西南大学学报 (自然科学版)  2017, Vol. 39 Issue (8): 57-64.  DOI: 10.13718/j.cnki.xdzk.2017.08.008
0
Article Options
  • PDF
  • Abstract
  • Figures
  • References
  • 扩展功能
    Email Alert
    RSS
    本文作者相关文章
    张建元
    韩艳
    张毅敏
    赵书芬
    刘承萍
    张昕
    欢迎关注西南大学期刊社
     

  • 无穷直线上K-解析函数的Riemann边值问题与Hilbert边值问题    [PDF全文]
    张建元1, 韩艳1, 张毅敏1, 赵书芬1, 刘承萍1, 张昕2     
    1. 昭通学院 数学与统计学院,云南 昭通 657000;
    2. 昭通市统计局,云南 昭通 657000
    摘要:首先引入了无穷直线上(分片)K-解析函数的Cauchy型K-积分的概念,利用K-对称变换的方法研究了Cauchy型K-积分的某些性质,然后借助函数在无穷直线上的指标与这些Cauchy型K-积分的性质,得到了在无穷直线上K-解析函数类中的Riemann边值问题的可解条件和解的表达式以及它们与指标之间的关系;进一步利用半平面内的K-对称扩张函数,把Hilbert边值问题转化为无穷直线X上的Riemann边值问题,又得到了Hilbert边值问题的可解条件和解的表达式.而解析函数和共轭解析函数都是K-解析函数的特例,所得结果推广了解析函数和共轭解析函数中的相应结论.
    关键词${{\hat H}_k}$类函数    直线上的Cauchy型K-积分    (分片)K-解析函数    K-对称变换    K-对称扩张函数    边界值公式    Riemann边值问题    Hilbert边值问题    指标    

    在文献[1-4]中,给出K-解析函数(变换)与K-积分的概念以及K-解析函数(在域内的值)可由其边界上的值来表示这一特性.文献[5]给出了一个定义在全平面互补的两个区域内的分片K-解析函数,只要它们(左右)的边界值(若存在)满足一定的条件,那么它也有此特性.由于以上各种情况的区域边界是周线,那么当区域边界是通过无穷的曲线,且它在其上边界值也满足一定的条件时,该函数是否也具有类似特性?不失一般性,设该无穷曲线是实轴,记为X,其正向与实轴相同,同时把以X为边界的上下半平面分别记为Z+Z-.本文将在无穷直线X上对其相应的Riemann边值问题及Hilbert边值问题作出如下的讨论与研究.

    1 基本概念

    z(k)≡x+iky(0≠k$\mathbb{R}$)为z=x+iyK-复数,K-对称变换[2, 6-8]ζ=z(k)一对一地把Z+变为K-平面Z+(k),边界X变为K-直线L:ξ=x(k),此时XL分别关于区域Z+Z+(k)是K保方(转)向,当k>0时,XL是保转向的;当k<0时,XL是反保转向的,X+对应L-,反之亦然.

    定义1[9] 设f(z)在$\overline {{Z^ + }} $上连续,若满足下列两个条件:

    (ⅰ)当z1z2$\overline {{Z^ + }} $,|z1(k)|,|z2(k)|≤R时,

    $ \left| {f\left( {{z_1}} \right) - f\left( {{z_2}} \right)} \right| < A{\left| {{z_1}\left( k \right) - {z_2}\left( k \right)} \right|^a} $

    (ⅱ)当z1z2$\overline {{Z^ + }} $,|z1(k)|,|z2(k)|>R时,

    $ \left| {f\left( {{z_1}} \right) - f\left( {{z_2}} \right)} \right| < A{\left| {\frac{1}{{{z_1}\left( k \right)}} - \frac{1}{{{z_2}\left( k \right)}}} \right|^a}\left( {0 < \alpha \le 1} \right) $

    则称$f\left(z \right) \in \hat H_k^\alpha \left({\overline {{Z^ + }} } \right)$,简记为$f\left(z \right) \in {{\hat H}_k}\left({\overline {{Z^ + }} } \right)$${{\hat H}_k}$.同样可在$\overline {{Z^ - }} $上定义$\hat H_k^\alpha $($\overline {{Z^ - }} $)以及在X上定义$\hat H_k^\alpha \left(X \right)$.

    显然,$f\left(x \right) \in {{\hat H}_k}\left(X \right)$,则

    $ \mathop {\lim }\limits_{x \to \pm \infty } f\left( x \right) = f\left( { \pm \infty } \right) = C\left( {存在} \right) $

    定义2[3-5, 8-13] 设$f\left(x \right) \in {{\hat H}_k}\left(X \right)$,称积分

    $ F\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2\pi i}}\int_{ - \infty }^{ + \infty } {\frac{{f\left( t \right)}}{{\left( {t - z} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} \;\;\;\;\;z \notin X $ (1)

    X上的Cauchy型K-积分,f(t)叫做它的密度.

    xX时,有

    $ F\left( x \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2\pi i}}\int_{ - \infty }^{ + \infty } {\frac{{f\left( t \right)}}{{\left( {t - x} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} \;\;\;\;\;z \in X $ (2)

    对于(1),(2) 式中的积分主值,有如下边界值性质.

