# Quantum Difference Langevin System with Nonlocalq-Derivative Conditions

Authors: Surang Sitho; Sorasak Laoprasittichok; Sotiris K. Ntouyas; Jessada Tariboon
Journal: International Journal of Mathematics and Mathematical Sciences
 (2016)

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## Abstract

We introduce a new class of boundary value problems for Langevin quantum difference systems. Some new existence and uniqueness results for coupled systems are obtained by using fixed point theorems. The existence and uniqueness of solutions are established by Banach’s contraction mapping principle, while the existence of solutions is derived by using Leray-Schauder’s alternative. The obtained results are well illustrated with the aid of examples.


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## Body

## 1. Introduction

Quantum calculus (q-calculus) has a rich history and the details of its basic notions, results, and methods can be found in the text [1]. Apart from the traditional treatment of quantum calculus, many interesting questions and problems, especially from theoretical point of view, either remained open or were partially answered. In recent years, the topic has attracted the attention of several researchers and a variety of new results can be found in the papers [2–12]. However, there are many aspects of boundary value problems of quantum difference equations that need attention. For instance, quantum difference Langevin systems with nonlocal q-derivative conditions are yet to be addressed.In this paper, we investigate the sufficient conditions for existence of solutions for quantum difference Langevin system of the form(1)DqDq+λ1xt=ft,xt,yt,t∈J,DpDp+λ2yt=gt,xt,yt,t∈J,x0=α1Dz1x0,yT=Du1xξ,y0=α2Dz2y0,xT=Du2yη,where J=[0,T], 0<p,q,u1,u2,z1,z2<1, are quantum numbers, λ1,λ2,α1,α2∈R are constants, f,g∈C(J×R2,R) are continuous functions, and ξ,η∈J are fixed points.The Langevin equation (first formulated by Langevin in 1908) is found to be an effective tool to describe the evolution of physical phenomena in fluctuating environments [13]. For some new developments on the fractional Langevin equation in physics, see, for example, [14–22].In this paper, we prove existence and uniqueness of solutions by using Banach’s contraction principle and existence of solutions via Leray-Schauder’s alternative.The paper is organized as follows. In Section2, we recall some preliminary results from quantum calculus needed in the sequel. Also two basic lemmas are proved. The main existence and uniqueness results are contained in Section 3. Finally, in Section 4, examples illustrating the obtained results are presented.


## 2. Preliminaries

Let us recall some basic concepts ofq-calculus [1, 23].Definition 1.
For0<q<1, we define the q-derivative of a real valued function f as (2)Dqft=ft-fqt1-qt,t∈J∖0,Dqf0=limt→0⁡Dqft.The higher orderq-derivatives are given by (3)Dq0ft=ft,Dqnft=DqDqn-1ft,n∈N.Forx≥0, we set Jx=xqn:n∈N∪0∪0 and define the definite q-integral of a function f:Jx→R by (4)Iqfx=∫0xfsdqs=∑n=0∞x1-qqnfxqnprovided that the series converges.Fora,b∈Jx, we set (5)∫abfsdqs=Iqfb-Iqfa=1-q∑n=0∞qnbfbqn-afaqn.Note that, fora,b∈Jx, we have a=xqn1, b=xqn2, for some n1,n2∈N, thus the definite integral ∫ab‍f(s)dqs is just a finite sum, so no question about convergence is raised.We note that(6)DqIqfx=fx;while, if f is continuous at x=0, then (7)IqDqfx=fx-f0.In q-calculus, the product rule and integration by parts formula are (8)Dqght=Dqgtht+gqtDqht,∫0xftDqgtdqt=ftgt0x-∫0xDqftgqtdqt.Further, the reversing order of integration is given by (9)∫0t∫0sfrdqrdqs=∫0t∫qrtfrdqsdqr.In the limit q→1 the above results correspond to their counterparts in standard calculus.Lemma 2.
