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Journal of Applied Mathematics / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 841349 | https://doi.org/10.1155/2012/841349

Gang Wang, Wenbin Liu, Jinyun Yang, Sinian Zhu, Ting Zheng, "The Existence of Solutions for a Fractional 2𝑚-Point Boundary Value Problems", Journal of Applied Mathematics, vol. 2012, Article ID 841349, 18 pages, 2012. https://doi.org/10.1155/2012/841349

The Existence of Solutions for a Fractional 2𝑚-Point Boundary Value Problems

Gang Wang,1 Wenbin Liu,1 Jinyun Yang,2 Sinian Zhu,1 and Ting Zheng1

1Department of Mathematics, China University of Mining and Technology, Xuzhou 221008, China

2School of Mathematics and Physical Sciences, Xuzhou Institute of Technology, Xuzhou 221008, China

Academic Editor: Yongkun Li
Received24 Jun 2011
Revised09 Oct 2011
Accepted11 Oct 2011
Published26 Jan 2012

Abstract

By using the coincidence degree theory, we consider the following 2𝑚-point boundary value problem for fractional differential equation 𝐷𝛼0+𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡),𝐷𝛼−10+𝑢(𝑡),𝐷𝛼−20+𝑢(𝑡))+𝑒(𝑡), 0<𝑡<1, 𝐼3−𝛼0+𝑢(𝑡)|𝑡=0=0,𝐷𝛼−20+∑𝑢(1)=𝑚−2𝑖=1𝑎𝑖𝐷𝛼−20+𝑢(𝜉𝑖∑),𝑢(1)=𝑚−2𝑖=1𝑏𝑖𝑢(𝜂𝑖), where 2<𝛼≤3,𝐷𝛼0+ and 𝐼𝛼0+ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively. A new result on the existence of solutions for above fractional boundary value problem is obtained.

1. Introduction

Fractional differential equations have been of great interest recently. This is because of the intensive development of the theory of fractional calculus itself as well as its applications. Apart from diverse areas of mathematics, fractional differential equations arise in a variety of different areas such as rheology, fluid flows, electrical networks, viscoelasticity, chemical physics, and many other branches of science (see [1–4] and references cited therein). The research of fractional differential equations on boundary value problems, as one of the focal topics has attained a great deal of attention from many researchers (see [5–13]).

However, there are few papers which consider the boundary value problem at resonance for nonlinear ordinary differential equations of fractional order. In [14], Hu and Liu studied the following BVP of fractional equation at resonance: 𝐷𝛼0+𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝑥(𝑡),𝑥(𝑡),0≤𝑡≤1,𝑥(0)=𝑥(1),𝑥(0)=𝑥(0)=0,(1.1) where 1<𝛼≤2, 𝐷𝛼0+ is the standard Caputo fractional derivative.

In [15], Zhang and Bai investigated the nonlinear nonlocal problem 𝐷𝛼0+𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),0<𝑡<1,𝑢(0)=0,𝛽𝑢(𝜂)=𝑢(1),(1.2) where 1<𝛼≤2, they consider the case 𝛽𝜂𝛼−1=1, that is, the resonance case.

In [16], Bai investigated the boundary value problem at resonance 𝐷𝛼0+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼−10+𝐼𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,2−𝛼0+𝑢(𝑡)|𝑡=0=0,𝐷𝛼−10+𝑢(1)=𝑚−2𝑖=0𝛽𝑖𝐷𝛼−10+𝑢𝜂𝑖(1.3) is considered, where 1<𝛼≤2 is a real number, 𝐷𝛼0+and𝐼𝛼0+ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and 𝑓∶[0,1]×𝑅2→𝑅 is continuous and 𝑒(𝑡)∈𝐿1[0,1],𝑚≥2,0<𝜉𝑖<1,𝛽𝑖∈𝑅,𝑖=1,2,…,𝑚−2 are given constants such that ∑𝑚−2𝑖=1𝛽𝑖=1.

In this paper, we study the 2𝑚-point boundary value problem𝐷𝛼0+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼−10+𝑢(𝑡),𝐷𝛼−20+𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,(1.4)𝐼3−𝛼0+𝑢(𝑡)|𝑡=0=0,𝐷𝛼−20+𝑢(1)=𝑚−2𝑖=1𝑎𝑖𝐷𝛼−20+𝑢𝜉𝑖,𝑢(1)=𝑚−2𝑖=1𝑏𝑖𝑢𝜂𝑖,(1.5) where 2<𝛼≤3, 𝑚≥2,0<𝜉1<⋯<𝜉𝑚<1,0<𝜂1<⋯<𝜂𝑚<1,𝑎𝑖,𝑏𝑖∈𝑅,𝑓∶[0,1]×𝑅3→𝑅, 𝑓 satisfies Carathéodory conditions, 𝐷𝛼0+ and 𝐼𝛼0+ are the standard Riemann-Liouville fractional derivative and fractional integral, respectively.

Setting Λ1=1𝛼(𝛼+1)1−𝑚−2𝑖=1𝑎𝑖𝜉𝑖𝛼+1,Λ2=1𝛼(𝛼−1)1−𝑚−2𝑖=1𝑎𝑖𝜉𝛼𝑖,Λ3=(Γ(𝛼))2Γ(2𝛼)1−𝑚−2𝑖=1𝑏𝑖𝜂𝑖2𝛼−1,Λ4=Γ(𝛼)Γ(𝛼−1)Γ(2𝛼−1)1−𝑚−2𝑖=1𝑏𝑖𝜂𝑖2𝛼−2.(1.6)

In this paper, we will always suppose that the following conditions hold:(C1): 𝑚−2𝑖=1𝑎𝑖𝜉𝑖=𝑚−2𝑖=1𝑎𝑖=1,𝑚−2𝑖=1𝑏𝑖𝜂𝑖𝛼−1=𝑚−2𝑖=1𝑏𝑖𝜂𝑖𝛼−2=1,(1.7)(C2): Λ=Λ1Λ4−Λ2Λ3≠0.(1.8)

We say that boundary value problem (1.4) and (1.5) is at resonance, if BVP 𝐷𝛼0+𝐼𝑢(𝑡)=0,3−𝛼0+𝑢(𝑡)|𝑡=0=0,D𝛼−20+𝑢(1)=𝑚−2𝑖=1𝑎𝑖𝐷𝛼−20+𝑢𝜉𝑖,𝑢(1)=𝑚−2𝑖=1𝑏𝑖𝑢𝜂𝑖(1.9) has 𝑢(𝑡)=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2, 𝑎,𝑏∈𝑅 as a nontrivial solution.

