The existence of solutions for nonlinear fractional multipoint boundary value problems at resonance

  • Na Xu1,

    Affiliated with

    • Wenbin Liu1Email author and

      Affiliated with

      • Lishun Xiao1

        Affiliated with

        Boundary Value Problems20122012:65

        DOI: 10.1186/1687-2770-2012-65

        Received: 17 January 2012

        Accepted: 18 May 2012

        Published: 28 June 2012

        Abstract

        A class of nonlinear fractional multipoint boundary value problems at resonance is considered in this article. The existence results are obtained by the method of the coincidence degree theory of Mawhin. An example is given to illustrate the results.

        MSC:34A08.

        Keywords

        coincidence degree fractional differential equation resonance multipoint boundary conditions

        1 Introduction

        The subject of fractional calculus has gained considerable popularity during the past decades, due mainly to its frequent appearance in a variety of different areas such as physics, aerodynamics, polymer rheology, etc. (see [13]). Many methods have been introduced for solving fractional differential equations (FDEs for short in the remaining), such as the Laplace transform method, the iteration method, the Fourier transform method, etc. (see [4]).

        Recently, there have been many works related to the existence of solutions for multipoint boundary value problems (BVPs for short in the remaining) at nonresonance of FDEs (see [511]). Motivated by the above articles and recent studies on FDEs (see [1219]), we consider the existence of solutions for a nonlinear fractional multipoint BVPs at resonance in this article.

        In [16], Zhang and Bai considered the following fractional three-point boundary value problems at resonance:
        { D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α ( n 1 ) u ( t ) , , D 0 + α 1 u ( t ) ) + e ( t ) , 0 < t < 1 , I 0 + n α u ( 0 ) = D 0 + α ( n 1 ) u ( 0 ) = = D 0 + α 2 u ( 0 ) = 0 , u ( 1 ) = σ u ( η ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equa_HTML.gif

        where n > 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq1_HTML.gif is a natural number; n 1 < α n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq2_HTML.gif is a real number; D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq3_HTML.gif and I 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq4_HTML.gif are the standard Riemann-Liouville derivative and integral respectively; f : [ 0 , 1 ] × R n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq5_HTML.gif is continuous; e ( t ) L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq6_HTML.gif; σ ( 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq7_HTML.gif η ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq8_HTML.gif are given constants such that σ η α 1 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq9_HTML.gif. In their article, they made the operator L u = D 0 + α u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq10_HTML.gif and got dim Ker L = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq11_HTML.gif. In [17], Bai discussed fractional m-point boundary value problems at resonance with the case of dim Ker L = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq11_HTML.gif.

        In 2010, Bai and Jiang studied the fractional differential equation of boundary value problems at resonance with the case of dim Ker L = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq12_HTML.gif respectively (see [18, 19]), and we can see that they obtained the results by the assumption that a specific algebraic expression is not equal to zero; for example,
        R = 1 α η α Γ ( α ) Γ ( α 1 ) Γ ( 2 α 1 ) [ 1 i = 1 m α i η i 2 α 2 ] 1 α 1 η α 1 Γ 2 ( α ) Γ ( 2 α ) [ 1 i = 1 m α i η i 2 α 1 ] 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equb_HTML.gif

        is referred to as a condition in [18]. We will show that the assumption like above R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq13_HTML.gif is not necessary.

        In this article, we will use the coincidence degree theory to study the existence of solutions for a nonlinear FDEs at resonance which is given by
        D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) , 0 < t < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ1_HTML.gif
        (1.1)
        with boundary conditions
        I 0 + 3 α u ( t ) | t = 0 = 0 , D 0 + α 1 u ( 1 ) = i = 1 m a i D 0 + α 1 u ( ξ i ) , D 0 + α 2 u ( 1 ) = i = 1 m b i D 0 + α 2 u ( η i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ2_HTML.gif
        (1.2)

        where 2 < α 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq14_HTML.gif; 0 < ξ 1 < < ξ m < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq15_HTML.gif; 0 < η 1 < < η m < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq16_HTML.gif; a i , b i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq17_HTML.gif; f : [ 0 , 1 ] × R 3 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq18_HTML.gif with satisfying Carathéodory conditions; D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq19_HTML.gif and I 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq20_HTML.gif are the standard Riemann-Liouville fractional derivative and fractional integral respectively.

        BVPs (1.1)-(1.2) being at resonance means that the associated linear homogeneous equation D 0 + α u ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq21_HTML.gif with boundary conditions (1.2) has u ( t ) = a t α 1 + b t α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq22_HTML.gif as a nontrivial solution, where 0 < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq23_HTML.gif, a , b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq24_HTML.gif.

        We will always suppose that the following conditions hold:
        i = 1 m a i = 1 , i = 1 m b i η i = 1 , i = 1 m b i = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ3_HTML.gif
        (C)

        The rest of this article is organized as follows: In Section 2, we give some definitions, lemmas and notations. In Section 3, we establish theorems of existence result for BVPs (1.1)-(1.2). In Section 4, we give an example to illustrate our result.

        2 Preliminaries

        We present here some necessary basic knowledge and definitions of the fractional calculus theory, which can be found in [13].

