# Existence and Uniqueness Results for a Class of Singular Fractional Boundary Value Problems with thep-Laplacian Operator via the Upper and Lower Solutions Approach

Authors: KumSong Jong; HuiChol Choi; KyongJun Jang; SunAe Pak
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we study the existence and uniqueness of positive solutions to a class of multipoint boundary value problems for singular fractional differential equations with thep-Laplacian operator. Here, the nonlinear source term f permits singularity with respect to its time variable t. Some fixed-point theorems such as the Leray-Schauder nonlinear alternative, the Schauder fixed-point theorem, and the Banach contraction mapping principle and the properties of the Gauss hypergeometric function are used to prove our main results. And by employing the upper and lower solutions technique, we derive a new approach to obtain the maximal and minimal solutions to the given problem. Finally, we present some examples to demonstrate our existence and uniqueness results.


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## Body

## 1. Introduction

In this paper, we consider the existence of positive solutions of the followingm-point boundary value problems for singular nonlinear fractional differential equations
(1)D0+βφpD0+αxt=ft,xt,0<t<1,x0=0,D0+γx1=∑i=1m−2ξiD0+γxηi,D0+αx0=0,φpD0+αx1=∑i=1m−2ζiφpD0+αxηi,where 1<α,β≤2,0<γ<1,0<ξi,ηi,ζi<1,i=1,2,⋯,m−2, and D0+α,D0+β,D0+γ are the standard Riemann–Liouville derivatives. Here, f∈C0,1×0,+∞,0,+∞, ft,x is singular at t=0,1 (i.e., limt→0+ft,⋅=+∞,limt→1−ft,⋅=+∞) and φp is defined as φps=sp−2s,p>1.Fractional calculus has a history of several hundred years, and many valuable results, which have contributed to the development of mathematical theories and their application to practice, have been created during its historical process (see [1]). Also, fractional differential equations are one of the powerful tools to model and solve scientific and technological problems arising in physics, chemistry, biology, and mechanics, and it has developed more and more in depth (see [2]). In particular, after Leibenson’s work dealt with the application of the integer-order differential equation with the p-Laplacian operator to the analysis of turbulent flow in porous media (see [3]), many valuable existence results for this equation have been achieved, and recently, the achievements obtained in this integer-order differential equation are more generalized to the fractional differential equation (see [4–6] and the references therein). However, due to the nonlinearity of the p-Laplacian operator, not much has been studied on the solutions to singular fractional differential equations with this operator and many researchers have been paid their attention to those equations. For instance, by using the fixed-point theorem of mixed monotone operators, Jong et al. [7] proved the existence of positive solutions to the boundary value problem (1), in which the nonlinear source term ft,x was singular with respect to its space variable x, and proposed a new approach by which the approximate solution of the given problem could be obtained. Unlikely in [7], this paper deals with the boundary value problem (1), in which the function ft,x permits singularity with respect to its time variable t.Many researchers have derived some important results for solutions to boundary value problems of fractional differential equations with singularity with respect to the time variable (see [8–29]). In [14], Henderson et al. established the existence and multiplicity of positive solutions for a system of nonlinear Riemann–Liouville fractional differential equations
(2)D0+αut+ft,ut,vt=0,t∈0,1,D0+βut+gt,ut,vt=0,t∈0,1,with the coupled multipoint boundary conditions
(3)u0=u′0=⋯=un−20=0,D0+p1utt=1=∑i=1NaiD0+q1vtt=ξi,v0=v′0=⋯=vm−20=0,D0+p2vtt=1=∑i=1MbiD0+q2utt=ηi,where α∈n−1,n,β∈m−1,m and the functions f,g can be singular at the points t=0 and/or t=1. They employed the Guo–Krasnosel’skii fixed-point theorem to prove that their problem has at least one positive solution. And Wu and Zhou [6] used the upper and lower solutions method to study the existence of positive solutions for the fractional-order eigenvalue problem with the p-Laplacian operator
(4)−DtβφpDtαxt=λft,xt,t∈0,1,x0=0,Dtαx0=0,Dtγx1=∑j=1m−2ajDtγxξj,where 1<α≤2,0<β,γ≤1 and f can be singular at t=0,1 and x=0 (for more detailed information about the upper and lower solutions method to solve integral and differential equations, see [30]).Taking the previous results together, to our best knowledge, very little is known about the existence and uniqueness of positive solutions ofp-Laplacian fractional boundary value problems with singularities with respect to their time variable.Motivated by the above works, in this paper, we first apply the Leray-Schauder nonlinear alternative to establish the existence of solutions to our problem (1) and then use the Schauder fixed-point theorem and upper and lower control functions to derive the upper and lower solutions method to obtain the maximal and minimal solutions. And we prove the uniqueness of solutions to the given problem by using some useful properties of the Gauss hypergeometric function F21a,b,c,s and the Banach contraction mapping principle.Throughout this paper, we suppose that(5)3<α+β≤4,α−γ−1>0,∑i=1m−2ξiηiα−γ−1<1,∑i=1m−2ζiηiβ−1<1.


