海区环境条件对声波在海中的传播规律有决定性影响，但复杂环境下声场所应满足的波动方程式很难求得相应的解析解，不便于进行理论分析，而利用抛物近似等数值方法又仅能适用于极为有限的条件，因此探索一种波动方程式近似解析解的方法，有实际的学术和应用意义。本文探讨利用逐次近似法，求取在复杂环境条件下波动方程式的近似解析解。

1 基本思路与公式

设在流体及弹性海底介质中的密度、纵波声速和弹性模量分别为ρ₁, ρ₂, c₁, c₂, λ, μ，并设水域中点声源位于(0, 0, z_s)处所发出声波，在流体和弹性海底介质中的声场势函数分别为ϕ₁, ϕ₂, ψ，若取声场的时间因子为exp(-iωt)，此时各类介质中声场所应满足的方程式将分别为：

$ {\nabla ^2}{\varphi _1} + k_1^2{\varphi _1} = - 4{\rm{ \mathsf{ π} }}\delta \left( {0,0,z - {z_s}} \right) $

(1)

$ {\nabla ^2}{\varphi _2} + k_2^2{\varphi _2} = 0 $

(2)

$ \nabla \times \nabla \times \mathit{\boldsymbol{\psi }} - {\chi ^2}\mathit{\boldsymbol{\psi }} = 0 $

(3)

$ \nabla \cdot \mathit{\boldsymbol{\psi }} = 0 $

(4)

其中k₁、k₂、χ分别为液态及海底介质中的纵波波数与弹性海底介质中的横波波数, 并均可为空间坐标(x, y, z)的某种函数。此时海底弹性介质中质点位移s将为$- {\rm{i}}\omega \mathit{\boldsymbol{s}} = \nabla {\phi _2} + \nabla \times \mathit{\boldsymbol{\psi }}$, 其应力张量为：

$ \frac{\mathit{\boldsymbol{T}}}{{ - {\rm{i}}\omega }} = \left[ {\begin{array}{*{20}{l}} {{\tau _{xx}}}&{{\tau _{xy}}}&{{\tau _{xz}}}\\ {{\tau _{yx}}}&{{\tau _{yy}}}&{{\tau _{yz}}}\\ {{\tau _{zx}}}&{{\tau _{zy}}}&{{\tau _{zz}}} \end{array}} \right] $

(5)

为书写简便，以下记：

$ \left\{ {\begin{array}{*{20}{l}} {{M_1} = \frac{{\partial {\varphi _2}}}{{\partial x}} + \frac{{\partial {\psi _z}}}{{\partial y}} - \frac{{\partial {\psi _y}}}{{\partial z}}}\\ {{M_2} = \frac{{\partial {\varphi _2}}}{{\partial y}} + \frac{{\partial {\psi _x}}}{{\partial z}} - \frac{{\partial {\psi _z}}}{{\partial x}}}\\ {{M_3} = \frac{{\partial {\varphi _2}}}{{\partial z}} + \frac{{\partial {\psi _y}}}{{\partial x}} - \frac{{\partial {\psi _x}}}{{\partial y}}} \end{array}} \right. $

(6)

则式(6)中个分量可以表示为：

$ \left\{ \begin{array}{l} {\tau _{xx}} = - \lambda k_2^2{\varphi _2} + 2\mu \frac{{\partial {M_1}}}{{\partial x}}\\ {\tau _{{\rm{yy}}}} = - \lambda k_2^2{\varphi _2} + 2\mu \frac{{\partial {M_2}}}{{\partial y}}\\ {\tau _{zz}} = - \lambda k_2^2{\varphi _2} + 2\mu \frac{{\partial {M_3}}}{{\partial z}}\\ {\tau _{yx}} = {\tau _{xy}} = \mu \left\{ {\frac{{\partial {M_1}}}{{\partial x}} + \frac{{\partial {M_2}}}{{\partial y}}} \right\}\\ {\tau _{zy}} = {\tau _{yz}} = \mu \left\{ {\frac{{\partial {M_2}}}{{\partial y}} + \frac{{\partial {M_3}}}{{\partial z}}} \right\}\\ {\tau _{zx}} = {\tau _{xz}} = \mu \left\{ {\frac{{\partial {M_3}}}{{\partial z}} + \frac{{\partial {M_1}}}{{\partial x}}} \right\} \end{array} \right. $

(7)

依照逐次近似法, 选取合适的小量ε, 将各坐标量改为：

$ X = \varepsilon x,Y = \varepsilon y,Z = z $

(8)

并将各声场势函数分别表示为以下ε的幂级数形式：

$ \left\{ {\begin{array}{*{20}{l}} {{\phi _1} = {{\rm{e}}^{{\rm{i}}\frac{{W\left( {X,Y} \right)}}{\varepsilon }}}\sum\limits_{m = 0}^\infty {{A_m}} (X,Y,Z){{({\rm{i}}\varepsilon )}^m}}\\ {{\phi _2} = {{\rm{e}}^{{\rm{i}}\frac{{W\left( {X,Y} \right)}}{\varepsilon }}}\sum\limits_{m = 0}^\infty {{B_m}} (X,Y,Z){{({\rm{i}}\varepsilon )}^m}}\\ {\vec \psi = {{\rm{e}}^{{\rm{i}}\frac{{W\left( {X,Y} \right)}}{\varepsilon }}}\sum\limits_{m = 0}^\infty {{\mathit{\boldsymbol{C}}_m}} (X,Y,Z){{({\rm{i}}\varepsilon )}^m}} \end{array}} \right. $

