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

Abstract

We discuss the existence of solutions about generalized antiperiodic boundary value problems for the fractional differential equation with p-Laplacian operator 𝜙_𝑝(^𝑐𝐷^𝛼₀₊𝑢(𝑡))=𝑓(𝑡,𝑢(𝑡),𝑢^′(𝑡)), 0<𝑡<𝑇, 1<𝛼≤2, 𝑢(0)+(−1)^𝜃𝑎𝑢(𝑇)=0, ^𝑐𝐷^𝛽₀₊𝑢(0)+(−1)^𝜃𝑏^𝑐𝐷^𝛽₀₊𝑢(𝑇)=𝜆, 0<𝛽<1, where ^𝑐𝐷^𝛼₀₊ is the Caputo fractional derivative, 𝜃=0,1, 𝑎>0, 𝑎≠1, 𝑏>0 and 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, 𝑝>1, 𝜙_𝑝⁻¹=𝜙_𝑞, 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.

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: ^𝑐𝐷^𝑝₀₊𝑢(𝑡)((𝑡𝑢(0)=𝑓𝑡,𝑢)),0<𝑡<1,+𝑢^′(0)=0,𝑢(1)+𝑢^′(1)=0, (1) where 1<𝑝≤2 is a real number, ^𝑐𝐷^𝑞 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: ^𝑐𝐷^𝑞𝑥(𝑡)((𝑡[]𝑥(0)(𝑇)=𝑓𝑡,𝑥)),𝑡∈0,𝑇,𝑇>0,3<𝑞≤4,=−𝑥,𝑥^′(0)=−𝑥^′(𝑇),𝑥^′′(0)=−𝑥^′′(𝑇),𝑥^′′′(0)=−𝑥^′′′(𝑇), (2) where ^𝑐𝐷^𝑞 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: ^𝑐𝐷^𝑞𝑥(𝑡)((𝑡[]𝑥(0)(𝑇)=𝑓𝑡,𝑥)),𝑡∈0,𝑇,𝑇>0,4<𝑞≤5,=−𝑥,𝑥^′(0)=−𝑥^′(𝑇),𝑥^′′(0)=−𝑥^′′(𝑇),𝑥^′′′(0)=−𝑥^′′′(𝑇),𝑥^(𝑖𝑣)(0)=−𝑥^(𝑖𝑣)(𝑇), (3) where ^𝑐𝐷^𝑞 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 ^𝑐𝐷^𝑞𝑥(𝑡)((𝑡[]𝑥(0)(𝑇),=𝑓𝑡,𝑥)),𝑡∈0,𝑇,𝑇>0,1<𝑞≤2,=−𝑥^𝑐𝐷^𝑝₀₊𝑥(0)=−^𝑐𝐷^𝑝₀₊𝑢(𝑇),0<𝑝<1, (4) where ^𝑐𝐷^𝑞 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 𝜙_𝑝^𝑐𝐷^𝛼₀₊𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)𝑢(0)+(),0<𝑡<𝑇,1<𝛼≤2,−1^𝜃𝑎𝑢(𝑇)=0,_𝑐𝐷^𝛽₀₊𝑢(0)+(−1)^𝜃𝑏^𝑐𝐷^𝛽₀₊𝑢(𝑇)=𝜆,0<𝛽<1, (5) where ^𝑐𝐷^𝛼₀₊ is the Caputo fractional derivative, 𝜃=0,1, 𝑎>0, 𝑎≠1, 𝑏>0, and 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, 𝑝>1, 𝜙_𝑝⁻¹=𝜙_𝑞, (1/𝑝)+(1/𝑞)=1.

If we take 𝑎=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 Section 2, 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 order 𝛼 with the lower limit 𝑡₀ for a function 𝑓 is defined as 𝐼^𝛼𝑓(𝑡)=1Γ(𝛼)^𝑡_𝑡0(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠,𝑡>𝑡₀,𝛼>0, (6) where Γ is the gamma function.