    定理1[5, 9-13](边界值公式) 若Xx轴,f(t)在X上满足条件$\hat H_k^\alpha \left(X \right)$,其中0<α≤1,则

    $ {F^ \pm }\left( x \right) = \frac{{ \pm 1}}{2}f\left( x \right) + \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2\pi i}}\int_{ - \infty }^{ + \infty } {\frac{{f\left( t \right)}}{{\left( {t - x} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} \;\;\;\;\;z \in X $ (3)
    $ \left\{ \begin{array}{l} {F^ + }\left( x \right) - {F^ - }\left( x \right) = f\left( x \right)\\ {F^ + }\left( x \right) + {F^ - }\left( x \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{\pi i}}\int_{ - \infty }^{ + \infty } {\frac{{f\left( t \right)}}{{\left( {t - x} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} \end{array} \right.\;\;\;\;\;\;\;\;x \in X $ (4)

    其中$F\left(x \right) = \frac{{{\text{sgn}}\left(k \right)}}{{2\pi i}}\int_{ -\infty }^{ + \infty } {\frac{{f\left(t \right)}}{{\left({t -x} \right)\left(k \right)}}{\text{d}}t\left(k \right)} $为积分(1) 的Cauchy主值.

    定理2[5, 9-13](边界值性质) 在定理1的条件下,F(x)满足条件$\hat H_k^\alpha \left(X \right)$F(∞)=0.

    引理1 若无穷曲线Xx轴,变换ζ=z(k)把X变为K-直线Lξ=x(k),则

    $ f\left( x \right) \in \mathop {{H_k}}\limits^ \wedge \left( x \right) \Leftrightarrow f\left( {\xi \left( {{k^{ - 1}}} \right)} \right) \in \mathop H\limits^ \wedge \left( L \right) $

    更一般地,若变换ζ=z(k)把$\overline {{Z^ \pm }} $变为$\overline {{Z^ \pm }} $(k),则

    $ f\left( z \right) \in \mathop {{H_k}}\limits^ \wedge \left( {\overline {{Z^ \pm }} } \right) \Leftrightarrow f\left( {\zeta \left( {{k^{ - 1}}} \right)} \right) \in \mathop H\limits^ \wedge \left( {\overline {{Z^ \pm }\left( k \right)} } \right) $

    我们只在$\overline {{Z^ + }} $上对定义1中(ⅱ)进行验证,其他情况可类似进行.

    $ g\left( \zeta \right) = f\left( {\zeta \left( {{k^{ - 1}}} \right)} \right) = f\left( z \right) $

    $f\left(z \right) \in {{\hat H}_k}\left({\overline {{Z^ + }} } \right)$,当z1z2$\overline {{Z^ + }} $,|ζ1|=|z1(k)|,|ζ2|=|z2(k)|>R时,有[1]

    $ \begin{array}{l} \left| {g\left( {{\zeta _1}} \right) - g\left( {{\zeta _2}} \right)} \right| = \left| {f\left( {{\zeta _1}\left( {{k^{ - 1}}} \right)} \right) - f\left( {{\zeta _2}\left( {{k^{ - 1}}} \right)} \right)} \right| \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;A{\left| {{{\left[ {{\zeta _1}\left( {{k^{ - 1}}k} \right)} \right]}^{ - 1}} - {{\left[ {{\zeta _2}\left( {{k^{ - 1}}k} \right)} \right]}^{ - 1}}} \right|^\alpha } = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;A{\left| {\zeta _1^{ - 1} - \zeta _2^{ - 1}} \right|^\alpha } \end{array} $

    $ g\left( \zeta \right) = f\left( {\zeta \left( {{k^{ - 1}}} \right)} \right) \in \mathop H\limits^ \wedge \left( {\overline {{Z^ \pm }\left( k \right)} } \right) $

    反之也成立.

    定理1-2的证明 由于K-对称变换ζ=z(k)分别把Z+Z-变为K-平面Z(k),边界XL分别关于相应的区域是K保方(转)向.于是X上Cauchy型K-积分(1) 变为L上Cauchy型积分

    $ F\left( {\zeta \left( {{k^{ - 1}}} \right)} \right) = \frac{1}{{2\pi i}}\int_{ - \infty }^{ + \infty } {\frac{{f\left( {\tau \left( {{k^{ - 1}}} \right)} \right)}}{{\tau - \zeta }}{\rm{d}}\tau } $

    $ G\left( \zeta \right) = \frac{1}{{2\pi i}}\int_{ - \infty }^{ + \infty } {\frac{{g\left( \tau \right)}}{{\tau - \zeta }}{\rm{d}}\tau } \;\;\;\;\;\;\;\zeta \notin L $

    $f\left(x \right) \in {{\hat H}_k}\left(X \right) \Leftrightarrow f\left({\xi \left({{k^{ -1}}} \right)} \right) \in \hat H\left(L \right)$,得G(ζ)解析[9].又G(ζ)=F(ζ(k-1))为解析⇔F(z)=G(z(k))(zX)为K-解析[1],由X上解析函数的普莱梅(J. Plemelj)公式及普利瓦洛夫(Privalov)定理[9]得定理1-2成立,且

    $ \left\{ \begin{array}{l} {G^ \pm }\left( \xi \right) = \frac{{ \pm 1}}{2}g\left( \xi \right) + \frac{1}{{2{\rm{\pi }}i}}\int_{ - \infty }^{ + \infty } {\frac{{g\left( \tau \right)}}{{\tau - \xi }}{\rm{d}}\tau } \\ G\left( \xi \right) \in H\left( L \right)\;\;\;\;\;\;\;\;\;G\left( \infty \right) = 0 \end{array} \right. $

    $ {F^ \pm }\left( x \right) = \frac{{ \pm 1}}{2}f\left( x \right) + \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2{\rm{\pi }}i}}\int_L {\frac{{f\left( t \right)}}{{\left( {t - x} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} ,F\left( x \right) \in {\mathop H\limits^ \wedge _k}\left( X \right),F\left( \infty \right) = 0 $

    此时ξ=x(k)∈L,即(3) 式成立,进一步可证(4) 式与(3) 式等价.

