβ2−1y2sds+μ2λ21+μ2Γα2∫011−sα2−1x2sds−μ21+μ2Γα2+β2∫011−sα2+β2−1y2sds.
By direct computation, the converse of the lemma can be easily verified.


## 3. Main Results

LetX be a Banach space of all continuous functions from 0,1⟶ℝ endowed with norm xi=supxit:t∈0,1 for i=1,2. Then, the product space X×X,x1;x2 is also a Banach space equipped with the norm x1;x2=x1+x2.In view of Lemma3, we define the operator U:X×X⟶X×X by Ux1,x2=U1x1,x2,U2x1,x2.Here,(14)U1x1,x2t=1Γα1+β1∫0tt−sα1+β1−1f1s,x1s,x2s,Ip2x2sds−λ1Γα1∫0tt−sα1−1x1sds+A11t∫01h1s;x1sds+A21t∫01g1s;x1sds+A31t∫01k1s;x1sds+A41tΓβ1−α1∫011−sβ1−α1−1f1s,x1s,x2sIp2x2sds+A51tΓβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds−μ11+μ1Γα1+β1∫011−sα1+β1−1f1s,x1s,x2s,Ip2x2sds,U2x1,x2t=1Γα2+β2∫0tt−sα2+β2−1f2s,x1s,x2s,Ip1x1sds−λ2Γα2∫0tt−sα2−1x2sds+A12t∫01h2s;x2sds+A22t∫01g2s;x2sds+A32t∫01k2s;x2sds+A42tΓβ2−α2∫011−sβ2−α2−1f2s,x1s,x2s,Ip1x1sds+A52tΓβ2∫011−sβ2−1f2s,x1s,x2s,Ip1x1sds+μ2λ21+μ2Γα2∫011−sα2−1x2sds−μ21+μ2Γα2+β2∫011−sα2+β2−1f2s,x1s,x2s,Ip1x1sds.For computational convenience, we set(15)r11=maxq111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+λ1Γα1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61,q21+q31Γp2+1×1Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1,r12=maxq121Γα2+β2+1+A42Γβ2−α2+1+A52Γβ2+1+μ21+μ2Γα2+β2+1+λ2Γα2+1+μ2λ21+μ2Γα2+1+A12q52+A22q42+A32q62,q22+q32Γp1+1×1Γα2+β2+1+A42Γβ2−α2+1+A52Γβ2+1+μ21+μ2Γα2+β2+1,Aij=maxt∈0,1Aijt,for i=1,2,…,5and j=1,2.Before introducing the main results, we impose some assumptions:H1f1,f2:0,1×ℝ3⟶ℝ and h1,g1,k1,h2,g2,k2:0,1×ℝ⟶ℝ are continuous functions.H2 There exist positive constants q11,q21,q31,q12,q22,q32 such that for all t∈0,1 and x1,x2,y1,y2,z1,z2∈ℝ, we have f1t,x1,y1,z1−f1t,x2,y2,z2≤q11x1−x2+q21y1−y2+q31z1−z2, f2t,x1,y1,z1−f2t,x2,y2,z2≤q12x1−x2+q22y1−y2+q32z1−z2.H3 There exist positive constants q41,q51,q61,q42,q52,q62 such that g1t,x1−g1t,x2≤q41x1−x2, g2t,x1−g2t,x2≤q42x1−x2, h1t,x1−h1t,x2≤q51x1−x2, h2t,x1−h2t,x2≤q52x1−x2, k1t,x1−k1t,x2≤q61x1−x2, k2t,x1−k2t,x2≤q62x1−x2,∀x1,x2∈ℝ.H4fit,x,y,z≤mit; hit,x≤ρit; kit,x≤ψit; git,x≤ϕit,∀t,x,y,z∈0,1×ℝ3 with mi,ρi,ϕi,ψi∈C0,1;ℝ+, for i=1,2.Theorem 2.
Suppose thatH1−H3 are satisfied.
