# Positive Solutions of Nonlocal Boundary Value Problem for High-Order Nonlinear Fractionalq-Difference Equations

Authors: Changlong Yu; Jufang Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the nonlinearq-difference equations of fractional order (Dqαu)(t)+f(t,u(t))=0,
0<t<1,
(Dqiu)(0)=0,
(Dqβu)(1)=a(Dqβu)(η),
0≤i≤n-2, where Dqα is the fractional q-derivative of the Riemann-Liouville type of order α,
n-1<α≤n,
α>2,
1≤β≤n-2, and 0≤a≤1. We obtain the existence and multiplicity results of positive solutions by using some fixed point theorems. Finally, we give examples to illustrate the results.


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## Body

## 1. Introduction

Theq-difference calculus or quantum calculus is an old subject that was initially developed by Jackson [1, 2]. It is rich in history and in applications as the reader can find in the work by Ernst [3]. For some recent existence results on q-difference equations, see [4–7] and the references therein.The fractionalq-difference calculus had its origin in the works by Al-Salam and Agarwal. Henceforth, fractional q-difference equations have gained considerable importance due to their application in various sciences, such as physics, chemistry, aerodynamics, biology, economics, control theory, mechanics, electricity, signal and image processing, biophysics, blood flow phenomena, and fitting of experimental data. It has been a significant development in difference equations involving fractional q-derivatives; see [8–11] and references therein. As well known, fractional differential equations boundary value problems is currently under strong research, see [12–21] and references therein. In particular, in recent years, fractional q-difference boundary value problem (BVP) was in its infancy, and many people begin to study the existence of positive solutions for this kind of BVP; see [22–27] and references therein. However, there are few related results available. Lots of work and development should be done in the future.Recently, in [16], Li et al. considered the BVP of nonlinear fractional difference equation
(1)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=0,D0+βu(1)=aD0+βu(ξ),
where 1<α≤2, 0≤β≤1, 0≤a≤1, ξ∈(0,1), and f:[0,1]×[0,∞)→[0,∞) satisfies Caratheodory type conditions.More recently, in [23], Ferreira considered the BVP of fractional q-difference equation
(2)(Dqαy)(x)=-f(x,y(x)),0<x<1,y(0)=(Dqy)(0)=0,(Dqy)(1)=β≥0,
where 2<α≤3 and f:[0,1]×R→R is a nonnegative continuous function.Motivated by the work above, in this paper, we will discuss the following BVP:(3)(Dqαu)(t)+f(t,u(t))=0,0<t<1,(Dqiu)(0)=0,(Dqβu)(1)=a(Dqβu)(η),0≤i≤n-2,
where n-1<α≤n(n>2), 1≤β≤n-2, 0<η<1, 0≤a≤1, and f:[0,1]×[0,∞)→[0,∞) satisfies Caratheodory type conditions. We discuss the existence of positive solutions for BVP(3) and obtain multiplicity results which extend and improve the known results by using some fixed point theorems.


## 2. Preliminary Results

In this section, we introduce definitions and preliminary facts which are used throughout this paper.Letq∈(0,1) and define
(4)[a]q=1-qa1-q,a∈R.Theq-analogue of the power function (a-b)n with n∈N0 is
(5)(a-b)(0)=1,(a-b)(n)=∏k=0n-1‍(a-bqk),n∈N,a,b∈R.
More generally, if α∈R, then
(6)(a-b)(α)=aα∏n=0n-1‍a-bqna-bqα+n.
Note that, if b=0, then a(α)=aα. The q-gamma function is defined by
(7)Γq(x)=(1-q)(x-1)(1-q)x-1,x∈R∖{0,-1,-2,…},
and satisfies Γq(x+1)=[x]qΓq(x).Then, let us recall some basic concepts ofq-calculus [28].Definition 1.
For0<q<1, we define the q-derivative of a real-value function f as
(8)(Dqf)(x)=f(x)-f(qx)(1-q)x,(Dqf)(0)=limx→0(Dqf)(x).
