Positive solutions for nonlocal fourth-order boundary value problems with all order derivatives

  • Yanping Guo1,

    Affiliated with

    • Fei Yang2 and

      Affiliated with

      • Yongchun Liang1Email author

        Affiliated with

        Boundary Value Problems20122012:29

        DOI: 10.1186/1687-2770-2012-29

        Received: 29 September 2011

        Accepted: 2 March 2012

        Published: 2 March 2012

        Abstract

        In this article, by the fixed point theorem in a cone and the nonlocal fourth-order BVP's Green function, the existence of at least one positive solution for the nonlocal fourth-order boundary value problem with all order derivatives

        u ( 4 ) ( t ) + A u ( t ) = λ f ( t , u ( t ) , u ( t ) , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equa_HTML.gif

        is considered, where f is a nonnegative continuous function, λ > 0, 0 < A < π2, p, qL[0, 1], p(s) ≥ 0, q(s) ≥ 0. The emphasis here is that f depends on all order derivatives.

        Keywords

        fourth-order boundary value problem fixed point theorem Green's function positive solution

        1 Introduction

        The deformation of an elastic beam in equilibrium state, whose two ends are simply supported, can be described by a fourth-order ordinary equation boundary value problem. Owing to its significance in physics, the existence of positive solutions for the fourth-order boundary value problem has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point index theory, the Krasnosel'skii's fixed point theorem and the method of upper and lower solutions, in reference [110].

        In recent years, there has been much attention on the question of positive solutions of the fourth-order differential equations with one or two parameters. By the Krasnosel'skii's fixed point theorem in cone [11], Bai [5] investigated the following fourth-order boundary value problem with one parameter
        u ( 4 ) ( t ) + β u ( t ) = λ f ( t , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equb_HTML.gif

        where λ > 0, 0 < β < π2, f: C([0, 1] × [0, ∞) × (-∞, 0], [0, ∞)) is continuous, p, qL[0, 1], p(s) ≥ 0, q(s) ≥ 0, 0 1 p ( s ) d s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq1_HTML.gif, 0 1 q ( s ) sin β s d s + 0 1 q ( s ) sin β ( 1 - s ) d s < sin β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq2_HTML.gif.

        By the fixed point index in cone, Ma [7] proved the existence of symmetric positive solutions for the nonlocal fourth-order boundary value problem
        u ( 4 ) ( t ) = h ( t ) f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equc_HTML.gif
        All the above works were done under the assumption that all order derivatives u', u″, u‴ are not involved explicitly in the nonlinear term f. In this article, we are concerned with the existence of positive solutions for the nonlocal fourth-order boundary value problem
        u ( 4 ) ( t ) + A u ( t ) = λ f ( t , u ( t ) , u ( t ) , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ1_HTML.gif
        (1.1)

        Throughout, we assume

        (H1) λ > 0, 0 < A < π2;

        (H2) f: [0, 1] × R4R+ is continuous, p, qL[0, 1], p(s) ≥ 0, q(s) ≥ 0, 0 1 p ( s ) d s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq1_HTML.gif, 0 1 q ( s ) sin A s d s + 0 1 q ( s ) sin A ( 1 - s ) d s < sin A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq3_HTML.gif.

        We will impose all order derivatives in f and make use of two continuous convex functionals which will ensure the existence of at least one positive solution to (1.1). Bai [5] applied Krasnoselskii's fixed point theorem. Ma [8] used fixed point index in cone and Leray-Schauder degree. In this article, to show the existence of positive solutions to (1.1), we define two positive continuous convex functionals. Then, using the new fixed point theorem [12] in a cone and the nonlocal fourth-order BVP's Green function, we give some new criteria for the existence of positive solutions to (1.1).

        2 The preliminary lemmas

        Let Y = C[0, 1] be the Banach space equipped with the norm
        | | u ( t ) | | 0 = max t [ 0 , 1 ] | u ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equd_HTML.gif

        Set λ1, λ2 be the roots of the polynomial P(λ) = λ2 + , namely λ1 = 0, λ2 = -A. By (H1), it is obviously that -π2< λ2< 0.

