Solvability of right focal boundary value problems with superlinear growth conditions

  • Minghe Pei1,

    Affiliated with

    • Sung Kag Chang2Email author and

      Affiliated with

      • Young Sun Oh3

        Affiliated with

        Boundary Value Problems20122012:60

        DOI: 10.1186/1687-2770-2012-60

        Received: 5 March 2012

        Accepted: 2 May 2012

        Published: 22 June 2012

        Abstract

        In this paper, we consider n th-order two-point right focal boundary value problems

        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equa_HTML.gif

        where f : [ 0 , 1 ] × R n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq1_HTML.gif is a L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq2_HTML.gif-Carathéodory ( 1 p < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq3_HTML.gif) function and satisfies superlinear growth conditions. The existence and uniqueness of solutions for the above right focal boundary value problems are obtained by Leray-Schauder continuation theorem and analytical technique. Meanwhile, as an application of our results, examples are given.

        MSC:34B15.

        Keywords

        right focal boundary value problem Leray-Schauder continuation theorem existence uniqueness

        1 Introduction

        In this paper, we shall discuss the existence and uniqueness of solutions of right focal boundary value problems for n th-order nonlinear differential equation
        u ( n ) ( t ) = f ( t , u ( t ) , u ( t ) , , u ( n 1 ) ( t ) ) , a.e. t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ1_HTML.gif
        (1.1)
        subject to the boundary conditions ( 1 m n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq4_HTML.gif)
        { u ( i ) ( 0 ) = 0 , i = 0 , 1 , , m 1 , u ( i ) ( 1 ) = 0 , i = m , m + 1 , , n 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ2_HTML.gif
        (1.2)
        where f : [ 0 , 1 ] × R n R = ( , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq5_HTML.gif satisfies the L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq2_HTML.gif-Carathéodory ( 1 p < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq3_HTML.gif) conditions, that is,
        1. (i)

          for each ( u 0 , u 1 , , u n 1 ) R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq6_HTML.gif, the function t [ 0 , 1 ] f ( t , u 0 , u 1 , , u n 1 ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq7_HTML.gif is measurable on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif;

           
        2. (ii)

          for a.e. t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq9_HTML.gif, the function ( u 0 , u 1 , , u n 1 ) f ( t , u 0 , u 1 , , u n 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq10_HTML.gif is continuous on R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq11_HTML.gif;

           
        3. (iii)

          for each r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq12_HTML.gif, there exists an α r L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq13_HTML.gif such that | f ( t , u 0 , u 1 , , u n 1 ) | α r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq14_HTML.gif for a.e. t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq9_HTML.gif and all ( u 0 , u 1 , , u n 1 ) R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq6_HTML.gif with j = 0 n 1 u j 2 r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq15_HTML.gif.

           

        As it is well known, the right focal boundary value problems have attracted many scholars’ attention. Among a substantial number of works dealing with right focal boundary value problems, we mention [116, 1825].

        Recently, using the Leray-Schauder continuation theorem, Hopkins and Kosmatov [16] have obtained sufficient conditions for the existence of at least one sign-changing solution for third-order right focal boundary value problems such as
        { u ( t ) = f ( t , u ( t ) , u ( t ) , u ( t ) ) , a.e. t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = u ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equb_HTML.gif
        and
        { u ( t ) = f ( t , u ( t ) , u ( t ) , u ( t ) ) , a.e. t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equc_HTML.gif

        where f : [ 0 , 1 ] × R 3 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq16_HTML.gif satisfies the L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq2_HTML.gif-Carathéodory ( 1 p < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq3_HTML.gif) conditions and the linear growth conditions.

        Motivated by [16], in this paper we study the solvability for general n th-order right focal boundary value problems (1.1), (1.2). The existence and uniqueness of sign-changing solutions for the problems are obtained by Leray-Schauder continuation theorem and analytical technique. We note that the nonlinearity of f in our problem allows up to the superlinear growth conditions.

        The rest of this paper is organized as follows. In Section 2, we give some lemmas which help to simplify the proofs of our main results. In Section 3, we discuss the existence and uniqueness of sign-changing solutions for n th-order right focal boundary value problems (1.1), (1.2) by Leray-Schauder continuation theorem and analytical technique, and give two examples to demonstrate our results. Our results improve and generalize the corresponding results in [16].

