# Existence Results for a Class ofp-Laplacian Fractional Differential Equations with Integral Boundary Conditions

Authors: SunAe Pak; KumSong Jong; KyuNam O; HuiChol Choi
Journal: Abstract and Applied Analysis
 (2020)

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## Abstract

In this paper, we investigate the existence and uniqueness of solutions for a class of integral boundary value problems of nonlinear fractional differential equations withp-Laplacian operator. We obtain some existence and uniqueness results concerned with our problem by using Schaefer’s fixed-point theorem and Banach contraction mapping principle. Finally, we present some examples to illustrate our main results.


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## Body

## 1. Introduction

This paper deals with the existence and uniqueness of solutions for the following fractional integral boundary value problem withp-Laplacian operator:
(1)Dc0+βφpDc0+αxt=ft,xt,t∈0,1,x0=∫01gsxsds,x1=0,φpDc0+αx0=φpDc0+αx1=∫01hsφpDc0+αxsds,where 1<α,β≤2,3<α+β<4, Dc0+α and Dc0+β are the Caputo fractional derivatives, f∈C0,1×R,R,g,h∈C0,1,R, and the p-Laplacian operator is defined as φps=sp−2s,p>1.In the past few decades, fractional differential equations have been widely applied to many fields in natural and social sciences, because they are important tools in mathematically describing many phenomena of science and engineering such as aerodynamics, control theory, signal and image processing, plasma dynamics, blood flow phenomena, and viscoelastic and non-Newtonian fluid mechanics (see [1–5] and their references). In 1983, Leibenson [2] proposed the following integer-order differential equation model with p-Laplacian operator to study the turbulent flow in a porous medium in his work:
(2)φpu′t′=ft,ut,u′t,t∈0,1,where φp is a p-Laplacian operator.Based on Leibenson’s results, many researchers have generalized his model into a variety of models with fractional order and obtained many valuable existence and uniqueness results forp-Laplacian fractional differential equations with two-point boundary conditions (see [6–14]), multipoint boundary conditions (see [15–18]), and nonlocal boundary conditions (see [19–23]). Since the 1st derivatives of unknown functions existed in Leibenson’s model, it became a special study to discuss the existence results for p-Laplacian fractional differential equations where the fractional orders were in the neighborhood of 1. So, Chai [7] used the fixed-point theorem on cones to investigate the existence and multiplicity of positive solutions for fractional differential equations with p-Laplacian operator:
(3)D0+βφpD0+αut=−ft,ut,0<t<1,u0=0,u1+σD0+γu1=0,D0+αu0=0,where 1<α≤2,0<β,γ≤1, and D0+α,D0+β, and D0+γ are the Riemann-Liouville fractional derivatives. Jong [15] established the existence and uniqueness of positive solutions for multipoint boundary value problems of nonlinear fractional differential equations with p-Laplacian operator by using the Banach contraction mapping principle
(4)D0+βφpD0+αut=ft,ut,0<t<1,u0=0,D0+γu1=∑i=1m−2ξiD0+γuηi,D0+αu0=0,φpD0+αu1=∑i=1m−2ζiφpD0+αuηi,where 1<α,β≤2,0<γ≤1, and f∈C0,1×0,+∞,0,+∞.Integral boundary value problems for differential equations have arisen in the study of various fields such as underground water flow, blood flow problems, and thermoelasticity. So, the study on the existence of solutions for thep-Laplacian integral boundary value problems has attracted the attention of many researchers recently (see [20–23]). Zhang et al. [23] considered the existence of symmetric positive solutions of the problem for the following nonlinear fourth-order p-Laplacian differential equations with integral boundary conditions:
(5)φpx″t″=ωtft,xt,t∈0,1,x0=x1=∫01gsxsds,φpx″0=φpx″1=∫01hsφpx″sds,where ω,g,h∈L0,1 are the nonnegative, symmetric functions and f∈C0,1×0,+∞,0,+∞. Zhang and Cui [22] investigated the existence of positive solutions for nonlinear fourth-order singular p-Laplacian differential equations with the integral boundary conditions
(6)φpx″t″=ft,xt,xt,t∈0,1,x0=∫01gsxsds,x1=0,φpx″0=φpx″1=∫01hsφpx″sds,where f∈C0,1×0,+∞×0,+∞,0,+∞, f may be singular at t=0,1,u=0, and g,h∈L0,1 are nonnegative. The existence results on solutions of problem (2) are established by employing upper and lower solution methods together with maximal principle. Jiang [20] used the generalized continuous theorem to investigate the existence of solutions to the integral boundary value problem of p-Laplacian multiterm fractional differential equations at resonance
(7)D0+βφpD0+αut=−ft,ut,D0+α−1ut,D0+αut,0<t<1,u0=D0+αu0=0,u1=∫01htutdt,where 0<β≤1,1<α≤2, and ∫01httα−1dt=1.Summarizing previous results, very few papers dealt with the existence of solutions for integral boundary value problems ofp-Laplacian fractional differential equations, especially, Zhang and Cui [22] who established the existence of positive solutions for fourth-order singular p-Laplacian differential equations under p≥2 and Jiang [20] who considered the existence of solutions for nonlinear multiterm fractional differential equations with 0<β≤1. Moreover, due to the nonlinearity of p-Laplacian operator φp, it is more difficult to study for the case 1<β≤2 rather than for the case 0<β≤1. Motivated by the above facts, this paper deals with the existence and uniqueness of solutions of problem (1) in which the fractional derivatives are Caputo fractional derivatives with 1<α,β≤2.The structure of this paper is organized as follows.In Section2, we recall some definitions and lemmas. In Section 3, we prove the existence and uniqueness of solutions for nonlinear integral boundary value problem with p-Laplacian operator by using Schaefer’s fixed-point theorem and Banach contraction mapping principle. Finally, we give two examples to illustrate our main results in Section 4.