(9)

式中：W为各势函数的水平相位项；A_m、B_m、C_m为各势函数不同阶垂直简正波的幅度项。由于考虑的是垂直简正波的水平相位, 因而在初步近似中对不同类型势函数其水平相位相同^[1]。

为书写简便计，以下记：

$ \left\{ {\begin{array}{*{20}{l}} {{f_x} = \frac{{\partial f}}{{\partial x}},{f_y} = \frac{{\partial f}}{{\partial y}}}\\ {\sigma = {{\left( {1 + f_x^2 + f_y^2} \right)}^{1/2}}}\\ {{w_x} = \frac{{\partial W}}{{\partial X}},{w_y} = \frac{{\partial W}}{{\partial Y}}} \end{array}} \right. $

(10)

函数W应满足一阶偏微分方程式:

$ {\left( {{w_x}} \right)^2} + {\left( {{w_y}} \right)^2} = {\xi ^2} $

(11)

该方程式不难利用一阶偏微分方程的全积分方法解出^[2]。将式(4)代入各自对应的波动方程式，并分别写出各阶ε的同幂次项，最后依次可得：

$ \begin{array}{*{20}{c}} { - {\xi ^2}{A_0} + {A_{0zz}} + k_1^2{A_0} = - 4{\rm{ \mathsf{ π} }}\delta \left( {0,0,Z - {Z_s}} \right)}\\ { - {\xi ^2}{A_1} + {A_{1zz}} + k_1^2{A_1} = - {\xi ^2}{A_0} - 2\left( {{w_x}{A_{0x}} + {w_y}{A_{0y}}} \right)}\\ { \cdots \cdots ,} \end{array} $

(12)

$ \begin{array}{*{20}{c}} { - {\xi ^2}{B_0} + {B_{0zz}} + k_2^2{B_0} = 0}\\ { - {\xi ^2}{B_1} + {B_{1zz}} + k_2^2{B_1} = - {\xi ^2}{B_0} - 2\left( {{w_x}{B_{0x}} + {w_y}{B_{0y}}} \right)}\\ { \cdots \cdots ,} \end{array} $

(13)

$ \begin{array}{*{20}{c}} { - {\xi ^2}{\mathit{\boldsymbol{C}}_0} + {\mathit{\boldsymbol{C}}_{0zz}} + {\chi ^2}{\mathit{\boldsymbol{C}}_0} = 0}\\ { - {\xi ^2}{\mathit{\boldsymbol{C}}_1} + {\mathit{\boldsymbol{C}}_{1zz}} + {\chi ^2}{\mathit{\boldsymbol{C}}_1} = - {\xi ^2}{\mathit{\boldsymbol{C}}_0} - 2\left( {{w_x}{\mathit{\boldsymbol{C}}_{0x}} + {w_y}{\mathit{\boldsymbol{C}}_{0y}}} \right)}\\ { \cdots \cdots ,} \end{array} $

(14)

$ \begin{array}{*{20}{c}} {{w_x}{C_{x0}} + {w_z}{C_{y0}} + {C_{z0z}} = 0}\\ { \cdots \cdots ,} \end{array} $

(15)

取海底界面为z=f(x, y)曲面，记n为海底界面上任意点指向海水方向的单位法线矢量，则有:

$ \mathit{\boldsymbol{n}} = \frac{{\left( {{f_x}\mathit{\boldsymbol{i}} + {f_y}\mathit{\boldsymbol{j}} - \mathit{\boldsymbol{k}}} \right)}}{\sigma } $

(16)

而声场在海底界面处所应满足的边界条件将分别为^[3]:

$ \nabla {\phi _1} \cdot \mathit{\boldsymbol{n}} = \left( {\nabla {\phi _2} + \nabla \times \mathit{\boldsymbol{\psi }}} \right) \cdot \mathit{\boldsymbol{n}} $

(17)

$ {\rm{i}}\omega {\rho _1}{\phi _1} = \left( {\mathit{\boldsymbol{T}} \cdot \mathit{\boldsymbol{n}}} \right) \cdot \mathit{\boldsymbol{n}} $

(18)

$ \left( {\mathit{\boldsymbol{T}} \cdot \mathit{\boldsymbol{n}}} \right) \times \mathit{\boldsymbol{n}} = 0 $

(19)

由各势函数所应满足的波动方程式，借助修正的W.K.B.方法，不难写出其零阶近似解(见附录)，只是对ϕ₁的解则不仅应该满足点源条件，还应满足在海面处其值为零的边界条件。而此时沿海底弹性界面传播的界面波将具有柱面波衰减规律。注意到当考虑海区有不完整水下声道时, 若声速梯度函数形式在声道轴上、下不同, 则ϕ₁在声道轴上、下的具体表示也将有所不同, 取声道轴所在深度为h, 此时各零阶垂直简正波的解可表示为^[4]：

