# Multiple Positive Solutions for a Class of Boundary Value Problem of Fractionalp,q-Difference Equations under p,q-Integral Boundary Conditions

Authors: Yongyang Liu; Yansheng Liu
Journal: Journal of Mathematics
 (2021)

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## Abstract

This paper is mainly concerned with a class of fractionalp,q-difference equations under p,q-integral boundary conditions. Multiple positive solutions are established by using the topological degree theory and Krein–Rutman theorem. Finally, two examples are worked out to illustrate the main results.


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## Body

## 1. Introduction

Theq-difference operator was first systematically studied by Jackson [1]. Then, q-calculus has been studied extensively. See [2–4] and references therein. q-calculus and q-difference equations have been used by many researchers to solve physical problems such as molecular problems and chemical physics [1, 5–7]. For example, in 1967, Floreanini and Vinet [3] studied the behaviors of hydrogen atoms by using Schrödinger equation and q-calculus. Diaz and Osler [8] investigated the q-field theory.In the last decades, the theory of quantum calculus based on two-parameterp,q-integer has been studied since it can be used efficiently in many fields such as difference equations, Lie group, hypergeometric series, and physical sciences. The p,q-calculus was first studied by Chakrabarti and Jagannathan [2] in the field of quantum algebra in 1991. Njionou Sadjang [9] systematically established the basic theory of p,q-calculus and some p,q-Taylor formula. Milovanovic and Gupta [10] developed the concept of p,q-beta and p,q-gamma functions. These basic concepts and theories promote the development of p,q-calculus. For detailed results on p,q-calculus, please see [9–13] and references therein.On the contrary, the research of fractional calculus in discrete settings was initiated in [8, 11, 14]. In 2020, Soontharanonl and Sitthiwirattham [15] introduced the fractional p,q-calculus, which has been found in a wide range of applications in many fields such as concrete mathematical models of quantum mechanics and fluid mechanics [7, 13, 15].As we all know, in recent decades, more and more researchers pay much attention to the fractional differential equations and have obtained substantial achievements, we refer the readers to see [16–34] and references therein. Although the results of discrete fractional calculus are similar to those of continuous fractional calculus, the theory of discrete fractional calculus remains much less developed than that of continuous fractional calculus [35, 36]. Therefore, it is very important to develop discrete calculus. In particular, the fractional p,q-difference equations involving p,q-integral boundary conditions have rarely been studied. In order to make up for this gap, the paper mainly studies the following boundary value problem of fractional p,q-difference equations under p,q-integral boundary conditions:(1)Dp,qαxt+ft,x=0,t∈0,1,x0=Dp,qx0=0,Dp,qx1=∫01htDp,qxtdp,qt,where 2<α<3,0<q<p≤1, f∈C0,1×R+,R+, and Dp,qα is fractional p,q-difference operator.It should be pointed out that the boundary conditions of BVP equation (1) are more extensive. Furthermore, two parameters in the discrete environment makes the boundary value problem more complex. In order to overcome these difficulties, we constructed a special cone. The existence and multiplicity of the positive solution for the BVP equation (1) are obtained by using the topological degree theory, Krein–Rutman theorem.This paper is structured as follows. In Section2, we introduce some definitions of p,q-fractional integral and differential operator together with some basic properties and lemmas. The main results are given and proved in Section 3. Finally, in Section 4, two examples are given to show the applicability of our main results.