Definition 2 (see [21]). Caputo's derivative of order 𝑞 with the lower limit 𝑡₀ for a function 𝑓 can be written as ^𝑐𝐷^𝑞𝑓(𝑡)=1Γ𝑛−𝑞^𝑡_𝑡0(𝑡−𝑠)^{𝑛−𝑞−1}𝑓^(𝑛)(𝑠)𝑑𝑠,𝑡>𝑡₀,𝑞>0,𝑛=𝑞+1. (7)

Lemma 3 (see [22]). Assume that 𝑢∈𝐶(0,1)∩𝐿(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)∩𝐿(0,1). Then 𝐼^𝛼𝑐₀₊𝐷^𝛼₀₊𝑢(𝑡)=𝑢(𝑡)+𝑐₀+𝑐₁𝑡+𝑐₂𝑡+⋯+𝑐_𝑛−1𝑡^𝑛−1, (8) where 𝑐_𝑖∈𝑅, 𝑖=0,1,2,…,𝑛−1, 𝑛=[𝛼]+1.

Lemma 4. Let 𝑦∈𝐶[0,1]. Then the fractional differential equation 𝜙_𝑝^𝑐𝐷^𝛼₀₊𝑢(𝑡(𝑡)𝑢(0)+())=𝑦,0<𝑡<𝑇,1<𝛼≤2,−1^𝜃𝑎𝑢(𝑇)=0,_𝑐𝐷^𝛽₀₊𝑢(0)+(−1)^𝜃𝑏^𝑐𝐷^𝛽₀₊𝑢(𝑇)=𝜆 (9) has a unique solution which is given by 𝑢(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑦(𝑠)−𝑎𝑑𝑠(−1)^𝜃+𝑎Γ(𝛼)^𝑇0(𝑇−𝑠)^𝛼−1𝑦(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑦(𝑠)−𝑑𝑠𝑎Γ2−𝛽1+(−1)^𝜃𝑎𝑏𝑇^−𝛽−Γ+𝑡2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑦(𝑠)+()𝑑𝑠−1^𝜃Γ2−𝛽𝑏𝑇^1−𝛽𝜆. (10)

Proof. From Lemma 3, we have 𝑢(𝑡)=𝐼^𝛼₀₊𝜙_𝑞𝑦(𝑡)+𝑐₀+𝑐₁𝑡=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝑦(𝑠)𝑑𝑠+𝑐₀+𝑐₁𝑡,_𝑐𝐷^𝛽₀₊𝑢(𝑡)=𝐼^{(𝛼−𝛽)}₀₊𝜙_𝑞𝑦(𝑡)+𝑐₁^𝑐𝐷^𝛽₀₊𝑡=1Γ𝛼−𝛽^𝑡₀(𝑡−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑦(𝑠)𝑑𝑠+𝑐₁𝑡^1−𝛽Γ.2−𝛽 (11) Thus, 𝑢(0)=𝑐₀,𝑢(𝑇)=1Γ(𝛼)^𝑇₀(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑦(𝑠)𝑑𝑠+𝑐₀+𝑐₁𝑇,_𝑐𝐷^𝛽₀₊𝑢(0)=0,_𝑐𝐷^𝛽₀₊𝑢(𝑇)=1Γ𝛼−𝛽^𝑇₀(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑦(𝑠)𝑑𝑠+𝑐₁𝑇^1−𝛽Γ.2−𝛽 (12) By ^𝑐𝐷^𝛽₀₊𝑢(0)+(−1)^𝜃𝑏^𝑐𝐷^𝛽₀₊𝑢(𝑇)=𝜆, we have 𝑐₁Γ=−2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑦(𝑠)+()𝑑𝑠−1^𝜃Γ2−𝛽𝑏𝑇^1−𝛽𝜆. (13) Using the boundary condition 𝑢(0)+(−1)^𝜃𝑎𝑢(𝑇)=0 and (13), we obtain 𝑐₀𝑎=−(−1)^𝜃+𝑎Γ(𝛼)^𝑇0(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑦(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑦(𝑠−)𝑑𝑠𝑎Γ2−𝛽1+(−1)^𝜃𝑎𝑏𝑇^−𝛽𝜆. (14) Thus, 𝑢(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝑦(𝑠)−𝑎𝑑𝑠(−1)^𝜃+𝑎Γ(𝛼)^𝑇0(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑦(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ𝛼−𝛽×^𝑇₀(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑦(𝑠)−𝑑𝑠𝑎Γ2−𝛽1+(−1)^𝜃𝑎𝑏𝑇^−𝛽𝜆+𝑡(−1)^𝜃−Γ2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑦(𝑠)+()𝑑𝑠−1^𝜃Γ2−𝛽𝑏𝑇^1−𝛽𝜆. (15)

