Existence and uniqueness of solutions for periodic-integrable boundary value problem of second order differential equation

  • Hongtu Hua1, 2Email author,

    Affiliated with

    • Fuzhong Cong1 and

      Affiliated with

      • Yi Cheng1, 2

        Affiliated with

        Boundary Value Problems20122012:89

        DOI: 10.1186/1687-2770-2012-89

        Received: 26 March 2012

        Accepted: 16 July 2012

        Published: 7 August 2012

        Abstract

        In this paper we deal with one kind of second order periodic-integrable boundary value problem. Using the lemma on bilinear form and Schauder’s fixed point theorem, we give the existence and uniqueness of solutions for the problem under Lazer type nonresonant condition.

        MSC:34B15, 34B16, 37J40.

        Keywords

        lemma on bilinear forms Schauder’s fixed point theorem existence and uniqueness periodic-integrable boundary value problems

        1 Introduction and main results

        In this paper, we consider the solutions to the following periodic-integrable boundary value problem (for short, PIBVP):
        ( p ( t ) x ) + f ( t , x ) = 0 , x ( 0 ) = x ( T ) , 0 T x ( s ) d s = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ1_HTML.gif
        (1.1)

        where p ( t ) C 1 ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq1_HTML.gif is a given T-periodic function in t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq2_HTML.gif, and p ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq3_HTML.gif; f C 1 ( R × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq4_HTML.gif is T-periodic in t.

        Throughout this paper, we assume

        (A1) there exist two constants m and M such that
        m f x ( t , x ) p ( t ) M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equa_HTML.gif

        for all t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq2_HTML.gif and x R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq5_HTML.gif;

        (A2) there exists N Z + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq6_HTML.gif such that
        4 π 2 T 2 N 2 < m M < 4 π 2 T 2 ( N + 1 ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equb_HTML.gif
        Recently, boundary value problems with integral conditions have been studied extensively [610]. As we all know Lazer type conditions are essential for the existence and uniqueness of periodic solutions of equations [14]. In [5] the existence of periodic solutions has been considered for the following second order equation:
        ( p ( t ) x ) + f ( x , t ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equc_HTML.gif

        Motivated by the above works, we will consider periodic-integrable boundary value problem (1.1). The main result obtained by us is the following theorem.

        Theorem 1 Assume that (A 1) and (A 2) are satisfied. Then PIBVP (1.1) has a unique solution.

        This paper is organized as follows. Section 2 deals with a linear problem. There, using the bilinear lemma developed by Lazer, one proves the uniqueness of solutions for linear equations. In Section 3, applying the result in Section 2 and Schauder’s fixed point theorem, we complete the proof of Theorem 1.

        2 Linear equation

        Consider the following linear PIBVP:
        ( p ( t ) x ) + q ( t ) x = 0 , x ( 0 ) = x ( T ) , 0 T x ( s ) d s = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ2_HTML.gif
        (2.1)

        here p ( t ) C 1 ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq1_HTML.gif is a given T-periodic function in t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq2_HTML.gif, and p ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq3_HTML.gif; q ( t ) C ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq7_HTML.gif is a T-periodic function. Assume that

        (L1) there exist two constants m and M such that
        m q ( t ) p ( t ) M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equd_HTML.gif

        for all t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq2_HTML.gif. Moreover, m and M suit (A2).

        Theorem 2 Assume that (L 1) and (A 2) are satisfied, then PIBVP (2.1) has only a trivial solution.

        In order to prove Theorem 2, let us give some following concepts.

        First, for any interval [ α , β ] [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq8_HTML.gif, define
        O α , β = { u ( t ) L 2 ( 0 , T ) : u ( t )  is absolutely continuous on  [ α , β ] , and  u ( t ) = 0  for any  t [ 0 , α ] [ β , T ] } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Eque_HTML.gif
        It is clear that O α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq9_HTML.gif is a linear space with the norm as follows:
        u = max t [ 0 , T ] | u ( t ) | + max t [ 0 , T ] | u ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equf_HTML.gif
        Define a bilinear form on O α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq9_HTML.gif as follows:
        H α , β ( u , v ) = 0 T [ p ( t ) u ( t ) v ( t ) q ( t ) u ( t ) v ( t ) ] d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equg_HTML.gif
        for any u ( t ) O α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq10_HTML.gif and v ( t ) O α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq11_HTML.gif. Let
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equh_HTML.gif

        where N suits assumption (L1), and a m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq12_HTML.gif, b m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq13_HTML.gif, c 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq14_HTML.gif, c k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq15_HTML.gif and d k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq16_HTML.gif are some constants. Then O α , β = X α , β Y α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq17_HTML.gif.

