International Journal of Differential Equations

International Journal of Differential Equations / 2011 / Article
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Fractional Differential Equations 2011

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Research Article | Open Access

Volume 2011 |Article ID 401803 | 14 pages | https://doi.org/10.1155/2011/401803

The Existence of Solutions for a Nonlinear Fractional Multi-Point Boundary Value Problem at Resonance

Academic Editor: Nikolai Leonenko
Received16 May 2011
Accepted16 Jun 2011
Published18 Aug 2011

Abstract

We discuss the existence of solution for a multipoint boundary value problem of fractional differential equation. An existence result is obtained with the use of the coincidence degree theory.

1. Introduction

In this paper, we study the multipoint boundary value problem š·š›¼0+ī€·š‘¢(š‘”)=š‘“š‘”,š‘¢(š‘”),š·š›¼āˆ’10+š‘¢(š‘”),š·š›¼āˆ’20+ī€øš¼š‘¢(š‘”)+š‘’(š‘”),0<š‘”<1,(1.1)3āˆ’š›¼0+š‘¢(0)=0,š·š›¼āˆ’20+š‘¢(0)=š‘›ī“š‘—=1š›½š‘—š·š›¼āˆ’20+š‘¢ī€·šœ‰š‘—ī€ø,š‘¢(1)=š‘šī“š‘–=1š›¼š‘–š‘¢ī€·šœ‚š‘–ī€ø,(1.2) where 2<š›¼ā‰¤3,0<šœ‰1<šœ‰2<ā‹Æ<šœ‰š‘›<1,š‘›ā‰„1,0<šœ‚1<ā‹Æ<šœ‚š‘š<1,š‘šā‰„2,š›¼š‘–,š›½š‘—āˆˆā„,š‘šī“š‘–=1š›¼š‘–šœ‚š‘–š›¼āˆ’1=š‘šī“š‘–=1š›¼š‘–šœ‚š‘–š›¼āˆ’2=1,š‘›ī“š‘—=1š›½š‘—šœ‰š‘—=0,š‘›ī“š‘—=1š›½š‘—=1,(1.3)š‘“āˆ¶[0,1]Ɨā„3ā†’ā„ satisfying the CarathĆ©odory conditions, š‘’āˆˆšæ1[0,1]. š·š›¼0+ and š¼š›¼0+ are the standard Riemann-Liouville derivative and integral, respectively. We assume, in addition, that š‘…=Ī“(š›¼)2Ī“(š›¼āˆ’1)Ī“(2š›¼)Ī“(š›¼+1)š‘›ī“š‘—=1š›½š‘—šœ‰š›¼š‘—īƒ©1āˆ’š‘šī“š‘–=1š›¼š‘–šœ‚š‘–2š›¼āˆ’1īƒŖāˆ’Ī“(š›¼)2Ī“(š›¼āˆ’1)Ī“(š›¼+2)Ī“(2š›¼āˆ’1)š‘›ī“š‘—=1š›½š‘—šœ‰š‘—š›¼+1īƒ©1āˆ’š‘šī“š‘–=1š›¼š‘–šœ‚š‘–2š›¼āˆ’2īƒŖā‰ 0,(1.4) where Ī“ is the Gamma function. Due to condition (1.3), the fractional differential operator in (1.1), (1.2) is not invertible.

Fractional differential equation can describe many phenomena in various fields of science and engineering. Many methods have been introduced for solving fractional differential equations, such as the popular Laplace transform method, the iteration method. For details, see [1, 2] and the references therein.

Recently, there are some papers dealing with the solvability of nonlinear boundary value problems of fractional differential equation, by use of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, etc.), see, for example, [3ā€“6]. But there are few papers that consider the fractional-order boundary problems at resonance. Very recently [7], Y. H. Zhang and Z. B. Bai considered the existence of solutions for the fractional ordinary differential equationš·š›¼0+ī‚€š‘¢(š‘”)=š‘“š‘”,š‘¢(š‘”),š·š›¼āˆ’(š‘›āˆ’1)0+š‘¢(š‘”),ā€¦,š·š›¼āˆ’10+ī‚š‘¢(š‘”)+š‘’(š‘”),0<š‘”<1,(1.5) subject to the following boundary value conditions: š¼š‘›āˆ’š›¼0+š‘¢(0)=š·š›¼āˆ’(š‘›āˆ’1)0+š‘¢(0)=ā‹Æ=š·š›¼āˆ’20+š‘¢(0)=0,š‘¢(1)=šœŽš‘¢(šœ‚),(1.6) where š‘›>2 is a natural number, š‘›āˆ’1<š›¼ā‰¤š‘› is a real number, š‘“āˆ¶[0,1]Ɨā„š‘›ā†’ā„ is continuous, and š‘’āˆˆšæ1[0,1],šœŽāˆˆ(0,āˆž), and šœ‚āˆˆ(0,1) are given constants such that šœŽšœ‚š›¼āˆ’1=1. š·š›¼0+ and š¼š›¼0+ are the standard Riemann-Liouville derivative and integral, respectively. By the conditions, the kernel of the linear operator is one dimensional.

Motivated by the above work and recent studies on fractional differential equations [8ā€“18], in this paper, we consider the existence of solutions for multipoint boundary value problem (1.1), (1.2) at resonance. Note that under condition (1.3), the kernel of the linear operator in (1.1), (1.2) is two dimensional. Our method is based upon the coincidence degree theory of Mawhin [18].

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

Let š‘Œ,š‘ be real Banach spaces, let šæāˆ¶dom(šæ)āŠ‚š‘Œā†’š‘ be a Fredholm map of index zero, and let š‘ƒāˆ¶š‘Œā†’š‘Œ,š‘„āˆ¶š‘ā†’š‘ be continuous projectors such that Im(š‘ƒ)=Ker(š‘ƒ),Ker(š‘„)=Im(šæ), and š‘Œ=Ker(šæ)āŠ•Ker(š‘ƒ),š‘=Im(šæ)āŠ•Im(š‘„). It follows that šæ|dom(šæ)āˆ©Ker(š‘ƒ)āˆ¶dom(šæ)āˆ©Ker(š‘ƒ)ā†’Im(šæ) is invertible. We denote the inverse of the map by š¾š‘ƒ. If Ī© is an open-bounded subset of š‘Œ such that dom(šæ)āˆ©Ī©ā‰ āˆ…, the map š‘āˆ¶š‘Œā†’š‘ will be called šæ-compact on Ī© if š‘„š‘(Ī©) is bounded and š¾š‘ƒ(š¼āˆ’š‘„)š‘āˆ¶Ī©ā†’š‘Œ is compact.

