# Coupled System of Nonlinear Fractional Langevin Equations with Multipoint and Nonlocal Integral Boundary Conditions

Authors: Ahmed Salem; Faris Alzahrani; Mohammad Alnegga
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

This research paper is about the existence and uniqueness of the coupled system of nonlinear fractional Langevin equations with multipoint and nonlocal integral boundary conditions. The Caputo fractional derivative is used to formulate the fractional differential equations, and the fractional integrals mentioned in the boundary conditions are due to Atangana–Baleanu and Katugampola. The existence of solution has been proven by two main fixed-point theorems: O’Regan’s fixed-point theorem and Krasnoselskii’s fixed-point theorem. By applying Banach’s fixed-point theorem, we proved the uniqueness result for the concerned problem. This research paper highlights the examples related with theorems that have already been proven.


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## Body

## 1. Introduction

Recently, many mathematical fields have been developed rapidly via fractional calculus. Different applications can be described by fractional equations involving fractional derivatives. Fractional calculus was an essential element in many recently published articles, such as a fractional biological population model, a fractional SISR-SI malaria disease model, a fractional Biswas–Milovic model, fractional wave equations, fractional reaction-diffusion equations, and nonlinear fractional shock wave equations. More recent published articles related with fractional calculus can be clearly found in [1–9]. Fractional differential equations have obtained a remarkable reputation among the mathematicians due to rapid development which is applicable in many fields such as mathematics, chemistry, and electronics. For more details, we refer to [10–16]. The coupled systems of fractional differential equations are mainly significant because such systems occur frequently in various scientific applications (see [17–19]).The Langevin equations (first formulated by Langevin in 1908) have been done with accuracy in order to have a full description of evolution of physical phenomena in fluctuating environment [20]. There is a clear progress on fractional Langevin equations in physics (see [21, 22]). New results on Langevin equations under the variety of boundary value conditions have been published [23–26].Different forms of fractional integral have been identified and employed in many different applications. Three of the fractional integrals will be used: Riemann–Liouville [10], Atangana and Baleanu [27, 28], and Ntouyas et al. [29, 30].Recent paper [31] has discussed the existence and uniqueness of solutions obtained from boundary value conditions for nonlinear fractional differential equations for Riemann–Liouville type under the generalized nonlocal integral boundary condition. In addition, the authors in [32] have studied existence and uniqueness of the solution for a certain class of ordinary differential equations of Atangana–Baleanu fractional derivative.In this paper, we modify the boundary value conditions of coupled systems of Langevin fractional differential equations of Caputo type into new boundary value conditions. So, we deal with the following coupled systems of nonlinear fractional Langevin equations ofα and β fractional orders:(1)Dcβ1Dcα1+λ1x1t=f1t,x1t,x2t,t∈0,1,Dcβ2Dcα2+λ2x2t=f2t,x1t,x2t,t∈0,1,supplemented by the following:(2)x10=0,Dcα1x10=Γα1+1Iρ1γ1x1η1,∑j=1m1aj1x1ξj1=μ1IABγ2x1η2,x20=0,Dcα1x20=Γα2+1Iρ2γ3x2η3,∑j=1m2aj2x2ξj2=μ2IABγ4x2η4,where Dc is the Caputo fractional derivative of order 0<αi≤1 and 1<βi≤2 for i=1,2. ABI and Iργ are Atangana–Baleanu, and Katugampola fractional integrals, respectively. ρi>0 and λi, μi∈ℝ for i=1,2, γk>0 for k=1,2,3,4, aji,∈ℝ for j=1,2,…,mi, and i=1,2. 0<ηk<ξ1i<ξ2i<⋯<ξmi<1 for i=1,2 and k=1,2,3,4,…f1,f2:0,1×ℝ2⟶ℝ are continuous functions.From the definitions of fractional integrals mentioned in the next section, it is worth pointing out that the fractional integral of Katugampola is a generalization for Riemann–Liouville fractional integralρ⟶1 and Hadamard fractional integral ρ⟶0. Also, the fractional integral of Atangana–Baleanu contains the Riemann–Liouville fractional integral and when γ=0, we recover the initial function and if γ=1, we obtain the ordinary integral. These motivate us to choose these fractional integrals in our boundary conditions. Furthermore, to the extent of our knowledge, this is the first paper that discusses the existence and uniqueness of the solutions to coupled systems of fractional Langevin equations involving the nonlocal integro-multipoint of Atangana–Baleanu type and the nonlocal integral of Katugampola type as boundary value conditions.The research article has been organized as follows. In the second section, we introduce some main concepts and essential lemma. In the third one, the main results show the existence and uniqueness of solutions to (1) and (2) by O’Regan’s fixed-point theorem, Krasnoselskii’s fixed-point theorem, and Banach’s fixed-point theorem, respectively. Under each one, examples have been considered in order to cover all theorems clearly.


