对多元复合函数偏导数求解的研究

Research on Solving Partial Derivatives of Multivariate Compound Functions

摘要: 本文主要讨论多元复合函数的偏导数问题，主要对中间变量是隐函数的情况进行讨论。文章主线是从隐函数的存在性问题出发，并结合“链式求导法则、矩阵求多元抽象复合函数二阶偏导数”等技巧和方法，获得了求解多元复合函数的一、二阶偏导数的理论性定理。

Abstract: In this article, we mainly discuss the partial derivatives of multivariate compound functions, and mainly discuss the case that the intermediate variables are implicit functions. The main line of this article starts from the existence of implicit function, and combines the techniques and methods of “chain derivative rule, using matrix to find the second partial derivative of multiple abstract composite function”. The theoretical theorems for solving the first and second partial derivatives of multivariate compound functions are obtained.

1. 引言

在现已知的多元函数变量代换的方法与应用中，都是仅对复合函数一阶偏导数的代换方法进行了讨论，然而对高阶可微函数的高阶偏导数(如：二阶偏导数)的代换求解公式极少提及。但当我们处理实际问题时，往往会有很多这方面的理论需求，如：多元复合函数的高精度近似计算、多元函数求极值问题等。另外，在对一般复合函数偏导数的计算过程中，常以链式求导法则为主，但在许多具体的问题，例如：函数 $\xi =\xi \left(z,r\right)$，， $z=\left(u,v\right)$， $r=\left(u,v\right)$， $z,r$ 为物理域中的变量，而 $\xi ,\eta$ 为计算域中的变量，利用计算域中的结果来讨论自变量变化问题，如果使用链式法则进行计算，求解过程就会极为复杂。故而，为了解决类似的这样一类问题，我们可从隐函数的存在唯一性定理出发，结合 [1] [2] 中的方法，可得到本文的结果。

2. 预备知识

为后期得到我们的结论，我们先给出如下的隐函数存在唯一性定理和隐函数组定理。

引理2.1 [3] 若函数 $F\left(x,y\right)$ 满足下列条件：

1) F在以 $P_0\left(x_0,y_0\right)$ 为内点的某一区域 $D\subset R^2$ 上连续；

2) $F\left(x_0,y_0\right)=0$ (通常称为初始条件)；

3) F在D内存在连续的偏导数 $F_y\left(x,y\right)$ ；

4) $F_y\left(x_0,y_0\right)\ne 0$。

则

i) 存在点 $p_0$ 的某领域 $U\left(P_0\right)\subset D$，在 $U\left(P_0\right)$ 上方程 $F\left(x,y\right)=0$ 唯一地决定了一个定义在某区间 $\left[x_0-\alpha ,x_0+\alpha \right]$ 上的隐函数 $y=f\left(x\right)$，使得当 $x\in \left[x_0-\alpha ,x_0+\alpha \right]$ 时， $\left(x,f\left(x\right)\right)\in U\left(P_0\right)$，且 $F\left(x,f\left(x\right)\right)\equiv 0$， $f\left(x_0\right)=y_0$。

ii) $y=f\left(x\right)$ 在 $\left[x_0-\alpha ,x_0+\alpha \right]$ 上连续。

引理2.2 [3] 若

1) $F\left(x,y,u,v\right)$ 与 $G\left(x,y,u,v\right)$ 在以点 $P_0\left(x_0,y_0,u_0,v_0\right)$ 为内点的区域 $V\subset R^4$ 上连续；

2) $F\left(x_0,y_0,u_0,v_0\right)=0$， $G\left(x_0,y_0,u_0,v_0\right)=0$ (初始条件)；

3) 在V上F，G具有一阶连续偏导数；

4) $J=\frac{\partial \left(F,G\right)}{\partial \left(u,v\right)}$ 在点 $P_0$ 不等于0。

i) 存在点 $P_0$ 的某一(四维)邻域 $U\left(P_0\right)\subset V$，在 $U\left(P_0\right)\subset V$ 上，方程组 $\{\begin{array}{l}F\left(x,y,u,v\right)=0\hfill \\ G\left(x,y,u,v\right)=0\hfill \end{array}$，唯一确定了定义在点 $Q_0\left(x_0,y_0\right)$ 的某一(二维空间)邻域 $U\left(Q_0\right)$ 上的两个二元隐函数 $u=f\left(x,y\right),v=g\left(x,y\right)$，使得 $u_0=f\left(x_0,y_0\right),v_0=g\left(x_0,y_0\right)$，且当 $\left(x,y\right)\in U\left(Q_0\right)$ 时