    注1 关于定理2有如下更一般的结论,命题:在定理1的条件下,F(z)分别满足条件${{\hat H}_k}\left({\overline {{Z^ + }} } \right), {{\hat H}_k}\left({\overline {{Z^ -}} } \right)$.

    注2 在定理1-2证明过程中知,(1) 式定义的函数F(z)(zX)分别在Z+Z-K-解析,但在X上边界值有跳跃:F+(x)-F-(x)=f(x),这样F(z)就是一个分片(区)K-解析函数.

    2 无穷直线上的Riemann边值问题

    求一个包括无穷远点在内的分片K-解析函数Φ(z),使它在X上满足条件[5, 9-14]

    $ {\mathit{\Phi }^ + }\left( t \right) = G\left( t \right){\mathit{\Phi }^ - }\left( t \right) + g\left( t \right)\;\;\;\;\;t \in X $ (5)

    其中G(t)≠0,g(t)∈${{\hat H}_k}\left(X \right)$的问题称为X上的Riemann边值问题,简称Riemann边值问题,记为R问题.在∞处,因G(t)≠0,g(t)∈${{\hat H}_k}\left(X \right)$G(∞)≠0,g(∞)皆存在.若要求Φ(z)在∞处有界,即当z$\overline {{Z^ \pm }} $z→∞时,Φ±(z)→Φ±(∞)存在,在XΦ±(+∞)=Φ±(-∞).

    $\kappa \triangleq {\text{sgn}}\left(k \right){\text{In}}{{\text{d}}_X}G\left(t \right) = {\left({2\pi i} \right)^{ -1}}{\text{sgn}}\left(k \right){\left[ {{\text{ln}}G\left(t \right)} \right]_X} = {\left({2\pi } \right)^{ -1}}{\text{sgn}}\left(k \right){\left[ {{\text{arg}}G\left(t \right)} \right]_X}$称为R问题的指标.

    $ {G_0}\left( t \right) = Q\left( t \right)G\left( t \right)\;\;\;\;\;\;\;Q\left( t \right) \buildrel \Delta \over = {\left[ {\left( {t + i} \right)\left( k \right)/\left( {t - i} \right)\left( k \right)} \right]^\kappa } $

    易证

    $ \begin{array}{*{20}{c}} {{G_0}\left( t \right) \in {{\mathop H\limits^ \wedge }_k}\left( X \right)}\\ {{G_0}\left( t \right) \ne 0,{G_0}\left( \infty \right) = G\left( \infty \right) \ne 0,{\rm{In}}{{\rm{d}}_X}{G_0}\left( t \right) = - {\mathop{\rm sgn}} \left( k \right)\kappa + {\rm{In}}{{\rm{d}}_X}G\left( t \right) = 0} \end{array} $

    对于齐次R问题:

    $ {\mathit{\Phi }^ + }\left( t \right) = G\left( t \right){\mathit{\Phi }^ - }\left( t \right)\;\;\;\;\;\;\;\;\;t \in X $ (6)

    改写:

    $ {\mathit{\Psi }^ + }\left( t \right) = {G_0}\left( t \right){\mathit{\Psi }^ - }\left( t \right) $

    其中

    $ \psi \left( z \right) = \left\{ \begin{array}{l} {\left[ {\left( {z + i} \right)\left( k \right)} \right]^\kappa }\phi \left( z \right)\;\;\;\;\;\;\;\;\;\;z \in {Z^ + }\\ {\left[ {\left( {z - i} \right)\left( k \right)} \right]^\kappa }\phi \left( z \right)\;\;\;\;\;\;\;\;\;\;z \in {Z^ - } \end{array} \right. $

    Ψ(z)分片K-解析,且Ψ(z)=O(|z|κ)(z→∞).

    $ \mathit{\Gamma }\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2{\rm{\pi }}i}}\int_{ - \infty }^{ + \infty } {\frac{{\ln {G_0}\left( t \right)}}{{\left( {t - z} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} $ (7)

    因IndXG0(t)=0,lnG0(t)单值$ \in {{\hat H}_k}\left(X \right)$,有

    $ {\mathit{\Gamma }^ \pm }\left( x \right) \in {\mathop H\limits^ \wedge _k}\left( X \right)\;\;\;\;\;\;\mathit{\Gamma }\left( { \pm \infty } \right) = 0 $

    再令

    $ X\left( z \right) = \left\{ \begin{array}{l} {\left[ {\left( {z + i} \right)\left( k \right)} \right]^{ - \kappa }}{e^{\mathit{\Gamma }\left( z \right)}}\;\;\;\;\;\;\;\;\;\;z \in {Z^ + }\\ {\left[ {\left( {z - i} \right)\left( k \right)} \right]^{ - \kappa }}{e^{\mathit{\Gamma }\left( z \right)}}\;\;\;\;\;\;\;\;\;\;z \in {Z^ - } \end{array} \right. $ (8)

    $ {X^ + }\left( t \right) = G\left( t \right){X^ - }\left( t \right) $ (9)

    X(z)分片K-解析,X(z)≠0(包括X±(t)≠0),在z=∞处有有限阶.