## 2. Preliminaries

The Riemann-Liouville fractional integral and the Caputo fractional derivative of orderα>0 of a function f:0,∞⟶R is given by
(8)I0+αft≔1Γα∫0tt−sα−1fsds,Dc0+αft≔1Γn−α∫0tt−sn−α−1fnsds,where n=α+1, provided that the right-hand side is pointwise defined on 0,∞ (see [4, 5]).Lemma 1 (see [1]).
Ifα>0 and f∈Ca,b, then Dca+αIa+αft=ft.Lemma 2 (see [9]).
Assume thatcD0+αu∈C0,1,α>0. Then, Ia+αcD0+αut=ut+c0+c1t+⋯cn−1tn−1,ci∈R,i=0,1,⋯,n−1,n−1<α≤n.For the sake of convenience, putσ1≔∫011−sgsds and σ2≔∫01hsds and assume that σ1≠1 and σ2≠1.Lemma 3.
Letσ∈C0,1. Then, the fractional boundary value problem
(9)Dc0+βzt=σt,0<t<1,z0=z1=∫01hszsdshas a unique solution which is given by
(10)zt=−11−σ2∫01∫01hsGβs,τστdτds−∫01Gβt,sσsds,where
(11)Gβt,s≔1Γβt1−sβ−1−t−sβ−1,0≤s≤t≤1,t1−sβ−1,0≤t≤s≤1.Proof.
In view of Lemma2, we have that
(12)zt=c0+c1t+I0+βσt.
By means of the property of the fractional integral of a continuous function, we obtain thatz0=c0.
Sincez0=z1=∫01hszsds, from (12), we obtain
(13)c0=∫01hszsds,c1=∫01hszsds−I0+βσtt=1−∫01hszsds=−I0+βσtt=1.
Hence, (12) can be written as
(14)zt=∫01hszsds+I0+βσt−I0+βσtt=1⋅t=∫01hszsds+1Γβ∫0tt−sβ−1σsds−∫01t1−sβ−1σsds.
Thus, we can easily get(15)zt=∫01hszsds−∫01Gβt,sσsds.
In the right side of (15), the term ∫01hszsds can be rewritten as
(16)∫01hszsds=∫01hs∫01hτzτdτ−∫01Gβs,τστdτds=∫01hsds∫01hszsds−∫01∫01hsGβs,τστdτds,so we get
(17)∫01hszsds=−11−σ2∫01∫01hsGβs,τστdτds.
Therefore, the unique solution of (9) is given by
(18)zt=−11−σ2∫01∫01hsGβs,τστdτds−∫01Gβt,sσsds.
Conversely, letz∈C0,1 be the function which is expressed by (10). Putting
(19)Aβ≔−11−σ2∫01∫01hsGβs,τστdτds,we get
(20)zt=Aβ−∫01Gβt,sσsds=Aβ−∫0tGβt,sσsds−∫t1Gβt,sσsds=Aβ−1Γβ∫0tt1−sβ−1−t−sβ−1σsds+∫t1t1−sβ−1σsds=Aβ−1Γβ∫01t1−sβ−1σsds+1Γβ∫0tt−sβ−1σsds.
Then, we have that(21)zt=Aβ−tI0+βσtt=1+I0+βσt.
Sinceσ∈C0,1, applying Dc0+β to both sides of (21) and using Lemma 1, we can obtain
(22)Dc0+βzt=Dc0+βI0+βσt=σt.
On the other hand, multiplying (10) by ht and integrating on 0,1, we have
(23)∫01zthtdt=−11−σ2∫01htdt∫01∫01hsGβs,τστdτds−∫01ht∫01Gβt,sσsdsdt=−σ21−σ2∫01∫01hsGβs,τστdτds−∫01∫01htGβt,sσsdsdt=−11−σ2∫01∫01hsGβs,τστdτds.
Hence, (10) can be written as
(24)zt=−∫01Gβt,sσsds+∫01hszsds.
SinceGβ0,s=Gβ1,s=0, we can also have that
(25)z0=z1=∫01hszsds
Therefore, we can know thatzt is a solution of problem (9) and (9) has a unique solution which is given by (10). The proof is completed.Lemma 4.
Letf∈C0,1×R,R, then p-Laplacian integral boundary value problem (1) has a unique solution which is given by
