Monotone and convex positive solutions for fourth-order multi-point boundary value problems

  • Yang Liu1, 2Email author,

    Affiliated with

    • Zhang Weiguo1 and

      Affiliated with

      • Shen Chunfang2

        Affiliated with

        Boundary Value Problems20112011:21

        DOI: 10.1186/1687-2770-2011-21

        Received: 27 December 2010

        Accepted: 5 September 2011

        Published: 5 September 2011

        Abstract

        The existence results of multiple monotone and convex positive solutions for some fourth-order multi-point boundary value problems are established. The nonlinearities in the problems studied depend on all order derivatives. The analysis relies on a fixed point theorem in a cone. The explicit expressions and properties of associated Green's functions are also given.

        MSC: 34B10; 34B15.

        Keywords

        multi-point boundary value problem positive solution cone fixed point

        1 Introduction

        Boundary value problems for second and higher order nonlinear differential equations play a very important role in both theory and applications. For example, the deformations of an elastic beam in the equilibrium state can be described as a boundary value problem of some fourth-order differential equations. Owing to its importance in application, the existence of positive solutions for nonlinear second and higher order boundary value problems has been studied by many authors. We refer to recent contributions of Ma [13], He and Ge [4], Guo and Ge [5], Avery et al. [6, 7], Henderson [8], Eloe and Henderson [9], Yang et al. [10], Webb and Infante [11, 12], and Agarwal and O'Regan [13]. For survey of known results and additional references, we refer the reader to the monographs by Agarwal [14] and Agarwal et al. [15].

        When it comes to positive solutions for nonlinear fourth-order ordinary differential equations, two point boundary value problems are studied extensively, see [1624]. Few papers deal with the multi-point cases. Furthermore, for nonlinear fourth-order equations, only the situation that the nonlinear term does not depend on the first, second and third order derivatives are considered, see [1623]. Few paper deals with the situation that lower order derivatives are involved in the nonlinear term explicitly. In fact, the derivatives are of great importance in the problem in some cases. For example, in the linear elastic beam equation (Euler-Bernoulli equation)
        ( EI u ( t ) ) = f ( t ) , t ( 0 , L ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equa_HTML.gif

        where u(t) is the deformation function, L is the length of the beam, f(t) is the load density, E is the Young's modulus of elasticity and I is the moment of inertia of the cross-section of the beam. In this problem, the physical meaning of the derivatives of the function u(t) is as follows: u(4)(t) is the load density stiffness, u'''(t) is the shear force stiffness, u''(t) is the bending moment stiffness and the u'(t) is the slope. If the payload depends on the shear force stiffness, bending moment stiffness or the slope, the derivatives of the unknown function are involved in the nonlinear term explicitly.

        In this paper, we are interested in the positive solution for fourth-order nonlinear differential equation
        x ( 4 ) ( t ) = f ( t , x ( t ) , x ( t ) , x ( t ) , x ( t ) ) , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ1_HTML.gif
        (1.1)
        subject to multi-point boundary condition
        x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = 0 , x ( 1 ) = i = 1 m - 2 β i x ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ2_HTML.gif
        (1.2)
        or
        x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ3_HTML.gif
        (1.3)

        where 0 < ξ1 < ξ2 < ⋯ < ξm-2< 1, β i > 0, 1 = 1, 2, ..., m - 2, i = 1 m - 2 β i > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq1_HTML.gif, and fC([0, 1] × R4, [0, +∞)).

        One can see that all lower order derivatives are involved in the nonlinear term explicitly and the BCs are the m-point cases. In this sense, the problems studied in this paper are more general than before. In the paper, multiple monotone and convex positive solutions for problems (1.1), (1.2) and (1.1), (1.3) are established. The results presented extend the study for fourth-order boundary value problems of nonlinear ordinary differential equations.

        This paper is organized as follows. In Section 2, we present some preliminaries and lemmas. Section 3 is devoted to the existence of at least three convex and increasing positive solutions for problem (1.1), (1.2). In Section 4, we prove that there exist at least three convex and decreasing positive solutions for problem (1.1), (1.3).

        2 Preliminaries and lemmas

        In this section, some preliminaries and lemmas used later are presented.

