# Some Properties of Solutions to Multiterm Fractional Boundary Value Problems withp-Laplacian Operator

Authors: KumSong Jong; HuiChol Choi; KyongJun Jang; SongGuk Jong; KyongSon Jon; Ok Ri
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study some properties of positive solutions to a class of multipoint boundary value problems for nonlinear multiterm fractional differential equations withp-Laplacian operator. Using the Banach contraction mapping principle, the existence, the uniqueness, the positivity, and the continuous dependency on m-point boundary conditions of the solutions to the given problem are investigated. Also, two examples are presented to demonstrate our main results.


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## Body

## 1. Introduction

This paper deals with a class of multipoint boundary value problems for nonlinear multiterm fractional differential equations withp-Laplacian operator:
(1)D0+βφpD0+αxt=ft,xt,D0+γxt,0<t<1,D0+γx0=0,D0+δx1=∑i=1m−2ξiD0+δxηi,D0+αx0=0,φpD0+αx1=∑i=1m−2ζiφpD0+αxηi,where 1<α,β≤2,0<γ,δ<1,0<ξi,ηi,ζi<1,i=1,2,⋯,m−2 and D0+α,D0+β,D0+γ,D0+δ are the standard Riemann-Liouville derivatives. And it is assumed that the nonlinear source term f satisfies f∈C0,1×0,+∞×0,+∞;0,+∞, and φp is defined as φps=sp−2s,p>1.Fractional differential equations have developed into a powerful tool to mathematically model and solve many real-world problems that arises in various fields such as physics, chemistry, biology, and mechanics (see [1, 2] and references therein). Especially, since the turbulent flow was one of the fundamental problems in the field of fluid dynamics, Leibenson [3] and Esteban and Vazquez [4] proposed a mathematical model for one-dimensional, polytropic, and turbulent flow of a gas in a porous medium as
(2)ut=∂∂x∂um∂x∂um∂xp−1,where ut,x was a scaled density at every point x and time t. According to them, if p=1, the equation refers to the laminar filtration, in which case, it is often known as the fast diffusion equation if 0<m<1 and the linear heat equation if m=1; otherwise, the porous media equation. On the other hand, the case m=1,p≠1 has been widely studied in non-Newtonian fluid dynamics (see [5] and the references therein). By some substitutions and tricks, this nonlinear problem was reduced into the following p-Laplacian equation to investigate the existence and properties of solutions to it [6, 7]:
(3)φpu′t′=ft,ut,u′t.Due to the significance of equation (3), many works have been studied to establish a lot of valuable existence and multiplicity results for various classes of higher-order and generalized p-Laplacian differential equations with integer-order derivatives [8–10]. For example, Guo et al. [8] dealt with the existence of at least three positive solutions to the quasilinear second-order differential equation
(4)φpu′t′+atft,ut=0,0<t<1,subject to one of the following m-point boundary conditions:
(5)φpu′0=∑i=1m−2aiφpu′ξi,u1=∑i=1m−2biuξior
(6)u0=∑i=1m−2aiuξi,φpu′1=∑i=1m−2biφpu′ξi,by using the five functional fixed point theorem. Generalizing the results of the above studies for integer-order differential equations, many researchers have obtained great outcomes of the existence of solutions to single-term fractional differential equations with p-Laplacian operator (see [11–15]). From the representative results, Lv [11] established the existence and multiplicity of positive solutions to m-point boundary value problems of p-Laplacian fractional differential equations with a parameter:
(7)D0+βφpD0+αut+φpλft,ut=0,0<t<1,1<α≤2,0<β≤1,u0=0,D0+γu1=∑i=1m−2ξiD0+γuηi,D0+αu0=0,0<γ≤1,by using the theory of the fixed-point index in a cone and the monotone iterative technique. In addition, Li and Qi [12] improved the works of Guo et al. [8] to show the existence of positive solutions for the following multipoint boundary value problems of nonlinear fractional differential equations with p-Laplacian operator:
(8)CD0+βφpCD0+αut+ft,ut=0,0<t<1,n−1<α≤n,l−1<β≤l,φpCD0+αu0i=0,i=1,2,⋯,l−1,φpCD0+αu1=∑i=1m−2biφpCD0+αuξi,ui0=0,i=1,2,⋯,n−1,u0=∑i=1m−2aiujξi,where l>1,n≥2,j∈0,α are integers and DC0+α,DC0+β are the Caputo fractional derivatives. It is noted that the notations D0+α and DC0+α appearing throughout the paper refer to the Riemann-Liouville and Caputo fractional derivatives, respectively, unless otherwise stated.However, since the differential equation studied by Leibenson [3] is a multiterm one, it is important to extend the research results for single-term fractional differential equations to the case of multiterm. To do this, the existence of solutions to nonlinear multiterm fractional differential equations with a p-Laplacian operator have been studied by employing useful techniques in a nonlinear functional analysis such as the Banach contraction mapping principle, the Schauder fixed-point theorem, and the Krasnoselskii fixed-point theorem (see [16–20]). In particular, it is of great significance to study the case where α,β,γ are in 0,1 or 1,2, when considering the following fractional differential equation:
(9)DβφpDαut=ft,ut,Dγut,generalized equation (3) because of the inclusion of first-order derivatives outside and inside of φp in it. Here, Dα,Dβ, and Dγ refer to fractional derivatives in several senses such as the Riemann-Liouville and the Caputo. Therefore, by using the coincidence degree theory, Chen et al. [16] studied two-point boundary value problems of fractional differential equations with p-Laplacian operator:
(10)DC0+βφpDC0+αxt=ft,xt,DC0+αxt,t∈0,1,0<α,β≤1,DC0+αx0=DC0+αx1=0,to obtain new results on the existence of solutions to them. Besides, according to Liu et al. [17], the following four-point boundary value problems of p-Laplacian fractional differential equations with mixed fractional derivatives:
(11)D0+αφpDC0+βut=ft,ut,DC0+βut,t∈0,1,1<α,β≤2,DC0+βu0=u′0=0,u1=r1uη,DC0+βu1=r2DC0+βuξ,were considered to establish a lower and upper solution method which was used to prove the existence of positive solutions to them. After analyzing the results for nonlinear multiterm fractional differential equations with p-Laplacian operator, we have found that most of these previous works only dealt with some kinds of p-Laplacian equations that had certain rules for the fractional derivatives in nonlinear source terms and were limited to the cases of two-point, three-point, and four-point boundary conditions.On the other hand, Dishlieva [21] and Li et al. [22] investigated the continuous dependency of solutions to their differential equations on initial conditions, barrier curves, and source terms in their studies but not on boundary conditions. This also implies that as far as we know, no works concerned with this issue for fractional differential equations with p-Laplacian operator can be done.Motivated by the analysis mentioned above, this paper is aimed at studying some properties of positive solutions tom-point boundary value problems of nonlinear multiterm fractional differential equations with p-Laplacian operator including the existence, the uniqueness, and the continuous dependency on boundary conditions.This study is organized as follows. In Section2, we give some necessary definitions and preliminary results which will be used to prove our main results. In Section 3, we prove the existence and uniqueness of positive solutions to p-Laplacian fractional boundary value problem (1) and continuous dependence of them on the perturbations with respect to the coefficients in m-point boundary conditions. Finally, in Section 4, we illustrate our results by giving two examples.Throughout the whole paper, it is also supposed that(12)3<α+β≤4,γ<δ,α−δ−1>0,∑i=1m−2ξiηiα−δ−1<1,∑i=1m−2ζiηiβ−1<1.


