Positive solutions for the third-order boundary value problems with the second derivatives

  • Yanping Guo1,

    Affiliated with

    • Yujing Liu2 and

      Affiliated with

      • Yonhchun Liang1Email author

        Affiliated with

        Boundary Value Problems20122012:34

        DOI: 10.1186/1687-2770-2012-34

        Received: 3 November 2011

        Accepted: 26 March 2012

        Published: 26 March 2012

        Abstract

        By using the fixed-point index theory in a cone and defining a linear operator, we obtain the existence of at least one positive solution for the third-order boundary value problem with integral boundary conditions

        u ( t ) + f ( t , u ( t ) , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 1 g ( t ) u ( t ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equa_HTML.gif

        where f : [0, 1] × R+ × R-→ R+ is a nonnegative function. The associated Green's function for the above problem is also used, and a new reproducing cone also used.

        Keywords

        fixed-point index theory Green's function positive solution boundary value problem

        1 Introduction

        By eigenvalue criteria, Webb [1] obtained the existence of multiple positive solutions of a Hammerstein integral equation of the form
        u ( t ) = 0 1 k ( t , s ) g ( s ) f ( s , u ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equb_HTML.gif
        where k can have discontinuities and gL1. Then, some articles have studied different BVPs by this way (see [25]). Webb [4] introduced an unified method to study existence of at least one nonzero solution for higher order boundary value problems
        u ( n ) ( t ) + g ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( n ) ( 0 ) = 0 , 0 k n - 2 , u ( 1 ) = 0 1 u ( s ) d A ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equc_HTML.gif
        In 2010, Hao [5] considered the existence of positive solutions of the n th-order BVP
        u ( n ) ( t ) + λ a ( t ) f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = = u ( n - 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equd_HTML.gif
        Guo [6] studied the existence of positive solutions for the there-point boundary problem with the first-order derivative.
        x + f ( t , x , x ) = 0 , 0 < t < 1 , x ( 0 ) = 0 , x ( 1 ) = α x ( η ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Eque_HTML.gif
        where f is a nonnegative continuous function. In 2011, Zhao [7] studied third-order differential equations:
        x + f ( t , x ( t ) ) = θ , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equf_HTML.gif
        subject to integral boundary condition of the form
        x ( 0 ) = θ , x ( 0 ) = θ , x ( 1 ) = 0 1 g ( t ) x ( t ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equg_HTML.gif

        where fC([0, 1] × P, P).

        In this article, we study the existence of positive solutions for the following boundary value problem
        u ( t ) + f ( t , u ( t ) , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 1 g ( t ) u ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ1_HTML.gif
        (1.1)

        The results are proved by applying the fixed point index theory in a cone and spectral radius of a linear operator. Unlike reference [7], the nonlinear part f involves the second-order derivative and just satisfies Caratheodory conditions.

        The following conditions are satisfied throughout this article:

        (H1) f : [0, 1] × R+ × R-R+ satisfies Caratheodory conditions, that is, f(·,u, v) is measurable for each fixed uR+, vR-, and f(t, ·,·) is continuous for a.e. t ∈ [0, 1]. For any r, r' > 0, there exists φ r , r ( t ) L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq1_HTML.gif, such that 0 f ( t , u , v ) φ r , r ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq2_HTML.gif, where (u, v) ∈ [0, r] × [-r', 0], a.e. t ∈ [0, 1];

        (H2) gL[0, 1] is nonnegative, b ∈ [0, 1), where b = 0 1 s g ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq3_HTML.gif.

        2 Preliminaries

        Lemma 2.1[7]. Let yL1[0, 1] and y ≥ 0, the problem
        u ( t ) + y ( t ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 1 g ( t ) u ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ2_HTML.gif
        (2.1)
        has a unique solution
        u ( t ) = 0 1 H ( t , s ) y ( t ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equh_HTML.gif
        where H ( t , s ) = G ( t , s ) + t 1 - b 0 1 G ( τ , s ) y ( τ ) d τ , b = 0 1 s g ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq4_HTML.gif,
        G ( t , s ) = 1 2 t ( 1 - s ) 2 - 1 2 ( t - s ) 2 , 0 s t 1 , 1 2 t ( 1 - s ) 2 , 0 t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equi_HTML.gif

        Lemma 2.2. Let yL1[0, 1] and y ≥ 0, the unique solution of the boundary value problem (2.1) satisfies the following conditions: u(t) ≥ 0, u"(t) ≤ 0, for t ∈ [0, 1].

