# Generalized Antiperiodic Boundary Value Problems for the Fractional Differential Equation with p-Laplacian Operator

Authors: Zhi-Wei Lv; Xu-Dong Zheng
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We discuss the existence of solutions about generalized antiperiodic boundary value problems for the fractional differential equation with p-Laplacian operatorϕp(cD0+αu(t))=f(t,u(t),u'(t)), 0<t<T, 1<α≤2, u(0)+(-1)θau(T)=0, cD0+βu(0)+(-1)θbcD0+βu(T)=λ, 0<β<1, where cD0+α is the Caputo fractional derivative, θ=0,1, a>0, a≠1, b>0 and ϕp(s)=|s|p-2s, p>1, ϕp-1=ϕq, 1/p+1/q=1. Our results are based on fixed point theorem and contraction mapping principle. Furthermore, three examples are also given to illustrate the results.


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## Body

## 1. Introduction

Fractional differential equations arise in various areas of science and engineering, such as physics, mechanics, chemistry, and engineering. The fractional order models become more realistic and practical than the classical integer models. Due to their applications, fractional differential equations have gained considerable attentions; one can see [1–14] and references therein.Anti-periodic boundary value problems occur in the mathematical modeling of a variety of physical processes. Anti-periodic problems constitute an important class of boundary value problems and have received considerable attention (see [15–19]).In [20], Zhang considered the existence and multiplicity results of positive solutions for the following boundary value problem of fractional differential equation:
(1)cD0+pu(t)=f(t,u(t)),0<t<1,u(0)+u′(0)=0,u(1)+u′(1)=0,
where 1<p≤2 is a real number, cDq is the Caputo fractional derivative, and f:[0,1]×[0,+∞)→[0,+∞) is continuous.In [15], the authors discussed some existence results for the following anti-periodic boundary value problem for fractional differential equations:
(2)cDqx(t)=f(t,x(t)),t∈[0,T],T>0,3<q≤4,x(0)=-x(T),x′(0)=-x′(T),x′′(0)=-x′′(T),x′′′(0)=-x′′′(T),
where cDq is the Caputo fractional derivative of order q; f is a given continuous function.In [16], the authors investigated the following anti-periodic boundary value problem for higher-order fractional differential equations:
(3)cDqx(t)=f(t,x(t)),t∈[0,T],T>0,4<q≤5,x(0)=-x(T),x′(0)=-x′(T),x′′(0)=-x′′(T),x′′′(0)=-x′′′(T),x(iv)(0)=-x(iv)(T),
where cDq is the Caputo fractional derivative of order q; f is a given continuous function.In [17], the authors investigated a class of anti-periodic boundary value problem of fractional differential equations
(4)cDqx(t)=f(t,x(t)),t∈[0,T],T>0,1<q≤2,x(0)=-x(T),cD0+px(0)=-cD0+pu(T),0<p<1,
where cDq is the Caputo fractional derivative of order q; f is a given continuous function.In this paper, we discuss the existence of solutions about generalized anti-periodic boundary value problems for the fractional differential equation with p-Laplacian operator(5)ϕp(cD0+αu(t))=f(t,u(t),u′(t)),0<t<T,1<α≤2,u(0)+(-1)θau(T)=0,cD0+βu(0)+(-1)θbcD0+βu(T)=λ,0<β<1,
where cD0+α is the Caputo fractional derivative, θ=0,1, a>0, a≠1, b>0, and ϕp(s)=|s|p-2s, p>1, ϕp-1=ϕq, (1/p)+(1/q)=1.If we takea=b=1, θ=λ=0 and p=2, then the problem (5) becomes the problem studied in [17]. In this paper, we let a≠1.This paper is organized as follows. In Section2, we present some background materials and preliminaries. Section 3 deals with some existence results. In Section 4, three examples are given to illustrate the results.


## 2. Background Materials and Preliminaries

Definition 1 (see [21]).
The fractional integral of orderα with the lower limit t0 for a function f is defined as
(6)Iαf(t)=1Γ(α)∫t0t‍(t-s)α-1f(s)ds,t>t0,α>0,
where Γ is the gamma function.Definition 2 (see [21]).
