Existence of positive solutions to a non-positive elastic beam equation with both ends fixed

  • Haixia Lu1,

    Affiliated with

    • Li Sun2, 3Email author and

      Affiliated with

      • Jingxian Sun2

        Affiliated with

        Boundary Value Problems20122012:56

        DOI: 10.1186/1687-2770-2012-56

        Received: 30 November 2011

        Accepted: 14 May 2012

        Published: 14 May 2012

        Abstract

        This article is concerned with the existence of nontrivial solutions for a non-positive fourth-order two-point boundary value problem (BVP) and the existence of positive solutions for a semipositive fourth-order two-point BVP. In mechanics, the problem describes the deflection of an elastic beam rigidly fixed at both ends. The method to show our main results is the topological degree and fixed point theory of nonlinear operator on lattice.

        Mathematics Subject Classification 2010: 34B18; 34B16; 34B15.

        Keywords

        lattice topological degree fixed point nontrivial solutions and positive solutions elastic beam equations

        1 Introduction

        The purpose of this article is to investigate the existence of nontrivial solutions and positive solutions of the following nonlinear fourth-order two-point boundary value problem (for short, BVP)
        u ( 4 ) ( t ) = λ f ( t , u ( t ) ) , 0 t 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ1_HTML.gif
        (P)

        where λ is a positive parameter, f : [0,1] × R1R1 is continuous.

        Fourth-order two-point BVPs are useful for material mechanics because the problems usually characterize the deflection of an elastic beam. The problem (P) describes the deflection of an elastic beam with both ends rigidly fixed. The existence and multiplicity of positive solutions for the elastic beam equations has been studied extensively when the non-linear term f : [0,1] × [0, +∞) → [0, +∞) is continuous, see for example [110] and references therein. Agarwal and Chow [1] investigated problem (P) by using of contraction mapping and iterative methods. Bai [3] applied upper and lower solution method and Yao [9] used Guo-Krasnosel'skii fixed point theorem of cone expansion-compression type. However, there are only a few articles concerned with the nonpositive or semipositive elastic beam equations. Yao [11] considered the existence of positive solutions of semipositive elastic beam equations by constructing control functions and a special cone and using fixed point theorem of cone expansion-compression type. In this article, we assume that f : [0,1] × R1R1, which implies the problem (P) is nonpositive (or semipositive particularly). By the topological degree and fixed point theory of superlinear operator on lattice (the definition of lattice will be given in Section 2), we obtain the existence of nontrivial solutions for the non-positive BVP (P) and the existence of positive solutions for the semipositive BVP (P).

        2 Preliminaries

        Let E be an ordered Banach space in which the partial ordering ≤ is induced by a cone PE, θ denote the zero element of E. P is called solid if int P ≠ ∅, i.e., P has nonempty interior. P is called a generating cone if E = P - P . For the concepts and properties about the cones we refer to [1214].

        We call E a lattice in the partial ordering ≤, if for arbitrary x, yE, sup{x, y} and inf{x, y} exist. For xE, let
        x + = sup{ x , θ } , x - = sup{ { x , θ } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equa_HTML.gif

        which are called the positive and the negative part of x, respectively. Take |x| = x+ + x - , then |x| ∈ P , and |x| is called the module of x. One can see [15] for the definition and the properties about the lattice.

        For convenience, we use the following notations:
        x + = x + , x - = - x - , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ2_HTML.gif
        (2.1)
        then
        x + P , x - ( - P ) , x = x + + x - . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equb_HTML.gif

        Remark 2.1 If E is a lattice, then P is a generating cone.

        Definition 2.1 [[16], Definition 3.2, p. 929]. Let DE and F : DE be a nonlinear operator. F is said to be quasi-additive on lattice, if there exists y0E such that
        F x = F x + + F x - + y 0 , x D , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ3_HTML.gif
        (2.2)

        where x+ and x - are defined by (2.1).

        Remark 2.2 We point out that the condition (2.2) appears naturally in the applications of nonlinear differential equations and integral equations.