The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions for BVP (1.4)-(1.5) under nonlinear growth restriction of 𝑓, basing on the coincidence degree theory due to Mawhin (see [17]).

Now, we will briefly recall some notation and an abstract existence result.

Let 𝑌,𝑍 be real Banach spaces, 𝐿∶dom𝐿⊂𝑌→𝑍 a Fredholm map of index zero,s and 𝑃∶𝑌→𝑌,𝑄∶𝑍→𝑍 continuous projectors such that 𝑌=Ker𝐿⊕Ker𝑃,𝑍=Im𝐿⊕𝑄,Im𝑃=Ker𝐿,Ker𝑄=Im𝐿.(1.10) It follows that 𝐿|dom𝐿∩Ker𝑃∶dom𝐿∩Ker𝑃→Im𝐿 is invertible. We denote the inverse of the map by 𝐾𝑝. If Ω is an open-bounded subset of 𝑌 such that dom𝐿∩Ω≠∅, the map 𝑁∶𝑌→𝑍 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾𝑝(𝐼−𝑄)𝑁∶Ω→𝑌 is compact.

The lemma that we used is [17, Theorem 2.4].

Lemma 1.1. Let 𝐿 be a Fredholm operator of index zero and let 𝑁 be L-compact on Ω. Assume that the following conditions are satisfied: (i)𝐿𝑥≠𝜆𝑁𝑥, for all (𝑥,𝜆)∈[dom𝐿⧵Ker𝐿∩𝜕Ω]×[0,1]; (ii)𝑁𝑥∉Im𝐿, for all 𝑥∈Ker𝐿∩𝜕Ω; (iii)deg(𝐽𝑄𝑁|Ker𝐿,Ker𝐿∩Ω,0)≠0,where 𝑄∶𝑍→𝑍 is a projection as above with Ker𝑄=Im𝐿 and 𝐽∶Im𝑄→Ker𝐿 is any isomorphism. Then the equation 𝐿𝑥=𝑁𝑥 has at least one solution in dom𝐿∩Ω.

2. Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory. These definitions can be found in the recent literature [1–16, 18].

Definition 2.1. The fractional integral of order 𝛼>0 of a function 𝑦∶(0,∞)→𝑅 is given by 𝐼𝛼0+1𝑦(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1,𝑦(𝑠)𝑑𝑠(2.1) provided the right side is pointwise defined on (0,∞), where Γ(⋅) is the Gamma function.

Definition 2.2. The fractional derivative of order 𝛼>0 of a function 𝑦∶(0,∞)→𝑅 is given by 𝐷𝛼0+1𝑦(𝑡)=𝑑Γ(𝑛−𝛼)𝑑𝑡𝑛𝑡0𝑦(𝑠)(𝑡−𝑠)𝛼−𝑛+1𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, provided the right side is pointwise defined on (0,∞).

Definition 2.3. We say that the map 𝑓∶[0,1]×𝑅𝑛→𝑅 satisfies Carathéodory conditions with respect to 𝐿1[0,1] if the following conditions are satisfied: (i)for each 𝑧∈𝑅𝑛, the mapping 𝑡→𝑓(𝑡,𝑧) is Lebesgue measurable; (ii)for almost every 𝑡∈[0,1], the mapping 𝑡→𝑓(𝑡,𝑧) is continuous on 𝑅𝑛; (iii) for each 𝑟>0, there exists 𝜌𝑟∈𝐿1([0,1],𝑅) such that for a.e. 𝑡∈[0,1] and every |𝑧|≤𝑟, we have 𝑓(𝑡,𝑧)≤𝜌𝑟(𝑡).

Lemma 2.4 (see [15]). Assume that 𝑢∈𝐶(0,1)∩𝐿1(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)∩𝐿1(0,1). Then 𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝑐1𝑡𝛼−1+𝑐1𝑡𝛼−2+⋯+𝑐𝑁𝑡𝛼−𝑁(2.3) for some 𝑐𝑖∈𝑅,𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

We use the classical Banach space 𝐶[0,1] with the norm ‖𝑥‖∞=max[]𝑡∈0,1||||,𝑥(𝑡)(2.4)𝐿[0,1] with the norm ‖𝑥‖1=10||||𝑥(𝑡)𝑑𝑡.(2.5)

Definition 2.5. For 𝑛∈𝑁, we denote by 𝐴𝐶𝑛[0,1] the space of functions 𝑢(𝑡) which have continuous derivatives up to order 𝑛−1 on [0,1] such that 𝑢(𝑛−1)(𝑡) is absolutely continuous: 𝐴𝐶𝑛[0,1]={𝑢∣[0,1]→𝑅 and (𝐷(𝑛−1))𝑢(𝑡) is absolutely continuous in [0,1]}.

Lemma 2.6 (see [15]). Given 𝜇>0 and 𝑁=[𝜇]+1 we can define a linear space 𝐶𝜇[]=0,1𝑢(𝑡)∣𝑢(𝑡)=𝐼𝛼0+𝑥(𝑡)+𝑐1𝑡𝜇−1+𝑐2𝑡𝜇−2+⋯+𝑐𝑁𝑡𝜇−(𝑁−1)[],,𝑡∈0,1(2.6) where 𝑥∈[0,1],𝑐𝑖∈𝑅,𝑖=1,2,…,𝑁−1. By means of the linear functional analysis theory, we can prove that with the ‖𝑢‖𝐶𝜇=‖‖𝐷𝜇0+𝑢‖‖∞‖‖𝐷+⋯+0𝜇−(𝑁−1)+𝑢‖‖∞+‖𝑢‖∞,(2.7)𝐶𝜇[0,1] is a Banach space.