        Definition 2.1 The map α is said to be a nonnegative continuous convex functional on cone P of a real Banach space E provided that α : P → [0, +∞) is continuous and
        α ( t x + ( 1 - t ) y ) t α ( x ) + ( 1 - t ) α ( y ) for all  x , y P and  t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equb_HTML.gif
        Definition 2.2 The map β is said to be a nonnegative continuous concave functional on cone P of a real Banach space E provided that β : P → [0, +∞) is continuous and
        β ( t x + ( 1 - t ) y ) t β ( x ) + ( 1 - t ) β ( y ) , for all  x , y P and  t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equc_HTML.gif
        Let γ, θ be nonnegative continuous convex functionals on P, α be a nonnegative continuous concave functional on P and ψ be a nonnegative continuous functional on P. Then for positive numbers a, b, c and d, we define the following convex sets:
        P ( γ , d ) = { x P | γ ( x ) < d } , P ( γ , α , b , d ) = { x P | b α ( x ) , γ ( x ) d } , P ( γ , θ , α , b , c , d ) = { x P | b α ( x ) , θ ( x ) c , γ ( x ) d } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equd_HTML.gif
        and a closed set
        R ( γ , ψ , a , d ) = { x P | a ψ ( x ) , γ ( x ) d } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Eque_HTML.gif
        Lemma 2.1 [25] Let P be a cone in Banach space E. Let γ, θ be nonnegative continuous convex functionals on P, α be a nonnegative continuous concave functional and ψ be a nonnegative continuous functional on P satisfying
        ψ ( λ x ) λ ψ ( x ) , f o r 0 λ 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equf_HTML.gif
        such that for some positive numbers l and d,
        α ( x ) ψ ( x ) , x l γ ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equg_HTML.gif

        for all x P ( γ , d ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq2_HTML.gif. Suppose T : P ( γ , d ) ¯ P ( γ , d ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq3_HTML.gif is completely continuous and there exist positive numbers a, b, c with a < b such that

        (S1) {xP (γ, θ, α, b, c, d)|α(x) > b} ≠ ∅ and α(Tx) > b for xP (γ, θ, α, b, c, d);

        (S2) α(Tx) > b for xP (γ, α, b, d) with θ(Tx) > c;

        (S3) 0 ∉ R(γ, ψ, a, d) and ψ(Tx) < a for xR(γ, ψ, a, d) with ψ(x) = a.

        Then T has at least three fixed points x1, x2, x 3 P ( γ , d ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq4_HTML.gif such that:
        γ ( x i ) d , i = 1 , 2 , 3 ; b < α ( x 1 ) ; a < ψ ( x 2 ) , α ( x 2 ) < b ; ψ ( x 3 ) < a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equh_HTML.gif

        3 Positive solutions for problem (1.1), (1.2)

        We begin with the fourth-order m-point boundary value problem
        x ( 4 ) ( t ) = y ( t ) , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ4_HTML.gif
        (3.1)
        x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = 0 , x ( 1 ) = i = 1 m - 2 β i x ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ5_HTML.gif
        (3.2)

        where 0 < ξ1 < ξ2 < ⋯ < ξm-2< 1, β i > 0, i = 1, 2, ..., m - 2.

        The following assumption will stand throughout this section:
        ( H 1 ) f C ( [ 0 , 1 ] × R 4 , [ 0 , + ) ) , i = 1 m - 2 β i > 1 , i = 1 m - 2 β i ξ i < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equi_HTML.gif
        Lemma 3.1 Denote ξ0 = 0, ξm -1= 1, β0 = βm -1= 0, and y(t) ∈ C[0, 1]. Problem (3.1), (3.2) has the unique solution
        x ( t ) = 0 1 G ( t , s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equj_HTML.gif
        where
        G ( t , s ) = - 1 6 t 3 + 1 2 s t 2 + 1 6 s 3 + k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s - 1 6 s 3 + 1 2 1 - k = i m - 2 β k ξ k s 2 k = 0 m - 1 β k - 1 , t s , 1 2 s 2 t + k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s - 1 6 s 3 + 1 2 1 - k = i m - 2 β k ξ k s 2 k = 0 m - 1 β k - 1 , t s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equk_HTML.gif

        for ξi-1sξ i , i = 1, 2, ..., m -1.

