Solving singular second-orderinitial/boundary value problems in reproducing kernel Hilbert space

  • Er Gao1Email author,

    Affiliated with

    • Songhe Song1 and

      Affiliated with

      • Xinjian Zhang1

        Affiliated with

        Boundary Value Problems20122012:3

        DOI: 10.1186/1687-2770-2012-3

        Received: 13 January 2011

        Accepted: 16 January 2012

        Published: 16 January 2012

        Abstract

        In this paper, we presents a reproducing kernel method for computing singular second-order initial/boundary value problems (IBVPs). This method could deal with much more general IBVPs than the ones could do, which are given by the previous researchers. According to our work, in the first step, the analytical solution of IBVPs is represented in the RKHS which we constructs. Then, the analytic approximation is exhibited in this RKHS. Finally, the n-term approximation is proved to converge to the analytical solution. Some numerical examples are displayed to demonstrate the validity and applicability of the present method. The results obtained by using the method indicate the method is simple and effective.

        Mathematics Subject Classification (2000) 35A24, 46E20, 47B32.

        1. Introduction

        Initial and boundary value problems of ordinary differential equations play an important role in many fields. Various applications of boundary to physical, biological, chemical, and other branches of applied mathematics are well documented in the literature. The main idea of this paper is to present a new algorithm for computing the solutions of singular second-order initial/boundary value problems (IBVPs) of the form:
        { p ( x ) u ( x ) + q ( x ) u ( x ) + r ( x ) u ( x ) = F ( x , u ) , a 1 u ( 0 ) + b 1 u ( 0 ) + c 1 u ( 1 ) = 0 , a 2 u ( 1 ) + b 2 u ( 1 ) + c 2 u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ1_HTML.gif
        (1.1)

        where u ( x ) W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq1_HTML.gif, for x ∈ [0, 1], p ≠ 0, p(x), q(x), r(x) ∈ C[0, 1]. a1, b1,c1, a2, b2, c2 arc real constants and satisfy that a1 u(0) + b1 u'(0) + c1 u (1) and a2 u(1) + b2u'(1) + c2u'(0) are linear independent. F(x, u) is continuous.

        Remark 1.1. We find that if
        b 1 = c 1 = b 2 = c 2 = 0 , a 1 0 , a 2 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ2_HTML.gif
        (1.2)
        the problems are two-point BVPs; if
        b 1 = c 1 = a 2 = b 2 = 0 , a 1 0 , c 2 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ3_HTML.gif
        (1.3)
        the problems are initial value problems; if
        b 1 = a 2 = 0 , a 1 = c 1 0 , b 2 = c 2 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ4_HTML.gif
        (1.4)
        the problems are periodic BVPs; if
        b 1 = a 2 = 0 , a 1 = - c 1 0 , b 2 = - c 2 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ5_HTML.gif
        (1.5)

        the problems are anti-periodic BVPs.

        Such problems have been investigated in many researches. Specially, the existence and uniqueness of the solution of (1.1) have been discussed in [15]. And in recent years, there are also a large number of special-purpose methods are proposed to provide accurate numerical solutions of the special form of (1.1), such as collocation methods [6], finite-element methods [7], Galerkin-wavelet methods [8], variational iteration method [9], spectral methods [10], finite difference methods [11], etc.

        On the other hands, reproducing kernel theory has important applications in numerical analysis, differential equation, probability and statistics, machine learning and precessing image. Recently, using the reproducing kernel method, Cui and Geng [1216] have make much effort to solve some special boundary value problems.

        According to our method, which is presented in this paper, some reproducing kernel Hilbert spaces have been presented in the first step. And in the second step, the homogeneous IBVPs is deal with in the RKHS. Finally, one analytic approximation of the solutions of the second-order BVPs is given by reproducing kernel method under the assumption that the solution to (1.1) is unique.

        2. Some RKHS

        In this section, we will introduce the RKHS W 2 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq2_HTML.gif and W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif. Then we will construct a RKHS H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif, in which every function satisfies the boundary condition of (1.1).