$ {A_0} = \left\{ \begin{array}{l} D{S_0}V_0^{1/3}{N_0},\;\;\;\;0 \le Z < {Z_s}\\ D{S_0}V_0^{1/3}{N_0} + P\left( {{V_0}} \right),\;\;\;\;{Z_s} \le Z < h\\ \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over D} {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over S} }_0}\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} _0^{1/3}{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over N} }_0} + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over P} \left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} }_0}} \right),\;\;\;\;h \le Z < f \end{array} \right. $

(20)

其中：

$ \begin{array}{*{20}{c}} {\left\{ {\begin{array}{*{20}{l}} {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over D} = D{{\rm{e}}^{{\rm{i}}\left( {{v_{0h}} - {v_{1h}}} \right)}}}\\ {{N_0} = {\rm{H}}_{1/3}^{(1)}\left( {{V_0}} \right){\rm{H}}_{1/3}^{(2)}\left( {{V_{00}}} \right) - {\rm{H}}_{1/3}^{(1)}\left( {{V_{00}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{V_0}} \right)}\\ {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over N} }_0} = {\rm{H}}_{1/3}^{(1)}\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} }_0}} \right){\rm{H}}_{1/3}^{(2)}\left( {{V_{00}}} \right) - {\rm{H}}_{1/3}^{(1)}\left( {{V_{00}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} }_0}} \right)} \end{array}} \right.}\\ {\left\{ \begin{array}{l} P\left( {{V_0}} \right) = \frac{{{\rm{ \mathsf{ π} i}}}}{2}{\left( {\frac{{{V_0}{V_{0s}}}}{{{Q_0}{Q_{0s}}}}} \right)^{1/2}}{N_{0s}}\\ \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over P} \left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} }_0}} \right) = \frac{{{\rm{ \mathsf{ π} i}}}}{2}{{\rm{e}}^{{\rm{i}}\left( {{v_{0h}} - {v_{1h}}} \right)}}{\left( {\frac{{{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} }_0}{V_{0s}}}}{{{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over Q} }_0}{Q_{0{\rm{s}}}}}}} \right)^{1/2}}{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over N} }_{0{\rm{s}}}} \end{array} \right.} \end{array} $

(21)

$ {S_0} = \frac{{V_0^{1/6}}}{{Q_0^{1/2}}},{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over S} }_0} = \frac{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} _0^{1/6}}}{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over Q} _0^{1/2}}} $

(22)

$ {Q_0} = \sqrt {k_1^2 - {\xi ^2}} ,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over Q} }_0} = \sqrt {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over k} _1^2 - {\xi ^2}} $

(23)

$ \left\{ {\begin{array}{*{20}{l}} {{V_0}(Z) = \int {{Q_0}} (Z){\rm{d}}Z}\\ {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} }_0}(Z) = \int {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over Q} }_0}} (Z){\rm{d}}Z} \end{array}} \right. $

(24)

并解出：

$ {B_0} = FS_0^\prime V{'}_0^{1/3}H_{1/3}^{(1)}\left( {V_0^\prime } \right),S_0^\prime = \frac{{V{'}_0^{1/6}}}{{Q{'}_0^{1/2}}} $

(25)

$ Q_0^\prime = \sqrt {k_2^2 - {\xi ^2}} ,V_0^\prime = \int {Q_0^\prime } {\rm{d}}Z $

(26)

$ {\mathit{\boldsymbol{C}}_0} = \mathit{\boldsymbol{G}}S_0^{\prime \prime }V_0^{\prime \prime 1/3}{\rm{H}}_{1/3}^{(1)}\left( {V_0^{\prime \prime }} \right),S_0^{\prime \prime } = \frac{{V_0^{\prime \prime 1/6}}}{{Q_0^{\prime \prime 1/2}}} $

(27)

$ Q_0^{\prime \prime } = \sqrt {{\chi ^2} - {\xi ^2}} ,V_0^{\prime \prime } = \int {Q_0^{\prime \prime }} {\rm{d}}Z $

(28)

$ {w_x}{G_{x0}} + {w_y}{G_{y0}} + Q_0^{\prime \prime }{G_{z0}} = 0 $

(29)

函数W应满足一阶偏微分方程式:

$ {\left( {\frac{{\partial W}}{{\partial X}}} \right)^2} + {\left( {\frac{{\partial W}}{{\partial Y}}} \right)^2} = {\xi ^2} $

(30)

该方程式不难利用一阶偏微分方程全积分方法解出^[5]。将式(10)~(14)所给出的零阶解代入边界条件各式后，可得求解常数$\overset\frown{D}$, F, G及ξ的方程组，其中ξ为边界条件矩阵系数行列式为零时的根。通常海水声速的水平梯度远小于其垂直声速梯度，故ε只是10^-2~10^-3左右量级，仅利用各势函数的零阶近似表达, 即可以获得对声场分布有一定参考价值的结果。在不过分影响所得规律的准确性条件下，为书写简便计，忽略含有S对z求偏导的各${\dot{S}}$项，且Hankel函数均用其渐近展开式表示, 并利用$G_{z}=-\frac{G_{x} w_{x}+G_{y} w_{y}}{Q^{\prime \prime}}$消去G_z, 可写出弹性介质中各应力张量表示, 及由边界条件形成的矩阵R的各项如下(忽略-iω因子)：