Note that limq→1-Dqf(x)=f′(x).Definition 2.
The higher-orderq-derivatives are defined inductively as
(9)(Dq0f)(x)=f(x),(Dqnf)(t)=Dq(Dqn-1f)(t),n∈N.Definition 3.
Theq-integral of a function f in the interval [0,b] is given by
(10)(Iqf)(x)=∫0x‍f(t)dqt=x(1-q)∑n=0∞‍f(xqn)qn,x∈[0,b].
If a∈[0,b] and f is defined in the interval [0,b], its integral from a to b is defined by
(11)∫ab‍f(t)dqt=∫0b‍f(t)dqt-∫0a‍f(t)dqt.
Similarly as done for derivatives, an operator Iqn can be defined; namely,
(12)(Iq0f)(x)=f(x),(Iqnf)(x)=Iq(Iqn-1f)(x),n∈N.
Observe that(13)DqIqf(x)=f(x),
and if f is continuous at x=0, then IqDqf(x)=f(x)-f(0).
We now point out three formulas (Diq denotes the derivative with respect to variable i):
(14)[a(t-s)](α)=aα(t-s)(α),tDq(t-s)(α)=[α]q(t-s)(α-1),xDq∫0x‍f(x,t)dqt=∫0x‍Dxqf(x,t)dqt+f(qx,x).Remark 4.
We note that ifα≥0 and a≤b≤t, then (t-a)(α)≥(t-b)(α) [19].Definition 5 (see [9]).
Letα≥0 and let f be a function defined on [0,1]. The fractional q-integral of the Riemann-Liouville type is (Iq0f)(x)=f(x) and
(15)(Iqαf)(x)=1Γq(α)∫0x‍(x-qt)(α-1)f(t)dqt,α>0,x∈[0,1].Definition 6 (see [11]).
The fractionalq-derivative of the Riemann-Liouville type of α≥0 is defined by (Dq0f)(x)=f(x) and
(16)(Dqαf)(x)=(DqmIqm-αf)(x),α>0,
where m is the smallest integer greater than or equal to α.Lemma 7 (see [9, 11]).
Letα,β≥0 and let f be a function defined on [0,1]. Then, the next formulas hold:(1)
(IqβIqαf)(x)=(Iqα+βf)(x),(2)
(DqαIqαf)(x)=f(x).Remark 8.
Assume thatg(t)∈[0,1] and α,β are two constants such that α>2≥β≥1. Then
(17)Dqβ∫0t‍(t-qs)(α-1)g(s)dqs=Γq(α)Γq(α-β)∫0t‍(t-qs)(α-β-1)g(s)dq.Proof.
From Lemma7, we can get
(18)(DqβIqαg)(t)=(DqβIqβIqα-βg)(t)=(Iqα-βg)(t),
so
(19)Dqβ1Γq(α)∫0t‍(t-qs)(α-1)g(s)dqs=1Γq(α-β)∫0t‍(t-qs)(α-β-1)g(s)dq;
that is, (17) holds. The proof is completed.Lemma 9 (see [22]).
Letα>0 and let p be a positive integer. Then, the following equality holds:
(20)(IqαDqpf)(x)=(DqpIqαf)(x)-∑k=0p-1‍xα-p+kΓq(α+k-p+1)(Dqkf)(0).Lemma 10.
Letg(t)∈L[0,1]; then the unique solution of
(21)(Dqαu)(t)+g(t)=0,0<t<1,(Dqiu)(0)=0,(Dqβu)(1)=a(Dqβu)(η),0≤i≤n-2,
is
(22)u(t)=∫01‍G(t,qs)g(s)dqs,
where
(23)G(t,s)=1ΛΓq(α)×{(1-s)(α-β-1)tα-1-a(η-s)(α-β-1)tα-1-Λ(t-s)(α-1),0≤s≤min(η,t)≤1,(1-s)(α-β-1)tα-1-a(η-s)(α-β-1)tα-1,0≤t≤s≤η≤1,(1-s)(α-β-1)tα-1-Λ(t-s)(α-1),0≤η≤s≤t≤1,(1-s)(α-β-1)tα-1,0≤max(t,η)≤s≤1,
where Λ=1-aηα-β-1.Proof.