        Definition 2.1 The Riemann-Liouville fractional integral of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq25_HTML.gif of a function y : ( 0 , ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq26_HTML.gif is given by
        I 0 + α y ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equc_HTML.gif

        where Γ ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq27_HTML.gif is the Gamma function, provided the right side is pointwise defined on ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq28_HTML.gif.

        Definition 2.2 The Riemann-Liouville fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq25_HTML.gif of a function y : ( 0 , ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq26_HTML.gif is given by
        D 0 + α y ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t y ( s ) ( t s ) α n + 1 d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equd_HTML.gif

        where n = [ α ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq29_HTML.gif, provided the right side is pointwise defined on ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq28_HTML.gif.

        Definition 2.3 ([18])

        We say that the map f : [ 0 , 1 ] × R n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq30_HTML.gif satisfies Carathéodory conditions with respect to L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq31_HTML.gif if the following conditions are satisfied:
        1. (i)

          for each z R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq32_HTML.gif, the mapping t f ( t , z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq33_HTML.gif is Lebesgue measurable;

           
        2. (ii)

          for almost every t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif, the mapping t f ( t , z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq33_HTML.gif is continuous on R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq35_HTML.gif;

           
        3. (iii)

          for each E > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq36_HTML.gif, there exists a ρ E L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq37_HTML.gif such that, for a.e. t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif and every | u | E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq38_HTML.gif, we have f ( t , u ) ρ E ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq39_HTML.gif.

           

        Lemma 2.4 ([2])

        Assume y ( t ) C [ 0 , 1 ] L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq40_HTML.gif, 0 β α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq41_HTML.gif, then D 0 + β I 0 + α y ( t ) = I 0 + α β y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq42_HTML.gif. And, for all α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq43_HTML.gif, β > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq44_HTML.gif, we have that
        I 0 + α t β = Γ ( β + 1 ) Γ ( α + β + 1 ) t α + β , D 0 + α t β = Γ ( β + 1 ) Γ ( β α + 1 ) t β α . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Eque_HTML.gif

        Lemma 2.5 ([2])

        Let α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq25_HTML.gif, n = [ α ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq29_HTML.gifand assume that y , D 0 + α y L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq45_HTML.gif, then the following equality holds almost everywhere on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq46_HTML.gif,
        ( I 0 + α D 0 + α y ) ( t ) = y ( t ) i = 1 n ( ( I 0 + n α y ) ( t ) ) n i | t = 0 Γ ( α i + 1 ) t α i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equf_HTML.gif
        Now, we briefly recall some notations and an abstract existence result, which can be found in [20]. Let Y Z be real Banach spaces, L : dom L Y Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq47_HTML.gif be a Fredholm map of index zero, and P : Y Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq48_HTML.gif Q : Z Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq49_HTML.gif be continuous projectors such that
        Im P = Ker L , Ker Q = Im L , Y = Ker L Ker P , Z = Im Q Im L . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equg_HTML.gif

        It follows that L | dom L Ker P : dom L Ker P Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq50_HTML.gif is invertible. We denote the inverse by K p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq51_HTML.gif. If Ω is an open bounded subset of Y such that dom L Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq52_HTML.gif, the map N : Y Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq53_HTML.gif will be called L-compact on Ω if Q N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq54_HTML.gif is bounded and K p ( I Q ) N : Ω ¯ Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq55_HTML.gif is compact.

        Lemma 2.6 ([20])

        Let L be a Fredholm operator of index zero and N be L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq56_HTML.gif. The equation L x = N x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq57_HTML.gifhas at least one solution in dom L Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq58_HTML.gifif the following conditions are satisfied:
        1. (i)

          L x λ N x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq59_HTML.gif for each ( x , λ ) [ dom L Ker L Ω ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq60_HTML.gif;

           
        2. (ii)

          N x Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq61_HTML.gif for each x Ker L Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq62_HTML.gif;

           
        3. (iii)

          deg ( J Q N | Ker L , Ker L Ω , 0 ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq63_HTML.gif,

           

        where Q : Z Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq49_HTML.gifis a projection such that Ker Q = Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq64_HTML.gifand J : Im Q Ker L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq65_HTML.gifis a any isomorphism.

        In this article, we use the Banach space C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq66_HTML.gif with the norm u = max t [ 0 , 1 ] | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq67_HTML.gif.

        Lemma 2.7 ([16])

        Given μ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq68_HTML.gifand N = [ μ ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq69_HTML.gif, for any x C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq70_HTML.gif, c i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq71_HTML.gif ( i = 1 , 2 , , N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq72_HTML.gif), we can define a Banach space
        C μ [ 0 , 1 ] = { u ( t ) | u ( t ) = I 0 + μ x ( t ) + c 1 t μ 1 + c 2 t μ 2 + + c N 1 t μ ( N 1 ) , t ( 0 , 1 ) } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equh_HTML.gif

        with the norm defined by u C μ = D 0 + μ u + + D 0 + μ ( N 1 ) u + u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq73_HTML.gif.

