# Monotone Iterative Schemes for Positive Solutions of a Fractionalq-Difference Equation with Integral Boundary Conditions on the Half-Line

Authors: Min Jiang; Rengang Huang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we study the boundary value problem of a fractionalq-difference equation with nonlocal integral boundary conditions on the half-line. Using the properties of the Green function and monotone iterative method, the extremal solutions are obtained. Finally, an example is presented to illustrate our main results.


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## Body

## 1. Introduction

In this paper, we are concerned with the following fractionalq-differential equation with integral boundary value problem on the half-line:(1)Dqαut+ft,ut,Dqα−1ut,Dqα−2ut,…,Dqα−n−1ut=0,n−1<α≤n,u0=Dqju0=0,j=1,…,n−2,Dqα−1u+∞=∫0+∞gtutdqt,where t∈J=0,+∞, DqαandDqα−j are the fractional q-derivative of Riemann–Liouville type of order α and α−j, Dqj is the q-derivative of order j, 0<q<1,n∈ℕ,n≥3,f∈CJ×ℝn,J, and g is a given function and satisfies some conditions which will be given later.Theq-difference calculus was initially developed by Jackson [1, 2] and had proved to have important applications in many subjects, such as quantum mechanics, complex analysis, and hypergeometric series. The fractional q-difference calculus had its origin in the works by Al-Salam [3] and Agarwal [4]. Due to the development of the fractional differential equations, fractional q-differential equations, regarded as fractional analogue of q-difference equations, have been studied by several researchers, especially, about the existence of the solutions for the boundary value problems (see [5–22] and the references therein).In [21], Yang studied the coupled integral boundary value problem for systems of nonlinear fractional q-difference equations as follows:(2)Dqαut+λft,ut,vt=0,Dqβvt+λgt,ut,vt=0,t∈0,1,λ>0,Dqju0=Dqjv0=0,0≤j≤n−2,u1=μ∫01vsdqs,v1=ν∫01usdqs,where λ,μ,andν are three parameters with 0<μ<βq and 0<ν<αq, α,β∈n−1,n are two real numbers and n≥3, and DqαandDqβ are the fractional q-derivative of Riemann–Liouville type. By applying the nonlinear alternative of Leray–Schauder type and Krasnoselskii’s fixed point theorems, the authors obtain some sufficient conditions for the existence of positive solutions to the considered problem.In [14], Wang et al. investigated the following nonlinear fractional q-difference equation with q-integral boundary condition:(3)Dqαut+fut,Iqβut=0,t∈0,1,Dqju0=0,0≤j≤n−2,u1=μ∫01usdqs,where α∈n−1,n,n≥3, μ is a parameter with 0<μ<αq, β>0, and Dqα is the fractional q-derivative of Riemann–Liouville type. By applying the hybrid monotone method, the existence and uniqueness of the positive solution of the q-integral boundary problem are obtained.In [17], the authors investigate the existence of solutions for the following boundary value problem of nonlinear fractional q-difference equations on the half-line(4)Dqαut+ft,ut=0,u0=0,Dqα−1u+∞=∑i=1maiuξi,where 1<α<2, 0≤∑i=1maiξiα−1<Γqα. Dqα is the q-derivative of Riemann–Liouville type of order α. By means of Schauder fixed point theorem and Leggett–Williams fixed point theorem, some results on existence and multiplicity of solutions to the above boundary value problem are obtained.To the best of our knowledge, there are few papers that consider the boundary value of nonlinear fractionalq-difference equations with nonlocal conditions on the half-line, although the study of such problems is very important. In [17], the authors only proved the existence and multiplicity of solutions, for the uniqueness of positive solutions without being given. Moreover, how to seek the solutions? It is very important and helpful for computational purpose. This thought motivates the research of iterative schemes of positive solutions for problem (1). We should mention that the paper has some new features. Firstly, we consider the infinite interval problem for higher-order nonlinear fractional q-difference equation with integral boundary conditions. Secondly, the nonlinearity relies on the lower-order fractional q-derivative of unknown function. Our main purpose of this paper is by constructing a suitable Banach space, defining appropriate operators, and using the monotone iterative method, which is different from the method in [17] to obtain the existence and uniqueness of positive solutions.


## 2. Preliminaries onq-Calculus and Lemmas

Here we recall some definitions and fundamental results on fractionalq-integral and fractional q-derivative. For more information regarding fractional q-calculus, see [23].Letq∈0,1 and define(5)aq=1−qa1−q,a∈ℝ.Theq-analogue of the power function a−bn with n∈N0 is(6)a−b0=1,a−bn=∏k=0n−1a−bqk,n∈N,a,b∈ℝ.More generally, ifα∈R, then(7)a−bα=aα∏n=0∞a−bqna−bqα+n.Ifb=0, then aα=aα. The q-gamma function is defined by(8)Γqs=1−qs−11−qs−1,s∈R,0,−1,−2,…,and satisfies Γqs+1=sqΓqs. The q-derivative of a function u is defined by(9)Dqus=us−uqs1−qs,Dqu0=lims⟶0Dqus,and q-derivative of higher order by(10)Dq0us=usandDqnus=DqDqn−1us,n∈N.Theq-integral of a function f defined in the interval 0,b is given by(11)Iqus=∫0sutdqt=s1−q∑n=0∞usqnqn,s∈0,b.Ifa∈0,b and u is defined in the interval 0,b, its integral from a to b is defined by(12)∫abutdqt=∫0butdqt−∫0autdqt.Similar to that for derivatives, an operatorIqn can be defined, namely,(13)Iq0us=usandIqnus=IqIqn−1us,n∈N.The fundamental theorem of calculus applies to these operatorsIq and Dq, i.e.,(14)DqIqus=us,and if u is continuous at s=0, then(15)IqDqus=us−u0.The basic properties of the two operators can be found in the book [23].The following definition was considered first in [4].Definition 1 (see [4]).