Then, there exists a unique solution for systems (1) and (2) provided that r11+r12<1.Proof.
Definesup0≤t≤1fit,0,0,0=M0i, sup0≤t≤1git,0=M1i, sup0≤t≤1hit,0=M2i, sup0≤t≤1kit,0=M3i, for i=1,2.
LetBr=x1,x2∈X×X:x1,x2≤r with(16)r≥r21+r221−r11+r12,where(17)r21=M01Γα1+β1+1+M01A41Γβ1−α1+1+A11M21+A21M11+A31M31+A51M01Γβ1+1+μ1M011+μ1Γα1+β1+1,r22=M02Γα2+β2+1+M02A42Γβ2−α2+1+A12M22+A22M12+A32M32+A52M02Γβ2+1+μ2M021+μ2Γα2+β2+1.
We prove thatTBr⊆Br.
Forx1,x2∈Br,t∈0,1, we have(18)U1x1,x2t≤1Γα1+β1∫0tt−sα1+β1−1f1s,x1s,x2s,Ip2x2s−f1s,0,0,0+f1s,0,0,0ds+λ1Γα1∫0tt−sα1−1x1sds+A11t∫01h1s;x1s−h1s;0+h1s;0ds+A21t∫01g1s;x1s−g1s;0+g1s;0ds+A31t∫01k1s;x1s−k1s;0+k1s;0ds+A41tΓβ1−α1×∫011−sβ1−α1−1f1s,x1s,x2s,Ip2x2s−f1s,0,0,0+f1s,0,0,0ds+A51tΓβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2s−f1s,0,0,0+f1s,0,0,0ds+μ1λ11+μ1Γα1∫011−sα1−1x1sds+μ11+μ1Γα1+β1×∫011−sα1+β1−1f1s,x1s,x2s,Ip2x2s−f1s,0,0,0+f1s,0,0,0ds.
Consequently,(19)U1x1,x2≤1Γα1+β1+1q11x1+q21x2+q31Γp2+1x2+M01+λ1Γα1+1x1+A11q51x1+M21+A21q41x1+M11+A31q61x1+M31+A41Γβ1−α1+1q11x1+q21x2+q31Γp2+1x2+M01+A51Γβ1+1q11x1+q21x2+q31Γp2+1x2+M01+μ1λ11+μ1Γα1+1x1+μ11+μ1Γα1+β1+1q11x1+q21x2+q31Γp2+1x2+M01≤q111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+λ1Γα1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61x1+q21+q31Γp2+11Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1x2+M01Γα1+β1+1+M01A41Γβ1−α1+1+A11M21+A21M11+A31M31+A51M01Γβ1+1+μ1M011+μ1Γα1+β1+1≤r11r+r21.
In the same way, we obtain that(20)U2x1,x2≤r12r+r22.
Therefore, we have(21)Ux1,x2=U1x1,x2+U2x1,x2≤r11+r12r+r21+r22≤r.
Now, forx1,x2,x1′,x2′∈X×X and for t∈0;1, we get(22)U1x1,x2t−U1x1′,x2′≤1Γα1+β1∫0tt−sα1+β1−1f1s,x1s,x2s,Ip2x2s−f1s,x1′s,x2′s,Ip2x2′sds+λ1Γα1∫0tt−sα1−1x1s−x1′sds+A11t∫01h1s;x1s−h1s;x1′sds+A21t∫01g1s;x1s−g1s;x1′sds+A31t×∫01k1s;x1s−k1s;x1′sds+A41tΓβ1−α1∫011−sβ1−α1−1×f1s,x1s,x2s,Ip2x2s−f1s,x1′s,x2′s,Ip2x2′sds+A51tΓβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2s−f1s,x1′s,x2′s,Ip2x2′sds+μ1λ11+μ1Γα1×∫011−sα1−1x1s−x1′sds+μ11+μ1Γα1+β1∫011−sα1+β1−1f1s,x1s,x2s,Ip2x2s−f1s,x1′s,x2′s,Ip2x2′sds≤1Γα1+β1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+λ1Γα1+1x1−x1′+A11q51x1−x1′+A21q41x1−x1′+A31q61x1−x1′+A41Γβ1−α1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+A51Γβ1+1q11x1−x1′+q21+q31Γp2+1x2−x2′+μ1λ11+μ1Γα1+1x1−x1′+μ11+μ1Γα1+β1+1q11x1−x1′+q21+q31Γp2+1x2−x2′≤q111Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1+λ1Γα1+1+μ1λ11+μ1Γα1+1+A11q51+A21q41+A31q61x1−x1′+q21+q31Γp2+1×1Γα1+β1+1+A41Γβ1−α1+1+A51Γβ1+1+μ11+μ1Γα1+β1+1x2−x2′≤r11x1−x1′+x2−x2′.
