Application of the shooting method to second-order multi-point integral boundary-value problems

  • Huilan Wang1Email author,

    Affiliated with

    • Zigen Ouyang1 and

      Affiliated with

      • Liguang Wang1

        Affiliated with

        Boundary Value Problems20132013:205

        DOI: 10.1186/1687-2770-2013-205

        Received: 3 March 2013

        Accepted: 8 August 2013

        Published: 9 September 2013

        Abstract

        In this paper, we focus on the following second-order multi-point integral boundary-value problem:

        u ( t ) + a ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = i = 1 n α i 0 η i u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equa_HTML.gif

        where 0 < η 1 < η 2 < < η n < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq1_HTML.gif, α i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq2_HTML.gif for i = 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq3_HTML.gif and α n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq4_HTML.gif are given constants. The proof is based on the shooting method. By constructing a quadratic function and a sine function as the shooting objects and combining the integral mean value theorem with the comparison principle, we consider the existence of positive solutions to the BVP respectively under the case 0 < i = 1 n α i η i 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq5_HTML.gif and the case i = 1 n α i η i > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq6_HTML.gif. The method is concise and some new criteria are established.

        MSC:34B10, 34B15, 34B18.

        Keywords

        shooting method integral boundary-value problem positive solution

        1 Introduction

        For the study of nonlinear second-order multi-point boundary-value problem, many results have been obtained by using all kinds of fixed point theorems related to a completely continuous map defined in a Banach space. We refer the reader to [19] and the references therein. Some of the results are so classical that little work can exceed; however, most of these papers are concerned with problems with boundary conditions of restrictions either on the slope of solutions and the solutions themselves, or on the number of boundary points [2, 58, 10].

        In [8], Ma investigated the existence of positive solutions of the nonlinear second-order m-point boundary value problem
        u ( t ) + a ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ1_HTML.gif
        (1.1)
        u ( 0 ) = 0 , i = 1 m 2 α i u i ( η i ) = u ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ2_HTML.gif
        (1.2)

        where 0 < η 1 < η 2 < < η m 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq7_HTML.gif, α i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq2_HTML.gif for i = 1 , , m 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq8_HTML.gif, α m 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq9_HTML.gif, f C ( [ 0 , ) ; [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq10_HTML.gif, a C ( [ 0 , 1 ] ; [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq11_HTML.gif, and there exists a t 0 [ η m 2 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq12_HTML.gif such that a ( t 0 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq13_HTML.gif.

        Set
        f 0 = lim u 0 + f ( u ) u , f = lim u f ( u ) u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equb_HTML.gif

        The author obtained the existence of a positive solution to (1.1)-(1.2) under the case f 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq14_HTML.gif and f = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq15_HTML.gif (super-linear case) or the case f 0 = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq16_HTML.gif and f = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq17_HTML.gif (sub-linear case) when 0 < i = 1 m 2 α i η i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq18_HTML.gif.

        Recently, Tariboon [9] considered three-point boundary-value problem (1.1) with the integral boundary condition
        u ( 0 ) = 0 , u ( 1 ) = α 0 η u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ3_HTML.gif
        (1.3)

        where 0 < η < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq19_HTML.gif, α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq20_HTML.gif.

        Such a boundary condition might be more realistic in the mathematical models of thermal conductivity, groundwater flow, thermoelectric flexibility and plasma physics, because it describes the fluid properties in a certain continuous medium. Under the assumption that 0 < α η 2 < 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq21_HTML.gif, Tariboon and the author proved that problem (1.1)-(1.3) has at least one positive solution in the super-linear case or in the sub-linear one.

        However, the method used in the previous two papers is Krasnoselskii’s fixed point theorem in a cone, which relates to constructing a completely continuous cone map in a Banach space, and the proof is somewhat procedural.