        Lemma 2.8 ([16])

        E C μ [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq74_HTML.gifis a sequentially compact set if and only if E is uniformly bounded and equicontinuous. Here, a uniform bound means that there exists a constant M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq75_HTML.gifwith each u E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq76_HTML.gif, such that
        u C μ = D 0 + μ u + + D 0 + μ ( N 1 ) u + u < M , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equi_HTML.gif
        and equicontinuation means that there exists a δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq77_HTML.gifwith | t 1 t 2 | < δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq78_HTML.giffor any t 1 , t 2 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq79_HTML.gif, u E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq76_HTML.gifand ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq80_HTML.gif, such that
        | u ( t 1 ) u ( t 2 ) | < ε , | D 0 + α i u ( t 1 ) D 0 + α i u ( t 2 ) | < ε ( i = 1 , 2 , , N 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equj_HTML.gif
        In this article, let Z = L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq81_HTML.gif with the norm y 1 = 0 1 | y ( s ) | d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq82_HTML.gif and Y = C α 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq83_HTML.gif with the norm u Y = D 0 + α 1 u + D 0 + α 2 u + u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq84_HTML.gif. Define the operator L : dom L Y Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq85_HTML.gif by
        L u = D 0 + α u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ4_HTML.gif
        (2.1)
        where dom L = { u C α 1 [ 0 , 1 ] | D 0 + α u Z , u satisfies (1.2) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq86_HTML.gif. Define the operator N : Y Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq53_HTML.gif by
        N u ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ5_HTML.gif
        (2.2)

        Thus, BVP (1.1) can be written as L u = N u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq87_HTML.gif for each u dom L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq88_HTML.gif.

        3 Main results

        First, let us introduce the following notations for convenience, with setting p { 1 , 2 , , m 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq89_HTML.gif and q Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq90_HTML.gif with q p + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq91_HTML.gif,
        Λ 1 : = 1 p ( p + 1 ) ( 1 i = 1 m b i η i p + 1 ) , Λ 2 : = 1 q ( q 1 ) ( 1 i = 1 m b i η i q ) , Λ 3 : = 1 p ( 1 i = 1 m a i ξ i p ) , Λ 4 : = 1 q 1 ( 1 i = 1 m a i ξ i q 1 ) , Λ : = Λ 1 Λ 4 Λ 2 Λ 3 , μ : = 3 + 1 Γ ( α ) + 1 Γ ( α 1 ) , ω : = 2 + 1 Γ ( α ) , ρ : = 5 + 2 Γ ( α ) + 1 Γ ( α 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equk_HTML.gif

        Then, let us make some assumptions which will be used throughout the article.

        (H1) There exist functions h ( t ) , r ( t ) , s ( t ) , w ( t ) , e ( t ) L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq92_HTML.gif and a constant θ [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq93_HTML.gif such that for all ( x , y , z ) R 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq94_HTML.gif, t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif,
        | f ( t , x , y , z ) | h ( t ) | x | + r ( t ) | y | + s ( t ) | z | + w ( t ) | z | θ + e ( t ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equl_HTML.gif
        (H2) For any u dom L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq88_HTML.gif, t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif, there exists a constant A > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq95_HTML.gif such that if D 0 + α 1 u ( t ) > A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq96_HTML.gif, then either
        Λ 2 T 1 N u Λ 4 T 2 N u < 0 , or Λ 2 T 1 N u Λ 4 T 2 N u > 0 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equm_HTML.gif
        (H3) For any u dom L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq88_HTML.gif, t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif, there exists a constant B > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq97_HTML.gif such that if D 0 + α 2 u ( t ) > B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq98_HTML.gif, then either
        Λ 1 T 1 N u Λ 3 T 2 N u > 0 , or Λ 1 T 1 N u Λ 3 T 2 N u < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equn_HTML.gif

        Theorem 3.1 If conditions (C), (H 1)-(H 3) hold, then BVPs (1.1)-(1.2) have at least one solution provided that ρ ( h 1 + r 1 + s 1 ) < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq99_HTML.gif.

        In order to obtain our main result, we first present and prove Lemmas 3.2-3.8. Now, let us define operators T j : Z Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq100_HTML.gif ( j = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq101_HTML.gif) as follows:
        T 1 x ( t ) = 0 1 x ( s ) d s i = 1 m a i 0 ξ i x ( s ) d s , t [ 0 , 1 ] , T 2 x ( t ) = 0 1 ( 1 s ) x ( s ) d s i = 1 m b i 0 η i ( η i s ) x ( s ) d s , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equo_HTML.gif
        Lemma 3.2 If condition (C) holds and L is defined by (2.1), then
        Ker L = { a t α 1 + b t α 2 | a , b R } , Im L = { x Z | T j x = 0 , j = 1 , 2 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equp_HTML.gif
        Proof By (2.1) and Lemma 2.5, D 0 + α u ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq21_HTML.gif has a solution
        u ( t ) = D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 + D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) t α 2 + I 0 + 3 α u ( t ) | t = 0 Γ ( α 2 ) t α 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equq_HTML.gif

        Combining with the condition (1.2), we get Ker L = { a t α 1 + b t α 2 | a , b R } R 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq102_HTML.gif.