Analogously, we can also have(23)U2x1,x2t−U2x1′,x2′t≤r12x1−x1′+x2−x2′,which leads to(24)Ux1,x2−Ux1′,x2′≤r11+r12x1−x1′+x2−x2′.
Asr11+r12<1, the operator U is a contraction mapping. Then, we deduce that systems (1) and (2) have a unique solution.Theorem 3.
Assume thatH1,H4 hold.
Then, systems (1) and (2) have at least one solution on 0,1 if R<1,where(25)R=maxλ1Γα1+1+μ1λ11+μ1Γα1+1,λ2Γα2+1+μ2λ21+μ2Γα2+1.Proof.
We define a bounded closed and convex ballBr′=x1,x2∈X×X:x1,x2≤r′ with r′≥r2′/1−R,where(26)r2′=m1Γα1+β1+1+A11ρ1+A21ϕ1+A31ψ1+A41m1Γβ1−α1+1+A51m1Γβ1+1+μ1m11+μ1Γα1+β1+1+m2Γα2+β2+1+A12ρ2+A22ϕ2+A32ψ2+A42m2Γβ2−α2+1+A52m2Γβ2+1+μ2m21+μ2Γα2+β2+1.
Let us introduce the decompositionUx1,x2t=W1x1,x2t+W2x1,x2t,where(27)W1x1,x2t=T1x1,x2,R1x1,x2t,W2x1,x2t=T2x1,x2,R2x1,x2t,with(28)T1x1,x2t=1Γα1+β1∫0tt−sα1+β1−1f1s,x1s,x2s,Ip2x2s+A11t∫01h1s;x1sds+A21t∫01g1s;x1sds+A31t∫01k1s;x1sds+A41tΓβ1−α1∫011−sβ1−α1−1f1s,x1s,x2s,Ip2x2sds+A51tΓβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2sds−μ11+μ1Γα1+β1∫011−sα1+β1−1f1s,x1s,x2s,Ip2x2sds,T2x1t=−λ1Γα1∫0tt−sα1−1x1sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds,R1x1,x2t=1Γα2+β2∫0tt−sα2+β2−1f2s,x1s,x2s,Ip1x1s+A12t∫01h2s;x2sds+A22t∫01g2s;x2sds+A32t∫01k2s;x2sds+A42tΓβ2−α2∫011−sβ2−α2−1f2s,x1s,x2s,Ip1x1sds+A52tΓβ2∫011−sβ2−1f2s,x1s,x2s,Ip1x1sds−μ21+μ2Γα2+β2∫011−sα2+β2−1f2s,x1s,x2s,Ip1x1sds,R2x2t=−λ2Γα2∫0tt−sα2−1x2sds+μ2λ21+μ2Γα2∫011−sα2−1x2sds.
