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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。
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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，Y′nm(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) $$ \begin{array}{*{20}{c}} {\sum\limits_{n = 1}^\infty {{x^n}{P_n}\left( {\cos \psi } \right)} = {{\left( {1 - 2x\cos \psi + {x^2}} \right)}^{ - 1/2}} - 1,}\\ {且\; - 1 \le x \le 1} \end{array} $$ (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) m≤n时，
$$ \begin{array}{*{20}{c}} {\frac{{3{m^2}}}{{2n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {2n + 3} \right)}} \le }\\ {\frac{{3{n^2}}}{{2n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {2n + 3} \right)}} \le {e^2}}\\ {\frac{{2{m^2}}}{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)\left( {2n + 3} \right)}} \le }\\ {\frac{{2{n^2}}}{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)\left( {2n + 3} \right)}} \le {e^2}} \end{array} $$ 在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。
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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。
Helmert Disturbing Potential and Its Integral Kernel Function with Ellipsoidal Harmonic Formula
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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。
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关键词:
- Neumann边值问题 /
- Helmert扰动位 /
- 第二类勒让德函数 /
- 椭球积分解 /
- 类椭球Hotine积分核函数
Abstract: Based on the theory of the Neumann boundary value problem of geodesy on the geocentric reference ellipsoid as the boundary surface. In the Helmert space, derive the ellipsoidal series expansion of harmonic functions outside the referential ellipsoidal, the relationship of second kinds of associa-ted Legendre functions and its derivatives recurrence formula, Helmert disturbing potential and its integral kernel function with ellipsoidal harmonic formula, in order to research the Neumann boundary value problem of geodesy in the ellipsoidal coordinates. -
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