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Helmert扰动位及其积分核函数的椭球实用公式

魏子卿 杨正辉

魏子卿, 杨正辉. Helmert扰动位及其积分核函数的椭球实用公式[J]. 武汉大学学报 ● 信息科学版, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327
引用本文: 魏子卿, 杨正辉. Helmert扰动位及其积分核函数的椭球实用公式[J]. 武汉大学学报 ● 信息科学版, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327
WEI Ziqing, YANG Zhenghui. Helmert Disturbing Potential and Its Integral Kernel Function with Ellipsoidal Harmonic Formula[J]. Geomatics and Information Science of Wuhan University, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327
Citation: WEI Ziqing, YANG Zhenghui. Helmert Disturbing Potential and Its Integral Kernel Function with Ellipsoidal Harmonic Formula[J]. Geomatics and Information Science of Wuhan University, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327

Helmert扰动位及其积分核函数的椭球实用公式

doi: 10.13203/j.whugis20180327
基金项目: 

国家自然科学基金 41674025

详细信息
    作者简介:

    魏子卿, 教授, 中国工程院院士, 主要从事大地边值问题的理论和方法研究。ziqingw@sina.com

    通讯作者: 杨正辉, 博士生。yangzhenghui.1986@163.com
  • 中图分类号: P223

Helmert Disturbing Potential and Its Integral Kernel Function with Ellipsoidal Harmonic Formula

Funds: 

The National Natural Science Foundation of China 41674025

More Information
    Author Bio:

    WEI Ziqing, professor, Academician of Chinese Academy of Engineering, mainly engages in the theory and methods of boundary value problem of geodesy. E-mail: ziqingw@sina.com

    Corresponding author: YANG Zhenghui, PhD candidate. E-mail: yangzhenghui.1986@163.com
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  • 被引次数: 0
出版历程
  • 收稿日期:  2018-08-30
  • 刊出日期:  2018-12-05

Helmert扰动位及其积分核函数的椭球实用公式

doi: 10.13203/j.whugis20180327
    基金项目:

    国家自然科学基金 41674025

    作者简介:

    魏子卿, 教授, 中国工程院院士, 主要从事大地边值问题的理论和方法研究。ziqingw@sina.com

    通讯作者: 杨正辉, 博士生。yangzhenghui.1986@163.com
  • 中图分类号: P223

摘要: 借助以地心参考椭球面为边界面的第二大地边值问题的理论,基于Helmert空间的Neumann边值条件,给定Helmert扰动位的椭球解表达式,并详细推导第二类勒让德函数及其导数的递推关系、Helmert扰动位函数的椭球积分解以及类椭球Hotine积分核函数的实用计算公式,便于后续椭球域第二大地边值问题的实际研究。

English Abstract

魏子卿, 杨正辉. Helmert扰动位及其积分核函数的椭球实用公式[J]. 武汉大学学报 ● 信息科学版, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327
引用本文: 魏子卿, 杨正辉. Helmert扰动位及其积分核函数的椭球实用公式[J]. 武汉大学学报 ● 信息科学版, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327
WEI Ziqing, YANG Zhenghui. Helmert Disturbing Potential and Its Integral Kernel Function with Ellipsoidal Harmonic Formula[J]. Geomatics and Information Science of Wuhan University, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327
Citation: WEI Ziqing, YANG Zhenghui. Helmert Disturbing Potential and Its Integral Kernel Function with Ellipsoidal Harmonic Formula[J]. Geomatics and Information Science of Wuhan University, 2018, 43(12): 1768-1774. doi: 10.13203/j.whugis20180327
  • 在地球重力场问题中,常用地球扰动位(地球实际引力位减去正常引力位)模型计算重力场元(大地水准面起伏、垂线偏差、重力扰动等)[1-2]。经典理论解决大地边值问题的思路都是以地面重力或者重力异常为边界值,用球近似加上一定的椭球改正进行边值问题的近似求解。因此,为满足厘米级精度的大地水准面要求,就必须考虑是否依旧在球近似下引入椭球改正,还是用椭球谐解椭球面大地边值问题[3-7]

    本文在文献[5]提出的以地心参考椭球面为边界面的第二大地边值问题的理论基础上,首先给定Helmert空间Neumann边值问题的椭球解表达式,然后继续推导第二类勒让德函数及其导数的计算公式,最后给定Helmert扰动位函数的椭球积分解及其类椭球Hotine积分核函数的实用公式。

    • 在椭球域坐标系下,作为Neumann外问题求解的第一步,是在Helmert空间求定扰动位Th,使得[5]

      $$ \Delta {T^h}\left( P \right) = 0,u > b $$ (1)
      $$ \frac{{\partial {T^h}\left( P \right)}}{{\partial u}} = - {\rm{ \mathsf{ δ} }}{g^{h * }},u = b $$ (2)
      $$ {T^h}\left( P \right) = O\left( {{u^{ - 3}}} \right),u \to \infty $$ (3)

      式(2)中, δgh*为参考椭球边界面上的Helmert扰动重力。

      Laplace方程(1)的椭球谐级数解可表示为[4]

      $$ {T^h}\left( {u,\mathit{\Omega }} \right) = \sum\limits_{n = 1}^\infty {\sum\limits_{m = - n}^n {T_{nm}^h\frac{{{Q_{nm}}\left( {i\frac{u}{E}} \right)}}{{{Q_{nm}}\left( {i\frac{b}{E}} \right)}}{Y_{nm}}\left( \mathit{\Omega } \right)} } $$ (4)