        2 Preliminary

        In this section, we give some lemmas which help to simplify the presentation of our main results.

        Let A C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq17_HTML.gif denote the space of absolutely continuous functions on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif, and C n 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq18_HTML.gif denote the Banach space of ( n 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq19_HTML.gif times continuously differentiable functions defined on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif with the norm u C n 1 = max { u ( i ) , i = 0 , 1 , , n 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq20_HTML.gif, where u ( i ) = sup t [ 0 , 1 ] | u ( i ) ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq21_HTML.gif. Let L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq22_HTML.gif be the usual Lebesgue space on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif with norm p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq23_HTML.gif, 1 p < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq3_HTML.gif.

        For 1 p < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq3_HTML.gif, we introduce the Sobolev space
        W n , p ( 0 , 1 ) = { u : [ 0 , 1 ] R | u ( i ) A C [ 0 , 1 ] , i = 0 , 1 , , n 1 , u ( n ) L p [ 0 , 1 ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equd_HTML.gif
        with the norm u = u C n 1 + u ( n ) p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq24_HTML.gif . Let us consider a special subspace
        W r n , p ( 0 , 1 ) = { u W n , p ( 0 , 1 ) : u satisfies ( 1.2 ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Eque_HTML.gif

        Then it is clear that W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq25_HTML.gif is closed in W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq26_HTML.gif and hence is itself a Banach space with the norm u = u C n 1 + u ( n ) p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq24_HTML.gif.

        Lemma 2.1 ([21])

        Let G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq27_HTML.gifbe the Green’s function of the differential equation ( 1 ) n m × u ( n ) ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq28_HTML.gifsubject to the boundary conditions (1.2). Then