Caputo's derivative of orderq with the lower limit t0 for a function f can be written as
(7)cDqf(t)=1Γ(n-q)∫t0t‍(t-s)n-q-1f(n)(s)ds,t>t0,q>0,n=[q]+1.Lemma 3 (see [22]).
Assume thatu∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L(0,1). Then
(8)I0+αcD0+αu(t)=u(t)+c0+c1t+c2t+⋯+cn-1tn-1,
where ci∈R, i=0,1,2,…,n-1, n=[α]+1.Lemma 4.
Lety∈C[0,1]. Then the fractional differential equation
(9)ϕp(cD0+αu(t))=y(t),0<t<T,1<α≤2,u(0)+(-1)θau(T)=0,cD0+βu(0)+(-1)θbcD0+βu(T)=λ
has a unique solution which is given by
(10)u(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds-a[(-1)θ+a]Γ(α)∫0T‍(T-s)α-1y(s)ds+aΓ(2-β)[(-1)θ+a]T-βΓ(α-β)∫0T‍(T-s)α-β-1y(s)ds-aΓ(2-β)[1+(-1)θa]bT-β+t[-Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1y(s)ds+(-1)θΓ(2-β)bT1-βλ].Proof.
From Lemma3, we have
(11)u(t)=I0+α(ϕq(y(t)))+c0+c1t=1Γ(α)∫0t‍(t-s)α-1ϕq(y(s))ds+c0+c1t,cD0+βu(t)=I0+(α-β)(ϕq(y(t)))+c1cD0+βt=1Γ(α-β)∫0t‍(t-s)α-β-1ϕq(y(s))ds+c1t1-βΓ(2-β).
Thus,
(12)u(0)=c0,u(T)=1Γ(α)∫0T‍(T-s)α-1ϕq(y(s))ds+c0+c1T,cD0+βu(0)=0,cD0+βu(T)=1Γ(α-β)∫0T‍(T-s)α-β-1ϕq(y(s))ds+c1T1-βΓ(2-β).
By cD0+βu(0)+(-1)θbcD0+βu(T)=λ, we have
(13)c1=-Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ϕq(y(s))ds+(-1)θΓ(2-β)bT1-βλ.
Using the boundary condition u(0)+(-1)θau(T)=0 and (13), we obtain
(14)c0=-a[(-1)θ+a]Γ(α)∫0T‍(T-s)α-1ϕq(y(s))ds+aΓ(2-β)[(-1)θ+a]T-βΓ(α-β)∫0T‍(T-s)α-β-1ϕq(y(s))ds-aΓ(2-β)[1+(-1)θa]bT-βλ.
Thus,
(15)u(t)=1Γ(α)∫0t‍(t-s)α-1ϕq(y(s))ds-a[(-1)θ+a]Γ(α)∫0T‍(T-s)α-1ϕq(y(s))ds+aΓ(2-β)[(-1)θ+a]T-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(y(s))ds-aΓ(2-β)[1+(-1)θa]bT-βλ+t[(-1)θΓ(2-β)bT1-β(-1)θ-Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(y(s))ds+(-1)θΓ(2-β)bT1-βλ].


## 3. Main Results

LetX=C1([0,1],R) denote the Banach space of continuous functions u(t) and u′(t) from [0,1]→R endowed with the norm defined by
(16)∥u∥1=max{∥u∥,∥u′∥},
where
(17)∥u∥=sup0≤t≤1|u(t)|,∥u′∥=sup0≤t≤1|u′(t)|.
Define an operator F:X→X as
(18)Fu(t)=1Γ(α)∫0t‍(t-s)α-1ϕq(f(s,u(s),u′(s)))ds-a[(-1)θ+a]Γ(α)×∫0T‍(T-s)α-1ϕq(f(s,u(s),u′(s)))ds+aΓ(2-β)[(-1)θ+a]T-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(f(s,u(s),u′(s)))ds-aΓ(2-β)[1+(-1)θa]bT-βλ+t[(-1)θΓ(2-β)bT1-β-Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(f(s,u(s),u′(s)))ds+(-1)θΓ(2-β)bT1-βλ].