Remark 2.7. If 𝜇 is a natural number, then 𝐶𝜇[0,1] is in accordance with the classical Banach space 𝐶𝑛[0,1].

Lemma 2.8 (see [15]). 𝑓⊂𝐶𝜇[0,1] is a sequentially compact set if and only if 𝑓 is uniformly bounded and equicontinuous. Here uniformly bounded means there exists 𝑀>0, such that for every 𝑢∈𝑓‖𝑢‖𝐶𝜇=‖‖𝐷𝜇0+𝑢‖‖∞‖‖𝐷+⋯+𝜇−(𝑁−1)0+𝑢‖‖∞+‖𝑢‖∞<𝑀,(2.8) and equicontinuous means that ∀𝜀>0, ∃𝛿>0, such that ||𝑢𝑡1𝑡−𝑢2||<𝜀,∀𝑡1,𝑡2∈[],||𝑡0,11−𝑡2||,||𝐷<𝛿,∀𝑢∈𝑓𝛼−𝑖0+𝑢𝑡1−𝐷𝛼−𝑖0+𝑢𝑡2||<𝜀,∀𝑡1,𝑡2∈[],||𝑡0,11−𝑡2||.<𝛿,∀𝑢∈𝑓,∀𝑖=1,2,…,𝑁−1(2.9)

Lemma 2.9 (see [1]). Let 𝛼>0,𝑛=[𝛼]+1. Assume that 𝑢∈𝐿1(0,1) with a fractional integration of order 𝑛−𝛼 that belongs to 𝐴𝐶𝑛[0,1]. Then the equality 𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)=𝑢(𝑡)−𝑛𝑖=1𝐼𝑛−𝛼0+𝑢(𝑡)𝑛−𝑖|𝑡=0𝑡Γ(𝛼−𝑖+1)𝛼−𝑖(2.10) holds almost everywhere on [0,1].

Definition 2.10 (see [16]). Let 𝐼𝛼0+(𝐿1(0,1)),𝛼>0 denote the space of functions 𝑢(𝑡), represented by fractional integral of order 𝛼 of a summable function: 𝑢=𝐼𝛼0+𝑣,𝑣∈𝐿1(0,1).

Let 𝑍=𝐿1[0,1], with the norm ∫‖𝑦‖=10|𝑦(𝑠)|𝑑𝑠, 𝑌=𝐶𝛼−1[0,1] defined by Lemma 2.6, with the norm ‖𝑢‖𝐶𝛼−1=‖𝐷0𝛼−1+𝑢‖∞+‖𝐷0𝛼−2+𝑢‖∞+‖𝑢‖∞, where 𝑌 is a Banach space.

Define 𝐿 to be the linear operator from dom𝐿⊂𝑌 to 𝑍 with dom𝐿=𝑢∈𝐶𝛼−1[]0,1∣𝐷𝛼0+𝑢∈𝐿1[]0,1,𝑢satisfies,(1.5)(2.11)𝐿𝑢=𝐷𝛼0+𝑢,𝑢∈dom𝐿,(2.12) we define 𝑁∶𝑌→𝑍 by setting 𝑁𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼−10+𝑢(𝑡)𝐷𝛼−20+𝑢(𝑡)+𝑒(𝑡).(2.13) Then boundary value problem (1.4) and (1.5) can be written as 𝐿𝑢=𝑁𝑢.

3. Main Results

Lemma 3.1. Let L be defined by (2.12), then Ker𝐿=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2∣𝑎,𝑏∈𝑅≅𝑅2,Im𝐿=𝑦∈𝑍∣10(1−𝑠)𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1𝑎𝑖𝜉𝑖0𝜉𝑖−𝑠𝑦(𝑠)𝑑𝑠=0,10(1−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1b𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1.𝑦(𝑠)𝑑𝑠=0(3.1)

Proof. In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that 𝐼𝛼0+=𝐷−𝛼0+ for 𝛼<0.
Let 𝐿𝑢=𝐷𝛼0+𝑢, by Lemma 2.9, 𝐷𝛼0+𝑢(𝑡)=0 has solution 𝑢(𝑡)=3𝑖=1𝐼3−𝛼0+𝑢(𝑡)3−𝑖|𝑡=0Γ𝑡(𝛼−𝑖+1)𝛼−𝑖=𝐼3−𝛼0+𝑢(𝑡)|𝑡=0𝑡Γ(𝛼)𝛼−1+𝐼3−𝛼0+𝑢(𝑡)|𝑡=0𝑡Γ(𝛼−1)𝛼−2+𝐼3−𝛼0+𝑢|(𝑡)𝑡=0𝑡Γ(𝛼−2)𝛼−3=𝐷𝛼−10+𝑢(𝑡)|𝑡=0𝑡Γ(𝛼)𝛼−1+𝐷𝛼−20+𝑢(𝑡)|𝑡=0𝑡Γ(𝛼−1)𝛼−2+𝐼3−𝛼0+𝑢|(𝑡)𝑡=0𝑡Γ(𝛼−2)𝛼−3.(3.2) Combine with (1.5), So, Ker𝐿=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2∣𝑎,𝑏∈𝑅≅𝑅2.(3.3) Let 𝑦∈𝑍 and let 𝑢𝑡=1Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠+𝑐1𝑡𝛼−1+𝑐2𝑡𝛼−2+𝑐3𝑡𝛼−3.(3.4)
Then 𝐷𝛼0+𝑢(𝑡)=𝑦(𝑡) a.e. 𝑡∈[0,1] and, if 10(1−𝑠)𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1𝑎𝑖𝜉𝑖0𝜉𝑖−𝑠𝑦(𝑠)𝑑𝑠=0,10(1−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1𝑏𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1𝑦(𝑠)𝑑𝑠=0(3.5) hold, then 𝑢(𝑡) satisfies the boundary conditions (1.5). That is, 𝑢∈dom𝐿 and we have {𝑦∈𝑍∣𝑦satisfies(3.4)}⊆Im𝐿.(3.6)
Let 𝑢∈dom𝐿. Then for 𝐷𝛼0+𝑢∈Im𝐿, we have 𝐼𝛼0+𝑦(𝑡)=𝑢(𝑡)−𝑐1𝑡𝛼−1−𝑐2𝑡𝛼−2−𝑐3𝑡𝛼−3,(3.7) where 𝑐1=𝐷𝛼−10+𝑢(𝑡)|𝑡=0Γ(𝛼),𝑐2=𝐷𝛼−20+𝑢(𝑡)|𝑡=0,Γ(𝛼−1)c3=𝐼3−𝛼0+𝑢(𝑡)|𝑡=0,Γ(𝛼−2)(3.8) which, due to the boundary value condition (1.5), implies that satisfies (3.5). In fact, from 𝐼3−𝛼0+𝑢(𝑡)|𝑡=0=0 we have 𝑐3=0, from 𝐷𝛼−20+∑𝑢(1)=𝑚−2𝑖=1𝑎𝑖𝐷𝛼−20+𝑢(𝜉𝑖), ∑𝑢(1)=𝑚−2𝑖=1𝑏𝑖𝑢(𝜂𝑖), we have 10(1−𝑠)𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1𝑎𝑖𝜉𝑖0𝜉𝑖−𝑠𝑦(𝑠)𝑑𝑠=0,10(1−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1𝑏𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1𝑦(𝑠)𝑑𝑠=0.(3.9) Hence, {𝑦∈𝑍∣𝑦satisfies(3.4)}⊇Im𝐿.(3.10) Therefore, {𝑦∈𝑍∣𝑦satisfies(3.4)}=Im𝐿.(3.11) The proof is complete.