        Let
        E = C [ a , b ] = { x ( t ) | x : [ a , b ] R 1  is continuous } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equc_HTML.gif
        and f : [a, b] × R1R1. Consider the Nemytskii operator
        ( F x ) ( t ) = f ( t , x ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equd_HTML.gif
        Set P = {xC[a, b] | x(t) ≥ 0}, then E = C[a, b] is a lattice in the partial ordering which is induced by P . For any xC[a, b], it is evident that
        x + ( t ) = max { x ( t ) , 0 } , x - ( t ) = min { x ( t ) , 0 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Eque_HTML.gif

        and hence |x|(t) = |x(t)|. By Remark 3.1 in [16], we know that there exists y0C[a, b] such that Fx = Fx+ + Fx - + y0, ∀xC[a, b].

        Suppose that B is a linear operator and A = BF . It follows that
        A x = A x + + A x - + B y 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equf_HTML.gif

        which means that A is quasi-additive on lattice.

        Definition 2.2 [[17], Definition, p. 261]. Let B : EE be a linear operator. B is said to be a u0- bounded linear operator, if there exists u0P\{θ}, such that for any xP\{θ}, there exist a natural number n and real numbers ζ, η > 0, such that
        ζ u 0 B n x η u 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equg_HTML.gif

        Lemma 2.1 [[18], Theorem 4.2.2, p. 122]. Let P be a generating cone and B a u0- bounded completely continuous linear operator. Then the spectral radius r(B) ≠ 0 and r - 1(B) is the only positive eigenvalue corresponding to positive eigenvectors and B has no other eigenvectors except those corresponding to r - 1(B).

        Let B : EE be a positive completely continuous linear operator, r(B) a spectral radius of B, B* the conjugated operator of B, and P* the conjugated cone of P. Since PE is a generating cone, according to the famous Krein-Rutman theorem (see [14]), if r(B) ≠ 0, then there exist φ ̄ P \ { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq1_HTML.gif, and g* ∈ P*\{θ}, such that
        B φ ̄ = r ( B ) φ ̄ , B * g * = r ( B ) g * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ4_HTML.gif
        (2.3)
        Fix φ ̄ P \ { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq2_HTML.gif, g* ∈ P*\{θ} such that (2.3) holds. For δ > 0, let
        P ( g * , δ ) = { x P , g * ( x ) δ | | x | | } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equh_HTML.gif

        then P (g*, δ) is also a cone in E.

        Definition 2.3 [[19], Definition, p. 528]. Let B be a positive linear operator. B is said to satisfy H condition, if there exist φ ̄ P \ { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq3_HTML.gif, g* ∈ P*\{θ} and δ > 0 such that (2.3) holds, and B maps P into P(g*, δ).

        Remark 2.3 Let B φ = a b k ( x , y ) φ ( y ) d y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq4_HTML.gif, where k(x, y) ∈ C([a, b] × [a, b]), k(x, y) ≥ 0, φC[a, b]. Suppose that (2.3) holds and there exists v(x) ∈ P\{θ} such that
        k ( x , y ) v ( x ) k ( τ , y ) , x , y , τ [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equi_HTML.gif

        and v(x)g*(x) ≢ 0, then B satisfies H condition (see [19]).

        Lemma 2.2 [[16], Theorem 3.1, p. 929]. Let P be a solid cone, A : EE be a completely continuous operator satisfying A = BF, where F is quasi-additive on lattice, B is a positive bounded operator satisfying H condition. Suppose that
        1. (i)
          there exist a 1 > r - 1(B), y 1P such that
          F x a 1 x - y 1 , x P ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ5_HTML.gif
          (2.4)
           
        2. (ii)
          there exist 0 ≤ a 2r - 1(B), y 2P such that
          F x a 2 x - y 2 , x ( - P ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ6_HTML.gif
          (2.5)
           

        Then there exists R0> 0 such that deg(I - A, T R , θ) = 0 for any R > R0 , where T R = {xC[0, 1] : ||x|| < R}.

        Lemma 2.3 [[16], Theorem 3.3, p. 932]. Let Ω be a bounded open subset of E, θ ∈ Ω, and A : Ω ̄ P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq5_HTML.gif a completely continuous operator. Suppose that A has no fixed point on ∂Ω. If
        1. (i)

          there exists a positive bounded linear operator B such that |Ax| ≤ B|x|, for all x ∈ ∂Ω;

           
        2. (ii)

          r(B) ≤ 1.

           

        Then deg(I - A, Ω, θ) = 1.

        3 Existence of nontrivial solutions for the non-positive BVP (P)

        In the sequel we always take E = C[0,1] with the norm ||u|| = max0≤t≤1|u(t)| and P = {uC[0, 1] | u(t) ≥ 0, 0 ≤ t ≤ 1}. Then P is a solid cone in E and E is a lattice under the partial ordering ≤ induced by P.