From (18), we conclude that
(19)F′u(t)=1Γ(α-1)∫0t‍(t-s)α-2ϕq(f(s,u(s),u′(s)))ds-Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(f(s,u(s),u′(s)))ds+(-1)θΓ(2-β)bT1-βλ.
Then (5) has a solution if and only if the operator F has a fixed point.Theorem 5.
Letf:[0,T]×R+×R+→R+ be continuous. Assume that f meets the following condition: there exist d≥0, γ≥0 such that
(20)f(t,x(t),x′(t))≤γϕp(|x(t)|),fort∈[0,T],∥x∥1≤d,(21)ϕq(γ)≤12(Tα-1Γ(α)+TαΓ(α+1)+aTα|1-a|Γ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1)+Tα-1Γ(2-β)Γ(α-β+1))-1.
Then the problem (5) has at least one solution on [0,T] for
(22)0≤λ≤d2(aΓ(2-β)|1-a|bT-β+Γ(2-β)bT-β+Γ(2-β)bT1-β)-1.Proof.
Fromf(t,u(t),u′(t))∈C([0,1]×R+×R+,R+), we know that T is continuous.
Let(23)Bd={u∣∥u∥1≤d,u∈X}.
For u∈Bd, we have
(24)|Fu(t)|≤1Γ(α)∫0t‍(t-s)α-1|ϕq(f(s,u(s),u′(s)))|ds+a|(-1)θ+a|Γ(α)×∫0T‍(T-s)α-1|ϕq(f(s,u(s),u′(s)))|ds+aΓ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u(s),u′(s)))|ds+|aΓ(2-β)[1+(-1)θa]bT-βλ|+t[Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u(s),u′(s)))|ds+|(-1)θΓ(2-β)bT1-βλ|]≤1Γ(α)∫0t‍(t-s)α-1ϕq(γ)|u(s)|ds+a|1-a|Γ(α)∫0T‍(T-s)α-1ϕq(γ)|u(s)|ds+aΓ(2-β)|1-a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(γ)|u(s)|ds+aΓ(2-β)|1-a|bT-βλ+t[Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(γ)|u(s)|ds+Γ(2-β)bT1-βλ]≤(Tαϕq(γ)Γ(α+1)+aTαϕq(γ)|1-a|Γ(α+1)+aTαΓ(2-β)ϕq(γ)|1-a|Γ(α-β+1)+TαΓ(2-β)ϕq(γ)Γ(α-β+1))∥u∥1+aΓ(2-β)|1-a|bT-βλ+Γ(2-β)bT-βλ,|F′u(t)|≤1Γ(α-1)×∫0t‍(t-s)α-2|ϕq(f(s,u(s),u′(s)))|ds+Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u(s),u′(s)))|ds+Γ(2-β)bT1-βλ≤1Γ(α-1)∫0t‍(t-s)α-2ϕq(γ)|u(s)|ds+Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ϕq(γ)|u(s)|ds+Γ(2-β)bT1-βλ≤(Tα-1ϕq(γ)Γ(α)+Tα-1Γ(2-β)ϕq(γ)Γ(α-β+1))∥u∥1+Γ(2-β)bT1-βλ.
This, together with (21) and (22), yields that
(25)∥Fu(t)∥1≤d.
Hence, F(Bd) is uniformly bounded.
Next we show thatF is equicontinuous.