The theorem that we used is Theoremā€‰ā€‰2.4 of [18].

Theorem 1.1. Let šæ be a Fredholm operator of index zero and š‘ be šæ-compact on Ī©. Assume that the following conditions are satisfied: (i)šæš‘„ā‰ šœ†š‘š‘„ for every (š‘„,šœ†)āˆˆ[(dom(šæ)ā§µKer(šæ))āˆ©šœ•Ī©]Ɨ(0,1),(ii)š‘š‘„āˆ‰Im(šæ) for every š‘„āˆˆKer(šæ)āˆ©šœ•Ī©,(iii)deg(š½š‘„š‘|Ker(šæ),Ī©āˆ©Ker(šæ),0)ā‰ 0, where š‘„āˆ¶š‘ā†’š‘ is a projection as above with Im(šæ)=Ker(š‘„), and š½āˆ¶Im(š‘„)ā†’Ker(šæ) is any isomorphism,
then the equation šæš‘„=š‘š‘„ has at least one solution in dom(šæ)āˆ©Ī©.

The rest of this paper is organized as follows. In Section 2, we give some notation and Lemmas. In Section 3, we establish an existence theorem of a solution for the problem (1.1), (1.2).

2. Background Materials and Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory, and these definitions can be found in the recent literature [1, 2].

Definition 2.1. The fractional integral of order š›¼>0 of a function š‘¦āˆ¶(0,āˆž)ā†’ā„ is given by š¼š›¼0+1š‘¦(š‘”)=ī€œĪ“(š›¼)š‘”0(š‘”āˆ’š‘ )š›¼āˆ’1š‘¦(š‘ )š‘‘š‘ ,(2.1) provided the right side is pointwise defined on (0,āˆž). And we let š¼00+š‘¦(š‘”)=š‘¦(š‘”) for every continuous š‘¦āˆ¶(0,āˆž)ā†’ā„.

Definition 2.2. The fractional derivative of order š›¼>0 of a function š‘¦āˆ¶(0,āˆž)ā†’ā„ is given by š·š›¼0+1š‘¦(š‘”)=ī‚€š‘‘Ī“(š‘›āˆ’š›¼)ī‚š‘‘š‘”š‘›ī€œš‘”0š‘¦(š‘ )(š‘”āˆ’š‘ )š›¼āˆ’š‘›+1š‘‘š‘ ,(2.2) where š‘›=[š›¼]+1, provided the right side is pointwise defined on (0,āˆž).

Lemma 2.3 (see [3]). Assume that š‘¢āˆˆš¶(0,1)āˆ©šæ1[0,1] with a fractional derivative of order š›¼>0 that belongs to š¶(0,1)āˆ©šæ1[0,1], then š¼š›¼0+š·š›¼0+š‘¢(š‘”)=š‘¢(š‘”)+š¶1š‘”š›¼āˆ’1+š¶2š‘”š›¼āˆ’2+ā‹Æ+š¶š‘š‘”š›¼āˆ’š‘,(2.3) for some š¶š‘–āˆˆā„,š‘–=1,2,ā€¦,š‘, where š‘ is the smallest integer greater than or equal to š›¼.

We use the classical space š¶[0,1] with the norm ā€–š‘„ā€–āˆž=maxš‘”āˆˆ[0,1]|š‘„(š‘”)|. Given šœ‡>0 and š‘=[šœ‡]+1, one can define a linear space š¶šœ‡[]ī€½0,1āˆ¶=š‘¢āˆ£š‘¢(š‘”)=š¼šœ‡0+š‘„(š‘”)+š‘1š‘”šœ‡āˆ’1+š‘2š‘”šœ‡āˆ’2+ā‹Æ+š‘š‘āˆ’1š‘”šœ‡āˆ’(š‘āˆ’1)[]ī€¾,š‘”āˆˆ0,1,(2.4) where š‘„āˆˆš¶[0,1] and š‘š‘–āˆˆā„,š‘–=1,2,ā€¦,š‘āˆ’1. By means of the linear function analysis theory, one can prove that with the norm ā€–š‘¢ā€–š¶šœ‡=ā€–š·šœ‡0+š‘¢ā€–āˆž+ā‹Æ+ā€–š·šœ‡āˆ’(š‘āˆ’1)0+š‘¢ā€–āˆž+ā€–š‘¢ā€–āˆž,š¶šœ‡[0,1] is a Banach space.

Lemma 2.4 (see [7]). š¹āŠ‚š¶šœ‡[0,1] is a sequentially compact set if and only if š¹ is uniformly bounded and equicontinuous. Here, uniformly bounded means that there exists š‘€>0 such that for every š‘¢āˆˆš¹, ā€–š‘¢ā€–š¶šœ‡=ā€–ā€–š·šœ‡0+š‘¢ā€–ā€–āˆžā€–ā€–š·+ā‹Æ+šœ‡āˆ’(š‘āˆ’1)0+š‘¢ā€–ā€–āˆž+ā€–š‘¢ā€–āˆž<š‘€,(2.5) and equicontinuous means that forallšœ€>0,āˆƒš›æ>0 such that ||š‘¢ī€·š‘”1ī€øī€·š‘”āˆ’š‘¢2ī€ø||ī€·<šœ€,āˆ€š‘”1,š‘”2āˆˆ[],||š‘”0,11āˆ’š‘”2||ī€ø,||š·<š›æ,āˆ€š‘¢āˆˆš¹š›¼āˆ’š‘–0+š‘¢ī€·š‘”1ī€øāˆ’š·š›¼āˆ’š‘–0+š‘¢ī€·š‘”2ī€ø||ī€·š‘”<šœ€,1,š‘”2āˆˆ[],||š‘”0,11āˆ’š‘”2||ī€ø.<š›æ,āˆ€š‘¢āˆˆš¹,āˆ€š‘–āˆˆ{0,ā€¦,š‘āˆ’1}(2.6)