Letu(t) be a solution of (21); in view of Lemma 7 and Lemma 9, (21) is equivalent to the integral equation
(24)u(t)=c1tα-1+c2tα-2+⋯+cntα-n-1Γq(α)∫0t‍(t-qs)(α-1)g(s)dqs,
where c1,c2,…,cn are some constants to be determined. The boundary conditions (Dqiu)(0)=0, 0≤i≤n-2, imply that c2=c3=⋯=cn=0. Thus,
(25)u(t)=c1tα-1-1Γq(α)∫0t‍(t-qs)(α-1)g(s)dqs.
By Remark 8, we have
(26)(Dqβu)(t)=c1Γq(α)Γq(α-β)tα-β-1-1Γq(α-β)×∫0t‍(t-qs)(α-β-1)g(s)dqs.
For (Dqβu)(1)=a(Dqβu)(η),
(27)c1=1(1-aηα-β-1)Γq(α)[∫01‍(1-qs)(α-β-1)g(s)dqs-a∫0η‍(η-qs)(α-β-1)g(s)dqs].
Hence,
(28)u(t)=tα-1ΛΓq(α)[∫01‍(1-qs)(α-β-1)g(s)dqs-a∫0η‍(η-qs)(α-β-1)g(s)dqs]-1Γq(α)∫0t‍(t-qs)(α-1)g(s)dqs=∫01‍G(t,qs)g(s)dqs.
The proof is complete.Remark 11.
For the special case wherea=0, it is easy to see that G(t,s) can be written as
(29)G(t,s)=1Γq(α)×{(1-s)(α-β-1)tα-1-(t-s)(α-1),0≤s≤t≤1,(1-s)(α-β-1)tα-1,0≤t≤s≤1.Lemma 12.
Green functionG in Lemma 10 satisfies the following conditions:(i)
G(t,qs)≥0 for t,s∈[0,1];(ii)
G(t,qs)≤G(1,qs) for t,s∈[0,1];(iii)
G(t,qs)≥tα-1G(1,qs) for t,s∈[0,1].Proof.
Let(30)g1(t,s)=(1-s)(α-β-1)tα-1-a(η-s)(α-β-1)tα-1-Λ(t-s)(α-1),0≤s≤min(η,t)≤1,g2(t,s)=(1-s)(α-β-1)tα-1-a(η-s)(α-β-1)tα-1,0≤t≤s≤η≤1,g3(t,s)=(1-s)(α-β-1)tα-1-Λ(t-s)(α-1),0≤η≤s≤t≤1,g4(t,s)=(1-s)(α-β-1)tα-1,0≤max(t,η)≤s≤1.
We first prove part (i). Fort,s∈[0,1], from Remark 4, for t≠0,
(31)g1(t,qs)=(1-qs)(α-β-1)tα-1-aηα-β-1×(1-qsη)(α-β-1)tα-1-Λ(1-qst)(α-1)tα-1=Λtα-1[(1-qs)(α-β-1)-(1-qst)(α-1)]+aηα-β-1tα-1[(1-qsη)(α-β-1)(1-qs)(α-β-1)-(1-qsη)(α-β-1)]≥Λtα-1[(1-qs)(α-β-1)-(1-qs)(α-1)]+aηα-β-1tα-1[(1-qs)(α-β-1)-(1-qs)(α-β-1)]≥0.
Since g1(t,qs)≥0, it is easy to know g2(t,qs)≥0, g3(t,qs)≥0, and g4(t,qs)≥0. Therefore, G(t,qs)≥0.