## 2. Preliminaries

For the sake of convenience of the readers, some necessary definitions and lemmas will be presented here.The Riemann-Liouville fractional integral and the Riemann-Liouville fractional derivative of orderα>0 of a function f:0,∞→R are defined as
(13)I0+αft≔1Γα∫0tt−sα−1fsds,D0+αft≔1Γn−αddtn∫0tt−sn−α−1fsds,where n=α+1, provided that the right-hand sides are pointwise defined on 0,∞ (see [23]).Lemma 1 (see [24]).
Assume thatu∈C0,1∩L0,1 with a fractional derivative of order α>0 that belongs to C0,1∩L0,1. Then,
(14)I0+αD0+αut=ut+C1tα−1+C2tα−2+⋯+CNtα−N,for some Ci∈R,i=1,2,⋯,N, where N is the smallest integer greater than or equal to α.
Since1<α≤2,0<γ<δ<1,α−δ>1, it is obvious that 1<α−γ<2. And putting A≔1−∑i=1m−2ξiηiα−γ−1 and B≔1−∑i=1m−2ζiηiβ−1, the following lemmas hold.Lemma 2 (see [11]).
Lety∈C0,1. Then, the fractional differential equation
(15)D0+α−γut+yt=0,0<t<1,u0=0,D0+δ−γu1=∑i=1m−2ξiD0+δ−γuηi,has a unique solution which is given by
(16)ut=∫01Gt,sysds,where Gt,s=G1t,s+G2t,s,
in which(17)G1t,s=1Γα−γtα−γ−11−sα−δ−1−t−sα−γ−1,0≤s≤t≤1,tα−γ−11−sα−δ−1,0≤t≤s≤1,G2t,s=tα−γ−1AΓα−γ∑s<ηiξiηiα−δ−11−sα−δ−1−ηi−sα−δ−1+∑s≥ηiξiηiα−δ−11−sα−δ−1,t,s∈0,1.Lemma 3 (see [11]).
If∑i=1m−2ξiηiα−δ−1, then the function Gt,s in Lemma 2 satisfies the following conditions:
(i)
Gt,s>0, for any t,s∈0,1(ii)
Gt,s≤G∗s,s, for any t,s∈0,1where
(18)G∗s,s=1AΓα−γ1−sα−δ−1.Lemma 4 (see [25]).
Letz∈C0,1. Then, the fractional differential equation
(19)D0+βvt+zt=0,v0=0,v1=∑i=1m−2ζivηi,has a unique solution which is given by
(20)vt=∫01Ht,szsds,where Ht,s=H1t,s+H2t,s, in which
(21)H1t,s=1Γβtβ−11−sβ−1−t−sβ−1,0≤s≤t≤1,tβ−11−sβ−1,0≤t≤s≤1,H2t,s=tβ−1BΓβ∑s<ηiζiηiβ−11−sβ−1−ηi−sβ−1+∑s≥ηiζiηiβ−11−sβ−1,t,s∈0,1.Lemma 5 (see [25]).
If∑i=1m−2ζiηiβ−1<1, then the function Ht,s in Lemma 4 satisfies the following conditions:
(i)
Ht,s>0, for any t,s∈0,1(ii)
Ht,s≤H∗s,s, for any t,s∈0,1where
(22)H∗s,s=1BΓβ1−sβ−1.
The following properties ofφp⋅ which will be used later can be found in [26]:
(i)
If1<p<2,xy>0, and x,y≥m>0, then
(23)∣φpx−φpy∣≤p−1mp−2∣x−y∣(ii)
Ifp>2,x,y≤M, then
(24)∣φpx−φpy∣≤p−1Mp−2∣x−y∣