        Proof. By Lemma 2.1, u(t) ≥ 0. By differential equations u'"(t) + y(t) = 0, t ∈ (0, 1), we get
        u ( t ) - u ( 0 ) = - 0 1 y ( s ) d s , u ( t ) = - 0 1 y ( s ) d s 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equj_HTML.gif
        Let X = C2[0, 1] with u = max 0 t 1 u ( t ) + max 0 t 1 u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq5_HTML.gif. Obviously, (X, ||·||) is a Banach space. Define the cone PX by
        P = u X u ( t ) 0 , u ( t ) 0 , P r = u P u < r , r > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equk_HTML.gif

        Obviously P is a cone in X, and P r is a bounded open subset in P.

        Definition 2.1[1]. Let P be a cone in a Banach space X. If for any xX and x+, x-P, writing x = x+ + x- shows that P is a reproducing cone.

        Lemma 2.3. P is a reproducing cone in X.

        Proof. Suppose uX, so u" ∈ C[0, 1] and
        u = u - - u + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ3_HTML.gif
        (2.2)
        where u- = min{u"(t), 0}, u+ = min{-u"(t), 0}. Obviously u+,u-C[0, 1] and u+ ≤ 0,u- ≤ 0. For (2.2), we get
        u ( t ) = 0 t u - ( s ) d s - 0 t u + ( s ) d s + u ( 0 ) , u ( t ) = 0 t d s 0 s u - ( τ ) d τ - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) t + u ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equl_HTML.gif
        If u(0) ≥ 0, u'(0)t ≥ 0, let
        u 1 = - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) t + u ( 0 ) , u 2 = - 0 t d s 0 s u - ( τ ) d τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equm_HTML.gif

        So u1 ≥ 0, u2 ≥ 0, then u1, u2P and u = u1 - u2.

        If u(0) ≤ 0, u'(0)t ≤ 0, let
        u 1 = - 0 t d s 0 s u + ( τ ) d τ , u 2 = - 0 t d s 0 s u - ( τ ) d τ - u ( 0 ) t - u ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equn_HTML.gif

        So u1 ≥ 0, u2 ≥ 0, then u1, u2P and u = u1 - u2.

        If u(0) ≥ 0, u'(0)t ≤ 0, let
        u 1 = - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) , u 2 = - 0 t d s 0 s u - ( τ ) d τ - u ( 0 ) t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equo_HTML.gif

        So u1 ≥ 0, u2 ≥ 0, then u1, u2P and u = u1 - u2.

        If u(0) ≤ 0, u'(0)t ≥ 0, let
        u 1 = - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) t , u 2 = - 0 t d s 0 s u - ( τ ) d τ - u ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equp_HTML.gif

        So u1 ≥ 0, u2 ≥ 0, then u1, u2P and u = u1 - u2.

        Then P is a reproducing cone in X.

        Lemma 2.4 (Krein-Rutman) [8]. Let K be a reproducing cone in a real Banach space X and let L : KK be a compact linear operator with L(K) ⊂ K. r(L) is the spectral radius of L. If r(L) > 0, then there is φ1K\{0} such that 1 = r(L)φ1.

        Lemma 2.5[9]. Let X be a Banach space, P be a cone in X and Ω(P) be a bounded open subset in P. Suppose that A : Ω ( P ) ¯ P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq6_HTML.gif is a completely continuous operator. Then the following results hold
        1. (1)

          If there exists u 0P\{0} such that uAu + λu 0, for any u Ω(P), λ ≥ 0, then the fixed-point index i(A, Ω(P), P) = 0.

           
        2. (2)

          If 0 ∈ Ω(P), Au ≠ λu, for any u Ω(P), λ ≥ 1, then the fixed-point index i(A, Ω(P), P) = 1.

           
        Define the operator A: XX, L: XX, by
        A u ( t ) = 0 1 H ( t , s ) f ( s , u ( s ) , u ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equq_HTML.gif
        L u ( t ) = 0 1 H ( t , s ) ( u ( s ) - u ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equr_HTML.gif

        So A : PP is completely continuous operator; L : P → P is a compact linear operator.