Letα≥0 and u be a function defined on 0,1. The fractional q-integral of the Riemann–Liouville type is Iq0us=us and(16)Iqαus=1Γqα∫0ss−qtα−1utdqt,α>0,s∈0,1.Definition 2 (see [24]).
The fractionalq-derivative of the Riemann–Liouville type of order α≥0 is defined by Dq0us=us and(17)Dqαus=DqmIqm−αus,α>0,where m is the smallest integer greater than or equal to α.Lemma 1 (see [4, 24]).
Letα,β≥0 and u be a function defined on 0,1. Then, the following formulas hold:(1)
IqβIqαus=Iqα+βus(2)
DqαIqαus=usLemma 2 (see [25]).
Letα>0 and n be a positive integer. Then, the following equality holds:(18)IqαDqnus=DqnIqαus−∑k=0n−1sα−n+kΓqα+k−n+1Dqku0.Lemma 3 (see [24]).
Forβ∈ℝ+,λ∈−1,∞, the following is valid:(19)Iqβs−aλ=Γqλ+1Γqβ+λ+1s−aβ+λ,0<a<x<b.In the following, we introduce the hypotheses that will be used in subsequent proof:(H1): gt satisfies ∫0+∞tα−1gtdqt=Λ<Γqα and ft,0,…,0≡0,∀t∈J=0,+∞.(H2): there exist nonnegative functionsa0t,ait∈Lq0,+∞ and constants λi≥0 satisfying(20)ft,u1,u2,…,un≤a0t+∑i=1naituiλi,∀t∈J,u∈ℝ,i=1,2,…,n∫0+∞a0tdqt=a0<+∞∫0+∞1+tα−1λ1a1tdqt=a1<+∞∫0+∞a2tdqt<+∞∫0+∞1+ti−2λiaitdqt=ai<+∞,i=3,…,n.(H3): there exist nonnegative functions bit∈Lq0,+∞ satisfying(21)ft,u1,u2,…,un−ft,v1,v2,…,vn≤∑i=1nbitui−vi,∀t∈J,ui,vi∈ℝ,i=1,2,…,n∫0+∞1+tα−1b1tdqt=b1<+∞∫0+∞1+ti−2bitdqt=bi<+∞,i=3,…,n∫0+∞b2tdqt<+∞∫0+∞ft,0,0,…,0dqt=κ<+∞.(H4): function ft,u1,u2,…,un is increasing with respect to the variables u1,u2,…,un,∀t∈J.Lemma 4.
Assume thatgt satisfies condition (H1), then fractional q-differential equation(22)Dqαut+ht=0,n−1<α≤n,u0=Dqju0=0,j=1,…,n−2,Dqα−1u+∞=∫0+∞gtutdqt,has a unique integral representation(23)ut=∫0+∞Gt,qshsdqs,where(24)Gt,qs=G1t,qs+G2t,qs,(25)G1t,qs=1Γqαtα−1−t−qsα−1,0≤s≤t≤+∞,tα−1,0≤t≤s≤+∞,G2t,qs=tα−1Γqα−Λ∫0+∞gtG1t,qsdqt.Proof.
In view of Definition2 and Lemma 1, we see that(26)IqαDqnIqn−αut=−Iqαht.
By Lemma2, we obtain(27)ut=−Iqαht+c1tα−1+c2tα−2+⋯+cntα−n,where ci,i=1,2,…,n are some constants. Since u0=Dqju0=0,j=1,…,n−2, we get(28)c2=c3=⋯=cn=0,which implies that(29)ut=−1Γaα∫0tt−qsα−1hsdqs+c1tα−1.
From (29) and Lemma 1, we have(30)Dqα−1ut=−∫0thsdqs+c1Γqα.
Hence,(31)Dqα−1u+∞=−∫0+∞hsdqs+c1Γqα.
The conditionDqα−1u+∞=∫0+∞gtutdqt together with (31) implies that(32)c1=1Γqα∫0+∞hsdqs+∫0+∞gtutdqt.
Submitting (32) to (29), we get(33)ut=−1Γqα∫0tt−qsα−1hsdqs+tα−1Γqα∫0+∞hsdqs+∫0+∞gtutdqt=1Γqα∫0+∞tα−1hsdqs−1Γqα∫0tt−qsα−1hsdqs+tα−1Γqα∫0+∞gtutdqt=∫0+∞G1t,qshsdqs+tα−1Γqα∫0+∞gtutdqt.
Equation (33) is multiplied by gt and q-integrating from 0 to +∞, and we obtain(34)∫0+∞gtutdqt=∫0+∞gt∫0+∞G1t,qshsdqsdqt+ΛΓqα∫0+∞gtutdqt.
From (34), we get(35)∫0+∞gtutdqt=ΓqαΓqα−Λ∫0+∞gt∫0+∞G1t,qshsdqsdqt.