        G ( t , s ) = ( 1 ) n m ( n 1 ) ! { i = 0 m 1 ( n 1 i ) t i ( s ) n i 1 , 0 s t 1 , i = m n 1 ( n 1 i ) t i ( s ) n i 1 , 0 t s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equf_HTML.gif
        and
        i t i G ( t , s ) 0 , ( t , s ) [ 0 , 1 ] × [ 0 , 1 ] , i = 0 , 1 , , m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equg_HTML.gif
        Lemma 2.2 Let g L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq29_HTML.gif. Then the solution of the differential equation
        u ( n ) ( t ) = g ( t ) , a.e. t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equh_HTML.gif
        subject to the boundary conditions (1.2) satisfies
        u ( j ) A j g p , j = 0 , 1 , , n 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ3_HTML.gif
        (2.1)
        where for p > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq30_HTML.gif ( 1 p + 1 q = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq31_HTML.gif),
        A j = { ( 1 ) ( n m ) ( n j 1 ) ! [ 0 1 ( i = 0 m j 1 ( n j 1 i ) ( s ) n j 1 i ) q d s ] 1 q , j = 0 , 1 , , m 1 , 1 ( n j 1 ) ! [ q ( n j 1 ) + 1 ] 1 q , j = m , m + 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ4_HTML.gif
        (2.2)
        and for p = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq32_HTML.gif,
        A j = { ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 i ) ( 1 ) n j 1 i , j = 0 , 1 , , m 1 , 1 ( n j 1 ) ! , j = m , m + 1 , , n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ5_HTML.gif
        (2.3)
        Proof Firstly, let us show the lemma for case p > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq30_HTML.gif. Since
        u ( t ) = ( 1 ) n m 0 1 G ( t , s ) g ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equi_HTML.gif
        we have that for j = 0 , 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq33_HTML.gif,
        u ( j ) ( t ) = ( 1 ) n m 0 1 j t j G ( t , s ) g ( s ) d s = : ( 1 ) n m 0 1 G j ( t , s ) g ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equj_HTML.gif
        where, for j = 0 , 1 , , m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq34_HTML.gif,
        G j ( t , s ) = ( 1 ) n m ( n 1 ) ! { i = j m 1 ( n 1 i ) i ! ( i j ) ! t i j ( s ) n i 1 , 0 s t 1 , i = m n 1 ( n 1 i ) i ! ( i j ) ! t i j ( s ) n i 1 , 0 t s 1 = ( 1 ) n m ( n j 1 ) ! { i = 0 m j 1 ( n j 1 i ) t i ( s ) n j 1 i , 0 s t 1 , i = m j n j 1 ( n j 1 i ) t i ( s ) n j 1 i , 0 t s 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equk_HTML.gif
        and for j = m , m + 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq35_HTML.gif,
        G j ( t , s ) = ( 1 ) n m ( n 1 ) ! { 0 , 0 s t 1 , i = j n 1 ( n 1 i ) i ! ( i j ) ! t i j ( s ) n i 1 , 0 t s 1 = ( 1 ) n m ( n j 1 ) ! { 0 , 0 s t 1 , ( t s ) n j 1 , 0 t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equl_HTML.gif
        It follows by Hölder’s inequality that, for each j = 0 , 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq33_HTML.gif,
        | u ( j ) ( t ) | 0 1 | G j ( t , s ) | | g ( s ) | d s g p G j ( t , ) q g p max t [ 0 , 1 ] G j ( t , ) q , t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equm_HTML.gif
        and consequently, for each j = 0 , 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq33_HTML.gif,
        u ( j ) g p max t [ 0 , 1 ] G j ( t , ) q , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ6_HTML.gif
        (2.4)
        But for j = m , m + 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq35_HTML.gif,
        max t [ 0 , 1 ] G j ( t , ) q q = max t [ 0 , 1 ] 0 1 | G j ( t , s ) | q d s = max t [ 0 , 1 ] 0 t | G j ( t , s ) | q d s + max t [ 0 , 1 ] t 1 | G j ( t , s ) | q d s = max t [ 0 , 1 ] t 1 | ( 1 ) n m ( n j 1 ) ! [ ( t s ) n j 1 ] | q d s = 1 [ ( n j 1 ) ! ] q max t [ 0 , 1 ] t 1 ( s t ) q ( n j 1 ) d s = 1 [ ( n j 1 ) ! ] q max t [ 0 , 1 ] ( 1 t ) q ( n j 1 ) + 1 q ( n j 1 ) + 1 = 1 [ ( n j 1 ) ! ] q [ q ( n j 1 ) + 1 ] = A j q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equn_HTML.gif
        It follows by (2.4) that for j = m , m + 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq35_HTML.gif,
        u ( j ) A j g p . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equo_HTML.gif
        For j = 0 , 1 , , m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq34_HTML.gif, by Lemma 2.1, G j ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq36_HTML.gif is nondecreasing in t, and thus
        max t [ 0 , 1 ] G j ( t , ) q q = max t [ 0 , 1 ] 0 1 [ G j ( t , s ) ] q d s 0 1 [ max t [ 0 , 1 ] G j ( t , s ) ] q d s = 0 1 [ G j ( 1 , s ) ] q d s = 0 1 [ ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 i ) ( s ) n j 1 i ] q d s = A j q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equp_HTML.gif
        Hence, by (2.4) we have for j = 0 , 1 , , m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq34_HTML.gif,
        u ( j ) A j g p . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equq_HTML.gif
        In summary,
        u ( j ) A j g p , j = 0 , 1 , , n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equr_HTML.gif
        Next, we show the lemma for the case p = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq32_HTML.gif. It is easy to see that for j = m , m + 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq35_HTML.gif,
        | u ( j ) ( t ) | 0 1 | G j ( t , s ) | | g ( s ) | d s = t 1 | ( 1 ) n m ( n j 1 ) ! [ ( t s ) n j 1 ] | | g ( s ) | d s = 1 ( n j 1 ) ! t 1 ( s t ) n j 1 | g ( s ) | d s ( 1 t ) n j 1 ( n j 1 ) ! t 1 | g ( s ) | d s 1 ( n j 1 ) ! g 1 = A j g 1 , t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equs_HTML.gif
        and thus for j = m , m + 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq35_HTML.gif,
        u ( j ) A j g 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equt_HTML.gif
        Also by Lemma 2.1, we have for j = 0 , 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq37_HTML.gif,
        G j ( t , s ) 0 , ( t , s ) [ 0 , 1 ] × [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equu_HTML.gif
        so that for each j = 0 , 1 , , m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq34_HTML.gif, G j ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq36_HTML.gif is nondecreasing in t, it follows that
        | u ( j ) ( t ) | 0 1 | G j ( t , s ) | | g ( s ) | d s 0 1 max t [ 0 , 1 ] G j ( t , s ) | g ( s ) | d s = 0 1 G j ( 1 , s ) | g ( s ) | d s = 0 1 [ ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 i ) ( s ) n j 1 i ] | g ( s ) | d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ7_HTML.gif
        (2.5)
        Let
        ϕ ( t ) = ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 i ) ( t ) n j 1 i , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equv_HTML.gif
        Then
        ϕ ( n m ) ( t ) = ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 i ) ( 1 ) n j 1 i ( n j 1 i ) ( n j 1 i 1 ) ( n j 1 i n + m + 1 ) t m j 1 i = ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 ) ! i ! ( m j 1 i ) ! ( 1 ) n j 1 i t m j 1 i = 1 ( m j 1 ) ! i = 0 m j 1 ( m j 1 ) ! i ! ( m j 1 i ) ! ( t ) m j 1 i = 1 ( m j 1 ) ! ( 1 t ) m j 1 0 , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equw_HTML.gif
        Since
        ϕ ( k ) ( 0 ) = 0 , k = n m 1 , n m 2 , , 2 , 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equx_HTML.gif
        we have for each k = n m 1 , n m 2 , , 2 , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq38_HTML.gif,
        ϕ ( k ) ( t ) 0 , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equy_HTML.gif
        in particular
        ϕ ( t ) 0 , t [ 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equz_HTML.gif
        so that ϕ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq39_HTML.gif is nondecreasing on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif. Hence by (2.5), we have
        | u ( j ) ( t ) | 0 1 [ ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 i ) ( 1 ) n j 1 i ] | g ( s ) | d s = ( 1 ) n m ( n j 1 ) ! i = 0 m j 1 ( n j 1 i ) ( 1 ) n j 1 i 0 1 | g ( s ) | d s = A j g 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equaa_HTML.gif
        Thus for j = 0 , 1 , , m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq34_HTML.gif,
        u ( j ) A j g 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equab_HTML.gif
        In summary,
        u ( j ) A j g 1 , j = 0 , 1 , , n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equac_HTML.gif