        Constructively, Agarwal [11] explored the solution of multi-point boundary value problems by converting BVPs to equivalent IVPs, which is called shooting method. After that Man Kam Kwong [4, 12] used the shooting method to consider second-order multi-point boundary value problems. In [12], Kwong studied the existence of a positive solution to the following three-point boundary value problem:
        u ( t ) + f ( u ( t ) ) = 0 , 0 < t < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ4_HTML.gif
        (1.4)
        u ( 0 ) = 0 , μ u ( 1 2 ) = u ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ5_HTML.gif
        (1.5)
        The principle of the shooting method used in [12] is converting BVP (1.4)-(1.5) into finding suitable initial slopes m > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq22_HTML.gif such that the solution of equation (1.4) with the initial value condition
        u ( 0 ) = 0 , u ( 0 ) = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ6_HTML.gif
        (1.6)
        vanishes for the first time after t > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq23_HTML.gif. Denote by u ( t , m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq24_HTML.gif the solution of (1.4)-(1.6) provided it exists. Then solving the boundary value problem is equivalent to finding m such that
        μ u ( 1 2 , m ) = u ( 1 , m ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equc_HTML.gif
        If we can find two solutions u ( t , m 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq25_HTML.gif and u ( t , m 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq26_HTML.gif of (1.4) such that
        u ( 1 , m 1 ) ( or  ) μ u ( 1 2 , m 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equd_HTML.gif
        and
        u ( 1 , m 2 ) ( or  ) μ u ( 1 2 , m 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Eque_HTML.gif

        where u ( t , m 1 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq27_HTML.gif, u ( t , m 2 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq28_HTML.gif for t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq29_HTML.gif, then there must exist a number m between m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq30_HTML.gif and m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq31_HTML.gif such that u ( t , m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq24_HTML.gif is the solution of (1.4)-(1.5). By constructing two sine functions as the shooting objects and combining with the comparison principle, the author obtained some better results than those via fixed point techniques for the existence of positive solutions to (1.4)-(1.5).

        In this paper, we try to employ the shooting method to establish the existence results of positive solutions for (1.1) with the more generalized multi-point integral boundary condition
        u ( 0 ) = 0 , i = 1 n α i 0 η i u ( s ) d s = u ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ7_HTML.gif
        (1.7)

        where 0 < η 1 < η 2 < < η n < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq1_HTML.gif, α i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq2_HTML.gif for i = 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq3_HTML.gif and α n > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq4_HTML.gif are given constants. Following the principle of the shooting method, there are two obstacles we encounter. The first one is that the boundary condition involves integral from 0 to η i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq32_HTML.gif ( i = 1 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq33_HTML.gif), so we transform the integral problem into a single-point problem by using the integral mean value theorem. The other difficulty is that we cannot obtain the existence results by constructing two sine functions as in [12] because of the particularity of η = 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq34_HTML.gif in [12]. Therefore, we construct a quadratic function and a sine function as the objective ones.

        The purpose of this article lies in two aspects. One is to explore the application of the shooting method in a more complicated multi-point integral boundary value problem, which demonstrates another way in studying BVPs. The other one is to establish new criteria for the existence of positive solutions to (1.1)-(1.7) under the case 0 < i = 1 n α i η i 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq5_HTML.gif and the case i = 1 n α i η i > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq6_HTML.gif.

        For the sake of convenience, we denote
        max 0 t 1 { a ( t ) } = a L , min 0 t 1 { a ( t ) } = a l , f ¯ x = lim u x sup f ( u ) u , f ̲ x = lim u x inf f ( u ) u , x { 0 , + } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equf_HTML.gif
        Let u ( t , m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq24_HTML.gif be the solution of (1.1)-(1.6) and define
        k ( m ) = i = 1 n α i 0 η i u ( s , m ) d s u ( 1 , m ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ8_HTML.gif
        (1.8)

        In this paper, we always assume:

        (H1) f C ( [ 0 , ) ; [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq10_HTML.gif, a C ( [ 0 , 1 ] ; [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq11_HTML.gif, a l > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq35_HTML.gif.

        Under the assumption, it is not difficult to prove that the initial problem (1.1)-(1.6) has at least one solution defined on [ 0 , 1 ] × [ 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq36_HTML.gif. In fact, after translating second-order differential equation (1.1) into one-order equations, one can draw the conclusion [13].

        Further, we introduce the comparison results derived from [4, 12], which evolved from the Sturm comparison theorem.