Combining (33) and (35), we have(36)ut=∫0+∞G1t,qshsdqs+tα−1Γqα−Λ∫0+∞gt∫0+∞G1t,qshsdqsdqt=∫0+∞G1t,qshsdqs+∫0+∞G2t,qshsdqs=∫0+∞Gt,qshsdqs.
The proof is completed.Lemma 5.
From Lemma4, by direct calculation, we have(37)Dqα−iut=∫0+∞Gi∗t,qshsdqs,i=1,2,…,n−1,where(38)Gi∗t,qs=Gi1∗t,qs+Gi2∗t,qs,Gi1∗t,qs=1Γqiti−1−t−qsi−1,0≤s≤t≤+∞,ti−1,0≤t≤s≤+∞,Gi2∗t,qs=Γqαti−1ΓqiΓqα−Λ∫0+∞gtG1t,qsdqt.Now, we shall prove a lemma which plays a key role in the next theorems.Lemma 6.
FunctionGt,qs defined in (24) satisfies the following conditions: 0≤Gt,qs≤tα−1/Γqα−Λ and 0≤Gt,qs/1+tα−1≤1/Γqα−Λ   for all 0≤t,s≤+∞.Proof.
From (24), it is obvious that, for all 0≤t,s≤+∞,(39)0≤G1t,qs≤tα−1Γqα,0≤G2t,qs≤tα−1Γqα−Λ∫0+∞gttα−1Γqαdqs=Λtα−1ΓqαΓqα−Λ.
The above two inequalities imply that(40)0≤Gt,qs=G1t,qs+G2t,qs≤tα−1Γqα+Λtα−1ΓqαΓqα−Λ=tα−1Γqα−Λ,0≤Gt,qs1+tα−1≤tα−1Γqα−Λ1+tα−1≤1Γqα−Λ.
The proof is completed.Lemma 7.
FunctionGi∗t,qs defined in Lemma 5 satisfies the following conditions:(41)0≤Gi∗t,qs≤Γqαti−1ΓqiΓqα−Λ,i=1,2,…,n−1,0≤Gi∗t,qs1+ti−1≤ΓqαΓqiΓqα−Λ,i=2,…,n−1,for all 0≤t,s≤+∞.Proof.
For0≤t,s≤+∞ and i=1,2,…,n−1, by Remark 2 and Lemma 5, it is easy to get(42)0≤Gi1∗t,qs≤ti−1Γqi,0≤Gi2∗t,qs≤Γqαti−1ΓqiΓqα−Λ∫0+∞gttα−1Γqαdqs=ti−1ΛΓqiΓqα−Λ.
This gives for,i=2,…,n−1,(43)0≤Gi∗t,qs≤ti−1Γqi+ti−1ΛΓqiΓqα−Λ=Γqαti−1ΓqiΓqα−Λ,0≤Gi∗t,qs1+ti−1≤ΓqαΓqiΓqα−Λ.
The proof is completed.
In the following, we will construct a suitable Banach space. Let(44)X=u|u,Dqα−1u,Dqα−iu∈C0,+∞,u1+tα−1∞,Dqα−1u∞<∞,Dqα−iu1+ti−1∞<∞,i=2,…,n−1,with the norm(45)u=maxu1+tα−1∞,Dqα−iu1+ti−1∞,Dqα−1u∞,i=2,…,n−1,where u∞=supt∈Jut,J=0,+∞.
Similar to the proof of Lemma 2.2 in [26], we can prove X is a Banach space. Note that the Arzela–Ascoli theorem fails to work in X. In order to proceed, we need the following compactness criterion.Lemma 8.
LetB⊂X be a bounded set. Then, B is relatively compact in X if the following conditions hold:(i)
u/1+tα−1|u∈B,Dqα−iu/1+ti−1|u∈B,Dqα−1u|u∈B,i=2,…,n−1 are equicontinuous on 0,T for any T∈0,+∞(ii)
For∀ε>0,∃Tε>0 such that for any t1,t2≥Tε,u∈B, the following inequalities hold:(46)ut21+t2α−1−ut11+t1α−1<ε,Dqα−iut21+t2i−1−Dqα−iut11+t1i−1<ε,Dqα−1ut2−Dqα−1ut1<ε,i=2,…,n−1.Proof.
The method is similar to the proof of Lemma 2.3 in [26], so we omit the details.Lemma 9.
If conditionH2 holds, then for u∈X, we have(47)∫0+∞ft,ut,Dqα−1ut,…,Dqα−n−1utdqt≤a0+∑i=1naiuλi.Proof.
Foru∈X, by condition H2, we have(48)∫0+∞ft,ut,Dqα−1ut,…,Dqα−n−1utdqt≤∫0+∞a0tdqt+∫0+∞a1tuλ1dqt+∫0+∞∑i=1naitDqα−iuλidqt=a0+∫0+∞1+tα−1λ1a1tuλ11+tα−1λ1dqt+∫0+∞a2tDqα−1uλ2dqt+∫0+∞∑i=2n−1ai+1t1+ti−1λi+1Dqα−iutλi+11+ti−1λi+1dqt≤a0+a1uλ1+⋯+anuλn=a0+∑i=1naiuλi.Lemma 10.