## 2. Preliminaries

In this section, we list some basic definitions and lemmas that will be used in this paper. For0<q<p≤1, we let(2)kp,q≔pk−qkp−q=pk−1kmqq/p,k∈N,1,k=0.Thep,q-analogue of the power function a−bp,qn with n∈N0\coloneq 0, 1, 2, …, is given by(3)a−bp,q0≕1,a−bp,qn≔∏k=0n−1apk−bqk,a,b∈R.Forα∈R,(4)a−bp,qα≔aα∏i=0∞1−b/aq/pi1−b/aq/pα+i,a≠0.By [15], we obtain(5)a−bp,qα=pα/2a−bq/pα=aα∏i=0∞1pα1−b/aq/pi1−b/aq/pα+i,a≠0.Note thataqα=ap,qα=aα and 0qα=0p,qα=0 for α>0. The p,q-gamma and p,q-beta functions are defined by(6)Γp,qx≔p−qp,qx−1p−qx−1=1−q/pp,qx−11−q/px−1,x∈R∖0,−1,−2,…,,x−1p,q!,x∈N,Bp,qx,y≔∫01tx−11−qtp,qy−1dp,qt=p12y−12x+y−2Γp,qxΓp,qyΓp,qx+y,respectively.Definition 1 (see [15]).
For0<q<p≤1 and f:0,T⟶R, we define the p,q-difference of f as(7)Dp,qft≔fpt−fqtp−qt,t≠0,where Dp,qf0=f′0, provided that f is differentiable at 0.Definition 2 (see [15]).
ForN−1<α<N,0<q<p≤1, and f:Ip,qT⟶R, the fractional p,q-difference is defined by(8)Dp,qαft=p−qtN−αp2N−αΓp,qN−α∑k=0∞qkpk+11−qpk+1p,qN−α−1Dp,qNfqkpk+N−αt,where Ip,qT=qk/pk+1T:k∈ℕ0∪0.Definition 3 (see [15]).
LetI be any closed interval of R containing a,b, and 0. Assuming that f:I⟶R is a given function, we define p,q-integral of f from a to b by(9)∫abftdp,qt=∫0bftdp,qt−∫0aftdp,qt,where(10)Ip,qfx=∫axftdp,qt=p−qx∑k=0∞qkpk+1fqkpk+1x,x∈I,provided that the series converges at x=a and x=b. f is called p,q-integrable on a,b.Definition 4 (see [15]).
Forα>0,0<q<p≤1, and f:0,T⟶R, the fractional p,q-integral is defined by(11)Ip,qαft≔1p2αΓp,qα∫0tt−qsp,qα−1fspα−1dp,qs,and Ip,q0ft=ft.Lemma 1 (see [15]).
Letf,g be p,q-differentiable. The properties of p,q-difference operator are as follows:(i)
Dp,qft+gt=Dp,qft+Dp,qgt(ii)
Dp,qαft=αDp,qft, for α∈RLemma 2 (see [15]).
For0<q<p≤1, α≥1, and a∈R, we have(i)
Dp,qt−ap,qα=αp,qpt−ap,qα−1(ii)
Dp,qa−tp,qα=−αp,qa−qtp,qα−1Lemma 3 (see [15]).
Forα,β≥0 and 0<q<p≤1, p,q-integral and p,q-difference operators have the following properties:(i)
Ip,qαIp,qβfx=Ip,qβIp,qαfx=Ip,qα+βfx(ii)
Dp,qIp,qfx=fx and Ip,qDp,qfx=fx−f0Lemma 4.