## 2. Basic Concepts and Relevant Lemmas

We will deduce the main outcomes by the following preliminary concepts in fractional calculus.Definition 1 (see [33]).
Forn∈ℕ, let f∈Cn0,∞, then the Caputo fractional derivative of order β for a continuous function f is defined by(3)Dcβfς=∫0ςς−xn−β+1Γn−βfnxdx,n−1<β≤n,provided the right-hand side exists.Definition 2 (see [33]).
The Riemann–Liouville fractional integral of orderω for a continuous function f:0,∞⟶ℝ is given by(4)Iωfτ=∫0ττ−xω−1fxΓωdx,ω>0,τ>0,provided the integral exists.Lemma 1 (see [33]).
IfDcβfx is a continuous function on 0,∞, then(5)IβDcβfx=fx+∑i=1nci−1xi−1,n−1<β≤n,where ci−1∈ℝ, i=1,2,…,n.Lemma 2 (see [33]).
Letγ>0. Then,(6)Iγxn=Γn+1Γn+γ+1xn+γ,n>−1.Definition 3 (see [34]).
The Katugampola fractional integral of orderβ for a function g defined on 0,∞ is given by the following formula:(7)Iρβgτ=ρ1−βΓβ∫0ττρ−xρβ−1xρ−1gxdx,β>0,ρ>0,provided the right-hand side exists.Definition 4 (see [35]).
TheAB “Atangana–Baleanu” fractional integral IABγ is defined by(8)IABγfθ=1−γBγfθ+γBγ∫0θθ−sγ−1fsdsΓγ,θ>0,provided the right-hand side exists whenever γ∈0,1 and f∈L10,∞. Bγ is called a normalization function satisfying B0=B1=1.Lemma 3.
Forh1,h2∈C0,1,ℝ, the coupled system(9)Dcβ1Dcα1+λ1x1t=h1t,t∈0,1,Dcβ2Dcα2+λ2x2t=h2t,t∈0,1,supplemented by(10)x10=0,Dcα1x10=Γα1+1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1,∑j=1m1aj1x1ξj1=μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2,x20=0,Dcα2x20=Γα2+1∫0η3ρ21−γ3η3ρ2−sρ2γ3−1sρ2−1x2sdsΓγ3,∑j=1m2aj2x2ξj2=μ21−γ4Bγ4x2η4+γ4Bγ4∫0η4η4−sγ4−1x2sdsΓγ4,has a solution given by(11)x1t=∫0tt−sα1+β1−1h1sdsΓα1+β1−λ1∫0tt−sα1−1x1sdsΓα1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+tα1+1Sα1+1Λ1x1,(12)x2t=∫0tt−sα2+β2−1h2sdsΓα2+β2−λ2∫0tt−sα2−1x2sdsΓα2+S2t∫0η1ρ21−γ1η3ρ2−sρ2γ1−1sρ2−1x2sdsΓγ3+tα2+1Sα2+1Λ2x2,where(13)Sit=tαi−SαiSαi+1tαi+1,Sαi=∑j=1miajiξjiαi,i=1,2,and for i=1,2(14)Λixi=μi1−γ2iBγ2ixiη2i+γ2iBγ2i∫0η2iη2i−sγ2i−1xisdsΓγ2i−∑j=1miaji∫0ξjiξji−sαi+βi−1hisdsΓαi+βi+λi∑j=1miaji∫0ξjiξji−sαi−1xisdsΓαi.Proof.
Clearly, by direct computation, both (11) and (12) are solutions to (1). Conversely, by using Lemma 1, the general solution of (1) can be given as(15)xit=∫0tt−sαi+βi−1hisdsΓαi+βi+λi∫0tt−sαi−1xisdsΓαi+c0itαiΓαi+1+c1itαi+1Γαi+2+c2i,i=1,2.
Fori = 1, both first and second boundary conditions, we obtain c21=0 and c01=Γα1+1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1s/Γγ1ds. Consequently, c11=Γα1+1/Sα1+1Λ1x1. By substituting c11 and c01 in (15), we get (11). Similarly, in case of i=2, the proof is done.Lemma 4.
∀
t
∈
0,1 and i=1,2, we have(16)S^i=maxt∈0,1Sit=max1−SαiSαi+1,αiαiαi+1αi+1Sαi+1Sαiαi.Proof.
For eachi=1,2, we have(17)Si′t=tαi−1αi−αi+1SαiSαi+1t.
IfSαi/Sαi+1≤0, then Si′t>0 for all t∈0,1 which means that Sit is increasing on 0,1 and so(18)Sit≤Si1=1−SαiSαi+1.
IfSαi/Sαi+1>0, then Si′t>0 for all t∈0,t0 and Si′t<0 for all t∈t0,1 where(19)t0=αiαi+1Sαi+1Sαi.
These mean thatSit is increasing on 0,t0 and decreasing on t0,1 and so(20)Sit≤Sit0=αiαiαi+1αi+1Sαi+1Sαiαi.
The proof is done.


## 3. Main Results

For eachi=1,2, consider Xi=xitxit∈C0,1,ℝ is the Banach space of all continuous functions from 0,1 to all real numbers introduced by the norm xi=supt∈0,1xit. Moreover, product space X1×X2,x1,x2 is a Banach space equipped with xi=x1,x2=x1+x2.For convenience, we simplify the following expressions:(21)Λ1=λ1Γα1+1+S^1η1ρ1−γ1Γγ1+1+1Sα1+1μ11−γ2Bγ2+η2γ2Bγ2Γγ2+λ1Sα1Γα1+1,(22)Λ2=λ2Γα2+1+S^2η3ρ2−γ3Γγ3+1+1Sα2+1μ21−γ4Bγ4+η4γ4Bγ4Γγ4λ2Sα2Γα2+1,(23)Λ3=1Γα1+β1+11+Sα1+β1Sα1+1,(24)Λ4=1Γα2+β2+11+Sα2+β2Sα2+1.Define the operatorT:X1×X2⟶X1×X2 by(25)Tx1,x2t=Ux1,x2t,Vx1,x2t,where(26)Ux1,x2t=∫0tt−sα1+β1−1f1s,x1,x2dsΓα1+β1−λ1∫0tt−sα1−1x1sdsΓα1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+tα1+1Sα1+1Λ1x1,Vx1,x2t=∫0tt−sα2+β2−1f2s,x1,x2dsΓα2+β2−λ2∫0tt−sα2−1x2sdsΓα2+S2t∫0η3ρ21−γ3η3ρ2−sρ2γ3−1sρ2−1x2sdsΓγ3+tα2+1Sα2+1Λ2x2,(27)Vx1,x2t=∫0tt−sα2+β2−1f2s,x1,x2dsΓα2+β2−λ2∫0tt−sα2−1x2sdsΓα2+S2t∫0η3ρ21−γ3η3ρ2−sρ2γ3−1sρ2−1x2sdsΓγ3+tα2+1Sα2+1Λ2x2.By splitting both (26) and (27), we have(28)U1x1,x2t=∫0tt−sα1+β1−1f1s,x1,x2dsΓα1+β1−tα1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2dsΓα1+β1,U2x1t=−λ1∫0tt−sα1−1x1sdsΓα1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+tα1+1Sα1+1μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1sdsΓα1,V1x1,x2t=∫0tt−sα2+β2−1f2s,x1,x2dsΓα2+β2−tα2+1Sα2+1∑j=1m2aj2∫0ξj2ξj2−sα2+β2−1f2s,x1,x2dsΓα2+β2,V2x2t=−λ2∫0tt−sα2−1x2sdsΓα2+S2t∫0η3ρ21−γ3η3ρ2−sρ2γ3−1sρ2−1x2sdsΓγ3+tα2+1Sα2+1μ21−γ4Bγ4x2η4+γ4Bγ4∫0η2η4−sγ4−1x2sdsΓγ4+λ2∑j=1m2aj2∫0ξj2ξj2−sα2−1x2sdsΓα2.Forthcoming theorems proof need more convenient dialog for the operatorT. So, it is required to rewrite it as follows:(29)Tx1,x2t=T1x1,x2t+T2x1,x2t,where(30)T1x1,x2t=U1x1,x2,V1x1,x2t,(31)T2x1,x2t=U2x1,V2x2t.