        Proof Let G(t, s) be the Green's function of problem x(4)(t) = 0 with boundary condition (3.2). We can suppose
        G ( t , s ) = a 3 t 3 + a 2 t 2 + a 1 t + a 0 , t s , ξ i - 1 s ξ i , i = 1 , 2 , , m - 1 , b 3 t 3 + b 2 t 2 + b 1 t + b 0 , t s , ξ i - 1 s ξ i , i = 1 , 2 , , m - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equl_HTML.gif
        Considering the definition and properties of Green's function together with the boundary condition (3.2), we have
        a 3 s 3 + a 2 s 2 + a 1 s + a 0 = b 3 s 3 + b 2 s 2 + b 1 s + b 0 , 3 a 3 s 2 + 2 a 2 s + a 1 = 3 b 3 s 2 + 2 b 2 s + b 1 , 6 a 3 s + 2 a 2 = 6 b 3 s + 2 b 2 , 6 a 3 - 6 b 3 = - 1 , 6 b 3 = 0 , 6 b 3 + 2 b 2 = 0 , a 1 = 0 , b 3 + b 2 + b 1 + b 0 = k = 0 i = 1 β k ( a 3 ξ k 3 + a 2 ξ k 2 + a 1 ξ k + a 0 ) + k = i m - 2 β k ( b 3 ξ k 3 + b 2 ξ k 2 + b 1 ξ k + b 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equm_HTML.gif
        A straightforward calculation shows that
        a 3 = - 1 6 , a 2 = s 2 , a 1 = 0 , a 0 = k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s - 1 6 s 3 + 1 2 1 - k = i m - 2 β k ξ k s 2 k = 0 m - 1 β k - 1 + 1 6 s 3 , b 3 = b 2 = 0 , b 1 = s 2 2 , b 0 = k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s - 1 6 s 3 + 1 2 1 - k = i m - 2 β k ξ k s 2 k = 0 m - 1 β k - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equn_HTML.gif

        These give the explicit expression of the Green's function and the proof of Lemma 3.1 is completed.

        Lemma 3.2 One can see that G(t, s) ≥ 0, t, s ∈ [0, 1].

        Proof For ξi-1sξ i , i = 1, 2, ..., m - 1,
        G ( t , s ) t = 1 2 t ( 2 s - t ) , t s , ξ i - 1 s ξ i , 1 2 s 2 , t s , ξ i - 1 s ξ i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equo_HTML.gif
        Then G ( t , s ) t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq5_HTML.gif, 0 ≤ t, s ≤ 1. Thus G(t, s) is increasing on t. By a simple computation, we see
        G ( 0 , s ) = 1 6 s 3 + k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s - 1 6 s 3 + 1 2 1 - k = i m - 2 β k ξ k s 2 k = 0 m - 1 β k - 1 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equp_HTML.gif

        These ensures that G(t, s) ≥ 0, t, s ∈ [0, 1].

        Lemma 3.3 Suppose x(t) ∈ C3[0, 1] and
        x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = 0 , x ( 1 ) = i = 1 m - 2 β i x ( ξ i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equq_HTML.gif
        Furthermore x(4)(t) ≥ 0 and there exist t0 such that x(4)(t0) > 0. Then x(t) has the following properties:
        1. (1)

          min 0 t 1 | x ( t ) | δ max 0 t 1 | x ( t ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq6_HTML.gif

           
        2. (2)

          max 0 t 1 | x ( t ) | γ max 0 t 1 | x ( t ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq7_HTML.gif

           
        3. (3)

          max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq8_HTML.gif

           

        where δ = 1 - i = 1 m - 2 β i ξ i i = 1 m - 2 β i ( 1 - ξ i ) , γ = i = 1 m - 2 β i ( 1 - ξ i ) ( i = 1 m - 2 β i - 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq9_HTML.gif are positive constants.

