# Iterative Positive Solutions to a Coupled Riemann-Liouville Fractionalq-Difference System with the Caputo Fractional q-Derivative Boundary Conditions

Authors: Yuan Meng; Xinran Du; Huihui Pang
Journal: Journal of Function Spaces
 (2023)

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## Abstract

This paper is devoted to the existence of positive solutions for a nonlinear coupled Riemann-Liouville fractionalq-difference system, with multistrip and multipoint mixed boundary conditions under Caputo fractional q-derivative. We obtain the existence of positive solutions and initial iterative solutions by the monotone iteration technique. Then, we also calculate the error limits of the numerical approximation solution by induction. In the end, two examples are given to illustrate the above research results, and in the second example, some graphs of the iterative solutions are also drawn to give a more intuitive sense of the iterative process.


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## Body

## 1. Introduction

Fractional calculus, which originated in 1695 [1], extended the traditional concept of integral calculus to the whole real field [2]. Over the next several hundred years, many mathematicians have studied the properties of fractional calculus. From these studies, it is not difficult to conclude that fractional calculus has some very nice properties and can be applied to various fields, such as random processes [3], non-Newtonian fluid mechanics [4], control theory [5], economics [6], and medical [7]. Especially in the modeling of dynamic problems, fractional differential equations are quite common. Therefore, fractional differential equations derived from fractional calculus have received more and more attention. The existence analysis stability analysis and numerical simulation of solutions have gradually become very important research directions [8–15].Quantum calculus has been around since the beginning of the twentieth century. It was first studied systematically by Jackson [16, 17]. Readers can refer to the works of Kac and Cheung for further research and applications [18]. Youm used q-calculus to study quantum mechanics in 2000 and came up with the nonexchange theories [19]. Quantum calculus is not only used in quantum mechanics but also plays an important role in economics [20], dynamic system and quantum model [21], heat and wave equation [22], sampling signal analysis theory [23], and so on. Because of its wide application, fractional q-difference equation has entered the field of view of researchers [24–40].In [25], the following q-difference boundary value problem (BVP) is discussed:
(1)Dqαut+ft,ut=0,t∈0,1,u0=u1=0,where 0<q<1, and 1<α⩽2. By means of the Guo–Krasnoselskii’s fixed point theorem, the authors proved the existence of nontrivial solutions to the fractional q-difference equation with Dirichlet boundary condition, which is one of the three well-known and fundamental boundary conditions. Thereafter, more and more researchers began to study the existence and uniqueness of solutions to fractional q-difference boundary value problem.In 2020, the authors [39] considered the following fractional q-difference system with four-point boundary conditions by using the monotone iterative approach:
(2)Dqαut+ft,vt=0,t∈0,1,Dqβvt+ft,ut=0,t∈0,1,u0=0,u1=γ1uη1,v0=0,v1=γ2vη2,where 1<β⩽α⩽2, 0<η1,η2<1, 0<γ1η1α−1<1, and 0<γ2η2β−1<1. As far as we all know, lots of the papers studied single fractional q-difference equation. This work attracted our attention to fractional q-difference system.Recently in [29], based on the Leray–Schauder alternative principle and Guo–Krasnoselskii’s fixed point theorem, the authors investigated the existence and uniqueness of nonnegative solutions of the nonlinear fractional q-difference boundary value problem:
(3)cDqσkt+wt,kt,cDqζkt=0,k0=k″0=0,k′r=λk″r,λ>0,where t∈J≔0,1 and 0<q<1, and w:0,1×ℝ2⟶ℝ is a given function with 0,1,2<σ<3,ζ,σ∈J. They presented the global convergence of the proposed method with the line searches. In addition, they built some algorithms to assist the proof process. The results of numerical experiments demonstrated the effectiveness of the proposed algorithm.In [27], the aim is to provide the conclusion of the existence and uniqueness of solutions for the following mixed fractional q-difference boundary value problem:
(4)Dqαut+ft,ut=0,t∈0,1,u0=cDqβu0=cDqβu1,where 0<β⩽1, 2<α<2+β, and Dqα,cDqβ are the Riemann–Liouville fractional q-derivative and Caputo fractional q-derivative of order α and β. Using the Guo–Krasnoselskii’s fixed point theorem and the Banach contraction mapping principle as well as the Schaefer’s fixed point theorem, the authors obtained the existence and uniqueness of solutions in cone.In the existing literatures, a large number of papers focus on the subject about equations or systems under the definition of a single type ofq-derivative, Riemann-Liouville, or Caputo derivative. In [27], the model of mixed q-derivative is studied for the first time, in which the form of the equation is concise and the boundary conditions are simple. However, as the models in the real world are extremely complex, such simple models cannot describe complex dynamic processes under some conditions. Therefore, we are more interested in the improvement of the model of mixed q-derivative which is more in line with the real needs. In [10], the authors study a system with single derivative, but it is a coupled system with the multistrip and multipoint mixed boundary conditions. We know that the multistrip and multipoint mixed boundary conditions highly summarize the characteristics of boundary conditions in the existing studies and have more research significance. In addition, the coupled derivative system can better depict the interaction between different factors in the real model.As a consequence, motivated by aforementioned work, we consider that there is no relevant research results about the coupled system under mixedq-derivative with complex boundary conditions, so we fill in this part of the gap. We investigate the following coupled nonlinear fractional q-difference system subject to the multistrip and multipoint mixed boundary conditions, where the Riemann-Liouville fractional q-derivative and Caputo fractional q-derivative are applied to the q-difference system and the boundary conditions separately:
(5)Dqα1ut+f1t,ut,vt=0,t∈0,1,Dqα2vt+f2t,ut,vt=0,t∈0,1,subject to the boundary conditions
(6)u0=0,v0=0,cDqα1−1u1=∑i=1mλ1iIqβ1ivξi+∑j=1nb1jvηj,cDqα2−1v1=∑i=1mλ2iIqβ2iuξi+∑j=1nb2juηj,where cDqα,Dqα are the Caputo fractional q-derivative and Riemann-Liouville fractional q-derivative of order α, respectively, and Iqβki is the Riemann-Liouville fractional q-integral of order βki, for k=1,2 and i=1,2,⋯,m. 1⩽α1,α2⩽2; λ1i,λ2i⩾0, β1i,β2i>0, and 0⩽ξi⩽1, for i=1,2,⋯,m; b1j,b2j⩾0, and 0⩽ηj⩽1, for j=1,2,⋯,n.At present, there are many theoretical methods to solve the boundary value problem of fractionalq-difference equations. The majority use all kinds of the fixed point theorem methods, such as the Guo–Krasnoselskii’s fixed point theorem [25–29], the Leray–Schauder alternative principle, and the Banach contraction mapping principle [30–34], and less use monotone iteration techniques [35–39]. All these methods can effectively study the existence of solutions, but the reason why we chose to use the monotone iterative method is that it has more advantages than other methods, which can not only prove the existence of positive solutions but also obtain numerical approximate solutions within certain limits of error.This article expands from the following aspects: in Section2, we introduce some fundamental definitions and lemmas. Also, we discuss some crucial results and their proofs. Section 3 contains the main conclusion: the existence results of monotone iterative positive solutions. In addition, two examples are given to prove our result.


## 2. Preliminaries

Here, we list some basic notion, lemmas, and some auxiliary results for the proof, which will be used in the next section.Forq∈0,1 and a∈ℝ, define
(7)aq=1−qa1−q.Theq-analogue of the power function is
(8)a−bq0=1,(9)a−bqk=∏i=0k−1a−bqi,k∈ℕ,a,b∈ℝ.Generally, ifα∈ℝ, there is
(10)a−bqα=aα∏i=0∞a−bqia−bqi+α.It is clear thataα=aα for b=0 and 0α=0 for α≥0.Theq-Gamma function is given by
(11)Γqx=1−qx−11−qx−1,x∈ℝ\0,−1,−2,⋯,then, we have Γqx+1=xqΓqx.Forx,y>0, we have
(12)Bqx,y=∫01tx−11−qty−1dqt,especially
(13)Bqx,y=ΓqxΓqyΓqx+y.Theq-derivative of a function f is defined by
(14)Dqfx=fqx−fxq−1x,(15)Dqf0=limx⟶0Dqfx,and the q-derivative of higher order by
(16)Dq0fx=fx,(17)Dqnfx=DqDqn−1fx,n∈ℕ.Theq-integral of a function f defined on the interval 0,b is given by
(18)Iqfx=∫0xfsdqs=x1−q∑k=0∞fxqkqk,x∈0,b.Ifa∈0,b and f is defined in the interval 0,b; then, its integral from a to b is defined by
(19)∫abfsdqs=∫0bfsdqs−∫0afsdqs,and similarly the q-integral of higher order is given by
(20)Iq0fx=fx,(21)Iqnfx=IqIqn−1fx,n∈ℕ.Definition 1 (see [41]).
Letα≥0 and f be a real function defined on a certain interval a,b. The Riemann-Liouville fractional q-integral of order α is defined by
(22)Iq0ft=ft,(23)Iqαft=1Γqα∫0tt−qsα−1fsdqs,α>0.Definition 2 (see [41]).
The fractionalq-derivative of the Riemann-Liouville type of order α≥0 of a continuous and differential function f on the interval a,b is given by
(24)Dq0ft=ft,(25)Dqαft=DqlIql−αft,α>0,where l is the smallest integer greater than or equal to α.Definition 3 (see [41]).
Letα≥0, and the Caputo fractional q-derivatives of f be defined by
(26)cDqαft=Iql−αDqlft,where l is the smallest integer greater than or equal to α.Lemma 4 (see [42]).
Letα,β≥0 and f:a,b⟶ℝ be a continuous function defined on a,b and its derivative exist. Then, the following formulas hold:
(27)DqαIqαft=ft,(28)IqαIqβft=Iqα+βft.Lemma 5 (see [42]).
Letα>0 and p be a positive integer. Then, the following equality holds:
(29)IqαDqpft=DqpIqαft−∑k=0p−1tα−p+kΓqα−p+k+1Dqkf0.Lemma 6 (see [18, 42]).
Letα>0 and n=α+1. Then, we have
(30)IqαcDqαft=ft+c0+c1t+c2t2+⋯+cn−1tn−1,where c0,c1,⋯,cn−1 are some constants.