## 3. Main Results

### 3.1. Existence and Uniqueness of Positive Solutions to Problem (1)

Definition 6.
A functionx∈u∣u∈C0,1,D0+αu∈C0,1,D0+βφpD0+αu∈C0,1 is called a solution of problem (1) if it satisfies the fractional differential equation and the boundary conditions of (1).Lemma 7.
If the functionx is a solution of problem (1), then ut≔D0+γxt is a solution of the integral equation
(25)ut=∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds,in C0,1 and conversely, if u∈C0,1 is a solution of the integral equation (25), then xt=I0+γut is a solution of problem (1), where q is a number such that 1/p+1/q=1.Proof.
Letx be a solution of the problem (1). Then, since x∈C0,1,D0+αx∈C0,1, by using Lemma 1, it can be easily seen that
(26)xt=I0+αD0+αxt+c1tα−1+c2tα−2.
The upper continuities of the functionsxt,I0+αD0+αxt at t=0 provide us that c2=0. Applying D0+γ on both sides, we have
(27)D0+γxt=D0+γI0+αD0+αxt+D0+γc1tα−1=I0+α−γD0+αxt+ΓαΓα−γc1tα−γ−1.
Combining withD0+αx∈C0,1,α−γ−1>0, we can see that D0+γx∈C0,1. Since x∈C0,1,D0+γx∈C0,1, Lemma 1 yields that
(28)xt=I0+γD0+γxt+c3tγ−1.
Therefore, it follows from the upper continuities of the functionsxt,I0+γD0+γxt at t=0 that c3=0. This implies that by putting ut≔D0+γxt, it holds that xt=I0+γut. Thus, we can know that u is a solution of the following problem:
(29)D0+βφpD0+α−γut=ft,I0+γut,ut,0<t<1,u0=0,D0+δ−γu1=∑i=1m−2ξiD0+δ−γuηi,D0+α−γu0=0,φpD0+α−γu1=∑i=1m−2ζiφpD0+α−γuηi.
Putwt≔D0+α−γut. Lemma 2 shows that
(30)ut=−∫01Gt,swsds.
Also, by applying Lemma4, the notation vt=φpwt gives us that
(31)vt=−∫01Ht,sfs,I0+γus,usds.
Sinceφp−1=φq, by equations (30) and (31), we obtain
(32)ut=∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds.
Conversely, let the functionu∈C0,1 be a solution of integral equation (25). From this, by putting xt≔I0+γut, it can be also obtained that x∈C0,1. Denote as follows:
(33)zt≔ft,I0+γut,ut,vt≔φq∫01Ht,szsds.
From the continuities of the functionsu,f, and H, it is obvious that z∈C0,1,v∈C0,1. The definition of the function Gt,s implies that
(34)ut=∫01Gt,svsds=−I0+α−γvt−Γα−δAΓα−γ∑i=1m−2ξiI0+α−δvtt=ηi−I0+α−δvtt=1tα−γ−1,so, we have
(35)D0+αxt=D0+α−γut=vt.
This means thatD0+αx∈C0,1. Also, combining the definition of the function Ht,s with the relation
(36)D0+βφpD0+αxt=D0+βφpvt=D0+β∫01Ht,szsds,we can get
(37)∫01Ht,szsds=−I0+βzt−1B∑i=1m−2ζiI0+βztt=ηi−I0+βztt=1tβ−1.
This yields that(38)D0+βφpD0+αxt=zt,and therefore, it holds that D0+βφpD0+αx∈C0,1. Since equations (15) and (19) have unique solutions by Lemmas 2 and 4, u, the solution of integral equation (25), satisfies the fractional differential equation and the boundary conditions of the problem (29). To sum up, it can be proved that x is a solution of problem (1).LetE be the Banach space C0,1 equipped with the norm u=max0≤t≤1ut and put P≔u∈E∣ut≥0,t∈0,1. Since ∀u∈P,∀t∈0,1,I0+γut=1/Γγ∫0tt−sγ−1usds≥0, it holds that I0+γu∈P. Therefore, define an operator T as
(39)Tut≔∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds,u∈P,then, it satisfies TP⊂P. This implies that integral equation (25) has a solution in P if and only if operator T has a fixed point. It is also obvious that the fixed point of operator T is a solution of integral equation (25) in P.PutM¯≔α−δAΓα−γp−1BΓβ. The following hypotheses will be used throughout the paper.(H1) There exista,b,c∈L0,1 such that
(40)Ma≔∫01atdt>0,Mb≔∫01btdt>0,Mc≔∫01ctdt>0,MbΓγ+1p−1+Mc<M¯,∀t,x,y∈0,1×0,R1×0,R2,ft,x,y<at+btxp−1+ctyp−1,where R2=Ma/M¯−Mb/Γγ+1p−1−Mcq−1 and R1=R2/Γγ+1(H2) There existL1,L2>0 such that for any t∈0,1 and any x1,y1,x2,y2∈0,R1×0,R2,
(41)ft,x1,y1−ft,x2,y2≤L1x1−x2+L2y1−y2(H3) There existm¯,μ>0 such that for any t,x,y∈0,1×0,R1×0,R2,
(42)ft,x,y≥m¯tμ−1Lemma 8.
Assume that the hypothesis (H1) is satisfied. Then, it holds that
(43)TBR2⊂BR2,where BR2=u∈Pu≤R2.Proof.
SinceTBR2⊂P, it is sufficient to prove that ∀u∈BR2,Tu≤R2. It can be easily evaluated that for any t∈0,1,
(44)I0+γut=1Γγ∫0tt−sγ−1usds≤R2Γγ∫0tt−sγ−1ds=R2Γγ+1⋅tγ≤R1.
By using Lemma5 and hypothesis (H1), we have
(45)Tut=∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds≤∫01Gt,sφqmaxs∈0,1H∗s,s∫01aτ+bτI0+γuτp−1+cτuτp−1dτds≤∫01Gtsφq1BΓβMa+MbΓγ+1p−1+McR2p−1ds.
Applying the properties of the functionGt,s indicated in Lemma 3, the following holds:
(46)Tut≤1BΓβMa+MbΓγ+1p−1+McR2p−1q−1∫01G∗s,sds=Ma+Mb/Γγ+1p−1+McR2p−1q−1Bq−1Γβq−1Aα−δΓα−γ.
The relation1/p+1/q=1 shows that p−1 is an inverse power of q−1. So, it can be rewritten as
(47)Tut≤Ma+MbΓγ+1p−1+McR2p−1/M¯q−1.
Since it is satisfied forR2 that
(48)Ma=M¯−MbΓγ+1p−1−McR2p−1,we can get
(49)Tut≤R2.
This concludes the lemma.Lemma 9.
If the hypothesis (H3) holds, then for any t∈0,1 and any u∈BR2,
(50)∫01Ht,sfs,I0+γus,usds≥K0tβ−1,where K0=m¯Γμ/BΓβ+μ∑i=1m−2ζiηiβ−1−ηiβ−1+μ.Proof.
It follows from Lemma4 and hypothesis (H3) that
(51)∫01Ht,sfs,I0+γus,usds≥m¯∫01H1t,ssμ−1ds+∫01H2t,ssμ−1ds.
By simple calculation, the integration terms in the right side of inequality (51) can be estimated as
(52)∫01H1t,ssμ−1ds=1Γβtβ−1∫011−sβ−1sμ−1ds−∫0tt−sβ−1sμ−1ds=1Γβtβ−1Bμ,β−tβ−1+μBμ,β=ΓμΓβ+μtβ−1−tβ−1+μ≥0,∫01H2t,ssμ−1ds=tβ−1BΓβ∑i=1m−2ζiηiβ−1∫011−sβ−1sμ−1ds−∫0ηiηi−sβ−1sμ−1ds=tβ−1BΓβ∑i=1m−2ζiηiβ−1Bμ,β−ηiβ−1+μBμ,β=ΓμBΓβ+μ∑i=1m−2ζiηiβ−1−ηiβ−1+μtβ−1.
These estimations provide us the conclusion (50).DenoteK1≔1/Γβ+11+1/B∑i=1m−2ζiηiβ−1−ηiβ. Then, it can be found in [25] that for any t∈0,1,
(53)∫01Ht,sds≤K1tβ−1.Theorem 10.
Suppose thatp>2 and hypotheses (H1)-(H3) are satisfied. If
(54)q−1K1Aα−δΓα−γK02−qL1Γγ+1+L2<1,then, integral equation (25) has a unique solution in BR2.Proof.