        Proof Since x(4)(t) ≥ 0, t ∈ [0, 1], then x'''(t) is increasing on [0, 1]. Considering x'''(1) = 0, we have x'''(t) ≤ 0, t ∈ [0, 1]. Thus x''(t) is decreasing on [0, 1]. Considering this together with the boundary condition x''(1) = 0, we conclude that x''(t) ≥ 0. Then x(t) is convex on [0, 1]. Taking into account that x'(0) = 0, we get that
        max 0 t 1 x ( t ) = x ( 1 ) , min 0 t 1 x ( t ) = x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equr_HTML.gif
        1. (1)
          From the concavity of x(t), we have
          ξ i ( x ( 1 ) - x ( 0 ) ) x ( ξ i ) - x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equs_HTML.gif
           
        Multiplying both sides with β i and considering the boundary condition, we have
        1 - i = 1 m - 2 β i ξ i x ( 1 ) i = 1 m - 2 β i ( 1 - ξ i ) x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ6_HTML.gif
        (3.3)
        Thus
        min 0 t 1 | x ( t ) | δ max 0 t 1 | x ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equt_HTML.gif
        1. (2)
          Considering the mean-value theorem together with the concavity of x(t), we have
          x ( 1 ) - x ( ξ i ) ( 1 - ξ i ) x ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ7_HTML.gif
          (3.4)
           
        Multiplying both sides with β i and considering the boundary condition, we have
        i = 1 m - 2 β i - 1 x ( 1 ) i = 1 m - 2 β i ( 1 - ξ i ) x ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ8_HTML.gif
        (3.5)
        which yields that x ( 1 ) i = 1 m - 2 β i ( 1 - ξ i ) ( i = 1 m - 2 β i - 1 ) | x ( 1 ) | = γ max 0 t 1 | x ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq10_HTML.gif.
        1. (3)
          For x ( t ) = x ( 0 ) + 0 t x ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq11_HTML.gif and x'(0) = 0, we get
          | x ( t ) | = | 0 t x ( s ) d s | 0 1 | x ( s ) | d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equu_HTML.gif
           
        For x ( t ) = x ( 1 ) - t 1 x ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq12_HTML.gif and x"(1) = 0, we get
        | x ( t ) | = | t 1 x ( s ) d s | 0 1 | x ( s ) | d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equv_HTML.gif
        Consequently
        max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equw_HTML.gif

        These give the proof of Lemma 3.3.