        Let Q1(t, s), Q2(t, s) be, respectively the Green's functions of the following problems
        - u ( t ) + λ 1 u ( t ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , - u ( t ) + λ 2 u ( t ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Eque_HTML.gif
        Then, carefully calculation yield
        Q 1 ( t , s ) = G 1 ( t , s ) + 0 1 G 1 ( s , x ) p ( x ) d x 1 - 0 1 p ( x ) d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equf_HTML.gif
        Q 2 ( t , s ) = G 2 ( t , s ) + sin A t + sin A ( 1 - t ) 0 1 G 2 ( s , x ) q ( x ) d x sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equg_HTML.gif
        G 1 ( t , s ) = s ( 1 - t ) , 0 s t 1 , t ( 1 - s ) , 0 t s 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equh_HTML.gif
        G 2 ( t , s ) = sin A s sin A ( 1 - t ) A sin A , 0 s t 1 , sin A t sin A ( 1 - s ) A sin A , 0 t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equi_HTML.gif
        Denote
        ω 1 = 1 1 - 0 1 p ( x ) d x , ω 2 ( t ) = sin A t + sin A ( 1 - t ) sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equj_HTML.gif
        Lemma 2.1. [5] Suppose that (H1) and (H2) hold. Then for any y(t) ∈ C[0, 1], the problem
        u ( 4 ) ( t ) + A u ( t ) = y ( t ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 p ( s ) u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 1 q ( s ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ2_HTML.gif
        (2.1)
        has a unique solution
        u ( t ) = 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) y ( τ ) d τ d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ3_HTML.gif
        (2.2)
        where
        Q 1 ( t , s ) = G 1 ( t , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x , Q 2 ( s , τ ) = G 2 ( s , τ ) + ω 2 ( s ) 0 1 G 2 ( τ , x ) q ( x ) d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equk_HTML.gif
        By (2.2), we get
        u ( t ) = 0 1 0 1 Q 2 ( s , τ ) y ( τ ) d τ d s - 0 1 0 1 s Q 2 ( s , τ ) y ( τ ) d τ d s ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ4_HTML.gif
        (2.3)
        u ( t ) = - 0 1 Q 2 ( t , τ ) y ( τ ) d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ5_HTML.gif
        (2.4)
        u ( t ) = - 0 1 Q 2 ( t , τ ) t y ( τ ) d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ6_HTML.gif
        (2.5)

        Lemma 2.2. [5] Assume that (H1) and (H2) hold. Then one has

        (i) Q i (t, s) ≥ 0, ∀t, s ∈ [0, 1]; Q i (t, s) > 0, ∀t, s ∈ (0, 1);

        (ii) G i (t, s) ≥ b i G i (t, t)G i (s, s), ∀t, s ∈ [0, 1];

        (iii) G i (t, s) ≤ c i G i (s, s), ∀t, s ∈ [0, 1].

        where b1 = 1, b 2 = A sin A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq4_HTML.gif; c1 = 1, c 2 = 1 sin A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq5_HTML.gif.

        Let
        d i = min 1 4 t 3 4 b i G i ( t , t ) , ( i = 1 , 2 ) ; ξ = min 1 4 t 3 4 ω 2 ( t ) max 1 4 t 3 4 ω 2 ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equl_HTML.gif

        Lemma 2.3. [5] Suppose that (H1) and (H2) hold and w2, d i , ξ i are given as above. Then

        one has

        (i) max 0 t 1 ω 2 ( t ) = ω 2 1 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq6_HTML.gif

        (ii) 0 < d i < 1 , 0 < ξ < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq7_HTML.gif