Lemma 3.2. The mapping 𝐿∶dom𝐿⊂𝑌→𝑍 is a Fredholm operator of index zero, and 𝑇𝑄𝑦(𝑡)=1𝑡𝑦(𝑡)𝛼−1+𝑇2𝑡𝑦(𝑡)𝛼−2,(3.12) where 𝑇11𝑦=ΛΛ4𝑄1𝑦−Λ2𝑄2𝑦,𝑇21𝑦=ΛΛ3𝑄1𝑦−Λ1𝑄2𝑦,(3.13) define by 𝐾𝑝∶Im𝐿→dom𝐿∩Ker𝑃 by 𝐾𝑝1𝑦(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠=𝐼𝛼0+𝑦(𝑡),𝑦∈Im𝐿,(3.14) and forall𝑦∈Im𝐿,‖𝐾𝑝𝑦‖𝐶𝛼−1≤((1/Γ(𝛼))+2)‖𝑦‖1.

Proof. Consider the continuous linear mapping 𝑄1∶𝑍→𝑍 and 𝑄2∶𝑍→𝑍 defined by 𝑄1𝑦=10(1−𝑠)𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1𝑎𝑖𝜉𝑖0𝜉𝑖𝑄−𝑠𝑦(𝑠)𝑑𝑠,2𝑦=10(1−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠−𝑚−2𝑖=1𝑏𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1𝑦(𝑠)𝑑𝑠.(3.15) Using the above definitions, we construct the following auxiliary maps 𝑇1∶𝑍→𝑍 and 𝑇2∶𝑍→𝑍: 𝑇11𝑦=ΛΛ4𝑄1𝑦−Λ2𝑄2𝑦,𝑇21𝑦=ΛΛ3𝑄1𝑦−Λ1Q2𝑦.(3.16) Since the condition (C2) holds, the mapping defined by 𝑇𝑄𝑦(𝑡)=1𝑡𝑦(𝑡)𝛼−1+𝑇2𝑡𝑦(𝑡)𝛼−2(3.17) is well defined.
Recall (C2) and note that 𝑇1𝑇1𝑦𝑡𝛼−1=1ΛΛ4𝑄1𝑇1𝑦𝑡𝛼−1−Λ2𝑄2𝑇1𝑦𝑡𝛼−1=1ΛΛ4Λ4Λ1Λ𝑄1Λ𝑦−1Λ2Λ𝑄2𝑦−Λ2Λ4Λ3Λ𝑄1Λ𝑦−2Λ3Λ𝑄2𝑦=𝑇1𝑦,(3.18) and similarly we can derive that 𝑇1𝑇2𝑦𝑡𝛼−2=0,𝑇2𝑇1𝑦𝑡𝛼−1=0,𝑇2𝑇2𝑦𝑡𝛼−2=𝑇2𝑦.(3.19) So, for 𝑦∈𝑍, it follows from the four relations above that 𝑄2𝑇𝑦=𝑄1𝑦𝑡𝛼−1+𝑇2𝑦𝑡𝛼−2=𝑇1𝑇1𝑦𝑡𝛼−1+𝑇2𝑦𝑡𝛼−2𝑡𝛼−1+𝑇2𝑇1𝑦𝑡𝛼−1+𝑇2𝑦𝑡𝛼−2𝑡𝛼−2=𝑇1𝑦𝑡𝛼−1+𝑇2𝑦𝑡𝛼−2=𝑄𝑦,(3.20) that is, the map 𝑄 is idempotent. In fact 𝑄 is a continuous linear projector.
Note that 𝑦∈Im𝐿 implies 𝑄𝑦=0. Conversely, if 𝑄𝑦=0, so Λ4𝑄1𝑦−Λ2𝑄2Λ𝑦=0,1𝑄2𝑦−Λ3𝑄1𝑦=0,(3.21) but ||||||Λ4−Λ2−Λ3Λ1||||||=Λ4Λ1−Λ2Λ3≠0,(3.22) then we must have 𝑄1𝑦=𝑄2𝑦=0; since the condition (C2) holds, this can only be the case if 𝑄1𝑦=𝑄2𝑦=0, that is, 𝑦∈Im𝐿. In fact Ker𝑄=Im𝐿, take 𝑦∈𝑍 in the form 𝑦=(𝑦−𝑄𝑦)+𝑄𝑦 so that 𝑦−𝑄𝑦∈Ker𝑄=Im𝐿,𝑄𝑦∈Im𝑄, thus, 𝑍=Im𝐿+Im𝑄, Let 𝑦∈Im𝐿∩Im𝑄 and assume that 𝑦=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2 is not identically zero on [0,1]. Then, since 𝑦∈Im𝐿, from (3.5) and the condition (C2), we have 𝑄1𝑦=10(1−𝑠)𝑎𝑠𝛼−1+𝑏𝑠𝛼−2𝑑𝑠−𝑚−2𝑖=1𝑎𝑖𝜉𝑖0𝜉𝑖−𝑠𝑎𝑠𝛼−1+𝑏𝑠𝛼−2𝑄𝑑𝑠=0,2𝑦=10(1−𝑠)𝛼−1𝑎𝑠𝛼−1+𝑏𝑠𝛼−2𝑑𝑠−𝑚−2𝑖=1𝑏𝑖𝜂𝑖0𝜂𝑖−𝑠𝛼−1𝑎𝑠𝛼−1+𝑏𝑠𝛼−2𝑑𝑠=0.