$ {\tau _{xx}} = F{U_x} - 2\mu \frac{{{w_x}}}{{{Q^{\prime \prime }}}}{P_y} $

(31)

$ {\tau _{yy}} = F{U_y} + 2\mu \frac{{{w_y}}}{{{Q^{\prime \prime }}}}{P_x} $

(32)

$ {\tau _{zz}} = F\left( {\lambda k_2^2 + 2\mu {{Q'}^2}} \right) - 2\mu {Q^{\prime \prime }}\kappa $

(33)

$ {\tau _{xy}} = \mu F{\xi ^2} + \mu {Q^{\prime \prime }}\kappa $

(34)

$ {\tau _{yz}} = \mu F{u_y} + \mu \frac{{{w_x}}}{{{Q^{\prime \prime }}}}{P_y} $

(35)

$ {\tau _{zx}} = \mu F{u_x} - \mu \frac{{{w_y}}}{{{Q^{\prime \prime }}}}{P_x} $

(36)

其中：

$ \left\{ \begin{array}{l} {U_x} = \lambda k_2^2 + 2\mu w_x^2,\;\;\;\;{U_y} = \lambda k_2^2 + 2\mu w_y^2\\ \begin{array}{*{20}{l}} {{u_x} = {{Q'}^2} + w_x^2,{u_y} = {{Q'}^2} + w_y^2}\\ {{P_x} = {G_y}{w_x}{w_y} + {G_x}\left( {{Q^{\prime \prime 2}} + w_x^2} \right)}\\ {{P_y} = {G_x}{w_x}{w_y} + {G_y}\left( {{Q^{\prime \prime 2}} + w_y^2} \right)}\\ {\kappa = {G_x}{w_y} - {G_y}{w_x}} \end{array} \end{array} \right. $

(37)

$ {\Re _{11}} = \mathit{\Omega }\sin \left( {v - {v_0}} \right) - {Q_{0f}}\cos \left( {v - {v_0}} \right) $

(38)

$ {\Re _{12}} = \left( {{f_x}{w_x} + {f_y}{w_y} - Q_{0f}^\prime } \right) $

(39)

$ {\Re _{13}} = - \frac{1}{{{Q^{\prime \prime }}}}\left( {{{\mathit{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \Omega } }}_2} - {{\bar \omega }_1}} \right) $

(40)

$ {\Re _{14}} = - \frac{1}{{{Q^{\prime \prime }}}}\left( {{{\mathit{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \Omega } }}_1} - {{\bar \omega }_1}} \right) $

(41)

$ {\Re _{21}} = {\omega ^2}{\rho _1}\sin \left( {v - {v_0}} \right) $

(42)

$ {\Re _{22}} = \lambda k_2^2{\sigma ^2} + 2\mu {\zeta _ - }{\mathit{\Omega }^\prime } $

(43)

$ {\Re _{23}} = - \frac{{2\mu {w_y}{\zeta _ - }}}{{{Q^{\prime \prime }}}}\left( {\mathit{\Omega }_1^{\prime \prime } - \bar \omega _1^\prime } \right) $

(44)

$ {\Re _{24}} = - \frac{{2\mu {w_x}{\zeta _ + }}}{{{Q^{\prime \prime }}}}\left( {\mathit{\Omega }_2^{\prime \prime } - \bar \omega _2^\prime } \right) $

(45)

$ {\Re _{31}} = 0 $

(46)

$ {\Re _{32}} = \mu \left( {{\zeta _ - }{\vartheta ^\prime } + {F_x}\vartheta _1^\prime } \right) $

(47)

$ {\Re _{33}} = - \frac{{\mu {w_y}}}{{{Q^{\prime \prime }}}}\left\{ {{\gamma _1}{Q^{\prime \prime 2}} + {\zeta _1}w_x^2} \right\} $

(48)

$ {\Re _{33}} = - \frac{{\mu {w_y}}}{{{Q^{\prime \prime }}}}\left\{ {{\gamma _1}{Q^{\prime \prime 2}} + {\zeta _1}w_y^2} \right\} $

(49)

$ {\Re _{41}} = 0 $

(50)

$ {\Re _{42}} = \mu \left\{ {{\zeta _ - }{\vartheta ^{\prime \prime }} + {F_y}\vartheta _1^{\prime \prime }} \right\} $

(51)

$ {\Re _{43}} = - \frac{{\mu {w_y}}}{{{Q^{\prime \prime }}}}\left\{ {{\gamma _3}{Q^{\prime \prime 2}} - \left( {{f_y}{\zeta _ - } - {F_y}} \right)w_x^2} \right\} $

(52)

$ {\Re _{44}} = \frac{{\mu {w_x}}}{{{Q^{\prime \prime }}}}\left\{ {{\gamma _4}{Q^{\prime \prime 2}} + {\zeta _2}w_y^2} \right\} $

(53)

其中：

$ \mathit{\Omega } = {f_x}{w_x} + {f_y}{w_y},{\mathit{\Omega }^\prime } = {f_x}w_x^2 + {f_y}w_y^2 - {{Q'}^2} $