         □

        Lemma 2.3 ([17] Leray-Schauder continuation theorem)

        Let X be a real Banach space and let Ω be a bounded open neighbourhood of 0 in X. Let T : Ω ¯ X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq40_HTML.gifbe a completely continuous operator such that for all λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq41_HTML.gif, and u Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq42_HTML.gif, u λ T u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq43_HTML.gif. Then the operator equation
        u = T u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equad_HTML.gif

        has a solution u Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq44_HTML.gif.

        3 Main results

        Now we are ready to establish our existence theorems of solutions for n th-order right focal boundary value problems (1.1), (1.2). The Leray-Schauder continuation theorem plays key roles in the proofs.

        Theorem 3.1 Let f : [ 0 , 1 ] × R n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq45_HTML.gifsatisfy L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq2_HTML.gif-Carathéodory’s conditions. Suppose that
        1. (i)
          there exist functions α j ( t ) , β j ( t ) , γ ( t ) L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq46_HTML.gif, j = 0 , 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq33_HTML.gif, and a constant σ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq47_HTML.gif such that
          | f ( t , u 0 , u 1 , , u n 1 ) | j = 0 n 1 α j ( t ) | u j | + j = 0 n 1 β j ( t ) | u j | σ + γ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ8_HTML.gif
          (3.1)
           
        for a.e. t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq48_HTML.gifand all ( u 0 , u 1 , , u n 1 ) R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq49_HTML.gif;(ii)
        a : = 1 j = 0 n 1 A j α j p > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ9_HTML.gif
        (3.2)
        where the constants A j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq50_HTML.gif, j = 0 , 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq33_HTML.gifare given in Lemma 2.2;(iii)
        a σ σ 1 ( σ σ 1 σ σ 1 1 σ ) + b 1 σ 1 γ p < 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ10_HTML.gif
        (3.3)

        where b : = j = 0 n 1 A j σ β j p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq51_HTML.gif.