        Theorem 1.1 Let u ( t , m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq24_HTML.gif, z ( t , m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq37_HTML.gif, Z ( t , m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq38_HTML.gif be the solution of the initial value problems, respectively,
        u ( t ) + F ( u ( t ) ) = 0 , u ( 0 ) = 0 , u ( 0 ) = m , Z ( t ) + G ( Z ( t ) ) = 0 , Z ( 0 ) = 0 , Z ( 0 ) = m , z ( t ) + g ( z ( t ) ) = 0 , z ( 0 ) = 0 , z ( 0 ) = m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equg_HTML.gif
        and suppose that F, G, g are nonnegative continuous functions on a certain interval I for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq39_HTML.gif and such that
        g ( ω ) F ( ω ) G ( ω ) , ω I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equh_HTML.gif
        If Z ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq40_HTML.gif does not vanish in [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq41_HTML.gif, then for 0 < η < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq19_HTML.gif, it yields
        z ( η ) z ( 1 ) u ( η ) u ( 1 ) Z ( η ) Z ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equi_HTML.gif

        The paper is arranged as follows. In the next section, we put forward the basic principle of the shooting method used in this paper, and show that BVP (1.1)-(1.7) has no positive solution when i = 1 n α i η i 2 > 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq42_HTML.gif. In Section 3, the general criteria are established for the existence of positive solutions to (1.1)-(1.7) under the case 0 < i = 1 n α i η i 2 < 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq43_HTML.gif. Moreover, we present the special results in the form of corollaries corresponding to the super-linear case or the sub-linear case. Finally, we come to the conclusion and an example is presented to illustrate our results.

        2 Preliminaries

        Lemma 2.1 If there exist two initial slopes m 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq44_HTML.gif and m 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq45_HTML.gif such that
        1. (i)

          the solution u ( t , m 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq25_HTML.gif of (1.1)-(1.6) remains positive in ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq46_HTML.gif and k ( m 1 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq47_HTML.gif;

           
        2. (ii)

          the solution u ( t , m 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq26_HTML.gif of (1.1)-(1.6) satisfies u ( t , m 2 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq28_HTML.gif for t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq29_HTML.gif and k ( m 2 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq48_HTML.gif; then multi-point boundary value problem (1.1)-(1.7) has a positive solution with the slope u ( 0 ) = m 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq49_HTML.gif between m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq30_HTML.gif and m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq31_HTML.gif.

           
        Proof Since the solutions of (1.1)-(1.6) depend on the initial value continuously, then from (1.8), it implies that k ( m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq50_HTML.gif is continuous on m. In view of the intermediate value theorem of continuous functions, there exists a number m 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq51_HTML.gif between m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq30_HTML.gif and m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq31_HTML.gif such that k ( m 0 ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq52_HTML.gif, that is,
        u ( t , m 0 ) = i = 1 n α i 0 η i u ( s , m 0 ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equj_HTML.gif

        Therefore, u ( t , m 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq53_HTML.gif is the solution of (1.1)-(1.7). □

        Lemma 2.2 Let i = 1 n α i η i 2 > 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq42_HTML.gif, then (1.1)-(1.7) has no positive solution.

        Proof Assume that (1.1)-(1.7) has a positive solution u.

        If u ( 1 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq54_HTML.gif, then i = 1 n α i 0 η i u ( s ) d s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq55_HTML.gif, the convexity of u implies that u ( η i ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq56_HTML.gif ( i = 1 , 2 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq57_HTML.gif) and
        u ( 1 ) 1 = i = 1 n α i 0 η i u ( s ) d s 1 2 i = 1 n α i η i u ( η i ) = 1 2 i = 1 n α i η i 2 u ( η i ) η i 1 2 i = 1 n α i η i 2 u ( η n ) η n > u ( η n ) η n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equk_HTML.gif

        which contradicts with the convexity of u.

        If u ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq58_HTML.gif, then i = 1 n α i 0 η i u ( s ) d s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq59_HTML.gif, that is, u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq60_HTML.gif for t [ 0 , η n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq61_HTML.gif. If there exists τ ( η n , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq62_HTML.gif such that u ( τ ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq63_HTML.gif, then u ( 0 ) = u ( η n ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq64_HTML.gif and u ( τ ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq63_HTML.gif, which contradicts with the convexity of u. Therefore u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq60_HTML.gif for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq39_HTML.gif.

        In the rest of this paper, we always assume:

        (H2) 0 < i = 1 n α i η i 2 < 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq65_HTML.gif.

         □

        3 Main results

        Theorem 3.1 Assume that (H1)-(H2) holds. Suppose 0 < i = 1 n α i η i 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq66_HTML.gif and there exists a constant A [ 0 , π 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq67_HTML.gif such that
        1. (i)

          f ¯ 0 < A 2 a L A 2 a l < f ̲ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq68_HTML.gif; or

           
        2. (ii)

          f ¯ < A 2 a L A 2 a l < f ̲ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq69_HTML.gif.