$\left(x,y,f\left(x,y\right),g\left(x,y\right)\right)\in U\left(P_0\right),F\left(x,y,f\left(x,y\right),g\left(x,y\right)\right)\equiv 0,G\left(x,y,f\left(x,y\right),g\left(x,y\right)\right)\equiv 0;$

ii) $f\left(x,y\right),g\left(x,y\right)$ 在 $U\left(Q_0\right)$ 上连续；

iii) $f\left(x,y\right),g\left(x,y\right)$ 在 $U\left(Q_0\right)$ 上有一阶连续偏导，且

为了得到我们的结果，我们先导出如下定理2.1：

定理2.1 设函数 $\xi =\xi \left(z,r\right),\eta =\eta \left(z,r\right)$ 关于变量具有连续的一阶偏导数，记

$J=\left(\begin{array}{cc}\frac{\partial \xi }{\partial z}& \frac{\partial \xi }{\partial r}\\ \frac{\partial \eta }{\partial z}& \frac{\partial \eta }{\partial r}\end{array}\right)$ (2.1)

若 $\left|J\right|\ne 0$，则存在唯一的函数 $z=z\left(\xi ,\eta \right),r=r\left(\xi ,\eta \right)$ 满足上述等式并且关于变量 $\xi ,\eta$ 具有连续的一阶偏导数

$\left(\begin{array}{cc}\frac{\partial z}{\partial \xi }& \frac{\partial z}{\partial \eta }\\ \frac{\partial r}{\partial \xi }& \frac{\partial r}{\partial \eta }\end{array}\right)=\frac{1}{\left|J\right|}\left(\begin{array}{cc}\frac{\partial \eta }{\partial r}& -\frac{\partial \eta }{\partial z}\\ -\frac{\partial \xi }{\partial r}& \frac{\partial \xi }{\partial z}\end{array}\right)$ (2.2)

其中 $\left|J\right|$ 矩阵J的行列式。

证明：假设存在两个不同的坐标系，中间变量为 $\left(z,r\right)$，自变量为 $\left(\xi ,\eta \right)$，两个坐标系存在以下的对应关系：

$\{\begin{array}{l}\xi =\xi \left(z,r\right)\\ \eta =\eta \left(z,r\right)\end{array}$ (2.3)

进一步，令

(2.4)

则由引理2.2可知，(2.4)所对应的Jacobi行列式不为0，即

$\frac{\partial \left(F,G\right)}{\partial \left(\xi ,\eta \right)}=\left|\begin{array}{cc}{F}_{\xi }& F_{\eta }\\ G_{\xi }& G_{\eta }\end{array}\right|\ne 0$

亦即

$\frac{\partial \left(F,G\right)}{\partial \left(z,r\right)}=\left|\begin{array}{cc}\frac{\partial F}{\partial z}& \frac{\partial F}{\partial r}\\ \frac{\partial G}{\partial z}& \frac{\partial G}{\partial r}\end{array}\right|=\left|\begin{array}{cc}-\frac{\partial \xi }{\partial z}& \frac{\partial \xi }{\partial r}\\ -\frac{\partial \eta }{\partial z}& \frac{\partial \eta }{\partial r}\end{array}\right|\ne 0$ (2.5)

进而对(2.3)式两边分别关于 $\xi ,\eta$ 求偏导，则有

$\frac{\partial \xi }{\partial z}\cdot \frac{\partial z}{\partial \xi }+\frac{\partial \xi }{\partial r}\cdot \frac{\partial r}{\partial \xi }=1$ (2.6)

(2.7)