If conditionH3 holds, then for u∈X, we have(49)∫0+∞ft,ut,Dqα−1ut,…,Dqα−n−1utdqt≤κ+∑i=1nbiu.Proof.
Foru∈X, by condition H3, we have(50)∫0+∞ft,ut,Dqα−1ut,…,Dqα−n−1utdqt≤∫0+∞ft,ut,Dqα−1ut,…,Dqα−n−1−ft,0,…,0dqt+∫0+∞ft,0,…,0dqt≤∫0+∞b1tutdqt+∫0+∞b2t|Dqα−1utdqt+⋯+∫0+∞bnt|Dqα−n−1utdqt=∫0+∞1+tα−1ut1+tα−1b1tdqt+∫0+∞b2tDqα−1utdqt+∫0+∞∑i=2n−1bi+1Dqα−iut1+ti−1dqt+∫0+∞ft,0,…,0dqt≤∑i=1nbiu+κ.
Define the coneP⊂X by (51)P=u∈X|ut≥0,Dqα−iut≥0,t∈J,i=1,2,…,n−1,and define the operator T:P↦P as follows:(52)Tut=∫0+∞Gt,qsfs,us,Dqα−1us,…,Dqα−n−1usdqs.
We also define(53)Dqα−iTut=∫0+∞Gi∗t,qsfs,us,Dqα−1us,…,Dqα−n−1usdqs.
It is easy to see problem (1) has a solution, if and only if the operator T has a fixed point.Lemma 11.
If conditions (H1) and (H2) hold, then the operator T:P↦P is completely continuous.Proof.
It is easy to seeT:P↦P. Since Gt,qs≥0 and f≥0, we know Tut≥0, for u∈P,t∈J. In the following, we divide the proof into four steps:Step 1: we show thatT is uniformly bounded. Let Ω be any bounded subset of P, i.e., there exists M>0 such that u≤M for each u∈Ω. We only need to show that Tu is bounded in P. For u∈Ω, by Lemmas 5 and 8, we have(54)Tu1+tα−1∞=supt∈JTu1+tα−1=supt∈J∫0+∞Gt,qs1+tα−1fs,us,Dqα−1us,…,Dqα−n−1usdqs≤1Γqα−Λa0+∑k=1nakuλk≤1Γqα−Λa0+∑k=1nakMλk.By Lemmas5, 7, and 9, we have(55)Dqα−iTu1+ti−1∞=supt∈JDqα−iTu1+ti−1=supt∈J∫0+∞Gi∗t,qs1+ti−1fs,us,Dqα−1us,…,Dqα−n−1usdqs≤ΓqαΓqiΓqα−Λa0+∑k=1nakuλk,i=2,3,…,n−1≤ΓqαΓqiΓqα−Λa0+∑k=1nakMλk,i=2,3,…,n−1,(56)Dqα−1Tu∞=supt∈J∫0+∞G1∗t,qsfs,us,Dqα−1us,…,Dqα−n−1usdqs≤ΓqαΓqα−Λa0+∑k=1nakuλk≤ΓqαΓqα−Λa0+∑k=1nakMλk.From (54)–(56), we obtain(57)Tu≤a0+∑k=1nakMλkmax1Γqα−Λ,ΓqαΓqα−Λ,ΓqαΓqiΓqα−Λ,i=2,3,…,n−1,which means thatTu is uniformly bounded.Step 2: we show that the operatorT is equicontinuous for all u∈Ω on any compact interval I of J. For any given T>0,t1,t2∈0,T and t2>t1,u∈Ω, we have(58)Tut21+t2α−1−Tut11+t1α−1=∫0+∞Gt2,qs1+t2α−1−Gt1,qs1+t1α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs≤∫0+∞G1t2,qs1+t2α−1−G1t1,qs1+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs+∫0+∞G1t2,qs1+t2α−1−G1t1,qs1+t1α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs+∫0+∞G2t2,qs1+t2α−1−G2t1,qs1+t1α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs.By Lemmas5 and 6, we have(59)∫0+∞G2t2,qs1+t2α−1−G2t1,qs1+t1α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs≤∫0+∞1Γqα−Λ2t2α−1−t1α−11+t1α−11+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs⟶0.On the other hand, for allu∈Ω, t1⟶t2, we have(60)∫0+∞G1t2,qs1+t2α−1−G1t1,qs1+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs≤∫0t1t2α−1−t1α−1−t2−qsα−1+t1−qsα−1Γqα1+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs+∫t1t2t2α−1−t1α−1−t2−qsα−1Γqα1+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs+∫t2+∞t2α−1−t1α−1Γqα1+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs⟶0.Similar to (60), for all u∈Ω,t1⟶t2, we