           

        Then problem (1.1)-(1.7) has a positive solution.

        Proof (i) Since f ¯ 0 < A 2 a L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq70_HTML.gif, we can choose a positive number m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq71_HTML.gif such that
        f ( u ) u A 2 a L , 0 < u m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equl_HTML.gif
        We claim that there exists a positive number m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq30_HTML.gif small enough such that 0 < u ( t , m 1 ) m 1 t m 1 < m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq72_HTML.gif for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq39_HTML.gif. The claim is based on the convexity of the function u ( t , m 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq25_HTML.gif and the Sturm comparison theorem (see [12]). Hence,
        a ( t ) f ( u ( t , m 1 ) ) a L A 2 a L u ( t , m 1 ) = A 2 u ( t , m 1 ) , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equm_HTML.gif
        Let
        Z ( t ) = sin A t , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ9_HTML.gif
        (3.1)
        then
        Z ( t ) + A 2 Z ( t ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equn_HTML.gif
        From (1.8), (3.1) and combining the integral mean value theorem with Theorem 1.1, we have
        k ( m 1 ) = i = 1 n α i 0 η i u ( s , m 1 ) d s u ( 1 , m 1 ) = i = 1 n α i η i u ( ξ i , m 1 ) u ( 1 , m 1 ) u ( ξ ¯ , m 1 ) i = 1 n α i η i u ( 1 , m 1 ) sin A ξ ¯ i = 1 n α i η i sin A sin A η n i = 1 n α i η i sin A < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ10_HTML.gif
        (3.2)

        where ξ i [ η i 1 , η i ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq73_HTML.gif and ξ ¯ { ξ 1 , , ξ n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq74_HTML.gif such that u ( ξ ¯ , m 1 ) = max 1 i n u ( ξ i , m 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq75_HTML.gif.

        The second inequality in (i) means that there exists a number M large enough such that
        f ( u ) u A 2 a l , u M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equo_HTML.gif
        For this M, there exist two numbers δ and M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq76_HTML.gif such that
        0 < δ < 1 η n , M = 2 ( 1 δ ) η n i = 1 n α i η i 2 2 i = 1 n α i η i 2 × M A 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ11_HTML.gif
        (3.3)
        and there exists another number m 2 M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq77_HTML.gif such that u ( t , m 2 ) M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq78_HTML.gif for t [ δ , 1 δ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq79_HTML.gif. Set
        z ( t ) = { M t M A 2 2 t 2 , t [ 0 , 1 δ ] , M ( 1 δ ) M A 2 2 ( 1 δ ) 2 , t [ 1 δ , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ12_HTML.gif
        (3.4)
        In view of (H2) and (3.3), it is not difficult to verify that
        M > M A 2 2 ( 1 δ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equp_HTML.gif
        which implies from (3.4) that z ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq80_HTML.gif for t ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq81_HTML.gif. Thus, by the convexity of u ( t , m 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq26_HTML.gif and Theorem 1.1, we have
        k ( m 2 ) = i = 1 n α i 0 η i u ( s , m 2 ) d s u ( 1 , m 2 ) i = 1 n α i η i u ( η i , m 2 ) 2 u ( 1 , m 2 ) = i = 1 n α i η i 2 u ( η i , m 2 ) η i 2 u ( 1 , m 2 ) i = 1 n α i η i 2 u ( η n , m 2 ) η n 2 u ( 1 , m 2 ) i = 1 n α i η i 2 z ( η n ) 2 η n z ( 1 ) = i = 1 n α i η i 2 [ M M A 2 2 η n ] 2 ( 1 δ ) [ M M A 2 2 ( 1 δ ) ] i = 1 n α i η i 2 [ M M A 2 2 η n ] 2 [ M M A 2 2 ( 1 δ ) ] = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ13_HTML.gif
        (3.5)

        By Lemma 2.1 and (3.2)-(3.5), there exists a number m 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq51_HTML.gif between m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq30_HTML.gif and m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq31_HTML.gif such that u ( t , m 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq53_HTML.gif is the positive solution of (1.1)-(1.7). The proof for (i) is complete.

        Now, we prove for (ii).