## 2. Preliminaries

For the convenience of the readers, we will give some necessary definitions and lemmas here.The Riemann-Liouville fractional integral and the Riemann-Liouville fractional derivative of orderα>0 of a function f:0,∞⟶R are given by
(6)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 [1]).Lemma 1 (see [31]).
Assume thatu∈C0,1∩L0,1 with a fractional derivative of order α>0 that belongs to C0,1∩L0,1. Then,
(7)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.Lemma 2 (see [32]).
(Schauder fixed-point theorem). LetD be a nonempty, closed, bounded, and convex subset of a Banach space X, and suppose A:D⟶D is a compact operator. Then, A has a fixed point.Lemma 3 (see [33]).
LetX be a Banach space with D a closed and convex subset of X. Assume U is a relatively open subset of D, with 0∈U, and A:U¯⟶D is a compact map. Then, either,
(i)
A has a fixed point in U¯, or(ii)
there is a pointx∈∂U and λ∈0,1, with x=λAxLemma 4.
Letz:0,1⟶0,+∞ be a continuous function such that
(8)∃σ0,σ1∈0,1;limt→0+tσ0zt=z0>0,limt→1−1−tσ1zt=z1>0,and define the function K as
(9)Kt≔0,t=0,∫0tt−sβ−1zsds,t∈0,1.
Then,K is continuous on 0,1.Proof.
From the assumption (5), we can see easily that
(10)∃δ0>0;∀t0<t<δ0,12z0<tσ0zt<32z0,(11)∃δ1>0;∀t1−δ1<t<1,12z1<1−tσ1zt<32z1.
Put as follows(12)δ≔min14,δ0,δ1,Mz≔maxδ≤t≤1−δzt.
Then, we divide the proof of this lemma into the following three cases:
Ift∈0,δ, the definition of an improper integral
(13)Kt=limt′→0+∫t′tt−sβ−1zsds,yields
(14)∫t′tt−sβ−1zsds=∫t′tt−sβ−1s−σ0sσ0zsds≤32z0∫t′tt−sβ−1s−σ0ds≤32z0∫011−sβ−1s−σ0tβ−σ0ds=32z0Bβ,1−σ0tβ−σ0.
Take the limitt′⟶0+ to obtain
(15)Kt≤32z0Bβ,1−σ0tβ−σ0.
This implieslimt→0+Kt=0=K0.
Ift∈δ,1−δ, we can get
(16)Kt=limt′→0+∫t′δt−sβ−1zsds+∫δtt−sβ−1zsds.
In a similar way above, the first term of the right side in Equation (17) can be evaluated as
(17)limt′→0+∫t′δt−sβ−1zsds≤32z0Bβ,1−σ0tβ−σ0.
And for the second term of the right side in Equation (17), it can be easily seen that
(18)∫δtt−sβ−1zsds≤Mz∫δtt−sβ−1ds≤Mz∫0tt−sβ−1ds=1βMztβ.
Combining these two inequalities above, we can find(19)Kt≤32z0Bβ,1−σ0tβ−σ0+1βMztβ.
For the caset∈1−δ,1, if t<1, then we have
(20)Kt=limt′→0+∫t′δt−sβ−1zsds+∫δ1−δt−sβ−1zsds+∫1−δtt−sβ−1zsds.
Similarly to that given above, we obtain(21)limt′→0+∫t′δt−sβ−1zsds≤32z0Bβ,1−σ0tβ−σ0,∫δ1−δt−sβ−1zsds≤1βMztβ.
By simple calculation, we can get(22)∫1−δtt−sβ−1zsds=∫1−δtt−sβ−11−s−σ11−sσ1zsds≤32z1∫1−δtt−sβ−11−s−σ1ds≤32z1∫011−sβ−11−s−σ1ds=32β−σ1z1.
So for anyt∈1−δ,1, it holds that
(23)Kt≤32z0Bβ,1−σ0tβ−σ0+1βMztβ+32β−σ1z1.
Ift=1, the definition of an improper integral also implies K1=limt→1−Kt. Taking the limit t⟶1− on both sides of the inequality (11), we can obtain
(24)K1≤32z0Bβ,1−σ0+1βMz+32β−σ1z1<+∞.
Therefore, it follows directly from the inequalities (7) and (10), (11), (12), (13) that K is well-defined on 0,1. Since z∈C0,1, it is obvious that K is continuous on 0,1. Combining this with the continuity of K at t=0,1, we can prove that K is a continuous function on 0,1.Lemma 5.
Letz:0,1⟶0,+∞ be a continuous function such that satisfies (6). Then, the boundary value problem
(25)D0+βvt+zt=0,v0=0,v1=∑i=1m−2ζivηi,has a unique solution which is given by
(26)vt=∫01Ht,szsds,where
(27)Ht,s=H1t,s+H2t,s,in which
(28)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,where
(29)B=1−∑i=1m−2ζiηiβ−1.Remark 6.
In Lemma5, a function v∈C0,1 with a fractional derivative of order β that belongs to C0,1 (i.e., D0+βu∈C0,1) is said to be a solution of the boundary value problem (14) if it satisfies the fractional differential equation and the boundary conditions of (12).Proof.
As we can see in the proof of Lemma4, we have
(30)∫01tσ01−tσ1ztdt<+∞,∫01t−σ01−t−σ1dt=B1−σ0,1−σ1<+∞.
So, we can get(31)∫01ztdt=∫01t−σ01−t−σ1tσ01−tσ1ztdt<+∞.
This impliesz∈L0,1.
Also, Lemma4 asserts that
(32)I0+βzt=1Γβ∫0tt−sβ−1zsds∈C0,1.
Sincez∈C0,1∩L0,1 and I0+βz∈C0,1, it follows from Lemma 1 that a solution of the boundary value problem (14), v∈C0,1, satisfies
(33)vt+c1tβ−1+c2tβ−2=−I0+βzt.
For the rest of the proof, it is easy to see that doing as in the proof of Lemma4 in [12] leads to a conclusion of this lemma.Lemma 7 (see [4]).
If∑i=1m−2ζiηiβ−1<1, then the function Ht,s in Lemma 5 satisfies the following conditions:
(i)
Ht,s>0, for t,s∈0,1(ii)
Ht,s≤H∗s,s, for t,s∈0,1where
(34)H∗s,s=1BΓβ1−sβ−1.
As we can see in the proof of Lemma4 in [4], it holds that
(35)vt=∫01Ht,szsds=−I0+βzt−1B∑i=1m−2ζiI0+βztt=ηi−I0+βztt=1tβ−1.Lemma 8 (see [34]).
Lety∈C0,1. Then, the boundary value problem
(36)D0+αut+yt=0,0<t<1,u0=0,D0+γu1=∑i=1m−2ξiD0+γuηi,has a unique solution which is given by
(37)ut=∫01Gt,sysds,where
(38)Gt,s=G1t,s+G2t,s,in which
(39)G1t,s=tα−11−sα−γ−1−t−sα−1Γα,0≤s≤t≤1,tα−11−sα−γ−1Γα,0≤t≤s≤1,G2t,s=tα−1AΓα∑s<ηiξiηiα−γ−11−sα−γ−1−ηi−sα−γ−1+∑s≥ηiξiηiα−γ−11−sα−γ−1,t,s∈0,1,where
(40)A=1−∑i=1m−2ξiηiα−γ−1.Lemma 9 (see [34]).
If∑i=1m−2ξiηiα−γ−1<1, then the function Gt,s in Lemma 8 satisfies the following conditions:
(i)
Gt,s>0, for t,s∈0,1(ii)
Gt,s≤G∗s,s, for t,s∈0,1where
(41)G∗s,s=1AΓα1−sα−γ−1.
The following useful properties ofφp⋅ which will be used later can be found in [5]:
(i)
If1<p<2,xy>0, and x,y≥m>0, then(42)φpx−φpy≤p−1mp−2x−y(ii)
Ifp>2,x,y≤M, then(43)φpx−φpy≤p−1Mp−2x−y.


## 3. Main Results

In this section, we will prove the existence and uniqueness of positive solutions for the boundary value problem (1) and derive the upper and lower solutions method by using some fixed-point theorems.


