# The Iterative Positive Solution for a System of Fractionalq-Difference Equations with Four-Point Boundary Conditions

Authors: Chuanzhi Bai; Dandan Yang
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this work, we investigate the following system of fractionalq-difference equations with four-point boundary problems: Dqαut+ft,vt=0,0<t<1;Dqβvt+gt,ut=0,0<t<1;u0=0,u1=γ1uη1;andv0=0,v1=γ2uη2, where Dqα and Dqβ are the fractional Riemann–Liouville q-derivative of order α and β, respectively, 0<q<1, 1<β≤α≤2, 0<η1,η2<1, 0<γ1η1α−1<1, and 0<γ2η2β−1<1. By virtue of monotone iterative approach, the iterative positive solutions are obtained. An example to illustrate our result is given.


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## Body

## 1. Introduction

In [1, 2], Jackson studied the q-difference calculus firstly; since then, many authors have investigated this subject duo to applications of the q-difference calculus in quantum mechanics, particle physics, hypergeometric series, and complex analysis [3, 4]. The extension of q-difference calculus is the fractional q-difference calculus, which was originally investigated by Al-Salam [5] and Agarwal [6]. In the past decade, in many works concerning nonlinear fractional q-difference boundary value problem, the results of the existence and the uniqueness of solutions have been given. In [7], Ferreira considered the existence of positive solutions to the nonlinear fractional q-difference equation:(1)Dqαut+ft,ut=0,0<t<1,1<α≤2,u0=u1=0.In [8], Ferreira studied the existence of positive solutions to the nonlinear fractional q-difference equation:(2)Dqαut+ft,ut=0,0<t<1,2<α≤3,u0=Dqu0=0,Dqu1=β≥0.By using a fixed-point theorem in partially ordered sets, Garzi and Agheli [9] studied the existence and uniqueness of a positive and nondecreasing solution to the fractional q-difference equation:(3)Dqαut+ft,ut=0,0<t<1,3<α≤4,u0=Dqu0=Dq2u0=0,Dq2u1=βDq2uη,where 0<η<1 and 1−βηα−3>0.In [10], Guo and Kang obtained the existence and uniqueness of a positive solution for the fractional q-difference equation of the form(4)Dqαut+ft,ut,ut+gt,ut=0,0<t<1,1<α≤2,u0=0,u1=βuη,by virtue of fixed-point theorems for the mixed monotone operator. Here, 1<α≤2 and 0<βηα−1<1.Recently, by using the monotone iterative approach, in [11], Wang investigated the iterative positive solutions of the following fractional q-difference equations with three-point boundary conditions:(5)Dqαux+λhxfux=0,0<x<1,2<α≤3,u0=Dqu0=Dqu1=0.It should be noted that the existence of positive solutions of problem (5) had been studied by Li et al. [12] by means of a fixed-point theorem in cones. The novel idea of [11] is to find the positive solution.Motivated by the above mentioned works, in this paper, we consider the following system of fractionalq-difference equations with four-point boundary conditions:(6)Dqαut+ft,vt=0,0<t<1,Dqβvt+gt,ut=0,0<t<1,u0=0,u1=γ1uη1,v0=0,v1=γ2uη2,where Dqα and Dqβ are the fractional Riemann–Liouville q-derivative of order α and β, respectively, 0<q<1, 1<β≤α≤2, 0<η1,η2<1, 0<γ1η1α−1<1, and 0<γ2η2β−1<1.By using the monotone iterative approach, in this paper, we will construct two convergent monotone iterative schemes for seeking one coupled positive solution and obtain the coupled positive solution of problem (6). To the best of our knowledge, there is no paper to study the iterative coupled positive solutions for the coupled system of fractional q-difference boundary value problems. It is noted that we may investigate the approximate solutions of problem (6) by numerical approximation algorithms, which will be presented as another paper. For the latest development of numerical approximation algorithms of some boundary value problems, see [13–17] and the references therein.