        In view of f ¯ < A 2 a L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq82_HTML.gif, we can choose a number N large enough such that
        f ( u ) u A 2 a L , u N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equq_HTML.gif
        For this N, there exist a number ϵ small enough and a number m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq30_HTML.gif large enough such that 0 < ϵ < η 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq83_HTML.gif and u ( t , m 1 ) N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq84_HTML.gif for t [ ϵ , 1 ϵ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq85_HTML.gif. Therefore
        a ( t ) f ( u ( t , m 1 ) ) a L A 2 a L u ( t , m 1 ) = A 2 u ( t , m 1 ) , t [ ϵ , 1 ϵ ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equr_HTML.gif

        Obviously, ϵ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq86_HTML.gif as m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq87_HTML.gif. Thus u ( t , m 1 ) N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq84_HTML.gif approximately for t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq39_HTML.gif as m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq87_HTML.gif.

        Let Z ( t ) = sin A t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq88_HTML.gif, t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq39_HTML.gif. Similar to (3.2), we obtain
        k ( m 1 ) = i = 1 n α i 0 η i u ( s , m 1 ) d s u ( 1 , m 1 ) = i = 1 n α i η i u ( ξ i , m 1 ) u ( 1 , m 1 ) u ( ξ ¯ , m 1 ) i = 1 n α i η i u ( 1 , m 1 ) sin A ξ ¯ i = 1 n α i η i sin A sin A η n i = 1 n α i η i sin A < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equs_HTML.gif

        where ξ i [ η i 1 , η i ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq73_HTML.gif and ξ ¯ { ξ 1 , , ξ n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq74_HTML.gif such that u ( ξ ¯ , m 1 ) = max 1 i n u ( ξ i , m 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq89_HTML.gif.

        Since f ̲ 0 > A 2 a l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq90_HTML.gif, then there exist two positive numbers m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq31_HTML.gif and σ small enough such that
        f ( u ) u A 2 a l , σ u m 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equt_HTML.gif
        By the convexity of u ( t , m 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq26_HTML.gif, for these σ and m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq31_HTML.gif, there exists a positive number τ small enough such that
        0 < τ < η 1 , σ u ( t , m 2 ) m 2 , t [ τ , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equu_HTML.gif
        which yields
        a ( t ) f ( u ( t , m 2 ) ) a l A 2 a l u ( t , m 2 ) A 2 σ , t [ τ , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equv_HTML.gif
        Let
        m = 2 η n i = 1 n α i η i 2 2 i = 1 n α i η i 2 A 2 σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ14_HTML.gif
        (3.6)
        and
        z ( t ) = m t A 2 σ 2 t 2 , t [ τ , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ15_HTML.gif
        (3.7)
        From (3.6) and (3.7), we have m > A 2 σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq91_HTML.gif and z ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq80_HTML.gif for t ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq81_HTML.gif. Thus
        k ( m 2 ) = i = 1 n α i 0 η i u ( s , m 2 ) d s u ( 1 , m 2 ) i = 1 n α i η i u ( η i , m 2 ) 2 u ( 1 , m 2 ) i = 1 n α i η i 2 u ( η i , m 2 ) η i 2 u ( 1 , m 2 ) i = 1 n α i η i 2 u ( η n , m 2 ) η n 2 u ( 1 , m 2 ) i = 1 n α i η i 2 z ( η n ) 2 η n z ( 1 ) i = 1 n α i η i 2 [ m A 2 σ 2 η n ] 2 [ m A 2 σ 2 ] = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equw_HTML.gif

        By Lemma 2.1, the proof for (ii) is complete. □

        Theorem 3.2 Assume that (H1)-(H2) holds. Suppose i = 1 n α i η i > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq92_HTML.gif and there exists a constant A [ 0 , π 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq93_HTML.gif such that
        sin A sin η n A = i = 1 n α i η i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equx_HTML.gif
        Then problem (1.1)-(1.7) has a positive solution under the case
        1. (i)

          f ¯ 0 < A 2 a L A 2 a l < f ̲ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq68_HTML.gif; or

           
        2. (ii)

          f ¯ < A 2 a L A 2 a l < f ̲ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq94_HTML.gif.

           
        Proof Note the computation of k ( m 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq95_HTML.gif in Theorem 3.1. In (3.2), if we substitute i = 1 n α i η i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq96_HTML.gif with
        sin A sin η n A , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equy_HTML.gif

        then k ( m 1 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq47_HTML.gif, and all the steps in the following are the same as in Theorem 3.1. □

        Now, let us consider the special super-linear case or the sub-linear case. It is not difficult to verify the following corollaries.