have(61)∫0+∞G1t1,qs1+t2α−1−G1t1,qs1+t1α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs≤∫0t1t1α−1−t1−qsα−1Γqαt2α−1−t1α−11+t1α−11+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs+∫t1+∞t1α−1Γqαt2α−1−t1α−11+t1α−11+t2α−1fs,us,Dqα−1us,…,Dqα−n−1usdqs⟶0.Together (59)–(61), which implies that Tut/1+tα−1 is equicontinuous on any compact intervals of 0,∞.Note that, from (53) and Lemma 5, we have(62)Dqα−1Tut=∫0+∞G1∗t,qsfs,us,Dqα−1us,…,Dqα−n−1usdqs,G1∗t,qs=G11∗t,qs+G12∗t,qs,where(63)G11∗t,qs=0,0≤s≤t≤+∞,1,0≤t≤s≤+∞,G12∗t,qs=ΓqαΓqα−Λ∫0+∞gtG1t,qsdqt.which meansG1∗t,qs does not rely on t. Hence, Dqα−1Tu is equicontinuous on any compact intervals of J.In the following, we show thatDqα−iTu, i=2,3,…,n−1, is equicontinuous on any compact intervals of J. In fact, for any given T>0,t1,t2∈0,T and t2>t1,u∈Ω, we have(64)Dqα−iTut21+t2i−1−Dqα−iTut11+t1i−1=∫0+∞Gi∗t2,qs1+t2i−1−Gi∗t1,qs1+t1i−1fs,us,Dqα−1us,…,Dqα−n−1usdqs≤∫0+∞Gi1∗t2,qs1+t2i−1−Gi1∗t1,qs1+t2i−1fs,us,Dqα−1us,…,Dqα−n−1usdqs+∫0+∞Gi1∗t1,qs1+t2i−1−Gi1∗t1,qs1+t1i−1fs,us,Dqα−1us,…,Dqα−n−1usdqs+∫0+∞Gi2∗t2,qs1+t2i−1−Gi2∗t1,qs1+t1i−1fs,us,Dqα−1us,…,Dqα−n−1usdqs.Similar to (59)–(61), we obtain(65)Dqα−iTut21+t2i−1−Dqα−iTut11+t1i−1⟶0,ast1⟶t2.Inequalities (59)–(64) imply that the operator T is equicontinuous on any compact intervals of J.Step 3: we show thatT is equiconvergent at ∞. We need to prove that, for ∀ε>0,∃Tε>0, such that, for any t1,t2>Tε, the following inequalities hold:(66)Tut21+t2α−1−Tut11+t1α−1<ε,Dqα−1Tut2−Dqα−1Tut1<ε,Dqα−iTut21+t2i−1−Dqα−iTut11+t1i−1<ε,i=2,3,…,n−1.Note that, since the functionG1∗t,s does not rely on t, it is easy to infer that(67)Dqα−1Tut2−Dqα−1Tut1<ε,i.e.,Dqα−1Tut is equiconvergent at +∞.By (H1) and Lemma 9, for any ε>0, there exists a>0 such that(68)∫a+∞fs,us,Dqα−1us,…,Dqα−n−1usdqs<ε.On the other hand, since(69)limt⟶+∞tγ1+tγ=1,γ∈α−1,i−1,i=2,…,n−1.There exists sufficiently largeT1>0 such that for any t1,t2>T1, we have(70)t2α−11+t2α−1−t1α−11+t1α−1<ε,andt2i−11+t2i−1−t1i−11+t1i−1<ε,i=2,…,n−1.Also becauselimt⟶+∞t−aθ−1/1+tθ−1=1,θ∈α,i,i=2,…,n−1, there exists sufficiently large T2>a such that, for any t1,t2>T3 and 0≤qs<s≤a, we have(71)t2−qsα−11+t2α−1−t1−qsα−11+t1α−1<εandt2−qsi−11+t2i−1−t1−qsi−11+t1i−1<ε,i=2,…,n−1.ChooseT>maxT1,T2, for any t1,t2>T, and let t1⟶t2. For convenience, we denote Fs:=fs,us,Dqα−1us,…,Dqα−n−1us, and we obtain(72)Tut21+t2α−1−Tut11+t1α−1=∫0+∞Gt2,qs1+t2α−1−Gt1,qs1+t1α−1Fsdqs≤∫0+∞G1t2,qs1+t2α−1−G1t1,qs1+t2α−1Fsdqs+∫0+∞G1t2,qs1+t2α−1−G1t1,qs1+t1α−1Fsdqs+∫0+∞G2t2,qs1+t2α−1−G2t1,qs1+t2α−1Fsdqs≤1Γqα∫0t1t2α−11+t2α−1−t1α−11+t1α−1+t2−qsα−11+t2α−1−t1−qsα−11+t1α−1Fsdqs+1Γqα∫t1t2t2α−11+t2α−1−t1α−11+t1α−1+t2−qsα−11+t2α−1Fsdqs+1Γqα∫t2+∞t2α−11+t2α−1−t1α−11+t1α−1Fsdqs+1Γqα−Λ2t2α−11+t2α−1−t1α−11+t1α−1∫0+∞Fsdqs.By (68) and (72), we