### 3.1. The Existence Results for Problem (1)

Definition 10.
A functionx∈uu∈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).
The following hypothesis concerned with the functionft,x, which permits singularity with respect to time variable, will be used in this article.(H1).
There existσ0,σ1∈0,1 such that for any x∈0,+∞,
(44)limt→0+tσ0ft,x=f0x>0,limt→1−1−tσ1ft,x=f1x>0.
LetE be the Banach space C0,1 equipped with the norm u=max0≤t≤1ut and put P≔u∈Eut≥0,t∈0,1.Lemma 11.
Assume that the hypothesis (H1) holds. Then, the function x∈P is a solution of the problem (1) if and only if x is a solution of the integral equation
(45)xt=∫01Gt,sφq∫01Hs,τfτ,xτdτds,in P, where q is a real number such that 1/p+1/q=1.Proof.
Suppose thatx∈P is a solution of the problem (1). Putting wt≔D0+αxt, by using Lemma 8, we have
(46)xt=−∫01Gt,swsds.
Also puttingvt≔φpwt, by Lemma 5, we obtain
(47)vt=−∫01Ht,sfs,xsds.
Sinceφp−1=φq, Equations (47) and(48) yields
(48)xt=∫01Gt,sφq∫01Hs,τfτ,xτdτds.
It is obvious thatx∈P and x is a solution of the integral Equation (46).
Conversely, suppose thatx∈P is a solution of the integral Equation (46). Put as follows:
(49)zt≔ft,xt,vt≔φq∫01Ht,szsds.
Then, Equation (36) implies that v∈C0,1. From the definition of Gt,s, we also have
(50)xt=∫01Gt,svsds=−I0+αvt−Γα−γAΓα∑i=1m−2ξiI0+α−γvtt=ηi−I0+α−γvtt=1tα−1.
So, we can get(51)D0+αxt=−vt,and it follows that D0+αx∈C0,1. Combining Equation (36) with
(52)D0+βφpD0+αxt=D0+βφp−vt=D0+β−∫01Ht,szsds,we can see
(53)D0+βφpD0+αxt=zt.
This yieldsD0+βφpD0+αx∈C0,1.
Since the boundary value problems (14) and (16) have unique solutions by Lemma 5 and Lemma 8, we can find that the function x satisfies the fractional differential equation and the boundary conditions of the problem (1). This completes the proof.
Define the operatorT on P as follows:
(54)Txt≔∫01Gt,sφq∫01Hs,τfτ,xτdτds.
Then, the functionx∈P is a solution of the integral Equation (46) if and only if the operator T has a fixed point x in P.Lemma 12.
If the hypothesis (H1) holds, then
(55)TP⊂P.Proof.
The conclusion of this lemma easily follows from Lemma4, so we omit the details.
For convenience we define the functionFt,x on 0,1×0,+∞ as follows:
(56)Ft,x≔f0x,t=0∧∀x∈0,+∞,tσ01−tσ1ft,x,∀t,x∈0,1×0,+∞,f1x,t=1∧∀x∈0,+∞.
Obviously, we can see thatF∈C0,1×0,+∞. In fact, using the hypothesis (H1), for any x∈0,+∞, it holds that
(57)limt→0+Ft,x=limt→0+tσ01−tσ1ft,x=limt→0+tσ0ft,x=f0x=F0,x,limt→1−Ft,x=limt→1−tσ01−tσ1ft,x=limt→1−1−tσ1ft,x=f1x=F1,x,and F∈C0,1×0,+∞.Lemma 13.
If the hypothesis (H1) holds, then the operator T:P⟶P is completely continuous.Proof.
We first prove that the operatorT is continuous on P. For this, choose any x∈P and any sequence xnn=1∞⊂P convergent to x. Then, there exists L1>0 such that ∀n,xn≤L1. From the continuity of the function F, we can put
(58)M1≔max0≤t≤10≤x≤L1Ft,x.
From the definition of the functionF, we can see that for any 0<t<1,
(59)ft,xnt≤t−σ01−t−σ1M1.
Sincexn∈C0,1, we have
(60)limn→∞ft,xnt=ft,xt.
A simple calculation provides that(61)∫01t−σ01−t−σ1M1dt=M1⋅B1−σ0,1−σ1<+∞,and therefore, by using the Lebesgue dominated convergence theorem, we know
(62)limn→∞∫01ft,xntdt=∫01ft,xtdt.
Combining this withH∈C0,1×0,1, we can obtain that for any s∈0,1,
(63)limn→∞φq∫01Hs,τfτ,xnτdτ=φq∫01Hs,τfτ,xτdτ.
SinceG∈C0,1×0,1, we can get that for any t∈0,1,
(64)limn→∞∫01Gt,sφq∫01Hs,τfτ,xnτdτds=∫01Gt,sφq∫01Hs,τfτ,xτdτds.
This gives(65)∀t∈0,1,limn→∞Txnt=Txt.
Next, we show that for any bounded setΩ⊂P, TΩ is relatively compact. To do this, by the Arzela–Ascoli theorem, it is sufficient to prove that TΩ is uniformly bounded and equicontinuous. Denote Ω=u∈P:u≤L2.
From Lemma7 and Lemma 9, we see that for any x∈Ω and any t∈0,1,
(66)Txt=∫01Gt,sφq∫01Hs,τfτ,xτdτds≤∫01G∗s,sds⋅φq∫01H∗τ,τfτ,xτdτ.
PutM2≔max0≤t≤10≤x≤L2Ft,x. Since x≤L2, we have
(67)∫01H∗τ,τfτ,xτdτ=∫01H∗τ,ττ−σ01−τ−σ1Fτ,xτdτ≤M2BΓβ∫011−τβ−1⋅τ−σ01−τ−σ1dτ=M2BΓβ⋅B1−σ0,β−σ1
Therefore, we know(68)Tut≤1Aα−γΓαφqM2⋅B1−σ0,β−σ1BΓβ.
This means thatTΩ is uniformly bounded.
By using the inequality (20), we can get that for any t1,t2∈0,1,
(69)Txt2−Txt1=∫01Gt2,s−Gt1,sφq∫01Hs,τfτ,xτdτds≤∫01Gt2,s−Gt1,sφq∫01H∗τ,τfτ,xτdτds≤φqM2⋅B1−σ0,β−σ1BΓβ∫01Gt2,s−Gt1,sds.
Combining this inequality with the uniform continuity ofGt,s on 0,1×0,1, it holds that for any ε>0, there exists δ>0 such that
(70)∀t1,t2∈0,1t2−t1<δ,∀s∈0,1,Txt2−Txt1<ε.
This implies thatTΩ is equicontinuous. The proof is completed.PutM¯=α−γAΓαp−1BΓβ and list more hypotheses to be used in this paper.(H2).
There exists a nondecreasing functiong∈C0,+∞,0,+∞ such that tσ01−tσ1ft,x≤gx for any t,x∈0,1×0,+∞.(H3).
There existsR>0 such that Rp−1/gR≥B1−σ0,β−σ1/M¯.Theorem 14.
Assume that the hypotheses (H1)-(H3) hold. Then, the problem (1) has at least one solution in BR, where BR=u∈Pu≤R.Proof.
PuttingUR=u∈Pu<R, it is obvious that U¯R=BR. By using Lemma 13, we can know that the operator T:U¯R⟶P is completely continuous.
Assume that there exist a pointx∈∂UR and a number λ∈0,1, with x=λTx. By the hypothesis (H2), we see that for any t∈0,1,
(71)xt=λTxt=λ∫01Gt,sφq∫01Hs,τfτ,xτdτds≤λ∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1gxτdτds≤λgxq−1∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds.
Using Lemma7 and Lemma 9, we have
(72)∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds≤∫01G∗s,sφq∫01H∗τ,ττ−σ01−τ−σ1dτds=∫011AΓα1−sα−γ−1φq∫011BΓβ1−τβ−1⋅τ−σ01−τ−σ1dτds=∫011AΓα1−sα−γ−1B1−σ0,β−σ1BΓβq−1ds=B1−σ0,β−σ1M¯q−1.
Combining these two inequalities, we obtain(73)x≤λgxq−1B1−σ0,β−σ1M¯q−1.
Now takingp−1th power on both sides of the above inequality, since x=R, p−1q−1=1, and g is a nondecreasing function, we get
(74)Rp−1≤λgR⋅B1−σ0,β−σ1M¯<gR⋅B1−σ0,β−σ1M¯.
This is a contradiction to the hypothesis (H3). Therefore, by Lemma 3, the operator T has at least one fixed point in BR. The proof is completed.