### 3.1. Existence via O’Regan’s Theorem

The first main result counts on O’Regan’s theorem which proves the existence of the solutions for (1) and (2).Lemma 5 (see [36]).
(O’Regan’s theorem)
Let0∈V⊂D where V be an open subset of a closed and convex set D in Banach space X. Assume G:V¯⟶D, where G=G1+G2, GV¯ is bounded, G1:V¯⟶X is completely continuous and G2:V¯⟶X is so called nonlinear contraction (i.e., there exists a nonnegative nondecreasing function ω:0,∞⟶0,∞ such that ωy<y,∀y>0 and 2y2−G2y1≤ωy2−y1,∀y2,y1∈V¯). Then, either(i)
Ghas a fixed pointy∈V¯ or(ii)
There exists a pointy∈∂Vandθ∈0,1withy=θGy.Theorem 1.
Letf1,f2:0,1×ℝ2⟶ℝ be continuous functions. Assume the following conditions hold:
E
1: there exist nonnegative functions Θ1,Θ2∈C0,1,ℝ and nondecreasing functions φi,ϕi:0,∞⟶0,∞ for each i=1,2 such that(32)f1t,x1,x2≤Θ1tφ1x1+φ2x2,for allt,x1,x2∈0,1×ℝ2,f2t,x1,x2≤Θ2tϕ1x1+ϕ2x2,for allt,x1,x2∈0,1×ℝ2.
E
2: there exists(33)Δ=maxΛ1,Λ2<1,where Λi for each i=1,2 are defined in (21) and (22), respectively.
Then, (1) and (2) have at least one solution on 0,1.Proof.
ConsiderVr=x1,x2∈X1×X2:x1,x2≤r with fix radius r as(34)r≥Θ1φ1r+φ2rΛ3+Θ2φ1r+φ2rΛ41−Δ,where Λi for each i=1,2,3,4, are defined in (21)–(24), respectively.
From (29) which are defined in both (30) and (31) as two separate operators, first of all, we will show that T is uniformly bounded on Vr¯. Indeed, T1x1,x2 is bounded since ∀x1,x2∈Vr¯, then we have(35)U1x1,x2t≤∫0tt−sα1+β1−1f1s,x1,x2dsΓα1+β1+tα1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2dsΓα1+β1≤Θ1tφ1x1+φ2x2Λ3.
By taking the norm over both sides, we obtain(36)U1x1,x2≤Θ1φ1r+φ2rΛ3.
Similarly,(37)V1x1,x2≤Θ2ϕ1r+ϕ2rΛ4.
Therefore,T1 is uniformly bounded:(38)T1x1,x2≤Θ1φ1r+φ2rΛ3+Θ2ϕ1r+ϕ2rΛ4.
T
2 is bounded, i.e., ∀x1,x2∈V¯r, and we have(39)U2x1≤λ1∫0tt−sα1−1x1sdsΓα1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+tα1+1Sα1+1μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1sΓα1dx≤Λ1x1.
By taking the norm over both sides, it yields(40)U2x1≤Λ1x1.
Similarly, we can deduce that(41)V2x2≤Λ2x2,which implies that(42)T2x1,x2≤Λ1x1+Λ2x2≤Δr.
Thus,(43)Tx1,x2≤Θ1φ1r+φ2rΛ3+Θ2ϕ1r+ϕ2rΛ4+Δr≤r.
Now, we shall show thatT1 is completely continuous, ∀t1,t2∈0,1, then we have(44)U1x1,x2t2−U1x1,x2t1≤Θ1φ1r+φ2r∫0t2t2−sα1+β1−1dsΓα1+β1−∫0t1t1−sα1+β1−1dsΓα1+β1+t2α1+1−t1α1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1dsΓα1+β1≤Θ1φ1r+φ2rΓα1+β1+1t2α1+β1−t1α1+β1+Sα1+β1Sα1+1t2α1+1−t1α1+1.
Similarly,(45)V1x1,x2t2−V1x1,x2t1≤Θ2ϕ1r+ϕ2rΓα2+β2+1t2α2+β2−t1α2+β2+Sα2+β2Sα2+1t2α2+1−t1α2+1,which shows the independence of the pair x1,x2 and t2−t1⟶0. We conclude that T1 is equicontinuous. By Arzelá–Ascoli theorem, T1Vr¯ is relatively compact. Thus, T1 is completely continuous.
We shall show thatT2 is a contraction mapping. Indeed, ∀t∈0,1 and for any x1,x1′,x2,x2′∈X×Y, we have(46)U2x2−U2x1≤Λ1x2−x1,V2x2′−V2x1′≤Λ2x2′−x1′.
So, we can write(47)T2x2,x2′−T2x1,x1′≤Δx2−x1+Δx2′−x1′=Δx2−x1+x2′−x1′,which clarifies that T2 is a contraction mapping via 3.11. The last step is to show the first case of Lemma 5. By the way of contradiction, so we suppose ∃θ∈0,1andx1,x2∈∂V¯r such that x1,x2=θTx1,x2. Then, we have x1,x2=r, where(48)x1≤Θ1φ1r+φ2rΛ3+Λ1x1,x2≤Θ2ϕ1r+ϕ2rΛ4+Λ2x2,consequently,(49)x1+x2≤Θ1φ1r+φ2rΛ3+Θ2ϕ1r+ϕ2rΛ4+Δr,equivalently,(50)r≤Θ1φ1r+φ2rΛ3+Θ2φ1r+φ2rΛ41−Δ,which clearly contradicts E2. Hence, the operator T has at least one fixed point x1,x2∈V¯r. This ensures that (1) and (2) has a solution on 0,1.Example 1.
Consider the following coupled system of the nonlinear fractional Langevin equation subject to both the nonlocal integro-multipoint of Atangana–Baleanu type and the nonlocal integral of Katugampola type as boundary value conditions.(51)DC3/2DC1/2+1101xt=ft,x,y,0<t<1,DC4/3DC1/3+1110yt=gt,x,y,0<t<1,x0=0,DC1/2x0=Γ323I1/4x1109,14x15+16x27+34x59=1105I1/10x1119,y0=0,DC1/3y0=Γ432I5/9y1103,17y15+58y29+34y411=4115IAB3/11y1135.
Obviously,(52)α1=12,α2=13,β1=32,β2=43,λ1=1101,λ2=1110,ρ1=3,ρ2=2,μ1=1105,μ2=1115,γ1=14,γ2=110,γ3=59,γ4=311,η1=1109,η2=1119,η3=1103,η4=1135,a1=14,a2=16,a3=34,b1=17,b2=58,b3=34,ξ11=15,ξ21=27,ξ31=59,ξ12=15,ξ22=29,ξ32=411,B110=99100,B311=98100,ft,x,y=t+25t+1xt23xt+2+yt2yt+2⟹ft,x,y≤t+25t+1xt22+yt2,gt,x,y=t+24t+3xt2xt+3+yt27yt+3⟹gt,x,y=t+24t+3xt3+yt23,Θ1t=t+25t+1,Θ2t=t+24t+3,φ1x=xt22,φ2y=yt2,ϕ1x=xt3,ϕ2y=yt23.
From the shown data above, we have(53)Λ1=0.0883,Λ2=0.1502,Λ3=0.8559,Λ4=1.1138,Δ=0.1502.
Clearly, by applying Theorem1 we have(54)supr∈0,∞rΘ1φ1r+φ2rΛ3+Θ2φ1r+φ2rΛ4∼2.5292≥11−Δ∼1.1767.
Hence, our example possess at least one solution on0,1.