        Suppose x Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq103_HTML.gif, then there exists u dom L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq88_HTML.gif such that x = L u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq104_HTML.gif, i.e., u Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq105_HTML.gif, x = D 0 + α u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq106_HTML.gif. By Lemma 2.5, we have
        I 0 + α x ( t ) = u ( t ) D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) t α 2 I 0 + 3 α u ( t ) | t = 0 Γ ( α 2 ) t α 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equr_HTML.gif
        Then in view of condition (C), (1.2) and Lemma 2.4, x satisfies
        { 0 1 x ( s ) d s i = 1 m a i 0 ξ i x ( s ) d s = 0 , 0 1 ( 1 s ) x ( s ) d s i = 1 m b i 0 η i ( η i s ) x ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ6_HTML.gif
        (3.1)
        On the other hand, suppose x Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq107_HTML.gif and it satisfies (3.1), let u ( t ) = I 0 + α x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq108_HTML.gif, then u dom L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq88_HTML.gif, D 0 + α u ( t ) = x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq109_HTML.gif, i.e., x Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq110_HTML.gif. Therefore, we obtain that
        Im L = { x Z | T j x = 0 , j = 1 , 2 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equs_HTML.gif

         □

        Lemma 3.3 If condition (C) holds, then there exist two constants q Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq90_HTML.gifand p { 1 , 2 , , m 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq89_HTML.gifwith q p + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq91_HTML.gifsuch that Λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq111_HTML.gif.

        Proof From i = 1 m a i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq112_HTML.gif, we obtain that for any nonnegative integer l, there exists k l 1 { l m + 1 , , ( l + 1 ) m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq113_HTML.gif such that i = 1 m a i ξ i k l 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq114_HTML.gif. If else, we obtain that i = 1 m a i ξ i k l 1 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq115_HTML.gif, k l 1 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq116_HTML.gif, l m + 1 , , ( l + 1 ) m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq117_HTML.gif.

        If l = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq118_HTML.gif, we have
        ( 1 1 1 ξ 1 ξ 2 ξ m ξ 1 m ξ 2 m ξ m m ) ( a 1 a 2 a m ) = ( 1 1 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equt_HTML.gif
        It is equal to
        ( 1 ξ 1 1 ξ 2 1 ξ m ξ 1 ( 1 ξ 1 ) ξ 2 ( 1 ξ 2 ) ξ m ( 1 ξ m ) ξ 1 m 1 ( 1 ξ 1 ) ξ 2 m 1 ( 1 ξ 2 ) ξ m m 1 ( 1 ξ m ) ) ( a 1 a 2 a m ) = ( 0 0 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equu_HTML.gif

        Since the determinant of coefficients is not equal to zero, we have that a i = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq119_HTML.gif ( i = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq120_HTML.gif), which is a contradiction to condition (C).

        If l Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq121_HTML.gif, we get
        ( 1 1 1 ξ 1 l m + 1 ξ 2 l m + 1 ξ m l m + 1 ξ 1 l m + m ξ 2 l m + m ξ m l m + m ) ( a 1 a 2 a m ) = ( 1 1 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equv_HTML.gif

        Similarly, we can deduce that the determinant of coefficients is not equal to zero, so we have that a i = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq119_HTML.gif ( i = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq120_HTML.gif), which is a contradiction to condition (C). Thus, there exists k l 1 { l m + 1 , , ( l + 1 ) m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq122_HTML.gif such that i = 1 m a i ξ i k l 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq114_HTML.gif.

        Similarly, from i = 1 m b i = i = 1 m b i η i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq123_HTML.gif, we have that there exists a constant p { 1 , 2 , , m 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq124_HTML.gif such that
        i = 1 m b i η i p + 1 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ7_HTML.gif
        (3.2)
        Let
        S = { ( k l 1 ) Z + | ( p + 1 ) ( 1 i = 1 m b i η i k l ) ( 1 i = 1 m a i ξ i p ) k l ( 1 i = 1 m a i ξ i k l 1 ) = 1 i = 1 m b i η i p + 1 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equw_HTML.gif
        we shall prove that S is a finite set. If else, there exists a strict increasing sequence { k l n } n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq125_HTML.gif such that
        ( p + 1 ) ( 1 i = 1 m b i η i k l n ) ( 1 i = 1 m a i ξ i p ) k l n ( 1 i = 1 m a i ξ i k l n 1 ) = 1 i = 1 m b i η i p + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equx_HTML.gif
        Since i = 1 m b i η i p + 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq126_HTML.gif, we have i = 1 m a i ξ i p 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq127_HTML.gif. Thus,
        1 i = 1 m b i η i p + 1 = lim k l n + ( p + 1 ) ( 1 i = 1 m b i η i k l n ) ( 1 i = 1 m a i ξ i p ) k l n ( 1 i = 1 m a i ξ i k l n 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equy_HTML.gif

        which is a contradiction to (3.2). Therefore, there exists two constants p { 1 , 2 , , m 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq89_HTML.gif and q Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq90_HTML.gif with q p + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq91_HTML.gif such that Λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq111_HTML.gif. □