## 2. Preliminaries

Letq∈0,1, the q-derivative of a function f is defined by(7)Dqfx=fqx−fxq−1x,Dqf0=limx⟶0Dqfx,and q-derivatives of higher order by(8)Dq0fx=fx,Dqnfx=DqDqn−1fx,n∈N.Theq-integral of a function f defined in the interval 0,b is given by(9)Iqfx=∫0xfsdqs=x1−q∑k=0∞fxqkqk,x∈0,b.Similar to the derivatives, the operatorIqn is given by(10)Iq0fx=fx,Iqnfx=IqIqn−1fx,n∈N.Define(11)aq=1−qa1−q,a∈ℝ.Theq-analogue of the power function a−bn with n∈ℕ0 is(12)a−b0=1,a−bn=∏k=0n−1a−bqk,n∈ℕ,a,b∈ℝ.Moreover, ifα∈ℝ, then(13)a−bα=aα∏n=0∞a−bqna−bqα+n.Remark 1.
Ifb=0, then aα=aα. If α>0 and a≤b≤t, then t−aα≥t−bα.
Theq-gamma function [18] is defined by(14)Γqx=1−qx−11−qx−1,x∈ℝ\0,−1,−2,…,and satisfies Γqx+1=xqΓqx.Definition 1.
We sayu∗,v∗ is a solution of system (6), if u∗,v∗ satisfies the first and second equations of (6) and boundary conditions of (6).Definition 2.
(see [19]). Let α>0 and f be a function defined on 0,1. The fractional q-integral of the Riemann–Liouville type is(15)Iqαfx=1Γqα∫0xx−qtα−1ftdqt,x∈0,1.Definition 3.
(see [19]). The fractional q-derivative of the Riemann–Liouville type is defined by(16)Dqαfx=DqnIqn−αfx,α>0,where n is the smallest integer greater than or equal to α.Lemma 1 (see [19]).
Letα,β≥0 and f be a function defined on 0,1. Then, the following formulas hold:(1)
IqβIqαfx=Iqα+βfx,(2)
DqαIqαfx=fx.Lemma 2 (see [13]).
Letα>0 and n be a positive integer. Then, the following equality holds:(17)IqαDqnfx=DqnIqαfx−∑k=0n−1xα−n+kΓqα+k−n+1Dqkf0.By Lemmas1 and 2, Guo and Kang in [10] obtained the following lemma.Lemma 3.
For anyg∈C0,1, the boundary value problem(18)Dqαut+gt=0,0<t<1,u0=0,u1=γ1uη1,has a unique solution:(19)ut=∫01G1t,qsgsdqs,where(20)G1t,qs=tα−11−qsα−1−tα−1γ1η1−qsα−1−t−qsα−11−γ1η1α−1Γqα1−γ1η1α−1,0≤qs≤t≤1,qs≤η1,tα−11−qsα−1−t−qsα−11−γ1η1α−1Γqα1−γ1η1α−1,0≤η1≤qs≤t≤1,tα−11−qsα−1−tα−1γ1η−qsα−1Γqα1−γ1η1α−1,0≤t≤qs≤1,tα−11−qsα−1Γqα1−γ1η1α−1,0≤t≤qs≤1,η1≤qs,is the Green function of BVP (18).Similarly, we have the following.Lemma 4.
For anyh∈C0,1, the boundary value problem(21)Dqβvt+ht=0,0<t<1,v0=0,v1=γ2vη2,has a unique solution:(22)vt=∫01G2t,qshsdqs,where(23)G2t,qs=tβ−11−qsβ−1−tβ−1γ2η2−qsβ−1−t−qsβ−11−γ2η2β−1Γqβ1−γ2η2β−1,0≤qs≤t≤1,qs≤η2,tβ−11−qsβ−1−t−qsβ−11−γ2η2β−1Γqβ1−γ2η2β−1,0≤η2≤qs≤t≤1,tβ−11−qsβ−1−tβ−1γ2η2−qsβ−1Γqβ1−γ2η2β−1,0≤t≤qs≤1,tβ−11−qsβ−1Γqβ1−γ2η2β−1,0≤t≤qs≤1,η2≤qs,is the Green function of BVP (21).Lemma 5.
(see [10]). For G1t,qs and G2t,qs defined as in Lemmas 3 and 4, respectively, we have(i)
G1t,qs and G2t,qs are two continuous functions(ii)
M1qs1−qsα−1/Γqα1−γ1η1α−1tα−1≤G1t,qs≤1−qsα−1/Γqα1−γ1η1α−1tα−1,∀0≤t,s≤1, where 0<M1=min1−γ1η1α−1,γ1η1α−21−η1,γ1η1α−1<1(iii)
M2qs1−qsβ−1/Γqβ1−γ2η2β−1tβ−1≤G2t,qs≤1−qsβ−1/Γqβ1−γ2η2β−1tβ−1,∀0≤t,s≤1, where 0<M2=min1−γ2η2β−1,γ2η2β−21−η2,γ2η2β−1<1