$\frac{\partial \eta }{\partial z}\cdot \frac{\partial z}{\partial \xi }+\frac{\partial \eta }{\partial r}\cdot \frac{\partial r}{\partial \xi }=0$ (2.8)

$\frac{\partial \eta }{\partial z}\cdot \frac{\partial z}{\partial \eta }+\frac{\partial \eta }{\partial r}\cdot \frac{\partial r}{\partial \eta }=1$ (2.9)

结合(2.6)~(2.9)，易得

$\left(\begin{array}{cc}\frac{\partial \xi }{\partial z}& \frac{\partial \xi }{\partial r}\\ \frac{\partial \eta }{\partial z}& \frac{\partial \eta }{\partial r}\end{array}\right)\left(\begin{array}{c}\frac{\partial z}{\partial \xi }\\ \frac{\partial r}{\partial \xi }\end{array}\right)=J\left(\begin{array}{c}\frac{\partial z}{\partial \xi }\\ \frac{\partial r}{\partial \xi }\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)$ (2.10)

$\left(\begin{array}{cc}\frac{\partial \xi }{\partial z}& \frac{\partial \xi }{\partial r}\\ \frac{\partial \eta }{\partial z}& \frac{\partial \eta }{\partial r}\end{array}\right)\left(\begin{array}{c}\frac{\partial z}{\partial \eta }\\ \frac{\partial r}{\partial \eta }\end{array}\right)=J\left(\begin{array}{c}\frac{\partial z}{\partial \eta }\\ \frac{\partial r}{\partial \eta }\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)$ (2.11)

于是有

$\left(\begin{array}{c}\frac{\partial z}{\partial \xi }\\ \frac{\partial r}{\partial \xi }\end{array}\right)=J^{-1}\left(\begin{array}{c}1\\ 0\end{array}\right)=\frac{1}{\left|J\right|}\left(\begin{array}{c}\frac{\partial \eta }{\partial r}\\ -\frac{\partial \xi }{\partial r}\end{array}\right)$ (2.12)

$\left(\begin{array}{c}\frac{\partial z}{\partial \xi }\\ \frac{\partial r}{\partial \xi }\end{array}\right)=J^{-1}\left(\begin{array}{c}1\\ 0\end{array}\right)=\frac{1}{\left|J\right|}\left(\begin{array}{c}-\frac{\partial \eta }{\partial z}\\ \frac{\partial \xi }{\partial z}\end{array}\right)$ (2.13)

由(2.12)和(2.13)可知(2.2)成立，证毕。

3. 多元复合函数的一阶、二阶偏导数

遍及本节，我们有下述假设：

假设 $\omega =f\left(z,r\right)$，且中间变量与自变量之间满足如下关系：

$\{\begin{array}{l}\xi =\xi \left(z,r\right)\hfill \\ \eta =\eta \left(z,r\right)\hfill \end{array}$

进而对上式两边分别关于求偏导，由定理2.1，得到：

$\{\begin{array}{l}\frac{\partial z}{\partial \xi }=\frac{1}{\left|J\right|}\frac{\partial \eta }{\partial r}\hfill \\ \frac{\partial r}{\partial \xi }=-\frac{1}{\left|J\right|}\frac{\partial z}{\partial \eta }\hfill \\ \frac{\partial \eta }{\partial z}=-\frac{1}{\left|J\right|}\frac{\partial r}{\partial \xi }\hfill \\ \frac{\partial \eta }{\partial r}=\frac{1}{\left|J\right|}\frac{\partial z}{\partial \xi }\hfill \end{array}$ (3.1)

其中

$J=\left(\begin{array}{cc}\frac{\partial z}{\partial \xi }& \frac{\partial r}{\partial \xi }\\ \frac{\partial z}{\partial \eta }& \frac{\partial r}{\partial \eta }\end{array}\right),\left|J\right|=\frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }-\frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \eta }$ (3.2)

接下来，我们首先导出多元复合函数一阶偏导数定理。

定理3.1 设 $w=f\left(u,v\right)$，函数 $x=x\left(u,v\right),y=y\left(u,v\right)$ 关于变量 $u,v$ 具有连续的一阶偏导数，记：