Letf:J→R be a continuous function and 0<p, q<1. Then we have the following: (i)
(10)Dp∫0tfsdqs=11-pt∫pttfsdqs,t≠0,limt→0⁡Dp∫0tfsdqs=f0;(ii)
(11)Dp∫0t∫0rfsdqsdqr=∫0ptfsdqs+∫pttt-qs1-ptfsdqs,t≠0,limt→0⁡Dp∫0t∫0rfsdqsdqr=0.Proof.
To prove (i), using the definition ofp-derivative, we have (12)Dp∫0tfsdqs=11-pt∫0tfsdqs-∫0ptfsdqs=11-pt∫pttfsdqs,t≠0.For t→0, we obtain (13)limt→0⁡Dp∫0tfsdqs=limt→0⁡Dpt1-q∑n=0∞qnftqn=limt→0⁡1-q1-p∑n=0∞qnftqn-p∑n=0∞qnfptqn=f0.Next, we will show that (ii) holds. From the reversing order of integration, the double q-integral can be reduced to a single integral as (14)∫0t∫0rfsdqsdqr=∫0tt-qsfsdqs.Taking the p-derivative to the both sides of the above equation, it follows that (15)Dp∫0t∫0rfsdqsdqr=Dp∫0tt-qsfsdqs=11-pt∫0tt-qsfsdqs+∫0ptqs-ptfsdqs=11-pt∫0tt-qsfsdqs-∫0ptt-qsfsdqs+∫0ptt-ptfsdqs=∫0ptfsdqs+∫pttt-qs1-ptfsdqs.Since (16)∫pttt-qs1-ptfsdqs=11-p∫pttfsdqs-q1-pt∫pttsfsdqs=1-q1-p∑n=0∞qntftqn-ptfptqn-q1-q1-pt∑n=0∞qnt2qnftqn-pt2qnfptqn,it is easy to see that (17)limt→0⁡Dp∫0t∫0rfsdqsdqr=0.This completes the proof.Let(18)A=1+α1λ11+α2λ21-Ω1Ω2,with (19)Ω1=T+α11+α1λ1,Ω2=T+α21+α2λ2.Lemma 3.
LetA≠0 and the functions ϕ,ψ∈C(J,R). Then x,y∈C(J,R) are solutions of the problem(20)DqDq+λ1xt=ϕt,t∈J,DpDp+λ2yt=ψt,t∈J,x0=α1Dz1x0,yT=Du1xξ,y0=α2Dz2y0,xT=Du2yη,if and only if(21)xt=-λ1∫0txsdqs+∫0tt-qsϕsdqs+t1+α1λ1+α11+α2λ2Aλ11-u1ξ∫u1ξξxsdqs-∫0u1ξϕsdqs-∫u1ξξξ-qs1-u1ξϕsdqs-λ2∫0Tysdps+∫0TT-psψsdps+Ω2-λ1∫0Txsdqs+∫0TT-qsϕsdqs+λ21-u2η∫u2ηηysdps-∫0u2ηψsdps-∫u2ηηη-ps1-u2ηψsdps,yt=-λ2∫0tysdps+∫0tt-psψsdps+t1+α2λ2+α21+α1λ1AΩ1λ11-u1ξ∫u1ξξxsdqs-∫0u1ξϕsdqs-∫u1ξξξ-qs1-u1ξϕsdqs-λ2∫0Tysdps+∫0TT-psψsdps-λ1∫0Txsdqs+∫0TT-qsϕsdqs+λ21-u2η∫u2ηηysdps-∫0u2ηψsdps-∫u2ηηη-ps1-u2ηψsdps.Proof.