For convenience, we denote(31)l1=1Γqα2∑i=1mλ1iΓqβ1i∫0ξiξi−qsβ1i−1sα2−1dqs+∑j=1nb1jηjα2−1,l2=1Γqα1∑i=1mλ2iΓqβ2i∫0ξiξi−qsβ2i−1sα1−1dqs+∑j=1nb2jηjα1−1.In the forthcoming analysis, we always need the following assumptions:F11⩽αk⩽2,βki>0, for k=1,2 and i=1,2,⋯,mF20⩽ηj,ξi⩽1,λ1i,λ2i⩾0,b1j,b2j⩾0, for i=1,2,⋯,m, j=1,2,⋯,nF31−l1l2>0, where l1,l2 are defined by (12)F4fk:0,1×0,+∞2⟶0,+∞ is continuous (k=1,2)Subject to BVP (5) and (6), we consider a corresponding linear differential system as follows and establish the expression of the corresponding Green’s functions.Lemma 7.
Assume thatF1-F3 hold. For h1,h2∈C0,1, the fractional differential system
(32)Dqα1ut+h1t=0,t∈0,1,Dqα2vt+h2t=0,t∈0,1,with boundary condition (6) has an integral representation
(33)ut=∫01K1t,qsh1sdqs+∫01H1t,qsh2sdqs,vt=∫01K2t,qsh2sdqs+∫01H2t,qsh1sdqs,where
(34)K1t,qs=g1t,qs+l1tα1−1Γqα11−l1l2∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1g1τ,qsdτ+∑j=1nb2jg1ηj,qs,H1t,qs=tα1−1Γqα11−l1l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1g2τ,qsdqτ+∑j=1nb1jg2ηj,qs,(35)K2t,qs=g2t,qs+l2tα2−1Γqα21−l1l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1g2τ,qsdτ+∑j=1nb1jg2ηj,qs,H2t,qs=tα2−1Γqα21−l1l2∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1g1τ,qsdqτ+∑j=1nb2jg1ηj,qs,and for k=1,2,
(36)gkt,qs=1Γqαktαk−1−t−qsαk−1,0⩽qs⩽t⩽1,tαk−1,0⩽t⩽qs⩽1.Proof.
From Lemma5, we can reduce (32) to the following equivalent integral equations:
(37)ut=−Iqα1h1t+c11tα1−1+c12tα1−2,vt=−Iqα2h2t+c21tα2−1+c22tα2−2,where c11,c12,c21,c22 are constants.
Fromu0=v0=0, we obtain c12=c22=0. By using Lemma 6, we get
(38)cDqα1−1ut=−cDqα1−1Iqα1h1t+c11cDqα1−1tα1−1=−Iqh1t+c11α1−1qIq2−α1tα1−2,cDqα2−1vt=−cDqα2−1Iqα2h2t+c21cDqα2−1tα2−1=−Iqh2t+c21α2−1qIq2−α2tα2−2,
Then, from (12), we get
(39)cDqα1−1u1=−Iqh11+c11α1−1qIq2−α11=−∫01h1sdqs+c11Γqα1,cDqα2−1v1=−Iqh21+c21α2−1qIq2−α21=−∫01h2sdqs+c21Γqα2,
From the rest of the condition of (6), it can be obtained that
(40)c11=1Γqα1∑i=1mλ1iIqβ1ivξi+∑j=1nb1jvηj+∫01h1sdqs,c21=1Γqα2∑i=1mλ2iIqβ2iuξi+∑j=1nb2juηj+∫01h2sdqs,
Further, we can reduce (37) to
(41)ut=tα1−1Γα1∑i=1mλ1iΓqβ1i∫0ξiξi−qsβ1i−1vsdqs+∑j=1nb1jvηj+∫01g1t,qsh1sdqs,vt=tα2−1Γα2∑i=1mλ2iΓqβ2i∫0ξiξi−qsβ2i−1usdqs+∑j=1nb2juηj+∫01g2t,qsh2sdqs,where gkt,qsk=1,2 is introduced by (36). Then, we can get
(42)∑i=1mλ1iΓqβ1i∫0ξiξi−qsβ1i−1vsdqs+∑j=1nb1jvηj=1Γqα2∑i=1mλ1iΓqβ1i∫0ξiξi−qsβ1i−1sα2−1dqs+∑j=1nb1jηjα2−1·∑i=1mλ2iΓqβ2i∫0ξiξi−qsβ2i−1usdqs+∑j=1nb2juηj+∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1∫01g2τ,qsh2sdqsdqτ+∑j=1nb1j∫01g2ηj,qsh2sdqs,(43)∑i=1mλ2iΓqβ2i∫0ξiξi−qsβ2i−1usdqs+∑j=1nb2juηj=1Γqα1∑i=1mλ2iΓqβ2i∫0ξiξi−qsβ2i−1sα1−1dqs+∑j=1nb2jηjα1−1·∑i=1mλ1iΓqβ1i∫0ξiξi−qsβ1i−1vsdqs+∑j=1nb1jvηj+∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1∫01g1τ,qsh1sdqsdqτ+∑j=1nb2j∫01g1ηj,qsh1sdqs.