$J=\left(\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right)$ (3.3)

若，则

${\left(\begin{array}{c}\frac{\partial w}{\partial x}\\ \frac{\partial w}{\partial y}\end{array}\right)}^{\text{T}}=\frac{1}{\left|J\right|}\left(\begin{array}{cc}\frac{\partial f}{\partial u}& \frac{\partial f}{\partial v}\end{array}\right)J^{-1}$ (3.4)

证明：依据链式求导法则，有

$\frac{\partial w}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\left(\begin{array}{cc}\frac{\partial f}{\partial u}& \frac{\partial f}{\partial v}\end{array}\right)\left(\begin{array}{c}\frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial x}\end{array}\right)$ (3.5)

$\frac{\partial w}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=\left(\begin{array}{cc}\frac{\partial f}{\partial u}& \frac{\partial f}{\partial v}\end{array}\right)\left(\begin{array}{c}\frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial y}\end{array}\right)$ (3.6)

进一步，通过(3.1)式，可以得到

$\frac{\partial w}{\partial x}=\frac{1}{\left|J\right|}\left(\begin{array}{cc}\frac{\partial f}{\partial u}& \frac{\partial f}{\partial v}\end{array}\right)\left(\begin{array}{c}\frac{\partial y}{\partial v}\\ -\frac{\partial y}{\partial u}\end{array}\right)$ (3.7)

$\frac{\partial w}{\partial y}=\frac{1}{\left|J\right|}\left(\begin{array}{cc}\frac{\partial f}{\partial u}& \frac{\partial f}{\partial v}\end{array}\right)\left(\begin{array}{c}-\frac{\partial x}{\partial v}\\ \frac{\partial x}{\partial u}\end{array}\right)$ (3.8)

结合(3.7)~(3.8)，易得(3.4)，证毕。

进一步，我们导出多元复合函数的二阶偏导数定理。

由文献 [1] 中矩阵求多元抽象复合函数二阶偏导数的方法，可以知道

$\left(\frac{{\partial }^2}{\partial z^2}\right)=\left(\begin{array}{cc}\frac{\partial \xi }{\partial z}& \frac{\partial \eta }{\partial z}\end{array}\right)\left[\begin{array}{cc}\frac{{\partial }^2}{\partial {\xi }^2}& \frac{{\partial }^2}{\partial \xi \partial \eta }\\ \frac{{\partial }^2}{\partial \eta \partial \xi }& \frac{{\partial }^2}{\partial {\eta }^2}\end{array}\right]\left(\begin{array}{c}\frac{\partial \xi }{\partial z}\\ \frac{\partial \eta }{\partial z}\end{array}\right)+\left(\begin{array}{cc}\frac{\partial }{\partial \xi }& \frac{\partial }{\partial \eta }\end{array}\right)\left(\begin{array}{c}\frac{{\partial }^2\xi }{\partial z^2}\\ \frac{{\partial }^2\eta }{\partial z^2}\end{array}\right)$ (3.9)

$\left(\frac{{\partial }^2}{\partial {\xi }^2}\right)=\left(\begin{array}{cc}\frac{\partial z}{\partial \xi }& \frac{\partial \gamma }{\partial \xi }\end{array}\right)\left[\begin{array}{cc}\frac{{\partial }^2}{\partial z^2}& \frac{{\partial }^2}{\partial z\partial \gamma }\\ \frac{{\partial }^2}{\partial z\partial \gamma }& \frac{{\partial }^2}{\partial {\gamma }^2}\end{array}\right]\left(\begin{array}{c}\frac{\partial z}{\partial \xi }\\ \frac{\partial \gamma }{\partial \xi }\end{array}\right)+\left(\begin{array}{cc}\frac{\partial }{\partial z}& \frac{\partial }{\partial \gamma }\end{array}\right)\left(\begin{array}{c}\frac{{\partial }^{2}z}{\partial {\xi }^2}\\ \frac{{\partial }^2\gamma }{\partial {\xi }^2}\end{array}\right)$ (3.10)