To prove this theorem, operatorsT0,T1 are defined as
(55)T0ut≔φq∫01Ht,sfs,I0+γus,usds,T1ut≔∫01Gt,susds.
Obviously, we can see that for anyu∈BR2, Tut=T1oT0ut, and Tu0=0. Because p>2, it is satisfied that 1<q<2. So, by employing Lemma 8, Lemma 9, and property (23), we can obtain that for any u,v∈BR2 and any t∈0,1,
(56)T0ut−T0vt=φq∫01Ht,sfs,I0+γus,usds−φq∫01Ht,sfs,I0+γvs,vsds≤q−1K0tβ−1q−2∫01Ht,sfs,I0+γus,usds−∫01Ht,sfs,I0+γvs,vsds≤q−1K0tβ−1q−2∫01Ht,sL1I0+γus−I0+γvs∣+L2∣us−vsds≤q−1K0tβ−1q−2∫01Ht,sL1Γγ+1u−v∣+L2u−vds≤q−1K0tβ−1q−2L1Γγ+1+L2u−v∫01Ht,sds.
Substituting inequality (53) into (56), we have
(57)T0ut−T0vt≤q−1K0q−2K1L1Γγ+1+L2u−vtβ−1q−1.
Thus, it holds that(58)Tut−Tvt=∫01Gt,sT0usds−∫01Gt,sT0vsds≤q−1K0q−2K1L1Γγ+1+L2u−v∫01Gt,ssβ−1q−1ds.
Sinceβ−1q−1>0, by Lemma 3, we can get
(59)∫01Gt,ssβ−1q−1ds≤∫01Gt,sds≤∫01G∗s,sds=1Aα−δΓα−γ.
This yields that(60)Tu−Tv≤q−1K1Aα−δΓα−γK02−qL1Γγ+1+L2u−v.
Combining inequalities (54) and (60) with Lemma 8, the operator T:BR2→BR2 is a contraction mapping and the Banach contraction mapping principle provides us that it has a unique fixed point in BR2. So, integral equation (25) has a unique solution in BR2.Lemma 11.
Assume that the hypothesis (H1) holds. Then, for any u∈BR2 and any t∈0,1, it is satisfied that
(61)∫01Ht,sfs,I0+γus,usds≤M0,where
(62)M0=Ma+Mb/Γγ+1p−1+McR2p−1BΓβ.Proof.
In a similar way to the proof of Lemma8, it can be easily proved that
(63)∫01Ht,sfs,I0+γus,usds≤maxs∈0,1H∗s,s∫01fs,I0+γus,usds≤1BΓβ∫01as+bsI0+γusp−1+csusp−1ds≤1BΓβ∫01asds+R1p−1∫01bsds+R2p−1∫01csds=M0.Theorem 12.
Suppose that1<p<2 and hypotheses (H1) and (H2) hold. If
(64)q−1R22−pβM¯L1Γγ+1+L2<1,then, integral equation (25) has a unique solution in BR2.Proof.
In the case1<p<2, due to 1/p+1/q=1, it holds that q>2. Similar to Theorem 10, defining operators T0,T1 with Lemma 8, Lemma 11, and property (24) give us that for any u,v∈BR2 and any t∈0,1,
(65)T0ut−T0vt=φq∫01Ht,sfs,I0+γus,usds−φq∫01Ht,sfs,I0+γvs,vsds≤q−1M0q−2∫01Ht,sfs,I0+γus,us−fs,I0+γvs,vsds≤q−1M0q−2∫01Ht,sL1∣I0+γus−I0+γvs∣+L2∣us−vs∣ds≤q−1M0q−2∫01Ht,sL1Γγ+1u−v+L2u−vds≤q−1M0q−2L1Γγ+1+L2u−v∫01H∗s,sds=q−1M0q−2BΓβ+1L1Γγ+1+L2u−v.
So, we have(66)Tut−Tvt=∫01Gt,sT0usds−∫01Gt,sT0vsds≤∫01Gt,sT0us−T0vsds≤q−1M0q−2BΓβ+1L1Γγ+1+L2u−v∫01Gt,sds≤q−1M0q−2α−δABΓα−γΓβ+1L1Γγ+1+L2u−v.
Obviously, forM0, the following holds:
(67)M0q−2α−δABΓα−γΓβ+1=1βM¯q−1⋅M¯R2p−1q−2=R22−pβM¯.
Therefore, by employing inequality (64), equality (67), and Lemma 8, the operator T:BR2→BR2 is a contraction mapping. It follows from the Banach contraction mapping principle that the operator T has a unique fixed point in BR2. This completes the proof.Remark 13.
Theorems10 and 12 show the existence and uniqueness of solutions to integral equation (25) in the cases p>2 and 1<p<2, respectively. The Banach contraction mapping principle, employed in the proofs of these theorems, guarantees that the iterative sequences can be constructed to converge to the exact solution of integral equation (25). In other words, the iterative sequence unn=0∞ initialized at any point u0∈BR2 and defined by
(68)un+1=Tun,n=0,1,2,⋯,converges to the exact solution of integral equation (25), u, in BR2. This provides the possibility to obtain the approximate solution un of integral equation (25) by using Theorems 10 and 12.Theorem 14.
Problem (1) has a unique solution in Ω if and only if the solution to integral equation (25) uniquely exists in BR2, where Ω=I0+γBR2. In particular, if hypothesis (H3) is satisfied, then the solution to problem (1) is positive.Proof.
By Lemma7, for x, a solution to problem (1), and u satisfying integral equation (25), the following relation holds:
(69)xt=I0+γut,t∈0,1.
This proves the first part of the theorem.
For the second part, it is necessary that if hypothesis (H3) holds, then(70)∀t∈0,1,xt>0.
To prove this, we must see if the following holds:(71)∀t∈0,1,ut>0
Sinceu∈BR2, using Lemma 9, it can be easily obtained that
(72)ut≥K0q−1∫01Gt,ssβ−1q−1ds=K0q−1∫01G1t,ssβ−1q−1ds+∫01G2t,ssβ−1q−1ds.
Considering thatG1t,s≥0 and β−1q−1>0, we can get
(73)∀t∈0,1,∫01G1t,ssβ−1q−1ds≥0..
From the definition of functionG2t,s, we have
(74)∫01G2t,ssβ−1q−1ds=tα−γ−1AΓα−γ∑i=1m−2ξiηiα−δ−1∫011−sα−δ−1sβ−1q−1ds−∫0ηiηi−sα−δ−1sβ−1q−1ds=Bα−δ,β−1q−1+1AΓα−γ∑i=1m−2ξiηiα−δ−1−ηiα−δ+β−1q−1tα−γ−1.
Denote as follows:(75)K2≔Bα−δ,β−1q−1+1AΓα−γ∑i=1m−2ζiηiα−δ−1−ηiα−δ+β−1q−1.
SinceK0,K2>0 and ut≥K0q−1K2tα−γ−1, it is obvious that (71) holds. Therefore, we can see that
(76)xt=I0+γut≥I0+γK0q−1K2tα−γ−1=K0q−1K2⋅Γα−γΓαtα−1.
The proof is completed.Remark 15.
Theorems10, 12, and 14 show new results that differ from some of the references listed in Section 1. One of these is the result of the existence, uniqueness, and positivity of solutions to p-Laplacian fractional differential equations with multipoint boundary conditions where the nonlinear source terms contain a more generalized form of fractional derivatives than those considered in [16, 17, 20]. The other is the possibility of constructing an iterative sequence that converges to a unique solution of a given problem, which is different from [16, 18, 19]. Combining Theorem 14 with Theorems 10 and 12, the iterative sequence xnn=0∞ given by
(77)xn=I0+γun,n=0,1,2,⋯,converges to the exact solution of problem (1), x, in Ω, where unn=0∞ is constructed by (68). This gives an opportunity to get the approximate solution, xn, for problem (1), although there might be some difficulties in its numerical computation.