Forx1,x2∈Br′, we have(29)T1x1,x2t+T2x1t≤1Γα1+β1∫0tt−sα1+β1−1f1s,x1s,x2s,Ip2x2sds+A11t×∫01h1s;x1sds+A21t∫01g1s;x1sds+A31t∫01k1s;x1sds+A41tΓβ1−α1∫011−sα1+β1−1f1s,x1s,x2s,Ip2x2sds+A51tΓβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2sds+μ11+μ1Γα1+β1∫011−sα1+β1−1f1s,x1s,x2s,Ip2x2sds+λ1Γα1∫01t−sα1−1x1sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds≤1Γα1+β1∫0tt−sα1+β1−1m1sds+A11∫0tρ1sds+A21∫01ϕ1sds+A31∫01ψ1sds+A41Γβ1−α1∫011−sβ1−α1−1m1sds+A51Γβ1∫011−sβ1−1m1sds+μ11+μ1Γα1+β1∫011−sα1+β1−1m1sds+λ1Γα1∫0tt−sα1−1x1sds+μ1λ11+μ1Γα1∫011−sα1−1x1sds≤mΓα1+β1+1+A11ρ1+A21ϕ1+A31ψ1+A41m1Γβ1−α1+1+A51m1Γβ1+1+μ1m11+μ1Γα1+β1+1+μ1λ11+μ1Γα1+1x1+λ1Γα1+1x1.
In a similar manner, we have(30)R1x1,x2t+R2x1t≤m2Γα2+β2+1+A12ρ2+A22ϕ2+A32ψ2+A42m2Γβ2−α2+1+A52m2Γβ2+1+μ2m21+μ2Γα2+β2+1+μ2λ21+μ2Γα2+1x2+λ2Γα2+1x2.
Further, we obtain(31)W1x1,x2t+W2x1,x2≤Rr′+r2′≤r′.
Hence,W1x1,x2t+W2x1,x2t∈Br′.
Forx1,x2,x1′,x2′∈Br′ and t∈0,1, we have(32)T2x1−T2x1′≤λ1Γα1+1+μ1λ11+μ1Γα1+1x1−x1′,R2x2−R2x2′≤λ2Γα2+1+μ2λ21+μ2Γα2+1x2−x2′.
Therefore,(33)W2x1,x2−W2x1′,x2′≤Rx1−x1′+Rx2−x2′≤Rx1−x1′,x2−x2′,
AsR<1, then W2 is a contraction.
Next, we prove thatW1 is compact and continuous. The continuity of f1,f2,h1,h2,g1,g2,k1,k2 implies that the operator W1 is continuous. Moreover, W1 is uniformly bounded on Br′.
Suppose that0≤t1<t2≤1. We have(34)T1x1,x2t2−T1x1,x2t1≤1Γα1+β1∫0t2t2−sα1+β1−1f1s,x1s,x2s,Ip2x2sds−∫0t1t1−sα1+β1−1f1s,x1s,x2s,Ip2x2sds+A11t2−A11t1×∫01h1s;x1sds+|A21t2−A21t1|∫01g1s;x1sds+A31t2−A31t1||∫01k1s;x1sds+A41t2−A41t1Γβ1−α1∫011−sβ1−α1−1f1s,x1s,x2s,Ip2x2sds+A51t2−A51t1Γβ1∫011−sβ1−1f1s,x1s,x2s,Ip2x2sds≤m1Γα1+β1+1t2α1+β1−t1α1+β1+ρ1A11t2−A11t1+ϕ1A21t2−A21t1+ψ1A31t2−A31t1m1Γβ1−α1+1A41t2−A41t1+m1Γβ1+1A51t2−A51t1.
Similarly, we obtain that(35)R1x1,x2t2−R1x1,x2t1≤m2Γα2+β2+1t2α2+β2−t1α2+β2+ρ2A12t2−A12t1+ϕ2A22t2−A22t1+ψ2A32t2−A32t1m2Γβ2−α2+1A42t2−A42t1+m2Γβ2+1A52t2−A52t1.
Therefore, the operatorW1 is equicontinuous. Thus, W1 is relatively compact on Br′. Then by Arzela–Ascoli theorem, the operator W1 is compact on Br′. In conclusion, all terms of Krasnoselskii’s theorem have been applied perfectly. Hence, (1) and (2) have at least one solution on Br′.