        Lemma 2.6[7]. Assume that (H2) holds, then choose δ 0 , 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq7_HTML.gif, for all t ∈ [δ, 1 - δ],v, s ∈ [0, 1], we have
        G ( t , s ) ρ G ( v , s ) , H ( t , s ) ρ H ( v , s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equs_HTML.gif

        where ρ = 4δ2(1 - δ).

        Note: r(L) is the spectral radius of L. h = min t [ δ , 1 - δ ] δ 1 - δ H ( t , s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq8_HTML.gif, where δ 0 , 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq7_HTML.gif. By Lemma 2.6, obviously h > 0.

        Lemma 2.7. Suppose conditions (H1), (H2) hold, then r(L) > 0.

        Proof. Take u(t) ≡ 1, then u"(t) = 0, for any t ∈ [δ, 1 - δ] we get
        L u ( t ) δ 1 - δ H ( t , s ) d s h > ( 0 ) . L 2 u ( t ) δ 1 - δ H ( t , s ) L u ( s ) d s h δ 1 - δ H ( t , s ) d s h 2 > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equt_HTML.gif
        Repeating the process gives
        L k u ( t ) h k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equu_HTML.gif

        So, we get L k h k , r ( L ) = lim k L k 1 k h > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq9_HTML.gif. The proof is completed.

        By Lemma 2.4, then there is φ1P\{0} such that 1 = r(L)φ1.

        3 Main results

        In the following, we use the notation:
        f ¯ ( u , v ) = sup t [ 0 , 1 ] \ E f ( t , u , v ) , f ( u , v ) = inf t [ 0 , 1 ] \ E f ( t , u , v ) , f = max { lim u sup { sup v R f ¯ ( u , v ) u v } , lim v sup { sup u R + f ¯ ( u , v ) u v } } , f 0 d = max { lim u 0 + inf { inf v [ d , 0 ] f ( u , v ) u v } , lim v 0 inf { inf u [ 0 , d ] f ( u , v ) u v } } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equv_HTML.gif

        where E is a fixed subset of [0, 1] of measure zero, d > 0.

        Lemma 3.1. Suppose
        0 f < μ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ4_HTML.gif
        (3.1)

        where μ = 1/r(L), then there exists R0 > 0 such that i(A, P r , P) = 1 for each r > R0.

        Proof. Let ε > 0 satisfy fμ - ε, then there exist r1 > 0 such that
        f ( t , u , v ) ( μ - ε ) ( u - v ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equw_HTML.gif

        for all u > r1 or v < -r1 and a.e. t ∈ [0, 1].

        By (H1), there exists φ r 1 L [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq10_HTML.gif such that
        0 f ( t , u , v ) φ r 1 ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equx_HTML.gif
        for all (u, v) ∈ [0, r1] × [-r1, 0] and a.e. t ∈ [0, 1]. Hence, we have
        f ( t , u , v ) ( μ - ε ) ( u - v ) + φ r 1 ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ5_HTML.gif
        (3.2)

        for all u ∈ R+, vR- and a.e. t ∈ [0, 1].

        Since 1 μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq11_HTML.gif is the spectrum radius of L. It follows from 1 μ - ε I - L - 1 = n = 0 ( μ - ε ) n + 1 L n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq12_HTML.gif, (I/(μ - ε) - L)-l exists, let
        C = 0 1 H ( t , s ) φ r 1 ( s ) d s , R 0 = 1 μ - ε I - L - 1 C μ - ε 3 2 - 1 2 t 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equy_HTML.gif
        For r > R0, by Lemma 2.5 we will prove
        A u λ u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equz_HTML.gif

        for each u ∈ ∂P r and λ ≥ 1.

        In fact, if not, there exist u0∂P r and λ0 ≥ 1 such that Au0 = λ0u0.