Let š‘=šæ1[0,1] with the norm ā€–š‘”ā€–1=āˆ«10|š‘”(š‘ )|š‘‘š‘ . š‘Œ=š¶š›¼āˆ’1[0,1]={š‘¢āˆ£š‘¢(š‘”)=š¼š›¼āˆ’10+š‘„(š‘”)+š‘š‘”š›¼āˆ’2,š‘”āˆˆ[0,1]}, where š‘„āˆˆš¶[0,1],š‘āˆˆā„, with the norm ā€–š‘¢ā€–š¶š›¼āˆ’1=ā€–š·š›¼āˆ’10+š‘¢ā€–āˆž+ā€–š·š›¼āˆ’20+š‘¢ā€–āˆž+ā€–š‘¢ā€–āˆž, and š‘Œ is a Banach space.

Definition 2.5. By a solution of the boundary value problem (1.1), (1.2), we understand a function š‘¢āˆˆš¶š›¼āˆ’1[0,1] such that š·š›¼āˆ’10+š‘¢ is absolutely continuous on (0,1) and satisfies (1.1), (1.2).

Definition 2.6. We say that the map š‘“āˆ¶[0,1]Ɨā„ā†’ā„ satisfies the CarathĆ©odory conditions with respect to šæ1[0,1] if the following conditions are satisfied:(i)for each š‘§āˆˆā„, the mapping š‘”ā†’š‘“(š‘”,š‘§) is Lebesgue measurable,(ii)for almost every š‘”āˆˆ[0,1], the mapping š‘§ā†’š‘“(š‘”,š‘§) is continuous on ā„,(iii)for each š‘Ÿ>0, there exists šœŒš‘Ÿāˆˆšæ1([0,1],ā„) such that, for a.e., š‘”āˆˆ[0,1] and every |š‘§|ā‰¤š‘Ÿ, we have |š‘“(š‘”,š‘§)|ā‰¤šœŒš‘Ÿ(š‘”).

Define šæ to be the linear operator from dom(šæ)āˆ©š‘Œ to š‘ with ī€½dom(šæ)=š‘¢āˆˆš¶š›¼āˆ’1[]0,1āˆ£š·š›¼0+š‘¢āˆˆšæ1[]ī€¾,0,1,š‘¢satisļ¬es(1.2)šæš‘¢=š·š›¼0+š‘¢,š‘¢āˆˆdom(šæ).(2.7) We define š‘āˆ¶š‘Œā†’š‘ by setting ī€·š‘š‘¢(š‘”)=š‘“š‘”,š‘¢(š‘”),š·š›¼āˆ’10+š‘¢(š‘”),š·š›¼āˆ’20+ī€øš‘¢(š‘”)+š‘’(š‘”).(2.8) Then boundary value problem (1.1), (1.2) can be written as šæš‘¢=š‘š‘¢.(2.9)

Lemma 2.7. Let condition (1.3) and (1.4) hold, then šæāˆ¶dom(šæ)āˆ©š‘Œā†’š‘ is a Fredholm map of index zero.