$ {{\mathit{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \Omega } }}_1} = {w_x}\left( {{f_y}{w_y} - {Q^{\prime \prime }}} \right),{{\mathit{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \Omega } }}_2} = {w_y}\left( {{f_x}{w_x} - {Q^{\prime \prime }}} \right) $

$ \mathit{\Omega }_1^{\prime \prime } = \left( {{f_x} - {f_y}} \right)w_x^2,\mathit{\Omega }_2^{\prime \prime } = \left( {{f_x} - {f_y}} \right)w_y^2 $

$ {{\bar \omega }_1} = {f_x}\left( {{Q^{\prime \prime 2}} + w_y^2} \right),{{\bar \omega }_2} = {f_y}\left( {{Q^{\prime \prime 2}} + w_x^2} \right) $

$ \bar \omega _1^\prime = \left( {{f_x} - 1} \right){Q^{\prime \prime 2}},\bar \omega _2^\prime = \left( {{f_y} + 1} \right){Q^{\prime \prime 2}} $

$ {\vartheta ^\prime } = {f_x}{Q^{\prime 2}} + w_x^2,\vartheta _1^\prime = {f_x}w_x^2 + {f_y}w_y^2 - {Q^{\prime 2}} $

$ {\vartheta ^{\prime \prime }} = {f_y}{Q^{\prime \prime 2}} + w_y^2,\vartheta _1^{\prime \prime } = {f_x}w_x^2 + {f_y}w_y^2 - {Q^{\prime \prime 2}} $

$ {\gamma _1} = f_x^2 - {f_x} - {\zeta _ + },{\gamma _2} = 4{f_x} + {f_y} - {f_x}{f_y} $

$ {\gamma _3} = {f_x}{f_y} - {f_x} - 4{f_y},{\gamma _4} = f_y^2 - {f_y} - {\zeta _ + } $

$ {\zeta _1} = f_x^2 + 2{f_x} - {f_x}{f_y} - 1, $

$ {\zeta _2} = f_y^2 + 2{f_y} - {f_x}{f_y} - 1 $

2 仿真算例

为进一步描述在不规则海区可能出现的声场分布特点，试考虑给定的如下海区环境条件：令ε=0.01, ω=200π, 取z坐标垂直向下，海面为Z=0平面，声源深度Z_s=300 m, 声道轴所在深度h=1 000 m, 海水密度为ρ₁=1 g/cm³, 水中声速随深度的变化规律如下：

海水中的声速:

$ {c_1} = \left\{ {\begin{array}{*{20}{l}} {\frac{{1500}}{{{c_{11}}}},}&{0 \le Z < 1000{\rm{m}}}\\ {\frac{{1500}}{{\sqrt {1.1} }}{c_{12}},}&{Z > 1000{\rm{m}}} \end{array}} \right. $

其中：

$ {c_{11}}(Z) = \sqrt {1 + {{10}^{ - 4}}Z[1 + 10\alpha (1000 - Z)]} $

$ \alpha = \sin \frac{{30}}{{1 + {{(X - 1)}^2}{{(Y - 2)}^2}/{{10}^4}}} $

$ {c_{12}}(Z) = \left[ {1 + 0.4 \times {{10}^{ - 4}}(Z - 1000)} \right] $

海底界面方程:

$ Z = 1200 + 1000\tanh (X/100 + Y/70) $

弹性介质密度:

$ {\rho _2} = 1.5(1 + 0.0005Z) $

弹性模量：

$ \lambda = 2 \times {10^4}\left( {1 + {{10}^{ - 5}}X + {{10}^{ - 5}}Y} \right), $

$ \mu = 3 \times {10^6}\left( {1 + {{10}^{ - 5}}X - {{10}^{ - 6}}Y} \right), $

按海区环境条件，可直接写出各简正波的零阶近似如下，对水中声场：

$ Z < 1000, $

$ \begin{array}{*{20}{c}} {{V_0} = \int {\sqrt {{{\left( {\frac{{\rm{ \mathsf{ π} }}}{{7.5}}} \right)}^2}{c_{11}} - {\xi ^2}{\rm{d}}Z} } = }\\ {\int {\sqrt {a + bZ + c{Z^2}} } {\rm{d}}Z = \frac{{(2cZ + b)}}{{4c}}\sqrt {a + bZ + c{Z^2}} + }\\ {\left\{ {\begin{array}{*{20}{l}} { - \frac{{4ac - {b^2}}}{{{{(2\sqrt { - c} )}^3}}}\arcsin \frac{{2cZ + b}}{{\sqrt {{b^2} - 4ac} }}}&{c < 0}\\ {\frac{{4ac - {b^2}}}{{{{(2\sqrt c )}^3}}}\ln (2\sqrt {cR} + 2cZ + b)}&{c > 0} \end{array}} \right.} \end{array} $

其中：

$ a = {\left( {\frac{{\rm{ \mathsf{ π} }}}{{7.5}}} \right)^2} - {\xi ^2}, $

$ b = {\left( {\frac{{\rm{ \mathsf{ π} }}}{{7.5}}} \right)^2}\left( {{{10}^{ - 4}} + \alpha } \right), $

$ c = - {\left( {\frac{{\rm{ \mathsf{ π} }}}{{7.5}}} \right)^2} \times {10^{ - 3}}\alpha ; $