### 3.2. Continuous Dependence of a Solution on Boundary Conditions

In this subsection, the continuous dependence of a unique positive solutionx to problem (1) on boundary conditions will be established. To do this, denote a unique solution to the problem
(78)D0+βφpD0+αxt=ft,xt,D0+γxt,0<t<1,D0+γx0=0,D0+δx1=∑i=1m−2ξ~iD0+δxηi,D0+αx0=0,φpD0+αx1=∑i=1m−2ζ~iφpD0+αxηi,by x~.As can be seen above, the positive solutionx to problem (1) is given by
(79)xt=I0+γut,t∈0,1,where u is a solution of integral equation (25). Also, the positive solution x~ to problem (78) is expressed by
(80)x~t=I0+γu~t,t∈0,1,where u~ is a solution of the integral equation
(81)u~t=∫01G~t,sϕq∫01H~s,τfτ,I0+γu~τ,u~τdτds,in which
(82)A~=1−∑i=1m−2ξ~iηiα−δ−1,B~=1−∑i=1m−2ζ~iηiβ−1,G~2t,s=tα−γ−1A~Γα−γ∑s<ηiξ~iηiα−δ−11−sα−δ−1−ηi−sα−δ−1+∑s≥ηiξ~iηiα−δ−11−sα−δ−1,H~2t,s=tβ−1B~Γβ∑s<ηiζ~iηiβ−11−sβ−1−ηi−sβ−1+∑s≥ηiζ~iηiβ−11−sβ−1,G~t,s=G1t,s+G~2t,s,H~t,s=H1t,s+H~2t,s,M¯~=α−δA~Γα−γp−1B~Γβ,R~2=Ma/M¯~−MbΓγ+1p−1−Mcq−1.Then, the continuous dependence of a unique positive solution to the problem (1) on boundary conditions can be formulated by
(83)limλξ→0λζ→0x~=x,where
(84)λξ≔max∣ξ~1−ξ1∣,∣ξ~2−ξ2∣,⋯,∣ξ~m−2−ξm−2∣,λζ≔max∣ζ~1−ζ1∣,∣ζ~2−ζ2∣,⋯,∣ζ~m−2−ζm−2∣.Considering relations (79) and (80), it is sufficient to prove that
(85)limλξ→0λζ→0u~=u.The solutions of integral equations (25) and (81), u and u~, uniquely exist in BR2,BR~2, respectively. Since A,B,M¯,andR2 are continuous at ξ1,ξ2,⋯,ξm−2,ζ1,ζ2,⋯,ζm−2, it holds that
(86)limλξ→0λζ→0A~=A,limλξ→0λζ→0B~=B,limλξ→0λζ→0M¯~=M¯,limλξ→0λζ→0R~2=R2.Define a functionG¯2t,s as follows:
(87)G¯2t,s≔1Γα−γtα−δ−11−sα−δ−1−t−sα−δ−1,0≤s≤t≤1,tα−δ−11−sα−δ−1,0≤t≤s≤1.Then, it can be easily seen that(88)G2t,s=1A∑i=1m−2ξiG¯2ηi,s⋅tα−γ−1,G~2t,s=1A~∑i=1m−2ξ~iG¯2ηi,s⋅tα−γ−1.And it can be found in [27] that
(89)H2t,s=1B∑i=1m−2ζiH1ηi,s⋅tβ−1,H~2t,s=1B~∑i=1m−2ζ~iH1ηi,s⋅tβ−1.For a fixedt, the functions G¯2t,s and H1t,s have maximum values at s=t, i.e.,
(90)maxs∈0,1G¯2t,s=tα−δ−11−tα−δ−1Γα−γ,maxs∈0,1H1t,s=tβ−11−tβ−1Γβ.Lemma 16.
Lety∈C0,1. Then, the following holds:
(91)∀t∈0,1,∣v~t−vt∣≤yΓα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1⋅tα−γ−1,where
(92)vt=∫01Gt,sysds,v~t=∫01G~t,sysds.Proof.
Using relation (88), we can see that for any t∈0,1,
(93)∣v~t−vt∣≤∫01∣G~t,s−Gt,s∣ysds≤y∫011A~∑i=1m−2ξ~iG¯2ηi,s−1A∑i=1m−2ξiG¯2ηi,stα−γ−1ds≤y∑i=1m−2ξ~iA~−ξiA∫01G¯2ηi,sds⋅tα−γ−1.
By applying (90), we obtain
(94)∣v~t−vt∣≤yΓα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1⋅tα−γ−1.Lemma 17.
Letz∈C0,1. Then, it holds that
(95)∀t∈0,1,∣w~t−wt∣≤z∣Γβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅tβ−1,where
(96)wt=∫01Ht,szsds,w~t=∫01H~t,szsds.Proof.
The proof is similar to Lemma16, so it is omitted.Theorem 18.
Suppose that all the assumptions of Theorem10 are satisfied. Then, it holds that
(97)limλξ→0λζ→0u~=u..Proof.
It can be obtained that for anyt∈0,1,
(98)∣u~t−ut∣≤∫01G~t,sφq∫01H~s,τfτ,I0+γu~τ,u~τdτds−∫01G~t,sφq∫01Hs,τfτ,I0+γuτ,uτdτds+∫01G~t,sφq∫01Hs,τfτ,I0+γuτ,uτdτds−∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds≤∫01G~t,sφq∫01H~s,τfτ,I0+γu~τ,u~τdτ−φq∫01Hs,τfτ,I0+γuτ,uτdτds+∫01G~t,s−Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds.
Lemma9 indicates that for u∈BR2,u~∈BR~2,
(99)∫01Hs,τfτ,I0+γuτ,uτdτ≥K0sβ−1,∫01H~s,τfτ,I0+γu~τ,u~τdτ≥K~0sβ−1,where K~0=m¯~Γμ~/B~Γβ+μ~∑i=1m−2ζ~iηiβ−1−ηiβ−1+μ~. And by Lemma 11, we know that
(100)∫01Hs,τfτ,I0+γuτ,uτdτ≤M0.
Considering Lemma17 and inequality (53), it is easy to see that
(101)∫01H~s,τfτ,I0+γu~τ,u~τdτ−∫01Hs,τfτ,I0+γuτ,uτdτ≤∫01H~s,τfτ,I0+γu~τ,u~τdτ−∫01H~s,τfτ,I0+γuτ,uτdτ+∫01H~s,τfτ,I0+γuτ,uτdτ−∫01Hs,τfτ,I0+γuτ,uτdτ≤∫01H~s,τ∣fτ,I0+γu~τ,u~τ−fτ,I0+γuτ,uτ∣dτ+∫01H~s,τ−Hs,τfτ,I0+γuτ,uτdτ≤maxL1,L~1Γγ+1+maxL2,L~2u~−u∫01H~s,τdτ+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅sβ−1≤K~1maxL1,L~1Γγ+1+maxL2,L~2u~−u+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅sβ−1,where
(102)K~1=1Γβ+11+1B~∑i=1m−2ζ~iηiβ−1−ηiβ,and C=maxt∈0,1x∈0,R1y∈0,R2∣ft,x,y∣.
Therefore, combining Lemma16 with 1<q<2, it follows that
(103)∣u~t−ut∣≤q−1minK02−q,K~02−qK~1maxL1,L~1Γγ+1+maxL2,L~2u~−u+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅∫01G~∗s,ssβ−1q−1ds+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1≤q−1K~1maxL1,L~1/Γγ+1+maxL2,L~2u~−u+C/Γβ∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1A~α−δΓα−γminK02−q,K~02−q+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1⋅
Since(104)q−1K1Aα−δΓα−γK02−qL1Γγ+1+L2<1,and A,B,K0,K1,L1,L2 are continuous at ξ1,ξ2,⋯,ξm−2,ζ1,ζ2,⋯,ζm−2, we can see that for sufficient small λξ and λζ,
(105)Λ1≔q−1K~1A~α−δΓα−γminK02−q,K~02−qmaxL1,L~1Γγ+1+maxL2,L~2<1.
So, putting(106)Λ2≔q−1CA~α−δΓβΓα−γminK02−q,K~02−q,Λ3≔M0Γα−γ,we have
(107)u~−u≤Λ2∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1+Λ3∑i=1m−2ξ~i/A~−ξi/Aηiα−δ−11−ηiα−δ−11−Λ1.
Finally, the fact that(108)limλξ→0λζ→0ξ~iA~=ξiA,limλξ→0λζ→0ζ~iB~=ζiB,i=1,2,⋯,m−2proves the conclusion of this theorem.Theorem 19.
Suppose that all the assumptions of Theorem12 hold. Then, it is satisfied that
(109)limλξ→0λζ→0u~=u.Proof.
Similar to Theorem18, we have that for any t∈0,1,
(110)∣u~t−ut∣≤∫01G~t,sφq∫01H~s,τfτ,I0+γu~τ,u~τdτ−φq∫01Hs,τfτ,I0+γuτ,uτdτds+∫01G~t,s−Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds.
By Lemma11, we know that for u∈BR2,u~∈BR~2,
(111)∫01Hs,τfτ,I0+γuτ,uτdτ≤M0,∫01Hs,τfτ,I0+γu~τ,u~τdτ≤M~0,where
(112)M~0=Ma+Mb/Γγ+1p−1+McR~2p−1B~Γβ.From Lemma 17 and the inequality (53), we obtain
(113)∫01H~s,τfτ,I0+γu~τ,u~τdτ−∫01Hs,τfτ,I0+γuτ,uτdτ≤∫01H~s,τ∣fτ,I0+γu~τ,u~τ−fτ,I0+γuτ,uτ∣dτ+∫01H~s,τ−Hs,τfτ,I0+γuτ,uτdτ≤maxL1,L~1Γγ+1+maxL2,L~2u~−u∫01H~∗s,sdτ+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅sβ−1≤maxL1,L~1Γγ+1+maxL2,L~2u~−uB~Γβ+1+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1,where
(114)C=maxt∈0,1x∈0,R1y∈0,R2ft,x,y.
Sinceq>2, using Lemma 16, we can get
(115)∣u~t−ut∣≤maxL1,L~1Γγ+1+maxL2,L~2u~−uB~Γβ+1+CΓβ∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1⋅q−1maxM0q−2,M~0q−2∫01G~∗s,sds+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1≤maxL1,L~1/Γγ+1+maxL2,L~2u~−u/B~Γβ+1+C/Γβ∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1A~α−δΓα−γ⋅q−1maxM0q−2,M~0q−2+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1.
From the condition(116)q−1M0q−2α−δABΓα−γΓβ+1L1Γγ+1+L2=q−1R22−pβM¯L1Γγ+1+L2<1,it holds that for sufficiently small λξ and λζ,
(117)Λ4≔q−1maxM0q−2,M~0q−2A~α−δΓα−γB~Γβ+1maxL1,L~1Γγ+1+maxL2,L~2<1.
Denote as follows:(118)Λ5≔q−1maxM0q−2,M~0q−2CA~α−δΓβΓα−γ,Λ6≔M0Γα−γ.
This gives us that(119)u~−u≤Λ5∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1+Λ6∑i=1m−2ξ~i/A~−ξi/Aηiα−δ−11−ηiα−δ−11−Λ4.
Since(120)limλξ→0λζ→0ξ~iA~=ξiA,limλξ→0λζ→0ζ~iB~=ζiB,i=1,2,⋯,m−2,the proof is completed.Remark 20.
It follows from Theorems18 and 20 that limλξ→0λζ→0x~=x. Therefore, it follows that the unique positive solution to problem (1) is continuously dependent on the perturbations with respect to the coefficients in m-point boundary conditions. This is a new result different from the achievements of Dishlieva [21] and Li et al. [22] who studied the continuous dependency of solutions on initial conditions, barrier curves, and source terms in differential equations but not on boundary conditions.