Next, we prove part (ii). Fixs∈[0,1], and
(32)tDqg1(t,qs)=[α-1]q(1-qs)(α-β-1)tα-2-aηα-β-1[α-1]q(1-qsη)(α-β-1)tα-2-Λ[α-1]q(1-qst)(α-2)tα-2=[α-1]qtα-2{qstΛ[qst(1-qst)(α-2)(1-qs)(α-β-1)-(1-qst)(α-2)]+aηα-β-1×[(1-qsη)(α-β-1)(1-qs)(α-β-1)-(1-qsη)(α-β-1)]}≥[α-1]qtα-2{Λ[(1-qs)(α-2)(1-qs)(α-β-1)-(1-qs)(α-2)]}≥0.
That is, g1(t,qs) is increasing function of t. By the same way, we can conclude that g2(t,qs), g3(t,qs), and g4(t,qs) are increasing functions of t for fixed s∈[0,1]. Thus, G(t,qs)≤G(1,qs) for t,s∈[0,1].
Finally, we prove part (iii). Suppose that0≤qs≤min{t,η}≤1; then
(33)G(t,qs)G(1,qs)=(1-qs)(α-β-1)tα-1-a(η-qs)(α-β-1)tα-1-Λ(t-qs)(α-1)(1-qs)(α-β-1)-a(η-qs)(α-β-1)-Λ(1-qs)(α-1)=tα-1[(1-qs)(α-β-1)-a(η-qs)(α-β-1)-Λ(1-qs/t)(α-1)](1-qs)(α-β-1)-a(η-qs)(α-β-1)-Λ(1-qs)(α-1)≥tα-1.
For other circumstances, we also getG(t,qs)≥tα-1G(1,qs) and this completes the proof.Remark 13.
Let0<τ<1; then 0<τα-1<1 and
(34)mint∈[τ,1]G(t,qs)≥τα-1G(1,qs)fors∈[0,1].Lemma 14 (see [29]).
LetX be a Banach space with C⊂X being closed and convex. Assume that U is a relatively open subset of C with 0∈U and T:U¯→C is complete continuous. Then either(i)
T has a fixed point in U¯, or(ii)
there existu∈∂U and λ∈(0,1) with u=λTu.Lemma 15 (see Krasnoselskii’s [30]).
LetE be a Banach space, and K∈E is a cone in E. Assume that Ω1 and Ω2 are open subsets of E with 0∈Ω1 and Ω¯1⊂Ω2. Let T:K∩(Ω¯2∖Ω1)→K be a completely continuous operator. In addition, suppose that either(H1)
∥Tu∥≤∥u∥, for all u∈K∩∂Ω1 and ∥Tu∥≥∥u∥, for all u∈K∩∂Ω2 or(H2)
∥Tu∥≤∥u∥, for all u∈K∩∂Ω2 and ∥Tu∥≥∥u∥, for all u∈K∩∂Ω1holds. Then T has a fixed point in K∩(Ω¯2∖Ω1).Lemma 16 (see [31]).
LetP be a cone in a real Banach space E, Pc={x∈P:∥x∥<c}, θ is a nonnegative continuous concave functional on P such that θ≤∥x∥, for all x∈Pc¯, and P(θ,b,d)={x∈P:b≤θ(x),∥x∥≤d}. Suppose that T:Pc¯→Pc¯ is completely continuous and there exist positive constants 0<a<b<d≤c such that(H1)
{x∈P(θ,b,d):θ(x)>b}≠∅ and θ(Tx)>b for x∈P(θ,b,d),(H2)
∥Tx∥<a for x∈Pa¯,(H3)
θ(Tx)>b for x∈P(θ,b,d) with ∥Tx∥>d.Then T has at least three fixed points x1,x2, and x3 with
(35)∥x∥<a,b<θ(x2),a<∥x3∥withθ(x3)<b.Remark 17.
Ifd=c, then (H1) implies (H3).