Combining (42) and (43), it can be seen that
(44)∑i=1mλ1iΓqβ1i∫0ξiξi−qsβ1i−1vsdqs+∑j=1nb1jvηj=11−l1l2l1∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1∫01g1τ,qsh1sdqsdqτ+∑j=1nb2j∫01g1ηj,qsh1sdqs+∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1∫01g2τ,qsh2sdqsdqτ+∑j=1nb1j∫01g2ηj,qsh2sdqs,∑i=1mλ2iΓqβ2i∫0ξiξi−qsβ2i−1usdqs+∑j=1nb2juηj=11−l1l2l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1∫01g2τ,qsh2sdqsdqτ+∑j=1nb1j∫01g2ηj,qsh2sdqs+∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1∫01g1τ,qsh1sdqsdqτ+∑j=1nb2j∫01g1ηj,qsh1sdqs,where lkk=1,2 is defined by (31). From (41) and (44), we have
(45)ut=∫01g1t,qsh1sdqs+tα1−1Γqα11−l1l2∫01l1∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1g1τ,qsdqτ+∑j=1nb2jg1ηj,qsh1sdqs+∫01∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1g2τ,qsdqτ+∑j=1nb1jg2ηj,qsh2sdqs=∫01K1t,qsh1sdqs+∫01H1t,qsh2sdqs,where K1t,qs and H1t,qs are introduced by (34). Similarly, we also have
(46)vt=∫01g2t,qsh2sdqs+tα2−1Γqα21−l1l2∫01l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1g2τ,qsdqτ+∑j=1nb1jg2ηj,qsh2sdqs+∫01∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1g1τ,qsdqτ+∑j=1nb2jg1ηj,qsh1sdqs=∫01K2t,qsh2sdqs+∫01H2t,qsh1sdqs,where K2t,qs and H2t,qs are also given by (35).
This completes the proof of the lemma.Lemma 8.
Assume thatF1 holds. Then, the functions gkt,s, for k=1,2, defined by (35) have the following properties:
(47)0⩽gkt,qs⩽1Γqαktαk−1,fort,qs∈0,1.Proof.
For0⩽qs⩽t⩽1, we have
(48)gkt,qs=1Γqαktαk−1−t−qsαk−1⩾0,(49)gkt,qs=1Γqαktαk−1−t−qsαk−1⩽1Γqαktαk−1.
For0⩽t⩽qs⩽1, we have
(50)0⩽gkt,qs=1Γqαktαk−1.
This completes the proof of the lemma.For convenience, we denote(51)ϱ1=1Γqα11+l1Γqα11−l1l2∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1τα1−1dqτ+∑j=1nb2jηjα1−1,(52)ϱ2=1Γqα21+l2Γqα21−l1l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1τα2−1dqτ+∑j=1nb1jηjα2−1,(53)ρ1=1Γqα1Γqα21−l1l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1τα2−1dqτ+∑j=1nb1jηjα2−1,(54)ρ2=1Γqα1Γqα21−l1l2∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1τα1−1dqτ+∑j=1nb2jηjα1−1.Lemma 9.
Assume thatF1-F3 hold. Then, for t,s∈0,1×0,1, the functions Kit,s and Hit,s, for i=1,2, defined by (33) and (34) satisfy the following results:
(55)0⩽K1t,qs⩽ϱ1tα1−1,0⩽K2t,qs⩽ϱ2tα2−1,(56)0⩽H1t,qs⩽ρ1tα1−1,0⩽H2t,qs⩽ρ2tα2−1.Proof.
(1)
According toF3, Lemma 8, and the definition of Kit,qsk=1,2, we obtain(57)0⩽K1t,qs=g1t,qs+l1tα1−1Γqα11−l1l2∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1g1τ,qsdqτ+∑j=1nb2jg1ηj,qs⩽tα1−1Γqα1+l1tα1−1Γqα11−l1l2∑i=1mλ2iΓqβ2iΓqα1∫0ξiξi−qτβ2i−1τα1−1dqτ+1Γqα1∑j=1nb2jηjα1−1=tα1−1Γqα11+l1Γqα11−l1l2∑i=1mλ2iΓqβ2i∫0ξiξi−qτβ2i−1τα1−1dqτ+∑j=1nb2jηjα1−1=ϱ1tα1−1,where ϱ1 is defined by (51). Similarly, we get
(58)0⩽K2t,qs⩽ϱ2tα2−1,where ϱ2 is defined by (52).
(2)
According toF3, Lemma 8, and the definition of Hkt,sk=1,2, we can also obtain(59)0⩽H1t,qs=tα1−1Γqα11−l1l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1g2τ,qsdqτ+∑j=1nb1jg2ηj,qs⩽tα1−1Γqα11−l1l2∑i=1mλ1iΓqβ1iΓqα2∫0ξiξi−qτβ1i−1τα2−1dqτ+1Γqα2∑j=1nb1jηjα2−1=tα1−1Γqα1Γqα21−l1l2∑i=1mλ1iΓqβ1i∫0ξiξi−qτβ1i−1τα2−1dqτ+∑j=1nb1jηjα2−1=ρ1tα1−1,where ρ1 is defined by (53). Analogously, we get
(60)0⩽H2t,s⩽ρ2tα2−1,where ρ2 is defined by (54).