        Lemma 2.4. If y(t) ∈ C[0, 1] and y(t) ≥ 0, then the unique solution u(t) of problem (2.1)

        satisfies
        min 1 4 t 3 4 u ( t ) d 1 | | u | | 0 , min 1 4 t 3 4 ( - u ( t ) ) d 2 ξ c 2 | | u | | 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equm_HTML.gif
        Proof. By (2.2) and (iii) of Lemma 2.2, we get
        u ( t ) = 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) y ( τ ) d τ d s 0 1 0 1 c 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s = 0 1 0 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s = 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equn_HTML.gif
        So,
        | | u | | 0 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equo_HTML.gif
        Using (ii) of Lemma 2.2, we have
        min 1 4 t 3 4 u ( t ) = min 1 4 t 3 4 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) y ( τ ) d τ d s min 1 4 t 3 4 0 1 0 1 [ b 1 G 1 ( t , t ) G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x ] Q 2 ( s , τ ) y ( τ ) d τ d s = 0 1 0 1 d 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s d 1 0 1 0 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) y ( τ ) d τ d s = d 1 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) y ( τ ) d τ d s d 1 | | u | | 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equp_HTML.gif
        By (2: 4) and (iii) of Lemma 2.2, we get
        max 1 4 t 3 4 ( - u ( t ) ) = max 1 4 t 3 4 0 1 Q 2 ( t , τ ) y ( τ ) d τ 0 1 c 2 G 2 ( τ , τ ) + max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ c 2 max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equq_HTML.gif
        So,
        | | u | | 0 c 2 max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equr_HTML.gif
        Using (ii) of Lemma 2.2, we have
        min 1 4 t 3 4 ( - u ( t ) ) = min 1 4 t 3 4 0 1 Q 2 ( t , τ ) y ( τ ) d τ min 1 4 t 3 4 0 1 b 2 G 2 ( t , t ) G 2 ( τ , τ ) + ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ 0 1 b 2 G 2 ( t , t ) G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ = 0 1 d 2 G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ d 2 min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x y ( τ ) d τ d 2 c 2 min 1 4 t 3 4 ω 2 ( t ) max 1 4 t 3 4 ω 2 ( t ) | | u | | 0 d 2 ξ c 2 | | u | | 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equs_HTML.gif

        The proof is completed.

        Let X be a Banach space and KX a cone. Suppose α, β: ×R+ are two continuous convex functionals satisfying α(λu) = |λ|α(u), β(λu) = |λ|β(u), for uX, λR, and ||u|| ≤ M max{α(u), β(u)}, for uX and α(u) ≤ α(v) for u, vK, uv, where M > 0 is a constant.

        Theorem 2.1. [12] Let r2> r1> 0, L > 0 be constants and
        Ω i = { u X : α ( u ) < r i , β ( u ) < L } , i = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equt_HTML.gif
        two bounded open sets in X. Set
        D i = { u X : α ( u ) = r i } , i = 1 , 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equu_HTML.gif

        Assume T: KK is a completely continuous operator satisfying

        (A1) α(Tu) < r1, uD1K; α(Tu) > r2, uD2K;

        (A2) β(Tu) < L, uK;

        (A3) there is a p ∈ (Ω2K) \ {0} such that α(p) ≠ 0 and α(u + λp) ≥ α(u), for all uK and λ ≥ 0.

        Then T has at least one fixed point in ( Ω 2 \ Ω ̄ 1 ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq8_HTML.gif.

        3 The main results

        Let X = C4[0, 1] be the Banach space equipped with the norm | | u | | = max t [ 0 , 1 ] | u ( t ) | + max t [ 0 , 1 ] | u ( t ) | + max t [ 0 , 1 ] | u ( t ) | + max t [ 0 , 1 ] | u ( t ) | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq9_HTML.gif and K = u X : u ( t ) 0 , u ( t ) 0 , min 1 4 t 3 4 u ( t ) d 1 | | u | | 0 , min 1 4 t 3 4 ( - u ( t ) ) d 2 ξ c 2 | | u | | 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq10_HTML.gif is a cone in X.

        Define two continuous convex functionals α ( u ) = max t [ 0 , 1 ] | u ( t ) | + max t [ 0 , 1 ] | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq11_HTML.gif and β ( u ) = max t [ 0 , 1 ] | u ( t ) | + max t [ 0 , 1 ] | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq12_HTML.gif, for each uX, then ||u|| ≤ 2 max{α(u), β(u)} and α(λu) = |λ|α(u), β(λu) = |λ|β(u), for uX, λR; α(u) ≤ α(v) for u, vK, uv.