    由(6),(9) 式得:Φ+(t)/X+(t)=Φ-(t)/X-(t),故F(z)=Φ(z)/X(z)在全平面K-解析,在z=∞处有κ阶.当κ≥0时,因∞至多为Φ(z)的0阶极,Φ(z)/X(z)在∞至多有κ阶极,由(广义)刘维尔定理得Φ(z)/X(z)=Pκ(z),即Φ(z)=Pκ(z)X(z);当κ<0时,Φ(∞)/X(∞)=0,由刘维尔定理得Φ(z)≡0.于是得如下结论.

    定理3 对于齐次R问题,若实数轴上R问题的指标为κ,则当κ≥0时,其一般解为Φ(z)=Pκ(z)X(z)(其中Pκ(z)为z(k)的κ次任复系数多项式,P0(z)≡CC是任意常数);当κ<0时,Φ(z)≡0.

    对于非齐次R问题,由(5),(9) 式得:

    $ \frac{{{\mathit{\Phi }^ + }\left( t \right)}}{{{X^ + }\left( t \right)}} = \frac{{{\mathit{\Phi }^ - }\left( t \right)}}{{{X^ - }\left( t \right)}} + \frac{{g\left( t \right)}}{{{X^ + }\left( t \right)}} $

    因一般情况下(除κ≤0外)

    $ \frac{{g\left( t \right)}}{{{X^ + }\left( t \right)}} = {\left[ {\left( {t + i} \right)\left( k \right)} \right]^\kappa }g\left( t \right){e^{ - \mathit{\Gamma } + \left( z \right)}} \notin {\mathop H\limits^ \wedge _k} $

    若令

    $ Y\left( z \right) = \left\{ \begin{array}{l} {e^{\mathit{\Gamma }\left( z \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;z \in {Z^ + }\\ {\left[ {\frac{{\left( {z + i} \right)\left( k \right)}}{{\left( {z - i} \right)\left( k \right)}}} \right]^\kappa }{e^{\mathit{\Gamma }\left( z \right)}}\;\;\;\;\;\;\;\;\;\;\;z \in {Z^ - } \end{array} \right. $ (10)

    $ \frac{{{\mathit{\Phi }^ + }\left( t \right)}}{{{Y^ + }\left( t \right)}} = \frac{{{\mathit{\Phi }^ - }\left( t \right)}}{{{Y^ - }\left( t \right)}} + \frac{{g\left( t \right)}}{{{Y^ + }\left( t \right)}} $ (11)

    $ \frac{{g\left( t \right)}}{{{Y^ + }\left( t \right)}} \in {\mathop H\limits^ \wedge _k} $

    $ \psi \left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2{\rm{\pi }}i}}\int_{ - \infty }^{ + \infty } {\frac{{g\left( t \right)}}{{{Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} $

    $ \frac{{{\mathit{\Phi }^ + }\left( t \right)}}{{{Y^ + }\left( t \right)}} - {\psi ^ + }\left( t \right) = \frac{{{\mathit{\Phi }^ - }\left( t \right)}}{{{Y^ - }\left( t \right)}} - {\psi ^ - }\left( t \right) $

    从而$F\left(z \right) = \frac{{\mathit{\Phi }\left(z \right)}}{{Y\left(z \right)}} -\psi \left(z \right)$在全平面K-解析,除在z=-i处可能有极外.当κ≥0时,F(z)在z=-i处有κ阶极,为了使它在此处有界,作[(z+i)(k)]κF(z),但在z=∞处它有κ阶,由(广义)刘维尔定理得[(z+i)(k)]κF(z)=Pκ(z).由(7),(8),(10) 式得非齐次R问题的一般解为:

    $ \mathit{\Phi }\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)Y\left( z \right)}}{{2{\rm{\pi }}i}}\int_{ - \infty }^{ + \infty } {\frac{{g\left( t \right)}}{{{Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}{\rm{d}}t\left( k \right) + X\left( z \right){P_\kappa }\left( z \right)} $ (12)

    其中Pκ(z)为z(k)的κ次任意复系数多项式,P0(z)≡CC是任意常数.

    κ<0时,F(z)在全平面有界,由刘维尔定理得F(z)=C,即

    $ \mathit{\Phi }\left( z \right) = Y\left( z \right)\left[ {\mathit{\Psi }\left( z \right) + C} \right]\;\;\;\;\;\;\;\;C\;为常数 $ (13)

    但它在z=-i处有极,为了消除极点,需且只需[1, 9, 15, 16] Ψ(-i)=-CΨk(n)(-i)=0,即

    $ \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2{\rm{\pi }}i}}\int_{ - \infty }^{ + \infty } {\frac{{g\left( t \right)}}{{{Y^ + }\left( t \right)\left( {t + i} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} = - C $ (14)
    $ \int_{ - \infty }^{ + \infty } {\frac{{g\left( t \right)}}{{{Y^ + }\left( t \right){{\left[ {\left( {t + i} \right)\left( k \right)} \right]}^{n + 1}}}}{\rm{d}}t\left( k \right)} = 0\;\;\;\;\;n = 1,2, \cdots , - \kappa - 1 $ (15)

    也就是当且仅当g(t)满足条件(14),(15) 时,非齐次R问题有唯一解(13),于是得如下结论.