(26)xt=∫01Gαt,sφq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+1−t1−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγdsdς,where q is the number that satisfies 1/p+1/q=1.Proof.
The proof of this lemma is divided into two steps.(Step 1)
Letr∈C0,1 and consider the boundary value problem
(27)Dc0+αxt=rt,0<t<1,x0=∫01gsxsds,x1=0.
By using Lemma2, we have that
(28)xt=d0+d1t+I0+αrt.
From (28) and boundary condition of (27) we can obtain
(29)d0=x0=∫01gsxsds,d1=−d0−I0+αrtt=1=−∫01gsxsds−I0+αrtt=1.
Then, (28) can be written as
(30)xt=d0+d1t+I0+αrt=∫01gsxsds+−∫01gsxsds−I0+αrt∣t=1t+I0+αrt=I0+αrt−I0+αrtt=1t+1−t∫01gsxsds=1Γα∫0tt−sα−1rsds−∫01t1−sα−1rsds+1−t∫01gsxsds=−∫01Gαt,srsds+1−t∫01gsxsds.
In a similar way to the proof of Lemma3, it can be easily seen that
(31)∫01gsxsds=−11−σ1∫01∫01gsGαs,τrτdτds.
Hence, we have that the unique solution of (27) is given by
(32)xt=−∫01Gαt,srsds−1−t1−σ1∫01∫01gsGαs,τrτdτds.
Conversely, letxt be the function which is expressed by (32). Then, from the definition of Gαt,s, (32) can be written as
(33)xt=−tI0+αrtt=1+I0+αrt−1−t1−σ1∫01∫01gsGαs,τrτdτds.
Sincer∈C0,1 and 1<α≤2, applying Dc0+α to both sides of (33), we can obtain
(34)Dc0+αxt=Dc0+αI0+αrt=rt.
On the other hand, multiplying (32) by gt and integrating on 0,1, we have that
(35)∫01gtxtdt=−∫01gt∫01Gαt,srsdsdt−11−σ1∫011−tgtdt∫01∫01gsGαs,τrτdτds=−∫01∫01Gαt,sgtrsdsdt−σ11−σ1∫01∫01gsGαs,τrτdτds=−1+σ11−σ1∫01∫01Gαt,sgtrsdsdt=−11−σ1∫01∫01Gαt,sgtrsdsdt.So, we can rewrite (32) as
(36)xt=−∫01Gαt,srsds+1−t∫01gtxtdt.Since Gα0,s=Gα1,s=0, we can get
(37)x0=∫01gtxtdt,x1=0.Therefore, we can know that xt is the solution of (27).
(Step 2)
Now letxt be the solution of (1). Putting yt≔Dc0+αxt, then we have that
(38)xt=−∫01Gαt,sysds−1−t1−σ1∫01∫01gsGαs,τyτdτds.Also denoting φpyt by zt, then by Lemma 3, we can see that that
(39)zt=−11−σ2∫01∫01hsGβs,τfτ,xτdτds−∫01Gβt,sfs,xsds.Since it is well-known that φp−1=φq, combining (38) and (39) yields
(40)xt=∫01Gαt,sφp−111−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+1−t1−σ1∫01∫01Gας,sgςφp−111−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγdsdς.
The proof is completed.Remark 5.
From the definition ofGαt,s and Gβt,s, it is easy to know that those functions are continuous in 0,1×0,1.Lemma 6 (see [24]).
Schaefer’s fixed-point theorem. LetX be the Banach space and T:X⟶X be completely continuous operator. If the set E≔u∈X∣u=ρTu,0<ρ<1 is bounded, then T has at least one fixed point in X.The basic properties of thep-Laplacian operator which will be used in the following studies are listed below (see [10]).
(i)
If1<p≤2,xy>0, and x,y≥m>0, then
(41)φpx−φpy≤p−1mp−2x−y
(ii)
Ifp>2 and x,y≤M, then
(42)φpx−φpy≤p−1Mp−2x−y