        Remark Lemma 3.3 ensures that
        max { max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | } γ max 0 t 1 | x ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equx_HTML.gif
        Let Banach space E = C3[0, 1] be endowed with the norm
        x = max { max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | } , x E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equy_HTML.gif
        Define the cone PE by
        P = { x E | x ( t ) 0 , x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = 0 , x ( 1 ) = i = 1 m - 2 β i x ( ξ i ) , x t  is convex on [0,1] } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equz_HTML.gif
        Let the nonnegative continuous concave functional α, the nonnegative continuous convex functionals γ, θ and the nonnegative continuous functional ψ be defined on the cone by
        γ ( x ) = max 0 t 1 | x ( t ) | , θ ( x ) = ψ ( x ) = max 0 t 1 | x ( t ) | , α ( x ) = min 0 t 1 | x ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equaa_HTML.gif
        By Lemma 3.3, the functionals defined above satisfy
        δ θ ( x ) α ( x ) θ ( x ) = ψ ( x ) , x γ γ ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ9_HTML.gif
        (3.6)
        Denote
        m = 0 1 G ( 0 , s ) d s , N = 0 1 G ( 1 , s ) d s , λ = min { m , δ γ } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equab_HTML.gif
        Assume that there exist constants 0 < a, b, d with a < b < λd such that
        ( A 1 ) f ( t , u , v , w , p ) d , ( t , u , v , w , p ) [ 0,1 ] × [ 0, γ d ] × [ 0, d ] × [ 0, d ] × [ d , 0 ] , ( A 2 ) f ( t , u , v , w , p ) > b / m , ( t , u , v , w , p ) [ 0, 1 ] × [ b , b / δ ] × [ 0, d ] × [ 0, d ] × [ d , 0 ] , ( A 3 ) f ( t , u , v , w , p ) < a / N , ( t , u , v , w , p ) [ 0, 1 ] × [ 0, a ] × [ 0, d ] × [ 0, d ] × [ d , 0 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equac_HTML.gif
        Theorem 3.1 Under assumptions (A1)-(A3), problem (1.1), (1.2) has at least three positive solutions x1, x2, x3 satisfying
        max 0 t 1 | x i ( t ) | d , i = 1 , 2 , 3 ; b < min 0 t 1 | x 1 ( t ) | ; a < max 0 t 1 | x 2 ( t ) | , min 0 t 1 | x 2 ( t ) | < b ; max 0 t 1 | x 3 ( t ) | a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equad_HTML.gif
        Proof Problem (1.1, 1.2) has a solution x = x(t) if and only if x solves the operator equation
        x ( t ) = 0 1 G ( t , s ) f ( s , x ( s ) , x ( s ) , x ( s ) , x ( s ) ) d s = ( T x ) ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equae_HTML.gif
        Then
        ( T x ) ( t ) = - t 1 f ( s , x , x , x , x ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equaf_HTML.gif
        For x P ( γ , d ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq2_HTML.gif, considering Lemma 3.3 and assumption (A1), we have f(t, x(t), x'(t), x''(t), x'''(t)) ≤ d. Thus
        γ ( T x ) = | ( T x ) ( 0 ) | = | - 0 1 f ( s , x , x , x , x ) d s | = 0 1 | f ( s , x , x , x , x ) | d s d . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equag_HTML.gif
        Hence T : P ( γ , d ) ¯ P ( γ , d ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq3_HTML.gif. An application of the Arzela-Ascoli theorem yields that T is a completely continuous operator. The fact that the constant function x(t) = b/δP(γ, θ, α, b, c, d) and α(b/δ) > b implies that
        { x P ( γ , θ , α , b , c , d ) | α ( x ) > b } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equah_HTML.gif
        For xP(γ, θ, α, b, c, d), we have bx(t) ≤ b/δ and |x'''(t)| < d. From assumption (A2), we see
        f ( t , x , x , x , x ) > b / m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equai_HTML.gif
        Hence, by definition of α and the cone P, we can get
        α ( T x ) = ( T x ) ( 0 ) = 0 1 G ( 0 , s ) f ( s , x , x , x , x ) d s b m 0 1 G ( 0 , s ) d s > b m m = b , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equaj_HTML.gif

        which means α(Tx) > b, ∀xP(γ, θ, α, b, b/δ, d). This ensures that condition (S1) of Lemma 2.1 is fulfilled.

        Second, with (3.4) and b < λd, we have
        α ( T x ) δ θ ( T x ) > δ × b δ = b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equak_HTML.gif

        for all xP(γ, α, b, d) with θ ( T x ) > b δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq13_HTML.gif.

        Thus, condition (S2) of Lemma 2.1 holds. Finally we show that (S3) also holds. We see ψ(0) = 0 < a and 0 ∉ R(γ, ψ, a, d). Suppose that xR(γ, ψ, a, d) with ψ(x) = a, then by the assumption of (A3),
        ψ ( T x ) = max 0 t 1 | ( T x ) ( t ) | = 0 1 G ( 1 , s ) f ( s , x , x , x , x ) d s < a N 0 1 G ( 1 , s ) d s = a , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equal_HTML.gif
        which ensures that condition (S3) of Lemma 2.1 is fulfilled. Thus, an application of Lemma 2.1 implies that the fourth-order m-point boundary value problem (1.1, 1.2) has at least three positive convex increasing solutions x1, x2, x3 with the properties that
        max 0 t 1 | x i ( t ) | d , i = 1 , 2 , 3 ; b < min 0 t 1 | x 1 ( t ) | ; a < max 0 t 1 | x 2 ( t ) | , min 0 t 1 | x 2 ( t ) | < b ; max 0 t 1 | x 3 ( t ) | a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equam_HTML.gif