        From p ( t ) C 1 ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq1_HTML.gif and p ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq3_HTML.gif, we can obtain that there exist two constants M 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq18_HTML.gif and M 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq19_HTML.gif such that
        0 M 1 p ( t ) M 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equi_HTML.gif
        for all t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq2_HTML.gif. Then from assumptions (L1) and (A2), we have
        H α , β ( x , x ) = 0 T p ( t ) ( x 2 ( t ) q ( t ) p ( t ) x 2 ( t ) ) d t 0 T p ( t ) ( x 2 ( t ) M x 2 ( t ) ) d t 2 π 2 M 1 T i = N + 1 ( i 2 ( N + 1 ) 2 ) ( a i 2 + b i 2 ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equj_HTML.gif
        for all x X α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq20_HTML.gif, and
        H α , β ( y , y ) = 0 T p ( t ) ( y 2 ( t ) q ( t ) p ( t ) y 2 ( t ) ) d t 0 T p ( t ) ( y 2 ( t ) m y 2 ( t ) ) d t 2 π 2 M 2 T k = 1 N ( k 2 N 2 ) ( c k 2 + d k 2 ) m M 2 T c 0 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equk_HTML.gif

        for all y Y α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq21_HTML.gif. Thus, H α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq22_HTML.gif is positive definite on X α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq23_HTML.gif and negative definite on Y α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq24_HTML.gif. By the lemma in [1], we assert that if H α , β ( u , v ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq25_HTML.gif for all u O α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq26_HTML.gif, then v 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq27_HTML.gif.

        For every x on [ α , β ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq28_HTML.gif with x ( α ) = x ( β ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq29_HTML.gif, we introduce an auxiliary function
        x α , β ( t ) = { x ( t ) , t [ α , β ] , 0 , t [ 0 , α ] [ β , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equl_HTML.gif

        The following lemma is very useful in our proofs.

        Lemma 1 If p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq30_HTML.gif, q ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq31_HTML.gif are continuous and satisfy (L 1) and (A 2), then the following two points boundary value problem
        ( p ( t ) x ) + q ( t ) x = 0 , x ( α ) = x ( β ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ3_HTML.gif
        (2.2)

        has only a trivial solution.

        Proof It is clear that 0 is a solution of two points boundary value problem (2.2). If v ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq32_HTML.gif is a solution of problem (2.2), then v α , β ( t ) O α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq33_HTML.gif. For any u O α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq34_HTML.gif, we have
        α β [ ( p ( t ) v ( t ) ) u ( t ) + q ( t ) u ( t ) v ( t ) ] d t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equm_HTML.gif
        by using (2.2). Integrating the first terms by parts, we derive
        H α , β ( u , v α , β ) = 0 T [ p ( t ) u ( t ) v α , β ( t ) q ( t ) u ( t ) v α , β ( t ) ] d t = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equn_HTML.gif

        By assumption (L1), H α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq22_HTML.gif is positive definite on X α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq23_HTML.gif and negative definite on Y α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq24_HTML.gif. These show v α , β ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq35_HTML.gif for t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq36_HTML.gif, that is, v ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq37_HTML.gif for t [ α , β ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq38_HTML.gif. The proof of Lemma 1 is ended. □

        Proof of Theorem 2 It is clear that PIBVP (2.1) has at least one solution, for example, x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq39_HTML.gif. Assume that PIBVP (2.1) possesses a nontrivial solution x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq40_HTML.gif. The proof is divided into three parts.

        Case 1: x ( 0 ) = x ( T ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq41_HTML.gif. By Lemma 1 ( α = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq42_HTML.gif and β = T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq43_HTML.gif), PIBVP (2.1) has only a trivial solution. This contradicts x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq40_HTML.gif.

        Case 2: x ( 0 ) = x ( T ) = η > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq44_HTML.gif. Denote
        S = { t [ 0 , T ] : x ( t ) = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equo_HTML.gif
        Take
        a = inf t S t and b = sup t S t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equp_HTML.gif
        From 0 2 π x ( s ) d s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq45_HTML.gif, there are at least two points in the set S, which implies that 0 < a < b < 2 π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq46_HTML.gif and x ( a ) = x ( b ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq47_HTML.gif. By Lemma 1 ( α = a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq48_HTML.gif and β = b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq49_HTML.gif) the two points boundary value problem
        ( p ( t ) x ) + q ( t ) x = 0 , x ( a ) = x ( b ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ4_HTML.gif
        (2.3)
        only has a trivial solution. Hence we obtain x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq50_HTML.gif, t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq51_HTML.gif. By the definitions of a and b, one has
        x ( t ) > 0 for all  t [ 0 , a ) ( b , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equq_HTML.gif
        From 0 T x ( s ) d s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq52_HTML.gif, we get
        0 a x ( s ) d s + b T x ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equr_HTML.gif

        This contradicts 0 a x ( s ) d s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq53_HTML.gif and b T x ( s ) d s > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq54_HTML.gif.