3. Main Results

Let 𝑋=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 ‖𝑢‖₁=max‖𝑢‖,𝑢^′, (16) where ‖𝑢‖=sup_0≤𝑡≤1|𝑢(𝑡)|,𝑢^′=sup_0≤𝑡≤1𝑢^′(𝑡). (17) Define an operator 𝖥∶𝑋→𝑋 as (𝑡)=1𝖥𝑢Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)−𝑎𝑑𝑠(−1)^𝜃Γ(𝛼)×+𝑎_𝑇0(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)−𝑑𝑠𝑎Γ2−𝛽1+(−1)^𝜃𝑎𝑏𝑇^−𝛽𝜆−Γ+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+()𝑑𝑠−1^𝜃Γ2−𝛽𝑏𝑇^1−𝛽𝜆. (18) From (18), we conclude that 𝖥^′𝑢(𝑡)=1Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)−Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+()𝑑𝑠−1^𝜃Γ2−𝛽𝑏𝑇^1−𝛽𝜆. (19) Then (5) has a solution if and only if the operator 𝖥 has a fixed point.

Theorem 5. Let 𝑓∶[0,𝑇]×𝑅₊×𝑅₊→𝑅₊ be continuous. Assume that 𝑓 meets the following condition: there exist 𝑑≥0, 𝛾≥0 such that 𝑓𝑡,𝑥(𝑡),𝑥^′(𝑡)≤𝛾𝜙_𝑝(|𝑥(𝑡)|),𝑓𝑜𝑟𝑡∈[0,𝑇],‖𝑥‖₁𝜙≤𝑑,_𝑞𝛾≤12𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼||Γ()+1−𝑎𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+𝑇1−𝑎𝛼−𝛽+1^𝛼Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼−1Γ2−𝛽Γ𝛼−𝛽+1⁻¹. (20)(21) Then the problem (5) has at least one solution on [0,𝑇] for 𝑑0≤𝜆≤2𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^1−𝛽−1. (22)

