# Existence of Solutions for Fractional Differential Equations withp-Laplacian Operator and Integral Boundary Conditions

Authors: Jingli Xie; Lijing Duan
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we investigate a class of integral boundary value problems of fractional differential equations with ap-Laplacian operator. Existence of solutions is obtained by using the fixed point theorem, and an example is given to show the applicability of our main result.


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## Body

## 1. Introduction

In this paper, we consider the nonlinear fractional differential equations with ap-Laplacian operator and integral boundary conditions
(1)Dc0+βϕpDc0+αut+ft,ut=0,t∈0,1,u1=λ∫01usds,u′1=0,Dc0+αu1=bcD0+αuξ,where 1<α≤2, 0<β≤1, 0<ξ, b, λ<1, Dc0+α and Dc0+β are the Caputo fractional derivative.ϕps=sp−2s is the p-Laplacian operator such that 1/p+1/q=1, p>1, and ϕp−1s=ϕqs, and ft,u:0,1×0,∞⟶0,∞ is a given continuous function.In recent years, boundary value problems of fractional differential equations have significantly been discussed by some researchers because fractional calculus theory and methods have been widely used in various fields of natural sciences and social sciences. In the field of physical mechanics, fractional calculus not only provides suitable mathematical tools for the study of soft matter but also provides new research ideas and plays an irreplaceable role in the modeling of soft matter [1–3]. Some nonlinear analysis tools such as coincidence degree theory [4, 5], upper and lower solution method [6–8], fixed point theorems [9–11], and variational methods [12–14] have been widely used to discuss existence of solutions for boundary value problems of fractional differential equations.On the other hand, it is well known that differential equation models withp-Laplacian operators are often used to simulate practical problems such as tides caused by celestial gravity and elastic deformation of beams and rich results of fractional differential equations with a p-Laplacian operator have been obtained [15–18]. In particular, in [15], by using the fixed point theorem, Yan et al. studied the existence of solutions for boundary value problems of fractional differential equations with a p-Laplacian operator:
(2)−D0+αϕpD0+βut=ft,ut,u0=u1=u′0=u′1,D0+βu0=0,D0+βu1=bD0+βuη,where 1<α≤2, 3<β≤4, 0<η<1, 0<b<η1−α/p−1, 1/p+1/q=1, 1<p, ϕp−1s=ϕqs,D0+α, D0+β is the standard Riemann-Liouville derivative, and ft,u:0,1×0,∞⟶0,∞ is a given countinuous function.Moreover, during the last decade, the integral boundary value problem of fractional differential equations is also a hot issue for scholars and some good results have been achieved [19–23]. In [24], by using the method of the upper and lower solutions and Schauder’s and Banach’s fixed points theorem, Abdo et al. obtained the existence and uniqueness of a positive solution of the fractional differential equations with integral boundary equations:
(3)Dc0+αut=ft,ut,t∈0,1,u0=λ∫01usds+d,where 0<α≤1, λ≥0, d>0, Dc0+α is the standard Caputo derivative, and f:0,1×0,∞⟶0,∞ is a given countinuous function.In [25], Bai and Qiu discuss the existence of positive solutions for boundary value problems of fractional differential equations:
(4)Dc0+αut+ft,ut=0,t∈0,1,u0=u′1=u″0=0,where 2<α≤3, f∈C0,1×0,+∞,0,+∞,and Dc0+α is the standard Caputo derivative.Motivated by the works mentioned above, we concentrate on the solutions for the nonlinear fractional differential equation (1). We obtain the existence result of the fractional differential equations with integral boundary equations by using the Schauder fixed point theorem and other mathematical analysis techniques.The rest of this paper is organized as follows. In Section2, we give some notations and lemmas. Section 3 is devoted to study existence of solutions for boundary value problems of fractional differential equations. Finally, we provide an example to illustrate our results.