    定理4 若实数轴上R问题的指标为κ,则非齐次R问题在Φ(∞)有界的附加条件下,当κ≥0时,问题有一般解(12);当κ=-1时,问题有唯一解(13),其中C由(14) 式给出;当κ<0时,当且仅当g(t)满足条件(14),(15) 时,R问题有唯一解(13).

    下面讨论非齐次R问题在Φ(∞)=0的附加条件下的解.要求Φ(∞)=0,必有g(∞)=0,从而Ψ(∞)=0.上述F(z)在z=∞处也为零.因此当κ≥0时,

    $ {\left[ {\left( {z + i} \right)\left( k \right)} \right]^\kappa }F\left( z \right) = {P_{\kappa - 1}}\left( z \right) $

    故问题有一般解

    $ \mathit{\Phi }\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)Y\left( z \right)}}{{2{\rm{\pi }}i}}\int_{ - \infty }^{ + \infty } {\frac{{g\left( t \right)}}{{{Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} + X\left( z \right){P_{\kappa - 1}}\left( z \right) $ (16)

    其中Pκ-1(z)为z(k)的κ-1次任意复系数多项式.当κ=0时,P-1(z)≡0;当κ≤-1时,F(z)≡0.故

    $ \mathit{\Phi }\left( z \right) = Y\left( z \right)\mathit{\Psi }\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)Y\left( z \right)}}{{2{\rm{\pi }}i}}\int_{ - \infty }^{ + \infty } {\frac{{g\left( t \right)}}{{{Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} $ (17)

    Φ(z)在z=-i处有-κ阶极,为了消除极点,应使[1, 9, 15-16] Ψk(n)(-i)=0,n=0,1,2,…,-κ-1.即

    $ \int_{ - \infty }^{ + \infty } {\frac{{g\left( t \right){\rm{d}}t\left( k \right)}}{{{Y^ + }\left( t \right){{\left[ {\left( {t + i} \right)\left( k \right)} \right]}^n}}}} = 0\;\;\;\;\;n = 1,2, \cdots , - \kappa $ (18)

    于是有如下结论.

    定理5 若实数轴上R问题的指标为κ,则非齐次R问题在Φ(∞)=0的附加条件下,当κ≥0时,问题有一般解(16);当κ≤-1时,当且仅当g(t)满足-κ个条件(18) 时,R问题有唯一解(17).

    3 半平面内的Hilbert边值问题

    把满足条件:

    $ z\left( k \right) \cdot {z^ * }\left( k \right) = {\left| {z\left( k \right)} \right|^2} = {\left| {{z^ * }\left( k \right)} \right|^2} $

    的点zz*称为是关于XK-对称点[6-9].显然

    $ \begin{array}{*{20}{c}} {{z^ * } = z\left( { - 1} \right)}&{{{\left( {z\left( { - 1} \right)} \right)}^ * } = {z^ * }\left( { - 1} \right) = {z^{ * * }} = z}&{{x^ * } = x\left( {x \in \mathbb{R}} \right)} \end{array} $

    命题1 若zZ±,则${z^*} \in {Z^ \mp }$.

    定义3[6-9] 设Φ(z)在Z+(或Z-)内有定义,Φ(z)在Z-(或Z+)的K-对称扩张函数为:

    $ \mathit{\bar \Phi }\left( z \right) = \overline {\mathit{\Phi }\left( {z\left( { - 1} \right)} \right)} \;\;\;\;\;\;\;\;\;z \in {Z^ - }\left( {或\;{Z^ + }} \right) $

    显然,若Φ(z)在Z++Z-内有定义,则Φ(z)在Z-+Z+内也有定义.

    对于边界值(若存在),有

    $ \begin{array}{*{20}{c}} {{{\mathit{\bar \Phi }}^ - }\left( x \right) = \overline {{\mathit{\Phi }^ + }\left( x \right)} }&{{{\mathit{\bar \Phi }}^ + }\left( x \right) = \overline {{\mathit{\Phi }^ - }\left( x \right)} }&{x \in X} \end{array} $ (19)

    Z+内的Hilbert边值问题[9-11](简称H问题)Ⅰ:求一个在Z+内的K-解析,在Z++X上连续的函数Φ(z),使在X上满足边界值条件:

    $ {\mathop{\rm Re}\nolimits} \left\{ {\left[ {a\left( t \right) + ib\left( t \right)} \right]{\mathit{\Phi }^ + }\left( t \right)} \right\} = c\left( t \right)或\left( {a + ib} \right){\mathit{\Phi }^ + }\left( t \right) + \left( {a - ib} \right)\overline {{\mathit{\Phi }^ + }\left( t \right)} = 2c\;\;\;\;\;\;t \in X $ (20)

    其中$a\left(t \right), b\left(t \right), c\left(b \right) \in {{\hat H}_k}\left(X \right)$是给定的实函数,且a2(t)+b2(t)≠0于X上(包括t=∞).

    c(t)≡0时,此问题称为齐次H问题,当c(t)≠0时,此问题称为非齐次H问题.