        Together with (3.2) implies
        u 0 ( t ) A u 0 ( t ) 0 1 H ( t , s ) [ ( μ ε ) u 0 ( s ) + φ r 1 ( t ) ] d s 0 1 H ( t , s ) ( μ ε ) [ u 0 ( s ) v 0 ( s ) ) + φ r 1 ( s ) ] d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equaa_HTML.gif
        So
        u 0 ( t ) ( μ - ε ) L u 0 ( t ) + C , u 0 ( t ) λ 0 u 0 ( t ) = ( A u 0 ( t ) ) ( μ - ε ) ( L u 0 ( t ) ) - C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equab_HTML.gif
        Then
        1 μ - ε I - L u 0 ( t ) C μ - ε 3 2 - 1 2 t 2 , 1 μ - ε I - L u 0 ( t ) C μ - ε 3 2 - 1 2 t 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equac_HTML.gif
        So
        C μ - ε 3 2 - 1 2 t 2 - 1 μ - ε I - L u 0 ( t ) P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equad_HTML.gif
        Then
        u 0 ( t ) ( I μ ε L ) 1 C μ ε ( 3 2 1 2 t 2 ) , u 0 ' ' ( t ) [ ( I μ ε L ) 1 C μ ε ( 3 2 1 2 t 2 ) ] ' ' , u 0 ( t ) R 0 < r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equae_HTML.gif

        This is a contradiction. By Lemma 2.5 (2), we get that i(A, P r , P) = 1 for each r > R0. The proof is completed.

        Lemma 3.2. Suppose there exists d > 0 such that
        μ < f 0 d . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ6_HTML.gif
        (3.3)

        Then there exists ρ0 > 0 and dρ0 such that for each ρ∈ (0, ρ0], if uAu for u ∈ ∂, then i(A, P ρ , P) = 0.

        Proof. Let ε > 0 satisfy f 0 d μ + ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq13_HTML.gif, there exist dρ0 > 0 such that
        f ( t , u , v ) ( μ + ε ) ( u - v ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ7_HTML.gif
        (3.4)

        for u ∈ [0, ρ0],v ∈ [-ρ0,0] and a.e. t ∈ [0, 1].

        Let ρ ∈ (0,ρ0], by Lemma 2.5 (1), we prove that: uAu + λφ1 for all u∂Pρ, λ > 0, where φ1P\{0} is the eigenfunction of L corresponding to the eigenvalue 1 μ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq11_HTML.gif. In fact, if not, there exist u0∂P ρ , λ0 > 0 such that u0 = Au0 + λ0φ1. This implies
        u 0 λ 0 φ 1 a n d u 0 λ 0 φ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equaf_HTML.gif

        Let: λ * = sup λ | u 0 λ φ 1 , u 0 " λ φ 1 " http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq14_HTML.gif.

        So 0 < λ0 < λ* < ∞ and u 0 λ * φ 1 , u 0 λ * φ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq15_HTML.gif. Then, u0- λ*φ1P.

        For L(P) ⊂ P, we get
        μ L u 0 λ * μ L φ 1 = λ * φ 1 , μ ( L u 0 ) λ * μ ( L φ 1 ) = λ * φ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equag_HTML.gif
        By (3.4), we get
        A u 0 = 0 1 H ( t , s ) f ( s , u 0 ( s ) , u 0 ' ' ( s ) ) d s ( μ + ε ) L u 0 . ( A u 0 ) ( μ + ε ) ( L u 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equah_HTML.gif
        So, we know
        u 0 = A u 0 + λ 0 φ 1 ( μ + ε ) L u 0 + λ 0 φ 1 ( λ * + λ 0 ) φ 1 . ( u 0 ) = ( A u 0 ) + λ 0 φ 1 ( μ + ε ) ( L u 0 ) + λ 0 φ 1 ( λ * + λ 0 ) φ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equai_HTML.gif

        which contradicts the definition of λ*.

        Lemma 3.3. Suppose there is ρ1 > 0 such that
        f ( t , u , v ) d 1 ρ 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ8_HTML.gif
        (3.5)

        for u ∈ [0, ρ1] and v ∈ [-ρ1, 0] a.e. t ∈ [0, 1], where d 1 = 1 0 1 H ( t , s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq16_HTML.gif, if Auu for u P ρ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, then i ( A , P ρ 1 , P ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq18_HTML.gif.

        Proof. Suppose u P ρ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, by Lemma 2.2, we get
        A u = max 0 t 1 A u ( t ) - min 0 t 1 ( A u ( t ) ) = max 0 t 1 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s + max 0 t 1 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s d 1 ρ 1 max 0 t 1 0 1 H ( t , s ) d s + max 0 t 1 0 1 H ( t , s ) d s ρ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equaj_HTML.gif

        That is Au ≠ λu for each u P ρ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, λ > 1. If Auu for u P ρ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, by Lemma 2.5, then i ( A , P ρ 1 , P ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq18_HTML.gif.