## 3. Main Results

In this section, we establish the existence and uniqueness of solutions of problem (1) by using Schaefer’s fixed-point theorem and the Banach contraction mapping principle.Let us consider the Banach spaceX=C0,1 endowed with the norm u≔max0≤t≤1ut.Define an operatorT:X⟶X by
(43)Txt≔∫01Gαt,sφq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfτ,xτdτds+1−t1−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+1−t1−σ1.Then, Equation (26) is equivalent to the operator equation
(44)x=Tx,x∈C0,1.From Lemma4, the existence of solutions for the problem (1) refers to the existence of fixed points of Equation (44). Therefore, it is sufficient to prove the existence of fixed points of (44).Lemma 7.
The operatorT is completely continuous.Proof.
Sincef∈C0,1×R,R and φq is continuous, we can know that T:X⟶X is continuous.
LetΩ⊂X be a bounded subset, then for any u∈Ω, there exists M0>0 such that u≤M0.
We will show thatTΩ is relatively compact in X. Since f is a continuous function, there exists Mf>0 such that ft,ut≤Mf,t∈0,1,u∈Ω. Then, we have
(45)Txt≤∫01Gαt,s⋅φq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+1−t1−σ1∫01∫01Gας,sgς⋅φq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγdsdς≤∫01Gαt,s⋅φq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+1−t1−σ1∫01∫01Gας,sgς⋅φq11−σ2⋅∫01∫01∣hγGβγ,τfτ,xτ∣dτdγ+∫01∣Gβs,γfγ,xγ∣dγdsdς.And since t,s∈0,1×0,1, evaluating the upper bound of Gβt,s gives
(46)Gβt,s≤1Γβ.So, we obtain
(47)11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγ≤MfΓβ1+h1−σ2≕M1.
Then, we can get easily that(48)Txt≤∫01Gαt,s⋅φqM1ds+1−t1−σ1∫01∫01Gας,sgς⋅φqM1dsdς≤φqM1Γα1+g1−σ1.
In view of the definition of thep-Laplacian operator, we have that
(49)Tx≤M1q−1Γα1+g1−σ1.
This shows thatTΩ is uniformly bounded in X. For any x∈Ω,0≤t1<t2≤1, we have that
(50)∣Txt2−Txt1∣=∫01Gαt2,sφq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+1−t21−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγdsdς−∫01Gαt1,sφq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds−1−t11−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγdsdς=∫01Gαt2,s−Gαt1,sφq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+t1−t21−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγdsdς≤M1q−1⋅∫01Gαt2,s−Gαt1,sds+M1q−1⋅t2−t11−σ1gΓα.
And since(51)Gαt2,s−Gαt1,s=1Γαt21−sα−1−t2−sα−1−t11−sα−1+t1−sα−1,s≤t1,t21−sα−1−t2−sα−1−t11−sα−1,t1≤s≤t2,t21−sα−1−t11−sα−1,s≥t2,the integral term ∫01Gαt2,s−Gαt1,sds can be divided into the following three parts:
(52)∫0t1Gαt2,s−Gαt1,sds=1Γα∫0t1∣t21−sα−1−t2−sα−1−t11−sα−1−t1−sα−1∣ds≤1Γα∫0t1t2−t11−sα−1+t2−sα−1−t1−sα−1ds≤1Γαt2−t1∫0t11−sα−1ds+1Γα∫0t1t2−sα−1−t1−sα−1ds≤t2−t1Γα+1+t2α−t1α−t2−t1αΓα+1.
In a similar way to this, we can obtain(53)∫t1t2∣Gαt2,s−Gαt1,s∣ds≤t2−t1Γα+1+t2−t1αΓα+1,∫t21∣Gαt2,s−Gαt1,s∣ds≤t2−t1Γα+1.
These inequalities yield(54)∫01∣Gαt2,s−Gαt1,s∣ds=∫0t1∣Gαt2,s−Gαt1,s∣ds+∫t1t2∣Gαt2,s−Gαt1,s∣ds+∫t21∣Gαt2,s−Gαt1,s∣ds≤3t2−t1+t2α−t1αΓα+1.Therefore, we get
(55)Txt2−Txt1≤M1q−1⋅3t2−t1+t2α−t1αΓα+1+M1q−1⋅t2−t11−σ1gΓα=M1q−1Γα3α+g1−σ1⋅t2−t1+M1q−1Γα+1⋅t2α−t1α.
This shows thatTΩ is equicontinuous in X. By using the Arzela-Ascoli theorem, we can see that TΩ is relatively compact in X. As a consequence of the above discussion, the operator T:X⟶X is completely continuous. The proof is completed.Denote as follows:(56)A≔1Γα1+g1−σ1p−11Γβ1+h1−σ2.In this article, the following hypotheses will be used.(H1).