Assumeh≥0, A=1−∫01httα−2dp,qt>0, and α∈2,3. If g∈C0,1, then the following boundary value problem,(12)Dp,qαxt+gt=0,t∈0,1,x0=Dp,qx0=0,Dp,qx1=∫01htDp,qxtdp,qt,has a unique solution(13)xt=∫01Gt,qsgspα−2dp,qs,where(14)Gt,qs=G0t,qs+tα−1A∫01htG1t,qsdp,qt,G0t,qs=1pα−1/2Γp,qα1p2α−2/α−2tα−11−qsα−2−t−qsα−1,0≤s≤t≤1,1p2α−2/α−2tα−11−qsα−2,0≤t≤s≤1,G1t,qs=1pα−1/2Γp,qα1p2α−2/α−2tα−21−qsα−2−t−qsα−2,0≤s≤t≤1;1p2α−2/α−2tα−21−qsα−2,0≤t≤s≤1.Proof.
According toDp,qαxt=−gt, we have(15)xt=C1tα−1+C2tα−2+C3tα−3−Ip,qαgt.
Fromx0=Dp,qx0=0, one can easily obtain C2=C3=0. Hence,(16)xt=C1tα−1−Ip,qαgt,Dp,qx1=C1Γp,qαΓp,qα−1−1pα−1/2Γp,qα−1∫011−qsα−2gspα−2dp,qs=∫01htDp,qxtdp,qt=∫01htC1Γp,qαΓp,qα−1tα−2−1pα−1/2Γp,qα−1∫0tt−qsα−2gspα−2dp,qsdp,qt.
Based on the hypothesis in Lemma4, we can deduce that(17)C1=1Apα−1/2Γp,qα∫011−qsα−2gspα−2dp,qs−1Apα−1/2Γp,qα∫01htdp,qt∫0tt−qsα−2gspα−2dp,qs.
Thus,(18)xt=C1tα−1−Ip,qαgt=1Apα−1/2Γp,qα∫01t1−qsα−2gspα−2dp,qs−1Apα−1/2Γp,qα∫01htdp,qt∫0ttα−1t−qsα−2gspα−2dp,qs−1pα/2Γp,qα∫0tt−qsα−1gspα−1dp,qs=1Apα−1/2Γp,qα∫01tα−21−qsα−2gspα−2dp,qs−1Apα−1/2Γp,qα∫01htdp,qt.−1pα−1/2Γp,qα∫0t1p2α−2/α−2t−qsα−1gspα−2dp,qs+1pα−1/2Γp,qα∫01tα−1p2α−2/α−21−qsα−1gspα−2dp,qs−1pα−1/2Γp,qα∫01tα−1p2α−2/α−21−qsα−1gspα−2dp,qs=∫01G0spα−2dp,qs+tα−1Apα−1/2Γp,qα∫01httα−2dp,qt∫011p2α−2/α−21−qsα−2gspα−1dp,qs−tα−1Apα−1/2Γp,qα∫01htdp,qt∫0t1p2α−2/α−2t−qsα−2gspα−2dp,qs=tα−1Apα−1/2Γp,qα∫01∫011p2α−2/α−2httα−21−qsα−2dp,qt−∫s11p2α−2/α−2htt−qsα−2dp,qtgspα−2dp,qs=∫01G0t,qsgspα−2dp,qs+tα−1A∫01∫01G1t,qsgspα−2dp,qs=∫01Gt,qsgspα−2dp,qs.
This completes the proof.Lemma 5.
The functionsGii=0,1 have the following properties:(1)
Git,qs≥0, for t,s∈0,1(2)
tα−1G01,qs≤G0t,qs≤G01,qs, for t,s∈0,1Proof.
(1)
On the one hand, for0≤s≤t≤1, we know(19)G0t,qs=1pα−1/2Γp,qα1p2α−2/α−2tα−11−qsα−2−t−qsα−1.Thus, fort≠0, it is easy to see that(20)G0t,qs=1pα−1/2Γp,qα1p2α−2/α−2tα−11−qsα−2−t−qsα−1=1pα−1/2Γp,qα1p2α−2/α−2tα−11−qsα−2−tα−11−qstα−1≥tα−1pα−1/2Γp,qα1−qsα−2−1−qsα−1≥0.Similarly, for0≤s≤t≤1, we know(21)G1t,qs=1pα−1/2Γp,qα1p2α−2/α−2tα−21−qsα−2−t−qsα−2.Thus, fort≠0, it is also easy to see that(22)G1t,qs=1pα−1/2Γp,qα1p2α−2/α−2tα−21−qsα−2−t−qsα−2=1pα−1/2Γp,qα1p2α−2/α−2tα−21−qsα−2−tα−21−qstα−2≥tα−2pα−1/2Γp,qα1−qsα−2−1−qsα−2=0.On the other hand, for0≤t≤s≤1, it is easy to see that, from Lemma 4, the conclusion is obviously established. Therefore, Git,qs≥0, for t,s∈0,1.(2)