Simplifying the first two equations of problem (20) and applying the double quantum integral, we obtain(22)xt=-λ1∫0txsdqs+∫0tt-qsϕsdqs+C1t+C2,yt=-λ2∫0tysdps+∫0tt-psψsdps+C3t+C4,where Ci, i=1,2,3,4∈R. From the conditions x(0)=α1Dz1x(0), y(0)=α2Dz2y(0), Lemma 2, and α1λ1, α2λ2≠-1, we have (23)C2=α1C11+α1λ1,C4=α2C31+α2λ2.Using the coupled nonlocal boundary conditions and Lemma 2, we get the following system:(24)-λ2∫0Tysdps+∫0TT-psψsdps+T1+α2λ2+α21+α2λ2C3=-λ11-u1ξ∫u1ξξxsdqs+∫0u1ξϕsdqs+∫u1ξξξ-qs1-u1ξϕsdqs+C1,-λ1∫0Txsdqs+∫0TT-qsϕsdqs+T1+α1λ1+α11+α1λ1C1=-λ21-u2η∫u2ηηysdps+∫0u2ηψsdps+∫u2ηηη-ps1-u2ηψsdps+C3.Solving system (24) for constants C1 and C3, we have(25)C1=1Aλ11-u1ξ∫u1ξξxsdqs-∫0u1ξϕsdqs-∫u1ξξξ-qs1-u1ξϕsdqs-λ2∫0Tysdps+∫0TT-psψsdps+Ω2-λ1∫0Txsdqs+∫0TT-qsϕsdqs+λ21-u2η∫u2ηηysdps-∫0u2ηψsdps-∫u2ηηη-ps1-u2ηψsdps,C3=1AΩ1λ11-u1ξ∫u1ξξxsdqs-∫0u1ξϕsdqs-∫u1ξξξ-qs1-u1ξϕsdqs-λ2∫0Tysdps+∫0TT-psψsdps-λ1∫0Txsdqs+∫0TT-qsϕsdqs+λ21-u2η∫u2ηηysdps-∫0u2ηψsdps-∫u2ηηη-ps1-u2ηψsdps.Substituting all values of constants Ci, i=1,2,3,4, in (22), we obtain the solutions of system (20) as in (21). The converse follows by direct computation. This completes the proof.


## 3. Main Results

In this section, we are going to prove the existence and uniqueness of solutions for the Langevin quantum difference system (1) with coupled boundary q-derivative conditions by using fixed point theorems. Let X=C([0,T],R) be the Banach space of all continuous functions from [0,T] to R with the norm defined by x=supt∈Jxt. Obviously (X,·) is a Banach space. In addition the product space (X×X,(x,y)) is a Banach space with norm (x,y)=x+y.In view of Lemma3, we define an operator G:X×X→X×X by (26)Gx,yt=G1x,ytG2x,yt,where(27)G1x,yt=-λ1∫0txsdqs+∫0tt-qsfs,xs,ysdqs+t1+α1λ1+α11+α2λ2Aλ11-u1ξ∫u1ξξxsdqs-∫0u1ξfs,xs,ysdqs-∫u1ξξξ-qs1-u1ξfs,xs,ysdqs-λ2∫0Tysdps+∫0TT-psgs,xs,ysdps+Ω2-λ1∫0Txsdqs+∫0TT-qsfs,xs,ysdqs+λ21-u2η∫u2ηηysdps-∫0u2ηgs,xs,ysdps-∫u2ηηη-ps1-u2ηgs,xs,ysdps,G2x,yt=-λ2∫0tysdps+∫0tt-psgs,xs,ysdps+t1+α2λ2+α21+α1λ1AΩ1λ11-u1ξ∫u1ξξxsdqs-∫0u1ξfs,xs,ysdqs-∫u1ξξξ-qs1-u1ξfs,xs,ysdqs-λ2∫0Tysdps+∫0TT-psgs,xs,ysdps-λ1∫0Txsdqs+∫0TT-qsfs,xs,ysdqs+λ21-u2η∫u2ηηysdps-∫0u2ηgs,xs,ysdps-∫u2ηηη-ps1-u2ηgs,xs,ysdps.In addition, we set constants(28)M1=T2A+u1ξ1+q+1-u1qξ+T2Ω2δA1+q,M2=T2+Ω2u2η1+p+1-u2pΩ2ηδA1+p,M3=λ1TA+Ω2λ1Tδ+λ1δA,M4=λ2Tδ+Ω2λ2δA,M5=T2+Ω1u1ξ1+q+1-u1qΩ1ξρA1+q,M6=T2A+u2η1+p+1-u2pη+T2Ω1ρA1+p,M7=λ1Tρ+Ω1λ1ρA,M8=λ2TA+Ω1λ2Tρ+λ2ρA,where (29)δ=T1+α1λ1+α11+α2λ2,ρ=T1+α2λ2+α21+α1λ1.Theorem 4.