For any0≤t1≤t2≤1, u∈Bd, we have
(26)|Fu(t2)-Fu(t1)|≤|1Γ(α)∫0t2‍(t2-s)α-1ϕq(f(s,u(s),u′(s)))ds-1Γ(α)∫0t1‍(t1-s)α-1ϕq(f(s,u(s),u′(s)))ds|+[∫0T‍Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(f(s,u(s),u′(s)))ds+Γ(2-β)bT1-βλ∫0T‍](t2-t1)≤1Γ(α)∫0t1‍|(t2-s)α-1-(t1-s)α-1|ϕq(γ)|u(s)|ds+1Γ(α)∫t1t2‍(t1-s)α-1ϕq(γ)|u(s)|ds+[Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ϕq(γ)|u(s)|ds+Γ(2-β)bT1-βλ∫0T‍](t2-t1)≤ϕq(γ)dΓ(α)∫0t1‍|(t2-s)α-1-(t1-s)α-1|ds+ϕq(γ)dΓ(α)∫t1t2‍(t1-s)α-1ds+[Tα-1ϕq(γ)dΓ(2-β)Γ(α-β+1)+Γ(2-β)bT1-βλTα-1ϕq(γ)dΓ(2-β)Γ(α-β+1)](t2-t1),|F′u(t2)-F′u(t1)|≤|1Γ(α-1)∫0t2‍(t2-s)α-2ϕq(f(s,u(s),u′(s)))ds-1Γ(α-1)∫0t1‍(t1-s)α-2ϕq(f(s,u(s),u′(s)))ds|≤1Γ(α-1)∫0t1‍|(t2-s)α-1-(t1-s)α-1|ϕq(γ)|u(s)|ds+1Γ(α-1)∫t1t2‍(t1-s)α-2ϕq(γ)|u(s)|ds≤ϕq(γ)dΓ(α-1)∫0t1‍|(t2-s)α-2-(t1-s)α-2|ds+ϕq(γ)dΓ(α-1)∫t1t2‍(t1-s)α-2ds.
Thus, we conclude that F is equicontinuous on Bd, and
(27)F:Bd⟶Bdiscompletelycontinuous.
By Schauder fixed point theorem we know that there exists a solution for the boundary value problem (5).Theorem 6.
Letf:[0,1]×R×R→R be continuous. Assume that f meets the following condition: there exist l1>0, l2>0 such that
(28)|ϕq(f(t,u,u′))-ϕq(f(t,v,v′))|≤l1|u-v|+l2|u′-v′|,(29)(Tα-1Γ(α)+TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))(l1+l2)<1.
Then the problem (5) has a unique solution on [0,1] for any λ≥0.Proof.
From (18) and (19), we have, for u1, u2∈X,
(30)|Fu2(t)-Fu1(t)|≤1Γ(α)∫0t‍(t-s)α-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+a|(-1)θ+a|Γ(α)×∫0T‍(T-s)α-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+aΓ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+t[∫0T‍Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds∫0T‍]≤1Γ(α)∫0t‍(t-s)α-1[|u2′(s)-u1′(s)||u2′(s)-u1′(s)|l1|u2(s)-u1(s)|+l2|u2′(s)-u1′(s)|]ds+a|(-1)θ+a|Γ(α)×∫0T‍(T-s)α-1[|u2′(s)-u1′(s)|l1|u2(s)-u1(s)|+l2|u2′(s)-u1′(s)|]ds+aΓ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1[|u2′(s)-u1′(s)|l1|u2(s)-u1(s)|+l2|u2′(s)-u1′(s)|]ds+t[Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1[|u2′(s)-u1′(s)|l1|u2(s)-u1(s)|+l2|u2′(s)-u1′(s)|]dsΓ(2-β)T1-βΓ(α-β)]≤([Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ds]1Γ(α)∫0t‍(t-s)α-1ds+aΓ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1ds+a|(-1)θ+a|Γ(α)∫0T‍(T-s)α-1ds+t[Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ds])×(l1+l2)∥u2-u1∥1≤(TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+TαΓ(2-β)Γ(α-β+1))×(l1+l2)∥u2-u1∥1,|F′u2(t)-F′u1(t)|≤1Γ(α-1)∫0t‍(t-s)α-2|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds≤1Γ(α-1)∫0t‍(t-s)α-2[l1|u2(s)-u1(s)|+l2|u2′(s)-u1′(s)|]ds+Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1[l1|u2(s)-u1(s)|+l2|u2′(s)-u1′(s)|]ds≤(1Γ(α-1)∫0t‍(t-s)α-2ds+Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ds)×(l1+l2)∥u2-u1∥1≤(Tα-1Γ(α)+Tα-1Γ(2-β)Γ(α-β+1))(l1+l2)∥u2-u1∥1.
Thus,
(31)∥Fu2-Fu1∥1≤(Tα-1Γ(α)+TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))(l1+l2)∥u2-u1∥1.