There exist nonnegative functionsa,b∈C0,1 such that
(i)
∣ft,x∣≤at+btxp−1,t∈0,1,x∈R
(ii)
Ab<1Theorem 8.
Assume that the hypothesis (H1) holds, then problem (1) has at least one solution.Proof.
Consider the following set:(57)E≔x∈X∣x=ρTx,0<ρ<1.
For anyx∈E, it can be easily seen that
(58)x=ρTx<Tx.
On the other hand, from the condition (i) of the hypothesis (H1) and (49), we have that
(59)Tx≤1Γα1+g1−σ11Γβ1+h1−σ2a+bxp−1q−1.
From (58), we get
(60)x<1Γα1+g1−σ11Γβ1+h1−σ2a+bxp−1q−1.
Since1/p+1/q=1 implies p−1q−1=1, we obtain
(61)xp−1<Aa+bxp−1.
Therefore, we have that(62)xp−11−Ab<Aa,and by using the condition (ii) of the hypothesis (H1), we can see that
(63)x<Aa1−Abq−1.
So, we can know thatE is bounded. In view of Schaefer’s fixed-point theorem (Lemma 6), the operator T:X⟶X has at least one fixed point which is the solution of the problem (1). The proof is completed.Here, putr≔Aa/1−Abq−1 and list more hypotheses to obtain the uniqueness results for our problem.(H2).
There exists a real numberL>0 such that
(64)ft,x−ft,y∣<L∣x−y,t∈0,1,x,y∈−r,r.
For the readers’ convenience, denote as follows:(65)M0≔1+h1−σ2a+brp−1,θ≔Lq−1M0q−2ΓαΓβ1+g1−σ11+h1−σ2.Theorem 9.
Let1<p<2. Assume that the hypotheses (H1) and (H2) are satisfied and 0<θ<1, then problem (1) has a unique solution.Proof.
PutB=x∈Xx≤r.
Firstly, we will show thatTB⊂B. In fact, from the hypothesis (H1) and (49), for any x∈B, we have that
(66)Tx≤1Γα1+g1−σ11Γβ1+h1−σ2a+bxp−1q−1=1Γα1+g1−σ1p−11Γβ1+h1−σ2q−1a+bxp−1q−1≤Aq−1a+brp−1q−1=Aq−1a+Aab1−Abq−1=Aa1−Abq−1=r.
Next, we will prove the uniqueness of solutions for problem (1).
For anyx∈B and any t∈0,1, by the hypothesis (H1), we get
(67)1∣1−σ2∣∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγ≤h∣1−σ2∣∫01∫01Gβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγ≤h∣1−σ2∣∫01∫01Gβγ,τaτ+bτrp−1dτdγ+∫01Gβs,γaγ+bγrp−1dγ≤h∣1−σ2∣+1a+brp−1=M0.
Since1<p<2, we can see q>2. Hence, from (H2) and one basic property of p-Laplacian operator (42), for any x,y∈B and any t∈0,1, we have that
(68)Txt−Tyt=∫01Gαt,sφq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds+1−t1−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγdsdς−∫01Gαt,sφq11−σ2∫01∫01hγGβγ,τfτ,yτdτdγ+∫01Gβs,γfγ,yγdγds−1−t1−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,yτdτdγ+∫01Gβs,γfγ,yγdγdsdς≤∫01Gαt,sφq11−σ2∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγds−φq11−σ2∫01∫01hγGβγ,τfτ,yτdτdγ+∫01Gβs,γfγ,yγdγds+1−t1−σ1∫01∫01Gας,sgςφq11−σ2⋅∫01∫01hγGβγ,τfτ,xτdτdγ+∫01Gβs,γfγ,xγdγ−φq11−σ2⋅∫01∫01hγGβγ,τfτ,yτdτdγ+∫01Gβs,γfγ,yγdγdsdς≤q−1M0q−21−σ2∫01Gαt,s∫01∫01hγGβγ,τfτ,xτ−fτ,yτdτdγds+q−1M0q−2∫01Gαt,s∫01Gβs,γfγ,xγ−fγ,yγdγds+1−tq−1M0q−21−σ1⋅1−σ2⋅∫01∫01Gας,sgς∫01∫01hγGβγ,τfτ,xτ−fτ,yτdτdγdsdς+1−tq−1M0q−21−σ1∫01∫01Gας,sgς∫01Gβs,γfγ,xγ−fγ,yγdγdsdς≤Lq−1M0q−2∣1−σ2∣∫01Gαt,s∫01∫01hγGβγ,τxτ−yτdτdγds+Lq−1M0q−2∫01Gαt,s∫01Gβs,γxγ−yγdγds+L1−tq−1M0q−21−σ1⋅1−σ2∫01∫01Gας,sgς∫01∫01hγGβγ,τxτ−yτdτdγdsdς+L1−tq−1M0q−21−σ1∫01∫01Gας,sgς∫01Gβs,γxγ−yγdγdsdς≤Lq−1M0q−2ΓαΓβh1−σ2+1+g1−σ1h1−σ2+g1−σ1x−y=θx−y.
This means that(69)Tx−Ty≤θx−y.
Since0<θ<1, we can see that T:B→B is a contraction mapping. By means of the Banach contraction mapping principle, we can prove that T has a unique fixed point in B. That is, problem (1) has a unique solution. The proof is completed.