### 3.2. Derivation of the Upper and Lower Solutions Method

We define the upper control functionKt,x and the lower control function kt,x on 0,1×0,R as follows:
(75)Kt,x≔max0≤ξ≤xft,ξ,kt,x≔minx≤ξ≤Rft,ξ.Then, we know that the functionsKt,x and kt,x are nondecreasing in x and for any t,x∈0,1×0,R,
(76)kt,x≤ft,x≤Kt,x.Definition 15.
The functionsx¯∈BR and x¯∈BR are said to be a upper solution and a lower solution of the integral Equation (46) in BR, respectively, if
(77)x¯t≥∫01Gt,sφq∫01Hs,τKτ,x¯τdτds,x¯t≤∫01Gt,sφq∫01Hs,τkτ,x¯τdτds.Lemma 16.
Assume that the hypotheses (H1)-(H3) hold and there exist a upper solution x¯ and a lower solution x¯ of the integral Equation (46) in BR. Then, the problem (1) has at least one solution in S, where S=u∈BR∀t∈0,1,x¯t≤ut≤x¯t.Proof.
It is obvious thatS is a nonempty, closed, bounded, and convex subset of the Banach space E and the operator T:BR⟶P is completely continuous. In a similar way to the proof of the Theorem 14, for any x∈BR, we can get
(78)∀t∈0,1,Txt≤gR⋅B1−σ0,β−σ1M¯q−1.
By the hypothesis (H3), we have
(79)∀t∈0,1,Txt≤R.
That is,TBR⊂BR.
Now we proveTS⊂S. In fact, since the functions Kt,x and kt,x are nondecreasing in x, by using the definition of upper and lower solution and the inequality (21), we obtain
(80)Txt≤∫01Gt,sφq∫01Hs,τKτ,xτdτds≤∫01Gt,sφq∫01Hs,τKτ,x¯τdτds≤x¯t,Txt≥∫01Gt,sφq∫01Hs,τkτ,xτdτds≥∫01Gt,sφq∫01Hs,τkτ,x¯τdτds≥x¯t.
Therefore, Lemma2 assures that the problem (1) has at least one solution in S.
SinceF∈C0,1×0,+∞, we can put
(81)ρ¯≔min0≤t≤10≤x≤RFt,x.
Also, putρ¯≔gR. Then, by the hypothesis (H2), we know that for any t,x∈0,1×0,R,
(82)ρ¯≤tσ01−tσ1ft,x≤ρ¯.
And by using the hypothesis (H3), we have
(83)ρ¯≤M¯B1−σ0,β−σ1Rp−1.
Denote as follows:(84)ω¯t≔ρ¯q−1∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds,ω¯t≔ρ¯q−1∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds.Theorem 17.
If the hypotheses (H1)-(H3) hold, then the problem (1) has at least one solution in S∗, where S∗≔u∈BR∀t∈0,1,ω¯t≤ut≤ω¯t.Proof.
From the inequality (23), similar to the proof of Theorem 14, we can see that for any t∈0,1,
(85)ω¯t≤ρ¯q−1∫01G∗s,sφq∫01H∗τ,ττ−σ01−τ−σ1dτds=ρ¯q−1B1−σ0,β−σ1M¯q−1≤R.
By using the estimation (22), we know that for any t,x∈0,1×0,R,
(86)tσ01−tσ1kt,x=minx≤ξ≤Rtσ01−tσ1ft,ξ≥ρ¯,tσ01−tσ1Kt,x=max0≤ξ≤xtσ01−tσ1ft,ξ≤ρ¯.
Since for anyt∈0,1, 0≤ω¯t≤ω¯t≤R, we have
(87)tσ01−tσ1kt,ω¯t≥ρ¯,tσ01−tσ1Kt,ω¯t≤ρ¯.
In other words, it holds that for anyt∈0,1,
(88)kt,ω¯t≥ρ¯t−σ01−t−σ1,Kt,ω¯t≤ρ¯t−σ01−t−σ1.
Using the inequalites (24), we obtain
(89)ω¯t≥∫01Gt,sφq∫01Hs,τ⋅Kτ,ω¯τdτds,ω¯t≤∫01Gt,sφq∫01Hs,τ⋅kτ,ω¯τdτds.
This implies that the functionsω¯t,ω¯t are upper and lower solutions of the integral Equation (46) in BR, respectively. Therefore, by Lemma 16, the problem (1) has at least one solution in S∗.Theorem 18.
Assume that the functionft,x is nondecreasing in x and the hypotheses (H1)-(H3) are satisfied. Then, the problem (1) has a maximal solution xmaxt and a minimal solution xmint in S∗ and the following estimation holds:
(90)ω¯t≤xmint≤xmaxt≤ω¯t.Proof.
Putx¯0≔ω¯,x¯0≔ω¯. Theorem 17 shows that the functions x¯0,x¯0 are upper and lower solutions of the integral Equation (46). Now, we construct the iterative sequences x¯n,x¯n as follows:
(91)x¯nt=∫01Gt,sφq∫01Hs,τfτ,x¯n−1τdτds,x¯nt=∫01Gt,sφq∫01Hs,τfτ,x¯n−1τdτds.
Since the functionft,x is nondecreasing in x, we know that if
(92)x¯n−1≤x¯n,x¯n≤x¯n,x¯n≤x¯n−1,then
(93)x¯n≤x¯n+1,x¯n+1≤x¯n+1,x¯n+1≤x¯n.
Considering that the functionsx¯0,x¯0 are upper and lower solutions of the integral Equation (46), we also have
(94)x¯0≤x¯1,x¯1≤x¯1,x¯1≤x¯0.
So, it is obvious that for anyn∈N,
(95)ω¯=x¯0≤x¯n≤x¯n+1≤x¯n+1≤x¯n≤x¯0=ω¯.
This implies that for the sequencesx¯n,x¯n, passing to the limit as n⟶∞, the limits exist. And denoting
(96)xmax=limn→∞x¯n,xmin=limn→∞x¯n,we can get
(97)ω¯≤xmin≤xmax≤ω¯.
Taking the limitn⟶∞ on both sides of the inequality (25), by the Lebesgue dominated convergence theorem, we obtain
(98)xmint=∫01Gt,sφq∫01Hs,τfτ,xminτdτds,xmaxt=∫01Gt,sφq∫01Hs,τfτ,xmaxτdτds.
Therefore, it holds that the functionsxmin, xmax are solutions to the problem (1).
On the other hand, for any solution to the problem (1) x∈S∗, it can be easily seen that
(99)x¯0≤x≤x¯0,xt=∫01Gt,sφq∫01Hs,τfτ,xτdτds,x¯1t=∫01Gt,sφq∫01Hs,τfτ,x¯0τdτds,x¯1t=∫01Gt,sφq∫01Hs,τfτ,x¯0τdτds.
Since the functionft,x is nondecreasing in x, it is obvious that
(100)x¯1≤x≤x¯1.
By induction, we know that for anyn∈N,
(101)x¯n≤x≤x¯n.
Taking the limitn⟶∞ on both sides of the inequality (26), we can see
(102)xmin≤x≤xmax.
This completes the proof.Remark 19.
If maximal and minimal solutions of the problem (1) exist in S∗ and they are equal to each other, then (1) has a unique solution in S∗.Remark 20.
Ifρ¯>0, then the problem (1) has a positive solution in S∗.