进而我们可以得到如下关于多元复合函数求二阶偏导数的定理。

定理3.2 设 $\omega =f\left(z,r\right),\xi =\xi \left(z,r\right),\eta =\eta \left(z,r\right)$，且 $f,\xi ,\eta$ 均有连续的二阶偏导数。若记：

$B=\left(\begin{array}{cccc}{\left(\frac{\partial r}{\partial \eta }\right)}^2& -\frac{\partial r}{\partial \xi }\frac{\partial r}{\partial \eta }& -\frac{\partial r}{\partial \xi }\frac{\partial r}{\partial \eta }& {\left(\frac{\partial r}{\partial \xi }\right)}^2\\ -\frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \eta }& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \eta }& -\frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \xi }\\ -\frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \eta }& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \xi }& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }& -\frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \xi }\\ {\left(\frac{\partial z}{\partial \eta }\right)}^2& -\frac{\partial z}{\partial \xi }\frac{\partial z}{\partial \eta }& -\frac{\partial z}{\partial \xi }\frac{\partial z}{\partial \eta }& {\left(\frac{\partial z}{\partial \xi }\right)}^2\end{array}\right)$ (3.11)

且满足 $\left|J\right|\ne 0$，则

(3.12)

(3.13)

证明：对(2.6)和(2.7)两边分别关于 $\xi ,\eta$ 求二阶偏导，易得

$\frac{\partial \xi }{\partial z}\cdot \frac{{\partial }^{2}z}{\partial {\xi }^2}+\frac{\partial z}{\partial \xi }\left[\frac{{\partial }^2\xi }{\partial z^2}\cdot \frac{\partial z}{\partial \xi }+\frac{{\partial }^2\xi }{\partial z\partial r}\cdot \frac{\partial r}{\partial \xi }\right]+\frac{\partial \xi }{\partial r}\cdot \frac{{\partial }^{2}r}{\partial {\xi }^2}+\frac{\partial r}{\partial \xi }\left[\frac{{\partial }^2\xi }{\partial r\partial z}\cdot \frac{\partial z}{\partial \xi }+\frac{{\partial }^2\xi }{\partial r^2}\cdot \frac{\partial r}{\partial \xi }\right]=0$

$\frac{\partial \xi }{\partial z}\cdot \frac{{\partial }^{2}z}{\partial \xi \partial \eta }+\frac{\partial z}{\partial \xi }\left[\frac{{\partial }^2\xi }{\partial z^2}\cdot \frac{\partial z}{\partial \eta }+\frac{{\partial }^2\xi }{\partial z\partial r}\cdot \frac{\partial r}{\partial \eta }\right]+\frac{\partial \xi }{\partial r}\cdot \frac{{\partial }^{2}r}{\partial \xi \partial \eta }+\frac{\partial r}{\partial \xi }\left[\frac{{\partial }^2\xi }{\partial r\partial z}\cdot \frac{\partial z}{\partial \eta }+\frac{{\partial }^2\xi }{\partial r^2}\cdot \frac{\partial r}{\partial \eta }\right]=0$

$\frac{\partial \xi }{\partial z}\cdot \frac{{\partial }^{2}z}{\partial \eta \partial \xi }+\frac{\partial z}{\partial \eta }\left[\frac{{\partial }^2\xi }{\partial z^2}\cdot \frac{\partial z}{\partial \xi }+\frac{{\partial }^2\xi }{\partial z\partial r}\cdot \frac{\partial r}{\partial \xi }\right]+\frac{\partial \xi }{\partial r}\cdot \frac{{\partial }^{2}r}{\partial \xi \partial \eta }+\frac{\partial r}{\partial \eta }\left[\frac{{\partial }^2\xi }{\partial r\partial z}\cdot \frac{\partial z}{\partial \xi }+\frac{{\partial }^2\xi }{\partial r^2}\cdot \frac{\partial r}{\partial \xi }\right]=0$