        4 Positive solutions for problem (1.1), (1.3)

        The following assumption will stand throughout this section:
        ( H 2 ) , f C ( [ 0 , 1 ] × R 4 , [ 0 , + ) ) , i = 1 m - 2 β i > 1 , i = 1 m - 2 β i ξ i + 1 - i = 1 m - 2 β i > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equan_HTML.gif
        Lemma 4.1 Denote ξ0 = 0, ξm-1= 1, β0 = βm-1= 0, the Green's function of problem
        x ( 4 ) ( t ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ10_HTML.gif
        (4.1)
        x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ11_HTML.gif
        (4.2)
        is
        H ( t , s ) = - 1 6 t 3 + 1 2 s t 2 - 1 2 s 2 t + k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s + 1 2 ξ k s 2 + k = i m - 2 1 6 β k s 3 k = 0 m - 1 β k - 1 , t s , ξ i - 1 s ξ i , - 1 6 s 3 + k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s + 1 2 ξ k s 2 + k = i m - 2 1 6 β k s 3 k = 0 m - 1 β k - 1 , t s , ξ i - 1 s ξ i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equao_HTML.gif

        for i = 1, 2, ..., m - 1.

        Proof Suppose that
        H ( t , s ) = a 3 t 3 + a 2 t 2 + a 1 t + a 0 t s , ξ i - 1 s ξ i , i = 1 , 2 , , m - 1 , b 3 t 3 + b 2 t 2 + b 1 t + b 0 t s , ξ i - 1 s ξ i , i = 1 , 2 , , m - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equap_HTML.gif
        Considering the definition and properties of Green's function together with the boundary condition (4.2), we have
        a 3 s 3 + a 2 s 2 + a 1 s + a 0 = b 3 s 3 + b 2 s 2 + b 1 s + b 0 , 3 a 3 s 2 + 2 a 2 s + a 1 = 3 b 3 s 2 + 2 b 2 s + b 1 , 6 a 3 s + 2 a 2 = 6 b 3 s + 2 b 2 , 6 a 3 - 6 b 3 = - 1 , b 3 = 0 , 6 b 3 + 2 b 2 = 0 , 3 b 3 + 2 b 2 + b 1 = 0 , a 0 = k = 0 i = 1 β k ( a 3 ξ k 3 + a 2 ξ k 2 + a 1 ξ k + a 0 ) + k = i m - 2 β k ( b 3 ξ k 3 + b 2 ξ k 2 + b 1 ξ k + b 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equaq_HTML.gif
        Consequently
        a 3 = - 1 6 , a 2 = 1 2 s , a 1 = - 1 2 s 2 , b 3 = b 2 = b 1 = 0 , a 0 = k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s + 1 2 ξ k s 2 + k = i m - 2 1 6 β k s 3 k = 0 m - 1 β k - 1 , b 0 = k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s + 1 2 ξ k s 2 + k = i m - 2 1 6 β k s 3 k = 0 m - 1 β k - 1 - 1 6 s 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equar_HTML.gif

        The proof of Lemma 4.1 is completed.

        Lemma 4.2 One can see that H(t, s) ≥ 0, t, s ∈ [0, 1].

        Proof For ξi-1sξ i , i = 1, 2, ..., m - 1,
        H ( t , s ) t = - 1 2 ( t - s ) 2 , t s , ξ i - 1 s ξ i , 0 , t s , ξ i - 1 s ξ i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equas_HTML.gif
        Then H ( t , s ) t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq14_HTML.gif, 0 ≤ t, s ≤ 1, which implies that H(t, s) is decreasing on t. The fact that
        H ( 1 , s ) = k = 0 i - 1 β k 1 6 ξ k 3 - 1 2 ξ k 2 s + 1 2 ξ k s 2 + k = i m - 2 1 6 β k s 3 k = 0 m - 1 β k - 1 - 1 6 s 3 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equat_HTML.gif

        ensures that H(t, s) ≥ 0, t, s ∈ [0, 1].

        Lemma 4.3 If x(t) ∈ C3[0, 1],
        x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equau_HTML.gif
        and x(4)(t) ≥ 0, there exists t0 such that x(4)(t0) > 0, then
        1. (1)

          min 0 t 1 | x ( t ) | δ 1 max 0 t 1 | x ( t ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq15_HTML.gif

           
        2. (2)

          max 0 t 1 | x ( t ) | γ 1 max 0 t 1 | x ( t ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq16_HTML.gif

           
        3. (3)

          max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq17_HTML.gif

           

        where δ 1 = 1 - i = 1 m - 2 β i ( 1 - ξ i ) i = 1 m - 2 β i ξ i , γ 1 = i = 1 m - 2 β i ξ i i = 1 m - 2 β i - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq18_HTML.gif are positive constants.