$ Z > 1000,\;令\;u = 0.96 + 0.4 \times {10^{ - 4}}Z $

$ \begin{array}{l} {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over V} }_0} = \int {\sqrt {\frac{{0.1930}}{{{{\left( {0.96 + 0.4 \times {{10}^{ - 4}}Z} \right)}^2}}} - {\xi ^2}} } {\rm{d}}Z = \\ 5 \times {10^4}\int {\frac{{u{\rm{d}}u}}{{\sqrt {0.1930 - {\xi ^2}{u^2}} }}} = \\ - \frac{{2.5 \times {{10}^4}}}{{{\xi ^2}}}\sqrt {0.1930 - {\xi ^2}{{\left( {0.96 + 0.4 \times {{10}^{ - 4}}Z} \right)}^2}} \end{array} $

对海底介质中声场，因海底介质密度随深度依线性变化，故有：

$ \begin{array}{l} k_2^2 = \frac{{{\rho _2}{\omega ^2}}}{{\lambda + 2\mu }} - \frac{3}{4}{\left( {\frac{{\partial {\rho _2}}}{{\partial z}}/{\rho _2}} \right)^2} = \\ \frac{{0.0984\left( {1 + 5 \times {{10}^{ - 4}}Z} \right)}}{{1 + {{10}^{ - 2}}X - {{10}^{ - 3}}Y}} - \frac{{1.875 \times {{10}^{ - 7}}}}{{{{(1 + 0.0005Z)}^2}}} \end{array} $

$ \begin{array}{l} {\chi ^2} = \frac{{{\rho _2}{\omega ^2}}}{\mu } - \frac{3}{4}{\left( {\frac{{\partial {\rho _2}}}{{\partial z}}/{\rho _2}} \right)^2} = \\ \frac{{0.1974\left( {1 + 5 \times {{10}^{ - 4}}Z} \right)}}{{1 + {{10}^{ - 2}}X - {{10}^{ - 3}}Y}} - \frac{{1.875 \times {{10}^{ - 7}}}}{{{{(1 + 0.0005Z)}^2}}} \end{array} $

记：

$ \beta = \frac{{0.0984}}{{1 + {{10}^{ - 2}}X - {{10}^3}Y}}, $

$ \eta = \frac{{0.1974}}{{1 + {{10}^{ - 2}}X - {{10}^{ - 3}}Y}}, $

$ {u^\prime } = 1 + 5 \times {10^{ - 4}}Z; $

$ \begin{array}{l} V_0^\prime = 2000\int {\sqrt {\beta {u^{\prime 3}} - {\xi ^2}{u^{\prime 2}} - 1.875 \times {{10}^{ - 7}}} } \frac{{{\rm{d}}{u^\prime }}}{{{u^\prime }}} \approx \\ 2000\int {\left\{ {\sqrt {\beta {u^{\prime 3}} - {\xi ^2}{u^{\prime 2}}} - \frac{{0.938 \times {{10}^{ - 7}}}}{{\sqrt {{\beta ^2}{u^{\prime 3}} - {\xi ^2}{u^{\prime 2}}} }}} \right\}} \frac{{{\rm{d}}{u^\prime }}}{{{u^\prime }}} = \\ \frac{{4000}}{{3\beta }}{\left( {\beta {u^\prime } - {\xi ^2}} \right)^{\frac{3}{2}}} + \frac{{1.875 \times {{10}^{ - 4}}\sqrt {\beta {u^\prime } - {\xi ^2}} }}{{{\xi ^2}{u^\prime }}} + \\ \frac{{1.875 \times {{10}^{ - 4}}\beta }}{{{\xi ^3}}}\arctan \frac{{\sqrt {\beta {u^\prime } - {\xi ^2}} }}{\xi } \end{array} $

同理：

$ \begin{array}{l} V_0^{\prime \prime } = \frac{{4000}}{{3\eta }}{\left( {\eta {u^\prime } - {\xi ^2}} \right)^{3/2}} + \frac{{1.875 \times {{10}^{ - 4}}\sqrt {\eta {u^\prime } - {\xi ^2}} }}{{{\xi ^2}{u^\prime }}} + \\ \frac{{1.875 \times {{10}^{ - 4}}\eta }}{{{\xi ^3}}}\arctan \frac{{\sqrt {\eta {u^\prime } - {\xi ^2}} }}{\xi } \end{array} $

由于仿真算例所给定的环境条件比较复杂, 利用边界条件所建立的联立方程组直接解算ξ将十分困难, 为简便计, 以下将利用数值计算求得近似解, 为了能保证必要的解算结果的准确度, 将对不同的X, Y坐标值，依次进行逐段近似。

对W取二次函数近似，设

$ W = {a_1}X + {a_2}Y + {a_3}{X^2} + {a_4}{Y^2} $

则有:

$ {w_x} = {a_1} + 2{a_3}X, $

$ {w_y} = {a_2} + 2{a_4}Y, $

$ {\xi ^2} = {\left( {{a_1} + 2{a_3}X} \right)^2} + {\left( {{a_2} + 2{a_4}Y} \right)^2} $