        2.1. The RKHS W 2 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq2_HTML.gif

        Inner space W 2 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq2_HTML.gif is defined as W 2 1 [ 0 , 1 ] = { u ( x ) | u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq5_HTML.gif is absolutely continuous real valued functions, u'L2[0, 1]}. The inner product in W 2 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq2_HTML.gif is given by
        ( f , h ) W 2 1 = f ( 0 ) h ( 0 ) + 0 1 f ( t ) h ( t ) d t , f , h W 2 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ6_HTML.gif
        (2.1)
        and the norm | | u | | W 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq6_HTML.gif is denoted by | | u | | W 2 1 = ( u , u ) W 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq7_HTML.gif. From [17, 18], W 2 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq2_HTML.gif is a reproducing kernel Hilbert space and the reproducing kernel is
        K 1 ( t , s ) = 1 + min { t , s } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ7_HTML.gif
        (2.2)

        2.2. The RKHS W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif

        Inner space W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif is defined as W 2 3 [ 0 , 1 ] = { u ( x ) | u , u , u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq8_HTML.gif is absolutely continuous real valued functions, u"'L2[0, 1]}.

        From [15, 1719], it is clear that W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif become a reproducing kernel Hilbert space if we endow it with suitable inner product.

        Zhang and Lu [18] and Long and Zhang [19] give us a clue to relate the inner product with the boundary conditions (1.1). Set L = D3, and
        γ 1 f = a 1 f ( 0 ) + b 1 f ( 0 ) + c 1 f ( 1 ) , γ 2 f = a 2 f ( 1 ) + b 2 f ( 1 ) + c 2 f ( 0 ) , γ 3 f = a 3 f ( 0 ) + b 3 f ( 0 ) + c 3 f ( 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ8_HTML.gif
        (2.3)

        where a3, b3, c3 is random but satisfying that γ3 is linearly independent of γ1 and γ2.

        It is easy to know that γ1, γ2, γ3 are linearly independent in Ker L. Then from [18, 19], it is easy to know one of the inner products of W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif
        ( f , h ) W 2 3 = i = 1 3 γ i f γ i h + 0 1 f ( t ) h ( t ) d t , f , h W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ9_HTML.gif
        (2.4)

        and its corresponding reproducing kernel K2(t, s).

        2.3. The RKHS H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif

        Inner space H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif is defined as H 2 3 [ 0 , 1 ] = { u ( x ) | u , u , u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq9_HTML.gif are absolutely continuous real valued functions, u"'L2[0, 1], and, a1 u(0) + b1 u'(0) + c1 u(1) = 0, a2 u(1) + b2u'(1) + c2u'(0) = 0}.

        It is clear that H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif is the complete subspace of W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif, so H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif is a RKHS. If P, which is the orthogonal projection from W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif to H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif, is found, we can get the reproducing kernel of H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif obviously. Under the assumptions of Section 2, note
        P f ( t ) = ( γ 3 f ) e 3 ( t ) + 0 1 G ( t , τ ) f ( τ ) d τ , f W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ10_HTML.gif
        (2.5)

        Theorem 2.1. Under the assumptions above, P is the orthogonal projection from W 2 3 [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq10_HTML.gif to H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif.

        Proof. For all f W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq11_HTML.gif, We have
        ( γ 1 ( P f ) ) ( t ) = ( γ 2 ( P f ) ) ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equa_HTML.gif
        That means P f H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq12_HTML.gif. At the same time, for any f , h W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq13_HTML.gif
        ( P f , h ) = ( γ 3 f ) e 3 ( t ) + 0 1 G ( t , τ ) L f ( τ ) d τ , h = ( γ 3 f ) ( γ 3 h ) + 0 1 L 0 1 G ( t , τ ) L f ( τ ) d τ L h ( t ) d t = ( γ 3 f ) ( γ 3 h ) + 0 1 L f ( t ) L h ( t ) d t ( f , P h ) = f , ( γ 3 h ) e 3 ( t ) + 0 1 G ( t , τ ) L h ( τ ) d τ = ( γ 3 f ) ( γ 3 h ) + 0 1 L f ( t ) L 0 1 G ( t , τ ) L h ( τ ) d τ d t = ( γ 3 f ) ( γ 3 h ) + 0 1 L f ( t ) L h ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equb_HTML.gif
        P is self-conjugate. And
        P ( P f ) = P ( γ 3 f ) e 3 ( t ) + 0 1 G ( t , τ ) L f ( τ ) d τ = ( γ 3 f ) e 3 ( t ) + 0 1 G ( t , τ ) L ( γ 3 f ) e 3 ( τ ) + 0 1 G ( τ , s ) L f ( s ) d s , d τ = ( γ 3 f ) e 3 ( t ) + 0 1 G ( t , τ ) L f ( τ ) d τ = P f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equc_HTML.gif

        P is idempotent.