### 3.2. Existence via Krasnoselskii’s Theorem

The second outcome relies on Krasnoselskii’s theorem in order to prove solution existence for equations (1) and (2).Lemma 6 (see [37]) (Krasnoselskii’s theorem).
LetΩ be a bounded, closed, convex, and nonempty subset of Banach space X. Let T1 and T2 be two operators on Ω as the following:(i)
T1x1+T2x2∈Ω,∀x1,x2∈Ω,(ii)
T1 is compact and continuous,(iii)
T2 is a contraction mapping.
Then, there existsx3∈Ω satisfying x3=T1x3+T2x3.Theorem 2.
Assume the following conditions holdA1f1t,x1,x2≤Φt,∀t,x1,x2∈0,1×ℝ2andΦ∈C0,1,ℝ+.A2f2t,x1,x2≤Ψt,∀t,x1,x2∈0,1×ℝ2andΨ∈C0,1,ℝ+.Then, the boundary value problem (1) and (2) has at least one solution on 0,1 if Δ<1 where Δ is defined in (33).Proof.
SetΦ=supt∈0,1Φt, Ψ=supt∈0,1Ψt and(55)ΦΛ3+ΨΛ41−Δ<r.
Consider(56)Br=x1,x2∈X1×X2:x1,x2≤r,is a subset of Banach space X1×X2. Our claim is to prove that for any point x1,x2 of Br, it implies T1x1,x2∈Br. Indeed, given x1,x2 of Br arbitrarily, we have(57)U1x1,x2t+U2x1t≤∫0tt−sα1+β1−1f1s,x1,x2dsΓα1+β1+λ1∫0tt−sα1−1x1sdsΓα1+tα1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2dsΓα1+β1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+tα1+1Sα1+1μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1sdsΓα1≤Φ∫0tt−sα1+β1−1dsΓα1+β1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1dsΓα1+β1+x1λ1∫0tt−sα1−1dsΓα1+S^1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1dsΓγ1+1Sα1+1μ11−γ2Bγ2+γ2Bγ2∫0η2η2−sγ2−1dsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1dsΓα1≤ΦΛ3+x2Λ1.
Similarly,(58)V1x1,x2t+V2x2t≤ΨΛ4+y2Λ2.
This shows thatT1x1,x2+T2x1,x2≤r.
Hence,T1x1,x2+T2x1,x2∈Br. T2 satisfies the contraction principle as it has been shown in (47). f is a continuous function which helps us to say that T1 is continuous. Moreover, T1 is uniformly bounded on Br(59)T1x1,x2≤ΦΛ3+ΨΛ4.
The last step in this proof is to show compactness of the operatorT1. Indeed, ∀t1,t2∈0,1 with t1<t2, we have(60)U1t2−U1t1≤∫0t2t2−sα1+β1−1f1s,x1,x2dsΓα1+β1−∫0t1t1−sα1+β1−1f1s,x1,x2dsΓα1+β1+t2α1+1−t1α1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2dsΓα1+β1≤Φt2α1+β1−t1α1+β1Γα1+β1+1+t2α1+1−t1α1+1Γα1+β1+1Sα1+β1Sα1+1,V1t2+V1t1≤∫0t2t2−sα2+β2−1f2s,x1,x2dsΓα2+β2−∫0t1t1−sα2+β2−1f2s,x1,x2dsΓα2+β2+t2α2+1−t1α2+1Sα2+1∑j=1m2aj2∫0ξj2ξj2−sα2+β2−1f2s,x1,x2dsΓα2+β2≤Ψt2α2+β2−t1α2+β2Γα2+β2+1+t2α2+1−t1α2+1Γα2+β2+1Sα2+β2Sα2+1,which is not dependent on the pair x1,x2 and the quantity t2−t1⟶0 which ensures that T1 is equicontinuous. So, T1 is relatively compact on Br. T1 is a compact operator on Br since T1 satisfies the Arzelá–Ascoli theorem. In conclusion, all terms of Krasnoselskii’s theorem have been applied perfectly. Hence, (1) and (2) possess at least one solution on the given period.Example 2.
Consider the following coupled system of nonlinear fractional Langevin equation subject to both the nonlocal integro-multipoint of Atangana–Baleanu type and the nonlocal integral of Katugampola type as boundary value conditions.(61)DC5/3DC1/3+1100xt=ft,x,y0<t<1,DC8/7DC1/7+1108yt=gt,x,y0<t<1,x0=0,DC1/3x0=Γ433I3/10x1107,13x14+14x35+14x56=1102IAB1/10x1120,y0=0,DC1/7y0=Γ871/3I5/12y1108,18y13+38y23+14y45=4110IAB1/12y1140.
Clearly,(62)α1=13,α2=17,β1=53,β2=87,λ1=1100,λ2=1108,ρ1=3,ρ2=13,μ1=1102,μ2=1110,γ1=310,γ2=110,γ3=512,γ4=112,η1=1107,η2=1120,η3=1108,η4=1140,a1=13,a2=14,a3=14,b1=18,b2=38,b3=14,ξ11=14,ξ21=35,ξ31=56,ξ12=13,ξ22=23,ξ32=45,B110=9991000,B112=9981000,ft,x,y=t+12xt20t+22+213xt+1+ytsin4yt24t+32+18yt+5,gt,x,y=t+13xt1+18t+237yt+5+ytcos2t2cosyt1+14t+2212yt+2,ft,x,y≤t+12260t+22+26+1192t+32+8,gt,x,y≤t+13126t+23+7+1168t+22+12,Φt=t+12260t+22+26+1192t+32+8,Ψt=t+13126t+23+7+1168t+22+12.
After calculating, we findΛ1≈0.0644, Λ2≈0.5194,Λ3≈0.8791, and Λ4≈1.6832. According to Theorem 2, we see that Δ=0.5194 which indicates that our example has at least one solution on 0,1.