(3.23) So 𝑎Λ1+𝑏Λ2=0,𝑎Λ3+𝑏Λ4=0,(3.24) but ||||||Λ1Λ2Λ3Λ4||||||=Λ1Λ4−Λ2Λ3≠0,(3.25) we derive 𝑎=𝑏=0, which is a contradiction. Hence, Im𝐿∩Im𝑄={0}; thus 𝑍=Im𝐿⊕Im𝑄.
Now, dimKer𝐿=2=codimIm𝐿 and so 𝐿 is a Fredholm operator of index zero.
Let 𝑃∶𝑌→𝑌 be defined by 1𝑃𝑢(𝑡)=Γ𝐷(𝛼)𝛼−10+𝑢(𝑡)|𝑡=0𝑡𝛼−1+1Γ𝐷(𝛼−1)𝛼−20+𝑢(𝑡)|𝑡=0𝑡𝛼−2[].,𝑡∈0,1(3.26) Note that 𝑃 is a continuous linear projector and Ker𝑃=𝑢∈𝑌∣𝐷𝛼−10+𝑢(0)=𝐷𝛼−20+.𝑢(0)=0(3.27) It is clear that 𝑌=Ker𝑃⊕Ker𝐿.
Note that the projectors 𝑃 and 𝑄 are exact. Define by 𝐾𝑝∶Im𝐿→dom𝐿∩Ker𝑃 by 𝐾𝑝1𝑦(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1𝑦(𝑠)𝑑𝑠=𝐼𝛼0+𝑦(𝑡),𝑦∈Im𝐿.(3.28) Hence we have 𝐷𝛼−10+𝐾𝑝𝑦𝑡=𝑡0𝑦(𝑠)𝑑𝑠,𝐷𝛼−20+𝐾𝑝𝑦𝑡=𝑡0(𝑡−𝑠)𝑦(𝑠)𝑑𝑠,(3.29) then ‖‖𝐾𝑝𝑦‖‖∞≤1Γ(𝛼)‖𝑦‖1,‖‖𝐷𝛼−10+𝐾𝑝𝑦‖‖∞≤‖𝑦‖1,‖‖𝐷𝛼−20+𝐾𝑝𝑦‖‖∞≤‖𝑦‖1,(3.30) and thus ‖‖𝐾𝑝𝑦‖‖𝐶𝛼−1≤1Γ(𝛼)+2‖𝑦‖1.(3.31)
In fact, if 𝑦∈Im𝐿, then 𝐿𝐾𝑝𝑦(𝑡)=𝐷𝛼0+𝐼𝛼0+𝑦(𝑡)=𝑦(𝑡).(3.32) Also, if 𝑢∈dom𝐿∩Ker𝑃, then 𝐾𝑝𝐿𝑢(𝑡)=𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝑐1𝑡𝛼−1+𝑐2𝑡𝛼−2+𝑐3𝑡𝛼−3,(3.33) where 𝑐1=𝐷𝛼−10+𝑢(𝑡)|𝑡=0Γ(𝛼),𝑐2=𝐷𝛼−20+𝑢(𝑡)|𝑡=0Γ(𝛼−1),𝑐3=𝐼3−𝛼0+𝑢(𝑡)|𝑡=0,Γ(𝛼−2)(3.34) and from the boundary value condition (1.5) and the fact that 𝑢∈dom𝐿∩Ker𝑃, 𝑃𝑢=0, 𝐷𝛼−10+𝑢(𝑡)|𝑡=0=𝐷𝛼−20+𝑢(𝑡)|𝑡=0=𝐼3−𝛼0+𝑢(𝑡)|𝑡=0=0, we have 𝑐1=𝑐2=𝑐3=0, thus 𝐾𝑝𝐿𝑢(𝑡)=𝑢(𝑡).(3.35) This shows that 𝐾𝑝=[𝐿|dom𝐿∩Ker𝑃]−1. The proof is complete. Using (3.16), we write 𝑇𝑄𝑁𝑢(𝑡)=1𝑡𝑁𝑢𝛼−1+𝑇2𝑡𝑁𝑢𝛼−2,𝐾𝑝1(𝐼−𝑄)𝑁𝑢(𝑡)=Γ(𝛼)𝑡0(𝑡−𝑠)𝛼−1[]𝑁𝑢(𝑠)−𝑄𝑁𝑢(𝑠)𝑑𝑠.(3.36)

By Lemma 2.8 and a standard method, we obtain the following lemma.

Lemma 3.3 (see [16]). For every given 𝑒∈𝐿1[0,1],𝐾𝑝(𝐼−𝑄)𝑁∶𝑌→𝑌 is completely continuous.