Proof. From 𝑓(𝑡,𝑢(𝑡),𝑢^′(𝑡))∈𝐶([0,1]×𝑅₊×𝑅₊,𝑅₊), we know that 𝑇 is continuous.
Let 𝐵_𝑑=𝑢∣‖𝑢‖₁≤𝑑,𝑢∈𝑋. (23) For 𝑢∈𝐵_𝑑, we have |(𝑡≤1𝖥𝑢)|Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+𝑎𝑑𝑠(−1)^𝜃Γ(𝛼)×+𝑎_𝑇0(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+𝑑𝑠𝑎Γ2−𝛽1+(−1)^𝜃𝑎𝑏𝑇^−𝛽𝜆Γ+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+()𝑑𝑠−1^𝜃Γ2−𝛽𝑏𝑇^1−𝛽𝜆≤1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝛾|𝑢(𝑠+𝑎)|𝑑𝑠|1−𝑎|Γ(𝛼)^𝑇₀(𝑇−𝑠)^𝛼−1𝜙_𝑞𝛾|𝑢(𝑠+)|𝑑𝑠𝑎Γ2−𝛽|1−𝑎|𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝛾|𝑢(𝑠+)|𝑑𝑠𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽𝜆Γ+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝛾|𝑢(𝑠+Γ)|𝑑𝑠2−𝛽𝑏𝑇^1−𝛽𝜆≤𝑇^𝛼𝜙_𝑞𝛾Γ()+𝛼+1𝑎𝑇^𝛼𝜙_𝑞𝛾||Γ()+1−𝑎𝛼+1𝑎𝑇^𝛼Γ2−𝛽𝜙_𝑞𝛾||Γ+𝑇1−𝑎𝛼−𝛽+1_𝛼Γ2−𝛽𝜙_𝑞𝛾Γ‖𝑢‖𝛼−𝛽+1¹+𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽Γ𝜆+2−𝛽𝑏𝑇^−𝛽𝖥𝜆,_′𝑢(𝑡≤1)Γ()×𝛼−1_𝑡0(𝑡−𝑠)^𝛼−2𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+Γ𝑑𝑠2−𝛽𝑏𝑇^1−𝛽𝜆≤1Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝜙_𝑞𝛾|𝑢(𝑠+Γ)|𝑑𝑠2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝛾|𝑢(𝑠+Γ)|𝑑𝑠2−𝛽𝑏𝑇^1−𝛽𝜆≤𝑇^𝛼−1𝜙_𝑞𝛾Γ(𝛼)+𝑇^𝛼−1Γ2−𝛽𝜙_𝑞𝛾Γ𝑢‖𝛼−𝛽+1‖₁+Γ2−𝛽𝑏𝑇^1−𝛽𝜆. (24) This, together with (21) and (22), yields that ‖𝖥𝑢(𝑡)‖₁≤𝑑. (25) Hence, 𝖥(𝐵_𝑑) is uniformly bounded.
Next we show that 𝖥 is equicontinuous.
For any 0≤𝑡₁≤𝑡₂≤1, 𝑢∈𝐵_𝑑, we have 𝖥𝑢𝑡₂−𝖥𝑢𝑡₁≤1Γ(𝛼)^𝑡2₀𝑡₂−𝑠^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)−1𝑑𝑠Γ(𝛼)^𝑡1₀𝑡₁−𝑠^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+Γ𝑑𝑠2−𝛽𝑏𝑇^1−𝛽𝜆𝑡₂−𝑡₁≤1Γ(𝛼)^𝑡1₀𝑡₂−𝑠^𝛼−1−𝑡₁−𝑠^𝛼−1𝜙_𝑞𝛾|𝑢(𝑠+1)|𝑑𝑠Γ(𝛼)^𝑡2_𝑡1𝑡₁−𝑠^𝛼−1𝜙_𝑞𝛾|𝑢(𝑠+Γ)|𝑑𝑠2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝛾𝑢(𝑠+Γ|)|𝑑𝑠2−𝛽𝑏𝑇^1−𝛽𝜆𝑡₂−𝑡₁≤𝜙_𝑞𝛾𝑑Γ(𝛼)^𝑡1₀𝑡₂−𝑠^𝛼−1−𝑡₁−𝑠^𝛼−1+𝜙𝑑𝑠_𝑞𝛾𝑑Γ(𝛼)^𝑡2_𝑡1𝑡₁−𝑠^𝛼−1+𝑇𝑑𝑠_𝛼−1𝜙_𝑞𝛾𝑑Γ2−𝛽Γ+Γ𝛼−𝛽+12−𝛽𝑏𝑇^1−𝛽𝜆𝑡₂−𝑡₁,𝖥^′𝑢𝑡₂−𝖥^′𝑢𝑡₁≤1Γ(𝛼−1)^𝑡2₀𝑡₂−𝑠^𝛼−2𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠−1)𝑑𝑠Γ(𝛼−1)^𝑡1₀𝑡₁−𝑠^𝛼−2𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)≤1𝑑𝑠Γ(𝛼−1)^𝑡1₀𝑡₂−𝑠^𝛼−1−𝑡₁−𝑠^𝛼−1𝜙_𝑞𝛾|𝑢(𝑠+1)|𝑑𝑠Γ(𝛼−1)^𝑡2_𝑡1𝑡₁−𝑠^𝛼−2𝜙_𝑞𝛾|𝑢(𝑠≤𝜙)|𝑑𝑠_𝑞𝛾𝑑Γ(𝛼−1)^𝑡1₀𝑡₂−𝑠^𝛼−2−𝑡₁−𝑠^𝛼−2+𝜙𝑑𝑠_𝑞𝛾𝑑Γ(𝛼−1)^𝑡2_𝑡1𝑡₁−𝑠^𝛼−2𝑑𝑠. (26) Thus, we conclude that 𝖥 is equicontinuous on 𝐵_𝑑, and 𝖥∶𝐵_𝑑⟶𝐵_𝑑iscompletelycontinuous. (27) By Schauder fixed point theorem we know that there exists a solution for the boundary value problem (5).