      式中,${Q_{nm}}\left( {i\frac{u}{E}} \right)$为第二类勒让德函数;Tnmh为Helmert扰动位函数的位系数。

      将式(4)代入边界条件式(2)中,得:

      $$ \begin{array}{*{20}{c}} {\sum\limits_{n = 1}^\infty {\sum\limits_{m = - n}^n {\frac{1}{{{Q_{nm}}\left( {i\frac{b}{E}} \right)}}\frac{{{\rm{d}}{Q_{nm}}\left( {i\frac{u}{E}} \right)}}{{{\rm{d}}u}}\left| {_{u = b}} \right.T_{nm}^h{Y_{nm}}\left( \mathit{\Omega } \right)} } = }\\ { - {\rm{ \mathsf{ δ} }}{g^{h * }}\left( \mathit{\Omega } \right)} \end{array} $$ (5)

      将参考椭球边界面上的Helmert扰动重力δgh*(Ω)展开为面球谐函数级数和的形式:

      $$ \begin{array}{*{20}{c}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( \mathit{\Omega } \right) = }\\ {\sum\limits_{n = 1}^\infty {\sum\limits_{m = - n}^n {\int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( {\mathit{\Omega '}} \right){{Y'}_{nm}}\left( {\mathit{\Omega '}} \right){\rm{d}}\mathit{\Omega '}{{\rm{Y}}_{{\rm{nm}}}}\left( \mathit{\Omega } \right)} } } } \end{array} $$ (6)

      将式(6)代入式(5)中,整理得到Helmert扰动位系数Tnmh为:

      $$ T_{nm}^h = \frac{{\int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( \mathit{\Omega } \right){{Y'}_{nm}}\left( {\mathit{\Omega '}} \right){\rm{d}}\mathit{\Omega '}} }}{{\frac{1}{{{Q_{nm}}\left( {i\frac{b}{E}} \right)}}\frac{{{\rm{d}}{Q_{nm}}\left( {i\frac{u}{E}} \right)}}{{{\rm{d}}u}}\left| {_{u = b}} \right.}} $$ (7)

      将式(7)代入式(4),可得到Neumann边值问题(1)~(3)的椭球解表达式:

      $$ \begin{array}{*{20}{c}} {{T^h}\left( {u,\mathit{\Omega }} \right) = \int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( {\mathit{\Omega '}} \right)} \cdot }\\ {\sum\limits_{n = 1}^\infty {\sum\limits_{m = - n}^n {{\alpha _{nm}}\left( u \right){{Y'}_{nm}}\left( {\mathit{\Omega '}} \right){Y_{nm}}\left( \mathit{\Omega } \right){\rm{d}}\mathit{\Omega '}} } } \end{array} $$ (8)

      式中,

      $$ {\alpha _{nm}}\left( u \right) = {Q_{nm}}\left( {i\frac{u}{E}} \right)/\left[ {\frac{{{\rm{d}}{Q_{nm}}\left( {i\frac{u}{E}} \right)}}{{{\rm{d}}u}}\left| {_{u = b}} \right.} \right] $$ (9)
    • 式(9)中,系数αnm(u)由第二类勒让德函数${Q_{nm}}\left( {i\frac{u}{E}} \right)$及其导数${\rm{d}}{Q_{nm}}\left( {i\frac{u}{E}} \right)/{\rm{d}}u$组成,故这里进行相应部分的公式推导计算。

      对于|z|>1,第二类勒让德函数定义为[8]

      $$ \begin{array}{*{20}{c}} {{Q_{nm}}\left( z \right) = \frac{{\sqrt {\rm{ \mathsf{ π} }} {{\rm{e}}^{{\rm{i}}m\pi }}}}{{{2^{n + 1}}}}\frac{{\mathit{\Gamma }\left( {n + m + 1} \right)}}{{\mathit{\Gamma }\left( {n + \frac{3}{2}} \right)}}{{\left( {{z^2} - 1} \right)}^{\frac{m}{2}}} \cdot }\\ {{z^{ - \left( {n + m + 1} \right)}}{{\left( {1 - \frac{1}{{{z^2}}}} \right)}^{ - \frac{{n + m + 1}}{2}}} \cdot }\\ {F\left( {\frac{{n + m + 1}}{2},\frac{{n - m + 1}}{2},\frac{{2n + 3}}{2};\frac{1}{{1 - {z^2}}}} \right)} \end{array} $$ (10)

      式中,Γ(z)为伽玛函数;F(a, b, c; z)为超几何函数;eimπ=(-1)m;

      $$ \mathit{\Gamma }\left( {n + m + 1} \right) = \left( {n + m} \right)! $$ (11)
      $$ \mathit{\Gamma }\left( {\frac{{2n + 3}}{2}} \right) = \frac{{\left( {2n + m} \right)!!}}{{{2^{n + 1}}}}\sqrt {\rm{ \mathsf{ π} }} $$ (12)

      则式(10)可改写为:

      $$ \begin{array}{*{20}{c}} {{Q_{nm}}\left( z \right) = {{\left( { - 1} \right)}^m}\frac{{\left( {n + m} \right)!}}{{\left( {2n + 1} \right)!!}}{{\left( {{z^2} - 1} \right)}^{ - \frac{{m + 1}}{2}}} \cdot }\\ {F\left( {\frac{{n + m + 1}}{2},\frac{{n - m + 1}}{2},\frac{{2n + 3}}{2};\frac{1}{{1 - {z^2}}}} \right)} \end{array} $$ (13)