## 3. Main Result

In this section, we will consider the question of positive solutions for BVP (3). At first, we prove some lemmas required for the main result.LetE=C[0,1] be the Banach space endowed with the norm ∥u∥=supt∈[0,1]|u(t)|. Let τ=qn for a given n∈N, and define the cone P⊂E by
(36)P={u∈E:u(t)≥0,mint∈[τ,1]u(t)≥τα-1∥u∥}.Let the nonnegative continuous concave functionalθ on the cone P be defined by
(37)θ(x)=minτ≤t≤1|u(t)|.In this paper, we assume thatf:[0,1]×[0,+∞)→[0,+∞] satisfies the following conditions of Caratheodory type:(D1)
f(t,u) is Lebesgue measurable with respect to t on [0,1];(D2)
f(t,u) is continuous with respect to u on [0,+∞).Theorem 18.
Assume that the conditions(D1) and (D2) hold. Suppose further that there exists a real-valued function h(t)∈L[0,1] such that |f(t,u)-f(t,v)|≤h(t)|u-v| for almost every t∈[0,1] and all u,v∈[0,+∞). If
(38)0<∫01‍G(1,qs)h(s)dqs<1,
then there exist unique positive solutions of BVP (3) on [0,1].Proof.
Consider the operatorT:P→P defined by
(39)Tu(t):=∫01‍G(t,qs)f(s,u(s))dqs<1.
For any u,v∈P, we have
(40)|Tu(t)-Tv(t)|=|∫01‍G(t,qs)(f(s,u(s))-f(s,v(s)))dqs|≤∫01‍G(t,qs)|(f(s,u(s))-f(s,v(s)))|dqs≤∫01‍G(t,qs)h(s)|u(s)-v(s)|dqs≤∫01‍G(1,qs)h(s)|u(s)-v(s)|dqs≤∥u-v∥∫01‍G(1,qs)h(s)dqs<∥u-v∥.
This implies that T is a contraction mapping. By the Banach contraction mapping principle, we deduce that T has a unique fixed point which is obviously a solution of BVP (3). The proof is complete.Corollary 19.
Assume that the conditions(D1) and (D2) hold. Suppose further that there exists a positive constant L∈(0,1/A) with |f(t,u)-f(t,v)|≤L|u-v|,t∈[0,1],(u,v)∈[0,+∞), where
(41)A=∫01‍G(1,qs)dqs;
then there exists a unique positive solution of BVP (3) on [0,1].Corollary 20.
Assume that the conditions(D1) and (D2) hold. Suppose further that there exists a real-valued function h(t)∈L[0,1] such that |f(t,u)-f(t,v)|≤h(t)|u-v|, t∈[0,1], (u,v)∈[0,+∞). If
(42)0<∫01‍h(s)dqs<ΛΓq(α),
then there exists a unique positive solution of BVP (3) on [0,1].Corollary 21.
Assume that the conditions(D1) and (D2) hold. Suppose further that there exists a positive constant L∈(0,ΛΓq(α)) with |f(t,u)-f(t,v)|≤L|u-v|, t∈[0,1], (u,v)∈[0,+∞); then there exists a unique positive solution of BVP (3) on [0,1].Next, we discuss multiple solutions of BVP (3).Lemma 22.
Assume that the conditions(D1) and (D2) hold. Suppose further that there exist two nonnegative real-valued functions m,n∈L[0,1] such that |f(t,u)|≤n(t)+m(t)u for almost every t∈[0,1] and all u,v∈[0,+∞). Then the operation T:P→P defined by (39) is completely continuous.Proof.
We will divide the proof into three parts.(I)
We show thatT:P→P is continuous.
For anyun,u∈P, n=1,2,…, with limn→∞∥un-u∥=0, we have
(43)limn→∞un(t)=u(t),t∈[0,1].
Thus
(44)limn→∞f(t,un(t))=f(t,u(t)),t∈[0,1].