Proof. It is clear that Ker(šæ)={š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2āˆ£š‘Ž,š‘āˆˆā„}ā‰…ā„2.
Let š‘”āˆˆš‘ and 1š‘¢(š‘”)=ī€œĪ“(š›¼)š‘”0(š‘”āˆ’š‘ )š›¼āˆ’1š‘”(š‘ )š‘‘š‘ +š‘1š‘”š›¼āˆ’1+š‘2š‘”š›¼āˆ’2,(2.10) then š·š›¼0+š‘¢(š‘”)=š‘”(š‘”)a.e.,š‘”āˆˆ(0,1) and, if ī€œ10(1āˆ’š‘ )š›¼āˆ’1š‘”(š‘ )š‘‘š‘ āˆ’š‘šī“š‘–=1š›¼š‘–ī€œšœ‚š‘–0ī€·šœ‚š‘–ī€øāˆ’š‘ š›¼āˆ’1š‘”(š‘ )š‘‘š‘ =0,š‘›ī“š‘—=1š›½š‘—ī€œšœ‰š‘—0ī€·šœ‰š‘—ī€øāˆ’š‘ š‘”(š‘ )š‘‘š‘ =0(2.11) hold. Then š‘¢(š‘”) satisfies the boundary conditions (1.2), that is, š‘¢āˆˆdom(šæ), and we have {š‘”āˆˆš‘āˆ£š‘”satisļ¬es(2.11)}āŠ†Im(šæ).(2.12) Let š‘¢āˆˆdom(šæ), then for š·š›¼0+š‘¢āˆˆIm(šæ), we have š‘¢(š‘”)=š¼š›¼0+š·š›¼0+š‘¢(š‘”)+š‘1š‘”š›¼āˆ’1+š‘2š‘”š›¼āˆ’2+š‘3š‘”š›¼āˆ’3,(2.13) which, due to the boundary value condition (1.2), implies that š·š›¼0+š‘¢ satisfies (2.11). In fact, from š¼3āˆ’š›¼0+š‘¢(0)=0, we have š‘3=0, from āˆ‘š‘¢(1)=š‘šš‘–=1š›¼š‘–š‘¢(šœ‚š‘–), we have ī€œ10(1āˆ’š‘ )š›¼āˆ’1š·š›¼0+š‘¢(š‘ )š‘‘š‘ āˆ’š‘šī“š‘–=1š›¼š‘–ī€œšœ‚š‘–0ī€·šœ‚š‘–ī€øāˆ’š‘ š›¼āˆ’1š·š›¼0+š‘¢(š‘ )š‘‘š‘ =0,(2.14) and from š·š›¼āˆ’20+āˆ‘š‘¢(0)=š‘›š‘—=1š›½š‘—š·š›¼āˆ’20+š‘¢(šœ‰š‘—), we have š‘›ī“š‘—=1š›½š‘—ī€œšœ‰š‘—0ī€·šœ‰š‘—ī€øš·āˆ’š‘ š›¼0+š‘¢(š‘ )š‘‘š‘ =0.(2.15) Hence, Im(šæ)āŠ†{š‘”āˆˆš‘āˆ£š‘”satisļ¬es(2.11)}.(2.16) Therefore, Im(šæ)={š‘”āˆˆš‘āˆ£š‘”satisļ¬es(2.11)}.(2.17) Consider the continuous linear mapping š‘„1āˆ¶š‘ā†’š‘ and š‘„2āˆ¶š‘ā†’š‘ defined by š‘„1ī€œš‘”=10(1āˆ’š‘ )š›¼āˆ’1š‘”(š‘ )š‘‘š‘ āˆ’š‘šī“š‘–=1š›¼š‘–ī€œšœ‚š‘–0ī€·šœ‚š‘–ī€øāˆ’š‘ š›¼āˆ’1š‘„š‘”(š‘ )š‘‘š‘ ,2š‘”=š‘›ī“š‘—=1š›½š‘—ī€œšœ‰š‘—0ī€·šœ‰š‘—ī€øāˆ’š‘ š‘”(š‘ )š‘‘š‘ .(2.18) Using the above definitions, we construct the following auxiliary maps š‘…1,š‘…2āˆ¶š‘ā†’š‘: š‘…11š‘”=š‘…īƒ¬Ī“(š›¼āˆ’1)Ī“(š›¼+1)š‘›ī“š‘—=1š›½š‘—šœ‰š›¼š‘—š‘„1š‘”(š‘”)āˆ’Ī“(š›¼)Ī“(š›¼āˆ’1)īƒ©Ī“(2š›¼āˆ’1)1āˆ’š‘šī“š‘–=1š›¼š‘–šœ‚š‘–2š›¼āˆ’2īƒŖš‘„2īƒ­,š‘…š‘”(š‘”)21š‘”=āˆ’š‘…īƒ¬Ī“(š›¼)Ī“(š›¼+2)š‘›ī“š‘—=1š›½š‘—šœ‰š‘—š›¼+1š‘„1š‘”(š‘”)āˆ’(Ī“(š›¼))2īƒ©Ī“(2š›¼)1āˆ’š‘šī“š‘–=1š›¼š‘–šœ‚š‘–2š›¼āˆ’1īƒŖš‘„2īƒ­.š‘”(š‘”)(2.19) Since the condition (1.4) holds, the mapping š‘„āˆ¶š‘ā†’š‘ defined by ī€·š‘…(š‘„š‘¦)(š‘”)=1ī€øš‘”š‘”(š‘”)š›¼āˆ’1+ī€·š‘…2ī€øš‘”š‘”(š‘”)š›¼āˆ’2(2.20) is well defined.
Recall (1.4) and note that š‘…1ī€·š‘…1š‘”š‘”š›¼āˆ’1ī€ø=1š‘…īƒ¬Ī“(š›¼āˆ’1)Ī“(š›¼+1)š‘›ī“š‘—=1š›½š‘—šœ‰š›¼š‘—š‘„1ī€·š‘…1š‘”š‘”š›¼āˆ’1ī€øāˆ’Ī“(š›¼)Ī“(š›¼āˆ’1)īƒ©Ī“(2š›¼āˆ’1)1āˆ’š‘šī“š‘–=1š›¼š‘–šœ‚š‘–2š›¼āˆ’2īƒŖš‘„2ī€·š‘…1š‘”š‘”š›¼āˆ’1ī€øīƒ­=š‘…1š‘”1š‘…īƒ¬ī€·š›¼Ī“(š›¼āˆ’1)Ī“2ī€øĪ“(š›¼+1)Ī“(2š›¼)š‘›ī“š‘—=1š›½š‘—šœ‰š›¼š‘—īƒ©1āˆ’š‘šī“š‘–=1š›¼š‘–šœ‚š‘–2š›¼āˆ’1īƒŖāˆ’ī€·š›¼Ī“(š›¼āˆ’1)Ī“2ī€øīƒ©Ī“(2š›¼āˆ’1)Ī“(š›¼+2)1āˆ’š‘šī“š‘–=1š›¼š‘–šœ‚š‘–2š›¼āˆ’2īƒŖš‘›ī“š‘—=1š›½š‘—šœ‰š‘—š›¼+1īƒ­=š‘…1š‘”,(2.21) and similarly we can derive that š‘…1ī€·š‘…2š‘”š‘”š›¼āˆ’2ī€øš‘…=0,2ī€·š‘…1š‘”š‘”š›¼āˆ’1ī€øš‘…=0,2ī€·š‘…2š‘”š‘”š›¼āˆ’2ī€ø=š‘…2š‘”.(2.22) So, for š‘”āˆˆš‘, it follows from the four relations above that š‘„2š‘”=š‘…1ī€·š‘…1š‘”š‘”š›¼āˆ’1+š‘…2š‘”š‘”š›¼āˆ’2ī€øš‘”š›¼āˆ’1+š‘…2ī€·š‘…1š‘”š‘”š›¼āˆ’1+š‘…2š‘”š‘”š›¼āˆ’2ī€øš‘”š›¼āˆ’2=š‘…1ī€·š‘…1š‘”š‘”š›¼āˆ’1ī€øš‘”š›¼āˆ’1+š‘…1ī€·š‘…2š‘”š‘”š›¼āˆ’2ī€øš‘”š›¼āˆ’1+š‘…2ī€·š‘…1š‘”š‘”š›¼āˆ’1ī€øš‘”š›¼āˆ’2+š‘…2ī€·š‘…2š‘”š‘”š›¼āˆ’2ī€øš‘”š›¼āˆ’2=š‘…1š‘”š‘”š›¼āˆ’1+š‘…2š‘”š‘”š›¼āˆ’2=š‘„š‘”,(2.23) that is, the map š‘„ is idempotent. In fact, š‘„ is a continuous linear projector.
Note that š‘”āˆˆIm(šæ) implies š‘„š‘”=0. Conversely, if š‘„š‘”=0, then we must have š‘…1š‘”=š‘…2š‘”=0; since the condition (1.4) holds, this can only be the case if š‘„1š‘”=š‘„2š‘”=0, that is, š‘”āˆˆIm(šæ). In fact, Im(šæ)=Ker(š‘„).
Take š‘”āˆˆš‘ in the form š‘”=(š‘”āˆ’š‘„š‘”)+š‘„š‘”, so that š‘”āˆ’š‘„š‘”āˆˆIm(šæ)=Ker(š‘„) and š‘„š‘”āˆˆIm(š‘„). Thus, š‘=Im(šæ)+Im(š‘„). Let š‘”āˆˆIm(šæ)āˆ©Im(š‘„) and assume that š‘”(š‘ )=š‘Žš‘ š›¼āˆ’1+š‘š‘ š›¼āˆ’2 is not identically zero on [0,1], then, since š‘”āˆˆIm(šæ), from (2.11) and the condition (1.4), we derive š‘Ž=š‘=0, which is a contradiction. Hence, Im(šæ)āˆ©Im(š‘„)={0}; thus, š‘=Im(šæ)āŠ•Im(š‘„).
Now, dimKer(šæ)=2=codimIm(šæ), and so šæ is a Fredholm operator of index zero.