Assume thatf,g:[0,T]×R2→R are continuous functions and there exist positive constants mi,ni=1,2 such that, for all t∈J and xi,yi∈R,i=1,2, (30)ft,x1,y1-ft,x2,y2≤m1x1-x2+m2y1-y2,gt,x1,y1-gt,x2,y2≤n1x1-x2+n2y1-y2.In addition, assume that(31)B1<12,U1<12,where (32)B1=m1+m2M1+n1+n2M2+M3+M4,U1=m1+m2M5+n1+n2M6+M7+M8.Then problem (1) has a unique solution on J.Proof.
Definesupt∈Jft,0,0=N1<∞ and supt∈Jgt,0,0=N2<∞ and choose a real number R such that (33)R≥max⁡M1N1+M2N21/2-B1,M5N1+M6N21/2-U1.Note that from (31) the constant R>0.
Firstly, we will show thatGBR⊂BR, where BR=x,y∈X×X:x,y≤R. For (x,y)∈BR, we have(34) G 1 x , y ≤ sup t ∈ J ⁡ - λ 1  ∫ 0 t x s d q s +  ∫ 0 t t - q s f s , x s , y s d q s + t 1 + α 1 λ 1 + α 1 1 + α 2 λ 2 A = · λ 1 1 - u 1 ξ  ∫ u 1 ξ ξ x s d q s -  ∫ 0 u 1 ξ f s , x s , y s d q s -  ∫ u 1 ξ ξ ξ - q s 1 - u 1 ξ f s , x s , y s d q s = - λ 2  ∫ 0 T y s d p s +  ∫ 0 T T - p s g s , x s , y s d p s + Ω 2 - λ 1  ∫ 0 T x s d q s +  ∫ 0 T T - q s f s , x s , y s d q s = + λ 2 1 - u 2 η  ∫ u 2 η η y s d p s -  ∫ 0 u 2 η g s , x s , y s d p s -  ∫ u 2 η η η - p s 1 - u 2 η g s , x s , y s d p s = ≤ m 1 x + m 2 y + N 1 T 2 A + u 1 ξ 1 + q + 1 - u 1 q ξ + T 2 Ω 2 δ A 1 + q + n 1 x + n 2 y + N 2 = · T 2 + Ω 2 u 2 η 1 + p + 1 - u 2 p Ω 2 η δ A 1 + p + λ 1 T A + Ω 2 λ 1 T δ + λ 1 δ A x + λ 2 T δ + Ω 2 λ 2 δ A y = ≤ m 1 + m 2 M 1 + n 1 + n 2 M 2 + M 3 + M 4 R + M 1 N 1 + M 2 N 2 = B 1 R + M 1 N 1 + M 2 N 2 ≤ R 2 ,and in a similar way(35) G 2 x , y ≤ m 1 x + m 2 y + N 1 T 2 + Ω 1 u 1 ξ 1 + q + 1 - u 1 q Ω 1 ξ ρ A 1 + q + n 1 x + n 2 y + N 2 T 2 A + u 2 η 1 + p + 1 - u 2 p η + T 2 Ω 1 ρ A 1 + p + λ 1 T ρ + Ω 1 λ 1 ρ A x  + λ 2 T A + Ω 1 λ 2 T ρ + λ 2 ρ A y ≤ m 1 + m 2 M 5 + n 1 + n 2 M 6 + M 7 + M 7 R + M 5 N 1 + M 6 N 2 = U 1 R + M 1 N 1 + M 2 N 2 ≤ R 2 .This shows that GBR⊂BR.