So, we can obtain that
(45)sups∈[0,1]|f(s,un(s))-f(s,u(s))|⟶0,asn⟶∞.
On the other hand, we have
(46)|(Tun)(t)-(Tu)(t)|=|∫01‍G(t,qs)(f(s,un(s))-f(s,u(s)))dqs|≤sups∈[0,1]|f(s,un(s))-f(s,u(s))|×∫01‍G(t,qs)dqs.
It implies that
(47)∥Tun-Tu∥≤sups∈[0,1]|f(s,un(s))-f(s,u(s))|×∫01‍G(t,qs)dqs.
Therefore, ∥Tun-Tu∥→0 as n→∞. This means that T is continuous.(II)
We will prove thatT maps bounded sets into bounded sets in P.
For anyξ>0, there exists a positive constant l>0 such that for each u∈Bξ={u∈P:∥u∥≤ξ}, we have ∥Tu∥<l. By the definition of T, for each t∈[0,1], we get
(48)|Tu(t)|=|∫01‍G(t,qs)f(s,u(s))dqs|≤∫01‍G(t,qs)(n(s)+m(s)u(s))dqs≤∫01‍G(1,qs)n(s)dqs+∥u∥∫01‍G(1,qs)m(s)dqs≤∫01‍G(1,qs)n(s)dqs+ξ∫01‍G(1,qs)m(s)dqs:=l.
That is, ∥Tu∥≤l.(III)
We will show thatT maps bounded sets into equicontinuous sets of P.
LetBξ⊂P be a bounded set, t1,t2∈[0,1] and t1<t2. For any u∈Bξ, we have
(49)|Tu(t2)-Tu(t1)|=|∫01‍(G(t2,qs)-G(t1,qs))f(s,u(s))dqs|≤∫01‍|G(t2,qs)-G(t1,qs)||f(s,u(s))|dqs≤∫01‍|G(t2,qs)-G(t1,qs)|(n(s)+m(s)u(s))dqs≤∫01‍|G(t2,qs)-G(t1,qs)|(n(s)+ξm(s))dqs.
Since G(t,qs) is continuous on [0,1]×[0,1], then G(t,qs) is uniformly continuous in [0,1]×[0,1]. Hence, for any ε>0, there exists δ>0, whenever |t2-t1|<δ, and we have
(50)|G(t2,qs)-G(t1,qs)|<ε1+∫01‍(n(s)+ξm(s))dqs.
So |Tu(t2)-Tu(t1)|<ε; that is, {Tu:u∈Bξ} is equicontinuous.
By Arzela-Ascoli theorem, we can conclude thatT is completely continuous. This completes the proof.Remark 23.
Iff:[0,1]×[0,∞)→[0,∞) is continuous, T is also completely continuous.Theorem 24.
Assume that all the assumptions of Lemma22 hold. If
(51)0<∫01‍G(1,qs)m(s)dqs<1,
then BVP (3) has at least one positive solution.Proof.
LetU={u∈P:∥u∥<r}, where
(52)r=∫01‍G(1,qs)n(s)dqs1-∫01‍G(1,qs)m(s)dqs>0
and T:U¯→P,Tu(t):=∫01‍G(t,qs)f(s,u(s))dqs. From Lemma 22, T is completely continuous.
Assume that there existu∈P and λ∈(0,1) such that u=λTu; we claim that ∥u∥≠r:
(53)u(t)=λ∫01‍G(t,qs)f(s,u(s))dqs≤λ∫01‍G(t,qs)(n(s)+m(s)u(s))dqs≤λ(∫01‍G(t,qs)n(s)dqs+∥u∥∫01‍G(t,qs)m(s)dqs),
and then
(54)∥u∥≤λ(∫01‍G(1,qs)n(s)dqs+r∫01‍G(1,qs)m(s)dqs)<r.