      令$z = i\frac{u}{E}$,则1-z2=1/e2,$e = E/\sqrt {{u^2} + {E^2}} $为共焦椭球的第一偏心率, ${e_0} = E/\sqrt {{b^2} + {E^2}} $为参考椭球的第一偏心率,则式(13)可变为:

      $$ \begin{array}{*{20}{c}} {{Q_{nm}}\left( {i\frac{u}{E}} \right) = {{\left( { - 1} \right)}^{m - \left( {n + 1} \right)/2}}\frac{{\left( {n + m} \right)!}}{{\left( {2n + 1} \right)!!}}{e^{n + 1}} \cdot }\\ {F\left( {\frac{{n + m + 1}}{2},\frac{{n - m + 1}}{2},\frac{{2n + 3}}{2};{e^2}} \right)} \end{array} $$ (14)

      利用F(a, b, c; z)超几何函数可以展开为高斯超几何级数和的形式[9]

      $$ \begin{array}{*{20}{c}} {F\left( {a,b,c;z} \right) = }\\ {\frac{{\mathit{\Gamma }\left( c \right)}}{{\mathit{\Gamma }\left( a \right)\mathit{\Gamma }\left( b \right)}}\sum\limits_{k = 0}^\infty {\frac{{\mathit{\Gamma }\left( {a + k} \right)\mathit{\Gamma }\left( {b + k} \right)}}{{\mathit{\Gamma }\left( {c + k} \right)}}\frac{{{z^k}}}{{k!}}} } \end{array} $$ (15)

      则第二类勒让德函数${Q_{nm}}\left( {i\frac{u}{E}} \right)$的递推关系表达式为:

      $$ {Q_{nm}}\left( {i\frac{u}{E}} \right) = {\left( { - 1} \right)^{m - \frac{{n + 1}}{2}}}\frac{{\left( {n + m} \right)!{e^{n + 1}}}}{{\left( {2n + 1} \right)!!}}\sum\limits_{k = 0}^\infty {{\alpha _{nmk}}{e^{2k}}} $$ (16)

      式中,

      $$ \begin{array}{*{20}{c}} {{\alpha _{nmk}} = \frac{1}{{{2^k}k!}}\frac{{\prod\limits_{l = 0}^{k - 1} {\left[ {{{\left( {n + 2l + 1} \right)}^2} - {m^2}} \right]} }}{{\prod\limits_{l = 0}^{k - 1} {\left( {2n + 2l + 3} \right)} }} = }\\ {\frac{{{{\left( {n + 2k + 1} \right)}^2} - {m^2}}}{{2k\left( {2n + 2k + 3} \right)}}{a_{nm,k - 1}},k \ge 1} \end{array} $$ (17)

      其中,

      $$ {\alpha _{nm0}} = 1 $$ (18)
      $$ {\alpha _{nm1}} = \frac{{{{\left( {n + 1} \right)}^2} - {m^2}}}{{2\left( {2n + 3} \right)}} $$ (19)
      $$ {\alpha _{nm2}} = \frac{{{{\left( {n + 3} \right)}^2} - {m^2}}}{{4\left( {2n + 5} \right)}}{\alpha _{nm1}} $$ (20)

      在式(16)中, ${Q_{nm}}\left( {i\frac{u}{E}} \right)$对u进行求导得:

      $$ \begin{array}{*{20}{c}} {\frac{{{\rm{d}}{Q_{nm}}\left( {i\frac{u}{E}} \right)}}{{{\rm{d}}u}} = {{\left( { - 1} \right)}^{m - \left( {n + 1} \right)/2}}\frac{{\left( {n + m} \right)!}}{{\left( {2n + 1} \right)!!}} \cdot }\\ {{e^n}\sum\limits_{k = 0}^\infty {\left( {2k + n + 1} \right){\alpha _{nmk}}{e^{2k}}\frac{{{\rm{d}}e}}{{{\rm{d}}u}}} } \end{array} $$ (21)
      $$ \frac{{{\rm{d}}e}}{{{\rm{d}}u}} = - \left( {1 - {e^2}} \right)\frac{e}{u} $$ (22)

      将式(22)代入式(21),得到第二类勒让德函数导数的递推关系:

      $$ \begin{array}{*{20}{c}} {\frac{{{\rm{d}}{Q_{nm}}\left( {i\frac{u}{E}} \right)}}{{{\rm{d}}u}} = {{\left( { - 1} \right)}^{m - \left( {n + 1} \right)/2}}\frac{{\left( {n + m} \right)!}}{{\left( {2n + 1} \right)!!}} \cdot }\\ {{{\left( {1 - e} \right)}^2}\frac{{{e^{n + 1}}}}{u}\sum\limits_{k = 0}^\infty {\left( { - 2k - n - 1} \right){\alpha _{nmk}}{e^{2k}}} } \end{array} $$ (23)
      $$ \begin{array}{*{20}{c}} {\frac{{{\rm{d}}{Q_{nm}}\left( {i\frac{u}{E}} \right)}}{{{\rm{d}}u}}\left| {_{u = b}} \right. = {{\left( { - 1} \right)}^{m - \left( {n + 1} \right)/2}}\frac{{\left( {n + m} \right)!}}{{\left( {2n + 1} \right)!!}} \cdot }\\ {{{\left( {1 - {e_0}} \right)}^2}\frac{{e_0^{n + 1}}}{b}\sum\limits_{k = 0}^\infty {\left( { - 2k - n - 1} \right){\alpha _{nmk}}e_0^{2k}} } \end{array} $$ (24)
    • 将式(16)和式(24)结合式(17)~(20),得:

      $$ \begin{array}{*{20}{c}} {{a_{nm}}\left( u \right) = }\\ {b\frac{{{e^{n + 1}}\sum\limits_{k = 0}^\infty {{\alpha _{nmk}}{e^{2k}}} }}{{e_0^{n + 1}\sum\limits_{k = 0}^\infty {{\alpha _{nmk}}{{\left( {1 - {e_0}} \right)}^2}\left( { - 2k - n - 1} \right)e_0^{2k}} }}} \end{array} $$ (25)