        Case 3: x ( 0 ) = x ( T ) = η < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq55_HTML.gif. This case is similar to Case 2.

        Thus, we complete the proof of Theorem 2. □

        Theorem 3 If p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq30_HTML.gif, q ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq31_HTML.gif are continuous and satisfy (L 1) and (A 2), then the following PIBVP
        ( p ( t ) x ) + q ( t ) x = h ( t ) , x ( 0 ) = x ( T ) , 0 T x ( s ) d s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ5_HTML.gif
        (2.4)

        has a unique solution.

        Proof Let x 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq56_HTML.gif and x 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq57_HTML.gif be two linear independent solutions of the following linear equation:
        ( p ( t ) x ) + q ( t ) x = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equs_HTML.gif
        Assume that x ( t ) = c 1 x 1 ( t ) + c 2 x 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq58_HTML.gif is a solution of PIBVP (2.1), where c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq59_HTML.gif and c 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq60_HTML.gif are constants. Then by the boundary value conditions of (2.1),
        { ( x 1 ( 0 ) x 1 ( T ) ) c 1 + ( x 2 ( 0 ) x 2 ( T ) ) c 2 = 0 , 0 T x 1 ( s ) d s c 1 + 0 T x 2 ( s ) d s c 2 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equt_HTML.gif
        By Theorem 3, PIBVP (2.1) has only a trivial solution, which shows
        | x 1 ( 0 ) x 1 ( T ) x 2 ( 0 ) x 2 ( T ) 0 T x 1 ( s ) d s 0 T x 2 ( s ) d s | 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ6_HTML.gif
        (2.5)
        Let x ( t ) = c 3 x 1 ( t ) + c 4 x 2 ( t ) + x 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq61_HTML.gif be a solution of PIBVP (2.4), where x 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq62_HTML.gif is a solution of the equation
        ( p ( t ) x ) + q ( t ) x = h ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equu_HTML.gif
        From the boundary value conditions, we have
        { ( x 1 ( 0 ) x 1 ( T ) ) c 3 + ( x 2 ( 0 ) x 2 ( T ) ) c 4 = x 0 ( 0 ) x 0 ( T ) , 0 T x 1 ( s ) d s c 3 + 0 T x 2 ( s ) d s c 4 = 0 T x 0 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equv_HTML.gif

        From (2.5) constants c 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq63_HTML.gif, c 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq64_HTML.gif are unique. Thus, PIBVP (2.4) has only one solution. □

        3 Nonlinear equations

        Let us prove Theorem 1. Rewrite (1.1) as follows:
        ( p ( t ) x ) + h ( t , x ) x = f ( t , 0 ) , x ( 0 ) = x ( T ) , 0 T x ( s ) d s = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ7_HTML.gif
        (3.1)
        where
        h ( t , x ) = 0 1 f x ( t , θ x ) d θ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equw_HTML.gif
        Define
        O = { u ( t ) L 2 ( 0 , T ) : u ( t )  is absolutely continuous on[0,T] , u ( 0 ) = u ( T )  and  0 T u ( s ) d s = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equx_HTML.gif
        Fix y O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq65_HTML.gif, introduce an auxiliary PIBVP
        ( p ( t ) x ) + h ( t , y ) x = f ( t , 0 ) , x ( 0 ) = x ( T ) , 0 T x ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ8_HTML.gif
        (3.2)

        To prove the main result, we need the following Lemma 2.

        Lemma 2 If f satisfies (A 1) and (A 2), then for any given y O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq65_HTML.gif, PIBVP (3.2) has only one solution, denoted as x y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq66_HTML.gif and x y M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq67_HTML.gif.

        Proof From condition (A2), it follows that
        m h ( t , y ) p ( t ) M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equy_HTML.gif

        By Theorem 3, PIBVP (3.2) has only one solution x y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq66_HTML.gif. If x y M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq68_HTML.gif does not hold, there would exist a sequence { y m ( t ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq69_HTML.gif such that x y m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq70_HTML.gif, m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq71_HTML.gif. Choose a subsequence of { h ( t , y m ) } m = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq72_HTML.gif, without loss of generality, express as itself, such that the sequences are weakly convergent in L 2 ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq73_HTML.gif. Denote the limit as h 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq74_HTML.gif. It is obvious that h 0 ( t ) L 2 ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq75_HTML.gif.