Next, we will prove that the operatorG is contractive. For any (x1,y1),(x2,y2)∈X×X, and t∈J, we have (36) G 1 x 2 , y 2 - G 1 x 1 , y 1 ≤ λ 1 ∫ 0 T x 2 s - x 1 s d q s + ∫ 0 T T - q s f s , x 2 s , y 2 s - f s , x 1 s , y 1 s d q s + δ A λ 1 1 - u 1 ξ ∫ u 1 ξ ξ x 2 s - x 1 s d q s + ∫ 0 u 1 ξ f s , x 2 s , y 2 s - f s , x 1 s , y 1 s d q s + ∫ u 1 ξ ξ ξ - q s 1 - u 1 ξ f s , x 2 s , y 2 s - f s , x 1 s , y 1 s + λ 2 ∫ 0 T y 2 s - y 1 s d p s + ∫ 0 T T - p s g s , x 2 s , y 2 s - g s , x 1 s , y 1 s d p s + Ω 2 λ 1 ∫ 0 T x 2 s - x 1 s d q s + ∫ 0 T T - q s f s , x 2 s , y 2 s - f s , x 1 s , y 1 s d q s + λ 2 1 - u 2 η ∫ u 2 η η y 2 s - y 1 s d p s  + ∫ 0 u 2 η g s , x 2 s , y 2 s - g s , x 1 s , y 1 s d p s + ∫ u 2 η η η - p s 1 - u 2 η g s , x 2 s , y 2 s - g s , x 1 s , y 1 s d p s ≤ m 1 x 2 - x 1 + m 2 y 2 - y 1 T 2 A + u 1 ξ 1 + q + 1 - u 1 q ξ + T 2 Ω 2 δ A 1 + q + n 1 x 2 - x 1 + n 2 y 2 - y 1 T 2 + Ω 2 u 2 η 1 + p + 1 - u 2 p Ω 2 η δ A 1 + p + λ 1 T A + Ω 2 λ 1 T δ + λ 1 δ A x 2 - x 1 + λ 2 T δ + Ω 2 λ 2 δ A y 2 - y 1 ≤ m 1 M 1 + n 1 M 2 + M 3 x 2 - x 1 + m 2 M 1 + n 2 M 2 + M 4 y 2 - y 1 .Thus,(37)G1x2,y2-G1x1,y1≤B1x2-x1+y2-y1.Similarly,(38)G2x2,y2-G2x1,y1≤U1x2-x1+y2-y1.It follows from (37) and (38) that (39)Gx2,y2-Gx1,y1≤B1+U1x2-x1+y2-y1.From (31), therefore, G is a contraction mapping. So, by Banach’s fixed point theorem, the operator G has a unique fixed point, which is the unique solution of problem (1). This completes the proof.In the next result, we prove the existence of solutions for problem (1) by Leray-Schauder alternative.Lemma 5 ((Leray-Schauder alternative) (see [24])).
LetG be a nonmed linear space and let F:G→G be a completely continuous operator (i.e., a map that is restricted to any bounded set in G is compact). Let (40)EF=x∈G:x=κFx,0<κ<1.Then either the set E(F) is unbounded or F has at least one fixed point.For convenience, we set constants(41)E1=M1+M5L1+M2+M6R1+M3+M7,E2=M1+M5L2+M2+M6R2+M4+M8,(42)E∗=min⁡1-E1,1-E2.Theorem 6.
Assume thatf,g:[0,T]×R2→R are continuous functions and there exist real constants Li,Ri≥0(i=1,2) and L0>0, R0>0, such that ∀xi∈R(i=1,2); we have (43)ft,x1,x2≤L0+L1x1+L2x2,gt,x1,x2≤R0+R1x1+R2x2.If E1<1, E2<1, M3+M4≠1, and M7+M8≠1, then there exists at least one solution for problem (1) on J.Proof.