### 3.3. Uniqueness via Banach Fixed-Point Theorem

The last result in this paper is about uniqueness criteria for the solution of (1) and (2), which can be achieved by Banach’s fixed-point theorem.Theorem 3.
Assume thatf1,f2:0,1×ℝ2⟶ℝ are continuous functions and there exist positive real constants l1,l2,j1, and j2 such that ∀t,x1,x2∈0,1×ℝ2,
E
3
f
1
t
,
x
1
,
x
1
′
−
f
t
,
x
2
,
x
2
′
≤
l
1
x
2
−
x
1
+
l
2
x
2
′
−
x
1
′
E
4
f
2
t
,
x
1
,
x
1
′
−
g
t
,
x
2
,
x
2
′
≤
j
1
x
2
−
x
1
+
j
2
x
2
′
−
x
1
′
Then (1) and (2) have a unique solution on 0,1 if N1+N2<1 where(63)N1=Λ3l1+l2+Λ1,N2=Λ4j1+j2+Λ2,and Λi for each i=1,2,3,4 are defined in (21)–(24).Proof.
SetF1=supt∈0,1f1t,0,0 and F2=supt∈0,1f2t,0,0 and r0>0 such that(64)Λ3F1+Λ4F21−N1+N2≤r0.
First of all, we will show thatTBr0⊆Br0 where(65)Br0=x1,x2∈X1×X2:x1,x2≤r0.
For allx1,x2∈Br0, we have(66)Ux1,x2t≤∫0tt−sα1+β1−1f1s,x1,x2−f1s,0,0+f1s,0,0dsΓα1+β1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2−f1s,0,0+f1s,0,0dsΓα1+β1+λ1∫0tt−sα1−1x1sdsΓα1+S^1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+1Sα1+1μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1sdsΓα1≤l1x1+l2x2+F1∫0tt−sα1+β1−1dsΓα1+β1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1dsΓα1+β1+x1λ1∫0tt−sα1−1dsΓα1+S^1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1dsΓγ1+1Sα1+1μ11−γ2Bγ2+γ2Bγ2∫0η2η2−sγ2−1dsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1dsΓα1=Λ3l1x1+l2x2+F1+Λ1x1=Λ3l1+Λ1x1+Λ3l2x2+Λ3F1≤Λ3l1+Λ1r0+Λ3l2r0+Λ3F1=N1r0+Λ3F1.
Likewise,(67)Vx1,x2t≤Λ4j2+Λ2r0+Λ4j1r0+Λ4F2=N2r0+Λ4F2.
Therefore, we obtain(68)Tx1,x2=Ux1,x2+Vx1,x2≤N1+N2r0+Λ3F1+Λ4F2<r0.
Next, we shall show that the operatorT is a contraction operator on 0,1. Indeed, for any distinct two pairs x1,x2,x1′,x2′∈X1×X2, we see that(69)Ux1,x2t−Ux1′,x2′t≤∫0tt−sα1+β1−1f1s,x1,x2−f1s,x1′,x2′dsΓα1+β1+tα1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2−f1s,x1′,x2′dsΓα1+β1+λ1∫0tt−sα1−1x1s−x1′sdsΓα1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1s−x1′sdsΓγ1+tα1+1Sα1+1μ11−γ2Bγ2x1η2−x1′η2+γ2Bγ2∫0η2η2−sγ2−1x1s−x1′sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1s−x1′sdsΓα1,≤Λ3l1x1−x1′+l2x2−x2′+Λ1x1−x1′=l1Λ3+Λ1x1−x1′+l2Λ3x2−x2′,obviously, it gives us(70)Ux1,x2t−Ux1′,x2′≤N1x1−x1′+x2−x2′.
In a similar technique, we can also have(71)Vx2,x2′t−Vx1,x1′≤N2x1−x1′+x2−x2′.
From (70) and (71), we obtain(72)Tx2,x2′−Tx1,x1′≤N1+N2x1−x1′+x2−x2′.
Since the sum of bothN1 and N2 is strictly less than one, we can say that the operator T satisfies a contraction criteria. Banach’s fixed-point theorem ensures that the operator T has a unique fixed point. Hence (1) and (2) have a unique solution on 0,1.Example 3.
Consider the previous example of the coupled system of nonlinear fractional Langevin equation subject to both the nonlocal integro-multipoint of Atangana–Baleanu type the and nonlocal integral of Katugampola type as boundary value conditions.(73)DC5/3DC1/3+1100xt=ft,x,y,0<t<1,DC8/7DC1/7+1108yt=gt,x,y,0<t<1,x0=0,DC1/3x0=Γ433I3/10x1107,13x14+14x35+14x56=1102IAB1/10x1120,y0=0,DC1/7y0=Γ871/3I5/12y1108,18y13+38y23+14y45=4110IAB1/12y1140.
Clearly,(74)α1=13,α2=17,β1=53,β2=87,λ1=1100,λ2=1108,ρ1=3,ρ2=13,μ1=1102,μ2=1110,γ1=310,γ2=110,γ3=512,γ4=112,η1=1107,η2=1120,η3=1108,η4=1140,a1=13,a2=14,a3=14,b1=18,b2=38,b3=14,ξ11=14,ξ21=35,ξ31=56,ξ12=13,ξ22=23,ξ32=45,B110=9991000,B112=9981000,ft,x,y=t+12xt20t+22+213xt+1+ytsin4yt24t+32+18yt+5,gt,x,y=t+13xt1+18t+237xt+5+ytcos2t2cosyt1+14t+2212yt+2,ft,x,y≤t+12260t+22+26+1192t+32+8,gt,x,y≤t+13126t+23+7+1168t+22+12,Φt=t+12260t+22+26+1192t+32+8,Ψt=t+13126t+23+7+1168t+22+12.
Based on Theorem3, we can rewrite ft,x,y and gt,x,y as follows:(75)ft,x2,y2−ft,x1,y1≤11066x2−x1+11736y2−y1,gt,x2,y2−gt,x1,y1≤11015x2−x1+1684y2−y1.
Obviously,l1=1/1066,l2=1/1736,j1=1/1015,j2=1/684,N1≈0.0657,andN2≈0.5235. We conclude that our example has a unique solution on 0,1 since N1+N2≈0.5892<1.