Firstly, for0≤s≤t≤1, one can easily obtain that(23)Dp,qtG0t,qs=Dp,qt1pα−1/2Γp,qα1p2α−2/α−2tα−11−qsα−2−t−qsα−1=1pα−1/2Γp,qα1p2α−2/α−2α−1tα−21−qsα−2−α−1t−qsα−2=1pα−1/2Γp,qαα−1tα−21−qsα−2−α−1tα−21−qstα−2≥0.
For0≤t≤s≤1 and 2<α<3, we know that Gt,qs=tα−11−qsα−2 is obviously an increasing function with respect to t.
Therefore, fort,s∈0,1, G0t,qs is an increasing function with respect to t. Then, G0t,qs≤G01,qs.
Secondly, for0≤s≤t≤1, we have(24)G0t,qsG01,qs=1/p2α−2/α−2tα−11−qsα−2−t−qsα−11/p2α−2/α−21−qsα−2−1−qsα−1=tα−11/p2α−2/α−21−qsα−2−1−qs/tα−11/p2α−2/α−21−qsα−2−1−qsα−1≥tα−11/p2α−2/α−21−qsα−2−1−qsα−11/p2α−2/α−21−qsα−2−1−qsα−1=tα−1.
For0≤t≤s≤1, we have(25)G0t,qsG01,qs=tα−1.
Therefore,tα−1G01,qs≤G0t,qs, for t,s∈0,1.Lemma 6.
From Lemma5, the following conclusions are established:(26)tα−1θ1qs≤Gt,qs≤θ1qsfort,s∈0,1,Gt,qs≤tα−1θ2qsfort,s∈0,1,where(27)θ1s=G01,s+1A∫01htG1t,sdp,qs,θ2s=1Γp,qα1−sα−2+1A∫01htG1t,sdp,qs.Lemma 7 (see [37]).
LetΩ be a bounded open set in a Banach space E, and T:Ω⟶E is a continuous compact operator. If there exists x0∈E∖0 such that(28)x−Tx≠μx0,∀x∈∂Ω,μ≥0,then the topological degree degI−T,Ω,0=0.Lemma 8 (see [37]).
LetΩ be a bounded open set in a Banach space E with 0∈Ω, and T:Ω⟶E is a continuous compact operator. If(29)Tx≠μx,∀x∈∂Ω,μ≥1,then the topological degree degI−T,Ω,0=1.
LetE≔C0,1,x≔maxt∈0,1xt, and P≔x∈E:xt≥tα−1x,∀t∈0,1. Then, E,⋅ is a real Banach space and P is a cone on E. From Lemma 4, we can define operator T:E⟶E as follows:(30)Txt≔∫01Gt,qsfspα−2,xspα−2dp,qs,x∈E,where G is determined in Lemma 4. Obviously, T is a completely continuous operator.In addition, from Lemma4, we can obtain that the solution of BVP equation (12) is equivalent to(31)xt=λ∫01Gt,qsgspα−2dp,qs,t∈0,1.For our purposes, we need to define the operatorL by(32)Lxt=∫01Gt,qsxspα−2dp,qs,t∈0,1,x∈E.It is easy to prove thatL:E⟶E is a linear completely continuous operator and LP⊂P. Obviously, we know that L has a spectral radius, denoted by rL, that is not equal to 0. From Krein–Rutman theorem, we know that L has a positive eigenfunction φ1 corresponding to its first eigenvalue λ1=rL−1, i.e., φ1=λ1Lφ1.