        So P is the orthogonal projection from W 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq3_HTML.gif to H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif.

        The proof of the Theorem 2.1 is complete.

        Now, H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif is a RKHS if endowed the inner product with the inner product below
        ( f , h ) H 2 3 = γ 3 f γ 3 h + 0 1 f ( t ) h ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ11_HTML.gif
        (2.6)

        and the corresponding reproducing kernel K3(t, s) is given in Appendix 4.

        3. The reproducing kernel method

        In this section, the representation of analytical solution of (1.1) is given in the reproducing kernel space H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif.

        Note Lu = p(x)u"(x) + q(x)u'(x) + r(x)u(x) in (1.1). It is clear that L : H 2 3 [ 0 , 1 ] W 2 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq14_HTML.gif is a bounded linear operator.

        Put φ i (x) = K1(x i , x), Ψ i (x) = L*φ i (x), where L* is the adjoint operator of L. Then
        Ψ i ( x ) = ( L * φ i ( y ) , K 3 ( x , y ) ) = ( φ i ( y ) , L y K 3 ( x , y ) ) = ( L y K 3 ( x , y ) , φ i ( x ) ) ¯ = L y K 3 ( x , y ) | y = x i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ12_HTML.gif
        (3.1)

        Lemma 3.1. Under the assumptions above, if { x i } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq15_HTML.gif is dense on [0, 1] then { Ψ i ( x ) } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq16_HTML.gif is the complete basis H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif.

        The orthogonal system { Ψ i ¯ ( x ) } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq17_HTML.gif of H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif can be derived from Gram-Schmidt orthogonalization process of { Ψ i ( x ) } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq16_HTML.gif, and
        Ψ i ¯ ( x ) = j = 1 i β i j Ψ j ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equd_HTML.gif

        Then

        Theorem 3.1. If { x i } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq15_HTML.gif is dense on [0, 1] and the solution of (1.1) is unique, the solution can be expressed in the form
        u ( x ) = i = 1 k = 1 i β i k F ( x k , u ( x k ) ) Ψ i ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ13_HTML.gif
        (3.2)
        Proof. From Lemma 3.1, { Ψ i ( x ) } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq16_HTML.gif is the complete system of H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif. Hence we have
        u ( x ) = i = 1 ( u ( x ) , Ψ ¯ i ( x ) ) Ψ ¯ i ( x ) = i = 1 k = 1 i β i j ( u ( x ) , Ψ i ( x ) ) Ψ ¯ i ( x ) = i = 1 k = 1 i β i k ( u ( x ) , L * φ k ( x ) ) Ψ ¯ i ( x ) = i = 1 k = 1 i β i k ( L u ( x ) , φ k ( x ) ) Ψ ¯ i ( x ) = i = 1 k = 1 i β i k ( F ( x , u ( x ) ) , φ k ( x ) ) Ψ ¯ i ( x ) = i = 1 k = 1 i β i k F ( x k , u ( x k ) ) Ψ ¯ i ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Eque_HTML.gif

        and the proof is complete.

        The approximate solution of the (1.1) is
        u n ( x ) = i = 1 n k = 1 i β i k F ( x k , u ( x k ) ) Ψ ¯ i ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ14_HTML.gif
        (3.3)
        If (1.1) is linear, that is F(x, u(x)) = F(x), then the approximate solution of (1.1) can be obtained directly from (3.3). Else, the approximate process could be modified into the following form:
        { u 0 ( x ) = 0 u n + 1 ( x ) = i = 1 n + 1 B i Ψ ¯ i ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ15_HTML.gif
        (3.4)

        where B i = k = 1 i β i k F ( x k , u n ( x k ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq18_HTML.gif.