This completes the proof of the lemma.LetX=uu∈C0,1 be a Banach space endowed with the norm
(61)uX=maxt∈0,1ut.Also, letY=vv∈C0,1 be a Banach space endowed with the norm
(62)vY=maxt∈0,1vt.Then, we introduce the product spaceX×Y,u,v endowed with the norm u,v=maxuX,vY and define a partial order over the product space
(63)u1v1⩾u2v2,while u1t⩾u2t, v1t⩾v2t, and t∈0,1. And we can know that X×Y,u,v is also a Banach space.Further, we define a coneP⊂X×Y by P=u,v∈X×Y:ut⩾0,vt⩾0,t∈0,1. For all u,v∈P, in view of Lemma 7 and F4, let T:P⟶P be the operator defined by
(64)Tu,vt=T1u,vtT2u,vt,where
(65)T1u,vt=∫01K1t,qsf1u,vsdqs+∫01H1t,qsf2u,vsdqs,(66)T2u,vt=∫01K2t,qsf2u,vsdqs+∫01H2t,qsf1u,vsdqs,and for convenience, we set
(67)f1u,vs≜f1s,us,vs,f2u,vs≜f2s,us,vs.Lemma 10.
The operatorT:P⟶P is completely continuous.Proof.
There are three steps to complete the whole proof.(a)
The operatorT:P⟶P is continuous
By the continuity of the functionsK1t,qs,K2t,qs,H1t,qs,H2t,qs,f1, and f2, the operator T is continuous.
(b)
The operatorT:P⟶P is uniformly bounded
LetΩ be any bounded subset of P. There exists a positive constant M satisfying the inequality
(68)maxt∈0,1f1u,vt,f2u,vt⩽M,∀u,v∈Ω.
For anyu,v∈Ω, combining Lemma 9 with (65) and (68), we get
(69)T1u,v=maxt∈0,1∫01K1t,qsf1u,vsdqs+∫01H1t,qsf2u,vsdqs⩽ϱ1tα1−1∫01f1u,vsdqs+ρ1tα1−1∫01f2u,vsdqs⩽Mϱ1tα1−1+Mρ1tα1−1⩽Mϱ1+ρ1.
So, foru,v∈Ω, T1 is uniformly bounded. Furthermore, we get T2 is also uniformly bounded. Thus, it follows from the above inequalities that the operator T is uniformly bounded.
(c)
The operatorT:P⟶P is equicontinous
For anyu,v∈Ω and t1,t2∈0,1, in view of Lemma 9, (65), and (68), we infer that
(70)T1u,vt2−T1u,vt1⩽∫01K1t2,qsf1u,vsdqs−∫01K1t1,qsf1u,vsdqs+∫01H1t2,qsf2u,vsdqs−∫01H1t1,qsf2u,vsdqs⩽M∫01K1t2,qs−K1t1,qsdqs+∫01H1t2,qs−H1t1,qsdqs⩽M∫01ϱ1t2α1−1−t1α1−1dqs+∫01ρ1t2α1−1−t1α1−1dqs=Mϱ1+ρ1t2α1−1−t1α1−1⩽Mϱ1+ρ1α1−1t2−t1⟶0ast2⟶t1.
Therefore,T1 is equicontinous for all u,v∈Ω. Similarly, we can get T2 is equicontinous for all u,v∈Ω. As a consequence, the operator T is equicontinous for all u,v∈Ω.
From the above steps, we get the operatorT:P⟶P is completely continuous. The proof is completed.


## 3. Existence Results of Monotone Iterative Positive Solutions

Theorem 11.
Assume thatF1-F4 hold. Let Ai,Li,Mi,i=1,2,l and γ<1/2 be positive constants satisfying
(71)γ=supϱ1L1+ρ1M1,ϱ1L2+ρ1M2,ϱ2M1+ρ2L1,ϱ2M2+ρ2L2,(72)l=maxA1ϱ1+A2ρ1,A2ϱ2+A1ρ2,
S1 For t∈0,1, fjt,x1,x2 is increasing in xi∈0,li=1,2 for j=1,2
S2max0⩽t⩽1fit,l,l⩽Ai, fit,0,0≡0, 0⩽t⩽1 for i=1,2
S3 For any xi,yi∈C0,1 with xi⩽yii=1,2, there exist nonnegative constants Li and Mii=1,2 such that
(73)0⩽f1t,y1,y2−f1t,x1,x2⩽L1y1−x1+L2y2−x2,t∈0,1,0⩽f2t,y1,y2−f2t,x1,x2⩽M1y1−x1+M2y2−x2,t∈0,1.
Then BVP (5) and (6) has iterative positive solutions u∗,v∗ and w∗,z∗ satisfying 0⩽u∗,v∗⩽l, which limn⟶∞un,vn=u∗,v∗,(74)un,vn=Tun−1,vn−1=T1un−1,vn−1T2un−1,vn−1,n=1,2,⋯,(75)u0tv0t=ltα1−1ltα2−1;
0⩽w∗,z∗X×Y⩽l, which limn⟶∞wn,zn=w∗,z∗,(76)wn,zn=Twn−1,zn−1=T1wn−1,zn−1T2wn−1,zn−1,n=1,2,⋯,(77)w0tz0t=00.
Moreover, the error estimate of the solution is given by(78)un−u∗⩽2γn21−2γu0−u1+v0−v1,(79)vn−v∗⩽2γn21−2γu0−u1+v0−v1,(80)w∗−wn⩽2γn21−2γw1−w0+z1−z0,(81)z∗−zn⩽2γn21−2γw1−w0+z1−z0.Proof.