Now we show that the operatorG:X×X→X×X is completely continuous. Let Br⊂X×X where Br=x,y∈X×X:x,y≤r. Then there exist positive constants P1 and P2 such that (44)ft,xt,yt≤P1,gt,xt,yt≤P2,∀x,y∈Br,and a positive real number r such that (45)r≥max⁡P1M1+P2M21-M3+M4,P1M5+P2M61-M7+M8.For any (x,y)∈Br, we have (46)G1x,y≤supt∈J⁡-λ1∫0txsdqs+∫0tt-qsfs,xs,ysdqs+t1+α1λ1+α11+α2λ2Aλ11-u1ξ∫u1ξξxsdqs-∫0u1ξfs,xs,ysdqs-∫u1ξξξ-qs1-u1ξfs,xs,ysdqs-λ2∫0Tysdps+∫0TT-psgs,xs,ysdps+Ω2-λ1∫0Txsdqs+∫0TT-qsfs,xs,ysdqs+λ21-u2η∫u2ηηysdps-∫0u2ηgs,xs,ysdps-∫u2ηηη-ps1-u2ηgs,xs,ysdps≤T2A+u1ξ1+q+1-u1qξ+T2Ω2δA1+qP1+T2+Ω2u2η1+p+1-u2pΩ2ηδA1+pP2+λ1TA+Ω2λ1Tδ+λ1δA+λ2Tδ+Ω2λ2δAr=M1P1+M2P2+M3+M4r.In the same way, we deduce that (47)G2x,y≤supt∈J⁡-λ2∫0tysdps+∫0tt-psgs,xs,ysdps+t1+α2λ2+α21+α1λ1AΩ1λ11-u1ξ∫u1ξξxsdqs-∫0u1ξfs,xs,ysdqs-∫u1ξξξ-qs1-u1ξfs,xs,ysdqs-λ2∫0Tysdps+∫0TT-psgs,xs,ysdps-λ1∫0Txsdqs+∫0TT-qsfs,xs,ysdqs+λ21-u2η∫u2ηηysdps-∫0u2ηgs,xs,ysdps-∫u2ηηη-ps1-u2ηgs,xs,ysdps≤T2+Ω1u1ξ1+q+1-u1qΩ1ξρA1+qP1+T2A+u2η1+p+1-u2pη+T2Ω1ρA1+pP2+λ1Tρ+Ω1λ1ρA+λ2TA+Ω1λ2Tρ+λ2ρAr=M5P1+M6P2+M7+M8r.Therefore, G is uniformly bounded.
Next, we show thatG is equicontinuous. Let t1,t2∈[0,T] with t2<t1. Setting supt∈Jft,xt,yt=f¯ and supt∈Jgt,xt,yt=g¯ and for any x,y∈Br, we get (48) G 1 x , y t 1 - G 1 x , y t 2 ≤ r λ 1 t 1 - t 2 + 1 A t 1 - t 2 1 + α 1 λ 1 + α 1 1 + α 2 λ 2 λ 1 + λ 2 T + λ 1 Ω 2 T + λ 2 Ω 2 + f ¯ t 1 - t 2 t 2 + t 1 - t 2 t 1 + t 2 1 + q + 1 A t 1 - t 2 1 + α 1 λ 1 + α 1 1 + α 2 λ 2 u 1 ξ + 1 - u 1 q ξ + T 2 Ω 2 1 + q  + g ¯ 1 A t 1 - t 2 1 + α 1 λ 1 + α 1 1 + α 2 λ 2 T 2 1 + p + u 2 η Ω 2 + 1 - u 2 p Ω 2 η .Similarly, we obtain (49) G 2 x , y t 1 - G 2 x , y t 2 ≤ r λ 2 t 1 - t 2 + 1 A t 1 - t 2 1 + α 2 λ 2 + α 2 1 + α 1 λ 1 λ 2 + λ 1 T + λ 2 Ω 1 T + λ 1 Ω 1  + f ¯ 1 A t 1 - t 2 1 + α 2 λ 2 + α 2 1 + α 1 λ 1 T 2 1 + q + u 1 ξ Ω 1 + 1 - u 1 q Ω 1 ξ + g ¯ t 1 - t 2 t 2 + t 1 - t 2 t 1 + t 2 1 + p + 1 A t 1 - t 2 1 + α 2 λ 2 + α 2 1 + α 1 λ 1 u 2 η + 1 - u 2 p η + T 2 Ω 1 1 + p .Then G is equicontinuous. So G is relatively compact on Br, and by the Arzelá-Ascoli theorem G is completely continuous on Br.