It follows from (29) that F is a contraction. Thus, the conclusion of the theorem follows from the contraction mapping principle.Theorem 7.
Let1<p<2. Assume that f meets the following condition: there exist d1>0, d2>0, d>0 such that
(32)|f(t,u,u′)|≤M,|f(t,u,u′)-f(t,v,v′)|≤d1|u-v|+d2|u′-v′|,(33)(Tα-1Γ(α)+TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))(q-1)M(q-2)(d1+d2)<1.
Then the problem (5) has unique solution on [0,1] for
(34)0≤λ≤(aΓ(2-β)|1-a|bT-β+Γ(2-β)bT-β+Γ(2-β)bT1-β)-1×[d-(Tα-1Γ(α)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(α+1)+aTα|1-a|Γ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))ϕq(M)].Proof.
Let(35)Bd={u∣∥u∥1≤d,u∈X},
where
(36)d≥(Tα-1Γ(α)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(α+1)+aTα|1-a|Γ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))ϕq(M)+aΓ(2-β)|1-a|bT-βλ+Γ(2-β)bT-βλ+Γ(2-β)bT1-βλ.
By (18) and (19), we have, for u∈Bd,
(37)|Fu(t)|≤1Γ(α)∫0t‍(t-s)α-1|ϕq(f(s,u(s),u′(s)))|ds+a|(-1)θ+a|Γ(α)×∫0T‍(T-s)α-1|ϕq(f(s,u(s),u′(s)))|ds+aΓ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u(s),u′(s)))|ds+|aΓ(2-β)[1+(-1)θa]bT-βλ|+t[∑Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u(s),u′(s)))|ds+|(-1)θΓ(2-β)bT1-βλ|∑]≤1Γ(α)∫0t‍(t-s)α-1ϕq(M)ds+a|1-a|Γ(α)∫0T‍(T-s)α-1ϕq(M)ds+aΓ(2-β)|1-a|T-βΓ(α-β)∫0T‍(T-s)α-β-1ϕq(M)ds+aΓ(2-β)|1-a|bT-βλ+t[Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(M)ds+Γ(2-β)bT1-βλ]≤(TαΓ(α+1)+aTα|1-a|Γ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))ϕq(M)+aΓ(2-β)|1-a|bT-βλ+Γ(2-β)bT-βλ,|F′u(t)|≤1Γ(α-1)∫0t‍(t-s)α-2ϕq(M)ds+Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1ϕq(M)ds+Γ(2-β)bT1-βλ≤(Tα-1Γ(α)+Tα-1Γ(2-β)Γ(α-β+1))ϕq(M)+Γ(2-β)bT1-βλ.
This, together with (36), yields that
(38)∥Fu(t)∥1≤d.
Hence,
(39)F:Bd⟶Bd.
In view of 1<p<2, we have q>2. Thus, by the following property of p-Laplacian operator:
ifp>2, |x|, |y|≤c, then |φp(x)-φp(y)|≤(p-1)c(p-2)|x-y|; we have, for u∈Bd, (40)|Fu2(t)-Fu1(t)|≤1Γ(α)∫0t‍(t-s)α-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+a|(-1)θ+a|Γ(α)×∫0T‍(T-s)α-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+aΓ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+t[Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds∫0T‍]≤(q-1)M(q-2)Γ(α)×∫0t‍(t-s)α-1|f(s,u2(s),u2′(s))-f(s,u1(s),u1′(s))|ds+a(q-1)M(q-2)|(-1)θ+a|Γ(α)×∫0T‍(T-s)α-1|f(s,u2(s),u2′(s))-f(s,u1(s),u1′(s))|ds+a(q-1)M(q-2)Γ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1|f(s,u2(s),u2′(s))-f(s,u1(s),u1′(