        Lemma 3.4 If the condition (C) holds and L is defined by (2.1), then L is a Fredholm operator of index zero. Define the linear operator K p : Im L dom L Ker P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq128_HTML.gifwith K P x = I 0 + α x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq129_HTML.gif, then it is the inverse of L. Furthermore, we have
        K p x Y ω x 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equz_HTML.gif
        Proof For each p { 1 , 2 , , m 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq89_HTML.gif and q Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq90_HTML.gif with q p + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq91_HTML.gif, define operator Q : Z Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq49_HTML.gif by
        Q x ( t ) = ( Q 1 x ( t ) ) t p 1 + ( Q 2 x ( t ) ) t q 2 , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ8_HTML.gif
        (3.3)
        where
        Q 1 x ( t ) = 1 Λ [ Λ 2 T 1 x ( t ) + Λ 4 T 2 x ( t ) ] , Q 2 x ( t ) = 1 Λ [ Λ 1 T 1 x ( t ) Λ 3 T 2 x ( t ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ9_HTML.gif
        (3.4)
        It is clear that dim Im Q 1 = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq130_HTML.gif. It follows from (3.4), the definition of T 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq131_HTML.gif and T 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq132_HTML.gif that
        Q 1 ( ( Q 1 x ) t p 1 ) = 1 Λ [ Λ 2 T 1 ( ( Q 1 x ) t p 1 ) + Λ 4 T 2 ( ( Q 1 x ) t p 1 ) ] = 1 Λ [ Λ 2 Λ 3 ( Q 1 x ) + Λ 1 Λ 4 ( Q 1 x ) ] = Q 1 x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ10_HTML.gif
        (3.5)
        similarly, we can derive that
        Q 1 ( ( Q 2 x ) t q 2 ) = 0 , Q 2 ( ( Q 1 x ) t p 1 ) = 0 , Q 2 ( ( Q 2 x ) t q 2 ) = Q 2 x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ11_HTML.gif
        (3.6)
        Hence, for each x Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq133_HTML.gif and t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif, it follows from the (3.3)-(3.6) that
        Q 2 x = Q 1 [ ( Q 1 x ) t p 1 + ( Q 2 x ) t q 2 ] t p 1 + Q 2 [ ( Q 1 x ) t p 1 + ( Q 2 x ) t q 2 ] t q 2 = ( Q 1 x ) t p 1 + ( Q 2 x ) t q 2 = Q x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equaa_HTML.gif

        Furthermore, Q is a continuous linear projector.

        For each x Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq110_HTML.gif, we have Q x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq134_HTML.gif, i.e., x Ker Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq135_HTML.gif. On the other hand, for each x Ker Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq135_HTML.gif, we have that
        { Λ 2 T 1 x + Λ 4 T 2 x = 0 , Λ 1 T 1 x Λ 3 T 2 x = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equab_HTML.gif
        However, the determinant of coefficients is as follows
        | Λ 2 Λ 4 Λ 1 Λ 3 | = Λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equac_HTML.gif

        then we have T j x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq136_HTML.gif ( j = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq137_HTML.gif), i.e., x Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq110_HTML.gif. Thus, Ker Q = Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq64_HTML.gif.

        Take any x Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq133_HTML.gif in the type x = ( x Q x ) + Q x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq138_HTML.gif, obviously, x Q x Ker Q = Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq139_HTML.gif and Q x Im Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq140_HTML.gif, so Z = Im L + Im Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq141_HTML.gif. For any x Im L Im Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq142_HTML.gif with x = a t p 1 + b t q 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq143_HTML.gif, by Lemma 3.2, we have
        0 1 ( a s p 1 + b s q 2 ) d s i = 1 m a i 0 ξ i ( a s p 1 + b s q 2 ) d s = 0 , 0 1 ( 1 s ) ( a s p 1 + b s q 2 ) d s i = 1 m b i 0 η i ( η i s ) ( a s p 1 + b s q 2 ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equad_HTML.gif
        That is,
        { a Λ 3 + b Λ 4 = 0 , a Λ 1 + b Λ 2 = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equae_HTML.gif
        but the determinant of coefficients is as follows
        | Λ 3 Λ 4 Λ 1 Λ 2 | = Λ 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equaf_HTML.gif

        we can deduce that a = b = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq144_HTML.gif. Hence, Im L Im Q = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq145_HTML.gif. Furthermore, we get Z = Im L Im Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq146_HTML.gif. Therefore, dim Ker L = dim Im Q = codim Im L = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq147_HTML.gif, which means that L is a Fredholm operator of index zero.

        Let operator P : Y Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq48_HTML.gif and
        P u ( t ) = D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 + D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) t α 2 , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ12_HTML.gif
        (3.7)
        It is easy to calculate that P u ( t ) = P 2 u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq148_HTML.gif; furthermore, P is a continuous linear projector. Obviously
        Ker P = { u Y | D 0 + α 1 u ( 0 ) = D 0 + α 2 u ( 0 ) = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equag_HTML.gif

        It is clear that Y = Ker L Ker P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq149_HTML.gif.

        For any x Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq110_HTML.gif, in view of the definition of operators Kp and L, we have L K P x = L I 0 + α x = D 0 + α I 0 + α x = x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq150_HTML.gif. On the other hand, if u dom L Ker P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq151_HTML.gif, we have D 0 + α 1 u ( 0 ) = D 0 + α 2 u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq152_HTML.gif, u dom L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq88_HTML.gif. Therefore, by Lemma 2.5 and definitions of operators K p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq51_HTML.gif and L, we know that ( K p L ) u = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq153_HTML.gif, which implies that K p = [ L | dom L Ker P ] 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq154_HTML.gif. By the definition of K p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq51_HTML.gif, we have
        K p x ( t ) = I 0 + α x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 x ( s ) d s , x Im L . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equah_HTML.gif
        It follows from Lemma 2.4 that
        D 0 + α 1 ( K p x ) ( t ) = 0 t ( t s ) x ( s ) d s , D 0 + α 2 ( K p x ) ( t ) = 0 t x ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equai_HTML.gif
        Then, we have
        K p x 1 Γ ( α ) x 1 , D 0 + α 1 ( K p x ) x 1 , D 0 + α 2 ( K p x ) x 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equaj_HTML.gif

        By the definition of the norm in space Y, we get K p x Y ω x 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq155_HTML.gif. □

        Lemma 3.5 Assume Ω Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq156_HTML.gifis an open bounded subset such that dom L Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq157_HTML.gif, and N is defined by (2.2), then N is L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq56_HTML.gif.