Theorem 6. Let 𝑓∶[0,1]×𝑅×𝑅→𝑅 be continuous. Assume that 𝑓 meets the following condition: there exist 𝑙₁>0, 𝑙₂>0 such that 𝜙_𝑞𝑓𝑡,𝑢,𝑢^′−𝜙_𝑞𝑓𝑡,𝑣,𝑣^′≤𝑙₁|𝑢−𝑣|+𝑙₂𝑢^′−𝑣^′𝑇_𝛼−1,Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝑙𝛼−𝛽+1¹+𝑙₂<1. (28)(29) 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, 𝖥𝑢₂(𝑡)−𝖥𝑢₁(𝑡≤1)Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+𝑎𝑑𝑠(−1)^𝜃Γ(𝛼)×+𝑎_𝑇0(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓s,𝑢₁(𝑠),𝑢^′₁(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)Γ𝑑𝑠+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)≤1𝑑𝑠Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑙₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑙₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠+𝑎)𝑑𝑠(−1)^𝜃Γ(𝛼)×+𝑎_𝑇0(𝑇−𝑠)^𝛼−1𝑙₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑙₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠+)𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑙₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑙₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠Γ)𝑑𝑠+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑙₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑙₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠≤1)𝑑𝑠Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}+𝑎𝑑𝑠(−1)^𝜃+𝑎Γ(𝛼)^𝑇0(𝑇−𝑠)^𝛼−1Γ𝑑𝑠+𝑡2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}×𝑙𝑑𝑠₁+𝑙₂𝑢₂−𝑢₁₁≤𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼Γ2−𝛽Γ×𝑙𝛼−𝛽+1₁+𝑙₂𝑢₂−𝑢₁₁,𝖥^′𝑢₂(𝑡)−𝖥^′𝑢₁(𝑡≤1)Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)≤1𝑑𝑠Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝑙₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑙₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠+Γ)𝑑𝑠2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑙₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑙₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠≤1)𝑑𝑠Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2+Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}×𝑙𝑑𝑠₁+𝑙₂𝑢₂−𝑢₁₁≤𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼−1Γ2−𝛽Γ𝑙𝛼−𝛽+1¹+𝑙₂𝑢₂−𝑢₁₁. (30) Thus, 𝖥𝑢₂−𝖥𝑢₁₁≤𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝑙𝛼−𝛽+1¹+𝑙₂𝑢₂−𝑢₁₁. (31) It follows from (29) that 𝖥 is a contraction. Thus, the conclusion of the theorem follows from the contraction mapping principle.

Theorem 7. Let 1<𝑝<2. Assume that 𝑓 meets the following condition: there exist 𝑑₁>0, 𝑑₂>0, 𝑑>0 such that 𝑓𝑡,𝑢,𝑢^′𝑓≤𝑀,𝑡,𝑢,𝑢^′−𝑓𝑡,𝑣,𝑣^′≤𝑑₁|𝑢−𝑣|+𝑑₂𝑢^′−𝑣^′,𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝑀𝛼−𝛽+1𝑞−1^(𝑞−2)𝑑₁+𝑑₂<1. (32)(33) Then the problem (5) has unique solution on [0,1] for 0≤𝜆≤𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^1−𝛽−1×𝑇𝑑−^𝛼−1Γ(𝛼)+𝑇^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼Γ()+𝛼+1𝑎𝑇^𝛼||Γ()+1−𝑎𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+𝑇1−𝑎𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝜙𝛼−𝛽+1^𝑞(𝑀). (34)