## 3. Main Result

In this paper, we will employ the Banach spaceC0,1, equipped with norm u=supt∈0,1ut for each u∈C0,1. Define two cones P1 and P2 in C0,1 as follows:(24)P1=u∈C0,1there exist two positive numbers0<a1<1<b1,such thata1tα−1≤ut≤b1tα−1,t∈0,1,P2=v∈C0,1there exist two positive numbers0<a2<1<b2,such thata2tβ−1≤vt≤b2tβ−1,t∈0,1.Now, we define the operatorsTi:C0,1⟶C0,1i=1,2 by (25)T1vt=∫01G1t,qsfs,vsdqs,T2ut=∫01G2t,qsgs,usdqs.From Lemmas3 and 4, BVP (6) can be transformed into the following system of integral equations:(26)ut=∫01G1t,qsfs,vsdqs,vt=∫01G2t,qsgs,usdqs,By (26), we know that u∗,v∗ is a solution of (6) if and only if u∗=T1v∗ and v∗=T2u∗.In order to facilitate our investigation, we make the following assumptions:(H1)f∈C0,1×0,∞,0,∞ is nondecreasing with respect to v, and there exists a positive constant σ1>1, such that(27)ft,rv≥rσ1ft,v,∀t∈0,1,v∈0,+∞,r∈0,1.(H2)g∈C0,1×0,∞,0,∞ is nondecreasing with respect to u, and there exists a positive constant 0<σ2<1, such that(28)gt,ru≥rσ2gt,u,∀t∈0,1,u∈0,+∞,r∈0,1.(H3)0<∫011−qsα−1fs,1dqs<+∞.(H4)0<∫011−qsβ−1gs,1dqs<+∞.Remark 2.
The conditions (H1) and (H2) imply that, for∀r>1, we have ft,rv≤rσ1ft,v and gt,ru≤rσ2gt,u.Theorem 1.
Assume that conditions (H1)–(H4) hold and there exist two positive constantsR1 and R2 such that(29)1Γqα1−γ1η1α−1∫011−qsα−1fs,1dqs≤R11−σ1,(30)1Γqβ1−γ2η2β−1∫011−qsβ−1gs,1dqs≤R21−σ2,then the fractional q-difference system (6) has one positive solution u∗,v∗, where u∗∈P1 and v∗∈P2. Moreover, for each t∈0,1, there exist constants 0<mi<1<nii=1,2, such that(31)u∗t∈m1tα−1,n1tα−1,v∗t∈m2tβ−1,n2tβ−1,which can be obtained by monotone iterative schemes un and vn generated by(32)unt=∫01G1t,qsfs,vn−1sdqs,vnt=∫01G2t,qsgs,un−1sdqs.i.e., un−u∗⟶0 and vn−v∗⟶0 as n⟶∞.Proof.
For anyv∈P2, we know that there exist two constants a1 and b1 with 0<a1<1<b1 such that(33)a1tβ−1≤vt≤b1tβ−1,t∈0,1.
From Lemma5 and condition (H1), we obtain(34)T1vt=∫01G1t,qsfs,vsdqs≥qM1tα−1Γqα1−γ1η1α−1∫01s1−qsα−1fs,vsdqs≥qM1tα−1Γqα1−γ1η1α−1∫01s1−qsα−1fs,a1sβ−1dqs≥qM1a1σ1tα−1Γqα1−γ1η1α−1∫01s1−qsα−1fs,sβ−1dqs≥c1tα−1,T1vt≤tα−1Γqα1−γ1η1α−1∫011−qsα−1fs,vsdqs≤tα−1Γqα1−γ1η1α−1∫011−qsα−1fs,b1sβ−1dqs≤b1σ1tα−1Γqα1−γ1η1α−1∫011−qsα−1fs,sβ−1dqs≤d1tα−1,where d1andc1 are two positive constants satisfying(35)d1>max1,b1σ1Γqα1−γ1η1α−1∫011−qsα−1fs,sβ−1dqs,0<c1<min1,qM1a1σ1Γqα1−γ1η1α−1∫01s1−qsα−1fs,sβ−1dqs.
Thus,T1 maps P2 into P1. For each u∈P1, there exist two constants a2 and b2 with 0<a2<1<b2 such that(36)a2tα−1≤ut≤b2tα−1,t∈0,1.
Similarly, by Lemma5 and condition (H2), we can get that(37)c2tβ−1≤T2ut≤d2tβ−1,where d2andc2 are two positive constants satisfying(38)d2>max1,b2σ2Γqβ1−γ2η2β−1∫011−qsβ−1gs,sα−1dqs,0<c2<min1,qM2a2σ2Γqβ1−γ2η2β−1∫01s1−qsβ−1gs,sα−1dqs,which implies that T2 maps P1 into P2. On the other hand, the proof of completely continuous T1 and T2 are as the same as in [12], and we omit it here.