        Proof In order to prove N is L-compact, we only need to prove that Q N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq54_HTML.gif is bounded and K p ( I Q ) N ( u ) : Ω ¯ Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq158_HTML.gif is compact. Since the function f satisfies Carathéodory conditions and u Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq159_HTML.gif, for each E > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq36_HTML.gif, there exists a ρ E ( t ) L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq160_HTML.gif such that, for a.e. t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif and every | u | E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq38_HTML.gif, we have f ρ E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq161_HTML.gif. By the definition of operators Q and K p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq51_HTML.gif on the interval [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq162_HTML.gif, it is easy to get that Q N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq54_HTML.gif and K p ( I Q ) N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq163_HTML.gif are bounded. Thus, there exists a constant r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq164_HTML.gif with each t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq165_HTML.gif, such that | Q N u ( t ) | r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq166_HTML.gif.

        For all 0 t 1 < t 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq167_HTML.gif, 2 < α 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq14_HTML.gif, u Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq168_HTML.gif, we have
        | K p ( I Q ) N u ( t 2 ) K p ( I Q ) N u ( t 1 ) | = 1 Γ ( α ) | 0 t 1 [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] ( I Q ) N u ( s ) d s + t 1 t 2 ( t 2 s ) α 1 ( I Q ) N u ( s ) d s | 1 Γ ( α ) { 0 t 1 [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] ( r + | ρ E ( s ) | ) d s + t 1 t 2 ( t 2 s ) α 1 ( r + | ρ E ( s ) | ) d s } 1 Γ ( α ) { 0 t 2 | ρ E ( s ) | ( t 2 s ) α 1 d s 0 t 1 | ρ E ( s ) | ( t 1 s ) α 1 d s } + r Γ ( α + 1 ) ( t 2 α t 1 α ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equak_HTML.gif
        and
        | D α 1 K p ( I Q ) N u ( t 2 ) D α 1 K p ( I Q ) N u ( t 1 ) | = | 0 t 2 ( I Q ) N u ( s ) d s 0 t 1 ( I Q ) N u ( s ) d s | r ( t 2 t 1 ) + t 1 t 2 | ρ E ( s ) | d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equal_HTML.gif

        Since t α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq169_HTML.gif is uniformly continuous on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq162_HTML.gif and ρ E ( t ) L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq160_HTML.gif, so K p ( I Q ) N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq170_HTML.gif and D α 1 K p ( I Q ) N ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq171_HTML.gif are equicontinuous. By Lemma 2.8, we get that K p ( I Q ) N : Y Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq172_HTML.gif is completely continuous. □

        Lemma 3.6 Suppose (H 1)-(H 3) hold, then the set Ω 1 = { u dom L Ker L : L u = λ N u , λ [ 0 , 1 ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq173_HTML.gifis bounded.

        Proof Taking any u Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq174_HTML.gif, then we have L u = λ N u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq175_HTML.gif, which yields λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq176_HTML.gif and N u Im L = Ker Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq177_HTML.gif, i.e., Q N u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq178_HTML.gif for all t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif. It follows from (H2) and (H3) that there exists t 0 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq179_HTML.gif such that | D 0 + α 1 u ( t 0 ) | + | D 0 + α 2 u ( t 0 ) | A + B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq180_HTML.gif. Then we can get that
        D 0 + α 1 u ( t ) = D 0 + α 1 u ( t 0 ) + t 0 t D 0 + α u ( s ) d s , D 0 + α 2 u ( t ) = D 0 + α 2 u ( t 0 ) + t 0 t D 0 + α 1 u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equam_HTML.gif
        Furthermore, we have that, with setting M = A + B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq181_HTML.gif,
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ13_HTML.gif
        (3.8)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ14_HTML.gif
        (3.9)
        By (3.7)-(3.9) and Lemma 2.4, we have that
        P u Y μ ( M + N u 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equan_HTML.gif
        As before, for any u Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq174_HTML.gif, we have ( I P ) ( u ) dom L Ker P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq182_HTML.gif and L P ( u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq183_HTML.gif. From Lemma 3.4 and for each λ ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq184_HTML.gif, we can get
        ( I P ) ( u ) Y = K p L ( I P ) ( u ) Y = K p ( L u ) Y ω N u 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equao_HTML.gif
        Furthermore, we have
        u Y ( I P ) ( u ) Y + P ( u ) Y ρ N u 1 + μ M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equap_HTML.gif
        By (H1) and the definition of N, we have
        u Y ρ [ h 1 u + r 1 D 0 + α 1 u s 1 D 0 + α 2 u + w 1 D 0 + α 2 u θ + D ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equaq_HTML.gif
        where D = e 1 + μ M / ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq185_HTML.gif. Since max { u , D 0 + α 1 u , D 0 + α 2 u } u Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq186_HTML.gif and ρ ( h 1 + r 1 + s 1 ) < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq99_HTML.gif hold true, we can get that
        u ρ 1 ρ h 1 [ r 1 D 0 + α 1 u + s 1 D 0 + α 2 u + w 1 D 0 + α 2 u θ + D ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equar_HTML.gif
        which yield that
        D 0 + α 1 u ρ 1 ρ h 1 ρ r 1 [ s 1 D 0 + α 2 u + w 1 D 0 + α 2 u θ + D ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equas_HTML.gif
        Furthermore, from the previous inequalities, we know that
        D 0 + α 2 u ρ 1 ρ h 1 ρ r 1 ρ s 1 ( w 1 D 0 + α 2 u θ + D ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equat_HTML.gif
        Since θ [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq93_HTML.gif, there exist constants m 1 , m 2 , m 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq187_HTML.gif such that
        D 0 + α 2 u m 1 , D 0 + α 1 u m 2 , u m 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equau_HTML.gif