Proof. Let 𝐵_𝑑=𝑢∣‖𝑢‖₁≤𝑑,𝑢∈𝑋, (35) where 𝑇𝑑≥^𝛼−1Γ(𝛼)+𝑇^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1Γ(^𝛼)+𝛼+1𝑎𝑇^𝛼||Γ()+1−𝑎𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+𝑇1−𝑎𝛼−𝛽+1^𝛼Γ2−𝛽Γ𝜙𝛼−𝛽+1^𝑞(𝑀)+𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽Γ𝜆+2−𝛽𝑏𝑇^−𝛽Γ𝜆+2−𝛽𝑏𝑇^1−𝛽𝜆. (36) By (18) and (19), we have, for 𝑢∈𝐵_𝑑, |(𝑡≤1𝖥𝑢)|Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+𝑎𝑑𝑠(−1)^𝜃Γ(𝛼)×+𝑎_𝑇0(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+𝑑𝑠𝑎Γ2−𝛽1+(−1)^𝜃𝑎𝑏𝑇^−𝛽𝜆Γ+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢(𝑠),𝑢^′(𝑠)+()𝑑𝑠−1^𝜃Γ2−𝛽𝑏𝑇^1−𝛽𝜆≤1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞(𝑀)+𝑎𝑑𝑠|1−𝑎|Γ(𝛼)^𝑇₀(𝑇−𝑠)^𝛼−1𝜙_𝑞(𝑀)+𝑑𝑠𝑎Γ2−𝛽|1−𝑎|𝑇^−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞(𝑀)+𝑑𝑠𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽𝜆Γ+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞(𝑀)Γ𝑑𝑠+2−𝛽𝑏𝑇^1−𝛽𝜆≤𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼||Γ()+1−𝑎𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+𝑇1−𝑎𝛼−𝛽+1^𝛼Γ2−𝛽Γ𝜙𝛼−𝛽+1^𝑞(𝑀)+𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽Γ𝜆+2−𝛽𝑏𝑇^−𝛽𝖥𝜆,_′𝑢(𝑡≤1)Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝜙_𝑞(𝑀)+Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ𝛼−𝛽×^𝑇₀(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞(𝑀)Γ𝑑𝑠+2−𝛽𝑏𝑇^1−𝛽𝜆≤𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼−1Γ2−𝛽Γ𝜙𝛼−𝛽+1^𝑞(𝑀)+Γ2−𝛽𝑏𝑇^1−𝛽𝜆. (37) This, together with (36), yields that ‖𝖥𝑢(𝑡)‖₁≤𝑑. (38) Hence, 𝖥∶𝐵_𝑑⟶𝐵_𝑑. (39) In view of 1<𝑝<2, we have 𝑞>2. Thus, by the following property of p-Laplacian operator:
if 𝑝>2, |𝑥|, |𝑦|≤𝑐, then |𝜑_𝑝(𝑥)−𝜑_𝑝(𝑦)|≤(𝑝−1)𝑐^(𝑝−2)|𝑥−𝑦|; we have, for 𝑢∈𝐵_𝑑, 𝖥𝑢₂(𝑡)−𝖥𝑢₁(𝑡≤1)Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+𝑎𝑑𝑠(−1)^𝜃Γ(𝛼)×+𝑎_𝑇0(𝑇−𝑠)^𝛼−1𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)Γ𝑑𝑠+𝑡2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_T0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)≤𝑀𝑑𝑠𝑞−1^(𝑞−2)Γ(𝛼)×^𝑡₀(𝑡−𝑠)^𝛼−1𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+𝑎𝑀𝑑𝑠𝑞−1^(𝑞−2)(−1)^𝜃Γ(𝛼)×+𝑎_𝑇0(𝑇−𝑠)^𝛼−1𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+𝑎𝑀𝑑𝑠𝑞−1^(𝑞−2)Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)𝑀𝑑𝑠+𝑡𝑞−1^(𝑞−2)Γ2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)≤𝑀𝑑𝑠𝑞−1^(𝑞−2)Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑑₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑑₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠+𝑎𝑀)𝑑𝑠𝑞−1^(𝑞−2)(−1)^𝜃+𝑎Γ(𝛼)×^𝑇₀(𝑇−𝑠)^𝛼−1𝑑₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑑₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠+𝑎𝑀)𝑑𝑠𝑞−1^(𝑞−2)Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑑₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑑₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠𝑀)𝑑𝑠+𝑡𝑞−1^(𝑞−2)Γ2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑑₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑑₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠≤1)𝑑𝑠Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1+𝑑𝑠𝑎Γ2−𝛽(−1)^𝜃+𝑎𝑇^−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}+𝑎𝑑𝑠(−1)^𝜃+𝛼Γ(𝛼)^𝑇0(𝑇−𝑠)^𝛼−1Γ𝑑𝑠+𝑡2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}×𝑀𝑑𝑠𝑞−1^(𝑞−2)𝑑₁+𝑑₂𝑢₂−𝑢₁₁≤𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼Γ2−𝛽Γ×𝑀𝛼−𝛽+1𝑞−1^(𝑞−2)𝑑₁+𝑑₂𝑢₂−𝑢₁₁,𝖥^′𝑢₂(𝑡)−𝖥^′𝑢₁(𝑡≤1)Γ()×𝛼−1_𝑡0(𝑡−𝑠)^𝛼−2𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝜙_𝑞𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝜙_𝑞𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)≤𝑀𝑑𝑠𝑞−1^(𝑞−2)Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)+𝑀𝑑𝑠𝑞−1^(𝑞−2)Γ2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑓𝑠,𝑢₂(𝑠),𝑢^′₂(𝑠)−𝑓𝑠,𝑢₁(𝑠),𝑢^′₁(𝑠)≤𝑀𝑑𝑠𝑞−1^(𝑞−2)Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝑑₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑑₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠+𝑀)𝑑𝑠𝑞−1^(𝑞−2)Γ2−𝛽𝑇^1−𝛽Γ×𝛼−𝛽_𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}𝑑₁𝑢₂(𝑠)−𝑢₁(𝑠)+𝑑₂𝑢^′₂(𝑠)−𝑢^′₁(𝑠≤1)𝑑𝑠Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2+Γ𝑑𝑠2−𝛽𝑇^1−𝛽Γ𝛼−𝛽^𝑇0(𝑇−𝑠)^{𝛼−𝛽−1}×𝑀𝑑𝑠𝑞−1^(𝑞−2)𝑑₁+𝑑₂𝑢₂−𝑢₁₁≤𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼−1Γ2−𝛽Γ×𝑀𝛼−𝛽+1𝑞−1^(𝑞−2)𝑑₁+𝑑₂𝑢₂−𝑢₁₁. (40) Thus, 𝖥𝑢₂−𝖥𝑢₁₁≤𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1^𝛼Γ2−𝛽Γ×𝑀𝛼−𝛽+1𝑞−1^(𝑞−2)𝑑₁+𝑑₂𝑢₂−𝑢₁₁. (41) 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: 𝜙₂^𝑐𝐷^3/2₀₊𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)𝑢(0)+(),0<𝑡<1,−1^𝜃2𝑢(1)=0,_𝑐𝐷^1/2₀₊𝑢(0)+(−1)^𝜃3√𝜋2^𝑐𝐷^1/2₀₊𝑢(1)=𝜆,𝜃=0,1, (42) where 3𝑇=1,𝛼=21,𝛽=2,3√𝑎=2,𝑏=𝜋2. (43) Let 𝑓(𝑡)𝑑=2,𝑝=2,𝑡,𝑢,𝑢^′(𝑡)=𝑡𝑢1005+|𝑢|+𝑢^′. (44) By computation, we deduce that 𝑑0≤𝜆≤2𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^1−𝛽−1=2Γ(3/2)3√+Γ()𝜋/23/23√+Γ()𝜋/23/23√𝜋/2⁻¹=2√𝜋/23√+√𝜋/2𝜋/23√+√𝜋/2𝜋/23√𝜋/2⁻¹=34,𝜙₂𝛾1=𝛾≤2𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼||Γ()+1−𝑎𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+𝑇1−𝑎𝛼−𝛽+1_𝛼Γ(2−𝛽)+𝑇Γ(𝛼−𝛽+1)^𝛼−1Γ(2−𝛽)Γ(𝛼−𝛽+1)⁻¹=121Γ()+13/2Γ()+25/2Γ()+()5/22Γ3/2Γ(2)+Γ(3/2)Γ(2)+Γ(3/2)Γ(2)⁻¹=1218+6𝜋3√𝜋⁻¹=3√𝜋<136+12𝜋5. (45) Thus, let 𝛾=1/9; we have 𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡=𝑡)𝑢1005+|𝑢|+𝑢^′<𝛾𝜙₂𝑢=1(||)9|𝑢|. (46) Hence, by Theorem 5, BVP (42) has at least one solution for 0≤𝜆≤3/4.