DenotePl=u,v∈X×Y:u,vX×Y⩽l, where l is introduced by (72). In the following, we first prove that T:Pl⟶Pl. For any u,v∈Pl, then for t∈0,1, we have
(82)0⩽ut⩽u⩽l,0⩽vt⩽v⩽l.
So, fort∈0,1 and i=1,2, by S1 and S2, we get
(83)0⩽fit,ut,vt⩽max0⩽t⩽1fit,l,l⩽Ai.
Consequently, fort∈0,1, in view of Lemma 9 and (83), we have
(84)T1u,v=max0⩽t⩽1∫01K1t,qsf1u,vsdqs+∫01H1t,qsf2u,vsdqs⩽max0⩽t⩽1ϱ1tα1−1∫01f1u,vsdqs+ρ1tα1−1∫01f2u,vsdqs⩽max0⩽t⩽1A1ϱ1+A2ρ1tα1−1⩽A1ϱ1+A2ρ1⩽l,(85)T2u,v=max0⩽t⩽1∫01K2t,qsf2u,vsdqs+∫01H2t,qsf1u,vsdqs⩽max0⩽t⩽1ϱ2tα2−1∫01f2u,vsdqs+ρ2tα2−1∫01f1u,vsdqs⩽max0⩽t⩽1A2ϱ2+A1ρ2tα2−1⩽A2ϱ2+A1ρ2⩽l.
As a result, we obtain(86)Tu,v⩽l,and thus T:Pl⟶Pl.
According to (75) and (77), it is obvious that u0,v0,w0,z0∈Pl. By using the completely continuous operator T, we define the sequences un,vn and wn,zn as un,vn=Tun−1,vn−1, wn,zn=Twn−1,zn−1, for n=1,2,⋯. Since T:Pl⟶Pl, we get that un,vn,wn,zn∈Pl, for n=1,2,⋯.
Fort∈0,1, by the definition of the iterative scheme, we have
(87)T1u0,v0t=∫01K1t,qsf1u0,v0sdqs+∫01H1t,qsf2u0,v0sdqs⩽∫01ϱ1tα1−1f1u0,v0sdqs+∫01ρ1tα1−1f2u0,v0sdqs⩽A1ϱ1+A2ρ1tα1−1⩽ltα1−1=u0t.
Similarly,(88)T2u0,v0t=∫01K2t,qsf2u0,v0sdqs+∫01H2t,qsf1u0,v0sdqs⩽A2ϱ2+A1ρ2tα2−1⩽ltα2−1=v0t.
From (87) and (88), we can get
(89)u1tv1t=T1u0,v0tT2u0,v0t⩽u0tv0t.
Thus, fort⩾1, from (89)and (S1), we do the second iteration
(90)u2tv2t=T1u1,v1tT2u1,v1t⩽T1u0,v0tT2u0,v0t=u1tv1t.
By induction, forn=0,1,2,⋯, we get
(91)un+1tvn+1t⩽untvnt,for0⩽t⩽1.
So we assert thatlimn⟶∞un,vn=u∗,v∗ and Tu∗,v∗=u∗,v∗, since T is completely continuous and un+1,vn+1=Tun,vn.
For the sequencewn,znn=1∞, we apply a similar argument. For t∈0,1, we have
(92)w1tz1t=T1w0,z0tT2w0,z0t=∫01K1t,qsf1w0,z0sds+∫01H1t,qsf2w0,z0sds,∫01K2t,qsf2w0,z0sds+∫01H2t,qsf1w0,z0sds⩾00=w0tz0t.
To sum up, for0⩽t⩽1, we have
(93)w2tz2t=T1w1,z1tT2w1,z1t⩾T1w0,z0tT2w0,z0t=w1tz1t.
Analogously, forn=1,2,⋯, we get
(94)wn+1tzn+1t⩾wntznt,for0⩽t⩽1.
Similarly, we can also assert thatlimn⟶∞wn,zn=w∗,z∗ and Tw∗,z∗=w∗,z∗, since T is completely continuous and wn+1,zn+1=Twn,zn.
Consequently, there existu∗,v∗ and w∗,z∗ in Pl which are nonnegative extremal solutions of BVP (5) and (6). Since fkt,0,0≡0,∀t∈0,1, for k=1,2; then, zero is not a solution of problem (5). It is obvious that u∗,v∗t>0, w∗,z∗t>0.
It follows from Lemma9 and (S3) that
(95)un−un+1=T1un−1,vn−1−T1un,vn=max0⩽t⩽1∫01K1t,qsf1un−1,vn−1sdqs+∫01H1t,qsf2un−1,vn−1sdqs−∫01K1t,qsf1un,vnsdqs−∫01H1t,qsf2un,vnsdqs⩽max0⩽t⩽1ϱ1tα1−1∫01f1un−1,vn−1s−f1un,vnsdqs+ρ1tα1−1∫01f2un−1,vn−1s−f2un,vnsdqs⩽ϱ1∫01L1un−1−un+L2vn−1−vndqs+ρ1∫01M1un−1−un+M2vn−1−vndqs⩽ϱ1L1un−1−un+L2vn−1−vn+ρ1M1un−1−un+M2vn−1−vn=ϱ1L1+ρ1M1un−1−un+ϱ1L2+ρ1M2vn−1−vn⩽γun−1−un+vn−1−vn.