        Therefore, Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq188_HTML.gif is bounded. □

        Lemma 3.7 Suppose (H 2) and (H 3) hold, then the set Ω 2 = { u Ker L : N u Im L } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq189_HTML.gifis bounded.

        Proof For any u Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq190_HTML.gif and a , b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq24_HTML.gif, then u ( t ) = a t α 1 + b t α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq191_HTML.gif and Q N u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq178_HTML.gif. By (H2), we get that | D 0 + α 1 u ( t ) | = | a Γ ( α ) | A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq192_HTML.gif, then we have | a | A / Γ ( α ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq193_HTML.gif. By (H3), we have that | D 0 + α 2 u ( t ) | = | a t Γ ( α ) + b Γ ( α 1 ) | B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq194_HTML.gif, thus | b | ( B + A ) / Γ ( α 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq195_HTML.gif. Therefore, Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq196_HTML.gif is bounded. □

        Lemma 3.8 If the first parts of (H 2) and (H 3) hold, then the set Ω 3 = { u Ker L : λ J 1 u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq197_HTML.gifis bounded.

        Proof Taking any u Ω 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq198_HTML.gif and a , b R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq24_HTML.gif, we have u ( t ) = a t α 1 + b t α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq22_HTML.gif. For all t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif, we define the isomorphism J 1 : Ker L Im Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq199_HTML.gif by
        J 1 ( a t α 1 + b t α 2 ) = a 2 Λ t p 1 + b 2 Λ t q 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equav_HTML.gif
        By the definition of the set Ω 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq200_HTML.gif, we can get that
        { λ a 2 + ( 1 λ ) [ Λ 2 T 1 N ( a t α 1 + b t α 2 ) + Λ 4 T 2 N ( a t α 1 + b t α 2 ) ] = 0 , λ b 2 + ( 1 λ ) [ Λ 1 T 1 N ( a t α 1 + b t α 2 ) Λ 3 T 2 N ( a t α 1 + b t α 2 ) ] = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ15_HTML.gif
        (3.10)
        If λ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq201_HTML.gif, we have
        T 1 N ( a t α 1 + b t α 2 ) = 0 , T 2 N ( a t α 1 + b t α 2 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equaw_HTML.gif
        By the first parts of (H2) and (H3), similar to the proof of Lemma 3.7, then
        | a | A Γ ( α ) , | b | B + A Γ ( α 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equax_HTML.gif

        Therefore, Ω 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq200_HTML.gif is bounded.

        If λ = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq202_HTML.gif, we have a = b = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq144_HTML.gif.

        If 0 < λ < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq203_HTML.gif, we get that | D α 1 u ( t ) | A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq204_HTML.gif and | D α 2 u ( t ) | B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq205_HTML.gif, similar to the proof of Lemma 3.7, Ω 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq200_HTML.gif is bounded. If else, we have that Λ 2 T 1 N u Λ 4 T 2 N u < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq206_HTML.gif and Λ 1 T 1 N u Λ 3 T 2 N u > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq207_HTML.gif. It contradicts (3.10), thus Ω 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq200_HTML.gif is bounded. □

        Remark 3.9 If the other parts of (H2) and (H3) hold, then the set Ω 3 = { u Ker L : λ J 1 u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq208_HTML.gif is bounded.

        Now with Lemmas 3.2-3.8 in hands, we can begin to prove our main result - Theorem 3.1.

        Proof of Theorem 3.1 Assume that Ω is a bounded open set of Y with i = 1 3 Ω ¯ Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq209_HTML.gif. By Lemma 3.5, N is L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq56_HTML.gif. Then by Lemmas 3.6 and 3.7, we have
        1. (i)

          L u λ N u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq210_HTML.gif for every ( u , λ ) [ dom L Ker L Ω ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq211_HTML.gif;

           
        2. (ii)

          N u Im L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq212_HTML.gif for every u Ker L Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq213_HTML.gif.