        Because the set
        S = { q ( t ) L 2 [ 0 , T ] : m q ( t ) p ( t ) M } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equz_HTML.gif
        is bounded convex in L 2 ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq73_HTML.gif, by the Mazur theorem, we have h 0 ( t ) S http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq76_HTML.gif. Hence,
        m h 0 ( t ) p ( t ) M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equaa_HTML.gif
        By the Arzela-Ascoli theorem, passing to a subsequence, we may assume that
        x m = x y m x y m x 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equab_HTML.gif

        and x m z ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq77_HTML.gif in C ( [ 0 , T ] , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq78_HTML.gif. Thus, x 0 ( 0 ) = x 0 ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq79_HTML.gif and 0 T x 0 ( s ) d s = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq80_HTML.gif.

        By
        x m ( t ) = x m ( 0 ) + 0 t x m ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equac_HTML.gif
        one has
        x 0 ( t ) = x 0 ( 0 ) + 0 t z ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equad_HTML.gif

        which implies z ( t ) = x 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq81_HTML.gif, for any t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq36_HTML.gif. Hence, x 0 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq82_HTML.gif.

        From PIBVP (3.2), we obtain
        ( p ( t ) x m ) + h ( t , y m ) x m = f ( t , 0 ) x y m , x m ( 0 ) = x m ( T ) , 0 T x m ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ9_HTML.gif
        (3.3)
        This shows that x 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq83_HTML.gif is a nontrivial solution of the following PIBVP:
        ( p ( t ) x ) + h 0 ( t ) x = 0 , x ( 0 ) = x ( T ) , 0 T x ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ10_HTML.gif
        (3.4)

        On the other hand, by Theorem 2, PIBVP (3.4) has only zero, which leads to a contradiction. The proof of Lemma 2 is completed. □

        Set
        B M = { x O : x M } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equae_HTML.gif

        Define an operator F : O O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq84_HTML.gif by F y = x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq85_HTML.gif. Applying Lemma 2, F : B M B M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq86_HTML.gif.

        Lemma 3 Operator F is completely continuous on O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq87_HTML.gif.

        Proof We first prove that F is continuous. Given any { y m } O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq88_HTML.gif such that y m y 0 O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq89_HTML.gif. Put u m = x y m x y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq90_HTML.gif. From the definition
        ( p ( t ) u m ) + h ( t , y m ) u m = [ h ( t , y 0 ) h ( t , y m ) ] x y 0 , u m ( 0 ) = u m ( T ) , 0 T u m ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ11_HTML.gif
        (3.5)
        We would prove that u m 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq91_HTML.gif in C 1 ( [ 0 , T ] , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq92_HTML.gif. If not, then there would be a c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq93_HTML.gif such that
        lim m sup u m c . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equaf_HTML.gif
        Utilizing Lemma 2 and Arzela-Ascoli theorem, passing to a subsequence, we may assume that u m u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq94_HTML.gif. Similar to the proof of Lemma 2, we have u m u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq95_HTML.gif. Then
        ( p ( t ) u 0 ) + h ( t , y 0 ) u 0 = 0 , u 0 ( 0 ) = u 0 ( T ) , 0 T u 0 ( s ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equ12_HTML.gif
        (3.6)
        Moreover,
        m h ( t , y 0 ) p ( t ) M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equag_HTML.gif

        Hence, from Theorem 2, u 0 ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq96_HTML.gif. This implies F is continuous. By Lemma 2, for any bounded subset D O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq97_HTML.gif, F ( D ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq98_HTML.gif is also bounded. Hence, applying the continuity of F and Arzela-Ascoli theorem, F ( D ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq98_HTML.gif is relatively compact. This shows F is completely continuous on O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq87_HTML.gif. The proof of Lemma 3 is completed. □

        Proof of Theorem 1 By Lemma 2, Lemma 3 and Schauder’s fixed point theorem, F has a fixed point in O http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq99_HTML.gif, that is, PIBVP (1.1) has a solution x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq100_HTML.gif.

        The following is to prove uniqueness. Let x 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq101_HTML.gif and x 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq102_HTML.gif be any two solutions of equation (1.1). Then x ( t ) = x 1 ( t ) x 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq103_HTML.gif is a solution of the equation
        ( p ( t ) x ) + x 0 1 f x ( t , x 2 + θ x ) d θ = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equah_HTML.gif
        Employing (A2), we have
        m 0 1 f x ( t , x 2 + θ x ) d θ p ( t ) M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_Equai_HTML.gif

        Hence by Theorem 3, x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-89/MediaObjects/13661_2012_Article_200_IEq104_HTML.gif. The uniqueness is proved. □

        Declarations

        Acknowledgements

        The authors are grateful to the referees for their useful comments. The research of F. Cong was partially supported by NSFC Grant (11171350) and Natural Science Foundation of Jilin Province of China (201115133).

        Authors’ Affiliations

        (1)
        Fundamental Department, Aviation University of Air Force
        (2)
        Institute of Mathematics, Jilin University

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