Example 9. Consider the following boundary value problem: 𝜙₂^𝑐𝐷^3/2₀₊𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)𝑢(0)+(),0<𝑡<1,−1^𝜃2𝑢(1)=0,_𝑐𝐷^1/2₀₊𝑢(0)+(−1)^𝜃√𝜋^𝑐𝐷^1/2₀₊𝑢(1)=𝜆,𝜃=0,1, (47) where 3𝑇=1,𝛼=21,𝛽=2,𝑎=2,𝑏=√𝜋,𝜆≥0. (48) Let 𝑓(𝑡)𝑝=2,𝑡,𝑢,𝑢^′(𝑡=1)(𝑡+cos𝑡+10)²𝑢(𝑡)+𝑢^′(𝑡)1+𝑢(𝑡)+𝑢^′(𝑡().)+101+sin𝑡 (49) By computation, we deduce that 𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)−𝑓𝑡,𝑣(𝑡),𝑣^′(𝑡)≤1|𝑢(𝑡)(𝑡+𝑢100−𝑣)|^′(𝑡)−𝑣^′(𝑡). (50) Let 𝑙₁=𝑙₂=1.100 (51) Thus, 𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝑙𝛼−𝛽+1¹+𝑙₂=1Γ()+13/2Γ()+()5/22Γ3/2Γ(2)+2Γ()+Γ()5/23/2Γ(2)+Γ(3/2)Γ(2)1+1100=1100√+1𝜋/23√+2√𝜋/4𝜋/21+23√+√𝜋/4𝜋/21+√𝜋/211+1100=1003+𝜋25√𝜋<1. (52) Hence, by Theorem 6, BVP (47) has a unique solution.