Let š‘ƒāˆ¶š‘Œā†’š‘Œ be defined by 1š‘ƒš‘¢(š‘”)=Ī“š·(š›¼)š›¼āˆ’10+š‘¢(0)š‘”š›¼āˆ’1+1Ī“š·(š›¼āˆ’1)š›¼āˆ’20+š‘¢(0)š‘”š›¼āˆ’2[],š‘”āˆˆ0,1.(2.24) Note that š‘ƒ is a continuous linear projector and ī€½Ker(š‘ƒ)=š‘¢āˆˆš‘Œāˆ£š·š›¼āˆ’10+š‘¢(0)=š·š›¼āˆ’20+ī€¾š‘¢(0)=0.(2.25) It is clear that š‘Œ=Ker(šæ)āŠ•Ker(š‘ƒ).

Note that the projectors š‘ƒ and š‘„ are exact. Define š¾š‘ƒāˆ¶Im(šæ)ā†’dom(šæ)āˆ©Ker(š‘ƒ) by š¾š‘ƒ1š‘”(š‘”)=ī€œĪ“(š›¼)š‘”0(š‘”āˆ’š‘ )š›¼āˆ’1š‘”(š‘ )š‘‘š‘ =š¼š›¼0+š‘”(š‘”).(2.26) Hence, we have š·š›¼āˆ’10+ī€·š¾š‘ƒš‘”ī€ø(ī€œš‘”)=š‘”0š‘”(š‘ )š‘‘š‘ ,š·š›¼āˆ’20+ī€·š¾š‘ƒš‘”ī€ø(ī€œš‘”)=š‘”0(š‘”āˆ’š‘ )š‘”(š‘ )š‘‘š‘ ,(2.27) then ā€–š¾š‘ƒš‘”ā€–āˆžā‰¤(1/Ī“(š›¼))ā€–š‘”ā€–1,ā€–š·š›¼āˆ’10+(š¾š‘ƒš‘”)ā€–āˆžā‰¤ā€–š‘”ā€–1,ā€–š·š›¼āˆ’20+(š¾š‘ƒš‘”)ā€–āˆžā‰¤ā€–š‘”ā€–1, and thus ā€–ā€–š¾š‘ƒš‘”ā€–ā€–š¶š›¼āˆ’1ā‰¤ī‚µ12+ī‚¶Ī“(š›¼)ā€–š‘”ā€–1.(2.28) In fact, if š‘”āˆˆIm(šæ), then (šæš¾š‘ƒ)š‘”=š·š›¼0+š¼š›¼0+š‘”=š‘”. Also, if š‘¢āˆˆdom(šæ)āˆ©Ker(š‘ƒ), then ī€·š¾š‘ƒī€øšæš‘”(š‘”)=š¼š›¼0+š·š›¼0+š‘”(š‘”)=š‘”(š‘”)+š‘1š‘”š›¼āˆ’1+š‘2š‘”š›¼āˆ’2+š‘3š‘”š›¼āˆ’3,(2.29) from boundary value condition (1.2) and the fact that š‘¢āˆˆdom(šæ)āˆ©Ker(š‘ƒ), we have š‘1=š‘2=š‘3=0. Thus, š¾š‘ƒ=ī‚€šæ||dom(šæ)āˆ©Ker(š‘ƒ)ī‚āˆ’1.(2.30) Using (2.19), we write ī€·š‘…š‘„š‘š‘¢(š‘”)=1ī€øš‘”š‘š‘¢š›¼āˆ’1+ī€·š‘…2ī€øš‘”š‘š‘¢š›¼āˆ’2,š¾š‘ƒ1(š¼āˆ’š‘„)š‘š‘¢(š‘”)=ī€œĪ“(š›¼)10(š‘”āˆ’š‘ )š›¼āˆ’1[]š‘š‘¢(š‘ )āˆ’š‘„š‘š‘¢(š‘ )š‘‘š‘ .(2.31) With arguments similar to those of [7], we obtain the following Lemma.

Lemma 2.8. š¾š‘ƒ(š¼āˆ’š‘„)š‘āˆ¶š‘Œā†’š‘Œ is completely continuous.