### 3.3. The Uniqueness Results for Problem (1)

The Gauss hypergeometric functionF21a,b;c;s is defined in the unit disk as the sum of the hypergeometric series
(103)F21a,b;c;s=∑k=0∞akbkckskk!,where z<1,a,b∈C,c∈C\Z0− and ak is the Pochhammer symbol. And it is known that this function has the following properties (see [35]):
(i)
F21a,b;c;s=F21b,a;c;s(ii)
IfRec−a−b>0, then F21a,b;c;1=ΓcΓc−a−b/Γc−aΓc−b(iii)
d/dsnsa+n−1F21a,b;c;s=ansa−1F21a+n,b;c;s for any n∈N(iv)
If0<Reb<Rec and s<1, then(104)2F1a,b;c;s=ΓcΓbΓc−b∫01tb−11−tc−b−11−st−adt,s<1.Denote the functionχs as follows:
(105)χs≔F211−σ0,σ1;β+1−σ0;s,s∈0,1.Lemma 21.
Put(106)K1=Γ1−σ0BΓβ+1−σ0∑i=1m−2ζiηiβ−1χ1−ηi1−σ0χηi,K2=Γ1−σ0Γβ+1−σ0χ1+1B∑i=1m−2ζiηiβ−1χ1−ηi1−σ0χηi.
Then, the following estimation holds(107)K1sβ−1≤∫01Hs,ττ−σ01−τ−σ1dτ≤K2sβ−1.Proof.
To estimate the function∫01Hs,ττ−σ01−τ−σ1dτ by using (15), we should calculate the fractional integral of order β of the function s−σ01−s−σ1. By the properties (i), (iv) of the Gauss hypergeometric function, we have
(108)1Γβ∫0ss−τβ−1τ−σ01−τ−σ1dτ=1Γβsβ−σ0∫011−uβ−1u−σ01−su−σ1du=1Γβsβ−σ0⋅Γ1−σ0ΓβΓβ+1−σ0F21σ1,1−σ0;β+1−σ0;s=Γ1−σ0Γβ+1−σ0sβ−σ0χs.
So, we obtain(109)∫01Hs,ττ−σ01−τ−σ1dτ=Γ1−σ0Γβ+1−σ0−s1−σ0χs−1B∑i=1m−2ζiηiβ−σ0χηi+1Bχ1⋅sβ−1=Γ1−σ0Γβ+1−σ0χ1−s1−σ0χs+1B∑i=1m−2ζiηiβ−1χ1−ηi1−σ0χηi⋅sβ−1.
By puttingds=s1−σ0χs, we can see
(110)∫01Hs,ττ−σ01−τ−σ1dτ=Γ1−σ0Γβ+1−σ0d1−ds+1B∑i=1m−2ζiηiβ−1d1−dηi⋅sβ−1.
The functionds is an increasing function on 0,1. In fact, it is obvious that d0=0 and ds≥0 for any s∈0,1, and by using the property (iii) of the Gauss hypergeometric function, we can get that for any s>0,
(111)d′s=1−σ0s−σ02F12−σ0;σ1;β+1−σ0;s>0.
The proof is completed.From the property (ii) of the Gauss hypergeometric function, we can calculateχ1 as
(112)χ1=Γβ+1−σ0Γβ−σ1ΓβΓβ+1−σ0−σ1.(H4).
There existsL>0 such that for any t∈0,1 and any x1,x2∈0,R,
(113)Ft,x1−Ft,x2≤Lx1−x2.Lemma 22.
Assume that the hypotheses (H1)-(H4) are satisfied. Then, the followings hold for any x,y∈BR:
(i)
If1<p<2, then
(114)Tx−Ty≤q−1ρ¯q−2K2q−1LAα−γΓα⋅x−y(ii)
Ifp>2, then(115)Tx−Ty≤q−1K2LAα−γΓαρ¯2−qK12−q⋅x−yProof.
From the definition of the operatorT, we can see
(116)Txt−Tyt≤∫01Gt,sφq∫01Hs,τfτ,xτdτ−φq∫01Hs,τfτ,yτdτds.
Sincex,y≤R, by using Lemma 21 and inequality (22), we have
(117)ρ¯K1sβ−1≤∫01Hs,τfτ,xτdτ≤ρ¯K2sβ−1,ρ¯K1sβ−1≤∫01Hs,τfτ,yτdτ≤ρ¯K2sβ−1.(i)
If1<p<2, then q>2. So, by the inequality (17) and the hypothesis (H4), we obtain that for any t∈0,1(118)Txt−Tyt≤q−1ρ¯K2q−2∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1⋅Fτ,xτ−Fτ,yτdτds≤q−1ρ¯K2q−2Lx−y∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1dτds≤q−1ρ¯q−2K2q−1Lx−y∫01Gt,ssβ−1q−1ds≤q−1ρ¯q−2K2q−1Lx−y∫01G∗s,sds≤q−1ρ¯q−2K2q−1LAα−γΓα⋅x−y(ii)
Ifp>2, then 1<q<2. Employing the inequality (18) and the hypothesis (H4), we have that for any t∈0,1(119)Txt−Tyt≤q−1ρ¯K1q−2∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1⋅Fτ,xτ−Fτ,yτdτds≤q−1ρ¯K1q−2Lx−y∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1dτds≤q−1ρ¯q−2K1q−2K2Lx−y∫01Gt,ssβ−1q−1ds≤q−1ρ¯q−2K1q−2K2Lx−y∫01G∗s,sds≤q−1K2LAα−γΓαρ¯2−qK12−q⋅x−yTheorem 23.
Assume that the hypotheses (H1)-(H4) are satisfied. Then, the problem (1) has a unique solution in S∗ if the following conditions hold
(i)
If1<p<2, then q−1ρ¯q−2K2q−1L/Aα−γΓα<1(ii)
Ifp>2, then q−1K2L/Aα−γΓαρ¯2−qK12−q<1Proof.
Letx≠y be two different solutions to the problem (1) in S∗. Using Lemma 22 we know that if 1<p<2, then
(120)x−y=Tx−Ty≤q−1ρ¯q−2K2q−1LAα−γΓα⋅x−y<x−y,and if p>2, then
(121)x−y=Tx−Ty≤q−1K2LAα−γΓαρ¯2−qK12−q⋅x−y<x−y.
This is a contradiction. The proof is completed.


## 3.1. The Existence Results for Problem (1)

Definition 10.
A functionx∈uu∈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).
The following hypothesis concerned with the functionft,x, which permits singularity with respect to time variable, will be used in this article.(H1).
There existσ0,σ1∈0,1 such that for any x∈0,+∞,
(44)limt→0+tσ0ft,x=f0x>0,limt→1−1−tσ1ft,x=f1x>0.
LetE be the Banach space C0,1 equipped with the norm u=max0≤t≤1ut and put P≔u∈Eut≥0,t∈0,1.Lemma 11.
Assume that the hypothesis (H1) holds. Then, the function x∈P is a solution of the problem (1) if and only if x is a solution of the integral equation
(45)xt=∫01Gt,sφq∫01Hs,τfτ,xτdτds,in P, where q is a real number such that 1/p+1/q=1.Proof.
Suppose thatx∈P is a solution of the problem (1). Putting wt≔D0+αxt, by using Lemma 8, we have
(46)xt=−∫01Gt,swsds.
Also puttingvt≔φpwt, by Lemma 5, we obtain
(47)vt=−∫01Ht,sfs,xsds.
Sinceφp−1=φq, Equations (47) and(48) yields
(48)xt=∫01Gt,sφq∫01Hs,τfτ,xτdτds.
It is obvious thatx∈P and x is a solution of the integral Equation (46).
Conversely, suppose thatx∈P is a solution of the integral Equation (46). Put as follows:
(49)zt≔ft,xt,vt≔φq∫01Ht,szsds.
Then, Equation (36) implies that v∈C0,1. From the definition of Gt,s, we also have
(50)xt=∫01Gt,svsds=−I0+αvt−Γα−γAΓα∑i=1m−2ξiI0+α−γvtt=ηi−I0+α−γvtt=1tα−1.
So, we can get(51)D0+αxt=−vt,and it follows that D0+αx∈C0,1. Combining Equation (36) with
(52)D0+βφpD0+αxt=D0+βφp−vt=D0+β−∫01Ht,szsds,we can see
(53)D0+βφpD0+αxt=zt.
This yieldsD0+βφpD0+αx∈C0,1.
Since the boundary value problems (14) and (16) have unique solutions by Lemma 5 and Lemma 8, we can find that the function x satisfies the fractional differential equation and the boundary conditions of the problem (1). This completes the proof.
Define the operatorT on P as follows:
(54)Txt≔∫01Gt,sφq∫01Hs,τfτ,xτdτds.
Then, the functionx∈P is a solution of the integral Equation (46) if and only if the operator T has a fixed point x in P.Lemma 12.
If the hypothesis (H1) holds, then
(55)TP⊂P.Proof.
The conclusion of this lemma easily follows from Lemma4, so we omit the details.
For convenience we define the functionFt,x on 0,1×0,+∞ as follows:
(56)Ft,x≔f0x,t=0∧∀x∈0,+∞,tσ01−tσ1ft,x,∀t,x∈0,1×0,+∞,f1x,t=1∧∀x∈0,+∞.
Obviously, we can see thatF∈C0,1×0,+∞. In fact, using the hypothesis (H1), for any x∈0,+∞, it holds that
(57)limt→0+Ft,x=limt→0+tσ01−tσ1ft,x=limt→0+tσ0ft,x=f0x=F0,x,limt→1−Ft,x=limt→1−tσ01−tσ1ft,x=limt→1−1−tσ1ft,x=f1x=F1,x,and F∈C0,1×0,+∞.Lemma 13.
If the hypothesis (H1) holds, then the operator T:P⟶P is completely continuous.Proof.
We first prove that the operatorT is continuous on P. For this, choose any x∈P and any sequence xnn=1∞⊂P convergent to x. Then, there exists L1>0 such that ∀n,xn≤L1. From the continuity of the function F, we can put
(58)M1≔max0≤t≤10≤x≤L1Ft,x.
From the definition of the functionF, we can see that for any 0<t<1,
(59)ft,xnt≤t−σ01−t−σ1M1.
Sincexn∈C0,1, we have
(60)limn→∞ft,xnt=ft,xt.
A simple calculation provides that(61)∫01t−σ01−t−σ1M1dt=M1⋅B1−σ0,1−σ1<+∞,and therefore, by using the Lebesgue dominated convergence theorem, we know
(62)limn→∞∫01ft,xntdt=∫01ft,xtdt.
Combining this withH∈C0,1×0,1, we can obtain that for any s∈0,1,
(63)limn→∞φq∫01Hs,τfτ,xnτdτ=φq∫01Hs,τfτ,xτdτ.
SinceG∈C0,1×0,1, we can get that for any t∈0,1,
(64)limn→∞∫01Gt,sφq∫01Hs,τfτ,xnτdτds=∫01Gt,sφq∫01Hs,τfτ,xτdτds.
This gives(65)∀t∈0,1,limn→∞Txnt=Txt.
Next, we show that for any bounded setΩ⊂P, TΩ is relatively compact. To do this, by the Arzela–Ascoli theorem, it is sufficient to prove that TΩ is uniformly bounded and equicontinuous. Denote Ω=u∈P:u≤L2.
From Lemma7 and Lemma 9, we see that for any x∈Ω and any t∈0,1,
(66)Txt=∫01Gt,sφq∫01Hs,τfτ,xτdτds≤∫01G∗s,sds⋅φq∫01H∗τ,τfτ,xτdτ.
PutM2≔max0≤t≤10≤x≤L2Ft,x. Since x≤L2, we have
(67)∫01H∗τ,τfτ,xτdτ=∫01H∗τ,ττ−σ01−τ−σ1Fτ,xτdτ≤M2BΓβ∫011−τβ−1⋅τ−σ01−τ−σ1dτ=M2BΓβ⋅B1−σ0,β−σ1
Therefore, we know(68)Tut≤1Aα−γΓαφqM2⋅B1−σ0,β−σ1BΓβ.
This means thatTΩ is uniformly bounded.
By using the inequality (20), we can get that for any t1,t2∈0,1,
(69)Txt2−Txt1=∫01Gt2,s−Gt1,sφq∫01Hs,τfτ,xτdτds≤∫01Gt2,s−Gt1,sφq∫01H∗τ,τfτ,xτdτds≤φqM2⋅B1−σ0,β−σ1BΓβ∫01Gt2,s−Gt1,sds.
Combining this inequality with the uniform continuity ofGt,s on 0,1×0,1, it holds that for any ε>0, there exists δ>0 such that
(70)∀t1,t2∈0,1t2−t1<δ,∀s∈0,1,Txt2−Txt1<ε.
This implies thatTΩ is equicontinuous. The proof is completed.PutM¯=α−γAΓαp−1BΓβ and list more hypotheses to be used in this paper.(H2).
There exists a nondecreasing functiong∈C0,+∞,0,+∞ such that tσ01−tσ1ft,x≤gx for any t,x∈0,1×0,+∞.(H3).
There existsR>0 such that Rp−1/gR≥B1−σ0,β−σ1/M¯.Theorem 14.
Assume that the hypotheses (H1)-(H3) hold. Then, the problem (1) has at least one solution in BR, where BR=u∈Pu≤R.Proof.
PuttingUR=u∈Pu<R, it is obvious that U¯R=BR. By using Lemma 13, we can know that the operator T:U¯R⟶P is completely continuous.
Assume that there exist a pointx∈∂UR and a number λ∈0,1, with x=λTx. By the hypothesis (H2), we see that for any t∈0,1,
(71)xt=λTxt=λ∫01Gt,sφq∫01Hs,τfτ,xτdτds≤λ∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1gxτdτds≤λgxq−1∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds.
Using Lemma7 and Lemma 9, we have
(72)∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds≤∫01G∗s,sφq∫01H∗τ,ττ−σ01−τ−σ1dτds=∫011AΓα1−sα−γ−1φq∫011BΓβ1−τβ−1⋅τ−σ01−τ−σ1dτds=∫011AΓα1−sα−γ−1B1−σ0,β−σ1BΓβq−1ds=B1−σ0,β−σ1M¯q−1.
Combining these two inequalities, we obtain(73)x≤λgxq−1B1−σ0,β−σ1M¯q−1.
Now takingp−1th power on both sides of the above inequality, since x=R, p−1q−1=1, and g is a nondecreasing function, we get
(74)Rp−1≤λgR⋅B1−σ0,β−σ1M¯<gR⋅B1−σ0,β−σ1M¯.
This is a contradiction to the hypothesis (H3). Therefore, by Lemma 3, the operator T has at least one fixed point in BR. The proof is completed.