        Proof It follows from the same methods as Lemma 3.3 that x(t) is convex on [0, 1]. Taking into account that x'(1) = 0, one can see that x(t) is decreasing on [0, 1] and
        max 0 t 1 x ( t ) = x ( 0 ) , min 0 t 1 x ( t ) = x ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equav_HTML.gif
        1. (1)
          From the concavity of x(t), we have
          ξ i ( x ( 1 ) - x ( 0 ) ) x ( ξ i ) - x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equaw_HTML.gif
           
        Multiplying both sides with β i and considering the boundary condition, we have
        i = 1 m - 2 β i ξ i x ( 1 ) 1 - i = 1 m - 2 β i ( 1 - ξ i ) x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ12_HTML.gif
        (4.3)
        Thus
        min 0 t 1 | x ( t ) | δ 1 max 0 t 1 | x ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equax_HTML.gif
        1. (2)
          Considering the mean-value theorem, we get
          ( 0 ) - x ( ξ i ) ξ i | x ( 0 ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equay_HTML.gif
           
        From the concavity of x similarly with above we know
        i = 1 m - 2 β i - 1 x ( 0 ) < i = 1 m - 2 β i ξ i | x ( 0 ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equ13_HTML.gif
        (4.4)
        Considering (4.3) together with (4.4) we have x ( 0 ) γ 1 | x ( 0 ) | = γ 1 max 0 t 1 | x ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq19_HTML.gif.
        1. (3)
          For x ( t ) = x ( 1 ) - t 1 x ( s ) d s , x ( t ) = x ( 1 ) - t 1 x ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq20_HTML.gif and x'(1) = 0, x''(1) = 0, we get
          | x ( t ) | = | t 1 x ( s ) d s | 0 1 | x ( s ) | d s , | x ( t ) | = | t 1 x ( s ) d s | 0 1 | x ( s ) | d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equaz_HTML.gif
           
        Thus
        max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | max 0 t 1 | x ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equba_HTML.gif
        Remark We see that
        max { max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | } γ 1 max 0 t 1 | x ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbb_HTML.gif
        Let Banach space E = C3[0, 1] be endowed with the norm
        x = max { max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | , max 0 t 1 | x ( t ) | } , x E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbc_HTML.gif
        Define the cone PE by
        P 1 = x E | x ( t ) 0 , x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 1 ) = 0 , x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) , x ( t ) is convex on [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbd_HTML.gif
        Denote
        m 1 = 0 1 H ( 1 , s ) d s , N 1 = 0 1 H ( 0 , s ) d s , λ 1 = min { m 1 , δ 1 γ 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Eqube_HTML.gif
        Assume that there exist constants 0 < a, b, d with a < b < λ1d such that
        ( A 4 ) f ( t , u , v , w , p ) d , ( t , u , v , w , p ) [ 0,1 ] × [ 0, γ 1 d ] × [ d , 0 ] × [ 0, d ] × [ d , 0 ] , ( A 5 ) f ( t , u , v , w , p ) > b / m 1 , ( t , u , v , w , p ) [ 0, 1 ] × [ b , b / δ 1 ] × [ d , 0 ] × [ 0, d ] × [ d , 0 ] , ( A 6 ) f ( t , u , v , w , p ) < a / N 1 , ( t , u , v , w , p ) [ 0, 1 ] × [ 0, a ] × [ d , 0 ] × [ 0, d ] × [ d , 0 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbf_HTML.gif
        Theorem 4.1 Under assumptions (A4)-(A6), problem (1.1), (1.3) has at least three positive solutions x1, x2, x3 with the properties that
        max 0 t 1 | x i ( t ) | d , i = 1 , 2 , 3 ; b < min 0 t 1 | x 1 ( t ) | ; a < max 0 t 1 | x 2 ( t ) | , min 0 t 1 | x 2 ( t ) | < b ; max 0 t 1 | x 3 ( t ) | a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbg_HTML.gif
        Proof Problem (1.1), (1.3) has a solution x = x(t) if and only if x solves the operator equation
        x ( t ) = 0 1 H ( t , s ) f ( s , x ( s ) , x ( s ) , x ( s ) , x ( s ) ) d s = ( T 1 x ) ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbh_HTML.gif
        Then
        ( T 1 x ) ( t ) = - t 1 f ( s , x , x , x , x ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbi_HTML.gif
        For x P 1 ( γ , d ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq21_HTML.gif, considering Lemma 4.3 and assumption (A4), we have
        f ( t , x ( t ) , x ( t ) , x ( t ) , x ( t ) ) d . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbj_HTML.gif
        Thus
        γ ( T 1 x ) = | ( T 1 x ) ( 0 ) | = | - 0 1 f ( s , x , x , x , x ) d s | = 0 1 | f ( s , x , x , x , x ) | d s d . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbk_HTML.gif