obtain(73)Tut21+t2α−1−Tut11+t1α−1≤1Γqα∫0at2α−11+t2α−1−t1α−11+t1α−1+t2−qsα−11+t2α−1−t1−qsα−11+t1α−1Fsdqs+1Γqα∫a+∞t2α−11+t2α−1−t1α−11+t1α−1+t2−qsα−11+t2α−1−t1−qsα−11+t1α−1Fsdqs+1Γqα∫a+∞t2α−11+t2α−1−t1α−11+t1α−1+t2−qsα−11+t2α−1Fsdqs+1Γqα∫a+∞t2α−11+t2α−1−t1α−11+t1α−1Fsdqs+1Γqα−Λ2t2α−11+t2α−1−t1α−11+t1α−1∫0+∞Fsdqs≤1Γqαmaxs∈0,a,u∈ΩFs∫0at2α−11+t2α−1−t1α−11+t1α−1+t2−qsα−11+t2α−1−t1−qsα−11+t1α−1dqs+9Γqα∫a+∞Fsdqs+εΓqα−Λ2∫0+∞Fsdqs≤2aεΓqαmaxs∈0,a,u∈ΩFs+9εΓqα+εΓqα−Λ2∫0+∞Fsdqs≤2aΓqαmaxs∈0,a,u∈ΩFs+9Γqα+a0+∑k=1nakMλkΓqα−Λ2ε.Similar to (72) and (73), we obtain(74)Dqα−iTut21+t2i−1−Dqα−iTut11+t1i−1≤2aΓqimaxs∈0,a,u∈ΩFs+9Γqi+a0+∑k=1nakMλkΓqαΛΓqiΓqα−Λ2ε.From (67), (73), and (74), we obtain T is equiconvergent at ∞.The above three steps and Lemma8 imply that the operator T:P⟶P is relatively compact.Step 4: we show that the operatorT:P⟶P is continuous. For um∈P, such that um⟶u as m⟶∞, and we need to show that Tum−Tu⟶0 as m⟶∞.By Lemma6, the continuity of function f, and the Lebesgue dominated convergence theorem, we obtain(75)Tum1+tα−1−Tu1+tα−1∞=supt∈J∫0+∞Gt,qs1+tα−1fs,ums,Dqα−1ums,…,Dqα−n−1ums−fs,us,Dqα−1us,…,Dqα−n−1usdqs≤1Γqα−Λ∫0+∞fs,ums,Dqα−1ums,…,Dqα−n−1ums−fs,us,Dqα−1us,…,Dqα−n−1usdqs⟶0,asm⟶∞.Similar to (75), by Lemmas 7 and 9, we get(76)Dqα−1Tum−Dqα−1Tu∞≤ΓqαΓqα−Λ∫0+∞fs,ums,Dqα−1ums,…,Dqα−n−1ums−fs,us,Dqα−1us,…,Dqα−n−1usdqs⟶0,asm⟶∞,(77)Dqα−iTum1+ti−1−Dqα−iTu1+ti−1∞≤ΓqαΓqiΓqα−Λ∫0+∞fs,ums,Dqα−1ums,…,Dqα−n−1ums−fs,us,Dqα−1us,…,Dqα−n−1usdqs⟶0,asm⟶∞,for alli=2,3,…,n−1.From (75)–(77), we have(78)Tum−Tu⟶0,asm⟶∞,which means that the operatorT is continuous.
In view of all the above arguments, we deduce that the operatorT:P⟶P is completely continuous. This completes the proof.


## 3. Main Results

In this section, we give the main results of this paper. For convenience, we denote(79)ρi=1ΓqiΓqα−Λ,i=1,2,…,n−1.Let(80)L=maxρ1,ρ1Γqα,ρ2Γqα,…,ρn−1Γqα.We have the following two theorems.Theorem 1.
Assume that (H1), (H2), and (H4) hold. There exists a positive R>0 such that problem (1) has two positive solutions u∗ and v∗ satisfying 0≤u∗≤R and 0≤v∗≤R. Moreover, limm⟶∞um=u∗ and limm⟶∞vm=v∗, um and vm can be obtained by the following monotone iterative schemes:(81)um=Tum−1,withu0t=Rtα−1,m=1,2,…,(82)vm=Tvm−1,withv0t=0,m=1,2,….
In addition, we have(83)v0t≤v1t≤⋯≤vmt≤⋯≤v∗t≤⋯≤u∗t≤⋯≤umt≤⋯≤u2t≤u1t,(84)Dqα−iv0t≤Dqα−iv1t≤⋯,≤Dqα−ivmt≤⋯≤Dqα−iv∗t≤⋯≤Dqα−iu∗t≤⋯≤Dqα−iumt≤⋯≤Dqα−iu2t≤Dqα−iu1t,i=2,3,…,n−1.Proof.
First, Lemma11 implies that TP⊂P for any u∈P and t∈J. Next, let 0≤λk<1, k=1,2,…,n. For the case λk≥1, the proof is similar to that of 0≤λk<1. Define PR=u∈P|u≤R, where R satisfies(85)R≥maxn+1La0,n+1La11/1−λ1,…,n+1Lan1/1−λ1.
For anyu∈PR, similar to (54)–(56), we obtain(86)Tu1+tα−1∞≤1Γqα−Λa0+∑k=1nakRλk=ρ1a0+∑k=1nakRλk≤La0+∑k=1nakRλk≤R,Dqα−iTu1+ti−1∞≤ρiΓqαa0+∑k=1nakRλk,i=2,3,…,n−1≤La0+∑k=1nakRλk≤R,i=2,3,…,n−1,Dqα−1Tu∞≤ΓqαΓqα−Λa0+∑k=1nakRλk=ρ1Γqαa0+∑k=1nakRλk≤La0+∑k=1nakRλk≤R.
Thus,(87)Tu=maxTu1+tα−1∞,Dqα−iTu1+ti−1∞,Dqα−1Tu∞,i=2,3,…,n−1≤R.
That isTPR⊂PR.
Denote thatu0t=Rtα−1, then u0t∈PR, and u1t=Tu0t, i.e.,(88)u1t=Tu0t=∫0+∞Gt,qsfs,u0s,Dqα−1u0s,…,Dqα−n−1u0sdqs≤tα−1Γqα−Λa0+∑k=1nakuλk≤tα−1La0+∑k=1nakRλk≤Rtα−1=u0t.