        A solution of BVP (P) is a fourth differentiable function u : [0,1] → R such that u satisfies (P). u is said to be a positive solution of BVP (P) if u(t) > 0, 0 < t < 1. Let G(t, s) be Green's function of homogeneous linear problem u(4)(t) = 0, u(0) = u(1) = u'(0) = u'(1) = 0. From Yao [11] we have
        G ( t , s ) = 1 6 t 2 ( 1 - s ) 2 [ ( s - t ) + 2 ( 1 - t ) s ] , 0 t s 1 , 1 6 s 2 ( 1 - t ) 2 [ ( t - s ) + 2 ( 1 - s ) t ] , 0 s t 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equj_HTML.gif

        and

        (G1) G(t, s) ≥ 0, 0 ≤ t, s ≤ 1;

        (G2) G(t, s) = G(s, t);

        (G3) G(t, s) ≥ p(t)G(τ; s), 0 ≤ t, s, τ ≤ 1, where p ( t ) = 2 3 min { t 2 , ( 1 - t ) 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq6_HTML.gif.

        Lemma 3.1 Let H ( t ) = 1 2 t 2 ( 1 - t ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq7_HTML.gif and q ( s ) = 2 3 s 2 ( 1 - s ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq8_HTML.gif. Then
        q ( s ) H ( t ) G ( t , s ) H ( t ) , 0 t , s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ7_HTML.gif
        (3.1)
        Proof. Since G(0, s) = G(1, s) = 0, 0 ≤ s ≤ 1, H(0) = H(1) = 0, then q(s)H(t) = G(t, s) = H(t) holds for t = 0 and t = 1. If 0 < ts ≤ 1 and t < 1, then
        G ( t , s ) = 1 6 t 2 ( 1 - s ) 2 [ 3 s ( 1 - t ) - t ( 1 - s ) ] 1 2 t 2 ( 1 - s ) 2 s ( 1 - t ) 1 2 t 2 ( 1 - t ) 2 s ( 1 - s ) < 1 2 t 2 ( 1 - t ) 2 = H ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equk_HTML.gif
        and
        G ( t , s ) H ( t ) = 1 6 t 2 ( 1 - s ) 2 [ 3 s ( 1 - t ) - t ( 1 - s ) ] 1 2 t 2 ( 1 - t ) 2 ( 1 - s ) 2 [ 3 s ( 1 - t ) - t ( 1 - t ) ] 3 ( 1 - t ) 2 = ( 1 - s ) 2 ( 3 s - t ) 3 ( 1 - t ) 2 s ( 1 - s ) 2 3 ( 1 - t ) 2 3 s 2 ( 1 - s ) 2 = q ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equl_HTML.gif

        Similarly, (3.1) holds for 0 ≤ st < 1 and t > 0. The proof is complete.    □

        It is well known that the problem (P) is equivalent to the integral equation
        u ( t ) = λ 0 1 G ( t , s ) f ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equm_HTML.gif
        Let
        ( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ8_HTML.gif
        (3.2)
        ( B u ) ( t ) = 0 1 G ( t , s ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ9_HTML.gif
        (3.3)

        Lemma 3.2 Let B be defined by (3.3). Then B is a u0- bounded linear operator.

        Proof. Let u 0 t = H t = 1 2 t 2 1 - t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq9_HTML.gif, t ∈ [0,1]. For any uP\{θ}, by Lemma 3.1

        we have
        0 1 q ( s ) H ( t ) u ( s ) d s B u ( t ) = 0 1 G ( t , s ) u ( s ) d s 0 1 H ( t ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equn_HTML.gif
        Take arbitrarily 0 < ε 0 < 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq10_HTML.gif, then
        u 0 ( t ) ε 0 1 - ε 0 q ( s ) u ( s ) d s B u ( t ) u u 0 ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equo_HTML.gif
        Let ζ = ε 0 1 - ε 0 q ( s ) u ( s ) d s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq11_HTML.gif, η = ||u|| > 0. Then
        ζ u 0 B u η u 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equp_HTML.gif

        This indicates that B : EE is a u0- bounded linear operator. □

        From Lemma 2.1 we have r(B) ≠ 0 and r - 1(B) is the only eigenvalue of B. Denote λ1 = r - 1(B).