        In the following, we denote
        B = 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s , D = 0 1 G 2 ( τ , τ ) + ω 2 1 2 0 1 G 2 ( τ , x ) q ( x ) d x d τ , F = 1 sin A 0 1 sin A τ d τ + A 0 1 0 1 G 2 ( τ , x ) q ( x ) d x d τ sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x , η 0 = 1 B + c 2 D , η 1 = 1 1 4 3 4 Q 2 1 2 , τ d τ , η 2 = 2 3 c 2 D + 4 F , θ = min d 1 2 , d 2 ξ 2 c 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equv_HTML.gif
        We will suppose that there are L > b > θb > c > 0 such that f(t, u, v, u0, v0) satisfies the following growth conditions:
        ( H 3 ) f ( t , u , v , u 0 , v 0 ) < c η 0 λ , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , c ] × [ - L , L ] × [ - c , 0 ] × [ - L , L ] , ( H 4 ) f ( t , u , v , u 0 , v 0 ) b η 1 λ , for ( t , u , v , u 0 , v 0 ) 1 4 , 3 4 × [ θ b , b ] × [ - L , L ] × [ - b , 0 ] × [ - L , L ] 1 4 , 3 4 × [ 0 , b ] × [ - L , L ] × [ - b , - θ b ] × [ - L , L ] , ( H 5 ) f ( t , u , v , u 0 , v 0 ) < L η 2 λ , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , b ] × [ - L , L ] × [ - b , 0 ] × [ - L , L ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equw_HTML.gif
        Let f 1 ( t , u , v , u 0 , v 0 ) = f 1 t , u * , v * , u 0 * , v 0 * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq13_HTML.gif, where
        u * = min { max ( u , 0 ) , b } , v * = min { max ( v , - L ) , L } , u 0 * = min { max ( u 0 , - b ) , 0 } , v 0 * = min { max ( v , - L ) , L } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equx_HTML.gif
        We denote
        ( T u ) ( t ) = λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ7_HTML.gif
        (3.1)
        ( T u ) ( t ) = λ t 1 0 1 Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s - 0 1 0 1 s Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ8_HTML.gif
        (3.2)
        ( T u ) ( t ) = - λ 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ9_HTML.gif
        (3.3)
        ( T u ) ( t ) = - λ 0 1 Q 2 ( t , τ ) t f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ10_HTML.gif
        (3.4)

        Lemma 3.1. Suppose that (H1) and (H2) hold. Then T: KK is completely continuous.

        Proof. For uK, by (3.1), (3.3) and Lemma 2.2, it is obviously that Tu ≥ 0, (Tu)″ ≤ 0. In view of c1 = 1, c2> 1, so
        | | T u | | 0 = max t [ 0 , 1 ] λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( t , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 [ c 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x ] Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s = λ 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equy_HTML.gif
        | | ( T u ) | | 0 = max t [ 0 , 1 ] - λ 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ λ 0 1 [ c 2 G 2 ( τ , τ ) + max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ max 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x × f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equz_HTML.gif
        By Lemma 2.3, (3.1) and (3.3), we have
        min 1 4 t 3 4 ( T u ) ( t ) = min 1 4 t 3 4 λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 [ b 1 G 1 ( t , t ) G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x ] Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s = λ 0 1 0 1 d 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x × Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s d 1 λ 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s d 1 | | T u | | 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equaa_HTML.gif
        min 1 4 t 3 4 ( - ( T u ) ( t ) ) = min 1 4 t 3 4 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ min 1 4 t 3 4 0 1 [ b 2 G 2 ( t , t ) G 2 ( τ , τ ) + ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ 0 1 [ b 2 G 2 ( t , t ) G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ = 0 1 [ d 2 G 2 ( τ , τ ) + min 1 4 t 3 4 ω 2 ( t ) 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d 2 min 1 4 t 3 4 ω 2 ( t ) 0 1 [ G 2 ( τ , τ ) + 0 1 G 2 ( τ , x ) q ( x ) d x ] f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d 2 c 2 min 1 4 t 3 4 ω 2 ( t ) max 1 4 t 3 4 ω 2 ( t ) | | ( T u ) | | 0 d 2 ξ c 2 | | ( T u ) | | 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equab_HTML.gif

        So we can get T(K) ⊂ K: Let BK is bounded, it is clear that T(B) is bounded. Using f1, Q1(t, s), Q2(t, s) is continuous, we show that T(B) is equicontinuous. By the Arzela-Ascoli theorem, a standard proof yields T: KK is completely continuous.