3. The Main Results

Assume that the following conditions on the function š‘“(š‘”,š‘„,š‘¦,š‘§) are satisfied:

(H1) there exists a constant š“>0, such that for š‘¢āˆˆdom(šæ)ā§µKer(šæ) satisfying |š·š›¼āˆ’10+š‘¢(š‘”)|+|š·š›¼āˆ’20+š‘¢(š‘”)|>š“ for all š‘”āˆˆ[0,1], we have š‘„1š‘š‘¢(š‘”)ā‰ 0orš‘„2š‘š‘¢(š‘”)ā‰ 0,(3.1)

(H2) there exist functions š‘Ž,š‘,š‘,š‘‘,š‘Ÿāˆˆšæ1[0,1] and a constant šœƒāˆˆ[0,1] such that for all (š‘„,š‘¦,š‘§)āˆˆā„3 and a.e., š‘”āˆˆ[0,1], one of the following inequalities is satisfied:||||||š‘¦||š‘“(š‘”,š‘„,š‘¦,š‘§)ā‰¤š‘Ž(š‘”)|š‘„|+š‘(š‘”)+š‘(š‘”)|š‘§|+š‘‘(š‘”)|š‘§|šœƒ||||||š‘¦||||š‘¦||+š‘Ÿ(š‘”),š‘“(š‘”,š‘„,š‘¦,š‘§)ā‰¤š‘Ž(š‘”)|š‘„|+š‘(š‘”)+š‘(š‘”)|š‘§|+š‘‘(š‘”)šœƒ||||||š‘¦||+š‘Ÿ(š‘”),š‘“(š‘”,š‘„,š‘¦,š‘§)ā‰¤š‘Ž(š‘”)|š‘„|+š‘(š‘”)+š‘(š‘”)|š‘§|+š‘‘(š‘”)|š‘„|šœƒ+š‘Ÿ(š‘”),(3.2)

(H3) there exists a constant šµ>0 such that for every š‘Ž,š‘āˆˆā„ satisfying š‘Ž2+š‘2>šµ, then either š‘Žš‘…1š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€ø+š‘š‘…2š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€ø<0,(3.3) or else š‘Žš‘…1š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€ø+š‘š‘…2š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€ø>0.(3.4)

Remark 3.1. š‘…1š‘(š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2) and š‘…2š‘(š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2) from (H3) stand for the images of š‘¢(š‘”)=š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2 under the maps š‘…1š‘ and š‘…2š‘, respectively.

Theorem 3.2. If (H1)ā€“(H3) hold, then boundary value problem (1.1)-(1.2) has at least one solution provided that ā€–š‘Žā€–1+ā€–š‘ā€–1+ā€–š‘ā€–1<1šœ,(3.5) where šœ=5+2/Ī“(š›¼)+1/Ī“(š›¼āˆ’1).