      为了满足适定性和类椭球Hotine核函数的表达形式[10-11],对式(25)进行改进:

      $$ \begin{array}{*{20}{c}} {{a_{nm}}\left( u \right) = \frac{b}{{n + 1}}{{\left( {\frac{e}{{{e_0}}}} \right)}^{n + 1}} \cdot }\\ {\left[ {1 - \frac{{\sum\limits_{k = 1}^\infty {\left[ {\frac{{2k}}{{n + 1}}{\alpha _{nmk}} - \left( {1 + \frac{{2k}}{{n + 1}}} \right){\alpha _{nm,k - 1}}} \right]{e^{2k}}} }}{{1 + \sum\limits_{k = 1}^\infty {\left( {1 + \frac{{2k}}{{n + 1}}} \right)\left( {{\alpha _{nmk}} - {\alpha _{nm,k - 1}}} \right)e_0^{2k}} }}} \right]} \end{array} $$ (26)

      将式(26)中第二项展开为关于e2项的级数形式为:

      $$ \begin{array}{*{20}{c}} {\frac{{\sum\limits_{k = 1}^\infty {\left[ {\frac{{2k}}{{n + 1}}{\alpha _{nmk}} - \left( {1 + \frac{{2k}}{{n + 1}}} \right){\alpha _{nm,k - 1}}} \right]{e^{2k}}} }}{{1 + \sum\limits_{k = 1}^\infty {\left( {1 + \frac{{2k}}{{n + 1}}} \right)\left( {{\alpha _{nmk}} - {\alpha _{nm,k - 1}}} \right)e_0^{2k}} }} = }\\ {\left( {\frac{2}{{n + 1}}{\alpha _{nm1}} - \frac{{n + 3}}{{n + 1}}} \right){e^2} + O\left( {{e^4}} \right)} \end{array} $$ (27)

      将式(27)代入式(25),舍弃O(e4)以上的高阶项,得到:

      $$ \begin{array}{*{20}{c}} {{\alpha _{nm}}\left( u \right) = }\\ {\frac{b}{{n + 1}}{{\left( {\frac{e}{{{e_0}}}} \right)}^{n + 1}}\left[ {1 - \left( {\frac{2}{{n + 1}}{\alpha _{nm1}} - \frac{{n + 3}}{{n + 1}}} \right){e^2}} \right]} \end{array} $$ (28)

      将式(28)代入式(8),得到Helmert扰动位函数的椭球解:

      $$ \begin{array}{*{20}{c}} {{T^h}\left( {u,\mathit{\Omega }} \right) = \int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( {\mathit{\Omega '}} \right)} \sum\limits_{n = 1}^\infty {\sum\limits_{m = - n}^n {\frac{b}{{n + 1}}{{\left( {\frac{e}{{{e_0}}}} \right)}^{n + 1}} \cdot } } }\\ {\left[ {1 - \left( {\frac{{2{\alpha _{nm1}}}}{{n + 1}} - \frac{{n + 3}}{{n + 1}}} \right){e^2}} \right]{{Y'}_{nm}}\left( {\mathit{\Omega '}} \right){Y_{nm}}\left( \mathit{\Omega } \right){\rm{d}}\mathit{\Omega '}} \end{array} $$ (29)

      代入如下关系式[7]

      $$ {P_n}\left( {\cos \psi } \right) = \frac{{4{\rm{ \mathsf{ π} }}}}{{2n + 1}}\sum\limits_{m = - n}^n {{{Y'}_{nm}}\left( {\mathit{\Omega '}} \right){Y_{nm}}\left( \mathit{\Omega } \right)} $$ (30)

      式中,Ynm(Ω′)和Ynm(Ω)为完全正常化球谐函数,则式(29)转换为:

      $$ \begin{array}{*{20}{c}} {{T^h}\left( {u,\mathit{\Omega }} \right) = }\\ {\frac{b}{{4{\rm{ \mathsf{ π} }}}}\sum\limits_{n = 1}^\infty {{{\left( {\frac{e}{{{e_0}}}} \right)}^{n + 1}}\frac{{2n + 1}}{{n + 1}}\left[ {1 - \frac{{2{\alpha _{nm1}} - \left( {n + 3} \right)}}{{n + 1}}{e^2}} \right]} \cdot }\\ {\int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( {\mathit{\Omega '}} \right){P_n}\left( {\cos \psi } \right){\rm{d}}\mathit{\Omega '}} = }\\ {\frac{b}{{4{\rm{ \mathsf{ π} }}}}\int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( {\mathit{\Omega '}} \right)} \sum\limits_{n = 1}^\infty {{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}} \frac{{2n + 1}}{{n + 1}} \cdot }\\ {{P_n}\left( {\cos \psi } \right){\rm{d}}\mathit{\Omega '} = \frac{b}{{4{\rm{ \mathsf{ π} }}}}\int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( {\mathit{\Omega '}} \right)} \cdot }\\ {\sum\limits_{n = 1}^\infty {{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}\frac{{2n + 1}}{{n + 1}}\left( {\frac{2}{{n + 1}}{\alpha _{nm1}} - \frac{{n + 3}}{{n + 1}}} \right)} \cdot }\\ {{e^2}{P_n}\left( {\cos \psi } \right){\rm{d}}\mathit{\Omega '}} \end{array} $$ (31)