Homoplastically, we have(96)vn−vn+1⩽γun−1−un+vn−1−vn,wn+1−wn⩽γwn−wn−1+zn−zn−1,zn+1−zn⩽γwn−wn−1+zn−zn−1.
Combining (95) and (96), we get
(97)u1−u2⩽γu0−u1+v0−v1,v1−v2⩽γu0−u1+v0−v1;u2−u3⩽γu1−u2+v1−v2⩽γ·2γu0−u1+v0−v1,v2−v3⩽γu1−u2+v1−v2⩽γ·2γu0−u1+v0−v1;⋯⋯
Then, we obtain(98)un−un+1⩽2γn2u0−u1+v0−v1,vn−vn+1⩽2γn2u0−u1+v0−v1.
Therefore, forn,p∈ℤ+, it comes that
(99)un−un+p⩽un−un+1+un+1−un+2+⋯+un+p−1−un+p⩽2γn2+2γn+12+⋯+2γn+p−12·u0−u1+v0−v1=2γn2·2γ0+2γ1+⋯+2γp−1·u0−u1+v0−v1=2γn2·1−2γp1−2γ·u0−u1+v0−v1.
For0<γ<1/2, we can obtain un−u∗⩽2γn/21−2γ·u0−u1+v0−v1 as p⟶∞, and similarly, un−u∗⩽2γn/21−2γ·u0−u1+v0−v1.
In the same manner, we can see thatw∗−wn⩽2γn/21−2γ·w1−w0+z1−z0, and z∗−zn⩽2γn/21−2γ·w1−w0+z1−z0.
The proof is completed.When the system degenerates into the following form and the boundary conditions remain unchanged, we still have the same conclusion:(100)Dqα1ut+f1t,ut=0,t∈0,1,Dqα2vt+f2t,vt=0,t∈0,1,Corollary 12.
Assume thatF1-F4 hold. Let Aii=1,2,L,M,l and γ<1/2 be positive constants satisfying
(101)γ=supϱ1L,ρ1M,ϱ2M,ρ2L,(102)l=maxA1ϱ1+A2ρ1,A2ϱ2+A1ρ2,
S1 For t∈0,1, fjt,x is increasing in x∈0,l for j=1,2
S2max0⩽t⩽1fit,l⩽Ai, fit,0≡0, 0⩽t⩽1 for i=1,2
S3 For any x,y∈C0,1 with x⩽y, there exist nonnegative constants L and M such that
(103)0⩽f1t,y−f1t,x⩽Ly−x,t∈0,1,(104)0⩽f2t,y−f2t,x⩽My−x,t∈0,1.Then, BVP (100) and (6) has iterative positive solutions u∗,v∗ and w∗,z∗ satisfying the same conclusion as Theorem 11.The absence of coupling in the BVP (100) and (6) results in a reduction of the value of the parameter γ, so the overall iteration accuracy is greatly improved. By comparing the error estimates and the plot of the solutions in the latter Example 3 with (139), we can verify that the BVP (100) and (6) has fewer iterations but higher accuracy.


## 4. Examples

Example 1.
Fort∈0,1, consider the following fractional differential system:
(105)D0.51.9ut+0.09t2+sin2t5ut+sint5vt=0,D0.51.8vt+0.11t+sint10ut+sin2t10vt=0,with the coupled integral and discrete mixed boundary conditions:
(106)u0=0,cD0.50.9u1=∑i=12λ1iI0.5β1ivξi+∑j=12b1jvηj,v1=0,cD0.50.8v1=∑i=12λ2iI0.5β2iuξi+∑j=12b2juηj.In this model, we set(107)λ11=0.25,λ21=0.2,β11=1.5,β21=1.4,ξ1=0.25,b11=0.33,b21=0.17,η1=0.33,(108)λ12=0.5,λ22=0.1,β12=2.5,β22=2.4,ξ2=0.75,b12=0.67,b22=0.83,η2=0.67.It is obvious thatF1 and F2 hold. By calculation, we get
(109)l1<0.753252,l2<0.681115,(110)ϱ1<2.106456,ϱ2<2.152105,(111)ρ1<1.586692,ρ2<1.465830.SettingA1=0.5,A2=0.3, we get l=1.529236. And for any ui,vi∈C0,1 with ui⩽vii=1,2, we have
(112)0⩽f1t,v1,v2−f1t,u1,u2⩽sin215u2−u1+sin15v2−v1,t∈0,1(113)0⩽f2t,v1,v2−f2t,u1,u2⩽sin110u2−u1+sin2110v2−v1,t∈0,1,so L1=0.039470,L2=0.198669,M1=0.099833,M2=0.009967. Thus, it can be obtained that γ<1/2. Then, all the hypotheses of Theorem 11 are satisfied. BVP (105) and (106) have monotone positive solutions u∗,v∗ and w∗,z∗, which can be approximated by the following iterative sequences:
(114)un+1tvn+1t=T1un,vntT2un,vnt=∫01K1t,qsf1un,vnsdqs+∫01H1t,qsf2un,vnsdqs∫01K2t,qsf2un,vnsdqs+∫01H2t,qsf1un,vnsdqs,(115)wn+1tzn+1t=∫01K1t,qsf1wn,znsds+∫01H1t,qsf2wn,znsds∫01K2t,qsf2wn,znsds+∫01H2t,qsf1wn,znsds,and the initial values are
(116)u0tv0t=ltα1−1ltα2−1=1.529236t0.91.529236t0.8,(117)w0tz0t=00.Example 2.