On the other hand, (H4) implies that the operator T is increasing, and then we have(89)u2t=Tu1t≤Tu0t=u1t,∀t∈J.
By the induction, we haveumt=Tumt,m=0,1,2,…. Then, the sequence umm=0∞⊂TΩR⊂PR and satisfies the following relation:(90)um+1t≤umt,∀t∈J,m=0,1,2,….
By Lemmas5 and 7, we have(91)Dqα−iu1t=Dqα−iTu0t=∫0+∞Gi∗t,qsfs,u0s,Dqα−1u0s,…,Dqα−n−1u0sdqs≤Γqαti−1ΓqiΓqα−Λfs,u0s,Dqα−1u0s,…,Dqα−n−1u0sdqs=1Γqα−ΛDqα−itα−1fs,u0s,Dqα−1u0s,…,Dqα−n−1u0sdqs≤1Γqα−Λa0+∑k=1nakRλkDqα−itα−1≤La0+∑k=1nakRλkDqα−itα−1≤RDqα−itα−1=Dqα−iu0t,i=1,2,…,n−1,Dqα−iu2t=Dqα−iTu1t≤Dqα−iTu0t=Dqα−iu1t,i=1,2,…,n−1.
By the induction, we have(92)Dqα−ium+1t≤Dqα−iumt,∀t∈J,i=1,2,…,n−1,m=0,1,2,….
In view of the complete continuity of the operatorT and um+1t=Tumt, umtm=0∞ is relatively compact. That is, umtm=0∞ has a convergent subsequence umktk=1∞ and there exists a u∗∈PR such that umk⟶u∗ as k⟶∞. This together with (90) implies limm⟶∞um=u∗. Since T is continuous and um+1t=Tumt, then we have u∗=Tu∗, that is, u∗ is a fixed point of operator T.
For the schemevmm=0∞, we use a similar discussion. For t∈J, by Lemma 5 and f∈CJ×ℝn,J, we have(93)v1t=Tv0=∫0+∞Gt,qsfs,v0s,Dqα−1v0sdqs,…,Dqα−n−1v0sdqs≥0=v0t.
Applying the monotonicity of functionf, we have(94)v2t=Tv1t≥Tv0t=v1t,∀t∈J.
By the induction, we have(95)vm+1t=Tvmtandvmt≤vm+1t,m=0,1,2,….
By Remark2 and Lemma 6, we get(96)Dqα−iv1t=Dqα−iTv0t=∫0+∞Gi∗t,qsfs,v0s,Dqα−1v0sdqs≥0=Dqα−iv0t,i=1,2,…,n−1.
By the induction, we have(97)Dqα−ivmt≤Dqα−ivm+1t,i=1,2,…,n−1,m=0,1,2,….
Using the complete continuity of the operatorT and the iterative scheme, it is easy to get(98)vm⟶v∗andTv∗=v∗.
Finally, we show thatu∗andv∗ are the maximal and minimal positive solutions of problem (1). Let wt be any positive solution of (1), then(99)v0t=0<wt≤Rtα−1=u0t,Dqα−iv0t≤Dqα−iwt≤Dqα−iu0t.
Noting thatT is increasing, then we have(100)v1t=Tv0t<Twt=wt≤Tu0t=u1t,for allt∈J.
Similarly, fort∈J and i=1,2,…,n−1, we can get(101)vmt≤wt≤umt,Dqα−ivmt≤Dqα−iwt≤Dqα−iumt.
Sincev∗=limm⟶∞vm and u∗=umm⟶∞lim, it follows that, for t∈J and i=1,2,…,n−1, we have(102)v0t≤v1t≤⋯≤vmt≤⋯≤v∗≤wt≤u∗≤⋯≤u1t≤u0t,Dqα−iv0t≤Dqα−iv1t≤⋯≤Dqα−ivmt≤⋯≤Dqα−iv∗t≤Dqα−iwt≤Dqα−iu∗t≤⋯≤Dqα−iu1t≤Dqα−iu0t.
This implies thatv∗t and u∗t are the minimal and maximal solutions of problem (1), and (53) and (54) hold.
On the other hand, we infer thatv∗t and u∗t are two positive solutions of (1). In fact, by the condition (H4), we get 0 function is not the solution of problem (1), which means v∗t>0, u∗t>0, and they can be constructed by means of two monotone iterative schemes in (81) and (82).Theorem 2.
Suppose conditions (H1) and (H3) are satisfied. If(103)A≔L∑k=1nbk<1,then problem (1) has a unique positive solution u˜t in P. Moreover, there is an iterative scheme umm=0∞ such that um⟶u˜ as m⟶∞ uniformly on any finite interval of J, where(104)um=Tum−1,m=1,2,…,hold. L is defined in (80). In addition, there is an error estimate for the approximation scheme(105)um−u˜=Am−11−Au1−u0,m=1,2,….Proof.
Chooser>Lκ/1−A, where κ is defined in (H3). Let Pr=u∈P|u≤r. For any u∈Pr, by Lemmas 5–7 and 10, we have(106)Tu1+tα−1∞≤L∑k=1nbkr+κ<Ar+Lκ,∀u∈Pr,Dqα−iTu1+ti−1∞≤L∑k=1nbkr+κ<Ar+Lκ,∀u∈Pr,i=2,3,…,Dqα−1Tu∞≤L∑k=1nbkr+κ<Ar+Lκ,∀u∈Pr.