Proof. Set Ī©1[]={š‘¢āˆˆdom(šæ)ā§µKer(šæ)āˆ£šæš‘¢=šœ†š‘š‘¢forsomešœ†āˆˆ0,1},(3.6) then for š‘¢āˆˆĪ©1,šæš‘¢=šœ†š‘š‘¢; thus, šœ†ā‰ 0,š‘š‘¢āˆˆIm(šæ)=Ker(š‘„), and hence š‘„š‘š‘¢(š‘”)=0 for all š‘”āˆˆ[0,1]. By the definition of š‘„, we have š‘„1š‘š‘¢(š‘”)=š‘„2š‘š‘¢(š‘”)=0. It follows from (H1) that there exists š‘”0āˆˆ[0,1] such that |š·š›¼āˆ’10+š‘¢(š‘”0)|+|š·š›¼āˆ’20+š‘¢(š‘”0)|ā‰¤š“. Now, š·š›¼āˆ’10+š‘¢(š‘”)=š·š›¼āˆ’10+š‘¢ī€·š‘”0ī€ø+ī€œš‘”š‘”0š·š›¼0+š·š‘¢(š‘ )š‘‘š‘ ,š›¼āˆ’20+š‘¢(š‘”)=š·š›¼āˆ’20+š‘¢ī€·š‘”0ī€ø+ī€œš‘”š‘”0š·š›¼āˆ’10+š‘¢(š‘ )š‘‘š‘ ,(3.7) so ||š·š›¼āˆ’10+||ā‰¤ā€–ā€–š·š‘¢(0)š›¼āˆ’10+ā€–ā€–š‘¢(š‘”)āˆžā‰¤||š·š›¼āˆ’10+š‘¢ī€·š‘”0ī€ø||+ā€–ā€–š·š›¼0+š‘¢ā€–ā€–1ā‰¤š“+ā€–šæš‘¢ā€–1ā‰¤š“+ā€–š‘š‘¢ā€–1,||š·š›¼āˆ’20+||ā‰¤ā€–ā€–š·š‘¢(0)š›¼āˆ’20+ā€–ā€–š‘¢(š‘”)āˆžā‰¤||š·š›¼āˆ’20+š‘¢ī€·š‘”0ī€ø||+ā€–ā€–š·š›¼āˆ’10+š‘¢ā€–ā€–āˆžā‰¤||š·š›¼āˆ’20+š‘¢ī€·š‘”0ī€ø||+||š·š›¼āˆ’10+š‘¢ī€·š‘”0ī€ø||+ā€–ā€–š·š›¼0+š‘¢ā€–ā€–1ā‰¤š“+ā€–šæš‘¢ā€–1ā‰¤š“+ā€–š‘š‘¢ā€–1.(3.8) Now by (3.8), we have ā€–š‘ƒš‘¢ā€–š¶š›¼āˆ’1=ā€–ā€–ā€–1š·Ī“(š›¼)š›¼āˆ’10+š‘¢(0)š‘”š›¼āˆ’1+1š·Ī“(š›¼āˆ’1)š›¼āˆ’20+š‘¢(0)š‘”š›¼āˆ’2ā€–ā€–ā€–š¶š›¼āˆ’1=ā€–ā€–ā€–1š·Ī“(š›¼)0š›¼āˆ’1š‘¢(0)š‘”š›¼āˆ’1+1š·Ī“(š›¼āˆ’1)0š›¼āˆ’2š‘¢(0)š‘”š›¼āˆ’2ā€–ā€–ā€–āˆž+ā€–ā€–š·š›¼āˆ’10+ā€–ā€–š‘¢(0)āˆž+ā€–ā€–š·š›¼āˆ’10+š‘¢(0)š‘”+š·š›¼āˆ’20+ā€–ā€–š‘¢(0)āˆžā‰¤ī‚µ12+ī‚¶||š·Ī“(š›¼)š›¼āˆ’10+||+ī‚µ1š‘¢(0)1+ī‚¶||š·Ī“(š›¼āˆ’1)š›¼āˆ’20+||ā‰¤ī‚µ1š‘¢(0)2+ī‚¶ī€·Ī“(š›¼)š“+ā€–š‘š‘¢ā€–1ī€ø+ī‚µ11+ī‚¶ī€·Ī“(š›¼āˆ’1)š“+ā€–š‘š‘¢ā€–1ī€ø.(3.9) Note that (š¼āˆ’š‘ƒ)š‘¢āˆˆIm(š¾š‘ƒ)=dom(šæ)āˆ©Ker(š‘ƒ) for š‘¢āˆˆĪ©1, then, by (2.28) and (2.30), ā€–ā€–(š¼āˆ’š‘ƒ)š‘¢š¶š›¼āˆ’1=ā€–ā€–š¾š‘ƒā€–ā€–šæ(š¼āˆ’š‘ƒ)š¶š›¼āˆ’1ā‰¤ī‚µ12āˆ’ī‚¶Ī“(š›¼)ā€–šæ(š¼āˆ’š‘ƒ)š‘¢ā€–1=ī‚µ12āˆ’ī‚¶Ī“(š›¼)ā€–šæš‘¢ā€–1ā‰¤ī‚µ12āˆ’ī‚¶Ī“(š›¼)ā€–š‘š‘¢ā€–1.(3.10) Using (3.9) and (3.10), we obtain ā€–š‘¢ā€–š¶š›¼āˆ’1=ā€–š‘ƒš‘¢+(š¼āˆ’š‘ƒ)š‘¢ā€–š¶š›¼āˆ’1ā‰¤ā€–š‘ƒš‘¢ā€–š¶š›¼āˆ’1ā€–+ā€–(š¼āˆ’š‘ƒ)š‘¢š¶š›¼āˆ’1ā‰¤ī‚µ12+ī‚¶ī€·Ī“(š›¼)š“+ā€–š‘š‘¢ā€–1ī€ø+ī‚µ11+ī‚¶ī€·Ī“(š›¼āˆ’1)š“+ā€–š‘š‘¢ā€–1ī€ø+ī‚µ12+ī‚¶Ī“(š›¼)ā€–š‘š‘¢ā€–1=ī‚µ25++1Ī“(š›¼)ī‚¶Ī“(š›¼āˆ’1)ā€–š‘š‘¢ā€–1+ī‚µ13++1Ī“(š›¼)ī‚¶š“Ī“(š›¼āˆ’1)=šœā€–š‘š‘¢ā€–1+š¶1,(3.11) where š¶1=(3+1/Ī“(š›¼)+1/Ī“(š›¼āˆ’1))š“ is a constant. This is for all š‘¢āˆˆĪ©1, ā€–š‘¢ā€–š¶š›¼āˆ’1ā‰¤šœā€–š‘š‘¢ā€–1+š¶1.(3.12) If the first condition of (H2) is satisfied, then we have ī€½maxā€–š‘¢ā€–āˆž,ā€–ā€–š·š›¼āˆ’10+š‘¢ā€–ā€–āˆž,ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–āˆžī€¾ā‰¤ā€–š‘¢ā€–š¶š›¼āˆ’1ī‚€ā‰¤šœā€–š‘Žā€–1ā€–š‘¢ā€–āˆž+ā€–š‘ā€–1ā€–ā€–š·š›¼āˆ’10+š‘¢ā€–ā€–āˆž+ā€–š‘ā€–1ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–āˆž+ā€–š‘‘ā€–1ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–šœƒāˆž+ā€–š‘Ÿā€–1+ā€–š‘’ā€–1ī‚+š¶1,(3.13) and consequently, ā€–š‘¢ā€–āˆžā‰¤šœ1āˆ’ā€–š‘Žā€–1šœī‚€ā€–š‘ā€–1ā€–ā€–š·š›¼āˆ’10+š‘¢ā€–ā€–āˆž+ā€–š‘ā€–1ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–āˆž+ā€–š‘‘ā€–1ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–šœƒāˆž+ā€–š‘Ÿā€–1+ā€–š‘’ā€–1ī‚+š¶11āˆ’ā€–š‘Žā€–1šœ,(3.14)ā€–ā€–š·š›¼āˆ’10+š‘¢ā€–ā€–āˆžā‰¤šœ1āˆ’ā€–š‘Žā€–1šœāˆ’ā€–š‘ā€–1šœī‚€ā€–š‘ā€–1ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–āˆž+ā€–š‘‘ā€–1ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–šœƒāˆž+ā€–š‘Ÿā€–1+ā€–š‘’ā€–1ī‚+š¶11āˆ’ā€–š‘Žā€–1šœāˆ’ā€–š‘ā€–1šœ,(3.15)ā€–ā€–š·š›¼āˆ’10+š‘¢ā€–ā€–āˆžā‰¤šœā€–š‘‘ā€–1ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–šœƒāˆž1āˆ’ā€–š‘Žā€–1šœāˆ’ā€–š‘ā€–1šœāˆ’ā€–š‘ā€–1šœ+šœī€·ā€–š‘Ÿā€–1+ā€–š‘’ā€–1ī€ø+š¶11āˆ’ā€–š‘Žā€–1šœāˆ’ā€–š‘ā€–1šœāˆ’ā€–š‘ā€–1šœ.(3.16) Note that šœƒāˆˆ[0,1) and ā€–š‘Žā€–1+ā€–š‘ā€–1+ā€–š‘ā€–1<1/šœ, so there exists š‘€1>0 such that ā€–š·š›¼āˆ’10+š‘¢ā€–āˆžā‰¤š‘€1 for all š‘¢āˆˆĪ©1. The inequalities (3.14) and (3.15) show that there exist š‘€2,š‘€3>0 such that ā€–š·š›¼āˆ’10+š‘¢ā€–āˆžā‰¤š‘€2,ā€–š‘¢ā€–āˆžā‰¤š‘€3 for all š‘¢āˆˆĪ©1. Therefore, for all š‘¢āˆˆĪ©1,ā€–š‘¢ā€–š¶š›¼āˆ’1=ā€–š‘¢ā€–āˆž+ā€–š·š›¼āˆ’10+š‘¢ā€–āˆž+ā€–š·š›¼āˆ’20+š‘¢ā€–āˆžā‰¤š‘€1+š‘€2+š‘€3, that is, Ī©1 is bounded given the first condition of (H2). If the other conditions of (H2) hold, by using an argument similar to the above, we can prove that Ī©1 is also bounded.