        Lemma 3.4. Suppose there is ρ2 > 0 such that
        f ( t , u , v ) d 2 ρ 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ9_HTML.gif
        (3.6)

        for u ∈ [0, ρ 2] and v ∈ [-ρ 2, 0] a.e. t ∈ [0, 1], where d 2 = 1 min t [ δ , 1 - δ ] 0 1 H ( t , s ) d s - max t [ δ , 1 - δ ] 0 1 H ( t , s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq19_HTML.gif. If Auu for u P ρ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, then i ( A , P ρ 2 , P ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq21_HTML.gif.

        Proof. For u P ρ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, t ∈ [δ, 1 - δ], by Lemma 2.2, we get
        A u + A u = 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s + 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s d 2 ρ 2 0 1 H ( t , s ) d s + 0 1 H ( t , s ) d s ρ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equak_HTML.gif

        This implies that uAu + λφ for each u P ρ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, λ > 0, where φP\{0} is the eigenfunction of L corresponding to r(L). Suppose uAu for u P ρ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, by Lemma 2.5, then i ( A , P ρ 2 , P ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq21_HTML.gif.

        Theorem 3.1. The boundary value problem (1.1) has at least one positive solution if one of the following conditions holds.

        (C1) There exists d > 0 such that (3.3) and (3.1) hold.

        (C2) There exists d > 0, ρ1> 0 such that (3.3) and (3.5) hold.

        (C3) There exists ρ2 > 0 such that (3.6) and (3.1) hold.

        (C4) There exists ρ1, ρ2 > 0 with 0 < ρ2 < ρ1d1/d2 such that (3.5) and (3.6) hold.

        Proof. When condition (C1) holds, by Lemma 3.1 and 0 ≤ f < μ, we get that there exists r > 0 such that i(A, P r , P) = 1. It follows from Lemma 3.2 and μ < f 0 b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq22_HTML.gif, then there exists 0 < ρ < min{r, d} such that either there exists u∂P ρ that i(A, P ρ , P) = 0 or u = Au. So BVP (1.1) has at least one positive solution uP with ρ ≤ ||u|| < r.

        When one of other conditions holds, the results can be proved similarly.

        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. Webb JRL, Lan KQ: Eigenvalue criteria for existence of multiple positive solutions of nonlinear boundary value problems of local and nonlocal type. Topolog Methods Nonlinear Anal J Juliusz Schauder Center 2006, 27: 91-115.MATHMathSciNet
        2. Jiang WH: The existence of positive solutions for second-order multi-point BVPs with the first derivative. J Comput Appl Math 2009, 225: 387-392. 10.1016/j.cam.2008.07.043MATHMathSciNetView Article
        3. Jiang WH: Eigenvalue criteria for existence of multiple positive solutions of high-order nonlinear BVPs. Nonlinear Anal 2008, 69: 295-303. 10.1016/j.na.2007.05.020MATHMathSciNetView Article
        4. Webb JRL: Nonlocal conjugate type boundary value problems of higher order. Nonlinear Anal 2009, 71: 1933-1940. 10.1016/j.na.2009.01.033MATHMathSciNetView Article
        5. Hao XN, Liu LS, Wu YH, Sun Q: Positive solutions for nonlinear nth-order singular eigenvalue problem with nonlocal conditions. Nonlinear Anal 2010, 73: 1653-1662. 10.1016/j.na.2010.04.074MATHMathSciNetView Article
        6. 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.2003.09.061MATHMathSciNetView Article
        7. Zhao JF, Wang PG, Ge WG: Existence and nonexistence of positive solutions for a class of third order BVP with integral boundary conditions in Banach spaces. Commun Nonlinear Sci Numer Simulat 2011, 16: 402-413. 10.1016/j.cnsns.2009.10.011MATHMathSciNetView Article
        8. Nussbaum RD: Eigenvectors of nonlinear positive operators and the linear Krein-Rutman theorem, Fixed Point Theory. Lecture Notes Math 1981, 886: 309-330. 10.1007/BFb0092191MathSciNetView Article
        9. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, New York; 1988.MATH

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        © Guo 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.