## 3. Main Results

In this section, we shall establish the existence and multiplicity results of BVP equation (1), which is based on the topological degree theory. For convenience, let λ1 be the first eigenvalue of the following eigenvalue problem:(33)Dp,qαxt+λxt=0,t∈0,1,x0=Dp,qx0=0,Dp,qx1=∫01htDp,qxtdp,qt.Now, let us list the following assumptions satisfied throughout the paper:(H1) liminfx⟶0ft,x/x>λ1 uniformly with respect to t∈0,1.(H2) limsupx⟶∞ft,x/x<λ1 uniformly with respect to t∈0,1.(H3) limsupx⟶0ft,x/x<λ1 uniformly with respect to t∈0,1.(H4) liminfx⟶∞ft,x/x>λ1 uniformly with respect to t∈0,1.(H5) There exist r∗>0 and a continuous function ϕr∗ such that(34)ft,x≥ϕr∗t,∀t∈0,1,x∈tα−1r∗,r∗,maxt∈0,1∫01tα−1G1,qsϕr∗spα−2dp,qs>r∗.(H6) There exist r∗>0 and a continuous function ψr∗ such that(35)ft,x≤ψr∗t,∀t∈0,1,x∈0,r∗,∫01G1,qsψr∗spα−2dp,qs<r∗.Now, we are in a position to give our main results.Theorem 1.
Under assumptions (H1) and (H2), BVP equation (1) admits at least one positive solution.Proof.
First, assumption (H1) implies that there exists r>0 such that(36)ft,x>λ1x,∀x∈0,r,t∈0,1.
We claim that, forμ≥0,(37)xt−Txt≠μφ1t,∀x∈∂Br∩P,t∈0,1.
Suppose, on the contrary, that there existx1∈∂Br∩P,μ1>0 such that(38)x1t−Tx1t=μ1φ1t,t∈0,1.
Without loss of generality, supposeμ1>0. Then, x1t≥μ1φ1t, for t∈0,1.
Let(39)μ∗=supμ:x1t≥μ1φ1t,t∈0,1.
Obviously,0<μ1≤μ∗<+∞ and x1t≥μ∗φ1t, for t∈0,1.
Thus,(40)x1t=Tx1t+μ1φ1t=∫01Gt,qsfspα−2,x1spα−2dp,qs+μ1φ1t≥λ1∫01Gt,qsx1spα−2dp,qs+μ1φ1t≥λ1μ∗∫01Gt,qsφ1spα−2dp,qs+μ1φ1t=μ∗+μ1φ1t.
It is a contradiction with the definition ofμ∗. According to Lemma 7, one obtains(41)degT,Br∩P,P=0.
On the contrary, we can chooseε0>0 such that 0<λ1−ε0L<1. Then, from (H2), there exists R>0 such that(42)ft,x≤λ1−ε0x,x≥R,t∈0,1.
Letm=maxt,x∈0,1×0,Rft,x. Thus, one can easily find that(43)ft,x≤λ1−ε0x+m,∀x≥0,t∈0,1.
ChooseR0>maxR,r,m∫01θ1qsdp,qs/1−λ1−ε0L. We claim that, for μ≥1,(44)Txt≠μxt,∀x∈∂BR0∩P,t∈0,1.
Suppose, on the contrary, that there existx2∈∂BR0∩P and μ2≥1 such that(45)Tx2t=μ2x2t,t∈0,1.
Hence,(46)x2t≤μ2xt=Tx2t=∫01Gt,qsfspα−2,x2spα−2dp,qs≤∫01Gt,qsλ1−ε0x2spα−2+mdp,q,noticing that 0<λ1−ε0L<1. We know that the inverse operator of I−λ1−ε0L exists, and(47)I−λ1−ε0L−1=∑n=0∞λ1−ε0nLn,which shows that I−λ1−ε0L−1P⊆P.
Thus,(48)x2t≤I−λ1−ε0L−1m∫01Gt,qsdp,qs.
In addition, byI−λ1−ε0L−1≤1/1−λ1−ε0L and Lemma 6, one can obtain(49)R0=x2t≤I−λ1−ε0L−1m∫01θ1qsdp,qs≤m∫01θ1qsdp,qs1−λ1−ε0L<R0,which is a contradiction. By Lemma 8, we obtain(50)degT,BR0∩P,P=1.
Therefore,(51)degT,BR0\B¯r∩P,P=degT,BR0∩P,P−degT,Br∩P,P=1−0=1.which means that BVP equation (1) has at least one positive solution.Theorem 2.
Under assumptions (H3) and (H4), BVP equation (1) admits at least one positive solution.Proof.
On the one hand, assumption (H3) implies that there exist ε∈0,λ1 and r1>0 such that(52)ft,x<λ1−ε,x<r1.
We claim that, forμ∈0,1,(53)xt≠μTxt,∀x∈∂Br1∩P,t∈0,1.
Suppose, on the contrary, that there existx1∈∂Br1∩P and μ∈0,1 such that(54)x1t=μ1Tx1t,t∈0,1.
Consequently, we have(55)x1t=μ1Tx1t<λ1−ε∫01Gt,qsx1spα−2dp,qs=λ1−εLx1t.