Thus, we have(107)Tu≤Ar+Lκ<Ar+1−Ar=r,∀u∈Pr,i.e., TPr⊂Pr.
In the following, we show thatT is a contraction. For any u1,u2∈Pr, by condition (H3) and Lemma 10, we obtain(108)Tu21+tα−1−Tu11+tα−1∞=supt∈J∫0+∞Gt,qs1+tα−1fs,u2s,Dqα−1u2s,…,Dqα−n−1u2s−fs,u1s,Dqα−1u1s,…,Dqα−n−1u1sdqs≤L∫0+∞b1s1+sα−1u2s−u1s1+sα−1+b2sDqα−1u2s−Dqα−1u1s+b3s1+sDqα−2u2s−Dqα−2u1s1+s+⋯+bi+1s1+si−1Dqα−iu2s−Dqα−iu1s1+si−1+⋯+bns1+sn−2Dqα−n−1u2s−Dqα−n−1u1s1+sn−2dqs≤Lb1u2−u11+tα−1∞+b2Dqα−1u2−Dqα−1u1∞+⋯+bnDqα−n−1u2−Dqα−n−1u1∞≤L∑k=1nbku2−u1∞.
In the same way, we have(109)Dqα−iTu21+ti−1−Dqα−iTu11+ti−1∞≤L∑k=1nbku2−u1∞,i=2,3,…,n−1,(110)Dqα−iu2−Dqα−1u1≤L∑k=1nbku2−u1∞.
From (108)–(110), we have(111)Tu2−Tu1≤L∑k=1nbku2−u1=Au2−u1,∀u1,u2∈Pr.
SinceA<1, then T is a contraction. The Banach fixed point theorem ensures that T has a unique fixed point u˜∈P. Hence, problem (1) has a unique positive solution u˜.
Moreover, for anyu0∈P,umm=0∞,um−u˜⟶0 as m⟶0, where um=Tum−1. By (111), we have(112)um−um−1≤Aum−1−um−2≤⋯≤Am−1u1−u0,(113)um−ul≤um−um−1+um−1−um−2+⋯+ul+1−ul≤Am−1u1−u0+Am−2u1−u0+⋯+Alu1−u0=Am−11−Am−l1−Au1−u0.
Letl⟶∞ on both sides of (113), we obtain(114)um−u˜≤Am−11−Au1−u0.Remark 1.
The functionf in problem (1) relies on lower-order fractional q-derivative of unknown function which is different from [17].Remark 2.
The results in Theorems1 and 2 extend the fractional q-derivative from the low-order to the high-order fractional derivative.Remark 3.
The two iterative sequences in Theorem1 begin with some simple functions which are helpful for computational purpose.


## 4. Example

Consider the following equation:(115)Dq5/2ut+ft,ut,Dq3/2ut,Dq1/2ut=0,u0=Dqu0=0,Dq3/2u+∞=∫0+∞e−tt−3/2utdqt,where n=3,α=5/2,q=1/2,gt=e−tt−3/2,λ1=0.1,λ2=0.2,λ3=0.4, and function(116)ft,ut,Dq3/2ut,Dq1/2ut=45+t2+e−tut0.11+t3/20.1+2tDq3/2u0.25+t22+e−2tDq1/2ut0.41+t0.4.By computation, we deduce that(117)ft,ut,Dq3/2ut,Dq1/2ut≤45+t2+e−tut0.11+t3/20.1+2tDq3/2u0.25+t22+e−2tDq1/2ut0.41+t0.4=a0t+a1tut0.1+a2tDq3/2ut0.2+a3tDq1/2ut0.4,a0=∫0+∞a0tdqt=∫0+∞45+t2dqt<112,a1=∫0+∞a1t1+t3/20.1dqt=∫0+∞e−t1+t3/20.11+t3/20.1dqt<∫0+∞1etdqt<1e,a2=∫0+∞a2tdqt=∫0+∞2t5+t22dqt<124,a3=∫0+∞a1t1+t0.4dqt=∫0+∞e−2t1+t0.40.41+t0.40.4dqt<∫0+∞1e2tdqt<12e,which means that hypothesis (H2) is satisfied.On the other hand,Λ=∫0+∞t3/2gtdqt=∫0+∞t3/2e−tt3/2dqt=∫0+∞e−tdqt<1/e<Γ1/25/2, and ft,0,0,0≡0. So condition (H1) is satisfied. Moreover, from the expression of the function f, it is obvious that f is increasing with respect to the variables u1,u2,u3,∀t∈J. Hence, condition (H4) is satisfied. By Theorem 1, it follows that problem (1) has two positive solutions, which can be given by the limits of two explicit monotone iterative schemes in (81) and (82).


## 5. Conclusions

In this paper, we study the boundary value problem of a fractionalq-difference equation with nonlocal integral boundary conditions on the half-line. We obtain some new results as follows: (1) the existence and uniqueness of positive solutions for a higher-order nonlinear fractional q-difference equation with integral boundary conditions on the half-line are obtained, and (2) the monotone iterative schemes of positive solutions for the problem which considered in our paper are obtained.



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