        Next, the convergence of u n (x) will be proved.

        Lemma 3.2. There exists a constant M, satisfied | u ( x ) | M | | u | | H 2 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equh_HTML.gif, for all u ( x ) H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq19_HTML.gif.

        Proof. For all x ∈ [0, 1] and u H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq20_HTML.gif, there are
        | u ( x ) | = | ( u ( ) , K 3 ( , x ) ) | | | K 3 ( , x ) | | H 2 3 | | u | | H 2 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equf_HTML.gif
        Since K 3 ( , x ) H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq21_HTML.gif, note
        M = max x [ 0 , 1 ] | | K 3 ( , x ) | | H 2 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equg_HTML.gif
        That is,
        | u ( x ) | M | | u | | H 2 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equh_HTML.gif
        .

        By Lemma 3.2, it is easy to obtain the following lemma.

        Lemma 3.3. If u n | | | | ū ( n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq22_HTML.gif, ||u n || is bounded, x n y(n → ∞) and F(x, u(x)) is continuous, then F ( x n , u n - 1 ( x n ) ) F ( y , ū ( y ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq23_HTML.gif.

        Theorem 3.2. Suppose that ||u n || is bounded in (3.3) and (1.1) has a unique solution. If { x i } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq15_HTML.gif is dense on [0, 1], then the n-term approximate solution u n (x) derived from the above method converges to the analytical solution u(x) of (1.1).

        Proof. First, we will prove the convergence of u n (x).

        From (3.4), we infer that
        u n + 1 ( x ) = u n ( x ) + B n + 1 Ψ ¯ n + 1 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equi_HTML.gif
        The orthonormality of { Ψ i ¯ } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq24_HTML.gif yield that
        | | u n + 1 | | 2 = | | u n | | 2 + ( B n + 1 ) 2 = = i = 1 n + 1 ( B i ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equj_HTML.gif
        That means ||un+1|| ≥ ||u n ||. Due to the condition that ||u n || is bounded, ||u n || is convergent and there exists a constant ℓ such that
        i = 1 ( B i ) 2 = . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equk_HTML.gif
        If m > n, then
        | | u m - u n | | 2 = | | u m - u m - 1 + u m - 1 - u m - 2 + + u n + 1 - u n | | 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equl_HTML.gif
        In view of (u m - um-1) ⊥ (um-1- um-2) ⊥ ··· ⊥ (un+1- u n ), it follows that
        | | u m - u n | | 2 = | | u m - u m - 1 | | 2 + | | u m - 1 - u m - 2 | | 2 + + | | u n + 1 - u n | | 2 = i = n + 1 m ( B i ) 2 0 as n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equm_HTML.gif

        The completeness of H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif shows that u n ū as n → ∞ in the sense of | | | | H 2 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq25_HTML.gif.

        Secondly, we will prove that ū is the solution of (1.1).

        Taking limits in (3.2), we get
        u ¯ ( x ) = i = 1 B i Ψ ¯ i ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equn_HTML.gif
        So
        L ū ( x ) = i = 1 B i L Ψ ¯ i ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equo_HTML.gif
        and
        ( L ū ) ( x ) = i = 1 B i ( L Ψ ¯ i , φ n ) = i = 1 B i ( Ψ ¯ i , L * φ n ) = i = 1 B i ( Ψ ¯ i , Ψ n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ16_HTML.gif
        (3.5)
        Therefore,
        i = 1 n β n j ( L υ ̄ ) ( x n ) = i = 1 B i Ψ ¯ i , i = 1 n β n j Ψ j = i = 1 B i ( Ψ ¯ i , Ψ ¯ n ) = B n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equp_HTML.gif
        If n = 1, then
        L ū ( x 1 ) = F ( x 1 , u 0 ( x 1 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equq_HTML.gif
        If n = 2, then
        β 21 L ū ( x 1 ) + β 22 L ū ( x 2 ) = β 21 F ( x 1 , u 0 ( x 1 ) ) + β 22 F ( x 2 , u 1 ( x 2 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equr_HTML.gif
        It is clear that
        ( L ū ) ( x 2 ) = F ( x 2 , u 1 ( x 2 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equs_HTML.gif
        Moreover, it is easy to see by induction that
        ( L ū ) ( x j ) = F ( x j , u j - 1 ( x j ) ) , j = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ17_HTML.gif
        (3.6)
        Since { x i } i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq15_HTML.gif is dense on [0, 1], for all Y ∈ [0, 1], there exists a subsequence { x n j } j = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq26_HTML.gif such that
        x n j Y as j . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ18_HTML.gif
        (3.7)
        It is easy to see that ( L ū ) ( x n j ) = F ( x n j , u n j - 1 ( x n j ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq27_HTML.gif. Let j → ∞, by the continuity of F(x, u(x)) and Lemma 3.3, we have
        ( L ū ) ( Y ) = F ( Y , ū ( Y ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equ19_HTML.gif
        (3.8)