## 3.2. Derivation of the Upper and Lower Solutions Method

We define the upper control functionKt,x and the lower control function kt,x on 0,1×0,R as follows:
(75)Kt,x≔max0≤ξ≤xft,ξ,kt,x≔minx≤ξ≤Rft,ξ.Then, we know that the functionsKt,x and kt,x are nondecreasing in x and for any t,x∈0,1×0,R,
(76)kt,x≤ft,x≤Kt,x.Definition 15.
The functionsx¯∈BR and x¯∈BR are said to be a upper solution and a lower solution of the integral Equation (46) in BR, respectively, if
(77)x¯t≥∫01Gt,sφq∫01Hs,τKτ,x¯τdτds,x¯t≤∫01Gt,sφq∫01Hs,τkτ,x¯τdτds.Lemma 16.
Assume that the hypotheses (H1)-(H3) hold and there exist a upper solution x¯ and a lower solution x¯ of the integral Equation (46) in BR. Then, the problem (1) has at least one solution in S, where S=u∈BR∀t∈0,1,x¯t≤ut≤x¯t.Proof.
It is obvious thatS is a nonempty, closed, bounded, and convex subset of the Banach space E and the operator T:BR⟶P is completely continuous. In a similar way to the proof of the Theorem 14, for any x∈BR, we can get
(78)∀t∈0,1,Txt≤gR⋅B1−σ0,β−σ1M¯q−1.
By the hypothesis (H3), we have
(79)∀t∈0,1,Txt≤R.
That is,TBR⊂BR.
Now we proveTS⊂S. In fact, since the functions Kt,x and kt,x are nondecreasing in x, by using the definition of upper and lower solution and the inequality (21), we obtain
(80)Txt≤∫01Gt,sφq∫01Hs,τKτ,xτdτds≤∫01Gt,sφq∫01Hs,τKτ,x¯τdτds≤x¯t,Txt≥∫01Gt,sφq∫01Hs,τkτ,xτdτds≥∫01Gt,sφq∫01Hs,τkτ,x¯τdτds≥x¯t.
Therefore, Lemma2 assures that the problem (1) has at least one solution in S.
SinceF∈C0,1×0,+∞, we can put
(81)ρ¯≔min0≤t≤10≤x≤RFt,x.
Also, putρ¯≔gR. Then, by the hypothesis (H2), we know that for any t,x∈0,1×0,R,
(82)ρ¯≤tσ01−tσ1ft,x≤ρ¯.
And by using the hypothesis (H3), we have
(83)ρ¯≤M¯B1−σ0,β−σ1Rp−1.
Denote as follows:(84)ω¯t≔ρ¯q−1∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds,ω¯t≔ρ¯q−1∫01Gt,sφq∫01Hs,ττ−σ01−τ−σ1dτds.Theorem 17.
If the hypotheses (H1)-(H3) hold, then the problem (1) has at least one solution in S∗, where S∗≔u∈BR∀t∈0,1,ω¯t≤ut≤ω¯t.Proof.
From the inequality (23), similar to the proof of Theorem 14, we can see that for any t∈0,1,
(85)ω¯t≤ρ¯q−1∫01G∗s,sφq∫01H∗τ,ττ−σ01−τ−σ1dτds=ρ¯q−1B1−σ0,β−σ1M¯q−1≤R.
By using the estimation (22), we know that for any t,x∈0,1×0,R,
(86)tσ01−tσ1kt,x=minx≤ξ≤Rtσ01−tσ1ft,ξ≥ρ¯,tσ01−tσ1Kt,x=max0≤ξ≤xtσ01−tσ1ft,ξ≤ρ¯.
Since for anyt∈0,1, 0≤ω¯t≤ω¯t≤R, we have
(87)tσ01−tσ1kt,ω¯t≥ρ¯,tσ01−tσ1Kt,ω¯t≤ρ¯.
In other words, it holds that for anyt∈0,1,
(88)kt,ω¯t≥ρ¯t−σ01−t−σ1,Kt,ω¯t≤ρ¯t−σ01−t−σ1.
Using the inequalites (24), we obtain
(89)ω¯t≥∫01Gt,sφq∫01Hs,τ⋅Kτ,ω¯τdτds,ω¯t≤∫01Gt,sφq∫01Hs,τ⋅kτ,ω¯τdτds.
This implies that the functionsω¯t,ω¯t are upper and lower solutions of the integral Equation (46) in BR, respectively. Therefore, by Lemma 16, the problem (1) has at least one solution in S∗.Theorem 18.
Assume that the functionft,x is nondecreasing in x and the hypotheses (H1)-(H3) are satisfied. Then, the problem (1) has a maximal solution xmaxt and a minimal solution xmint in S∗ and the following estimation holds:
(90)ω¯t≤xmint≤xmaxt≤ω¯t.Proof.
Putx¯0≔ω¯,x¯0≔ω¯. Theorem 17 shows that the functions x¯0,x¯0 are upper and lower solutions of the integral Equation (46). Now, we construct the iterative sequences x¯n,x¯n as follows:
(91)x¯nt=∫01Gt,sφq∫01Hs,τfτ,x¯n−1τdτds,x¯nt=∫01Gt,sφq∫01Hs,τfτ,x¯n−1τdτds.
Since the functionft,x is nondecreasing in x, we know that if
(92)x¯n−1≤x¯n,x¯n≤x¯n,x¯n≤x¯n−1,then
(93)x¯n≤x¯n+1,x¯n+1≤x¯n+1,x¯n+1≤x¯n.
Considering that the functionsx¯0,x¯0 are upper and lower solutions of the integral Equation (46), we also have
(94)x¯0≤x¯1,x¯1≤x¯1,x¯1≤x¯0.
So, it is obvious that for anyn∈N,
(95)ω¯=x¯0≤x¯n≤x¯n+1≤x¯n+1≤x¯n≤x¯0=ω¯.
This implies that for the sequencesx¯n,x¯n, passing to the limit as n⟶∞, the limits exist. And denoting
(96)xmax=limn→∞x¯n,xmin=limn→∞x¯n,we can get
(97)ω¯≤xmin≤xmax≤ω¯.
Taking the limitn⟶∞ on both sides of the inequality (25), by the Lebesgue dominated convergence theorem, we obtain
(98)xmint=∫01Gt,sφq∫01Hs,τfτ,xminτdτds,xmaxt=∫01Gt,sφq∫01Hs,τfτ,xmaxτdτds.
Therefore, it holds that the functionsxmin, xmax are solutions to the problem (1).
On the other hand, for any solution to the problem (1) x∈S∗, it can be easily seen that
(99)x¯0≤x≤x¯0,xt=∫01Gt,sφq∫01Hs,τfτ,xτdτds,x¯1t=∫01Gt,sφq∫01Hs,τfτ,x¯0τdτds,x¯1t=∫01Gt,sφq∫01Hs,τfτ,x¯0τdτds.
Since the functionft,x is nondecreasing in x, it is obvious that
(100)x¯1≤x≤x¯1.
By induction, we know that for anyn∈N,
(101)x¯n≤x≤x¯n.
Taking the limitn⟶∞ on both sides of the inequality (26), we can see
(102)xmin≤x≤xmax.
This completes the proof.Remark 19.
If maximal and minimal solutions of the problem (1) exist in S∗ and they are equal to each other, then (1) has a unique solution in S∗.Remark 20.
Ifρ¯>0, then the problem (1) has a positive solution in S∗.