        Theorem 3.1. Suppose that (H1)-(H5) hold. Then BVP (1.1) has at least one positive solution u(t) satisfying
        c < α ( u ) < b , β ( u ) < L . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equac_HTML.gif
        Proof. Take
        Ω 1 = { u X : α ( u ) < c , β ( u ) < L } , Ω 2 = { u X : α ( u ) < b , β ( u ) < L } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equad_HTML.gif
        two bounded open sets in X, and
        D 1 = { u X : α ( u ) = c } , D 2 = { u X : α ( u ) = b } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equae_HTML.gif

        By Lemma 3.1, T: KK is completely continuous. Let p = b 2 ( Ω 2 K ) \ { 0 } , α ( p ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq14_HTML.gif. It is easy to see that α(u + λp) ≥ α(u), for all uK and λ ≥ 0.

        Let uD1, we have
        | | T u | | 0 = max t [ 0 , 1 ] λ 0 1 0 1 Q 1 ( t , s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 c 1 G 1 ( s , s ) + ω 1 0 1 G 1 ( s , x ) p ( x ) d x Q 2 ( s , τ ) d τ d s × c η 0 λ = c η 0 0 1 0 1 Q 1 ( s , s ) Q 2 ( s , τ ) d τ d s = B c η 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equaf_HTML.gif
        | | ( T u ) | | 0 = max t [ 0 , 1 ] - λ 0 1 Q 2 ( t , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ < λ 0 1 c 2 G 2 ( τ , τ ) + ω 2 1 2 0 1 G 2 ( τ , x ) q ( x ) d x d τ × c η 0 λ c 2 c η 0 0 1 G 2 ( τ , τ ) + ω 2 1 2 0 1 G 2 ( τ , x ) q ( x ) d x d τ = c 2 D c η 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equag_HTML.gif
        Hence, for uD1K, α(u) = c, we get
        α ( T u ) = | | T u | | 0 + | | ( T u ) | | 0 < B c η 0 + c 2 D c η 0 = ( B + c 2 D ) c η 0 = c . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equah_HTML.gif
        Whereas for uD2K, α(u) = b, there is | | u | | 0 b 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq15_HTML.gif or | | u | | 0 b 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq16_HTML.gif, By Lemma 2.4, we get
        min 1 4 t 3 4 u ( t ) d 1 | | u | | 0 d 1 b 2 or min 1 4 t 3 4 ( - u ( t ) ) d 2 ξ c 2 | | u | | 0 d 2 ξ b 2 c 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equai_HTML.gif
        Therefore, using (H4) and (3.3), we have
        | ( T u ) 1 2 | = λ 0 1 Q 2 1 2 , τ f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ > λ 1 4 3 4 Q 2 1 2 , τ f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ λ × b η 1 λ 1 4 3 4 Q 2 1 2 , τ d τ = b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equaj_HTML.gif
        Hence,
        α ( T u ) ( T u ) 1 2 > b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equak_HTML.gif
        By (3.2), (3.4), and (H5), for uK, we have
        ( T u ) 0 = max t [ 0 , 1 ] | λ t 1 0 1 Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s λ 0 1 0 1 s Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | max t [ 0 , 1 ] | λ t 1 0 1 Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | + max t [ 0 , 1 ] | λ 0 1 0 1 s Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | λ | 0 1 0 1 ( 1 + s ) Q 2 ( s , τ ) f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ d s | λ × η 2 L λ | 0 1 0 1 ( 1 + s ) [ c 2 G 2 ( τ , τ ) + ω 2 ( 1 2 ) 0 1 G 2 ( τ , x ) q ( x ) d x ] d τ d s | η 2 L × 3 2 c 2 0 1 [ G 2 ( τ , τ ) + ω 2 ( 1 2 ) 0 1 G 2 ( τ , x ) q ( x ) d x ] d τ = 3 2 c 2 D η 2 L , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equal_HTML.gif
        ( T u ) 0 = max t [ 0 , 1 ] λ 0 1 Q 2 ( t , τ ) t f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) d τ 2 λ 0 1 sin A τ sin A + A 0 1 G 2 ( τ , x ) q ( x ) d x sin A - 0 1 q ( x ) sin A x d x - 0 1 q ( x ) sin A ( 1 - x ) d x × | f 1 ( τ , u ( τ ) , u ( τ ) , u ( τ ) , u ( τ ) ) | d τ < λ 2 F × η 2 L λ = 2 F η 2 L . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equam_HTML.gif
        So,
        β ( T u ) = ( T u ) 0 + ( T u ) 0 < 3 2 c 2 D η 2 L + 2 F η 2 L = 3 2 c 2 D + 2 F η 2 L = L . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equan_HTML.gif
        Theorem 2.1 implies there is ( Ω 2 \ Ω ̄ 1 ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq17_HTML.gif such that u = Tu. So, u(t) is a positive solution for BVP (1.1) satisfying
        c < α ( u ) < b , β ( u ) < L . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equao_HTML.gif