## 3.1. Existence via O’Regan’s Theorem

The first main result counts on O’Regan’s theorem which proves the existence of the solutions for (1) and (2).Lemma 5 (see [36]).
(O’Regan’s theorem)
Let0∈V⊂D where V be an open subset of a closed and convex set D in Banach space X. Assume G:V¯⟶D, where G=G1+G2, GV¯ is bounded, G1:V¯⟶X is completely continuous and G2:V¯⟶X is so called nonlinear contraction (i.e., there exists a nonnegative nondecreasing function ω:0,∞⟶0,∞ such that ωy<y,∀y>0 and 2y2−G2y1≤ωy2−y1,∀y2,y1∈V¯). Then, either(i)
Ghas a fixed pointy∈V¯ or(ii)
There exists a pointy∈∂Vandθ∈0,1withy=θGy.Theorem 1.
Letf1,f2:0,1×ℝ2⟶ℝ be continuous functions. Assume the following conditions hold:
E
1: there exist nonnegative functions Θ1,Θ2∈C0,1,ℝ and nondecreasing functions φi,ϕi:0,∞⟶0,∞ for each i=1,2 such that(32)f1t,x1,x2≤Θ1tφ1x1+φ2x2,for allt,x1,x2∈0,1×ℝ2,f2t,x1,x2≤Θ2tϕ1x1+ϕ2x2,for allt,x1,x2∈0,1×ℝ2.
E
2: there exists(33)Δ=maxΛ1,Λ2<1,where Λi for each i=1,2 are defined in (21) and (22), respectively.
Then, (1) and (2) have at least one solution on 0,1.Proof.
ConsiderVr=x1,x2∈X1×X2:x1,x2≤r with fix radius r as(34)r≥Θ1φ1r+φ2rΛ3+Θ2φ1r+φ2rΛ41−Δ,where Λi for each i=1,2,3,4, are defined in (21)–(24), respectively.
From (29) which are defined in both (30) and (31) as two separate operators, first of all, we will show that T is uniformly bounded on Vr¯. Indeed, T1x1,x2 is bounded since ∀x1,x2∈Vr¯, then we have(35)U1x1,x2t≤∫0tt−sα1+β1−1f1s,x1,x2dsΓα1+β1+tα1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2dsΓα1+β1≤Θ1tφ1x1+φ2x2Λ3.
By taking the norm over both sides, we obtain(36)U1x1,x2≤Θ1φ1r+φ2rΛ3.
Similarly,(37)V1x1,x2≤Θ2ϕ1r+ϕ2rΛ4.
Therefore,T1 is uniformly bounded:(38)T1x1,x2≤Θ1φ1r+φ2rΛ3+Θ2ϕ1r+ϕ2rΛ4.
T
2 is bounded, i.e., ∀x1,x2∈V¯r, and we have(39)U2x1≤λ1∫0tt−sα1−1x1sdsΓα1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+tα1+1Sα1+1μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1sΓα1dx≤Λ1x1.
By taking the norm over both sides, it yields(40)U2x1≤Λ1x1.
Similarly, we can deduce that(41)V2x2≤Λ2x2,which implies that(42)T2x1,x2≤Λ1x1+Λ2x2≤Δr.
Thus,(43)Tx1,x2≤Θ1φ1r+φ2rΛ3+Θ2ϕ1r+ϕ2rΛ4+Δr≤r.
Now, we shall show thatT1 is completely continuous, ∀t1,t2∈0,1, then we have(44)U1x1,x2t2−U1x1,x2t1≤Θ1φ1r+φ2r∫0t2t2−sα1+β1−1dsΓα1+β1−∫0t1t1−sα1+β1−1dsΓα1+β1+t2α1+1−t1α1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1dsΓα1+β1≤Θ1φ1r+φ2rΓα1+β1+1t2α1+β1−t1α1+β1+Sα1+β1Sα1+1t2α1+1−t1α1+1.
Similarly,(45)V1x1,x2t2−V1x1,x2t1≤Θ2ϕ1r+ϕ2rΓα2+β2+1t2α2+β2−t1α2+β2+Sα2+β2Sα2+1t2α2+1−t1α2+1,which shows the independence of the pair x1,x2 and t2−t1⟶0. We conclude that T1 is equicontinuous. By Arzelá–Ascoli theorem, T1Vr¯ is relatively compact. Thus, T1 is completely continuous.
We shall show thatT2 is a contraction mapping. Indeed, ∀t∈0,1 and for any x1,x1′,x2,x2′∈X×Y, we have(46)U2x2−U2x1≤Λ1x2−x1,V2x2′−V2x1′≤Λ2x2′−x1′.
So, we can write(47)T2x2,x2′−T2x1,x1′≤Δx2−x1+Δx2′−x1′=Δx2−x1+x2′−x1′,which clarifies that T2 is a contraction mapping via 3.11. The last step is to show the first case of Lemma 5. By the way of contradiction, so we suppose ∃θ∈0,1andx1,x2∈∂V¯r such that x1,x2=θTx1,x2. Then, we have x1,x2=r, where(48)x1≤Θ1φ1r+φ2rΛ3+Λ1x1,x2≤Θ2ϕ1r+ϕ2rΛ4+Λ2x2,consequently,(49)x1+x2≤Θ1φ1r+φ2rΛ3+Θ2ϕ1r+ϕ2rΛ4+Δr,equivalently,(50)r≤Θ1φ1r+φ2rΛ3+Θ2φ1r+φ2rΛ41−Δ,which clearly contradicts E2. Hence, the operator T has at least one fixed point x1,x2∈V¯r. This ensures that (1) and (2) has a solution on 0,1.Example 1.
Consider the following coupled system of the nonlinear fractional Langevin equation subject to both the nonlocal integro-multipoint of Atangana–Baleanu type and the nonlocal integral of Katugampola type as boundary value conditions.(51)DC3/2DC1/2+1101xt=ft,x,y,0<t<1,DC4/3DC1/3+1110yt=gt,x,y,0<t<1,x0=0,DC1/2x0=Γ323I1/4x1109,14x15+16x27+34x59=1105I1/10x1119,y0=0,DC1/3y0=Γ432I5/9y1103,17y15+58y29+34y411=4115IAB3/11y1135.
Obviously,(52)α1=12,α2=13,β1=32,β2=43,λ1=1101,λ2=1110,ρ1=3,ρ2=2,μ1=1105,μ2=1115,γ1=14,γ2=110,γ3=59,γ4=311,η1=1109,η2=1119,η3=1103,η4=1135,a1=14,a2=16,a3=34,b1=17,b2=58,b3=34,ξ11=15,ξ21=27,ξ31=59,ξ12=15,ξ22=29,ξ32=411,B110=99100,B311=98100,ft,x,y=t+25t+1xt23xt+2+yt2yt+2⟹ft,x,y≤t+25t+1xt22+yt2,gt,x,y=t+24t+3xt2xt+3+yt27yt+3⟹gt,x,y=t+24t+3xt3+yt23,Θ1t=t+25t+1,Θ2t=t+24t+3,φ1x=xt22,φ2y=yt2,ϕ1x=xt3,ϕ2y=yt23.
From the shown data above, we have(53)Λ1=0.0883,Λ2=0.1502,Λ3=0.8559,Λ4=1.1138,Δ=0.1502.
Clearly, by applying Theorem1 we have(54)supr∈0,∞rΘ1φ1r+φ2rΛ3+Θ2φ1r+φ2rΛ4∼2.5292≥11−Δ∼1.1767.
Hence, our example possess at least one solution on0,1.