## 3.1. Existence and Uniqueness of Positive Solutions to Problem (1)

Definition 6.
A functionx∈u∣u∈C0,1,D0+αu∈C0,1,D0+βφpD0+αu∈C0,1 is called a solution of problem (1) if it satisfies the fractional differential equation and the boundary conditions of (1).Lemma 7.
If the functionx is a solution of problem (1), then ut≔D0+γxt is a solution of the integral equation
(25)ut=∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds,in C0,1 and conversely, if u∈C0,1 is a solution of the integral equation (25), then xt=I0+γut is a solution of problem (1), where q is a number such that 1/p+1/q=1.Proof.
Letx be a solution of the problem (1). Then, since x∈C0,1,D0+αx∈C0,1, by using Lemma 1, it can be easily seen that
(26)xt=I0+αD0+αxt+c1tα−1+c2tα−2.
The upper continuities of the functionsxt,I0+αD0+αxt at t=0 provide us that c2=0. Applying D0+γ on both sides, we have
(27)D0+γxt=D0+γI0+αD0+αxt+D0+γc1tα−1=I0+α−γD0+αxt+ΓαΓα−γc1tα−γ−1.
Combining withD0+αx∈C0,1,α−γ−1>0, we can see that D0+γx∈C0,1. Since x∈C0,1,D0+γx∈C0,1, Lemma 1 yields that
(28)xt=I0+γD0+γxt+c3tγ−1.
Therefore, it follows from the upper continuities of the functionsxt,I0+γD0+γxt at t=0 that c3=0. This implies that by putting ut≔D0+γxt, it holds that xt=I0+γut. Thus, we can know that u is a solution of the following problem:
(29)D0+βφpD0+α−γut=ft,I0+γut,ut,0<t<1,u0=0,D0+δ−γu1=∑i=1m−2ξiD0+δ−γuηi,D0+α−γu0=0,φpD0+α−γu1=∑i=1m−2ζiφpD0+α−γuηi.
Putwt≔D0+α−γut. Lemma 2 shows that
(30)ut=−∫01Gt,swsds.
Also, by applying Lemma4, the notation vt=φpwt gives us that
(31)vt=−∫01Ht,sfs,I0+γus,usds.
Sinceφp−1=φq, by equations (30) and (31), we obtain
(32)ut=∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds.
Conversely, let the functionu∈C0,1 be a solution of integral equation (25). From this, by putting xt≔I0+γut, it can be also obtained that x∈C0,1. Denote as follows:
(33)zt≔ft,I0+γut,ut,vt≔φq∫01Ht,szsds.
From the continuities of the functionsu,f, and H, it is obvious that z∈C0,1,v∈C0,1. The definition of the function Gt,s implies that
(34)ut=∫01Gt,svsds=−I0+α−γvt−Γα−δAΓα−γ∑i=1m−2ξiI0+α−δvtt=ηi−I0+α−δvtt=1tα−γ−1,so, we have
(35)D0+αxt=D0+α−γut=vt.
This means thatD0+αx∈C0,1. Also, combining the definition of the function Ht,s with the relation
(36)D0+βφpD0+αxt=D0+βφpvt=D0+β∫01Ht,szsds,we can get
(37)∫01Ht,szsds=−I0+βzt−1B∑i=1m−2ζiI0+βztt=ηi−I0+βztt=1tβ−1.
This yields that(38)D0+βφpD0+αxt=zt,and therefore, it holds that D0+βφpD0+αx∈C0,1. Since equations (15) and (19) have unique solutions by Lemmas 2 and 4, u, the solution of integral equation (25), satisfies the fractional differential equation and the boundary conditions of the problem (29). To sum up, it can be proved that x is a solution of problem (1).LetE be the Banach space C0,1 equipped with the norm u=max0≤t≤1ut and put P≔u∈E∣ut≥0,t∈0,1. Since ∀u∈P,∀t∈0,1,I0+γut=1/Γγ∫0tt−sγ−1usds≥0, it holds that I0+γu∈P. Therefore, define an operator T as
(39)Tut≔∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds,u∈P,then, it satisfies TP⊂P. This implies that integral equation (25) has a solution in P if and only if operator T has a fixed point. It is also obvious that the fixed point of operator T is a solution of integral equation (25) in P.PutM¯≔α−δAΓα−γp−1BΓβ. The following hypotheses will be used throughout the paper.(H1) There exista,b,c∈L0,1 such that
(40)Ma≔∫01atdt>0,Mb≔∫01btdt>0,Mc≔∫01ctdt>0,MbΓγ+1p−1+Mc<M¯,∀t,x,y∈0,1×0,R1×0,R2,ft,x,y<at+btxp−1+ctyp−1,where R2=Ma/M¯−Mb/Γγ+1p−1−Mcq−1 and R1=R2/Γγ+1(H2) There existL1,L2>0 such that for any t∈0,1 and any x1,y1,x2,y2∈0,R1×0,R2,
(41)ft,x1,y1−ft,x2,y2≤L1x1−x2+L2y1−y2(H3) There existm¯,μ>0 such that for any t,x,y∈0,1×0,R1×0,R2,
(42)ft,x,y≥m¯tμ−1Lemma 8.
Assume that the hypothesis (H1) is satisfied. Then, it holds that
(43)TBR2⊂BR2,where BR2=u∈Pu≤R2.Proof.
SinceTBR2⊂P, it is sufficient to prove that ∀u∈BR2,Tu≤R2. It can be easily evaluated that for any t∈0,1,
(44)I0+γut=1Γγ∫0tt−sγ−1usds≤R2Γγ∫0tt−sγ−1ds=R2Γγ+1⋅tγ≤R1.
By using Lemma5 and hypothesis (H1), we have
(45)Tut=∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds≤∫01Gt,sφqmaxs∈0,1H∗s,s∫01aτ+bτI0+γuτp−1+cτuτp−1dτds≤∫01Gtsφq1BΓβMa+MbΓγ+1p−1+McR2p−1ds.
Applying the properties of the functionGt,s indicated in Lemma 3, the following holds:
(46)Tut≤1BΓβMa+MbΓγ+1p−1+McR2p−1q−1∫01G∗s,sds=Ma+Mb/Γγ+1p−1+McR2p−1q−1Bq−1Γβq−1Aα−δΓα−γ.
The relation1/p+1/q=1 shows that p−1 is an inverse power of q−1. So, it can be rewritten as
(47)Tut≤Ma+MbΓγ+1p−1+McR2p−1/M¯q−1.
Since it is satisfied forR2 that
(48)Ma=M¯−MbΓγ+1p−1−McR2p−1,we can get
(49)Tut≤R2.
This concludes the lemma.Lemma 9.
If the hypothesis (H3) holds, then for any t∈0,1 and any u∈BR2,
(50)∫01Ht,sfs,I0+γus,usds≥K0tβ−1,where K0=m¯Γμ/BΓβ+μ∑i=1m−2ζiηiβ−1−ηiβ−1+μ.Proof.
It follows from Lemma4 and hypothesis (H3) that
(51)∫01Ht,sfs,I0+γus,usds≥m¯∫01H1t,ssμ−1ds+∫01H2t,ssμ−1ds.
By simple calculation, the integration terms in the right side of inequality (51) can be estimated as
(52)∫01H1t,ssμ−1ds=1Γβtβ−1∫011−sβ−1sμ−1ds−∫0tt−sβ−1sμ−1ds=1Γβtβ−1Bμ,β−tβ−1+μBμ,β=ΓμΓβ+μtβ−1−tβ−1+μ≥0,∫01H2t,ssμ−1ds=tβ−1BΓβ∑i=1m−2ζiηiβ−1∫011−sβ−1sμ−1ds−∫0ηiηi−sβ−1sμ−1ds=tβ−1BΓβ∑i=1m−2ζiηiβ−1Bμ,β−ηiβ−1+μBμ,β=ΓμBΓβ+μ∑i=1m−2ζiηiβ−1−ηiβ−1+μtβ−1.
These estimations provide us the conclusion (50).DenoteK1≔1/Γβ+11+1/B∑i=1m−2ζiηiβ−1−ηiβ. Then, it can be found in [25] that for any t∈0,1,
(53)∫01Ht,sds≤K1tβ−1.Theorem 10.
Suppose thatp>2 and hypotheses (H1)-(H3) are satisfied. If
(54)q−1K1Aα−δΓα−γK02−qL1Γγ+1+L2<1,then, integral equation (25) has a unique solution in BR2.Proof.
To prove this theorem, operatorsT0,T1 are defined as
(55)T0ut≔φq∫01Ht,sfs,I0+γus,usds,T1ut≔∫01Gt,susds.
Obviously, we can see that for anyu∈BR2, Tut=T1oT0ut, and Tu0=0. Because p>2, it is satisfied that 1<q<2. So, by employing Lemma 8, Lemma 9, and property (23), we can obtain that for any u,v∈BR2 and any t∈0,1,
(56)T0ut−T0vt=φq∫01Ht,sfs,I0+γus,usds−φq∫01Ht,sfs,I0+γvs,vsds≤q−1K0tβ−1q−2∫01Ht,sfs,I0+γus,usds−∫01Ht,sfs,I0+γvs,vsds≤q−1K0tβ−1q−2∫01Ht,sL1I0+γus−I0+γvs∣+L2∣us−vsds≤q−1K0tβ−1q−2∫01Ht,sL1Γγ+1u−v∣+L2u−vds≤q−1K0tβ−1q−2L1Γγ+1+L2u−v∫01Ht,sds.
Substituting inequality (53) into (56), we have
(57)T0ut−T0vt≤q−1K0q−2K1L1Γγ+1+L2u−vtβ−1q−1.
Thus, it holds that(58)Tut−Tvt=∫01Gt,sT0usds−∫01Gt,sT0vsds≤q−1K0q−2K1L1Γγ+1+L2u−v∫01Gt,ssβ−1q−1ds.
Sinceβ−1q−1>0, by Lemma 3, we can get
(59)∫01Gt,ssβ−1q−1ds≤∫01Gt,sds≤∫01G∗s,sds=1Aα−δΓα−γ.
This yields that(60)Tu−Tv≤q−1K1Aα−δΓα−γK02−qL1Γγ+1+L2u−v.
Combining inequalities (54) and (60) with Lemma 8, the operator T:BR2→BR2 is a contraction mapping and the Banach contraction mapping principle provides us that it has a unique fixed point in BR2. So, integral equation (25) has a unique solution in BR2.Lemma 11.
Assume that the hypothesis (H1) holds. Then, for any u∈BR2 and any t∈0,1, it is satisfied that
(61)∫01Ht,sfs,I0+γus,usds≤M0,where
(62)M0=Ma+Mb/Γγ+1p−1+McR2p−1BΓβ.Proof.
In a similar way to the proof of Lemma8, it can be easily proved that
(63)∫01Ht,sfs,I0+γus,usds≤maxs∈0,1H∗s,s∫01fs,I0+γus,usds≤1BΓβ∫01as+bsI0+γusp−1+csusp−1ds≤1BΓβ∫01asds+R1p−1∫01bsds+R2p−1∫01csds=M0.Theorem 12.
Suppose that1<p<2 and hypotheses (H1) and (H2) hold. If
(64)q−1R22−pβM¯L1Γγ+1+L2<1,then, integral equation (25) has a unique solution in BR2.Proof.
In the case1<p<2, due to 1/p+1/q=1, it holds that q>2. Similar to Theorem 10, defining operators T0,T1 with Lemma 8, Lemma 11, and property (24) give us that for any u,v∈BR2 and any t∈0,1,
(65)T0ut−T0vt=φq∫01Ht,sfs,I0+γus,usds−φq∫01Ht,sfs,I0+γvs,vsds≤q−1M0q−2∫01Ht,sfs,I0+γus,us−fs,I0+γvs,vsds≤q−1M0q−2∫01Ht,sL1∣I0+γus−I0+γvs∣+L2∣us−vs∣ds≤q−1M0q−2∫01Ht,sL1Γγ+1u−v+L2u−vds≤q−1M0q−2L1Γγ+1+L2u−v∫01H∗s,sds=q−1M0q−2BΓβ+1L1Γγ+1+L2u−v.
So, we have(66)Tut−Tvt=∫01Gt,sT0usds−∫01Gt,sT0vsds≤∫01Gt,sT0us−T0vsds≤q−1M0q−2BΓβ+1L1Γγ+1+L2u−v∫01Gt,sds≤q−1M0q−2α−δABΓα−γΓβ+1L1Γγ+1+L2u−v.
Obviously, forM0, the following holds:
(67)M0q−2α−δABΓα−γΓβ+1=1βM¯q−1⋅M¯R2p−1q−2=R22−pβM¯.
Therefore, by employing inequality (64), equality (67), and Lemma 8, the operator T:BR2→BR2 is a contraction mapping. It follows from the Banach contraction mapping principle that the operator T has a unique fixed point in BR2. This completes the proof.Remark 13.
Theorems10 and 12 show the existence and uniqueness of solutions to integral equation (25) in the cases p>2 and 1<p<2, respectively. The Banach contraction mapping principle, employed in the proofs of these theorems, guarantees that the iterative sequences can be constructed to converge to the exact solution of integral equation (25). In other words, the iterative sequence unn=0∞ initialized at any point u0∈BR2 and defined by
(68)un+1=Tun,n=0,1,2,⋯,converges to the exact solution of integral equation (25), u, in BR2. This provides the possibility to obtain the approximate solution un of integral equation (25) by using Theorems 10 and 12.Theorem 14.
Problem (1) has a unique solution in Ω if and only if the solution to integral equation (25) uniquely exists in BR2, where Ω=I0+γBR2. In particular, if hypothesis (H3) is satisfied, then the solution to problem (1) is positive.Proof.
By Lemma7, for x, a solution to problem (1), and u satisfying integral equation (25), the following relation holds:
(69)xt=I0+γut,t∈0,1.
This proves the first part of the theorem.
For the second part, it is necessary that if hypothesis (H3) holds, then(70)∀t∈0,1,xt>0.
To prove this, we must see if the following holds:(71)∀t∈0,1,ut>0
Sinceu∈BR2, using Lemma 9, it can be easily obtained that
(72)ut≥K0q−1∫01Gt,ssβ−1q−1ds=K0q−1∫01G1t,ssβ−1q−1ds+∫01G2t,ssβ−1q−1ds.
Considering thatG1t,s≥0 and β−1q−1>0, we can get
(73)∀t∈0,1,∫01G1t,ssβ−1q−1ds≥0..
From the definition of functionG2t,s, we have
(74)∫01G2t,ssβ−1q−1ds=tα−γ−1AΓα−γ∑i=1m−2ξiηiα−δ−1∫011−sα−δ−1sβ−1q−1ds−∫0ηiηi−sα−δ−1sβ−1q−1ds=Bα−δ,β−1q−1+1AΓα−γ∑i=1m−2ξiηiα−δ−1−ηiα−δ+β−1q−1tα−γ−1.
Denote as follows:(75)K2≔Bα−δ,β−1q−1+1AΓα−γ∑i=1m−2ζiηiα−δ−1−ηiα−δ+β−1q−1.
SinceK0,K2>0 and ut≥K0q−1K2tα−γ−1, it is obvious that (71) holds. Therefore, we can see that
(76)xt=I0+γut≥I0+γK0q−1K2tα−γ−1=K0q−1K2⋅Γα−γΓαtα−1.
The proof is completed.Remark 15.
Theorems10, 12, and 14 show new results that differ from some of the references listed in Section 1. One of these is the result of the existence, uniqueness, and positivity of solutions to p-Laplacian fractional differential equations with multipoint boundary conditions where the nonlinear source terms contain a more generalized form of fractional derivatives than those considered in [16, 17, 20]. The other is the possibility of constructing an iterative sequence that converges to a unique solution of a given problem, which is different from [16, 18, 19]. Combining Theorem 14 with Theorems 10 and 12, the iterative sequence xnn=0∞ given by
(77)xn=I0+γun,n=0,1,2,⋯,converges to the exact solution of problem (1), x, in Ω, where unn=0∞ is constructed by (68). This gives an opportunity to get the approximate solution, xn, for problem (1), although there might be some difficulties in its numerical computation.