利用数值计算近似可得X=Y=0时, a₁≈0.24, a₂≈0.23;若取X=Y=5, 则求得a₁≈0.115, a₂≈0.225, a₃≈0.001, a₄≈0.002, 即此时由于海底系大陆坡, 故声传播开始向y方向偏转；若要求得更大范围声场的形式，则需要进行更多的数值计算。本文只满足于提出一种对复杂海区环境可用的分析方法，为保证文章篇幅的有限，将不再做更多的声场计算结果描述。

附录:不规则半声道海域声场分析

设声源位于(0, 0, z_s)处, 声道轴深度为z=h, 声道轴以上海水声速为c₀, 声道轴以下海水声速为c₁, 海面为绝对软界面, c₀, c₁均为x, y, z的函数。依修正的W.K.B.方法由逐次近似法，取声场势函数如下：

$ {\varphi _1} = {{\rm{e}}^{{\rm{i}}\frac{{W(x,y)}}{\varepsilon }}}\sum\limits_{m = 0}^\infty {{A_m}} (x,y,z){({\rm{i}}\varepsilon )^m} $

(A1)

A_m为不同阶局地简正波。不难直接写出其零阶形式解为：

$ A = \left\{ {\begin{array}{*{20}{l}} {{C_0}{{\left( {\frac{{{v_0}}}{{{Q_0}}}} \right)}^{1/2}}{l_0}}&{0 \le z < {z_s}}\\ {{{\left( {\frac{{{v_0}}}{{{Q_0}}}} \right)}^{1/2}}{E_1}}&{{z_s} < z < h}\\ {{{\left( {\frac{{{v_1}}}{{{Q_1}}}} \right)}^{1/2}}{E_2}}&{h < z} \end{array}} \right. $

(A2)

其中：

$ {Q_j} = \sqrt {k_j^2 - {\xi ^2}} ,{v_j} = \int {{Q_j}} {\rm{d}}z,{S_j} = \frac{{v_j^{1/6}}}{{Q_j^{1/2}}},j = 0,1; $

$ \begin{array}{*{20}{c}} {{l_0} = {\rm{H}}_{13}^{(1)}\left( {{v_0}} \right){\rm{H}}_{13}^{(2)}\left( {{v_{00}}} \right) - {\rm{H}}_{13}^{(1)}\left( {{v_{00}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_0}} \right)}\\ {{E_1} = {C_1}{\rm{H}}_{1/3}^{(1)}\left( {{v_0}} \right) + {C_2}{\rm{H}}_{1/3}^{(2)}\left( {{v_0}} \right)}\\ {{E_2} = {C_3}{\rm{H}}_{1/3}^{(1)}\left( {{v_0}} \right) + {C_4}{\rm{H}}_{1/3}^{(2)}\left( {{v_0}} \right)} \end{array} $

(A3)

由点源条件可得：

$ \begin{array}{*{20}{l}} {{C_0}\left\{ {{\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{00}}} \right) - {\rm{H}}_{1/3}^{(1)}\left( {{v_{00}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right)} \right\} = }\\ {{C_1}{\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right) + {C_2}{\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right)} \end{array} $

(A4)

$ \begin{array}{*{20}{l}} {{C_0}\left\{ {{\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{0s}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{00}}} \right) - {\rm{H}}_{1/3}^{(1)}\left( {{v_{00}}} \right){\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{0s}}} \right)} \right\} - }\\ {\left\{ {{C_1}{\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{0s}}} \right) + {C_2}{\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{{0_s}}}} \right)} \right\} = 2} \end{array} $

(A5)

由声道轴处声压与振速连续条件:

$ \begin{array}{*{20}{l}} {{S_{0h}}v_{0h}^{1/3}\left\{ {{C_1}{\rm{H}}_{1/3}^{(1)}\left( {{v_{0h}}} \right) + {C_2}{\rm{H}}_{1/3}^{(2)}\left( {{v_{0h}}} \right)} \right\} = }\\ {{S_{1h}}v_{1h}^{1/3}\left\{ {{C_3}{\rm{H}}_{1/3}^{(1)}\left( {{v_{1h}}} \right) + {C_4}{\rm{H}}_{1/3}^{(2)}\left( {{v_{1h}}} \right)} \right\}} \end{array} $

(A6)

$ \begin{array}{*{20}{l}} {{{\left( {{Q_{0h}}{v_{0h}}} \right)}^{1/2}}\left\{ {{C_1}{\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{0h}}} \right) + {C_2}{\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{0h}}} \right)} \right\} = }\\ {{{\left( {{Q_{1h}}{v_{1h}}} \right)}^{1/2}}\left\{ {{C_3}{\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{1h}}} \right) + {C_4}{\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{1h}}} \right)} \right\}} \end{array} $

(A7)

因$\dot{S}_{0 s}, \dot{S}_{0 h}, \dot{S}_{1 h}$均远小于1上式中一律忽略不计，式(A4)乘以H_-2/3⁽²⁾(v_0 s)减去式(A5)乘H_1/3⁽²⁾(v_0 s)，可得到：