        Thus, Theorem 3.1 is completed.

        4 Example

        Example 4.1. Consider the following boundary value problem
        u ( 4 ) ( t ) + π 2 9 u ( t ) = π 2 f ( t , u ( t ) , u ( t ) , u ( t ) , u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 1 s u ( s ) d s , u ( 0 ) = u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equ11_HTML.gif
        (4.1)
        where
        f ( t , u , v , u 0 , v 0 ) = 1 20 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 2 ] × [ - 16000 , 16000 ] × [ - 2 , 0 ] × [ - 16000 , 16000 ] , 1 20 ( 2 - u 0 ) ( 3 - u ) + 27 2 ( 3 - u 0 ) ( u - 2 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 2 , 3 ] × [ - 16000 , 16000 ] × [ - 2 , 0 ] × [ - 16000 , 16000 ] , 1 20 ( u + 2 ) ( u 0 + 3 ) - 27 2 ( u + 3 ) ( u 0 + 2 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 2 ] × [ - 16000 , 16000 ] × [ - 3 , - 2 ] × [ - 16000 , 16000 ] , 1 5 ( 3 - u ) ( u 0 + 3 ) + 135 2 ( u - 2 ) ( u 0 + 3 ) - 27 2 ( u + 3 ) ( u 0 + 2 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 2 , 3 ] × [ - 16000 , 16000 ] × [ - 3 , - 2 ] × [ - 16000 , 16000 ] , 27 2 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | , ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 3 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , 0 ] × [ - 16000 , 16000 ] , [ 0 , 1 ] × [ 0 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , - 3 ] × [ - 16000 , 16000 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equap_HTML.gif
        In this problem, we know that A = π 2 9 , λ = π 2 , p ( t ) = t , q ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq18_HTML.gif, then we can get b 1 = 1 , b 2 = 3 π 6 , c 1 = 1 , c 2 = 2 3 3 , ω 1 = 2 , ω 2 = 2 3 sin π 3 ( 1 + t ) 3 , d 1 = 3 16 , d 2 = 3 - 1 4 , ξ = 2 + 3 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq19_HTML.gifFurther more, we obtain
        B = 1944 3 - 972 π - 9 π 3 4 π 5 , D = 9 - 3 π 2 π 2 , F = 3 π . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equaq_HTML.gif

        then η 0 = 12 π 5 5832 3 - 2916 π - 27 π 3 + 36 3 π 3 - 12 π 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq20_HTML.gif, η 1 = π 2 3 6 + 3 3 - 9 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq21_HTML.gif, η 2 = 2 π 2 9 3 + 4 3 π - 3 π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq22_HTML.gif, θ = min d 1 2 , d 2 ξ 2 c 2 = 2 + 3 ( 3 - 3 ) 32 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_IEq23_HTML.gif, θb ≈ 3.06 > 3.