s))|ds+t[(q-1)M(q-2)Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|f(s,u2(s),u2′(s))-f(s,u1(s),u1′(s))|ds(q-1)M(q-2)Γ(2-β)T1-βΓ(α-β)]≤(q-1)M(q-2)Γ(α)∫0t‍(t-s)α-1[|u2′(s)-u1′(s)|d1|u2(s)-u1(s)|+d2|u2′(s)-u1′(s)|]ds+a(q-1)M(q-2)|(-1)θ+a|Γ(α)×∫0T‍(T-s)α-1[|u2′(s)-u1′(s)|d1|u2(s)-u1(s)|+d2|u2′(s)-u1′(s)|]ds+a(q-1)M(q-2)Γ(2-β)|(-1)θ+a|T-βΓ(α-β)×∫0T‍(T-s)α-β-1[|u2′(s)-u1′(s)|d1|u2(s)-u1(s)|+d2|u2′(s)-u1′(s)|]ds+t[(q-1)M(q-2)Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1[+d2|u2′(s)-u1′(s)|]dsd1|u2(s)-u1(s)|+d2|u2′(s)-u1′(s)|]ds(q-1)M(q-2)Γ(2-β)T1-βΓ(α-β)]≤(1Γ(α)∫0t‍(t-s)α-1ds+aΓ(2-β)|(-1)θ+a|T-βΓ(α-β)∫0T‍(T-s)α-β-1ds+a|(-1)θ+α|Γ(α)∫0T‍(T-s)α-1ds+t[Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ds])×(q-1)M(q-2)(d1+d2)∥u2-u1∥1≤(TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+TαΓ(2-β)Γ(α-β+1))×(q-1)M(q-2)(d1+d2)∥u2-u1∥1,|F′u2(t)-F′u1(t)|≤1Γ(α-1)×∫0t‍(t-s)α-2|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds+Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|ϕq(f(s,u2(s),u2′(s)))-ϕq(f(s,u1(s),u1′(s)))|ds≤(q-1)M(q-2)Γ(α-1)∫0t‍(t-s)α-2|f(s,u2(s),u2′(s))-f(s,u1(s),u1′(s))|ds+(q-1)M(q-2)Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1|f(s,u2(s),u2′(s))-f(s,u1(s),u1′(s))|ds≤(q-1)M(q-2)Γ(α-1)∫0t‍(t-s)α-2[|u2′(s)-u1′(s)|d1|u2(s)-u1(s)|+d2|u2′(s)-u1′(s)|]ds+(q-1)M(q-2)Γ(2-β)T1-βΓ(α-β)×∫0T‍(T-s)α-β-1[|u2′(s)-u1′(s)|d1|u2(s)-u1(s)|+d2|u2′(s)-u1′(s)|]ds≤(1Γ(α-1)∫0t‍(t-s)α-2ds+Γ(2-β)T1-βΓ(α-β)∫0T‍(T-s)α-β-1ds)×(q-1)M(q-2)(d1+d2)∥u2-u1∥1≤(Tα-1Γ(α)+Tα-1Γ(2-β)Γ(α-β+1))×(q-1)M(q-2)(d1+d2)∥u2-u1∥1.
Thus,
(41)∥Fu2-Fu1∥1≤(Tα-1Γ(α)+TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))×(q-1)M(q-2)(d1+d2)∥u2-u1∥1.
It follows from (33) that F is a contraction. Thus, the conclusion of the theorem follows from the contraction mapping principle.


## 4. Examples

Example 8.
Consider the following boundary value problem:(42)ϕ2(cD0+3/2u(t))=f(t,u(t),u′(t)),0<t<1,u(0)+(-1)θ2u(1)=0,cD0+1/2u(0)+(-1)θ3π2cD0+1/2u(1)=λ,θ=0,1,
where
(43)T=1,α=32,β=12,a=2,b=3π2.
Let
(44)d=2,p=2,f(t,u(t),u′(t))=t100u5+|u|+|u′|.
By computation, we deduce that
(45)0≤λ≤d2(aΓ(2-β)|1-a|bT-β+Γ(2-β)bT-β+Γ(2-β)bT1-β)-1=(2Γ(3/2)(3π/2)+Γ(3/2)(3π/2)+Γ(3/2)(3π/2))-1=(2(π/2)(3π/2)+π/2(3π/2)+π/2(3π/2))-1=34,ϕ2(γ)=γ≤12(Tα-1Γ(α)+TαΓ(α+1)+aTα|1-a|Γ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1)+Tα-1Γ(2-β)Γ(α-β+1))-1=12(1Γ(3/2)+1Γ(5/2)+2Γ(5/2)+2Γ(3/2)Γ(2)+Γ(3/2)Γ(2)+Γ(3/2)Γ(2))-1=12(18+6π3π)-1=3π36+12π<15.