Finally, it will be verified that the setE=x,y∈X×X∣(x,y)=κG(x,y),0<κ<1 is bounded. Let (x,y)∈E; then (x,y)=κG(x,y). For any t∈J, we have (50)xt=κG1x,yt,yt=κG2x,yt.Therefore, we obtain (51) x t = κ G 1 x , y t ≤ λ 1 x ∫ 0 T 1 d q s + L 0 + L 1 x + L 2 y ∫ 0 T T - q s d q s + δ A λ 1 x 1 - u 1 ξ ∫ u 1 ξ ξ 1 d q s + L 0 + L 1 x + L 2 y ∫ 0 u 1 ξ 1 d q s + L 0 + L 1 x + L 2 y ∫ u 1 ξ ξ ξ - q s 1 - u 1 ξ + λ 2 y ∫ 0 T 1 d p s + R 0 + R 1 x + R 2 y ∫ 0 T T - p s d p s + Ω 2 λ 1 T x ∫ 0 T 1 d q s + L 0 + L 1 x + L 2 y ∫ 0 T T - q s d q s + λ 2 y 1 - u 2 η ∫ u 2 η η 1 d p s + R 0 + R 1 x + R 2 y ∫ 0 u 2 η 1 d p s + R 0 + R 1 x + R 2 y ∫ u 2 η η η - p s 1 - u 2 η d p s = L 0 + L 1 x + L 2 y M 1 + R 0 + R 1 x + R 2 y M 2 + x M 3 + y M 4 , y t = κ G 2 x , y t ≤ λ 2 y ∫ 0 T 1 d p s + R 0 + R 1 x + R 2 y ∫ 0 T T - q s d p s + ρ A Ω 1 λ 1 x 1 - u 1 ξ ∫ u 1 ξ ξ 1 d q s + λ 2 y ∫ 0 T 1 d p s + L 0 + L 1 x + L 2 y ∫ 0 u 1 ξ 1 d q s + R 0 + R 1 x + R 2 y ∫ 0 T T - p s d p s + L 0 + L 1 x + L 2 y ∫ u 1 ξ ξ ξ - q s 1 - u 1 ξ d q s  + λ 1 x ∫ 0 T 1 d q s + L 0 + L 1 x + L 2 y ∫ 0 T T - q s d q s + λ 2 y 1 - u 2 η ∫ u 2 η η 1 d p s + R 0 + R 1 x + R 2 y ∫ 0 u 2 η 1 d p s + R 0 + R 1 x + R 2 y ∫ u 2 η η η - p s 1 - u 2 η d p s = L 0 + L 1 x + L 2 y M 5 + R 0 + R 1 x + R 2 y M 6 + x M 7 + y M 8 .So, we have (52)x+y≤M1+M5L0+M2+M6R0+M1+M5L1+M2+M6R1+M3+M7x+M1+M5L2+M2+M6R2+M4+M8y.Consequently, (53)x,y≤M1+M5L0+M2+M6R0E∗,for any t∈J, where E∗ is defined by (42), so that E is bounded. Thus, by Lemma 5, the operator G has at least one fixed point. Hence, problem (1) has at least one solution on J. The proof is completed.


## 4. Examples

In this section, we present examples to illustrate our result.Example 1.