LetPiR=uu∈Pi,u≤Ri=1,2. In the following, we will prove T1P2R1⊂P1R1 and T2P1R2⊂P2R2. In fact, for any v∈P2R1 and u∈P1R2, by conditions (29) and (30), we obtain(39)T1vt≤tα−1Γqα1−γ1η1α−1∫011−qsα−1fs,vsdqs≤tα−1Γqα1−γ1η1α−1∫011−qsα−1fs,R1dqs≤R1σ1Γqα1−γ1η1α−1∫011−qsα−1fs,1dqs≤R1,T2ut≤tβ−1Γqβ1−γ2η2β−1∫011−qsβ−1gs,usdqs≤tβ−1Γqβ1−γ2η2β−1∫011−qsβ−1gs,R2dqs≤R2σ2Γqβ1−γ2η2β−1∫011−qsβ−1gs,1dqs≤R2,which implies that T1v≤R1 and T2u≤R2. So, T1P2R1⊂P1R1 and T2P1R2⊂P2R2.
Takinge1t=tα−1 and e2t=tβ−1, then e1∈P1, e2∈P2, T1e2∈P1, and T2e1∈P2. Thus, there exist constants 0<mi<1<nii=1,2 such that(40)m1tα−1≤T1e2t≤n1tα−1,m2tβ−1≤T2e1t≤n2tβ−1.
Letl1 and l2 be two positive numbers satisfying 0<l1<l2<1, l2<l1σ2, and(41)l1l2−σ1≤m1,l2l1−σ2≤m2.
Set(42)u0t=l1e1t,v0t=l2e2t,(43)un=T1vn−1,vn=T2un−1,n=1,2,…
Obviously,u0t≤v0t by β≤α and 0<l1<l2<1, u0∈P1R2 and v0∈P2R1. By (H1) and (H2), we have(44)u1t=T1v0t=∫01G1t,qsfs,v0sdqs=∫01G1t,qsfs,l2e2sdqs≥l2σ1∫01G1t,qsfs,e2sdqs=l2σ1T1e2t≥l2σ1m1e1t≥l1e1t=u0t,v1t=T2u0t=∫01G2t,qsgs,u0sdqs=∫01G2t,qsgs,l1e1sdqs≥l1σ2∫01G2t,qsgs,e1sdqs=l1σ2T2e1t≥l1σ2m2e2t≥l2e2t=v0t.
From conditions (H1) and (H2), we know thatT1 and T2 are two nondecreasing operators. Thus, by induction, we can obtain(45)u0≤u1≤⋯≤un≤⋯,v0≤v1≤⋯≤vn≤⋯,un∈P1R1,vn∈P2R2,n=1,2,….
By the compactness of the operatorsT1 and T2, we have that un and vn are two sequentially compact sets. Therefore, there exist u∗∈P1R1 and v∗∈P2R2, such that un converges to u∗ and vn converges to v∗ as n⟶∞, respectively. Since the operators T1 and T2 are continuous, un=T1vn−1 and vn=T2un−1, and we obtain u∗=T1v∗ and v∗=T2u∗ as n⟶∞, which implies that system (6) has a positive solution u∗,v∗, and u∗∈m1tα−1,n1tα−1, v∗∈m2tβ−1,n2tβ−1, and ∀t∈0,1, where miandni are constants and 0<mi<1<nii=1,2, which can be achieved by the monotone scheme:(46)unt=∫01G1t,qsfs,vn−1sdqs,vnt=∫01G2t,qsgs,un−1sdqs,with initial values u0t and v0t defined as in (42).
In the following, we give an example to illustrate the existence of positive solutions of BVP (6).Example 1.
Consider the following system of fractionalq-difference with boundary conditions:(47)D1/35/3ut+18tv3/2t=0,0<t<1,D1/33/2vt+tu1/4t+u1/3t1+u1/4t=0,0<t<1,u0=0,u1=u34,v0=0,v1=54v12,where q=1/3, α=5/3, β=3/2, η1=3/4, η2=1/2, γ1=1, γ2=5/4, and(48)ft,v=18tv3/2,gt,u=tu1/4+u1/31+u1/4,
Obviously,ft,v and gt,u are nondecreasing with respect to v and u, respectively, and(49)0<γ1η1α−1=342/3<1,0<γ2η2β−1=54121/2<1.
Choosingσ1=2>1 and σ2=1/3<1, we have(50)ft,rv=18tr3/2v3/2≥r2ft,v,∀v∈0,+∞,r∈0,1,gt,ru=tr1/4u1/4+r1/3u1/31+r1/4u1/4,≥tr1/3u1/4+r1/3u1/31+u1/4=r1/3gt,u,∀u∈0,+∞,r∈0,1.
So, conditions (H1) and (H2) hold. Moreover, we can show that(51)0<∫011−13s2/3fs,1dqs=18∫011−13s2/3sdqs<∞,0<∫011−13s1/2gs,1dqs=32∫011−13s1/2sdqs<∞,which implies that (H3) and (H4) hold. Moreover, we know that there exist two positive constants R1 and R2 such that (29) and (30) hold, respectively. Thus, it follows from Theorem 1 that boundary value problem of fractional q-difference system (47) has one iterative positive solution u∗,v∗ which can be obtained with the aid of monotone iterative sequences.



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