           
        Finally, we will prove that (iii) of Lemma 2.6 is satisfied. We let I as the identity operator in the Banach space Y and H ( u , λ ) = ± λ J 1 ( u ) + ( 1 λ ) Q N ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq214_HTML.gif, according to Lemma 3.8 (or Remark 3.9) we know that for all u Ω Ker L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq215_HTML.gif, H ( u , λ ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq216_HTML.gif. By the homotopic property of degree, we have
        deg ( J Q N | Ker L , Ker L Ω , 0 ) = deg ( H ( , 0 ) , Ker L Ω , 0 ) = deg ( H ( , 1 ) , Ker L Ω , 0 ) = deg ( ± I , Ker L Ω , 0 ) 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equay_HTML.gif

        so (iii) of Lemma 2.6 is satisfied.

        Consequently, by Lemma 2.6, the equation L u = N u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq87_HTML.gif has at least one solution in dom L Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq58_HTML.gif. Namely, BVPs (1.1)-(1.2) have at least one solution in the space Y. □

        According to Theorem 3.1, we have the following corollary.

        Corollary 3.10 Suppose that (H 1) is replaced by the following condition,

        (H4) there exist functions h ( t ) , r ( t ) , s ( t ) , w ( t ) , e ( t ) L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq92_HTML.gifand a constant θ [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq93_HTML.gifsuch that for all ( x , y , z ) R 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq94_HTML.gif, t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq34_HTML.gif,

        | f ( t , x , y , z ) | h ( t ) | x | + r ( t ) | y | + s ( t ) | z | + w ( t ) | y | θ + e ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equaz_HTML.gif
        or
        | f ( t , x , y , z ) | h ( t ) | x | + r ( t ) | y | + s ( t ) | z | + w ( t ) | x | θ + e ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equba_HTML.gif

        and the others in Theorem 3.1 are not changed, then BVPs (1.1)-(1.2) have at least one solution.

        4 An example

        Example Consider the following boundary value problem for all t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq217_HTML.gif:
        { D 13 5 u ( t ) = 7 u ( t ) 200 + 9 D 0 + 8 5 u ( t ) 500 + D 0 + 3 5 u ( t ) 60 + 3 sin 1 7 u ( t ) 80 + cos 2 t + 1 , I 0 + 2 5 u ( 0 ) = 0 , D 0 + 8 5 u ( 1 ) = 2 5 D 0 + 8 5 u ( 1 3 ) + 3 5 D 0 + 8 5 u ( 1 2 ) , D 0 + 3 5 u ( 1 ) = 6 D 0 + 3 5 u ( 1 8 ) + 7 D 0 + 3 5 u ( 1 4 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equ16_HTML.gif
        (4.1)
        Let α = 13 / 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq218_HTML.gif, and a 1 = 2 / 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq219_HTML.gif, ξ 1 = 1 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq220_HTML.gif, a 2 = 3 / 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq221_HTML.gif, ξ 2 = 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq222_HTML.gif, b 1 = 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq223_HTML.gif, η 1 = 1 / 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq224_HTML.gif, b 2 = 7 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq225_HTML.gif, η 2 = 1 / 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq226_HTML.gif. We can get that the condition (C) holds, i.e., i = 1 2 a i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq227_HTML.gif, i = 1 2 b i η i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq228_HTML.gif, i = 1 2 b i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq229_HTML.gif. Moreover,
        f ( t , x , y , z ) = 7 200 x + 9 500 y + 1 60 z + 3 80 sin 1 / 7 x + cos 2 t + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equbb_HTML.gif
        Thus, we have
        | f ( t , x , y , z ) | 7 200 | x | + 9 500 | y | + 1 60 | z | + 3 80 | x | 1 / 7 + 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equbc_HTML.gif
        Taking h = 7 / 200 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq230_HTML.gif, r = 9 / 500 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq231_HTML.gif, s = 1 / 60 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq232_HTML.gif, w = 3 / 80 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq233_HTML.gif, e = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq234_HTML.gif, Λ 1 = ( 1 i = 1 2 b i η i 2 ) / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq235_HTML.gif, Λ 2 = ( 1 i = 1 2 b i η i 2 ) / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq236_HTML.gif, Λ 3 = 1 i = 1 2 a i ξ i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq237_HTML.gif, Λ 4 = 1 i = 1 2 a i ξ i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq238_HTML.gif, μ = 3 + Γ 1 ( 13 / 5 ) + Γ 1 ( 8 / 5 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq239_HTML.gif, ω = 2 + Γ 1 ( 13 / 5 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq240_HTML.gif, ρ = 5 + 2 Γ 1 ( 13 / 5 ) + Γ 1 ( 8 / 5 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq241_HTML.gif, we can calculate that (H1)-(H3) hold. Furthermore, we can get
        ρ ( h 1 + r 1 + s 1 ) = 0.52754 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_Equbd_HTML.gif

        By Corollary 3.10, the BVP (4.1) has at least one solution in C 8 / 5 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-65/MediaObjects/13661_2012_Article_165_IEq242_HTML.gif.

        Author’s contributions

        NX designed all the steps of proof in this research and also wrote the article. WBL suggested many good ideas in this article. LSX helped to draft the first manuscript and gave an example to illustrate our result. All authors read and approved the final manuscript.

        Declarations

        Acknowledgement

        The authors would like to acknowledge the anonymous referee for many helpful comments and valuable suggestions on this article. This work is sponsored by Fundamental Research Funds for the Central Universities (2012LWB44).

        Authors’ Affiliations

        (1)
        College of Sciences, China University of Mining and Technology

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