        Corollary 3.1 Assume that 0 < i = 1 n α i η i 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq97_HTML.gif and
        1. (i)

          f 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq14_HTML.gif, f = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq98_HTML.gif; or

           
        2. (ii)

          f 0 = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq16_HTML.gif, f = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq17_HTML.gif.

           

        Then problem (1.1)-(1.7) has a positive solution.

        Corollary 3.2 If i = 1 n α i η i > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq92_HTML.gif and there exists a constant A [ 0 , π 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq99_HTML.gif such that
        sin A sin η n A = i = 1 n α i η i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equz_HTML.gif
        Then, problem (1.1)-(1.7) has a positive solution under the case
        1. (i)

          f 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq14_HTML.gif, f = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq98_HTML.gif ; or

           
        2. (ii)

          f 0 = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq16_HTML.gif, f = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq17_HTML.gif.

           

        4 Conclusion and examples

        The tool which we used for the analysis in this article is the shooting method derived from [4, 12]; however, we considered a more general problem which involves integral boundary-value and multiplicity of boundary-point. The meaningful work that we have done lies in the following three aspects. The first one is that we transform the integral problem into a single-point value one by using the integral mean value theorem. The other one is that we construct a quadratic function and a sine function as the comparison functions because it does not take effect to construct two sine functions as in [12]. Finally, we established the new criteria for the existence of positive solutions to (1.1)-(1.7) under the case i = 1 n α i η i 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq100_HTML.gif and the case i = 1 n α i η i > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq101_HTML.gif. Obviously, (1.7) vanishes to (1.3) when n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq102_HTML.gif and the sup-linear case or the sub-linear case is sufficient for the conditions in Theorem 3.1 and Theorem 3.2, so some of our results are more general or better than those via fixed point techniques. However, in Theorem 3.2, whether the transcendental equation has a solution is somewhat difficult to verify. It can be seen that each method has its pros and cons.

        Example 4.1 Consider the BVP
        y ( t ) + ( 2 t + 1 ) ( 2 y 3 + 1 ) = 0 , 0 < t < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ16_HTML.gif
        (4.1)
        y ( 0 ) = 0 , y ( 1 ) = 1 4 0 1 4 y ( s ) d s + 83 60 0 3 4 y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equ17_HTML.gif
        (4.2)
        where
        a ( t ) = 2 t + 1 , f ( y ) = 2 y 3 + 1 , α 1 = 1 4 , η 1 = 1 4 , α 2 = 83 60 , η 2 = 3 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equaa_HTML.gif
        It is not difficult to see that
        a L = 3 , a l = 1 , f ¯ = 2 3 , f ̲ 0 = , i = 1 2 α i η i = 1.1 > 1 , i = 1 2 α i η i 2 < 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equab_HTML.gif
        In view of sin A sin 3 4 A = 1.1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq103_HTML.gif, Matlab software gives A = 1.5173 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq104_HTML.gif and A 2 = 2.3022 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq105_HTML.gif. Hence
        f ¯ = 2 3 < A 2 a L < A 2 a l < f ̲ 0 = . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Equac_HTML.gif
        Therefore, the condition (ii) of Theorem 3.2 is satisfied. A numerical simulation (Figure 1) for Example 4.1 demonstrates that BVP (4.1)-(4.2) has a positive solution y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq106_HTML.gif such that y ( 1 ) = 0.0027 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_IEq107_HTML.gif.
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-205/MediaObjects/13661_2013_Article_462_Fig1_HTML.jpg
        Figure 1

        Numerical simulation for Example 4.1.

        Declarations

        Acknowledgements

        The authors would like to thank the editors and the anonymous referees for their valuable suggestions on the improvement of this paper. First author was partially supported by the Scientific Research Fund of Hunan Provincial Educational Department (1200361), Project of Science and Technology Bureau of Hengyang, Hunan Province (2012KJ2). Second author was partially supported by the Doctor Foundation of University of South China ( No. 5-XQD-2006-9), the Foundation of Science and Technology Department of Hunan Province (No. 2009RS3019), the Natural Science Foundation of Hunan Province (No. 13JJ3074) and the Subject Lead Foundation of University of South China (No. 2007XQD13).

        Authors’ Affiliations

        (1)
        School of Mathematics and Physics, University of South China

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