        Now let us list the following conditions which will be used in this article:

        (H1) there exist constants α and β with α > β ≥ 0 satisfying
        lim inf u + f ( t , u ) u α , lim sup u - f ( t , u ) u β , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ10_HTML.gif
        (3.4)
        (H2) there exists a constant γ ≥ 0 satisfying
        lim sup u 0 f ( t , u ) u γ , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ11_HTML.gif
        (3.5)

        (H3) lim u + f ( t , u ) u = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq12_HTML.gif.

        Theorem 3.1 Suppose that (H1) and (H2) hold. Then for any λ λ 1 α , λ 1 ι http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq13_HTML.gif, BVP (P) has at least one nontrivial solution, where λ1 = r - 1(B) is the only eigenvalue of B, B is denoted by (3.3), ι = max{β, γ}.

        Proof. Let (Fu)(t) = f(t, u(t)). Then A = BF, where A is denoted by (3.2). By Remark 2.2, F is quasi-additive on lattice. Applying the Arzela-Ascoli theorem and a standard argument, we can prove that A : EE is a completely continuous operator.

        Now we show that λA = λBF has at least one nontrivial fixed point, which is the nontrivial solution of BVP (P).

        On account of (G3) we have that p ( t ) = 2 3 min { t 2 , ( 1 - t ) 2 } P \ { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq14_HTML.gif such that
        G ( t , s ) p ( t ) G ( τ , s ) , 0 t , s , τ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equq_HTML.gif

        Notice that B u t = 0 1 G ( t , s ) u ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq15_HTML.gif, where G(t, s) ≥ 0, G(t, s) ∈ C([0,1] × [0,1]). From Lemma 3.2 B is a u0- bounded linear operator. By Lemma 2.1 we have r(B) ≠ 0 and λ1 = r - 1(B) is the only eigenvalue of B. Then there exist φ ̄ P \ { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq16_HTML.gif and g* ∈ P*\{θ} such that (2.3) holds. Notice that λ > 0, from Remark 2.3, λB satisfies H condition.

        By (3.4) and (3.5), there exist r > 0, M > 0 and 0 < ε < min { α λ - λ 1 λ , λ 1 - β λ λ , λ 1 - γ λ λ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq17_HTML.gif such that
        f ( t , u ) ( α - ε ) u - M , t [ 0 , 1 ] , u 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ12_HTML.gif
        (3.6)
        f ( t , u ) ( β + ε ) u - M , t [ 0 , 1 ] , u 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ13_HTML.gif
        (3.7)
        | f ( t , u ) | ( γ + ε ) | u | , t [ 0 , 1 ] , | u | r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ14_HTML.gif
        (3.8)

        By (3.6) and (3.7), we have (2.4) and (2.5) hold, where a1 = α - ε, a2 = β + ε.

        Let B1 = λB. Then r - 1 B 1 = λ 1 λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq18_HTML.gif. Obviously, for any λ ( λ 1 α , λ 1 ι ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq19_HTML.gif, a1> r - 1(B1), a2< r - 1(B1). From Lemma 2.2 there exists R0> 0 such that for any R > max{R0, r},
        deg ( I - λ A , T R , θ ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ15_HTML.gif
        (3.9)
        Let B2 = λ(γ + ε)B. From (3.8) we have |λAu| ≤ B2|u|, also r ( B 2 ) = λ ( γ + ε ) λ 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq20_HTML.gif. Without loss of generality we assume that λA has no fixed point on ∂T r , where T r = {uC[0,1] | ||u|| < r}. By Lemma 2.3 we have
        deg ( I - λ A , T r , θ ) = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ16_HTML.gif
        (3.10)

        It is easy to see from (3.9) and (3.10) that λA has at least one nontrivial fixed point. Thus problem (P) has at least one nontrivial solution. □

        Remark 3.1 If α = +∞, β = γ = 0, then for any λ > 0 problem (P) has at least one nontrivial solution.

        Theorem 3.2 Suppose that (H1) holds. Assume f(t, 0) ≡ 0, ∀t ∈ [0,1] and
        lim u 0 f ( t , u ) u = ρ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ17_HTML.gif
        (3.11)

        Then for any λ ( λ 1 α , λ 1 β ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq21_HTML.gif and λ λ 1 ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq22_HTML.gif, BVP (P) has at least one nontrivial solution.