## 3.2. Existence via Krasnoselskii’s Theorem

The second outcome relies on Krasnoselskii’s theorem in order to prove solution existence for equations (1) and (2).Lemma 6 (see [37]) (Krasnoselskii’s theorem).
LetΩ be a bounded, closed, convex, and nonempty subset of Banach space X. Let T1 and T2 be two operators on Ω as the following:(i)
T1x1+T2x2∈Ω,∀x1,x2∈Ω,(ii)
T1 is compact and continuous,(iii)
T2 is a contraction mapping.
Then, there existsx3∈Ω satisfying x3=T1x3+T2x3.Theorem 2.
Assume the following conditions holdA1f1t,x1,x2≤Φt,∀t,x1,x2∈0,1×ℝ2andΦ∈C0,1,ℝ+.A2f2t,x1,x2≤Ψt,∀t,x1,x2∈0,1×ℝ2andΨ∈C0,1,ℝ+.Then, the boundary value problem (1) and (2) has at least one solution on 0,1 if Δ<1 where Δ is defined in (33).Proof.
SetΦ=supt∈0,1Φt, Ψ=supt∈0,1Ψt and(55)ΦΛ3+ΨΛ41−Δ<r.
Consider(56)Br=x1,x2∈X1×X2:x1,x2≤r,is a subset of Banach space X1×X2. Our claim is to prove that for any point x1,x2 of Br, it implies T1x1,x2∈Br. Indeed, given x1,x2 of Br arbitrarily, we have(57)U1x1,x2t+U2x1t≤∫0tt−sα1+β1−1f1s,x1,x2dsΓα1+β1+λ1∫0tt−sα1−1x1sdsΓα1+tα1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2dsΓα1+β1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+tα1+1Sα1+1μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1sdsΓα1≤Φ∫0tt−sα1+β1−1dsΓα1+β1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1dsΓα1+β1+x1λ1∫0tt−sα1−1dsΓα1+S^1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1dsΓγ1+1Sα1+1μ11−γ2Bγ2+γ2Bγ2∫0η2η2−sγ2−1dsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1dsΓα1≤ΦΛ3+x2Λ1.
Similarly,(58)V1x1,x2t+V2x2t≤ΨΛ4+y2Λ2.
This shows thatT1x1,x2+T2x1,x2≤r.
Hence,T1x1,x2+T2x1,x2∈Br. T2 satisfies the contraction principle as it has been shown in (47). f is a continuous function which helps us to say that T1 is continuous. Moreover, T1 is uniformly bounded on Br(59)T1x1,x2≤ΦΛ3+ΨΛ4.
The last step in this proof is to show compactness of the operatorT1. Indeed, ∀t1,t2∈0,1 with t1<t2, we have(60)U1t2−U1t1≤∫0t2t2−sα1+β1−1f1s,x1,x2dsΓα1+β1−∫0t1t1−sα1+β1−1f1s,x1,x2dsΓα1+β1+t2α1+1−t1α1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2dsΓα1+β1≤Φt2α1+β1−t1α1+β1Γα1+β1+1+t2α1+1−t1α1+1Γα1+β1+1Sα1+β1Sα1+1,V1t2+V1t1≤∫0t2t2−sα2+β2−1f2s,x1,x2dsΓα2+β2−∫0t1t1−sα2+β2−1f2s,x1,x2dsΓα2+β2+t2α2+1−t1α2+1Sα2+1∑j=1m2aj2∫0ξj2ξj2−sα2+β2−1f2s,x1,x2dsΓα2+β2≤Ψt2α2+β2−t1α2+β2Γα2+β2+1+t2α2+1−t1α2+1Γα2+β2+1Sα2+β2Sα2+1,which is not dependent on the pair x1,x2 and the quantity t2−t1⟶0 which ensures that T1 is equicontinuous. So, T1 is relatively compact on Br. T1 is a compact operator on Br since T1 satisfies the Arzelá–Ascoli theorem. In conclusion, all terms of Krasnoselskii’s theorem have been applied perfectly. Hence, (1) and (2) possess at least one solution on the given period.Example 2.
Consider the following coupled system of nonlinear fractional Langevin equation subject to both the nonlocal integro-multipoint of Atangana–Baleanu type and the nonlocal integral of Katugampola type as boundary value conditions.(61)DC5/3DC1/3+1100xt=ft,x,y0<t<1,DC8/7DC1/7+1108yt=gt,x,y0<t<1,x0=0,DC1/3x0=Γ433I3/10x1107,13x14+14x35+14x56=1102IAB1/10x1120,y0=0,DC1/7y0=Γ871/3I5/12y1108,18y13+38y23+14y45=4110IAB1/12y1140.
Clearly,(62)α1=13,α2=17,β1=53,β2=87,λ1=1100,λ2=1108,ρ1=3,ρ2=13,μ1=1102,μ2=1110,γ1=310,γ2=110,γ3=512,γ4=112,η1=1107,η2=1120,η3=1108,η4=1140,a1=13,a2=14,a3=14,b1=18,b2=38,b3=14,ξ11=14,ξ21=35,ξ31=56,ξ12=13,ξ22=23,ξ32=45,B110=9991000,B112=9981000,ft,x,y=t+12xt20t+22+213xt+1+ytsin4yt24t+32+18yt+5,gt,x,y=t+13xt1+18t+237yt+5+ytcos2t2cosyt1+14t+2212yt+2,ft,x,y≤t+12260t+22+26+1192t+32+8,gt,x,y≤t+13126t+23+7+1168t+22+12,Φt=t+12260t+22+26+1192t+32+8,Ψt=t+13126t+23+7+1168t+22+12.
After calculating, we findΛ1≈0.0644, Λ2≈0.5194,Λ3≈0.8791, and Λ4≈1.6832. According to Theorem 2, we see that Δ=0.5194 which indicates that our example has at least one solution on 0,1.