## 4. Examples

Example 1.
Consider the following system of fractional Langevin:(36)D4/3cD1/3c+1300x1t=1500+t2sinx1t+x2t1+x2t+1Γ15/2∫0tt−s13/2x2sds,t∈0,1,D8/7cD1/7c+1400x2t=1250+t212sinx1t+x2t2+2x2t+12Γ15/2∫0tt−s13/2x2sds,t∈0,1,x10+x11=1200∫01x1s300+x1sds,D1/3cx10+D1/3cx11=1200∫011s+23x1s30+x1sds,D2/3cx10+D2/3cx11=1200∫011s+42x1s30+x1sds,x20+x21=1200∫01x2s300+x2sds,D1/7cx20+D1/7cx21=1200∫011s+8x2s30+x2sds,D2/7cx20+D2/7cx21=1200∫011s+16x2s30+x2sds,where(37)β1=43,α1=13,β2=87,α2=17,p1=p2=152,λ1=1300,λ2=1400,μ1=1,μ2=1,σ11=σ21=σ31=σ12=σ22=σ32=1200,f1t,x,y,z=1500+t2sinxt+yt1+yt+zt,f2t,x,y,z=1250+t212sinxt+yt2+2yt+zt2,g1t,x=xt300+xt,g2t,x=xt300+xt,h1t,x=1t+23xt30+xt,h2t,x=1t+8xt30+xt,k1t,x=1t+42xt30+xt,k2t,x=1t+16xt30+xt.
Clearly,q11=q21=q12=q22=q31=q32=1/500, q41=q42=1/300, q51=q52=1/240, and q61=q62=1/480; furthermore, we have(38)r11+r12≈max0.007421,0.0128+max0.012255,0.007583≈0.026<1.
Thus, by Theorem2, system (36) has a unique solution.Example 2.
Consider the following problem:(39)D3/2cD1/2c+1600x1t=1t2+4t2x1t3x1t+1+x2t6x2t+10+1Γ4/3∫0tt−s1/3ds1+x22s,t∈0,1,cD4/3D1/3c+1700x2t=11+t4t4x1t4x1t+10+x2t4x2t+6+1Γ4/3∫0tt−s1/3ds1+x12s,t∈0,1,x10+x11=1300∫011s+100x1s300+x1sds,D1/2cx10+D1/2cx11=1300∫011s+203x1s30+x1sds,D1cx10+D1cx11=1300∫011s+402x1s30+x1sds,x20+x21=1300∫011s+200x2s300+x2sds,D1/3cx20+D1/3cx21=1300∫011s+8000x2s30+x2sds,D2/3cx20+D2/3cx21=1300∫011s+1600x2s30+x2sds,where(40)β1=32,α1=12,β2=43,α2=13,p1=p2=43,λ1=1600,λ2=1700,μ1=1,μ2=1,σ11=σ21=σ31=σ12=σ22=σ32=1300,andf1t,x,y,z=1t2+4t2xt3|xt|+1+yt6|yt|+10+11+z2t,f2t,x,y,z=11+t4t4xt4|xt|+10+yt4|yt|+6+11+z2t,g1t,x=1t+100xt300+xt,g2t,x=1t+200xt300+|xt|,h1t,x=1t+203xt30+xt,h2t,x=1t+8000xt30+xt,k1t,x=1t+402xt30+xt,k2t,x=1t+1600xt30+xt.
After calculating, we obtainR≈0.0029<1.
So, by Theorem3, problem (39) has at least one solution.


## 5. Conclusion

In this paper, we have investigated the existence and uniqueness results for a coupled system of nonlinear fractional Langevin equations supplemented with nonseparated integral boundary conditions by using the Banach contraction principle and Krasnoselskii’s fixed point theorem. Finally, we gave two examples to prove the validity of our results.



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