## 2. Preliminaries

In the section, we present some definitions and lemmas, which are required for building our theorems.Definition 1 (see [1]).
The fractional integral of orderαα>0of functionf:0,∞⟶Ris given by(5)I0+αft=1Γα∫0tt−sα−1fsds,whereΓαis the Gamma function, provided the right side is pointwise defined on0,+∞.Definition 2 (see [2]).
The Caputo fractional derivative of orderαα>0of functionf:0,∞⟶Ris given by(6)Dcαft=1Γn−α∫0tt−sn−α−1fnsds,wheret>0, n=α+1,Γαis the Gamma function.Lemma 3 (see [26]).
Forα>0, the solution of fractional differential equationDc0+αut=0is given byut=c0+c1t+⋯+cn−1tn−1, ci∈R, i=0,1,2,⋯,n−1, n=α+1, andαdenotes the integer part of the real numberα.Lemma 4 (see [1]).
Forα>0, then (i) I0+αDc0+αut=ut+c0+c1t+⋯+cn−1tn−1, ci∈R, n=α+1 and (ii) Dc0+αI0+αut=ut.Lemma 5 (see [1]).
LetXbe a Banach space andΩ⊂Xa convex, closed, and bounded set. IfT:Ω⟶Ωis a continuous operator such thatTΩ⊂X, TΩis relatively compact, thenThas at least one fixed point inΩ.
LetϕpDc0+αut=vt, then v1=bp−1vξ. We now consider the following equations:
(7)−Dc0+βvt=yt,t∈0,1,v1=bp−1vξ.Lemma 6.
Lety∈C0,1, then (7) has a unique solution(8)vt=∫01Ht,τyτdτ,where
(9)Ht,τ=1−τβ−11−bp−1Γβ−t−τβ−1Γβ,0≤τ≤t≤1,ξ≤τ,1−τβ−11−bp−1Γβ−t−τβ−1Γβ−bp−1ξ−τβ−11−bp−1Γβ,0≤τ≤t≤1,ξ≤τ,1−τβ−11−bp−1Γβ,0≤t≤τ≤1,ξ≤τ,1−τβ−11−bp−1Γβ−bp−1ξ−τβ−11−bp−1Γβ,0≤t≤τ≤1,ξ≤τ.Proof.
Supposev satisfies boundary value problem (7), by (i) of Lemma 4, we can obtain
(10)vt=−I0+βyt−c0=−1Γβ∫0tt−sβ−1ysds−c0.
Using the boundary conditionv1=bp−1vξ, we can obtain
(11)c0=bp−11−bp−1Γβ∫0ξξ−sβ−1ysds−11−bp−1Γβ∫011−sβ−1ysds.
Thus,(12)vt=−1Γβ∫0tt−sβ−1ysds−bp−11−bp−1Γβ∫0ξξ−sβ−1ysds+11−bp−1Γβ∫011−sβ−1ysds=∫01Ht,sysds.
From the above analysis, the equation(13)Dc0+βϕpDc0+αut+yt=0,u1=λ∫01usds,u′1=0,Dc0+αu1=bcD0+αuξis equivalent to