      在式(31)中,引入记号H(u, ψ)和K(u, ψ),则式(31)变为:

      $$ \begin{array}{*{20}{c}} {{T^h}\left( {u,\mathit{\Omega }} \right) = \frac{b}{{4{\rm{ \mathsf{ π} }}}}\int_{{\mathit{\Omega }_0}} {{\rm{ \mathsf{ δ} }}{g^{h * }}\left( {\mathit{\Omega '}} \right)} \cdot }\\ {\sum\limits_{n = 1}^\infty {\left[ {H\left( {u,\psi } \right) - {e^2}K\left( {u,\psi } \right)} \right]{\rm{d}}\mathit{\Omega '}} } \end{array} $$ (32)

      式中,

      $$ H\left( {u,\psi } \right) = \sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{n + 1}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}{P_n}\left( {\cos \psi } \right)} $$ (33)
      $$ \begin{array}{*{20}{c}} {H\left( {u,\psi } \right) = \sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{n + 1}}\left[ {\frac{{{{\left( {n + 1} \right)}^2} - {m^2}}}{{2n + 3}} - } \right.} }\\ {\left. {\left( {n + 3} \right)} \right]{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}{P_n}\left( {\cos \psi } \right)} \end{array} $$ (34)
    • 以下对于椭球Hotine积分核函数H(u, ψ)-e2K(u, ψ)公式进行推导,舍弃O(e2)以上的高阶项数。推导类椭球Hotine积分核函数部分中的H(u, ψ)函数。

      利用等式:

      $$ \frac{{2n + 1}}{{n + 1}} = 2 - \frac{1}{{n + 1}} $$ (35)

      并记$x = \frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}$,则式(33)可写成:

      $$ \begin{array}{*{20}{c}} {H\left( {b/x,\psi } \right) = }\\ {2x\sum\limits_{n = 1}^\infty {{x^n}{P_n}\left( {\cos \psi } \right)} - \sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{n + 1}}{P_n}\left( {\cos \psi } \right)} } \end{array} $$ (36)

      根据勒让德多项式及其生成函数之间的关系[12-13],有:

      $$ \begin{array}{*{20}{c}} {\sum\limits_{n = 1}^\infty {{x^n}{P_n}\left( {\cos \psi } \right)} = {{\left( {1 - 2x\cos \psi + {x^2}} \right)}^{ - 1/2}} - 1,}\\ {且\; - 1 \le x \le 1} \end{array} $$ (37)

      式(37)两端在0与x之间对x进行积分,得:

      $$ \begin{array}{*{20}{c}} {\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{n + 1}}{P_n}\left( {\cos \psi } \right)} = \ln \left[ {{{\left( {1 - 2x\cos \psi + {x^2}} \right)}^{ - 1/2}} + } \right.}\\ {\left. {x - \cos \psi } \right] + \ln \left( {1 - \cos \psi } \right) - x} \end{array} $$ (38)

      将式(37)、式(38)代入式(36),得:

      $$ \begin{array}{*{20}{c}} {H\left( {b/x,\psi } \right) = 2x{{\left( {1 - 2x\cos \psi + {x^2}} \right)}^{ - 1/2}} - }\\ {\ln \left[ {{{\left( {1 - 2x\cos \psi + {x^2}} \right)}^{ - 1/2}} + x - \cos \psi } \right] + }\\ {\ln \left( {1 - \cos \psi } \right) - x} \end{array} $$ (39)

      将变量x还原,有核函数H(u, ψ)为:

      $$ \begin{array}{*{20}{c}} {H\left( {u,\psi } \right) = 2\frac{{\sqrt {{b^2} + {E^2}} }}{l} - \frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }} - }\\ {\ln \left( {\frac{{l + \sqrt {{b^2} + {E^2}} - \sqrt {{u^2} + {E^2}} \cos \psi }}{{\sqrt {{u^2} + {E^2}} \left( {1 - \cos \psi } \right)}}} \right)} \end{array} $$ (40)

      式中,

      $$ l = {\left( {{b^2} + {u^2} - 2\sqrt {{b^2} + {E^2}} \sqrt {{u^2} + {E^2}} \cos \psi } \right)^{1/2}} $$

      式(40)适用于边界面外部和边界面上的任意一点。

      对于参考椭球边界面上的点(u=b),有:

      $$ \begin{array}{l} l\left| {_{u = b}} \right. = {l_0} = 2\sqrt {{b^2} + {E^2}} \sin \left( {\psi /2} \right)\\ H\left( {b,\psi } \right) = \csc \frac{\psi }{2} - \ln \left( {1 + \csc \frac{\psi }{2}} \right) - 1 \end{array} $$ (41)

      顾及式(37)、式(38)及$x = \frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}$,得:

      $$ \begin{array}{*{20}{c}} {H\left( {u,\psi } \right) = \sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{{\left( {n + 1} \right)}^2}}}\left[ {\frac{{{{\left( {n + 1} \right)}^2} - {m^2}}}{{2n + 3}} - } \right.} }\\ {\left. {\left( {n + 3} \right)} \right]{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}{P_n}\left( {\cos \psi } \right) = - }\\ {\sum\limits_{n = 1}^\infty {{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} + \sum\limits_{n = 1}^\infty {\frac{2}{{2n + 3}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} - }\\ {\sum\limits_{n = 1}^\infty {\frac{{\left( {2n + 1} \right){m^2}}}{{{{\left( {n + 1} \right)}^2}\left( {2n + 3} \right)}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} - }\\ {\sum\limits_{n = 1}^\infty {\frac{{3n + 1}}{{{{\left( {n + 1} \right)}^2}}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} } \end{array} $$ (42)