Thenth iteration of this inequality shows that(56)x1t<λ1−εnLnx1tn=1,2,…,.
Then,(57)x1<λ1−εnLnx1,i.e.,1<λ1−εnLn.
It means that(58)1≤λ1−εlimn⟶∞Lnn=λ1−εrL=λ1−ελ1<1,which is a contradiction. It follows from Lemma 8 that(59)degT,Br1∩P,P=1.
On the other hand, let(60)Lnxt=∫1/n1Gt,qsxspα−2dp,qs,t∈0,1,where n>1. It is easy to see that Ln:P⟶P is completely continuous operator and spectral radius rLn>0, denoted by λn=r−1Ln. We know limn⟶+∞λn=λ1.
It follows that there existN0,ε0 such that λN0<λ1+ε0, namely, rLN0>1/r−1L+ε0. From the Krein–Rutman theorem, there exists a φN0t∈E\θ such that(61)φN0t=r−1LN0∫1/N01Gt,qsφN0spα−2dp,qs,t∈0,1.
By the proof of Lemma5, we know(62)Gt,qs≥tα−1Gτ,qs.
Hence,(63)φN0t≥r−1LN0∫1/N01tα−1Gτ,qsφN0spα−2dp,qs=tα−1φN0τ,∀,t,τ∈0,1,which means that φN0t≥tα−1φN0, namely, φN0t∈P\θ.
Thus, from (H4), there exists R1>0 such that(64)ft,x>λ1+εx,t∈0,1,∀,x≥R1.
ChooseR¯0>maxr1,N0α−1R1. Thus,(65)xt≥tα−1x=tα−1R¯0>R1,t∈1N0,1,x∈∂BR¯0∩P.
Now, we prove that, forμ≥0,(66)xt−Txt≠μφN0t,∀x∈∂BR¯0∩P,t∈0,1.
Similar to the proof of Theorem1, this conclusion is clearly established. So, according to Lemma 7,(67)degT,BR¯0∩P,P=0.
Therefore,(68)degT,BR¯0\B¯r1∩P,P=degT,BR¯0∩P,P−degT,Br1∩P,P=0−1=−1,which means that BVP equation (1) has at least one positive solution.
Up to now, some existence results of BVP equation (1) have been obtained by using the topological degree theory and Krein–Rutman theorem. In the following, the multiple solutions will be considered for BVP equation (1).Theorem 3.
Suppose that (H2), (H3), and (H5) are satisfied. Then, BVP equation (1) has at least two positive solutions.Proof.
By (H5), we know(69)Txt=∫01Gt,qsfspα−2,xspα−2dp,qs≥∫01tα−1G1,qsϕr∗spα−2dp,qs,x∈∂Br∗∩P.
Consequently,(70)Tx≥maxt∈0,1∫01tα−2G1,qsϕr∗spα−2dp,qs>r∗=x.
So, similar to the previous proof of Theorem1, it is easy to know, for μ>0,(71)xt−Txt≠μφ1t,∀t∈0,1,x∈∂Br∗∩P.
By Lemma7, one can immediately obtain that(72)degT,Br∩P,P=0.
By the proof of Theorems1 and 2, we know that there exist r1∈0,r∗ and R1≥maxr∗,R0 such that(73)degT,Br1∩P,P=1,degT,BR1∩P,P=1.
Consequently,(74)degT,Br∗\B¯r1∩P,P=degT,Br∗∩P,P−degT,Br1∩P,P=0−1=−1,degT,BR1\B¯r∗∩P,P=degT,BR1∩P,P−degT,Br∗∩P,P=1−0=1,which means that BVP equation (1) has at least two positive solutions.Theorem 4.
Suppose that (H1), (H4), and (H6) are satisfied. Then, BVP equation (1) has at least two positive solutions.Proof.
By (H6), we know(75)Txt=∫01Gt,qsfspα−2,xspα−2dp,qs≤∫01G1,qsψr∗spα−2dp,qs,x∈∂Br∗∩P.
Therefore,(76)Tx≤∫01G1,qsψr∗spα−2dp,qs<r∗=x.
So, similar to the previous proof of Theorem1, it is easy to know, for μ≥1,(77)Txt≠μxt,∀x∈∂Br∗∩P,t∈0,1.
By Lemma8, one can immediately obtain that(78)degT,Br∗∩P,P=1.
By the proof of Theorems1 and 2, we know that there exist r¯1∈0,r∗ and R¯1≥maxr∗,R¯0 such that(79)degT,Br¯1∩P,P=0,degT,BR¯1∩P,P=0.
Consequently,(80)degT,Br∗\B¯r¯1∩P,P=degT,Br∗∩P,P−degT,Br¯1∩P,P=1−0=1,degT,BR¯1\B¯r∗∩P,P=degT,BR¯1∩P,P−degT,Br∗∩P,P=0−1=−1,which means that BVP equation (1) has at least two positive solutions.