        If we take c = 2, b = 40, L = 16000, then we get
        f ( t , u , v , u 0 , v 0 ) = 1 20 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | 0 . 7 < c η 0 λ 0 . 8 , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 2 ] × [ - 16000 , 16000 ] × [ - 2 , 0 ] × [ - 16000 , 16000 ] , f ( t , u , v , u 0 , v 0 ) = 27 2 ( u - u 0 ) + 1 2 | cos ( v + v 0 ) | 40 > b η 1 λ 38 , for ( t , u , v , u 0 , v 0 ) 1 4 , 3 4 × [ θ b , 40 ] × [ - 16000 , 16000 ] × [ - 40 , 0 ] × [ - 16000 , 16000 ] [ 1 4 , 3 4 ] × [ 0 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , - θ b ] × [ - 16000 , 16000 ] , f ( t , u , v , u 0 , v 0 ) 1080 . 5 < L η 2 λ 1146 , for ( t , u , v , u 0 , v 0 ) [ 0 , 1 ] × [ 0 , 40 ] × [ - 16000 , 16000 ] × [ - 40 , 0 ] × [ - 16000 , 16000 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equar_HTML.gif
        Then all the conditions of Theorem 3.1 are satisfied. Therefore, by Theorem 3.1 we know that boundary value problem (4.1) has at least one positive solution u(t) satisfying
        2 < α ( u ) < 40 , β ( u ) < 16000 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-29/MediaObjects/13661_2011_Article_153_Equas_HTML.gif

        Declarations

        Acknowledgements

        The project is supported by the Natural Science Foundation of China (10971045) and the Natural Science Foundation of Hebei Province (A2009000664, A2011208012). The research item financed by the talent training project funds of Hebei Province. The authors would like to thank the referee for helpful comments and suggestions.

        Authors’ Affiliations

        (1)
        College of Electrical Engineering and Information, Hebei University of Science and Technology
        (2)
        College of Sciences, Hebei University of Science and Technology

        References

        1. Bai ZB: The method of lower and upper solution for a bending of an elastic beam equation. J Math Anal Appl 2000, 248: 195-202. 10.1006/jmaa.2000.6887MathSciNetView Article
        2. Bai ZB: The method of lower and upper solutions for some fourth-order boundary value problems. Nonlinear Anal 2007, 67: 1704-1709. 10.1016/j.na.2006.08.009MathSciNetView Article
        3. Chai GQ: Existence of positive solutions for fourth-order boundary value problem with variable parameters. Nonlinear Anal 2007, 66: 870-880. 10.1016/j.na.2005.12.028MathSciNetView Article
        4. Zhao JF, Ge WG: Positive solutions for a higher-order four-point boundary value problem with a p -Laplacian. Comput Math Appl 2009, 58: 1103-1112. 10.1016/j.camwa.2009.04.022MathSciNetView Article
        5. Bai ZB: Positive solutions of some nonlocal fourth-order boundary value problem. Appl Math Comput 2010, 215: 4191-4197. 10.1016/j.amc.2009.12.040MathSciNetView Article
        6. Li YX: Positive solutions of fourth-order boundary value problems with two parameters. J Math Anal Appl 2003, 281: 477-484. 10.1016/S0022-247X(03)00131-8MathSciNetView Article
        7. Ma HL: Symmetric positive solutions for nonlocal boundary value problems of fourth-order. Nonlinear Anal 2008, 68: 645-651. 10.1016/j.na.2006.11.026MathSciNetView Article
        8. Ma RY: Existence of positive solutions of a fourth-order boundary value problem. Appl Math Comput 2005, 168: 1219-1231. 10.1016/j.amc.2004.10.014MathSciNetView Article
        9. Ma R, Wang H: On the existence of positive solutions of fourth-order ordinary differential equations. Appl Anal 1995, 59: 225-231. 10.1080/00036819508840401MathSciNetView Article
        10. Yao QL: Local existence of multiple positive solutions to a singular cantilever beam equation. J Math Anal Appl 2010, 363: 138-154. 10.1016/j.jmaa.2009.07.043MathSciNetView Article
        11. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, New York; 1988.
        12. Guo YP, Ge WG: Positive solutions for three-point boundary value problems with dependence on the first order derivatives. J Math Anal Appl 2004, 290: 291-301. 10.1016/j.jmaa.2003.09.061MathSciNetView Article

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