联立上面4个式子，可得

$\left(\begin{array}{cccc}{\left(\frac{\partial z}{\partial \xi }\right)}^2& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \xi }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \xi }& {\left(\frac{\partial r}{\partial \xi }\right)}^2\\ \frac{\partial z}{\partial \xi }\frac{\partial z}{\partial \eta }& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial r}{\partial \eta }\\ \frac{\partial z}{\partial \eta }\frac{\partial z}{\partial \xi }& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \xi }& \frac{\partial r}{\partial \eta }\frac{\partial z}{\partial \xi }& \frac{\partial r}{\partial \eta }\frac{\partial r}{\partial \xi }\\ {\left(\frac{\partial z}{\partial \eta }\right)}^2& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \eta }\frac{\partial z}{\partial \eta }& {\left(\frac{\partial r}{\partial \eta }\right)}^2\end{array}\right)\left(\begin{array}{c}\frac{{\partial }^2\xi }{\partial z^2}\\ \frac{{\partial }^2\xi }{\partial z\partial r}\\ \frac{{\partial }^2\xi }{\partial r\partial z}\\ \frac{{\partial }^2\xi }{\partial r^2}\end{array}\right)=-\frac{\partial \xi }{\partial z}\cdot \left(\begin{array}{c}\frac{{\partial }^{2}z}{\partial {\xi }^2}\\ \frac{{\partial }^{2}z}{\partial \xi \partial \eta }\\ \frac{{\partial }^{2}z}{\partial \eta \partial \xi }\\ \frac{{\partial }^{2}z}{\partial {\eta }^2}\end{array}\right)-\frac{\partial \xi }{\partial r}\cdot \left(\begin{array}{c}\frac{{\partial }^{2}r}{\partial {\xi }^2}\\ \frac{{\partial }^{2}r}{\partial \xi \partial \eta }\\ \frac{{\partial }^{2}r}{\partial \eta \partial \xi }\\ \frac{{\partial }^{2}r}{\partial {\eta }^2}\end{array}\right)$ (3.14)

记

$A=\left(\begin{array}{cccc}{\left(\frac{\partial z}{\partial \xi }\right)}^2& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \xi }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \xi }& {\left(\frac{\partial r}{\partial \xi }\right)}^2\\ \frac{\partial z}{\partial \xi }\frac{\partial z}{\partial \eta }& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial r}{\partial \eta }\\ \frac{\partial z}{\partial \eta }\frac{\partial z}{\partial \xi }& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \xi }& \frac{\partial r}{\partial \eta }\frac{\partial z}{\partial \xi }& \frac{\partial r}{\partial \eta }\frac{\partial r}{\partial \xi }\\ {\left(\frac{\partial z}{\partial \eta }\right)}^2& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \eta }\frac{\partial z}{\partial \eta }& {\left(\frac{\partial r}{\partial \eta }\right)}^2\end{array}\right)$ (3.15)

为求 $A^{-1}$，依据分块矩阵表示方法，A可写为：

$\begin{array}{l}A=\left(\begin{array}{cccc}{\left(\frac{\partial z}{\partial \xi }\right)}^2& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \xi }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \xi }& {\left(\frac{\partial r}{\partial \xi }\right)}^2\\ \frac{\partial z}{\partial \xi }\frac{\partial z}{\partial \eta }& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial r}{\partial \eta }\\ \frac{\partial z}{\partial \eta }\frac{\partial z}{\partial \xi }& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \xi }& \frac{\partial r}{\partial \eta }\frac{\partial z}{\partial \xi }& \frac{\partial r}{\partial \eta }\frac{\partial r}{\partial \xi }\\ {\left(\frac{\partial z}{\partial \eta }\right)}^2& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \eta }\frac{\partial z}{\partial \eta }& {\left(\frac{\partial r}{\partial \eta }\right)}^2\end{array}\right)\\ =\left(\begin{array}{cc}\frac{\partial z}{\partial \xi }J& \frac{\partial r}{\partial \xi }J\\ \frac{\partial z}{\partial \eta }J& \frac{\partial r}{\partial \eta }J\end{array}\right)=\left(\begin{array}{cc}\frac{\partial z}{\partial \xi }{E}_2& \frac{\partial r}{\partial \xi }{E}_2\\ \frac{\partial z}{\partial \eta }{E}_2& \frac{\partial r}{\partial \eta }{E}_2\end{array}\right)\left(\begin{array}{cc}J& 0\\ 0& J\end{array}\right)\end{array}$ (3.16)