Assume that the following conditions on the function 𝑓(𝑡,𝑥,𝑦,𝑧) are satisfied.(H1)There exist functions 𝑎(𝑡),𝑏(𝑡),𝑐(𝑡),𝑑(𝑡),𝑟(𝑡)∈𝐿1[0,1], and a constant 𝜃∈[0,1) such that for all (𝑥,𝑦,𝑧)∈𝑅3, 𝑡∈[0,1], one of the following inequalities is satisfied:||||||𝑦||𝑓(𝑡,𝑥,𝑦,𝑧)≤𝑎(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑥|𝜃+𝑟(𝑡),(3.37)||||||𝑦||||𝑦||𝑓(𝑡,𝑥,𝑦,𝑧)≤𝑎(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)𝜃+𝑟(𝑡),(3.38)||||||𝑦||𝑓(𝑡,𝑥,𝑦,𝑧)≤𝑎(𝑡)|𝑥|+𝑏(𝑡)+𝑐(𝑡)|𝑧|+𝑑(𝑡)|𝑧|𝜃+𝑟(𝑡).(3.39)(H2)There exists a constant 𝐴>0, such that for 𝑥∈dom𝐿⧵Ker𝐿 satisfying |𝐷𝛼−10+𝑥(𝑡)|+|𝐷𝛼−20+𝑥(𝑡)|>𝐴 for all 𝑡∈[0,1], we have 𝑄1𝑁𝑥(𝑡)≠0,or𝑄2𝑁𝑥(𝑡)≠0.(3.40)(H3)There exists a constant 𝐵>0 such that for every 𝑎,𝑏∈𝑅 satisfying 𝑎2+𝑏2>𝐵 then either 𝑎𝑇1𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2+𝑏𝑇2𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2<0,(3.41)

or 𝑎𝑇1𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2+𝑏𝑇2𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2>0.(3.42)

Remark 3.4. 𝑇1𝑁(𝑎𝑡𝛼−1+𝑏𝑡𝛼−2) and 𝑇2𝑁(𝑎𝑡𝛼−1+𝑏𝑡𝛼−2) from (H3) stand for the images of 𝑢(𝑡)=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2 under the maps 𝑇1𝑁 and 𝑇2𝑁, respectively.

Lemma 3.5. Suppose (H1)-(H2) hold, then the set Ω1={𝑥∈dom𝐿⧵Ker[]}𝐿∶𝐿𝑥=𝜆𝑁𝑥,𝜆∈0,1(3.43) is bounded.