        Then BVP (1.1), (1.2) has at least one solution in W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq26_HTML.gif.

        Proof We define a linear mapping L : W r n , p ( 0 , 1 ) W n , p ( 0 , 1 ) L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq52_HTML.gif, by setting for u W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq53_HTML.gif,
        ( L u ) ( t ) = u ( n ) ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equae_HTML.gif
        We also define a nonlinear mapping N : W r n , p ( 0 , 1 ) L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq54_HTML.gif by setting for y W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq55_HTML.gif,
        ( N u ) ( t ) = f ( t , u ( t ) , u ( t ) , , u ( n 1 ) ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equaf_HTML.gif
        Then, we note that N is a bounded continuous mapping by Lebesgue’s dominated convergence theorem. It is easy to see that the linear mapping L : W r n , p ( 0 , 1 ) L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq56_HTML.gif is a one-to-one mapping. Also, let the linear mapping K : L p [ 0 , 1 ] W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq57_HTML.gif for u L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq58_HTML.gif be defined by
        ( K u ) ( t ) = ( 1 ) n m 0 1 G ( t , s ) u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equag_HTML.gif

        where G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq27_HTML.gif is the Green’s function of BVP in Lemma 2.1.

        Then K satisfies that for u L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq58_HTML.gif, K u W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq59_HTML.gif and L K u = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq60_HTML.gif, and also for u W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq53_HTML.gif, K L u = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq61_HTML.gif. Furthermore, it follows easily by using Arzelà-Ascoli theorem that K N : W r n , p ( 0 , 1 ) W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq62_HTML.gif is a completely continuous operator.

        Here we also note that u W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq53_HTML.gif is a solution of BVP (1.1), (1.2) if and only if u W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq63_HTML.gif is a solution of the operator equation
        L u = N u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equah_HTML.gif
        which is equivalent to the operator equation
        u = K N u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equai_HTML.gif
        We now apply the Leray-Schauder continuation theorem to the operator equation u = K N u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq64_HTML.gif. To do this, it is sufficient to verify that the set of all possible solutions of the family of equations
        u ( n ) ( t ) = λ f ( t , u ( t ) , u ( t ) , , u ( n 1 ) ( t ) ) , 0 < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ11_HTML.gif
        (3.4)
        with boundary conditions
        { u ( i ) ( 0 ) = 0 , i = 0 , 1 , , m 1 , u ( i ) ( 1 ) = 0 , i = m , m + 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ12_HTML.gif
        (3.5)

        is, a priori, bounded in W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq25_HTML.gif by a constant independent of λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq65_HTML.gif.

        Suppose u ( t ) W r n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq66_HTML.gif is a solution of BVP (3.4), (3.5) for some λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq65_HTML.gif. Then from (3.4), (3.1) and (2.2) in Lemma 2.2, we obtain
        u ( n ) p = λ f ( t , u ( t ) , u ( t ) , , u ( n 1 ) ( t ) ) p f ( t , u ( t ) , u ( t ) , , u ( n 1 ) ( t ) ) p j = 0 n 1 α j u ( j ) p + j = 0 n 1 β j ( u ( j ) ) σ p + γ p j = 0 n 1 α j p u ( j ) + j = 0 n 1 β j p u ( j ) σ + γ p j = 0 n 1 A j α j p u ( n ) p + j = 0 n 1 A j σ β j p u ( n ) p σ + γ p = ( 1 a ) u ( n ) p + b u ( n ) p σ + γ p . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equaj_HTML.gif
        Consequently we obtain
        b u ( n ) p σ a u ( n ) p + γ p 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ13_HTML.gif
        (3.6)

        Now we have two cases to consider:

        Case 1. b = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq67_HTML.gif. In this case (3.6) becomes a u ( n ) p + γ p 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq68_HTML.gif, i.e. u ( n ) p γ p a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq69_HTML.gif. Thus from (2.1) in Lemma 2.2, we have that there exists a constant M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq70_HTML.gif which is independent of λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq65_HTML.gif such that
        u = max { u ( j ) , j = 0 , 1 , , n 1 } + u ( n ) p max { A j , j = 0 , 1 , , n 1 } u ( n ) p + u ( n ) p ( 1 + max { A j , j = 0 , 1 , , n 1 } ) γ p a = : M . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ14_HTML.gif
        (3.7)
        Now, let
        Ω = { u W r n , p ( 0 , 1 ) : u < M + 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equak_HTML.gif

        Then estimate (3.7) show that λ K N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq71_HTML.gif has no fixed point on Ω. Hence KN has a fixed point in Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq72_HTML.gif by the Leray-Schauder continuation theorem.

        Case 2. b > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq73_HTML.gif. When γ p = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq74_HTML.gif in (3.1), it is easy to see that BVP (1.1), (1.2) has the trivial solution u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq75_HTML.gif. Thus assume γ p > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq76_HTML.gif and let h ( t ) = b t σ a t + γ p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq77_HTML.gif, t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq78_HTML.gif. Then from (3.6), h ( u ( n ) p ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq79_HTML.gif. It is easy to see that h ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq80_HTML.gif has a unique positive solution ( a b σ ) 1 σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq81_HTML.gif, say ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq82_HTML.gif. By (3.3), we have h ( ρ ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq83_HTML.gif and thus h ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq84_HTML.gif has a minimum positive solution, say ρ ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq85_HTML.gif which is less than ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq82_HTML.gif and independent of λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq65_HTML.gif. Hence it follows that if u ( n ) p ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq86_HTML.gif, then
        u ( n ) p ρ ¯ < ρ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ15_HTML.gif
        (3.8)
        From (2.1) in Lemma 2.2, we get
        u = max { u ( j ) , j = 0 , 1 , , n 1 } + u ( n ) p ( 1 + max { A j , j = 0 , 1 , , n 1 } ) u ( n ) p . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ16_HTML.gif
        (3.9)
        Now, we let
        Ω = { u W r n , p ( 0 , 1 ) : u < M + 1 , u ( n ) p < ρ } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equal_HTML.gif

        where M = ( 1 + max { A j , j = 0 , 1 , , n 1 } ) ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq87_HTML.gif. Then estimates (3.8) and (3.9) show that λ K N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq71_HTML.gif has no fixed point on Ω. Consequently, KN has a fixed point in Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq72_HTML.gif by the Leray-Schauder continuation theorem. This completes the proof of the theorem. □

        Corollary 3.1 Let conditions (i) and (ii) of Theorem 3.1 hold. If b = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq67_HTML.gifor b > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq73_HTML.gifis small enough, then BVP (1.1), (1.2) has at least one solution in W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq26_HTML.gif.

        Corollary 3.2 Let conditions (i) and (ii) of Theorem 3.1 hold. If γ p > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq88_HTML.gifis small enough, then BVP (1.1), (1.2) has at least one solution in W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq26_HTML.gif.

        Remark 3.1 Theorem 3.1-3.4 in [16] are special cases of above Theorem 3.1.

        Next, we give some results on the uniqueness of solutions for BVP (1.1), (1.2).

        Theorem 3.2 Let f : [ 0 , 1 ] × R n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq45_HTML.gifsatisfy L p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq2_HTML.gif-Carathéodory’s conditions. Suppose that
        1. (i)
          there exist functions α j ( t ) , β j ( t ) L p [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq89_HTML.gif, j = 0 , 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq33_HTML.gif, and a constant σ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq90_HTML.gif such that
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ17_HTML.gif
          (3.10)
           
        for a.e. t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq91_HTML.gifand all ( u 0 , u 1 , , u n 1 ) , ( v 0 , v 1 , , v n 1 ) R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq92_HTML.gif;(ii)
        a : = 1 j = 0 n 1 A j α j p > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ18_HTML.gif
        (3.11)
        where the constants A j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq50_HTML.gif, j = 0 , 1 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq33_HTML.gifare given in Lemma 2.2;(iii)
        a σ σ 1 ( σ σ 1 σ σ 1 1 σ ) + b 1 σ 1 f ( t , 0 , , 0 ) p < 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ19_HTML.gif
        (3.12)

        where b : = j = 0 n 1 A j σ β j p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq51_HTML.gif.