其中 $E_2$ 是 $2\times 2$ 阶单位矩阵，J是(3.3)中所定义的Jacobi矩阵。

进一步，结合(3.1)式，我们有

$A^{-1}=\frac{1}{\left|J\right|}\left(\begin{array}{cc}{J}^{-1}& 0\\ 0& J^{-1}\end{array}\right)\left(\begin{array}{cc}\frac{\partial r}{\partial \eta }{E}_2& -\frac{\partial r}{\partial \xi }{E}_2\\ -\frac{\partial z}{\partial \eta }{E}_2& \frac{\partial z}{\partial \xi }{E}_2\end{array}\right),J^{-1}=\frac{1}{\left|J\right|}\left(\begin{array}{cc}\frac{\partial r}{\partial \eta }& -\frac{\partial r}{\partial \xi }\\ -\frac{\partial z}{\partial \eta }& \frac{\partial z}{\partial \xi }\end{array}\right)$ (3.17)

则可得到

$A^{-1}=\frac{1}{{\left|J\right|}^2}\left(\begin{array}{cccc}{\left(\frac{\partial r}{\partial \eta }\right)}^2& -\frac{\partial r}{\partial \xi }\frac{\partial r}{\partial \eta }& -\frac{\partial r}{\partial \xi }\frac{\partial r}{\partial \eta }& {\left(\frac{\partial r}{\partial \xi }\right)}^2\\ -\frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \eta }& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }& \frac{\partial r}{\partial \xi }\frac{\partial z}{\partial \eta }& -\frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \xi }\\ -\frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \eta }& \frac{\partial z}{\partial \eta }\frac{\partial r}{\partial \xi }& \frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \eta }& -\frac{\partial z}{\partial \xi }\frac{\partial r}{\partial \xi }\\ {\left(\frac{\partial z}{\partial \eta }\right)}^2& -\frac{\partial z}{\partial \xi }\frac{\partial z}{\partial \eta }& -\frac{\partial z}{\partial \xi }\frac{\partial z}{\partial \eta }& {\left(\frac{\partial z}{\partial \xi }\right)}^2\end{array}\right)$ (3.18)

结合(3.14)，易得

$\left(\begin{array}{c}\frac{{\partial }^2\xi }{\partial z^2}\\ \frac{{\partial }^2\xi }{\partial z\partial r}\\ \frac{{\partial }^2\xi }{\partial r\partial z}\\ \frac{{\partial }^2\xi }{\partial r^2}\end{array}\right)=-\frac{\partial \xi }{\partial z}\cdot A^{-1}\left(\begin{array}{c}\frac{{\partial }^{2}z}{\partial {\xi }^2}\\ \frac{{\partial }^{2}z}{\partial \xi \partial \eta }\\ \frac{{\partial }^{2}z}{\partial \eta \partial \xi }\\ \frac{{\partial }^{2}z}{\partial {\eta }^2}\end{array}\right)-\frac{\partial \xi }{\partial r}\cdot A^{-1}\left(\begin{array}{c}\frac{{\partial }^{2}r}{\partial {\xi }^2}\\ \frac{{\partial }^{2}r}{\partial \xi \partial \eta }\\ \frac{{\partial }^{2}r}{\partial \eta \partial \xi }\\ \frac{{\partial }^{2}r}{\partial {\eta }^2}\end{array}\right)$ (3.19)

(3.19)左边可化为

(3.20)

同理，对(2.8)和(2.9)两边分别关于 $\xi ,\eta$ 求偏导，可得

(3.21)

证毕。

致谢

作者衷心感谢张江卫师兄的帮助，感谢湖南省研究生科研创新项目资助(编号CX20200891)。