Proof. Take Ω1[]={𝑥∈dom𝐿⧵Ker𝐿∶𝐿𝑥=𝜆𝑁𝑥,𝜆∈0,1}.(3.44) Then for 𝑥∈Ω1, 𝐿𝑥=𝜆𝑁𝑥 thus 𝜆≠0,𝑁𝑥∈Im𝐿=Ker𝑄, and hence 𝑄𝑁𝑥(𝑡) for all 𝑡∈[0,1]. By the definition of 𝑄, we have 𝑄1𝑁𝑥(𝑡)=𝑄2𝑁𝑥(𝑡)=0. It follows from (H2) that there exists 𝑡0∈[0,1], such that |𝐷𝛼−10+𝑢(𝑡0)|+|𝐷𝛼−20+𝑢(𝑡0)|≤𝐴.
Now 𝐷𝛼−10+𝑥(𝑡)=𝐷𝛼−10+𝑥𝑡0+𝑡𝑡0𝐷𝛼0+𝐷𝑥(𝑠)𝑑𝑠,𝛼−20+𝑥(𝑡)=𝐷𝛼−20+𝑥𝑡0+𝑡𝑡0𝐷𝛼−10+𝑥(𝑠)𝑑𝑠,(3.45) and so ||𝐷𝛼−10+||≤‖‖𝐷𝑥(0)𝛼−10+‖‖𝑥(𝑡)∞≤||𝐷𝛼−10+𝑥𝑡0||+‖‖𝐷𝛼−10+𝑥‖‖1≤𝐴+‖𝐿𝑥‖1≤𝐴+‖𝑁𝑥‖1,||𝐷𝛼−20+𝑥||≤‖‖𝐷(0)𝛼−20+𝑥‖‖(𝑡)∞≤||𝐷𝛼−20+𝑥𝑡0||+‖‖𝐷𝛼−10+𝑥‖‖∞≤||𝐷𝛼−20+𝑥𝑡0||+||𝐷𝛼−10+𝑥𝑡0||+‖‖𝐷𝛼0+𝑥‖‖1≤𝐴+‖𝐿𝑥‖1≤𝐴+‖𝑁𝑥‖1.(3.46) Therefore, we have ‖𝑃𝑥‖𝐶𝛼−1=‖‖‖1𝐷Γ(𝛼)𝛼−10+𝑥(0)𝑡𝛼−1+1𝐷Γ(𝛼−1)𝛼−20+𝑥(0)𝑡𝛼−2‖‖‖𝐶𝛼−1=‖‖‖1𝐷Γ(𝛼)𝛼−10+𝑥(0)𝑡𝛼−1+1𝐷Γ(𝛼−1)𝛼−20+𝑥(0)𝑡𝛼−2‖‖‖∞+‖‖𝐷𝛼−10+‖‖𝑥(0)∞+‖‖𝐷𝛼−10+𝑥(0)𝑡+𝐷𝛼−20+‖‖𝑥(0)∞≤12+||𝐷Γ(𝛼)𝛼−10+||+1𝑥(0)1+||𝐷Γ(𝛼−1)𝛼−20+||≤1𝑥(0)2+Γ(𝛼)𝐴+‖𝑁𝑥‖1+11+Γ(𝛼−1)𝐴+‖𝑁𝑥‖1.(3.47) Note that (𝐼−𝑃)𝑥∈dom𝐿∩Ker𝑃 for all 𝑥∈Ω1. Then, by Lemma 3.2, we have ‖‖(𝐼−𝑃)𝑥𝐶𝛼−1=‖‖𝐾𝑝‖‖𝐿(𝐼−𝑃)𝑥𝐶𝛼−1≤12+Γ(𝛼)‖𝑁𝑥‖1,(3.48) so, we have ‖𝑥‖𝐶𝛼−1≤‖(𝐼−𝑃)𝑥‖𝐶𝛼−1+‖𝑃𝑥‖𝐶𝛼−1≤12+Γ(𝛼)𝐴+‖𝑁𝑥‖1+11+Γ(𝛼−1)𝐴+‖𝑁𝑥‖1+12+Γ(𝛼)‖𝑁𝑥‖1=2+1Γ(𝛼)Γ(𝛼−1)+5‖𝑁𝑥‖1+1+1Γ(𝛼)+1Γ(𝛼−1)𝐴Γ(𝛼−2)+3≤𝑚‖𝑁𝑥‖1+𝑛𝐴,(3.49) where 𝑚=((2/Γ(𝛼))+(1/Γ(𝛼−1))+5),𝑛=((1/Γ(𝛼))+(1/Γ(𝛼−1))+(1/(Γ(𝛼−2))+3),𝐴 is a constant. This is for all 𝑥∈Ω1. If the first condition of (H1) is satisfied, then, we have ‖𝑥‖𝐶𝛼−1=max‖𝑥‖∞,‖‖𝐷𝛼−10+𝑥‖‖∞,‖‖𝐷𝛼−20+𝑥‖‖∞≤𝑚‖𝑎‖1‖𝑥‖∞+‖𝑏‖1‖‖𝐷𝛼−10+𝑥‖‖∞+‖𝑐‖1‖‖𝐷𝛼−20+𝑥‖‖∞+‖𝑑‖1‖‖𝐷𝛼−20+𝑥‖‖𝜃∞,+𝐷(3.50) where 𝐷=‖𝑟‖1+‖𝑒‖1+𝑛/𝑚, and consequently, for ‖𝑥‖∞≤‖𝑥‖𝐶𝛼−1,‖‖𝐷𝛼−10+𝑥‖‖∞≤‖𝑥‖𝐶𝛼−1,‖‖𝐷𝛼−20+𝑥‖‖∞≤‖𝑥‖𝐶𝛼−1,(3.51) so ‖𝑥‖∞≤𝑚1−𝑚‖𝑎‖1‖𝑏‖1‖‖𝐷𝛼−10+𝑥‖‖∞+‖𝑐‖1‖‖𝐷𝛼−20+𝑥‖‖∞+‖𝑑‖1‖‖𝐷𝛼−20+𝑥‖‖𝜃∞,‖‖𝐷+𝐷𝛼−10+𝑥‖‖∞≤𝑚1−𝑚‖𝑎‖1−𝑚‖𝑏‖1‖𝑐‖1‖‖𝐷𝛼−20+𝑥‖‖∞+‖𝑑‖1‖‖𝐷𝛼−20+𝑥‖‖𝜃∞,‖‖𝐷+𝐷𝛼−20+𝑥‖‖∞≤𝑚1−𝑚‖𝑎‖1−𝑚‖𝑏‖1−𝑚‖𝑐‖1‖𝑑‖1‖‖𝐷𝛼−20+𝑥‖‖𝜃∞.+𝐷(3.52) But 𝜃∈[0,1) and ‖𝑎‖1+‖𝑏‖1+‖𝑐‖1≤1/𝑚, so there exists 𝐴1,𝐴2,𝐴3>0 such that ‖‖𝐷𝛼−20+𝑥‖‖∞≤𝐴1,‖‖𝐷𝛼−10+𝑥‖‖∞≤𝐴2,‖𝑥‖∞≤𝐴3.(3.53) Therefore, for all 𝑥∈Ω1, ‖𝑥‖𝐶𝛼−1=max‖𝑥‖∞,‖‖𝐷𝛼−10+𝑥‖‖∞,‖‖𝐷𝛼−20+𝑥‖‖∞𝐴≤max1,𝐴2,𝐴3,(3.54) we can prove that Ω1 is also bounded.
If (3.38) or (3.39) holds, similar to the above argument, we can prove that Ω1 is bounded too.

Lemma 3.6. Suppose (H3) holds, then the set Ω2={𝑥∈Ker𝐿∶𝑁𝑥∈Im𝐿}(3.55) is bounded.

Proof. Let Ω2={𝑥∈Ker𝐿∶𝑁𝑥∈Im𝐿},(3.56) for 𝑥∈Ω2,𝑥∈Ker𝐿={𝑥∈dom𝐿∶𝑥=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2,𝑎,𝑏∈𝑅,𝑡∈[0,1]} and 𝑄𝑁𝑥(𝑡)=0; thus 𝑇1𝑁(𝑎𝑡𝛼−1+𝑏𝑡𝛼−2)=𝑇2𝑁(𝑎𝑡𝛼−1+𝑏𝑡𝛼−2)=0. By (H3), 𝑎2+𝑏2≤𝐵, that is, Ω2 is bounded.

Lemma 3.7. Suppose (H3) holds, then the set Ω3={𝑥∈Ker[]}𝐿∶−𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈0,1(3.57) is bounded.