        Proof. Since f(t, 0) ≡ 0, ∀t ∈ [0,1], then = θ. By (3.11) we have that the Frechet derivative A θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq23_HTML.gif of A at θ exists and
        ( A θ u ) ( t ) = 0 1 G ( t , s ) ρ u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equr_HTML.gif
        Notice that λ λ 1 ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq24_HTML.gif, then 1 is not an eigenvalue of λ A θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq25_HTML.gif. By the famous Leray-Schauder theorem there exists r > 0 such that
        deg ( I - λ A , T r , θ ) = deg ( I - λ A θ , T r , θ ) = ( - 1 ) κ 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ18_HTML.gif
        (3.12)

        where κ is the sum of algebraic multiplicities for all eigenvalues of λ A θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq26_HTML.gif lying in the interval (0, 1). From the proof of Theorem 3.1 we have that (3.9) holds for any λ ( λ 1 α , λ 1 β ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq27_HTML.gif. By (3.9) and (3.12), λA has at least one nontrivial fixed point. Thus problem (P) has at least one nontrivial solution. □

        4 Existence of positive solutions for the semipositive BVP (P)

        In many real problems, the positive solution is more significant. In this section we will study this kind of question.

        Lemma 4.1 [[20], Theorem 1, p. 90]. Let D = [a, b]. Suppose
        1. (i)
          G(t, s) is a symmetric kernel. And there exist D 0D, mesD 0 ≠ 0 and δ > 0 such that
          G ( t , s ) δ G ( τ , s ) , t D 0 , s , τ D ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equs_HTML.gif
           
        2. (ii)

          f(t, u) is bounded from below and lim u + f ( t , u ) = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq28_HTML.gif uniformly holds for tD 0. Then for any λ* > 0, there exists R = R(λ*) > 0 such that if 0 < λ0 ≤ λ*, ||φ 0|| ≥ R and φ 0 = λ0 0, then φ 0(x) ≥ 0, where A is denoted by (3.2).

           

        Theorem 4.1 Suppose that (H3) holds. Then there exists λ* > 0 such that for any 0 < λ < λ* BVP (P) has at least one positive solution.

        Proof. By (H3) there exists a constant b > 0 such that
        f ( t , u ) - b , t [ 0 , 1 ] , u 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ19_HTML.gif
        (4.1)
        Let
        f 1 ( t , u ) = f ( t , u ) , u 0 , f ( t , - u ) , u < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equt_HTML.gif
        From (4.1) f1 is bounded from below. Let
        ( A 1 u ) ( t ) = 0 1 G ( t , s ) f 1 ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equu_HTML.gif

        Then A1 : EE is a completely continuous operator.

        From the proof of Theorems 2.7.3 and 2.7.4 in Sun [18], there exists R0> 0 such that for any R > R0,
        deg ( I - λ A 1 , T R , θ ) = 0 , λ > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ20_HTML.gif
        (4.2)
        Take 0 < r < R0. Let m = max0≤t≤1,|u|<r|f1(t, u)|, G = max0≤s,t≤1G(t, s), λ ̄ = r m G - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq29_HTML.gif. For any 0 < λ < λ ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq30_HTML.gif, u ∈ ∂T r , we have
        λ A 1 u = max 0 t 1 | 0 1 λ G ( t , s ) f 1 ( s , u ( s ) ) d s | < λ ̄ G m = r = u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equv_HTML.gif
        Thus
        deg ( I - λ A 1 , T r , θ ) = 1 , 0 < λ < λ ̄ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ21_HTML.gif
        (4.3)
        From (4.2) and (4.3), we have that for any 0 < λ < λ ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq31_HTML.gif, there exist uλ,C[0,1], ||uλ|| > r such that uλ = λA1uλ. In order to apply Lemma 4.1 we claim that
        lim λ 0 + , u λ = λ A 1 u λ , u λ > r u λ = + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ22_HTML.gif
        (4.4)
        In fact, if (4.4) doesn't hold, then there exist λ n > 0, u λ n C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq32_HTML.gif such that λ n → 0, r < | | u λ n | | < c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq33_HTML.gif (c > 0 is a constant) and
        u λ n = λ n A 1 u λ n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ23_HTML.gif
        (4.5)

        Since A1 is completely continuous, then { u λ n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq34_HTML.gif has a subsequence converging to u* ∈ C[0,1]. Assume, without loss of generality, that it is { u λ n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq34_HTML.gif. Taking n → +∞ in (4.5), we have u* = θ, which is a contradiction to | | u λ n | | > r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq35_HTML.gif. Hence (4.4) holds.