Example 10. Consider the following boundary value problem: 𝜙_4/3^𝑐𝐷^3/2₀₊𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)𝑢(0)+(),0<𝑡<1,−1^𝜃2𝑢(1)=0,_𝑐𝐷^1/2₀₊𝑢(0)+(−1)^𝜃√𝜋^𝑐𝐷^1/2₀₊𝑢(1)=𝜆,𝜃=0,1, (53) where 4𝑝=33,𝑇=1,𝛼=2,1𝛽=2,𝑎=2,𝑏=√𝜋. (54) Let 𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)=1(𝑡+cos𝑡+10)²𝑢(𝑡)+𝑢^′(𝑡)1+𝑢(𝑡)+𝑢^′(𝑡().)+101+sin𝑡 (55) Thus, 𝑓(𝑡)𝑞=4,𝑡,𝑢,𝑢^′(𝑡)≤1()=1001+1213𝑀100=𝑀,𝑞−1^(𝑞−2)=(4−1)𝑀⁴⁻²=313100²<1. (56) By Example 9, we know that 𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)−𝑓𝑡,𝑣(𝑡),𝑣^′(𝑡)≤1|𝑢(𝑡)(𝑡+𝑢100−𝑣)|^′(𝑡)−𝑣^′(𝑡). (57) Let 𝑑₁=𝑑₂=1.100 (58) It follows from Example 9 that 𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝑑𝛼−𝛽+1¹+𝑑₂<1. (59) On the other hand, we have 𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^1−𝛽−1=2Γ(3/2)√𝜋+Γ(3/2)√𝜋+Γ(3/2)√𝜋⁻¹=12. (60) Let 𝑑=2×3+𝜋25√𝜋. (61) Thus, 𝑇^𝛼−1Γ(𝛼)+𝑇^𝛼Γ()+𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+1−𝑎𝛼−𝛽+1𝑎𝑇^𝛼||Γ()+𝑇1−𝑎𝛼+1^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝑀𝛼−𝛽+1𝑞−1^(𝑞−2)𝑑₁+𝑑₂≤<1,0≤𝜆𝑎Γ2−𝛽|1−𝑎|𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^−𝛽+Γ2−𝛽𝑏𝑇^1−𝛽−1×𝑇𝑑−^𝛼−1Γ(𝛼)+𝑇^𝛼−1Γ2−𝛽Γ+𝑇𝛼−𝛽+1Γ(^𝛼)+𝛼+1𝑎𝑇^𝛼||Γ()+1−𝑎𝛼+1𝑎𝑇^𝛼Γ2−𝛽||Γ+𝑇1−𝑎𝛼−𝛽+1_𝛼Γ2−𝛽Γ𝜙𝛼−𝛽+1^𝑞(𝑀=1)22×3+𝜋25√𝜋−3+𝜋25√𝜋×13100³=3+𝜋50√𝜋2−13100³. (62) Hence, by Theorem 7, BVP (53) has a unique solution for 0≤𝜆≤((3+𝜋)/50√𝜋)[2−(13/100)³].

Acknowledgments

This research is supported by Henan Province College Youth Backbone Teacher Funds (2011GGJS-213) and the National Natural Science Foundation of China (11271336).