That is, u∉∂U. By Lemma 14, T has a fixed point u∈U¯. Therefore, BVP (3) has at least a positive solution. The proof is complete.In the following, we set(55)M=(∫01‍G(1,qs)dqs)-1,N=(∫τ1‍τα-1G(1,qs)dqs)-1.Theorem 25.
Assume that all the assumptions of Lemma22 hold. If there exist two positive constants r2>r1>0 such that
(56)f(t,u)≥Nr1for(t,u)∈[0,1]×[0,r1],(57)f(t,u)≤Mr2for(t,u)∈[0,1]×[0,r2],
then BVP (3) has at least one positive solution.Proof.
BecauseT:P→P is completely continuous, we just only show that u=Tu has a solution u(t)>0 for t∈[0,1].
LetΩ1={u∈E:∥u∥<r1}. For any u∈P∩∂Ω1, we know τα-1r1≤u(t)≤r1 on [τ,1]. Using (56) and (34), we have
(58)|Tu(t)|=|∫01‍G(t,qs)f(s,u(s))dqs|≥∫τ1‍G(t,qs)f(s,u(s))dqs≥Nr1∫τ1‍G(t,qs)τα-1dqs=r1=∥u∥,
which implies that ∥Tu∥≥∥u∥, u∈P∩∂Ω1.
LetΩ2={u∈E:∥u∥<r2}. For any u∈P∩∂Ω2, we get 0≤u(t)≤r2 on [0,1]. Using (57), we have
(59)|Tu(t)|=|∫01‍G(t,qs)f(s,u(s))dqs|≤Mr2∫01‍G(1,qs)dqs=r2=∥u∥;
that is, ∥Tu∥≤∥u∥, u∈P∩∂Ω2.
In view of Lemma15, T has a fixed point u0∈P∩(Ω¯2∖Ω1) which is the solution of BVP (3).Theorem 26.
Assume that all assumptions of Lemma22 hold. If there exist constants 0<A<B<C such that
(60)(G1)f(t,u)≤MA,(t,u)∈[0,1]×[0,A],(G2)f(t,u)≤MC,(t,u)∈[0,1]×[0,C],(G3)f(t,u)≥NB,(t,u)∈[τ,1]×[B,C],
hold, then BVP (3) has at least three positive solutions u1,u2, and u3 with
(61)∥u1∥<A,B<θ(u2)<∥u2∥≤C,A<∥u3∥,θ(u3)≤B.Proof.
First, ifu∈P¯C={u∈P:∥u∥<C}, then ∥u∥≤C. So 0≤u(t)≤C, t∈[0,1]. By (G2), we have
(62)|Tu(t)|=|∫01‍G(t,qs)f(s,u(s))dqs|≤MC∫01‍G(1,qs)dqs=C,
which implies that ∥Tu∥≤C,u∈P¯C. Hence, T:P¯C→P¯C. In view of Lemma 22, T:P¯C→P¯C is completely continuous.
By using the analogous argument, from(G1), we can get that if u∈P¯A, then ∥Tu∥<A.
Setu(t)=(B+C)/2, t∈[0,1], so u(t)=(B+C)/2∈P(θ,B,C), θ(u)=(B+C)/2>B. Therefore, {u∈P(θ,B,C)|θ(u)>B}≠∅.
On the other hand, ifu∈P(θ,B,C), then B≤u(t)≤C,t∈[τ,1]. By (G3), we have
(63)θ(Tu)=mint∈[τ,1]|Tu(t)|=mint∈[τ,1]|∫01‍G(t,qs)f(s,u(s))dqs|≥mint∈[τ,1]∫01‍G(t,qs)NBdqs,≥NB∫τ1‍τα-1G(1,qs)dqs=B,
which implies that θ(Tu)>B, for u∈P(θ,B,C).
By Lemma16, BVP(3) has at least three positive solutions u1, u2, and u3 with
(64)∥u1∥<A,B<θ(u2)<∥u2∥<C,A<∥u3∥,θ(u3)<B.
The proof is complete.


## 4. Example

Example 27.