(14)Dc0+αut=ϕq∫01Ht,τyτdτ,0<t<1,u′1=0,u1=λ∫01usds.Lemma 7.
Lety∈C0,1. Then (14) has a unique solution:(15)ut=1Γα∫0tt−sα−1ϕq∫01Hs,τyτdτds−1Γα∫011−sα−1ϕq∫01Hs,τyτdτds+1Γα−1∫011−sα−2ϕq∫01Hs,τyτdτds−1Γα−1∫01t1−sα−2ϕq∫01Hs,τyτdτds+λA,where
(16)A=11−λ1Γα+1∫011−sαϕq∫01Hs,τyτdτds−1Γα∫011−sα−1ϕq∫01Hs,τyτdτds+12Γα−1∫011−sα−2ϕq∫01Hs,τyτdτds.Proof.
By (i) of Lemma (4), we can obtain ut=1/Γα∫0tt−sα−1ϕq∫01Hs,τyτdτds−c1−c2t. Then, u′t=1/Γα−1∫0tt−sα−2ϕq∫01Hs,τyτdτds−c2, using the boundary condition u′1=0, we can obtain
(17)c2=1Γα−1∫011−sα−2ϕq∫01Hs,τyτdτds.
Another, because(18)u1=λ∫01usds,(19)u1=1Γα∫011−sα−1ϕq∫01Hs,τyτdτds−c1−c2,we have
(20)c1=1Γα∫011−sα−1ϕq∫01Hs,τyτdτds−1Γα−1∫011−sα−2ϕq∫01Hs,τyτdτds−λ∫01usds.
Now, we express∫01usds, let
(21)A=∫01utdt=∫01∫0tt−sα−1Γαϕq∫01Hs,τyτdτdsdt−∫01∫011−sα−1Γαϕq∫01Hs,τyτdτdsdt+∫01∫011−sα−2Γα−1ϕq∫01Hs,τyτdτdsdt−∫01∫01t1−sα−2Γα−1ϕq∫01Hs,τyτdτdsdt+∫01λAdt=∫011−sαΓα+1ϕq∫01Hs,τyτdτds−∫011−sα−1Γαϕq∫01Hs,τyτdτds+∫011−sα−2Γα−1ϕq∫01Hs,τyτdτds−∫011−sα−22Γα−1ϕq∫01Hs,τyτdτds+λA.
We obtain(22)A=11−λ1Γα+1∫011−sαϕq∫01Hs,τyτdτds−1Γα∫011−sα−1ϕq∫01Hs,τyτdτds+12Γα−1∫011−sα−2ϕq∫01Hs,τyτdτds.
Therefore,(23)ut=1Γα∫0tt−sα−1ϕq∫01Hs,τyτdτds−1Γα∫011−sα−1ϕq∫01Hs,τyτdτds+1Γα−1∫011−sα−2ϕq∫01Hs,τyτdτds−1Γα−1∫01t1−sα−2ϕq∫01Hs,τyτdτds+λA.
Reverse, if(24)ut=1Γα∫0tt−sα−1ϕq∫01Hs,τyτdτds−1Γα∫011−sα−1ϕq∫01Hs,τyτdτds+1Γα−1∫011−sα−2ϕq∫01Hs,τyτdτds−1Γα−1∫01t1−sα−2ϕq∫01Hs,τyτdτds+λA,by (ii) of Lemma (4), we can obtain that ut is a solution of (14).
The proof is completed.Lemma 8.
The functionsHis continuous on0,1×0,1and has the following properties:(1)
Ht,τ≤Hτ,τ,t,τ∈0,1(2)
∫01Hτ,τdτ≤1/1−bp−1Γβ+1Proof.
(1) For anyt,τ∈0,1, by (9), it is obvious that Ht,τ≤Hτ,τ. (2) For any t,τ∈0,1, by (9), we conclude that
(25)Hτ,τ≤1−τβ−11−bp−1Γβ.
Therefore,(26)∫01Hτ,τdτ≤11−bp−1Γβ+1.
This completes the proof.