## 3.3. The Uniqueness Results for Problem (1)

The Gauss hypergeometric functionF21a,b;c;s is defined in the unit disk as the sum of the hypergeometric series
(103)F21a,b;c;s=∑k=0∞akbkckskk!,where z<1,a,b∈C,c∈C\Z0− and ak is the Pochhammer symbol. And it is known that this function has the following properties (see [35]):
(i)
F21a,b;c;s=F21b,a;c;s(ii)
IfRec−a−b>0, then F21a,b;c;1=ΓcΓc−a−b/Γc−aΓc−b(iii)
d/dsnsa+n−1F21a,b;c;s=ansa−1F21a+n,b;c;s for any n∈N(iv)
If0<Reb<Rec and s<1, then(104)2F1a,b;c;s=ΓcΓbΓc−b∫01tb−11−tc−b−11−st−adt,s<1.Denote the functionχs as follows:
(105)χs≔F211−σ0,σ1;β+1−σ0;s,s∈0,1.Lemma 21.
Put(106)K1=Γ1−σ0BΓβ+1−σ0∑i=1m−2ζiηiβ−1χ1−ηi1−σ0χηi,K2=Γ1−σ0Γβ+1−σ0χ1+1B∑i=1m−2ζiηiβ−1χ1−ηi1−σ0χηi.
Then, the following estimation holds(107)K1sβ−1≤∫01Hs,ττ−σ01−τ−σ1dτ≤K2sβ−1.Proof.
To estimate the function∫01Hs,ττ−σ01−τ−σ1dτ by using (15), we should calculate the fractional integral of order β of the function s−σ01−s−σ1. By the properties (i), (iv) of the Gauss hypergeometric function, we have
(108)1Γβ∫0ss−τβ−1τ−σ01−τ−σ1dτ=1Γβsβ−σ0∫011−uβ−1u−σ01−su−σ1du=1Γβsβ−σ0⋅Γ1−σ0ΓβΓβ+1−σ0F21σ1,1−σ0;β+1−σ0;s=Γ1−σ0Γβ+1−σ0sβ−σ0χs.
So, we obtain(109)∫01Hs,ττ−σ01−τ−σ1dτ=Γ1−σ0Γβ+1−σ0−s1−σ0χs−1B∑i=1m−2ζiηiβ−σ0χηi+1Bχ1⋅sβ−1=Γ1−σ0Γβ+1−σ0χ1−s1−σ0χs+1B∑i=1m−2ζiηiβ−1χ1−ηi1−σ0χηi⋅sβ−1.
By puttingds=s1−σ0χs, we can see
(110)∫01Hs,ττ−σ01−τ−σ1dτ=Γ1−σ0Γβ+1−σ0d1−ds+1B∑i=1m−2ζiηiβ−1d1−dηi⋅sβ−1.
The functionds is an increasing function on 0,1. In fact, it is obvious that d0=0 and ds≥0 for any s∈0,1, and by using the property (iii) of the Gauss hypergeometric function, we can get that for any s>0,
(111)d′s=1−σ0s−σ02F12−σ0;σ1;β+1−σ0;s>0.
The proof is completed.From the property (ii) of the Gauss hypergeometric function, we can calculateχ1 as
(112)χ1=Γβ+1−σ0Γβ−σ1ΓβΓβ+1−σ0−σ1.(H4).
There existsL>0 such that for any t∈0,1 and any x1,x2∈0,R,
(113)Ft,x1−Ft,x2≤Lx1−x2.Lemma 22.
Assume that the hypotheses (H1)-(H4) are satisfied. Then, the followings hold for any x,y∈BR:
(i)
If1<p<2, then
(114)Tx−Ty≤q−1ρ¯q−2K2q−1LAα−γΓα⋅x−y(ii)
Ifp>2, then(115)Tx−Ty≤q−1K2LAα−γΓαρ¯2−qK12−q⋅x−yProof.
From the definition of the operatorT, we can see
(116)Txt−Tyt≤∫01Gt,sφq∫01Hs,τfτ,xτdτ−φq∫01Hs,τfτ,yτdτds.
Sincex,y≤R, by using Lemma 21 and inequality (22), we have
(117)ρ¯K1sβ−1≤∫01Hs,τfτ,xτdτ≤ρ¯K2sβ−1,ρ¯K1sβ−1≤∫01Hs,τfτ,yτdτ≤ρ¯K2sβ−1.(i)
If1<p<2, then q>2. So, by the inequality (17) and the hypothesis (H4), we obtain that for any t∈0,1(118)Txt−Tyt≤q−1ρ¯K2q−2∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1⋅Fτ,xτ−Fτ,yτdτds≤q−1ρ¯K2q−2Lx−y∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1dτds≤q−1ρ¯q−2K2q−1Lx−y∫01Gt,ssβ−1q−1ds≤q−1ρ¯q−2K2q−1Lx−y∫01G∗s,sds≤q−1ρ¯q−2K2q−1LAα−γΓα⋅x−y(ii)
Ifp>2, then 1<q<2. Employing the inequality (18) and the hypothesis (H4), we have that for any t∈0,1(119)Txt−Tyt≤q−1ρ¯K1q−2∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1⋅Fτ,xτ−Fτ,yτdτds≤q−1ρ¯K1q−2Lx−y∫01Gt,ssβ−1q−2∫01Hs,ττ−σ01−τ−σ1dτds≤q−1ρ¯q−2K1q−2K2Lx−y∫01Gt,ssβ−1q−1ds≤q−1ρ¯q−2K1q−2K2Lx−y∫01G∗s,sds≤q−1K2LAα−γΓαρ¯2−qK12−q⋅x−yTheorem 23.
Assume that the hypotheses (H1)-(H4) are satisfied. Then, the problem (1) has a unique solution in S∗ if the following conditions hold
(i)
If1<p<2, then q−1ρ¯q−2K2q−1L/Aα−γΓα<1(ii)
Ifp>2, then q−1K2L/Aα−γΓαρ¯2−qK12−q<1Proof.
Letx≠y be two different solutions to the problem (1) in S∗. Using Lemma 22 we know that if 1<p<2, then
(120)x−y=Tx−Ty≤q−1ρ¯q−2K2q−1LAα−γΓα⋅x−y<x−y,and if p>2, then
(121)x−y=Tx−Ty≤q−1K2LAα−γΓαρ¯2−qK12−q⋅x−y<x−y.
This is a contradiction. The proof is completed.