        Hence T 1 : P 1 ( γ , d ) ¯ P 1 ( γ , d ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq22_HTML.gif and T1 is a completely continuous operator obviously. The fact

        that the constant function x(t) = b/δ1P1(γ, θ, α, b, c, d) and α(b/δ1) > b implies that
        { x P 1 ( γ , θ , α , b , c , d | α ( x ) > b ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbl_HTML.gif

        This ensures that condition (S1) of Lemma 2.1 holds.

        For xP1(γ, θ, α, b, c, d), we have bx(t) ≤ b/δ1 and |x'''(t)| < d. From assumption (A4),
        f ( t , x , x , x , x ) > b / m 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbm_HTML.gif
        Hence, by definition of α and the cone P1, we can get
        α ( T 1 x ) = ( T 1 x ) ( 1 ) = 0 1 H ( 1 , s ) f ( s , x , x , x , x ) d s b m 1 0 1 H ( 1 , s ) d s > b m 1 m 1 = b , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbn_HTML.gif

        which means α(T1x) > b, ∀xP1(γ, θ, α, b, b/δ, d).

        Second, with (4.4) and b < λ1d, we have
        α ( T 1 x ) δ 1 θ ( T 1 x ) > δ 1 × b δ 1 = b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbo_HTML.gif

        for all xP1(γ, α, b, d) with θ ( T 1 x ) > b δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_IEq23_HTML.gif.

        Thus, condition (S2) of Lemma 2.1 holds. Finally we show that (S3) also holds. We see ψ(0) = 0 < a and 0 ∉ R(γ, ψ, a, d). Suppose that xR(γ, ψ, a, d) with ψ(x) = a, then by the assumption of (A6),
        ψ ( T 1 x ) = max 0 t 1 | ( T 1 x ) ( t ) | = 0 1 H ( 0 , s ) f ( s , x , x , x , x ) d s < a N 1 0 1 H ( 0 , s ) d s = a , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbp_HTML.gif
        which ensures that condition (S3) of Lemma 2.1 is satisfied. Thus, an application of Lemma 2.1 implies that the fourth-order m-point boundary value problem (1.1), (1.3) has at least three positive convex decreasing solutions x1, x2, x3 satisfying the conditions that
        max 0 t 1 | x i ( t ) | d , i = 1 , 2 , 3 ; b < min 0 t 1 | x 1 ( t ) | ; a < max 0 t 1 | x 2 ( t ) | , min 0 t 1 | x 2 ( t ) | < b ; max 0 t 1 | x 3 ( t ) | a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-21/MediaObjects/13661_2010_Article_18_Equbq_HTML.gif

        Declarations

        Acknowledgements

        We are indebted to the anonymous referee for a detailed reading and useful comments and suggestions, which allowed us to improve this work. This work was supported by the Anhui Provincial Natural Science Foundation (10040606Q50), National Natural Science Foundation of China (No.11071164), Shanghai Natural Science Foundation (No.10ZR1420800).

        Authors’ Affiliations

        (1)
        College of Science, University of Shanghai for Science and Technology
        (2)
        Department of Mathematics, Hefei Normal University

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