        Let D = [0,1], D0 = [t1, t2] ⊂ (0, 1) ⊂ D, δ = min t 1 t t 2 p ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq36_HTML.gif. By (G3)
        G ( t , s ) δ G ( τ , s ) , t D 0 , s , τ D . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equw_HTML.gif
        From (H3) and the definition of f1, we have
        lim u + f 1 ( t , u ) = lim u + f ( t , u ) = + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equx_HTML.gif
        By Lemma 4.1 there exists R = R ( λ ̄ ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq37_HTML.gif such that if 0 < λ λ ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq38_HTML.gif, ||uλ|| ≥ R and uλ = λA1uλ, then uλ(t) ≥ 0. By (4.4), there exists λ * < λ ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq39_HTML.gif such that if 0 < λ ≤ λ*, ||uλ|| ≥ r and uλ = λA1uλ, then ||uλ|| ≥ R. Thus uλ(t) ≥ 0. By the definitions of A1 and f1 we have
        u λ ( t ) = λ A 1 u λ ( t ) = λ 0 1 G ( t , s ) f 1 ( s , u ( s ) ) d s = λ 0 1 G ( t , s ) f ( s , u ( s ) ) d s = λ A u λ ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equy_HTML.gif

        And so u λ (t) is a positive solution of problem (P). □

        Remark 4.1 In Theorem 4.1, without assuming that f(t, u) ≥ 0 when u ≥ 0, we obtain the existence of positive solutions for the semipositive BVP (P).

        Remark 4.2 Noticing that, in this article, we only study the existence of positive solutions for BVP (P), which is irrelevant to the value of f(t, u) when u ≤ 0, we only need to suppose that f(t, u) is bounded from below when u ≥ 0. In fact, f(t, u) may be unbounded from below when u ≤ 0.

        5 Two examples

        In order to illustrate possible applications of Theorems 3.2 and 4.1, we give two examples.

        Example 5.1 Consider the fourth-order BVP
        u ( 4 ) ( t ) = λ [ sin u ( t ) + u ( t ) arctan u ( t ) + π u ( t ) ] , 0 t 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ24_HTML.gif
        (P1)
        In this example, f(t, u) = sinu + u arctanu + πu, then
        lim inf u + f ( t , u ) u = 3 π 2 , lim inf u - f ( t , u ) u = π 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ25_HTML.gif
        (5.1)
        lim u 0 f ( t , u ) u = 1 + π . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ26_HTML.gif
        (5.2)

        Take α = 3π/2, β = π/2, ρ = π +1. Then (5.1), (5.2) indicate (H1), (3.11) hold, repectively. Notice that α > ρ > β > 0 and f(t, 0) ≡ 0, ∀t ∈ [0,1], by Theorem 3.2 for any λ ( λ 1 α , λ 1 β ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq40_HTML.gif and λ λ 1 ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq41_HTML.gif, BVP (P1) has at least one nontrivial solution.

        Example 5.2 Consider the fourth-order BVP
        u ( 4 ) ( t ) = λ [ ( t + 1 ) u 2 - u 3 ] , 0 t 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ27_HTML.gif
        (P2)
        In this example, f ( t , u ) = t + 1 u 2 - u 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_IEq42_HTML.gif, then
        lim u + f ( t , u ) u = lim u + ( t + 1 ) u - 1 u 2 3 = + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-56/MediaObjects/13661_2011_Article_150_Equ28_HTML.gif
        (5.3)

        (5.3) means (H3) holds. By Theorem 4.1 there exists λ* > 0 such that for any 0 < λ < λ* BVP (P2) has at least one positive solution.

        Declarations

        Acknowledgements

        The authors would like to thank the referees for carefully reading this article and making valuable comments and suggestions. This work is supported by the Foundation items: NSFC (10971179, 11026203), NSF (BK2011202) of the Jiangsu Province, NSF (09XLR04) of the Xuzhou Normal University and NSF (2010KY07) of the Suqian College.

        Authors’ Affiliations

        (1)
        Department of Teachers Education, Suqian College
        (2)
        Department of Mathematics, Jiangsu Normal University
        (3)
        School of Mechanics and Civil Engineering, China University of Mining & Technology

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        © Lu et al; licensee Springer. 2012

        This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.