      式(42)由4项基本项组成,第一项计算式为:

      $$ \begin{array}{*{20}{c}} { - \sum\limits_{n = 1}^\infty {{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} = - x\left( {1 - 2x\cos \psi + } \right.}\\ {{{\left. {{x^2}} \right)}^{ - 1/2}} = \frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }} - \frac{{\sqrt {{b^2} + {E^2}} }}{l}} \end{array} $$ (43)

      又有:

      $$ \begin{array}{*{20}{c}} {\frac{2}{{2n + 3}} = \frac{1}{2}\left( {\frac{1}{{n + 1}} + \frac{1}{{n + 2}}} \right) - }\\ {\frac{1}{{2\left( {n + 1} \right)\left( {n + 2} \right)\left( {2n + 3} \right)}}} \end{array} $$

      其中,$\frac{1}{{2\left( {n + 1} \right)\left( {n + 2} \right)\left( {2n + 3} \right)}} \le {e^2}$,所以可以忽略O(e2)这一项。

      第二项计算式为:

      $$ \begin{array}{*{20}{c}} {\sum\limits_{n = 1}^\infty {\frac{2}{{2n + 3}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} = \sum\limits_{n = 1}^\infty {\frac{1}{2}\left( {\frac{1}{{n + 1}} + \frac{1}{{n + 2}}} \right){x^{n + 1}}{P_n}\left( {\cos \psi } \right)} = \frac{{\sqrt {{b^2} + {E^2}} + 3\sqrt {{u^2} + {E^2}} \cos \psi }}{{\sqrt {{b^2} + {E^2}} l}} + }\\ {\frac{{\sqrt {{b^2} + {E^2}} + 3\sqrt {{u^2} + {E^2}} \cos \psi }}{{2\sqrt {{b^2} + {E^2}} }}\ln \frac{{\sqrt {{u^2} + {E^2}} \left( {1 - \cos \psi } \right)}}{{l + \sqrt {{b^2} + {E^2}} - \sqrt {{u^2} + {E^2}} \cos \psi }} + \frac{{3\sqrt {{u^2} + {E^2}} l - 3{u^2} - {E^2} - {b^2}}}{{2{b^2}l}}} \end{array} $$ (44)

      由于,

      $$ \begin{array}{*{20}{c}} {\frac{{\left( {2n + 1} \right){m^2}}}{{{{\left( {n + 1} \right)}^2}\left( {2n + 3} \right)}} = \frac{1}{2}\left[ {\frac{{{m^2}}}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} - } \right.}\\ {\frac{{{m^2}}}{{4n\left( {n + 1} \right)\left( {n + 3} \right)}} + \frac{{{m^2}}}{{4\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} - }\\ {\frac{{3{m^2}}}{{2n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {2n + 3} \right)}} - }\\ {\left. {\frac{{2{m^2}}}{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)\left( {2n + 3} \right)}}} \right]} \end{array} $$ (45)

      mn时,

      $$ \begin{array}{*{20}{c}} {\frac{{3{m^2}}}{{2n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {2n + 3} \right)}} \le }\\ {\frac{{3{n^2}}}{{2n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {2n + 3} \right)}} \le {e^2}}\\ {\frac{{2{m^2}}}{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)\left( {2n + 3} \right)}} \le }\\ {\frac{{2{n^2}}}{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)\left( {2n + 3} \right)}} \le {e^2}} \end{array} $$

      O(e2)量级下,忽略式(45)中的后两项,第三项计算式为:

      $$ \begin{array}{*{20}{c}} {\sum\limits_{n = 1}^\infty {\frac{{\left( {2n + 1} \right){m^2}}}{{{{\left( {n + 1} \right)}^2}\left( {2n + 3} \right)}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} = \sum\limits_{n = 1}^\infty {\frac{1}{2}\left[ {\frac{{{m^2}}}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} - \frac{{{m^2}}}{{4n\left( {n + 1} \right)\left( {n + 3} \right)}} + } \right.} }\\ {\left. {\frac{{{m^2}}}{{4\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}} \right]{x^{n + 1}}{P_n}\left( {\cos \psi } \right) = \sum\limits_{n = 1}^\infty {\frac{1}{{48}}\left[ {\left( {\frac{{14}}{n} + \frac{{24}}{{n + 1}} - \frac{{30}}{{n + 2}} - \frac{5}{{n + 3}} + \frac{3}{{n + 1}}} \right) \cdot } \right.} }\\ {\left. {C{x^{n + 1}}\frac{{{\rm{d}}{P_n}\left( {\cos \psi } \right)}}{{{\rm{d}}\psi }} - \left( {48 - \frac{{60}}{{n + 2}} - \frac{{30}}{{n + 3}} - \frac{6}{{n - 1}}} \right)} \right]D{x^{n + 1}}{P_n}\left( {\cos \psi } \right) = }\\ {\frac{C}{{48}}\left( {14{K_8} + 24{K_9} - 30{K_{10}} - 5{K_{11}} - 3{K_{13}}} \right) - \frac{D}{8}\left( {\frac{8}{l} - 10{K_3} - 5{K_2} - {K_1}} \right)} \end{array} $$ (46)