Consider the following system of Langevin quantum difference equations subject to the coupled nonlocalq-derivatives boundary conditions:(54)D1/4D1/4+130xt=xe-t510+t2+ysin2⁡t14-t3yy+1,t∈0,4D2/9D2/9+135yt=2x125t+102xx+3+12ycos2⁡t5t+103,t∈0,4x0=211D1/4x0,y4=D3/11x1,y0=316D5/6y0,x4=D1/13y2.Hereq=1/4, p=2/9, λ1=1/30, λ2=1/35, α1=2/11, α2=3/16, z1=1/4, z2=5/6, u1=3/11, u2=1/13, T=4, ξ=1, η=2, f(t,x,y)=|x|e-t/5(10+t)2+ysin2(t)/(14-t)3(y/y+1), and g(t,x,y)=2x/12(5t+10)2(x/x+3)+12ycos2(t)/5(t+10)3.We have(55)ft,x1,y1-ft,x2,y2≤1500x1-x2+11000y1-y2,gt,x1,y1-gt,x2,y2≤1600x1-x2+12400y1-y2,α1λ1≃0.00606≠1,α2λ2≃0.00536≠1,Ω1=T+α11+α1λ1≃4.18072,Ω2=T+α21+α2λ2≃4.18650,A=1+α1λ11+α2λ21-Ω1Ω2≃16.69158,δ=T1+α1λ1+α11+α2λ2≃4.77698,ρ=T1+α2λ2+α21+α1λ1≃4.8004.Then, the assumption of Theorem 4 is satisfied with m1=1/500, m2=1/1000, n1=1/600, n2=1/2400, M1≃28.42756, M2≃5.85791, M3≃0.30263, M4≃0.06694, M5≃4.90542, M6≃29.33758, M7≃0.07269, M8≃0.25991, and (56)B1=m1+m2M1+n1+n2M2+M3+M4≃0.46701<12,U1=m1+m2M5+n1+n2M6+M7+M8≃0.40844<12.Therefore, we get that (57)B1+U1≃0.87545<1.Hence, by Theorem 4, problem (54) has a unique solution on [0,4].Example 2.
Consider the following system of Langevin quantum difference equations subject to the coupled nonlocalq-derivatives boundary conditions:(58)D1/3D1/3+132xt=12+xe-t815-t2+4ycos2⁡t610-t3yy+3,t∈0,5D1/4D1/4+136yt=34+2x2413+txx+2+ysin2⁡t4719-t,t∈0,5x0=-23D1/5x0,y5=D2/3x2,y0=35D1/4y0,x5=D2/5y3.Hereq=1/3, p=1/4, λ1=1/32, λ2=1/36, α1=-2/3, α2=3/5, z1=1/5, z2=1/4, u1=2/3, u2=2/5, T=5, ξ=2, η=3, f(t,x,y)=1/2+xe-t/8(15-t)2+(4ycos2(t)/6(10-t)3)(y/y+3), and g(t,x,y)=3/4+(2x/241(3+t)3)(x/x+2)+ysin2(t)/47(19-t).So that(59)ft,x1,x2≤12+1800x1+1750x2,gt,x1,x2≤34+2723x1+1658x2,α1λ1≃0.020833≠1,α2λ2≃0.035294≠1,Ω1=T+α11+α1λ1≃5.653061,Ω2=T+α21+α2λ2≃5.579546,A=1+α1λ11+α2λ21-Ω1Ω2≃32.278176,δ=T1+α1λ1+α11+α2λ2≃8.735291,ρ=T1+α2λ2+α21+α1λ1≃8.346811.Then, the assumptions of Theorem 6 are satisfied with L0=1/2, L1=1/800, L2=1/750, R0=3/4, R1=2/723, R2=1/658, M1≃47.738433, M2≃10.485993, M3≃0.400639, M4≃0.079530, M5≃8.503122, M6≃50.105354, M7≃0.086087, M8≃0.349103, and (60)E1=M1+M5L1+M2+M6R1+M3+M7≃0.724639<1,E2=M1+M5L2+M2+M6R2+M4+M8≃0.595706<1.Consequently all conditions in Theorem 6 are satisfied. Therefore, problem (1) has at least one solution on [0,5].



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*Source: 4928314-2016-06-05.xml*