## 4. Examples

The following examples are given to illustrate our main results:Example 1.
Consider the singular four-point boundary value problem(122)D0+1.5φ3D0+1.75xt=ln5+xt2t1−t,0<t<1,x0=0,D0+1.75x0=0,D0+0.4x1=0.1D0+0.4x0.3+0.1D0+0.4x0.7,φ3D0+1.75x1=0.1φ3D0+1.75x0.3+0.1φ3D0+1.75x0.7.
The problem (27) can be regarded as the singular boundary value problem (1) where ft,x=ln5+x/2t1−t,p=3,α=1.75,β=1.5,γ=0.4,η1=0.3,η2=0.7,ξ1=0.1,ξ2=0.1,ζ1=0.1,andζ2=0.1. Then, we have
(123)q=1.5<2,A=0.846122⋯,B=0.861562⋯,M¯=0.841503⋯.
We check that the hypotheses (H1)-(H3) are satisfied. Putting σ0=σ1=1/2, it can be easily seen that
(124)limt→0+t⋅ln5+x2t1−t=ln5+x2>0,limt→1−1−t⋅ln5+x2t1−t=ln5+x2>0,which means that the hypothesis (H1) holds. Put gx=2+x/2. Then, the function gx is nondecreasing in x. Since ln5+x/2≤gx, the hypothesis (H2) is satisfied. The hypothesis (H3) can be checked, while for R=3,
(125)Rp−1gR=187=2.57143⋯,B1−σ0,β−σ1M¯=2.3767⋯.
The constantsρ¯,ρ¯ are calculated as
(126)ρ¯=ln52=0.804719⋯,ρ¯=2+32=3.5.
Therefore, from Remark20, we can know that the boundary value problem (27) has a positive solution.
On the other hand, since the functionft,x=ln5+x/2t1−t is nondecreasing in x, we can construct iterative sequences x¯nn=0∞,x¯nn=0∞ initialized at
(127)x¯0t=ρ¯∫01Gt,sφ1.5∫01Hs,ττ1−τdτds,x¯0t=ρ¯∫01Gt,sφ1.5∫01Hs,ττ1−τdτds,and defined by
(128)x¯n+1t=∫01Gt,sφ1.5∫01Hs,τfτ,x¯nτdτds,x¯n+1t=∫01Gt,sφ1.5∫01Hs,τfτ,x¯nτdτds,n=0,1,2,⋯.
Then, the sequencesx¯nn=0∞,x¯nn=0∞ are convergent to the maximal and minimal solutions of the problem (27), respectively.
PuttingL=1/10 and Ft,x=ln5+x/2, it holds that for any t∈0,1 and any x1,x2≥0,
(129)Ft,x1−Ft,x2<Lx1−x2.
This implies that the hypothesis (H4) holds. By simple calculation, we can get
(130)K1=0.136124⋯,K2=2.39288⋯.
Checking the condition (ii) of Theorem23, we obtain
(131)q−1K2LAα−γΓαρ¯2−qK12−q=0.344342⋯<1.
Finally, we can find that the boundary value problem (27) has a unique solution and the iterative sequences (27) are convergent to the unique solution of the given problem.Example 2.
Consider the following singular four-point boundary value problem:(132)D0+1.5φ1.5D0+1.75xt=ln2+xt10t1−t,0<t<1,x0=0,D0+1.75x0=0,D0+0.4x1=0.1D0+0.4x0.3+0.1D0+0.4x0.7,φ1.5D0+1.75x1=0.1φ1.5D0+1.75x0.3+0.1φ1.5D0+1.75x0.7.
The problem (30) can be regarded as the boundary value problem (1) where ft,x=ln2+x/10t1−t,p=1.5,α=1.75,β=1.5,γ=0.4,η1=0.3,η2=0.7,ξ1=0.1,ξ2=0.1,ζ1=0.1,andζ2=0.1. Then, we can get
(133)q=3>2,A=0.846122⋯,B=0.861562⋯,M=0.782325⋯.
Now, we see whether the hypotheses (H1)-(H3) are satisfied. Putting σ0=σ1=1/2, we have
(134)limt→0+t⋅ln2+x10t1−t=ln2+x10>0,limt→1−1−t⋅ln2+x10t1−t=ln2+x10>0,which means that the hypothesis (H1) holds. Put gx=1+x/10. Then, the function gx is nondecreasing in x. Since ln2+x/10≤gx, the hypothesis (H2) is satisfied. The hypothesis (H3) can be checked, while for R=1,
(135)Rp−1gR=5,B1−σ0,β−σ1M¯=2.55648⋯.
The constantsρ¯,ρ¯ are obtained as
(136)ρ¯=ln210=0.0693147⋯,ρ¯=1+110=0.2.
Therefore, Remark20 provides that the boundary value problem (30) has a positive solution.
Since the functionft,x=ln2+x/10t1−t/ is nondecreasing in x, we can construct iterative sequences x¯nn=0∞,x¯nn=0∞ initialized at
(137)x¯0t=ρ¯∫01Gt,sφ3∫01Hs,ττ1−τdτds,x¯0t=ρ¯∫01Gt,sφ3∫01Hs,ττ1−τdτds,and defined by
(138)x¯n+1t=∫01Gt,sφ3∫01Hs,τfτ,x¯nτdτds,x¯n+1t=∫01Gt,sφ3∫01Hs,τfτ,x¯nτdτds,n=0,1,2,⋯.
Then, the sequencesx¯nn=0∞,x¯nn=0∞ are convergent to the maximal and minimal solutions of the problem (30), respectively.
PuttingL=1/20 and Ft,x=ln2+x/10, it holds that for any t∈0,1 and any x1,x2≥0,
(139)Ft,x1−Ft,x2<Lx1−x2
This assures that the hypothesis (H4) holds. A simple calculation gives
(140)K2=2.39288⋯.
Checking the condition (i) of Theorem23, we obtain
(141)q−1ρ¯q−2K2q−1LAα−γΓα=0.109084⋯<1.
Finally, we can find that the boundary value problem (30) has a unique solution and the iterative sequences (32) are convergent to the unique solution of the given problem.


## 5. Conclusion

In this paper, we have studied the existence and uniqueness results and upper and lower solution methods for multipoint boundary value problems for singular fractional differential equations with thep-Laplacian operator. For this purpose, we used some fixed-point theorems such as the Leray-Schauder nonlinear alternative and the Schauder fixed-point theorem to prove the existence of positive solutions to the problem (1). And by employing the upper and lower solutions technique, we derived a new approach to obtain the maximal and minimal solutions to the given problem. After that, some useful properties of the Gauss hypergeometric function were applied to the establishment of our uniqueness results. For the application of this work, two expressive examples were illustrated.



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