Proof. We define the isomorphism 𝐽∶Ker𝐿→Im𝑄 by 𝐽𝑎𝑡𝛼−1+𝑏𝑡𝛼−2=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2.(3.58)
If the first part of (H3) is satisfied, let Ω3[]={𝑥∈Ker𝐿∶−𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈0,1}.(3.59) For every 𝑥=𝑎𝑡𝛼−1+𝑏𝑡𝛼−2∈Ω3, 𝜆𝑎𝑡𝛼−1+𝑏𝑡𝛼−2𝑇=(1−𝜆)1𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2𝑡𝛼−1+𝑇2𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2𝑡𝛼−1.(3.60) If 𝜆=1, then 𝑎=𝑏=0, and if 𝑎2+𝑏2>𝐵, then by (H3) 𝜆𝑎2+𝑏2=(1−𝜆)𝑎𝑇1𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2+𝑏𝑇2𝑁𝑎𝑡𝛼−1+𝑏𝑡𝛼−2<0,(3.61) which, in either case, is a contradiction. If the other part of (H3) is satisfied, then we take Ω3[]={𝑥∈Ker𝐿∶𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈0,1},(3.62) and, again, obtain a contradiction. Thus, in either case ‖𝑥‖𝐶𝛼−1=‖‖𝑎𝑡𝛼−1+𝑏𝑡𝛼−2‖‖𝐶𝛼−1=‖‖𝑎𝑡𝛼−1+𝑏𝑡𝛼−2‖‖∞+‖𝑎Γ(𝛼)‖∞+‖𝑎Γ(𝛼)𝑡+𝑏Γ(𝛼−1)‖∞≤||𝑏||(1+2Γ(𝛼))|𝑎|+(1+Γ(𝛼−1))≤(2+2Γ(𝛼)+Γ(𝛼−1))𝐵,(3.63) for all 𝑥∈Ω3, that is, Ω3 is bounded.

Remark 3.8. Suppose the second part of (H3) holds, then the set Ω3[]}={𝑥∈Ker𝐿∶𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥=0,𝜆∈0,1(3.64) is bounded.

Theorem 3.9. If (C1)-(C2) and (H1)–(H3) hold, then the boundary value problem (1.4)-(1.5) has at least one solution.

Proof. Set Ω to be a bounded open set of 𝑌 such that ∪3𝑖=1Ω⊂Ω. It follows from Lemmas 3.2 and 3.3 that 𝐿 is a Fredholm operator of index zero, and the operator 𝐾𝑝(𝐼−𝑄)𝑁∶Ω→𝑌 is compact 𝑁, thus, is L-compact on Ω. By Lemmas 3.5 and 3.6, we get that the following two conditions are satisfied: (i)𝐿𝑥≠𝜆𝑁𝑥 for every (𝑥,𝜆)∈[dom𝐿⧵Ker𝐿∩𝜕Ω]×[0,1]; (ii)𝑁𝑥∉Im𝐿, for every 𝑥∈Ker𝐿∩𝜕Ω.
Finally, we will prove that (iii) of Lemma 1.1 is satisfied. Let 𝐻(𝑥,𝜆)=±𝜆𝐽𝑥+(1−𝜆)𝑄𝑁𝑥, where 𝐼 is the identity operator in the Banach space 𝑌. According to Lemma 3.7 (or Remark 3.8), we know that 𝐻(𝑥,𝜆)≠0, for all 𝑥∈𝜕Ω∩Ker𝐿, and thus, by the homotopy property of degree, deg𝑄𝑁|Ker𝐿⎡⎢⎢⎢⎣±||||||Λ,Ker𝐿∩Ω,0=deg(𝐻(⋅,0),Ker𝐿∩Ω,0)=deg(𝐻(⋅,1),Ker𝐿∩Ω,0)=deg(±𝐼,Ker𝐿∩Ω,0)=sgn4Λ−Λ2Λ−Λ3ΛΛ1Λ||||||⎤⎥⎥⎥⎦±Λ=sgn1Λ4−Λ2Λ3Λ=±1≠0.(3.65) Then by Lemma 1.1, 𝐿𝑥=𝑁𝑥 has at least one solution in dom𝐿∩Ω, so the boundary value problem (1.4) and (1.5) has at least one solution in the space 𝐶𝛼−1[0,1]. The proof is finished.

4. An Example

Let us consider the following boundary value problem:𝐷5/20+𝑢(𝑡)=𝑓𝑡,𝑢(𝑡),𝐷𝛼−10+𝑢(𝑡),𝐷𝛼−20+𝐼𝑢(𝑡)+𝑒(𝑡),0<𝑡<1,3−𝛼0+𝑢(𝑡)|𝑡=0=0,𝐷1/20+𝑢(1)=2𝐷1/20+𝑢23−𝐷1/20+𝑢13,𝑢(1)=645𝑢14−815𝑢19,(4.1) where 𝑓𝑡,𝑢(𝑡),𝐷𝛼−10+𝑢(𝑡),𝐷𝛼−20+=1𝑢(𝑡)140sin(𝑢(𝑡))+𝐷203/20+1𝑢(𝑡)+𝐷201/20+𝐷𝑢(𝑡)+5cos1/20+𝑢(𝑡)1/5.(4.2) Corresponding to the problem (1.4)-(1.5), we have that 𝑒(𝑡)=1+3sin2𝑡, 𝛼=5/2, 𝑎1=−1, 𝑎2=2, 𝜉1=1/3, 𝜉2=2/3, 𝑏1=64/5, 𝑏2=−81/5, 𝜂1=1/4, 𝜂2=1/9 and 1𝑓(𝑡,𝑥,𝑦,𝑧)=140sin𝑥+120𝑦+20𝑧+5cos(𝑧)1/5,(4.3) then there is 𝑎1+𝑎2=1,𝑎1𝜉1+𝑎2𝜉2𝑏=1,1𝜂13/2+𝑏2𝜂23/2=1,𝑏1𝜂11/2+𝑏2𝜂21/2Λ=1,1=4351−2𝑖=1𝑎𝑖𝜉𝑖7/2,Λ2=4151−2𝑖=1𝑎𝑖𝜉𝑖5/2,Λ3=(Γ(5/2))2241−2𝑖=1𝑏𝑖𝜂4𝑖,Λ4=16Γ(5/2)Γ(3/2)241−2𝑖=1𝑏𝑖𝜂3𝑖,Λ=Λ1Λ4−Λ2Λ3||||≤1≠0,𝑓(𝑡,𝑥,𝑦,𝑧)|140𝑥|+||𝑦||+120|20𝑧|+5|𝑧|1/5.(4.4) Again, taking 𝑎=1/40, 𝑏=𝑐=1/20, then ‖𝑎‖1+‖𝑏‖1+‖𝑐‖11=1/8,𝑚=1(2/Γ(𝛼