Fort∈0,1, consider the following fractional differential system:
(118)D0.52ut+0.09t2+0.03tut+0.14t2vt=0,D0.52vt+0.11t+0.07t2ut+0.01tvt=0,with the coupled integral and discrete mixed boundary conditions:
(119)u0=0,cD0.51u1=∑i=12λ1iI0.5β1ivξi+∑j=12b1jvηj,v1=0,cD0.51v1=∑i=12λ2iI0.5β2iuξi+∑j=12b2juηj.In this model, we set(120)λ11=0.25,λ21=0.2,β11=1,β21=1,ξ1=0.25,b11=0.33,b21=0.17,η1=0.33,(121)λ12=0.75,λ22=0.8,β12=1,β22=1,ξ2=0.75,b12=0.67,b22=0.83,η2=0.67.It is obvious thatF1 and F2 hold. By calculation, we get
(122)l1=0.77655,l2=0.84345,(123)ϱ1=2.8983919,ϱ2=2.8983919,(124)ρ1=2.2507462,ρ2=2.4446487.SettingA1=0.5,A2=0.3, we get l=2.1244199. And for any ui,vi∈C0,1 with ui⩽vii=1,2, we have
(125)0⩽f1t,v1,v2−f1t,u1,u2⩽0.03tu2−u1+0.14t2v2−v1,t∈0,1,(126)0⩽f2t,v1,v2−f2t,u1,u2⩽0.07t2u2−u1+0.01tv2−v1,t∈0,1,soL1=0.03,L2=0.14,M1=0.07,M2=0.01. Thus, it can be obtained that γ=0.4282824<1/2. Then, all the hypotheses of Theorem 11 are satisfied. BVP (118) and (119) have monotone positive solutions u∗,v∗ and w∗,z∗, which can be approximated by the following iterative sequences:
(127)un+1tvn+1t=T1un,vntT2un,vnt=∫01K1t,qsf1un,vnsdqs+∫01H1t,qsf2un,vnsdqs∫01K2t,qsf2un,vnsdqs+∫01H2t,qsf1un,vnsdqs,(128)wn+1tzn+1t=∫01K1t,qsf1wn,znsds+∫01H1t,qsf2wn,znsds∫01K2t,qsf2wn,znsds+∫01H2t,qsf1wn,znsds,and the initial values are
(129)u0tv0t=ltα1−1ltα2−1=2.1244199t2.1244199t,(130)w0tz0t=00.From Figure1(a), we can see that the images of the two functions u4 and u5 almost coincide, indicating that when the fifth iteration occurs, u5 is already very close to the extremal solution u∗. So we approximate that
(131)u∗t≈u5t.Figure 1
A plot of the solutions of Example2.
(a)
un(b)
vn(c)
wn(d)
znAnalogously, from Figures1(b)–1(d), we can also get that
(132)v∗t≈v5t,(133)w∗t≈w5t,(134)z∗t≈z5t.The error estimates are as follows(135)u5t−u∗t⩽2γ521−2γu0t−u1t+v0t−v1t<6.2023006,(136)v5t−v∗t⩽2γ521−2γu0t−u1t+v0t−v1t<6.2023006,(137)w∗t−w5t⩽2γ521−2γw1t−w0t+z1t−z0t<0.1821279,(138)z∗t−z5t⩽2γ521−2γw1t−w0t+z1t−z0t<0.1821279.Example 3.
Fort∈0,1, consider the following fractional differential system:
(139)D0.52ut+0.09t2+0.03tut=0D0.52vt+0.11t+0.01tvt=0,with the coupled integral and discrete mixed boundary conditions (119). Then, we can also get (122) by calculation. Setting A1=0.5,A2=0.3, we get l=2.1244199. And for any ui,vi∈C0,1 with u⩽v, we have
(140)0⩽f1t,v−f1t,u⩽0.03tv−u,t∈0,1(141)0⩽f2t,v−f2t,u⩽0.01tv−u,t∈0,1,so L=0.03,M=0.01. Thus, it can be obtained that γ=0.0869518<1/2. Then, all the hypotheses of Corollary 12 are satisfied. BVP (139) and (119) have monotone positive solutions u∗,v∗ and w∗,z∗, which are the same conclusion as (127)–(130) in Example 2.From Figure2, we can also get that
(142)u∗t≈u3t(143)v∗t≈v3t,(144)w∗t≈w3t,(145)z∗t≈z3t.Figure 2
A plot of the solutions of Example3.
(a)
un(b)
vn(c)
wn(d)
znThe error estimates are as follows(146)u3t−u∗t⩽2γ321−2γu0t−u1t+v0t−v1t<0.013027938,(147)v3t−v∗t⩽2γ321−2γu0t−u1t+v0t−v1t<0.013027938,(148)w∗t−w3t⩽2γ321−2γw1t−w0t+z1t−z0t<0.0003606834,(149)z∗t−z3t⩽2γ321−2γw1t−w0t+z1t−z0t<0.0003606834.



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