        Then BVP (1.1), (1.2) has at least one solution u ( t ) W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq93_HTML.gifand in particular has at most one solution u ( t ) W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq93_HTML.gifwith u ( n ) p < 1 2 ( a b ) 1 σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq94_HTML.gif.

        Proof We note that assumption (3.10) implies
        | f ( t , u 0 , u 1 , , u n 1 ) | j = 0 n 1 α j ( x ) | u j | + j = 0 n 1 β j ( x ) | u j | σ + | f ( t , 0 , , 0 ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equam_HTML.gif

        for a.e. x [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq95_HTML.gif and all ( u 0 , u 1 , , u n 1 ) R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq49_HTML.gif. Accordingly from Theorem 3.1, BVP (1.1), (1.2) has at least one solution in W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq26_HTML.gif.

        Now, suppose that u 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq96_HTML.gif, u 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq97_HTML.gif are two solutions of BVP (1.1), (1.2) with u i ( n 1 ) < 1 2 ( a b ) 1 σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq98_HTML.gif, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq99_HTML.gif. Let w ( t ) = u 1 ( t ) u 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq100_HTML.gif. Then w ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq101_HTML.gif satisfies the boundary condition (1.2) and
        | w ( n ) ( t ) | j = 0 n 1 α j ( t ) | w ( j ) ( t ) | + j = 0 n 1 β j ( t ) | w ( j ) ( t ) | σ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equan_HTML.gif
        Similarly to the proof of Theorem 3.1, we can show easily that
        w ( n ) p ( 1 a ) w ( n ) p + b w ( n ) p σ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equao_HTML.gif
        which gives
        b w ( n ) p σ a w ( n ) p 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equ20_HTML.gif
        (3.13)

        Now consider two cases. If b = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq67_HTML.gif, then w ( n ) p = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq102_HTML.gif from (3.13). Since w A 0 w ( n ) p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq103_HTML.gif, we have w ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq104_HTML.gif on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif, i.e., u 1 ( t ) u 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq105_HTML.gif on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif.

        If b > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq73_HTML.gif, let h ( t ) = b t σ a t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq106_HTML.gif. Then h ( w ( n ) p ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq107_HTML.gif from (3.13). It follows that h ( 0 ) = h ( ( a b ) 1 σ 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq108_HTML.gif and h ( t ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq109_HTML.gif on ( 0 , ( a b ) 1 σ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq110_HTML.gif. Since w ( n ) p u 1 ( n ) p + u 2 ( n ) p < ( a b ) 1 σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq111_HTML.gif, we get w ( n ) p = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq102_HTML.gif. Consequently, u 1 ( t ) u 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq105_HTML.gif on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq8_HTML.gif. This completes the proof of the theorem. □

        Corollary 3.3 Let conditions (i) and (ii) of Theorem 3.2 hold. If b = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq67_HTML.gif, then BVP (1.1), (1.2) has exactly one solution in W n , p ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq26_HTML.gif.

        Finally, we give two examples to which our results can be applicable.

        Example 3.1 Consider the boundary value problem
        { u = 1 16 t 1 4 + t 1 3 u 1 3 ( u ) 2 3 + 1 10 ( u ) 2 , a.e. t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = u ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equap_HTML.gif
        Let f ( t , u 0 , u 1 , u 2 ) = 1 16 t 1 4 + t 1 3 u 0 1 3 u 1 2 3 + 1 10 u 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq112_HTML.gif. Then it is easy to see that f satisfies L 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq113_HTML.gif-Carathéodory’s conditions. By the inequality A 1 p B 1 q A p + B q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq114_HTML.gif for any A , B > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq115_HTML.gif with p , q > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq116_HTML.gif and 1 p + 1 q = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq31_HTML.gif, we get
        | f ( t , u 0 , u 1 , u 2 ) | 1 16 t 1 4 + 1 3 t 1 3 | u 0 | + 2 3 t 1 3 | u 1 | + 1 10 u 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equaq_HTML.gif
        Let α 0 ( t ) = 1 3 t 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq117_HTML.gif, α 1 ( t ) = 2 3 t 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq118_HTML.gif, α 2 ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq119_HTML.gif, β 0 ( t ) = β 1 ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq120_HTML.gif, β 2 ( t ) = 1 10 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq121_HTML.gif, γ ( t ) = 1 16 t 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq122_HTML.gif, σ = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq123_HTML.gif. Then we have
        | f ( t , u 0 , u 1 , u 2 ) | j = 0 2 α j ( t ) | u j | + j = 0 2 β j ( t ) | u j | σ + γ ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equar_HTML.gif
        It is easy to compute that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equas_HTML.gif
        Consequently, we have
        a = 1 j = 0 2 A j α j p = 1 3 15 30 > 0 , b = j = 0 2 A j σ β j 2 = 1 10 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equat_HTML.gif
        and
        a σ σ 1 ( σ σ 1 σ σ 1 1 σ ) + b 1 σ 1 γ 2 = ( 1 3 15 30 ) 2 1 4 + 2 160 < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equau_HTML.gif