        At the same time, ū H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq28_HTML.gif. Clearly, u satisfies the boundary conditions of (1.1).

        That is, ū is the solution of (1.1).

        The proof is complete.

        In fact, u n (x) is just the orthogonal projection of exact solution ū(x) onto the space Span { Ψ ̄ i } i = 1 n ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq29_HTML.gif.

        4. Numerical example

        In this section, some examples are studied to demonstrate the validity and applicability of the present method. We compute them and compare the results with the exact solution of each example.

        Example 4.1. Consider the following IBVPs:
        x 2 ( 1 - x ) u ( x ) + 2 u ( x ) + 10 x u ( x ) + x 2 ( 1 - x ) ( u ( x ) + 1 ) 2 = f ( x ) , 0 < x < 1 , u ( 0 ) + u ( 0 ) + u ( 1 ) = 0 , u ( 1 ) + u ( 1 ) + u ( 0 ) = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equt_HTML.gif
        Where f ( x ) = 10 x e 10 ( x - x 2 ) 2 + 40 e 10 ( x - x 2 ) 2 ( 1 - 2 x ) ( x - x 2 ) + x 2 ( 1 - x ) ( e 20 ( x - x 2 ) 2 + 20 e 10 ( x - x 2 ) 2 ( 1 - 2 x ) 2 - 40 e 10 ( x - x 2 ) 2 ( x - x 2 ) + 400 e 10 ( x - x 2 ) 2 ( 1 - 2 x ) 2 ( x - x 2 ) 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq30_HTML.gif. The exact solution is u ( x ) = e 10 ( x - x 2 ) 2 - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq31_HTML.gif. Using our method, take a3 = 1, b3 = c3 = 0 and n = 21, 51, N = 5, x i = i - 1 n - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq32_HTML.gif. The numerical results are given in Tables 1 and 2.
        Table 1

        Numerical results for Example 4.1 (n = 21, N = 5)

        x

        True solution u(x)

        Approximate solution u11

        Absolute error

        Relative error

        0.08

        0.05566

        0.05530

        3.6E-4

        6.5E-3

        0.16

        0.19798

        0.19765

        3.3E-4

        1.7E-3

        0.24

        0.39473

        0.39443

        3.0E-4

        7.6E-4

        0.32

        0.60560

        0.60526

        3.4E-4

        5.6E-4

        0.40

        0.77891

        0.77839

        5.2E-4

        6.6E-4

        0.48

        0.86452

        0.86385

        6.7E-4

        7.7E-4

        0.56

        0.83516

        0.83457

        5.9E-4

        7.1E-4

        0.64

        0.70036

        0.70009

        2.7E-4

        3.8E-4

        0.72

        0.50144

        0.50146

        1.8E-5

        3.6E-4

        0.80

        0.29175

        0.29175

        3.2E-6

        1.1E-5

        0.88

        0.11797

        0.11771

        2.6E-4

        2.2E-3

        0.96

        0.01485

        0.01453

        3.3E-4

        2.2E-3

        Table 2

        Numerical results for Example 4.1 (n = 51, N = 5)

        x

        True solution u(x)