## 3.2. Continuous Dependence of a Solution on Boundary Conditions

In this subsection, the continuous dependence of a unique positive solutionx to problem (1) on boundary conditions will be established. To do this, denote a unique solution to the problem
(78)D0+βφpD0+αxt=ft,xt,D0+γxt,0<t<1,D0+γx0=0,D0+δx1=∑i=1m−2ξ~iD0+δxηi,D0+αx0=0,φpD0+αx1=∑i=1m−2ζ~iφpD0+αxηi,by x~.As can be seen above, the positive solutionx to problem (1) is given by
(79)xt=I0+γut,t∈0,1,where u is a solution of integral equation (25). Also, the positive solution x~ to problem (78) is expressed by
(80)x~t=I0+γu~t,t∈0,1,where u~ is a solution of the integral equation
(81)u~t=∫01G~t,sϕq∫01H~s,τfτ,I0+γu~τ,u~τdτds,in which
(82)A~=1−∑i=1m−2ξ~iηiα−δ−1,B~=1−∑i=1m−2ζ~iηiβ−1,G~2t,s=tα−γ−1A~Γα−γ∑s<ηiξ~iηiα−δ−11−sα−δ−1−ηi−sα−δ−1+∑s≥ηiξ~iηiα−δ−11−sα−δ−1,H~2t,s=tβ−1B~Γβ∑s<ηiζ~iηiβ−11−sβ−1−ηi−sβ−1+∑s≥ηiζ~iηiβ−11−sβ−1,G~t,s=G1t,s+G~2t,s,H~t,s=H1t,s+H~2t,s,M¯~=α−δA~Γα−γp−1B~Γβ,R~2=Ma/M¯~−MbΓγ+1p−1−Mcq−1.Then, the continuous dependence of a unique positive solution to the problem (1) on boundary conditions can be formulated by
(83)limλξ→0λζ→0x~=x,where
(84)λξ≔max∣ξ~1−ξ1∣,∣ξ~2−ξ2∣,⋯,∣ξ~m−2−ξm−2∣,λζ≔max∣ζ~1−ζ1∣,∣ζ~2−ζ2∣,⋯,∣ζ~m−2−ζm−2∣.Considering relations (79) and (80), it is sufficient to prove that
(85)limλξ→0λζ→0u~=u.The solutions of integral equations (25) and (81), u and u~, uniquely exist in BR2,BR~2, respectively. Since A,B,M¯,andR2 are continuous at ξ1,ξ2,⋯,ξm−2,ζ1,ζ2,⋯,ζm−2, it holds that
(86)limλξ→0λζ→0A~=A,limλξ→0λζ→0B~=B,limλξ→0λζ→0M¯~=M¯,limλξ→0λζ→0R~2=R2.Define a functionG¯2t,s as follows:
(87)G¯2t,s≔1Γα−γtα−δ−11−sα−δ−1−t−sα−δ−1,0≤s≤t≤1,tα−δ−11−sα−δ−1,0≤t≤s≤1.Then, it can be easily seen that(88)G2t,s=1A∑i=1m−2ξiG¯2ηi,s⋅tα−γ−1,G~2t,s=1A~∑i=1m−2ξ~iG¯2ηi,s⋅tα−γ−1.And it can be found in [27] that
(89)H2t,s=1B∑i=1m−2ζiH1ηi,s⋅tβ−1,H~2t,s=1B~∑i=1m−2ζ~iH1ηi,s⋅tβ−1.For a fixedt, the functions G¯2t,s and H1t,s have maximum values at s=t, i.e.,
(90)maxs∈0,1G¯2t,s=tα−δ−11−tα−δ−1Γα−γ,maxs∈0,1H1t,s=tβ−11−tβ−1Γβ.Lemma 16.
Lety∈C0,1. Then, the following holds:
(91)∀t∈0,1,∣v~t−vt∣≤yΓα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1⋅tα−γ−1,where
(92)vt=∫01Gt,sysds,v~t=∫01G~t,sysds.Proof.
Using relation (88), we can see that for any t∈0,1,
(93)∣v~t−vt∣≤∫01∣G~t,s−Gt,s∣ysds≤y∫011A~∑i=1m−2ξ~iG¯2ηi,s−1A∑i=1m−2ξiG¯2ηi,stα−γ−1ds≤y∑i=1m−2ξ~iA~−ξiA∫01G¯2ηi,sds⋅tα−γ−1.
By applying (90), we obtain
(94)∣v~t−vt∣≤yΓα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1⋅tα−γ−1.Lemma 17.
Letz∈C0,1. Then, it holds that
(95)∀t∈0,1,∣w~t−wt∣≤z∣Γβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅tβ−1,where
(96)wt=∫01Ht,szsds,w~t=∫01H~t,szsds.Proof.
The proof is similar to Lemma16, so it is omitted.Theorem 18.
Suppose that all the assumptions of Theorem10 are satisfied. Then, it holds that
(97)limλξ→0λζ→0u~=u..Proof.
It can be obtained that for anyt∈0,1,
(98)∣u~t−ut∣≤∫01G~t,sφq∫01H~s,τfτ,I0+γu~τ,u~τdτds−∫01G~t,sφq∫01Hs,τfτ,I0+γuτ,uτdτds+∫01G~t,sφq∫01Hs,τfτ,I0+γuτ,uτdτds−∫01Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds≤∫01G~t,sφq∫01H~s,τfτ,I0+γu~τ,u~τdτ−φq∫01Hs,τfτ,I0+γuτ,uτdτds+∫01G~t,s−Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds.
Lemma9 indicates that for u∈BR2,u~∈BR~2,
(99)∫01Hs,τfτ,I0+γuτ,uτdτ≥K0sβ−1,∫01H~s,τfτ,I0+γu~τ,u~τdτ≥K~0sβ−1,where K~0=m¯~Γμ~/B~Γβ+μ~∑i=1m−2ζ~iηiβ−1−ηiβ−1+μ~. And by Lemma 11, we know that
(100)∫01Hs,τfτ,I0+γuτ,uτdτ≤M0.
Considering Lemma17 and inequality (53), it is easy to see that
(101)∫01H~s,τfτ,I0+γu~τ,u~τdτ−∫01Hs,τfτ,I0+γuτ,uτdτ≤∫01H~s,τfτ,I0+γu~τ,u~τdτ−∫01H~s,τfτ,I0+γuτ,uτdτ+∫01H~s,τfτ,I0+γuτ,uτdτ−∫01Hs,τfτ,I0+γuτ,uτdτ≤∫01H~s,τ∣fτ,I0+γu~τ,u~τ−fτ,I0+γuτ,uτ∣dτ+∫01H~s,τ−Hs,τfτ,I0+γuτ,uτdτ≤maxL1,L~1Γγ+1+maxL2,L~2u~−u∫01H~s,τdτ+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅sβ−1≤K~1maxL1,L~1Γγ+1+maxL2,L~2u~−u+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅sβ−1,where
(102)K~1=1Γβ+11+1B~∑i=1m−2ζ~iηiβ−1−ηiβ,and C=maxt∈0,1x∈0,R1y∈0,R2∣ft,x,y∣.
Therefore, combining Lemma16 with 1<q<2, it follows that
(103)∣u~t−ut∣≤q−1minK02−q,K~02−qK~1maxL1,L~1Γγ+1+maxL2,L~2u~−u+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅∫01G~∗s,ssβ−1q−1ds+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1≤q−1K~1maxL1,L~1/Γγ+1+maxL2,L~2u~−u+C/Γβ∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1A~α−δΓα−γminK02−q,K~02−q+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1⋅
Since(104)q−1K1Aα−δΓα−γK02−qL1Γγ+1+L2<1,and A,B,K0,K1,L1,L2 are continuous at ξ1,ξ2,⋯,ξm−2,ζ1,ζ2,⋯,ζm−2, we can see that for sufficient small λξ and λζ,
(105)Λ1≔q−1K~1A~α−δΓα−γminK02−q,K~02−qmaxL1,L~1Γγ+1+maxL2,L~2<1.
So, putting(106)Λ2≔q−1CA~α−δΓβΓα−γminK02−q,K~02−q,Λ3≔M0Γα−γ,we have
(107)u~−u≤Λ2∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1+Λ3∑i=1m−2ξ~i/A~−ξi/Aηiα−δ−11−ηiα−δ−11−Λ1.
Finally, the fact that(108)limλξ→0λζ→0ξ~iA~=ξiA,limλξ→0λζ→0ζ~iB~=ζiB,i=1,2,⋯,m−2proves the conclusion of this theorem.Theorem 19.
Suppose that all the assumptions of Theorem12 hold. Then, it is satisfied that
(109)limλξ→0λζ→0u~=u.Proof.
Similar to Theorem18, we have that for any t∈0,1,
(110)∣u~t−ut∣≤∫01G~t,sφq∫01H~s,τfτ,I0+γu~τ,u~τdτ−φq∫01Hs,τfτ,I0+γuτ,uτdτds+∫01G~t,s−Gt,sφq∫01Hs,τfτ,I0+γuτ,uτdτds.
By Lemma11, we know that for u∈BR2,u~∈BR~2,
(111)∫01Hs,τfτ,I0+γuτ,uτdτ≤M0,∫01Hs,τfτ,I0+γu~τ,u~τdτ≤M~0,where
(112)M~0=Ma+Mb/Γγ+1p−1+McR~2p−1B~Γβ.From Lemma 17 and the inequality (53), we obtain
(113)∫01H~s,τfτ,I0+γu~τ,u~τdτ−∫01Hs,τfτ,I0+γuτ,uτdτ≤∫01H~s,τ∣fτ,I0+γu~τ,u~τ−fτ,I0+γuτ,uτ∣dτ+∫01H~s,τ−Hs,τfτ,I0+γuτ,uτdτ≤maxL1,L~1Γγ+1+maxL2,L~2u~−u∫01H~∗s,sdτ+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1⋅sβ−1≤maxL1,L~1Γγ+1+maxL2,L~2u~−uB~Γβ+1+CΓβ∑i=1m−2ζ~iB~−ζiBηiβ−11−ηiβ−1,where
(114)C=maxt∈0,1x∈0,R1y∈0,R2ft,x,y.
Sinceq>2, using Lemma 16, we can get
(115)∣u~t−ut∣≤maxL1,L~1Γγ+1+maxL2,L~2u~−uB~Γβ+1+CΓβ∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1⋅q−1maxM0q−2,M~0q−2∫01G~∗s,sds+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1≤maxL1,L~1/Γγ+1+maxL2,L~2u~−u/B~Γβ+1+C/Γβ∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1A~α−δΓα−γ⋅q−1maxM0q−2,M~0q−2+M0Γα−γ∑i=1m−2ξ~iA~−ξiAηiα−δ−11−ηiα−δ−1.
From the condition(116)q−1M0q−2α−δABΓα−γΓβ+1L1Γγ+1+L2=q−1R22−pβM¯L1Γγ+1+L2<1,it holds that for sufficiently small λξ and λζ,
(117)Λ4≔q−1maxM0q−2,M~0q−2A~α−δΓα−γB~Γβ+1maxL1,L~1Γγ+1+maxL2,L~2<1.
Denote as follows:(118)Λ5≔q−1maxM0q−2,M~0q−2CA~α−δΓβΓα−γ,Λ6≔M0Γα−γ.
This gives us that(119)u~−u≤Λ5∑i=1m−2ζ~i/B~−ζi/Bηiβ−11−ηiβ−1+Λ6∑i=1m−2ξ~i/A~−ξi/Aηiα−δ−11−ηiα−δ−11−Λ4.
Since(120)limλξ→0λζ→0ξ~iA~=ξiA,limλξ→0λζ→0ζ~iB~=ζiB,i=1,2,⋯,m−2,the proof is completed.Remark 20.
It follows from Theorems18 and 20 that limλξ→0λζ→0x~=x. Therefore, it follows that the unique positive solution to problem (1) is continuously dependent on the perturbations with respect to the coefficients in m-point boundary conditions. This is a new result different from the achievements of Dishlieva [21] and Li et al. [22] who studied the continuous dependency of solutions on initial conditions, barrier curves, and source terms in differential equations but not on boundary conditions.