      其中,

      $$ \begin{array}{*{20}{c}} {C = \frac{{2\cos \psi {{\left[ {\sin \theta \sin \theta '\sin \left( {\lambda - \lambda '} \right)} \right]}^2}}}{{{{\sin }^2}\psi }} + }\\ {\sin \theta \sin \theta '\cos \left( {\lambda - \lambda '} \right)} \end{array} $$
      $$ D = - \frac{{{{\left[ {\sin \theta \sin \theta '\sin \left( {\lambda - \lambda '} \right)} \right]}^2}}}{{{{\sin }^2}\psi }} $$
      $$ \begin{array}{*{20}{c}} {{K_1} = \sum\limits_{n = 1}^\infty {\frac{1}{{n - 1}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}{P_n}\left( {\cos \psi } \right)} = }\\ {\frac{{\sqrt {{b^2} + {E^2}} \left( {\sqrt {{u^2} + {E^2}} - l - \sqrt {{b^2} + {E^2}} \cos \psi } \right)}}{{{u^2} + {E^2}}} + }\\ {\frac{{\left( {{b^2} + {E^2}} \right)\cos \psi }}{{{u^2} + {E^2}}}\ln \frac{{2\sqrt {{u^2} + {E^2}} }}{{l + \sqrt {{u^2} + {E^2}} - \sqrt {{b^2} + {E^2}} \cos \psi }}} \end{array} $$
      $$ \begin{array}{*{20}{c}} {{K_2} = \sum\limits_{n = 1}^\infty {\frac{1}{{n + 2}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}{P_n}\left( {\cos \psi } \right)} = }\\ {\frac{{l - \sqrt {{u^2} + {E^2}} \left( {1 + \cos \psi } \right)}}{{\sqrt {{b^2} + {E^2}} }} + }\\ {\frac{{\sqrt {{u^2} + {E^2}} \cos \psi }}{{\sqrt {{b^2} + {E^2}} }}\ln \frac{{\sqrt {{u^2} + {E^2}} \left( {1 - \cos \psi } \right)}}{{1 + \sqrt {{b^2} + {E^2}} - \sqrt {{u^2} + {E^2}} \cos \psi }}} \end{array} $$
      $$ \begin{array}{*{20}{c}} {{K_3} = \sum\limits_{n = 1}^\infty {\frac{1}{{n + 3}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}} \cdot }\\ {{P_n}\left( {\cos \psi } \right) = \frac{{\sqrt {{b^2} + {E^2}} }}{{2\left( {{b^2} + {E^2}} \right)}} - }\\ {\frac{{3\sqrt {{u^2} + {E^2}} \left( {\sqrt {{u^2} + {E^2}} - l} \right)\cos \psi }}{{2\left( {{b^2} + {E^2}} \right)}} + }\\ {\frac{{\sqrt {{u^2} + {E^2}} \left( {1 - 3{{\cos }^2}\psi } \right)}}{{{b^2} + {E^2}}} \cdot }\\ {\ln \frac{{\sqrt {{u^2} + {E^2}} \left( {1 - \cos \psi } \right)}}{{l + \sqrt {{b^2} + {E^2}} - \sqrt {{u^2} + {E^2}} \cos \psi }}} \end{array} $$
      $$ \begin{array}{*{20}{c}} {{K_4} = \sum\limits_{n = 1}^\infty {\frac{1}{n}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}\frac{{{\rm{d}}{P_n}\left( {\cos \psi } \right)}}{{{\rm{d}}\cos \psi }}} = }\\ {\frac{{\left( {{b^2} + {E^2}} \right)\left( {\sqrt {{u^2} + {E^2}} + l} \right)\cos \psi }}{{\sqrt {{u^2} + {E^2}} l\left( {l + \sqrt {{u^2} + {E^2}} - \sqrt {{b^2} + {E^2}} \cos \psi } \right)}}} \end{array} $$
      $$ \begin{array}{*{20}{c}} {{K_5} = \sum\limits_{n = 1}^\infty {\frac{1}{{n + 1}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}\frac{{{\rm{d}}{P_n}\left( {\cos \psi } \right)}}{{{\rm{d}}\cos \psi }}} = }\\ {\sqrt {{u^2} + {E^2}} \cos \psi {K_4} - \frac{{\sqrt {{u^2} + {E^2}} }}{l} + 1} \end{array} $$
      $$ \begin{array}{*{20}{c}} {{K_6} = \sum\limits_{n = 1}^\infty {\frac{1}{{n + 2}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}\frac{{{\rm{d}}{P_n}\left( {\cos \psi } \right)}}{{{\rm{d}}\cos \psi }}} = }\\ {\frac{{\left( {{u^2} + {E^2}} \right)\cos \psi }}{{\sqrt {{b^2} + {E^2}} }}{K_8} - \sqrt {{u^2} + {E^2}} \cdot }\\ {\ln \frac{{\sqrt {{u^2} + {E^2}} \left( {1 - \cos \psi } \right)}}{{l + \sqrt {{b^2} + {E^2}} - \sqrt {{u^2} + {E^2}} \cos \psi }} - }\\ {\frac{{\left( {{u^2} + {E^2}} \right)\cos \psi + \sqrt {{b^2} + {E^2}} \sqrt {{u^2} + {E^2}} }}{{\sqrt {{b^2} + {E^2}} l}} + }\\ {\frac{{\sqrt {{u^2} + {E^2}} \cos \psi }}{{\sqrt {{b^2} + {E^2}} }}} \end{array} $$
      $$ \begin{array}{*{20}{c}} {{K_7} = \sum\limits_{n = 1}^\infty {\frac{1}{{n + 3}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}\frac{{{\rm{d}}{P_n}\left( {\cos \psi } \right)}}{{{\rm{d}}\cos \psi }}} = }\\ {\frac{{{{\left( {{u^2} + {E^2}} \right)}^{3/2}}{{\cos }^3}\psi }}{{{b^2} + {E^2}}}{K_4} - \frac{{\left( {{u^2} + {E^2}} \right)\cos \psi }}{{{b^2} + {E^2}}} \cdot }\\ {\ln \frac{{\sqrt {{u^2} + {E^2}} \left( {1 - \cos \psi } \right)}}{{l + \sqrt {{b^2} + {E^2}} - \sqrt {{u^2} + {E^2}} \cos \psi }} + }\\ {2\left( {{u^2} + {E^2}} \right){K_2} - }\\ {\frac{{{{\left( {{u^2} + {E^2}} \right)}^{3/2}}{{\cos }^2}\psi + b\left( {{u^2} + {E^2}} \right)\cos \psi }}{{\left( {{b^2} + {E^2}} \right)l}} \cdot }\\ {\frac{{\sqrt {{u^2} + {E^2}} }}{l} + \frac{{\left( {{u^2} + {E^2}} \right){{\cos }^2}\psi }}{{{b^2} + {E^2}}}} \end{array} $$
      $$ \begin{array}{*{20}{c}} {{K_8} = \sum\limits_{n = 1}^\infty {\frac{1}{{n - 1}}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}\frac{{{\rm{d}}{P_n}\left( {\cos \psi } \right)}}{{{\rm{d}}\cos \psi }}} = }\\ {\frac{{{b^2} + {E^2}}}{{\sqrt {{u^2} + {E^2}} l}} + \frac{{{b^2} + {E^2}}}{{{u^2} + {E^2}}} \cdot }\\ {\ln \frac{{2\sqrt {{u^2} + {E^2}} }}{{l + \sqrt {{u^2} + {E^2}} - \sqrt {{b^2} + {E^2}} \cos \psi }} + }\\ {\frac{{\left( {{b^2} + {E^2}} \right)\cos \psi }}{{\sqrt {{u^2} + {E^2}} }}{K_4}} \end{array} $$