## 3. Main Results

In this section, we will show the existence results for boundary value problem (1) by the Schauder fixed point theorem.LetI=0,1, U=ut∣ut∈CI, and definite the normu=maxt∈0,1ut, U,∙ is a Banach space.Theorem 9.
Assume that the following conditions (H1) and (H2) are satisfied:
(H1) ft,u:0,1×0,∞⟶0,∞ is continuous
(H2)There exists a constant k>0, satisfying ft,u≤Lϕpu,t∈0,1,u≤k, where
(27)0<L≤ϕp21−λΓα+22−λα2+α11−bp−1Γβ+1−1,then the problem (1) has at least one solution.Proof.
LetP=ut∣ut≤k,t∈0,1; thus, P⊂U is convex, bounded, and closed.
Define an operatorT:P⟶U by
(28)Tut=1Γα∫0tt−sα−1ϕq∫01Hs,τfτ,uτdτds−1Γα∫011−sα−1ϕq∫01Hs,τfτ,uτdτds+1Γα−1∫011−sα−2ϕq∫01Hs,τfτ,uτdτds−1Γα−1∫01t1−sα−2ϕq∫01Hs,τfτ,uτdτds+λ1−λ1Γα+1∫011−sαϕq∫01Hs,τfτ,uτdτds−1Γα∫011−sα−1ϕq∫01Hs,τfτ,uτdτds+12Γα−1∫011−sα−2ϕq∫01Hs,τfτ,uτdτds.
For anyu∈P, then by (H2), we have
(29)ft,u≤Lϕpu≤Lϕpk.
By Lemma (8), we conclude that
(30)Tut≤1Γα∫0tt−sα−1ϕq∫01Hτ,τfτ,uτdτds−1Γα∫011−sα−1ϕq∫01Hτ,τfτ,uτdτds+1Γα−1∫011−sα−2ϕq∫01Hτ,τfτ,uτdτds−1Γα−1∫01t1−sα−2ϕq∫01Hτ,τfτ,uτdτds+λ1−λ1Γα+1∫011−sαϕq∫01Hτ,τfτ,uτdτds−1Γα∫011−sα−1ϕq∫01Hτ,τfτ,uτdτds+12Γα−1∫011−sα−2ϕq∫01Hτ,τfτ,uτdτds=tαΓα+1ϕq∫01Hτ,τfτ,uτdτ−1Γα+1ϕq∫01Hτ,τfτ,uτdτ+1Γαϕq∫01Hτ,τfτ,uτdτ−tΓαϕq∫01Hτ,τfτ,uτdτ+λ1−λ1Γα+2ϕq∫01Hτ,τfτ,uτdτ−1Γα+1ϕq∫01Hτ,τfτ,uτdτ+12Γαϕq∫01Hτ,τfτ,uτdτ≤1Γα+1ϕq∫01Hτ,τfτ,uτdτ−1Γα+1ϕq∫01Hτ,τfτ,uτdτ+1Γαϕq∫01Hτ,τfτ,uτdτ+λ1−λ1Γα+2ϕq∫01Hτ,τfτ,uτdτ−1Γα+1ϕq∫01Hτ,τfτ,uτdτ+12Γαϕq∫01Hτ,τfτ,uτdτ=2α2+2α−λα2−3λα21−λΓα+2ϕq∫01Hτ,τfτ,uτdτ≤2−λα2+α21−λΓα+2ϕq∫01Hτ,τfτ,uτdτ≤2−λα2+α21−λΓα+2ϕq∫01Hτ,τLϕpuτdτ≤2−λα2+α21−λΓα+2ϕq∫01Hτ,τLϕpkdτ≤2−λα2+α21−λΓα+2ϕq∫01Hτ,τdτϕqLk≤2−λα2+α21−λΓα+2ϕq11−bp−1Γβ+1ϕqLk≤k.
Thus,TP⊆P. By (H2), we have
(31)ϕq∫01Hs,τfτ,uτdτds≤ϕqLϕpk1−bp−1Γβ+1≤Lϕpk1−bp−1Γβ+1q−1.
We express(32)M≔Lϕpk1−bp−1Γβ+1q−1.
For eacht1,t2∈0,1, t1<t2, we get
(33)Tut2−Tut1=1Γα∫0t2t2−sα−1ϕq∫01Hs,τfτ,uτdτds−1Γα∫0t1t1−sα−1ϕq∫01Hs,τfτ,uτdτds−1Γα−1∫0tt21−sα−2ϕq∫01Hs,τfτ,uτdτds−1Γα−1∫01t11−sα−2ϕq∫01Hs,τfτ,uτdτds≤Lϕpk1−bp−1Γβ+1q−1t2α−t1αΓα+1−t2−t1Γα≤Mt2α−1+t2α−2t1+t2α−3t12+⋯+t2t1α−2+t1α−1Γα+1−1Γαt2−t1.
Ast2⟶t1, the right-hand side of the previous inequality is independent of u and tends to zero; thus, TP is equicontinuous. From the Arzela-Ascoli Theorem, T is compact. Applying Schauder’s fixed point theorem, T has at least one fixed point u∈P. Therefore, the problem (1) has at least one positive solution u in P.


## 4. Applications

In this section, we will give an example to illustrate our main results.Example 1.
Consider the following equation:(34)Dc0+1/2ϕ4/3Dc0+3/2ut=ϕputt+52,t∈0,1,u1=12∫01usds,u′1=0,Dc0+αu1=12Dc0+αu12,where β=1/2, α=3/2, p=4/3, q=4, λ=ξ=b=1/2, and ft,u=ϕput/t+52. Since f is continuous and
(35)ft,u≤125ϕput,for t,u∈0,1×0,∞, we obtain
(36)L=125.
Then,(37)ϕp21−λΓα+22−λα2+α11−bp−1Γβ+1−1≈0.1534.
It is obvious that(38)L<ϕp21−λΓα+22−λα2+α11−bp−1Γβ+1−1≈0.1534.
By Theorem (9), we conclude that the problem (34) has at least one solution.



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