$ \begin{array}{l} \left\{ {{C_0}{\rm{H}}_{1/3}^{(2)}\left( {{v_{00}}} \right) - {C_1}} \right\}{L_{0s}} = - 2{\left( {{Q_{0s}}{v_{0s}}} \right)^{ - 1/2}}{\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right)\\ {L_{0s}} = {\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right){\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{0s}}} \right) - {\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{0s}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right)\\ {C_0}{\rm{H}}_{1/3}^{(2)}\left( {{v_{00}}} \right) = {C_1} - \frac{{{\rm{ \mathsf{ π} }}i}}{2}{\left( {\frac{{{v_{0s}}}}{{{Q_{0s}}}}} \right)^{1/2}}{\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right) \end{array} $

(A8)

式(A4)乘以H_-2/3⁽¹⁾(v_0 s)减去式(A5)乘H_1/3⁽¹⁾(v_0 s)，可得到：

$ \begin{array}{*{20}{c}} {\left\{ {{C_0}{\rm{H}}_{1/3}^{(1)}\left( {{v_{00}}} \right) - {C_2}} \right\}{L_{0s}} = 2{{\left( {{Q_{0s}}{v_{0s}}} \right)}^{ - 1/2}}{\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right)}\\ {{C_0}{\rm{H}}_{1/3}^{(1)}\left( {{v_{00}}} \right) = {C_2} + \frac{{{\rm{ \mathsf{ π} i}}}}{2}{{\left( {\frac{{{v_{0s}}}}{{{Q_{0s}}}}} \right)}^{1/2}}{\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right)} \end{array} $

(A9)

同理利用Q_0h=Q_1h, 由声道轴上、下声场与法线振速连续条件可得：

$ {C_0}{\left( {\frac{{{v_{0h}}}}{{{Q_{0h}}}}} \right)^{1/2}}{l_{0h}} + \frac{{{\rm{ \mathsf{ π} }}i}}{2}{\left( {\frac{{{v_{0h}}{v_{0s}}}}{{{Q_{0h}}{Q_{0s}}}}} \right)^{1/2}}{l_{0hs}} = {\left( {\frac{{{v_{1h}}}}{{{Q_{1h}}}}} \right)^{1/2}}{E_{2h}} $

(A10)

$ {C_0}{\left( {{Q_{0h}}{v_{0h}}} \right)^{1/2}}{L_{0h}} + \frac{{{\rm{ \mathsf{ π} }}i}}{2}{\left( {\frac{{{Q_{0h}}{v_{0h}}{v_{0s}}}}{{{Q_{0s}}}}} \right)^{1/2}}{L_{0hs}} = {\left( {{Q_{1h}}{v_{1h}}} \right)^{1/2}}{E_{41h}} $

(A11)

式中：

$ {l_{0hs}} = {\rm{H}}_{1/3}^{(1)}\left( {{v_{0h}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right) - {\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{0h}}} \right) $

$ {L_{0h}} = {\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{0h}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{00}}} \right) - {\rm{H}}_{1/3}^{(1)}\left( {{v_{00}}} \right){\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{0h}}} \right) $

$ {L_{0hs}} = {\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{0h}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right) - {\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right){\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{0h}}} \right) $

$ {E_{41h}} = {C_3}{\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{1h}}} \right) + {C_4}{\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{1h}}} \right) $

利用Q_0 h=Q_1 h，

联合式(A10)和式(A11)可得：

$ \begin{array}{*{20}{c}} {\left\{ {{C_0}{\rm{H}}_{1/3}^{(2)}\left( {{v_{00}}} \right) + \frac{{{\rm{ \mathsf{ π} }}i}}{2}{\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right){{\left( {\frac{{{v_{0s}}}}{{{Q_{0s}}}}} \right)}^{1/2}}} \right\} \cdot }\\ {\left. {{\rm{H}}_{1/3}^{(1)}\left( {{v_{0h}}} \right){\rm{H}}_{ - 2/3}^{(2)}\left( {{v_{1h}}} \right) - {\rm{H}}_{ - 2/3}^{(1)}\left( {{v_{0h}}} \right){\rm{H}}_{1/3}^{(2)}\left( {{v_{1h}}} \right)} \right\} = }\\ {{C_3}\frac{{ - 4i}}{{{\rm{ \mathsf{ π} }}\sqrt {{v_{0h}}{v_{1h}}} }}} \end{array} $

(A12)

可得：

$ {C_3} \approx \left\{ {{C_0}{\rm{H}}_{1/3}^{(2)}\left( {{v_{00}}} \right) + \frac{{{\rm{ \mathsf{ π} }}i}}{2}{\rm{H}}_{1/3}^{(2)}\left( {{v_{0s}}} \right)\sqrt {\frac{{{v_{0s}}}}{{{Q_{0s}}}}} } \right\}{{\rm{e}}^{{\rm{i}}\left( {{v_{oh}} - {v_{1h}}} \right)}} $

同理

$ {C_4} \approx - \left\{ {{C_0}{\rm{H}}_{1/3}^{(1)}\left( {{v_{00}}} \right) - \frac{{{\rm{ \mathsf{ π} }}i}}{2}{\rm{H}}_{1/3}^{(1)}\left( {{v_{0s}}} \right)\sqrt {\frac{{{v_{0s}}}}{{{Q_{0s}}}}} } \right\}{{\rm{e}}^{{\rm{i}}\left( {{v_{0h}} - {v_{1h}}} \right)}} $