## 4. Examples

Example 21.
Consider the following boundary value problem:(121)D0+1.5φ3D0+1.75xt=10t+0.2xt+0.1D0+0.2xt2,0<t<1,D0+0.2x0=0,D0+0.4x1=0.3D0+0.4x0.3+0.2D0+0.4x0.7,D0+1.75x0=0,φ3D0+1.75x1=0.2φ3D0+1.75x0.3+0.5φ3D0+1.75x0.7.
Problem (121) can be regarded as the boundary value problem (1) where ft,x,y=10t+0.2x+0.1y2,p=3,α=1.75,β=1.5,γ=0.2,δ=0.4,η1=0.3,η2=0.7,ξ1=0.3,ξ2=0.2,ζ1=0.2,andζ2=0.5. Then, we have
(122)q=1.5<2,A=0.626632⋯,B=0.472125⋯,M¯=0.236575⋯.
Sincex≤4+x2/16, denoting at=10.8,bt=0.0125,andct=0.1 yields that
(123)Ma=10.8,Mb=0.0125,Mc=0.1,ft,x,y<at+btxp−1+ctyp−1.
Puttingm¯=10,μ=1.5, we find that
(124)ft,x,y≥m¯tμ−1.
R2,R1 can be calculated as
(125)R2=9.41849⋯,R1=R2Γγ+1=10.2579⋯.
Also, putL1=0.2,L2=0.1⋅2,andR2=1.8837⋯. Then, we can see that for any t∈0,1 and any x1,y1,x2,y2∈0,R1×0,R2,
(126)ft,x1,y1−ft,x1,y2≤0.2x1−x2+0.1y12−y22≤L1x1−x2+L2y1−y2.
A simple calculation provides that(127)K0=2.48598⋯,K1=1.07439⋯,K=0.952215⋯.
By Theorems10 and 14, K<1 shows that fractional boundary value problem (121) has a unique positive solution in I0+γBR2. Besides, Theorem 18 shows that the unique positive solution to the problem (121) is continuously dependent on its boundary conditions.Example 22.
Consider the boundary value problem(128)D0+1.5φ1.5D0+1.75xt=0.1t+0.1xt2+0.1D0+0.2xt,0<t<1,D0+0.2x0=0,D0+0.4x1=0.3D0+0.4x0.3+0.2D0+0.4x0.7,D0+1.75x0=0,φ1.5D0+1.75x1=0.2φ1.5D0+1.75x0.3+0.5φ1.5D0+1.75x0.7.
Problem (128) can be regarded as the boundary value problem (1) where ft,x,y=0.1t+0.1x2+0.1y,p=1.5,α=1.75,β=1.5,γ=0.2,δ=0.4,η1=0.3,η2=0.7,ξ1=0.3,ξ2=0.2,ζ1=0.2,andζ2=0.5. Then, we can get
(129)q=3>2,A=0.626632⋯,B=0.472125⋯,M¯=0.362822⋯.
Denoteat=0.1,bt=0.1,ct=0.1 and we have
(130)Ma=0.1,Mb=0.1,Mc=0.1,R2=0.398248⋯,R1=0.433742⋯.
Also, it holds that for anyt,x,y∈0,1×0,R1×0,R2,
(131)ft,x,y=0.1t+0.1x2+0.1y<0.1+0.1x+0.1y.
Since∣ft,x1,y1−ft,x2,y2∣≤0.1∣x12−x22∣+0.1∣y1−y2∣≤0.1⋅2R1∣x1−x2∣+0.1⋅∣y1−y2∣ for any t∈0,1 and any x1,y1,x2,y2∈0,R1×0,R2, we can put
(132)L1=0.0867483⋯,L2=0.1.
By simple calculation, we have(133)K=0.45102⋯.
SinceK<1, using Theorems 12 and 14, boundary value problem (128) has a unique positive solution in I0+γBR2. Also, by employing Theorem 20, the continuous dependence of the unique positive solution to problem (128) on its boundary conditions is proved.Comparing the problems (121) and (128) considered in Examples 22 and 23, respectively, there exists two main differences between them. One is the value of p. If p=2, it is obvious that φps=s. As can be seen in the inequalities (23) and (24), some properties of φp⋅ depend on whether the value of p is greater than 2 or not. So, Examples 22 and 23 demonstrated the existence, the uniqueness, the positivity, and the continuous dependency on four-point boundary conditions of the solutions to the problems (121) and (128) in the cases of p>2 and 1<p<2 by using Theorems 10 and 18 and Theorems 12 and 20, respectively. The other difference lies in the nonlinear source term ft,x,y. The source term in problem (121) was nonlinear to the fractional derivative of the unknown function, whereas, in problem (128), the nonlinearity appeared for the unknown function itself. However, similar techniques were applied to these two nonlinear source terms in the problems (121) and (128) to illustrate our main results.


## 5. Conclusion

In this paper, we have proved the existence, the uniqueness, the positivity, and the continuous dependency on multipoint boundary conditions of solutions tom-point boundary value problems for nonlinear multiterm fractional differential equations with p-Laplacian operator by employing the Banach contraction mapping principle. For the application of this study, two examples have been illustrated.



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