    $ \mathit{\Omega }\left( z \right) = \left\{ \begin{array}{l} \mathit{\Phi }\left( z \right)\;\;\;\;\;\;\;z \in {Z^ + }\\ \mathit{\bar \Phi }\left( z \right)\;\;\;\;\;\;\;z \in {Z^ - } \end{array} \right. $

    利用(19) 式,改写(20) 式得Riemann边值问题[5, 9-14](简称R问题)Ⅱ:

    $ \left( {a + ib} \right){\mathit{\Omega }^ + }\left( t \right) + \left( {a - ib} \right){\mathit{\Omega }^ - }\left( t \right) = 2c\;或\;{\mathit{\Omega }^ + }\left( t \right) = G\left( t \right){\mathit{\Omega }^ - }\left( t \right) + g\left( t \right) $ (21)

    其中

    $ G\left( t \right) = - \left( {a - ib} \right)/\left( {a + ib} \right)\;\;\;\;\;g\left( t \right) = 2c/\left( {a + ib} \right) $

    于是H问题Ⅰ等价于在附加条件

    $ {\mathit{\Omega }^ - }\left( t \right) = \overline {{\mathit{\Omega }^ + }\left( t \right)} \;\;\;\;\;\;t \in X $ (22)

    下求R问题Ⅱ的有界解.

    命题2 若Ω(z)是R问题Ⅱ的有界解,则Ω(z),进而Ω0(z)=[Ω(z)+Ω(z)]/2皆是这样的解,且Ω0(z)满足条件(22).

    事实上,对(21) 式取共轭:

    $ \left( {a - ib} \right)\overline {{\mathit{\Omega }^ + }\left( t \right)} + \left( {a + ib} \right)\overline {{\mathit{\Omega }^ - }\left( t \right)} = 2c $

    $ {{\mathit{\bar \Omega }}^ + }\left( t \right) = \overline {{\mathit{\Omega }^ - }\left( t \right)} \;\;\;\;\;\;\;\;{{\mathit{\bar \Omega }}^ - }\left( t \right) = \overline {{\mathit{\Omega }^ + }\left( t \right)} $

    $ \begin{array}{*{20}{c}} {\left( {a + ib} \right)\overline {{\mathit{\Omega }^ + }\left( t \right)} + \left( {a - ib} \right)\overline {{\mathit{\Omega }^ - }\left( t \right)} = 2c}\\ {\mathit{\Omega }_0^ - \left( t \right) = \left[ {{\mathit{\Omega }^ - }\left( t \right) + {{\mathit{\bar \Omega }}^ - }\left( t \right)} \right]/2 = \left[ {\overline {{{\mathit{\bar \Omega }}^ + }\left( t \right)} + \overline {{\mathit{\Omega }^ + }\left( t \right)} } \right]/2 = \overline {\mathit{\Omega }_0^ + \left( t \right)} } \end{array} $

    $\kappa \triangleq {\pi ^{ -1}}{\text{sgn}}\left(k \right)\left[ {{\text{arg}}\left({a -ib} \right)} \right]_{ -\infty }^{ + \infty }$称为H问题Ⅰ(即R问题Ⅱ)的指标,因$a, b \in {{\hat H}_k}\left(X \right)$a(-∞)=a(+∞),b(-∞)=b(+∞),易得κ为一偶数.又|G(t)|=1,由(7) 式得

    $ \mathit{\Gamma }\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)}}{{2\pi }}\int_{ - \infty }^{ + \infty } {\frac{{\theta \left( t \right)}}{{\left( {t - z} \right)\left( k \right)}}{\rm{d}}t\left( k \right)} $

    其中

    $ \theta \left( t \right) = \arg \left\{ { - {{\left[ {\frac{{\left( {t + i} \right)\left( k \right)}}{{\left( {t - i} \right)\left( k \right)}}} \right]}^\kappa }\frac{{\left( {a - ib} \right)}}{{\left( {a + ib} \right)}}} \right\} = \arg \left\{ { - {{\left[ {\left( {t + i} \right)\left( k \right)} \right]}^{2\kappa }}{{\left( {a - ib} \right)}^2}} \right\} \in \mathbb{R} $

    $ \mathit{\bar \Gamma }\left( z \right) = \mathit{\Gamma }\left( z \right)\;\;\;\;\;\;{\mathit{\Gamma }^ - }\left( t \right) = \overline {{\mathit{\Gamma }^ + }\left( t \right)} $ (23)

    由于齐次H问题(c(t)≡0) 的相应问题是齐次R问题(g(t)≡0),对于后者由定理3知,当κ≥0时,其一般解为

    $ \mathit{\Omega }\left( z \right) = X\left( z \right)\left( {{C_0}{z^\kappa }\left( k \right) + {C_1}{z^{\kappa - 1}}\left( k \right) + \cdots + {C_\kappa }} \right) $ (24)

    其中C0C1,…,Cκ为任意复常数,X(z)由(8) 式给出.