Thus, let γ=1/9; we have
(46)f(t,u(t),u′(t))=t100u5+|u|+|u′|<γϕ2(|u|)=19|u|.
Hence, by Theorem 5, BVP (42) has at least one solution for 0≤λ≤3/4.Example 9.
Consider the following boundary value problem:(47)ϕ2(cD0+3/2u(t))=f(t,u(t),u′(t)),0<t<1,u(0)+(-1)θ2u(1)=0,cD0+1/2u(0)+(-1)θπcD0+1/2u(1)=λ,θ=0,1,
where
(48)T=1,α=32,β=12,a=2,b=π,λ≥0.
Let
(49)p=2,f(t,u(t),u′(t))=1(t+cost+10)2(|u(t)+u′(t)|1+|u(t)+u′(t)|+10(1+sint)).
By computation, we deduce that
(50)|f(t,u(t),u′(t))-f(t,v(t),v′(t))|≤1100(|u(t)-v(t)|+|u′(t)-v′(t)|).
Let
(51)l1=l2=1100.
Thus,
(52)(Tα-1Γ(α)+TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))(l1+l2)=(1Γ(3/2)+1Γ(5/2)+2Γ(3/2)Γ(2)+2Γ(5/2)+Γ(3/2)Γ(2)+Γ(3/2)Γ(2))(1100+1100)=(1(π/2)+1(3π/4)+2(π/2)1+2(3π/4)+(π/2)1+(π/2)1)(1100+1100)=3+π25π<1.
Hence, by Theorem 6, BVP (47) has a unique solution.Example 10.
Consider the following boundary value problem:(53)ϕ4/3(cD0+3/2u(t))=f(t,u(t),u′(t)),0<t<1,u(0)+(-1)θ2u(1)=0,cD0+1/2u(0)+(-1)θπcD0+1/2u(1)=λ,θ=0,1,
where
(54)p=43,T=1,α=32,β=12,a=2,b=π.
Let
(55)f(t,u(t),u′(t))=1(t+cost+10)2(|u(t)+u′(t)|1+|u(t)+u′(t)|+10(1+sint)).
Thus,
(56)q=4,|f(t,u(t),u′(t))|≤1100(1+12)=13100=M,(q-1)M(q-2)=(4-1)M4-2=3(13100)2<1.
By Example 9, we know that
(57)|f(t,u(t),u′(t))-f(t,v(t),v′(t))|≤1100(|u(t)-v(t)|+|u′(t)-v′(t)|).
Let
(58)d1=d2=1100.
It follows from Example 9 that
(59)(Tα-1Γ(α)+TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))(d1+d2)<1.
On the other hand, we have
(60)(aΓ(2-β)|1-a|bT-β+Γ(2-β)bT-β+Γ(2-β)bT1-β)-1=(2Γ(3/2)π+Γ(3/2)π+Γ(3/2)π)-1=12.
Let
(61)d=2×3+π25π.
Thus,
(62)(Tα-1Γ(α)+TαΓ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+aTα|1-a|Γ(α+1)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))(q-1)M(q-2)(d1+d2)<1,0≤λ≤(aΓ(2-β)|1-a|bT-β+Γ(2-β)bT-β+Γ(2-β)bT1-β)-1×[d-(Tα-1Γ(α)+Tα-1Γ(2-β)Γ(α-β+1)+TαΓ(α+1)+aTα|1-a|Γ(α+1)+aTαΓ(2-β)|1-a|Γ(α-β+1)+TαΓ(2-β)Γ(α-β+1))ϕq(M)]=12[2×3+π25π-3+π25π×(13100)3]=3+π50π[2-(13100)3].
Hence, by Theorem 7, BVP (53) has a unique solution for 0≤λ≤((3+π)/50π)[2-(13/100)3].



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*Source: 308024-2013-04-15.xml*