Let Ī©2={š‘¢āˆˆKer(šæ)āˆ£š‘š‘¢āˆˆIm(šæ)}.(3.17) For š‘¢āˆˆĪ©2,š‘¢āˆˆKer(šæ)={š‘¢āˆˆdom(šæ)āˆ£š‘¢=š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2,š‘Ž,š‘āˆˆā„,š‘”āˆˆ[0,1]}, and š‘„š‘(š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2)=0; thus, š‘…1š‘(š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2)=š‘…2š‘(š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2)=0. By (H3), š‘Ž2+š‘2ā‰¤šµ, that is, Ī©2 is bounded.
We define the isomorphism š½āˆ¶Im(š‘„)ā†’Ker(šæ) by š½ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€ø=š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2,š‘Ž,š‘āˆˆā„.(3.18)
If the first part of (H3) is satisfied, let Ī©3=ī€½š‘¢āˆˆKeršæāˆ¶āˆ’šœ†š½āˆ’1[]ī€¾š‘¢+(1āˆ’šœ†)š‘„š‘š‘¢=0,šœ†āˆˆ0,1.(3.19) For every š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2āˆˆĪ©3, šœ†ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€øš‘…=(1āˆ’šœ†)ī€ŗī€·1š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2š‘”ī€øī€øš›¼āˆ’1+ī€·š‘…2š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2š‘”ī€øī€øš›¼āˆ’2ī€».(3.20) If šœ†=1, then š‘Ž=š‘=0, and if š‘Ž2+š‘2>šµ, then by (H3), šœ†ī€·š‘Ž2+š‘2ī€ø=ī€ŗ(1āˆ’šœ†)š‘Žš‘…1š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€ø+š‘š‘…2š‘ī€·š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ī€øī€»<0,(3.21) which, in either case, obtain a contradiction. If the other part of (H3) is satisfied, then we take Ī©3=ī€½š‘¢āˆˆKeršæāˆ¶šœ†š½āˆ’1[]ī€¾š‘¢+(1āˆ’šœ†)š‘„š‘š‘¢=0,šœ†āˆˆ0,1,(3.22) and, again, obtain a contradiction. Thus, in either case, ā€–š‘¢ā€–š¶š›¼āˆ’1=ā€–š‘¢ā€–āˆž+ā€–ā€–š·š›¼āˆ’10+š‘¢ā€–ā€–āˆž+ā€–ā€–š·š›¼āˆ’20+š‘¢ā€–ā€–āˆž=ā€–ā€–š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ā€–ā€–š¶š›¼āˆ’1=ā€–ā€–š‘Žš‘”š›¼āˆ’1+š‘š‘”š›¼āˆ’2ā€–ā€–āˆž+ā€–š‘ŽĪ“(š›¼)ā€–āˆž+ā€–š‘ŽĪ“(š›¼)š‘”+š‘Ī“(š›¼āˆ’1)ā€–āˆž||š‘||ā‰¤(1+2Ī“(š›¼))|š‘Ž|+(1+Ī“(š›¼āˆ’1))ā‰¤(2+2Ī“(š›¼)+Ī“(š›¼āˆ’1))|š‘Ž|,(3.23) for all š‘¢āˆˆĪ©3, that is, Ī©3 is bounded.
In the following, we will prove that all the conditions of Theorem 1.1 are satisfied. Set Ī© to be a bounded open set of š‘Œ such that š‘ˆ3š‘–=1Ī©š‘–āŠ‚Ī©. by Lemma 2.8, the operator š¾š‘ƒ(š¼āˆ’š‘„)š‘āˆ¶Ī©ā†’š‘Œ is compact; thus, š‘ is šæ-compact on Ī©, then by the above argument, we have(i)šæš‘¢ā‰ šœ†š‘š‘„, for every (š‘¢,šœ†)āˆˆ[(dom(šæ)ā§µKeršæ)āˆ©šœ•Ī©]Ɨ(0,1),(ii)š‘š‘¢āˆ‰Im(šæ), for every š‘¢āˆˆKer(šæ)āˆ©šœ•Ī©.Finally, we will prove that (iii) of Theorem 1.1 is satisfied. Let š»(š‘¢,šœ†)=Ā±š¼š‘¢+(1āˆ’šœ†)š½š‘„š‘š‘¢, where š¼ is the identity operator in the Banach space š‘Œ. According to the above argument, we know that š»(š‘¢,šœ†)ā‰ 0,āˆ€š‘¢āˆˆšœ•Ī©āˆ©Ker(šæ),(3.24) and thus, by the homotopy property of degree, ī‚€||degš½š‘„š‘Ker(šæ)ī‚,Ī©āˆ©Ker(šæ),0=deg(š»(ā€¦,0),Ī©āˆ©Ker(šæ),0)=deg(š»(ā€¦,1),Ī©āˆ©Ker(šæ),0)=deg(Ā±š¼,Ī©āˆ©Ker(šæ),0)=Ā±1ā‰ 0,(3.25) then by Theorem 1.1, šæš‘¢=š‘š‘¢ has at least one solution in dom(šæ)āˆ©Ī©, so boundary problem (1.1), (1.2) has at least one solution in the space š¶š›¼āˆ’1[0,1]. The proof is finished.

Acknowledgments

This work is supported by the NSFC (no. 11061030, no. 11026060), the Fundamental Research Funds for the Gansu Universities, the nwnu-kjcxgc-03-69, nwnu-kjcxgc-03-61.

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Copyright Ā© 2011 Xiaoling Han and Ting Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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