    由(23) 式得

    $ \bar X\left( z \right) = X\left( z \right)\;\;\;\;\;\;\;\;\;\mathit{\bar \Omega }\left( z \right) = X\left( z \right)\left( {\overline {{C_0}} {z^\kappa }\left( k \right) + \overline {{C_1}} {z^{\kappa - 1}}\left( k \right) + \cdots + \overline {{C_\kappa }} } \right) $

    于是齐次H问题的一般解Φ(z)=[Ω(z)+Ω(z)]/2由(24) 给出,但其中系数Cj(0≤jκ)为任意实常数.当κ<0时,Ω(z)≡0,因此Φ(z)≡0.这样得如下结论.

    定理6 对于齐次H问题,若实数轴上H问题的指标为κ,则当κ≥0时,其一般解为Φ(z)=Pκ(z)X(z)(其中Pκ(z)为z(k)的κ次任意实系数多项式,X(z)由(8) 式给出);当κ<0时,Φ(z)≡0.

    对于非齐次H问题,只要求出其一个特解,再由上述定理便可得其一般解,下面求非齐次H问题的一个特解Φ0(z).

    κ≥0时,由(12) 式知

    $ {\mathit{\Omega }_0}\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)Y\left( z \right)}}{{\pi i}}\int_{ - \infty }^{ + \infty } {\frac{{c\left( t \right){\rm{d}}t\left( k \right)}}{{\left( {a + bi} \right){Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}} $

    R问题一个特解,其中Y(z)由(10) 式给出.注意到(23) 式,当zZ±时,

    $ \begin{array}{l} \bar Y\left( z \right) = \overline {Y\left( {z\left( { - 1} \right)} \right)} = {\left[ {\frac{{\left( {z - i} \right)\left( k \right)}}{{\left( {z + i} \right)\left( k \right)}}} \right]^\kappa }Y\left( z \right)\\ \overline {{Y^ + }\left( t \right)} = {{\bar Y}^ - }\left( t \right) = {\left[ {\frac{{\left( {t - i} \right)\left( k \right)}}{{\left( {t + i} \right)\left( k \right)}}} \right]^\kappa }{Y^ - }\left( t \right) \end{array} $

    $ {Y^ + }\left( t \right) = G\left( t \right){Y^ - }\left( t \right) $

    $ \overline {{Y^ + }\left( t \right)} = - {\left[ {\frac{{\left( {t - i} \right)\left( k \right)}}{{\left( {t + i} \right)\left( k \right)}}} \right]^\kappa }\frac{{\left( {a + ib} \right)}}{{\left( {a - ib} \right)}}{Y^ + }\left( t \right) $

    这样

    $ \overline {{\mathit{\Omega }_0}} \left( z \right) = {\left[ {\frac{{\left( {z - i} \right)\left( k \right)}}{{\left( {z + i} \right)\left( k \right)}}} \right]^\kappa }\frac{{{\mathop{\rm sgn}} \left( k \right)Y\left( z \right)}}{{\pi i}}\int_{ - \infty }^{ + \infty } {{{\left[ {\frac{{\left( {t + i} \right)\left( k \right)}}{{\left( {t - i} \right)\left( k \right)}}} \right]}^\kappa }\frac{{c\left( t \right){\rm{d}}t\left( k \right)}}{{\left( {a + bi} \right){Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}} $

    于是

    $ \begin{array}{l} {\mathit{\Phi }_0}\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)Y\left( z \right)}}{{2\pi i}}\left\{ {\int_{ - \infty }^{ + \infty } {\frac{{c\left( t \right){\rm{d}}t\left( k \right)}}{{\left( {a + bi} \right){Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}} + } } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\left. {{{\left[ {\frac{{\left( {z - i} \right)\left( k \right)}}{{\left( {z + i} \right)\left( k \right)}}} \right]}^\kappa }\int_{ - \infty }^{ + \infty } {{{\left[ {\frac{{\left( {t + i} \right)\left( k \right)}}{{\left( {t - i} \right)\left( k \right)}}} \right]}^\kappa }\frac{{c\left( t \right){\rm{d}}t\left( k \right)}}{{\left( {a + bi} \right){Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}} } \right\} \end{array} $ (25)

    κ≤-2时,由(14),(15) 式知当且仅当下列条件:

    $ \int_{ - \infty }^{ + \infty } {\frac{{c\left( t \right){\rm{d}}t\left( k \right)}}{{\left( {a + bi} \right){Y^ + }\left( t \right){{\left[ {\left( {t + i} \right)\left( k \right)} \right]}^n}}}} = 0\;\;\;\;\;\;\;n = 2,3, \cdots , - \kappa $ (26)

    满足时,R问题Ⅱ有唯一解:

    $ {\mathit{\Phi }_0}\left( z \right) = \frac{{{\mathop{\rm sgn}} \left( k \right)Y\left( z \right)}}{{\pi i}}\left[ {\int_{ - \infty }^{ + \infty } {\frac{{c\left( t \right){\rm{d}}t\left( k \right)}}{{\left( {a + bi} \right){Y^ + }\left( t \right)\left( {t - z} \right)\left( k \right)}}} + \int_{ - \infty }^{ + \infty } {\frac{{c\left( t \right){\rm{d}}t\left( k \right)}}{{\left( {a + bi} \right){Y^ + }\left( t \right)\left( {t + i} \right)\left( k \right)}}} } \right] $ (27)

    R问题Ⅱ解的唯一性,Φ0(z)=Φ0(z),从而(27) 式也就是非齐次H问题Ⅰ的唯一解.于是有以下结论.