      第四项计算式为:

      $$ \begin{array}{l} \sum\limits_{n = 1}^\infty {\frac{{3n + 1}}{{{{\left( {n + 1} \right)}^2}}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} = \sum\limits_{n = 1}^\infty {\left( { - \frac{1}{n}{x^{n + 1}}} \right){P_n}\left( {\cos \psi } \right)} + \sum\limits_{n = 1}^\infty {\frac{3}{{n + 1}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} + \\ \sum\limits_{n = 1}^\infty {\frac{1}{{n + 2}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} + \sum\limits_{n = 1}^\infty {\frac{2}{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}{x^{n + 1}}{P_n}\left( {\cos \psi } \right)} = - {K_9} + 3{K_{10}} + {K_2} \end{array} $$ (47)

      其中,

      $$ {K_9} = \sum\limits_{n = 1}^\infty {\frac{1}{n}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}} {P_n}\left( {\cos \psi } \right) $$
      $$ {K_{10}} = \sum\limits_{n = 1}^\infty {\frac{1}{n+1}{{\left( {\frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }}} \right)}^{n + 1}}} {P_n}\left( {\cos \psi } \right) $$

      将式(43)、(44)、(46)和(47)代入式(42),整理得到K(u, ψ)为:

      $$ \begin{array}{*{20}{c}} {K\left( {u,\psi } \right) = \frac{{\sqrt {{b^2} + {E^2}} }}{{\sqrt {{u^2} + {E^2}} }} - \frac{{\sqrt {{b^2} + {E^2}} }}{l} + \frac{{\sqrt {{b^2} + {E^2}} + 3\sqrt {{u^2} + {E^2}} \cos \psi }}{{\sqrt {{b^2} + {E^2}} l}} + \frac{{\sqrt {{b^2} + {E^2}} + 3\sqrt {{u^2} + {E^2}} \cos \psi }}{{2\left( {{b^2} + {E^2}} \right)}} \cdot }\\ {\ln \frac{{\sqrt {{u^2} + {E^2}} \left( {1 - \cos \psi } \right)}}{{l + \sqrt {{b^2} + {E^2}} - \sqrt {{u^2} + {E^2}} \cos \psi }}\frac{{3\sqrt {{u^2} + {E^2}} l - 3{u^2} - {E^2} - {b^2}}}{{2{b^2}l}} - \frac{C}{{48}}\left( {14{K_8} - 24{K_9} - } \right.}\\ {\left. {30{K_{10}} - 5{K_{11}} - 3{K_{13}}} \right) - \frac{D}{8}\left( {\frac{8}{l} - 10{K_3} - 5{K_2} - {K_1}} \right) - \left( { - {K_9} + 3{K_{10}} + {K_2}} \right)} \end{array} $$ (48)

      将式(40)和式(48)代入式(32)中,得到类椭球Hotine核函数H(u, ψ)-e2K(u, ψ)的计算公式,该式适用于参考椭球边界面外部和边界面上任意一点。

    • 本文在Neumann边值条件下,研究了Helmert空间扰动位函数的椭球积分解表达式,并详细地推导了相应的第二类勒让德函数及其导数的递推计算关系式、类椭球Hotine积分核函数等实用化公式,为解决椭球域下的相关位函数及其泛函的研究提供重要参考。

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