## 4. Examples

Example 1.
Consider the following boundary value problem:(81)D1,1/25/2xt+28x2tx2+1=0,0<t<1,x0=D1,1/2x0=0,D1,1/2x1=∫01t1/2D1,1/2xtd1,1/2t.
Conclusion: BVP equation (81) has at least two positive solutions.Proof.
BVP equation (81) can be regarded as a BVP of the form of equation (1), where(82)ft,x=28x2tx2+1,p=1,q=1/2, α=5/2, and ht=t1/2. Choose ϕr∗t=t2+2 and r∗=1/5. Obviously, f:0,1×1/5t3/2,1/5⟶0,+∞ is continuous and ft,x≥ϕr∗t for t,x∈0,1×1/5t3/2,1/5.
Consequently,(83)limx⟶0supt∈0,1ft,xx=28x2txx2+1=0<λ1,limx⟶+∞supt∈0,1ft,xx=28x2txx2+1=0<λ1.
From the definition of functionG, one can obtain that(84)Gt,s2=G0t,s2+tα−1A∫01htG1t,s2d1,1/2t,where(85)G0t,s2=11α/2Γ1,1/25/2t3/21−s21/2−t−s23/2,0≤s≤t≤1,t3/21−s21/2,0≤t≤s≤1,G1t,s2=11α/2Γ1,1/25/2t1/21−s21/2−t−s21/2,0≤s≤t≤1,t1/21−s21/2,0≤t≤s≤1.
Consequently,(86)maxt∈0,1∫01tα−1G1,s2ϕr∗sdp,qs≈0.235>15=r∗.
Therefore, by Theorem3, BVP equation (81) has at least two positive solutions.Example 2.
Consider the following boundary value problem:(87)D1,1/25/2xt+3x3/2+2t=0,0<t<1,x0=D1,1/2x0=0,D1,1/2x1=∫01t1/2D1,1/2xtd1,1/2t.
Conclusion: BVP equation (87) has at least two positive solutions.Proof.
BVP equation (87) can be regarded as a BVP of the form of equation (1), where(88)ft,x=3x3/2+2t,p=1,q=1/2, α=5/2, and ht=t1/2. Choose ψr∗t=2t+9, r∗=33. Obviously, f:0,1×0,33⟶0,+∞ is continuous and ft,x≤ψr∗t for t,x∈0,1×0,33.
Consequently,(89)limx⟶0inft∈0,1ft,xx=3x3/2+2tx=+∞>λ1,limx⟶+∞inft∈0,1ft,xx=3x3/2+2tx=+∞>λ1.
From the definition of functionG, one can obtain that(90)Gt,s2=G0t,s2+tα−1A∫01htG1t,s2d1,1/2t,where(91)G0t,s2=11α/2Γ1,1/25/2t1/21−s21/2−t−s21/2,0≤s≤t≤1;t3/21−s21/2,0≤t≤s≤1,G1t,s2=11α/2Γ1,1/25/2t1/21−s21/2−t−s21/2,0≤s≤t≤1;t1/21−s21/2,0≤t≤s≤1.
Thus,(92)∫01G1,s2ψr∗sdp,qs≈1.324<33=r∗.
Therefore, by Theorem4, BVP equation (87) has at least two positive solutions.


## 5. Conclusions

This paper is mainly concerned with a class of fractionalp,q-difference equations with p,q-integral boundary conditions. We first give the definition of fractional p,q-difference operator and fractional p,q-integral operator. Then, the existence and multiplicity of positive solutions for boundary value problems are obtained by using topological degree theory and Krein–Rutman theorem. Finally, two illustrative examples are given to show the practical usefulness of the analytical results.



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