## 4. Examples

The following examples are concerned with the illustration of Theorem8 and Theorem 9.Example 1.
Consider the following integral boundary value problem:(70)Dc0+1.8φ1.2Dc0+1.5xt=1+13sint+e−t10+tx0.2t,t∈0,1,x0=∫011+e−sxsds,x1=0,φ1.2Dc0+1.5x0=φ1.2Dc0+1.5x1=∫011100+s2φ1.2Dc0+1.5xsds.The problem (70) can be rated as the boundary value problem where α=1.5,β=1.8,p=1.2,ft,x=1+1/3sint+e−t/100+tx0.2,at=1+1/3sint,bt=e−t/10+t,gt=1+e−t, and ht=1/100+t2. By simple calculation, we have a=4/3,b=1/10,g=2,h=1/10,σ1=∫011−sgsds≈0.87≠1, and σ2=∫01hsds≈0.1≠1.So, testing whether the hypothesis (H1) holds or not, we get
(71)A=1Γα1+g1−σ1p−11Γβ1+h1−σ2≈2.13,Ab≈0.67<1.Therefore, by Theorem8, the problem (70) has at least one solution.Example 2.
Consider the boundary value problem(72)Dc0+1.7φ1.5Dc0+1.3xt=0.4t+0.1sinxt,t∈0,1,x0=∫01sxsds,x1=0,φ1.5Dc0+1.3x0=φ1.5Dc0+1.3x1=∫01110+s2φ1.5Dc0+1.3xsds.Comparing with the problem (1), it can be easily seen that α=1.3,β=1.7,p=1.5, ft,x=0.4t+0.1sinx,at=0.4t,bt=0.1,gt=t, and ht=1/10+t2.Sincea=0.4,b=0.1,g=1,h=0.1,σ1=∫01s1−sds≈0.17≠1, and σ2=∫011/10+s2ds≈0.1≠1, we obtain
(73)ft,x≤0.4t+0.1sinx≤0.4t+0.1⋅min1,x≤0.4t+0.1⋅x0.5,A=1Γα1+g1−σ1p−11Γβ1+h1−σ2≈1.91,Ab≈0.19<1.So, we can see that the hypothesis (H1) is satisfied. Calculating the radius r, we have
(74)r=Aa1−Abq−1≈0.9.Sincer<1, for any t∈0,1 and any x,y∈−r,r, we get
(75)ft,x−ft,y=0.1sinx−siny≤0.1x−y,M0=1+h1−σ2a+brp−1≈0.55,θ=Lq−1M0q−2ΓαΓβ1+g1−σ11+h1−σ2≈0.33<1.From the above discussion, we can see that the hypothesis (H2) holds. Therefore, it follows by Theorem 9 that the problem (72) has a unique solution.



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*Source: 7129796-2020-05-01.xml*