## 3.3. Uniqueness via Banach Fixed-Point Theorem

The last result in this paper is about uniqueness criteria for the solution of (1) and (2), which can be achieved by Banach’s fixed-point theorem.Theorem 3.
Assume thatf1,f2:0,1×ℝ2⟶ℝ are continuous functions and there exist positive real constants l1,l2,j1, and j2 such that ∀t,x1,x2∈0,1×ℝ2,
E
3
f
1
t
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1
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−
f
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2
−
x
1
+
l
2
x
2
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−
x
1
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E
4
f
2
t
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1
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−
g
t
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≤
j
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+
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−
x
1
′
Then (1) and (2) have a unique solution on 0,1 if N1+N2<1 where(63)N1=Λ3l1+l2+Λ1,N2=Λ4j1+j2+Λ2,and Λi for each i=1,2,3,4 are defined in (21)–(24).Proof.
SetF1=supt∈0,1f1t,0,0 and F2=supt∈0,1f2t,0,0 and r0>0 such that(64)Λ3F1+Λ4F21−N1+N2≤r0.
First of all, we will show thatTBr0⊆Br0 where(65)Br0=x1,x2∈X1×X2:x1,x2≤r0.
For allx1,x2∈Br0, we have(66)Ux1,x2t≤∫0tt−sα1+β1−1f1s,x1,x2−f1s,0,0+f1s,0,0dsΓα1+β1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2−f1s,0,0+f1s,0,0dsΓα1+β1+λ1∫0tt−sα1−1x1sdsΓα1+S^1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1sdsΓγ1+1Sα1+1μ11−γ2Bγ2x1η2+γ2Bγ2∫0η2η2−sγ2−1x1sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1sdsΓα1≤l1x1+l2x2+F1∫0tt−sα1+β1−1dsΓα1+β1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1dsΓα1+β1+x1λ1∫0tt−sα1−1dsΓα1+S^1∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1dsΓγ1+1Sα1+1μ11−γ2Bγ2+γ2Bγ2∫0η2η2−sγ2−1dsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1dsΓα1=Λ3l1x1+l2x2+F1+Λ1x1=Λ3l1+Λ1x1+Λ3l2x2+Λ3F1≤Λ3l1+Λ1r0+Λ3l2r0+Λ3F1=N1r0+Λ3F1.
Likewise,(67)Vx1,x2t≤Λ4j2+Λ2r0+Λ4j1r0+Λ4F2=N2r0+Λ4F2.
Therefore, we obtain(68)Tx1,x2=Ux1,x2+Vx1,x2≤N1+N2r0+Λ3F1+Λ4F2<r0.
Next, we shall show that the operatorT is a contraction operator on 0,1. Indeed, for any distinct two pairs x1,x2,x1′,x2′∈X1×X2, we see that(69)Ux1,x2t−Ux1′,x2′t≤∫0tt−sα1+β1−1f1s,x1,x2−f1s,x1′,x2′dsΓα1+β1+tα1+1Sα1+1∑j=1m1aj1∫0ξj1ξj1−sα1+β1−1f1s,x1,x2−f1s,x1′,x2′dsΓα1+β1+λ1∫0tt−sα1−1x1s−x1′sdsΓα1+S1t∫0η1ρ11−γ1η1ρ1−sρ1γ1−1sρ1−1x1s−x1′sdsΓγ1+tα1+1Sα1+1μ11−γ2Bγ2x1η2−x1′η2+γ2Bγ2∫0η2η2−sγ2−1x1s−x1′sdsΓγ2+λ1∑j=1m1aj1∫0ξj1ξj1−sα1−1x1s−x1′sdsΓα1,≤Λ3l1x1−x1′+l2x2−x2′+Λ1x1−x1′=l1Λ3+Λ1x1−x1′+l2Λ3x2−x2′,obviously, it gives us(70)Ux1,x2t−Ux1′,x2′≤N1x1−x1′+x2−x2′.
In a similar technique, we can also have(71)Vx2,x2′t−Vx1,x1′≤N2x1−x1′+x2−x2′.
From (70) and (71), we obtain(72)Tx2,x2′−Tx1,x1′≤N1+N2x1−x1′+x2−x2′.
Since the sum of bothN1 and N2 is strictly less than one, we can say that the operator T satisfies a contraction criteria. Banach’s fixed-point theorem ensures that the operator T has a unique fixed point. Hence (1) and (2) have a unique solution on 0,1.Example 3.
Consider the previous example of the coupled system of nonlinear fractional Langevin equation subject to both the nonlocal integro-multipoint of Atangana–Baleanu type the and nonlocal integral of Katugampola type as boundary value conditions.(73)DC5/3DC1/3+1100xt=ft,x,y,0<t<1,DC8/7DC1/7+1108yt=gt,x,y,0<t<1,x0=0,DC1/3x0=Γ433I3/10x1107,13x14+14x35+14x56=1102IAB1/10x1120,y0=0,DC1/7y0=Γ871/3I5/12y1108,18y13+38y23+14y45=4110IAB1/12y1140.
Clearly,(74)α1=13,α2=17,β1=53,β2=87,λ1=1100,λ2=1108,ρ1=3,ρ2=13,μ1=1102,μ2=1110,γ1=310,γ2=110,γ3=512,γ4=112,η1=1107,η2=1120,η3=1108,η4=1140,a1=13,a2=14,a3=14,b1=18,b2=38,b3=14,ξ11=14,ξ21=35,ξ31=56,ξ12=13,ξ22=23,ξ32=45,B110=9991000,B112=9981000,ft,x,y=t+12xt20t+22+213xt+1+ytsin4yt24t+32+18yt+5,gt,x,y=t+13xt1+18t+237xt+5+ytcos2t2cosyt1+14t+2212yt+2,ft,x,y≤t+12260t+22+26+1192t+32+8,gt,x,y≤t+13126t+23+7+1168t+22+12,Φt=t+12260t+22+26+1192t+32+8,Ψt=t+13126t+23+7+1168t+22+12.
Based on Theorem3, we can rewrite ft,x,y and gt,x,y as follows:(75)ft,x2,y2−ft,x1,y1≤11066x2−x1+11736y2−y1,gt,x2,y2−gt,x1,y1≤11015x2−x1+1684y2−y1.
Obviously,l1=1/1066,l2=1/1736,j1=1/1015,j2=1/684,N1≈0.0657,andN2≈0.5235. We conclude that our example has a unique solution on 0,1 since N1+N2≈0.5892<1.


## 4. Conclusion

In this research paper, we have proven the existence and uniqueness of solutions for the coupled system of nonlinear fractional Langevin equations with multipoint and nonlocal integral boundary value conditions by selecting0<α<1<β<2. Boundary value conditions have been chosen as two types of fractional integrals as we have shown in (2) for which have never been used together before in any article as far as we know. Existence of solutions have been shown by Krasnoselskii’s theorem and O’Regan’s theorem, and uniqueness solutions have been investigated by Banach’s fixed-point theorem. Examples have been supported in order to demonstrate all theorems very well. Results of this paper are not new in giving configuration, but also provide us new cases related with the choice of the parameters involving in the given problem. For example, the results associated with nonperiodic and nonlocal multipoint nonclassical integral boundary conditions follow by considering γ>1 of this problem. In case of aj1=aj2=0 for all j=1,2,⋯, the results change with respect to the boundary conditions:(76)x10=0,Dcα1x10=Γα1+1Iρ1γ1x1η1,μ1IABγ2x1η2=0,x20=0,Dcα2x20=Γα2+1Iρ2γ3x2η3,μ2IABγ4x2η4=0.Other results can be considered if we takeμ1=μ2=0, then the boundary conditions will be(77)x10=0,Dcα1x10=Γα1+1Iρ1γ1x1η1,∑j=1m1aj1x1ξj1=0,x20=0,Dcα2x20=Γα2+1Iρ2γ3x2η3,∑j=1m2aj2x2ξj2=0.In the future, in the case of obtaining theorems related with the existence and uniqueness of solutions under certain boundary conditions, both how can they be used to prove existence and uniqueness of solutions to the given problem and what are the conditions have to be considered.



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