    定理7 若实数轴上H问题的指标为κ,对于非齐次H问题,当κ≥0时,其一般解为

    $ {\mathit{\Phi }_0}\left( z \right) + {P_\kappa }\left( z \right)X\left( z \right) $

    其中Φ0(z)由(25) 式给出,Pκ(z)为z(k)的κ次任意实系数多项式;当κ<0时,当且仅当条件(26) 满足时有唯一解(27).

    4 结束语

    在前面部分,我们给出了无穷直线X上的${{\hat H}_k}$类函数、指标、Cauchy型K-积分及其边界值性质,利用它们分别得到无穷直线X上的Riemann边值问题有分片K-解析函数解的条件和解的表达式,即定理3~5.进一步利用K-对称扩张函数把Hilbert边值问题转化为无穷直线X上的Riemann边值问题,又得到了K-解析函数类F(D(k))中Hilbert边值问题的可解条件和解的表达式,即定理6~7.当k=±1时,K-解析函数[1-7, 15, 16]分别为解析函数[8-12]与共轭解析函数[13-14, 17],因此以上所得结果包含了(共轭)解析函数中的相应结论[9-14],它们是(共轭)解析函数理论的继续和应用.

    参考文献
    [1] 张建元. K-解析函数及其存在的条件[J]. 云南民族大学学报(自然科学版), 2007, 16(4): 298-302.
    [2] 张建元. K-共形映射[J]. 西南大学学报(自然科学版), 2010, 32(10): 119-125.
    [3] 张建元, 张毅敏, 姜锐武, 等. 复变函数的K-积分[J]. 云南师范大学学报(自然科学版), 2009, 29(1): 24-28.
    [4] 张毅敏, 张建元, 赵书芬. K-留数及其应用[J]. 云南师范大学学报(自然科学版), 2010, 30(2): 15-20.
    [5] 张建元, 赵书芬, 韩艳. K-解析函数的Riemann边值问题[J]. 应用数学和力学, 2014, 35(7): 805-814. DOI:10.3879/j.issn.1000-0887.2014.07.010
    [6] 张建元, 刘秀, 吴科. K-对称变换及其K保圆(周)性[J]. 西南民族大学学报(自然科学版), 2011, 37(2): 167-171.
    [7] 张建元. K-解析变换下曲线的转向[J]. 江西师范大学学报(自然科学版), 2011, 35(3): 292-296.
    [8] 钟玉泉. 复变函数论[M]. 3版. 北京: 高等教育出版社, 2004.
    [9] 路见可. 解析函数边值问题教程[M]. 武汉: 武汉大学出版社, 2009.
    [10] 赵桢. 奇异积分方程[M]. 北京: 北京师范大学出版社, 1984.
    [11] 侯宗义, 李明忠, 张万国. 奇异积分方程论及其应用[M]. 上海: 上海科学技术出版社, 1990.
    [12] 陈方权, 蒋绍惠. 解析函数论基础[M]. 北京: 北京师范大学出版社, 2008.
    [13] 张建元. 共轭解析函数的Riemann边值问题[J]. 北京工业大学学报, 1996, 22(3): 99-106.
    [14] 张建元. 一类复调和函数的Riemann边值问题[J]. 昭通师专学报, 1997, 19(2): 1-7.
    [15] 张建元, 张毅敏, 刘承萍, 等. K-解析函数的幂级数展开式[J]. 大理学院学报(综合版), 2009, 8(4): 14-18.
    [16] 张建元, 张毅敏, 熊绍武. K-解析函数的双边幂级数与孤立奇点[J]. 云南民族大学学报(自然科学版), 2009, 18(3): 198-201.
    [17] 王见定. 半解析函数、共轭解析函数[M]. 北京: 北京工业大学出版社, 1988.
    Riemann Boundary Value Problem and Hilbert Boundary Value Problem for K-Analytic Function on an Infinite Straight Line
    ZHANG Jian-yuan1, HAN Yan1, ZHANG Yi-min1, ZHAO Shu-fen1, LIU Cheng-ping1, ZHANG xin2     
    1. School of Mathematics and Statistics, Zhaotong University, Zhaotong Yunnan 657000, China;
    2. Zhaotong Bureau of Statistics, Zhaotong Yunnan 657000, China
    Abstract: In this paper, we first introduce the concept of K-analytic function of Cauchy type K-integral on an infinite straight line (fragmentation) and use the K-symmetry transformation method to study some properties of the Cauchy type K-integral. Then, with the help of the index that functions on the infinite straight line and the properties of the Cauchy type K-integral, we obtain the solvable conditions and its expression of Riemann boundary value problem of the K-analytic function on the infinite straight line as well as the relationship between them and the index. Finally, we use the K-symmetry expansion function in a half plane to transform the Hilbert boundary value problem into Riemann boundary value problems on the infinite straight line X, thus obtaining the solvable conditions and its expression of the Hilbert boundary value problem. Both the analytic function and the conjugate analytic function are special cases of the K-analytic function. The results obtained in this paper generalize the analytic function and the conjugate analytic function in the corresponding conclusions.
    Key words: ${{\hat H}_k}$-function    Cauchy type K-integral on a straight line    (Piecewise) K-analytic function    K-symmetry transformation    K-symmetry expansion function    boundary value formula    Riemann boundary value problem    Hilbert boundary value problem    index    
    X