        Thus by Theorem 3.1, the above boundary value problem has at least one solution in W 3 , 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq124_HTML.gif.

        Example 3.2 Consider the boundary value problem
        { u = 1 32 t 1 4 + 3 8 t 1 3 sin ( 4 u + u ) + 2 8 g ( u ) , a.e. t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equav_HTML.gif
        where
        g ( u 2 ) = { u 2 2 1 , u 2 2 , 1 2 u 2 2 , 0 u 2 2 , 1 2 u 2 2 , 2 u 2 0 , u 2 2 1 , u 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equaw_HTML.gif
        Let f ( t , u 0 , u 1 , u 2 ) = 1 32 t 1 4 + 3 8 t 1 3 sin ( 4 u 0 + u 1 ) + 2 8 g ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq125_HTML.gif. Then it is easy to see that f satisfies L 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq113_HTML.gif-Carathéodory’s conditions and
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equax_HTML.gif
        Let α 0 ( t ) = 3 2 t 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq126_HTML.gif, α 1 ( t ) = 3 8 t 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq127_HTML.gif, α 2 ( t ) = 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq128_HTML.gif, β 0 ( t ) = β 1 ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq120_HTML.gif, β 2 ( t ) = 2 16 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq129_HTML.gif. Then it is easy to compute that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equay_HTML.gif
        Consequently, we have
        a = 1 j = 0 2 A j α j p = 1 3 5 20 3 8 1 4 > 1 8 > 0 , b = 2 16 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equaz_HTML.gif
        Since f ( t , 0 , 0 , 0 ) 2 = 1 32 t 1 4 2 = 2 32 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq130_HTML.gif and σ = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq123_HTML.gif, we have
        a σ σ 1 ( σ σ 1 σ σ 1 1 σ ) + b 1 σ 1 f ( t , 0 , 0 , 0 ) 2 < ( 1 8 ) 2 ( 1 4 1 2 ) + 2 16 2 32 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equba_HTML.gif

        Thus by Theorem 3.2, the above boundary value problem has at least one solution u ( t ) W 3 , 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq131_HTML.gif and in particular has at most one solution u ( t ) W 3 , 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq132_HTML.gif with u 2 < 1 2 ( a b ) 1 σ 1 = 4 2 a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq133_HTML.gif.

        Also, since from the equation of the boundary value problem we have
        u 2 1 32 t 1 3 2 + 3 8 t 1 3 2 + 2 8 u 2 2 32 + 3 8 + 2 8 u 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equbb_HTML.gif
        it follows that
        u 2 2 32 + 3 8 1 2 8 0.518 < 2 2 < 4 2 a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_Equbc_HTML.gif

        Hence above boundary value problem has a unique solution u ( t ) W 3 , 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-60/MediaObjects/13661_2012_Article_172_IEq132_HTML.gif.

        Declarations

        Acknowledgement

        SKC was supported by Yeungnam University Research Grants 2012. YSO was supported by Daegu University Research Grants 2010.

        Authors’ Affiliations

        (1)
        Department of Mathematics, Beihua University
        (2)
        Department of Mathematics, Yeungnam University
        (3)
        Department of Mathematics Education, Daegu University

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