        Approximate solution u11

        Absolute error

        Relative error

        0.08

        0.05566

        0.05564

        2.3E-5

        4.IE-4

        0.16

        0.19798

        0.19796

        2.IE-5

        1.1E-4

        0.24

        0.39473

        0.39471

        2.0E-5

        4.9E-5

        0.32

        0.60560

        0.60557

        2.8E-5

        4.6E-5

        0.40

        0.77891

        0.77885

        5.6E-5

        7.IE-5

        0.48

        0.86452

        0.86444

        8.0E-5

        9.3E-5

        0.56

        0.83516

        0.83509

        6.6E-5

        7.9E-5

        0.64

        0.70036

        0.70035

        9.6E-6

        1.4E-5

        0.72

        0.50144

        0.50148

        4.3E-5

        8.6E-5

        0.80

        0.29175

        0.29180

        4.7E-5

        1.6E-5

        0.88

        0.11797

        0.11797

        2.2E-6

        1.9E-5

        Example 4.2. Consider the following IBVPs:
        { u ( x ) + u ( x ) + x ( 1 x ) ( u ( x ) 1 ) 3 = f ( x ) , 0 x 1 , π 2 u ( 0 ) + u ( 0 ) π 2 u ( 1 ) = 0 , π u ( 1 ) + 2 u ( 1 ) + 3 u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equu_HTML.gif
        where f(x) = π cos(πx) - sin(πx)(x2 + (-1 + x) * x * sin2(π* x)). The true solution is u(x) = sin(πx) + 1. Using our method, take a3 = 1, b3 = c3 = 0, and N = 5, n = 21, 51, x i = i - 1 n - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq32_HTML.gif. The numerical results are given in Figures 1, 2, 3, and 4.
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Fig1_HTML.jpg
        Figure 1

        The absolute error of Example 4.2 ( n = 21, N = 5).

        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Fig2_HTML.jpg
        Figure 2

        The relative error of Example 4.2 ( n = 21, N = 5).

        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Fig3_HTML.jpg
        Figure 3

        The absolute error of Example 4.2 ( n = 51, N = 5).

        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Fig4_HTML.jpg
        Figure 4

        The relative error of Example 4.2 ( n = 51, N = 5).

        Contributions

        Er Gao gives the main idea and proves the most of the theorems and propositions in the paper. He also takes part in the work of numerical experiment of the main results. Xinjian Zhang suggests some ideas for the prove of the main theorems. Songhe Song mainly accomplishes most part of the numerical experiments. All authors read and approved the final manuscript.

        Appendix A: The reproducing kernel of H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif

        The reproducing kernel of H 2 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_IEq4_HTML.gif is
        K 3 [ t , s ] = Λ 1 120 Δ + Λ 2 Δ 2 + Λ 3 120 Δ + Λ 4 40 Δ 2 + Λ 5 120 Δ + Λ 6 120 Δ + Λ 7 120 Δ 2 + { 1 120 ( s 5 5 s 4 t + 10 s 3 t 2 ) , t s , 1 120 ( t 5 5 t 4 s + 10 t 3 s 2 ) , t < s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equv_HTML.gif
        where
        Δ = a 3 ( 2 b 1 b 2 + b 2 c 1 c 1 c 2 ) + a 2 ( a 3 b 1 a 1 b 3 + 2 a 1 c 3 2 b 1 c 3 ) 2 ( a 1 + c 1 ) ( b 2 b 3 b 2 c 3 c 2 c 3 ) , Λ 1 = ( c 1 ( 2 c 2 c 3 + a 3 c 2 s 2 a 2 ( 1 + s ) ( b 3 2 c 3 a 3 s + b 3 s ) + b 2 ( 2 b 3 2 c 3 + a 3 ( 2 + s ) s ) ) t 3 ( 10 5 t + t 2 ) ) , Λ 2 = ( 2 b 1 b 2 + b 2 c 1 c 1 c 2 2 a 1 b 2 s 2 b 2 c 1 s + a 1 b 2 s 2 + b 2 c 1 s 2 ) + a 1 c 2 s 2 + c 1 c 2 s 2 a 2 ( 1 + s ) ( b 1 a 1 s + b 1 s ) ) × ( 2 b 1 b 2 + b 2 c 1 c 1 c 2 2 a 1 b 2 t 2 b 2 c 1 t + a 1 b 2 t 2 + b 2 c 1 t 2 + a 1 c 2 t 2 + c 1 c 2 t 3 a 2 ( 1 + t ) ( b 1 a 1 t + b 1 t ) ) , Λ 3 = c 1 s 3 ( 10 5 s + s 2 ) ( a 2 ( 1 + t ) ( b 3 2 c 3 a 3 t + b 3 t ) 2 c 2 c 3 + a 3 c 2 t 2 + b 2 ( 2 b 3 2 c 3 + a 3 ( 2 + t ) t ) , Λ 4 = c 1 ( 2 ( c 2 c 2 ( 2 c 3 + a 3 s 2 ) + a 2 ( 2 b 1 c 3 2 a 1 c 3 s a 3 b 1 s 2 + a 1 b 3 s 2 ) ) + b 2 ( 10 b 1 c 3 + 6 c 1 c 3 + a 3 c 1 s 10 a 1 c 3 s 10 c 1 c 3 s 5 a 3 b 1 s 2 3 a 3 c 1 s 2 + b 3 ( c 1 + 5 a 1 s 2 + 5 c 1 s 2 ) ) ) ( 2 c 2 c 3 + a 3 c s t 2 a 2 ( 1 + t ) ( b 3 2 c 3 a 3 t + b 3 t ) + b 2 ( 2 b 3 2 c 3 + a 3 ( 2 + t ) t ) ) , Λ 5 = ( 2 b 1 c 3 + 2 c 1 c 3 + a 3 c 1 s 2 a 1 c 3 s 2 c 1 c 3 s a 3 b 1 s 2 a 3 c 1 s 2 + b 3 ( a 1 s 2 + c 1 ( 1 + s 2 ) ) ) t 3 ( 5 b 2 ( 4 + t ) + a 2 ( 10 5 t + t 2 ) ) , Λ 6 = s 3 ( 5 b 2 ( 4 + s ) + a 2 ( 10 5 s + s 2 ) ) ( 2 b 1 c 3 + 2 c 1 c 3 + a 3 c 1 t 2 a 1 c 3 t 2 c 1 c 3 t a 3 b 1 t 3 a 3 c 1 t 2 + b 3 ( a 1 t 2 + c 1 ( 1 + t 2 ) ) ) , Λ 7 = ( 6 a 2 2 ( a 1 s ( 2 c 3 + b 3 s ) + b 1 ( 2 c 3 a 3 s 2 ) ) + 3 a 2 ( 2 c 1 c 2 ( 2 c 3 + a 3 s 2 ) + b 2 ( b 3 c 1 + 20 b 1 c 3 + 6 c 1 c 3 + a 3 c 1 s 20 a 1 c 3 s 10 c 1 c 3 s 10 a 3 b 1 s 2 + 10 a 1 b 3 s 2 3 a 3 c 1 s 2 + 5 b 3 c 1 s 2 ) ) + 5 b 2 ( 3 c 1 c 2 ( 2 c 3 + a 3 s 2 ) + b 2 ( 2 b 3 c 1 + 16 b 1 c 3 + 10 c 1 c 3 + 2 a 3 c 1 s 16 a 1 c 3 s 16 c 1 c 3 s 8 a 3 b 1 s 2 + 8 a 1 b 3 s 2 5 a 3 c 1 s 2 + 8 b 3 c 1 s 2 ) ) ) ( 2 b 1 c 3 + 2 c 1 c 3 + a 3 c 1 t 2 a 1 c 3 t 2 c 1 c 3 t a 3 b 1 t 2 a 3 c 1 t 2 + b 3 ( a 1 t 2 + c 1 ( 1 + t 2 ) ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-3/MediaObjects/13661_2011_Article_106_Equw_HTML.gif

        Declarations

        Acknowledgements

        The work is supported by NSF of China under Grant Numbers 10971226.

        Authors’ Affiliations

        (1)
        Department of Mathematics and Systems Science, College of Science, National University of Defense Technology

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