Consider the following BVP:(65)(D1/24.5u)(t)+etu1+et+t2+1=0,0<t<1,(D1/2iu)(0)=0,(D1/21.5u)(1)=1100(D1/21.5u)(910),0≤i≤3.
Letf(t,u)=etu/(1+et)+t2+1, (t,u)∈[0,1]×[0,+∞), h(t)=et/(1+et), so |f(t,u)-f(t,v)|≤h(t)|u-v|, for (t,u),(t,v)∈[0,1]×[0,+∞).
By simple calculation, we get(66)0<∫01‍G(1,qs)h(s)dqs≤1ΛΓq(α)=0.8471⋯<1.
All conditions of Theorem 18 are satisfied. Thus, BVP (65) has a unique positive solution.Example 28.
Consider the following BVP:(67)(D1/24.5u)(t)+14(sint+1)u1+u2+t2+1=0,0<t<1,(D1/2iu)(0)=0,(D1/21.5u)(1)=1100(D1/21.5u)(910),0≤i≤3.
Letn(t)=t2+1, m(t)=(1/4)(sint+1). It is easy to check that f(t,u)≤n(t)+m(t)u for (t,u)∈[0,1]×[0,+∞). Since
(68)0<∫01‍G(1,qs)m(s)dqs≤1ΛΓq(α)·12=0.4235⋯<1,
by Theorem 24, BVP (67) has at least one positive solution.Example 29.
Consider the following BVP:(69)(D1/22.5u)(t)+u4+sin2t5+1=0,0<t<1,(D1/2iu)(0)=0,(D1/2u)(1)=12(D1/2u)(34),0≤i≤1.
Letf(t,u)=u/4+sin2t/5+1, (t,u)∈[0,1]×[0,+∞). By calculation, we get
(70)M=(∫01‍G(1,qs)dqs)-1≥ΛΓq(α)=1.4809…,N=(∫τ1‍τα-1G(1,qs)dqs)-1≤2.414….
Choosing r1=1/3, r2=1, we have
(71)f(t,u)=u4+sin2t5+1≤1.45≤Mr2,(t,u)∈[0,1]×[0,r2],f(t,u)=u4+sin2t5+1≥1≥Nr1,(t,u)∈[0,1]×[0,r1].
By Theorem 25, BVP (69) has at least one positive solution u such that 1/3≤∥u∥≤1.Example 30.
Consider the following BVP:(72)(D1/22.5u)(t)+f(t,u)=0,0<t<1,(D1/2iu)(0)=0,(D1/2u)(1)=12(D1/2u)(34),0≤i≤1.
Here,
(73)f(t,u)={t4+4u4,(t,u)∈[0,1]×[0,1],t+74+u,(t,u)∈[0,1]×[1,+∞).
By Example29, we have
(74)M=(∫01‍G(1,qs)dqs)-1≥ΛΓq(α)=1.4809…,N=(∫τ1‍τα-1G(1,qs)dqs)-1≤2.414….
Choosing A=1/2, B=1, C=5, we have
(75)f(t,u)=t4+4u4≤0.5≤MA,(t,u)∈[0,1]×[0,12],f(t,u)=t+74+u≤7≤MC,(t,u)∈[0,1]×[0,5],f(t,u)=t+74+u≥2.875≥NB,(t,u)∈[12,1]×[1,5].
By Theorem 26, BVP (72) has at least three solutions u1, u2, and u3 such that
(76)∥u1∥<12,1<θ(u2),12<∥u3∥,θ(u3)<1.


## 5. Conclusion

In this paper, we obtain the existence and multiplicity results of positive solutions of BVP for high-order fractionalq-difference equations by some fixed point theorems, which enrich the theories for fractional q-difference equations, and provide the theoretical guarantee for the  application of fractional q-difference equations in every field. In the future, we will use bifurcation theory, critical point theory, variational method, and other methods to continue our works in this area.



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