6.2.4 微分表达式中的变量代换与坐标变换

6.2.4.1 一元函数

已知一个函数 $y\left(x\right)$ 和一个含自变量、函数 $y\left(x\right)$ 及其导数的微分表达式 $F$ :

$$\begin{array}{}\text{(6.58a)}& y=f\left(x\right),\end{array}$$$$\begin{array}{}\text{(6.58b)}& F=F\left(x,y,\frac{\mathrm{d}y}{\mathrm{d}x},\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2},\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3},\cdots \right).\end{array}$$

若作变量代换, 则可按如下方法计算导数.

情况 1a 用变量 $t$ 代替变量 $x$ ,有

$$\begin{array}{}\text{(6.59a)}& x=\phi \left(t\right),\end{array}$$

则

$$\begin{array}{}\text{(6.59b)}& \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{\phi }^{\prime }\left(t\right)}\frac{\mathrm{d}y}{\mathrm{d}t},\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{1}{{\left[{\phi }^{\prime }\left(t\right)\right]}^3}\left\{{\phi }^{\prime }\left(t\right)\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}-{\phi }^{\prime \prime }\left(t\right)\frac{\mathrm{d}y}{\mathrm{d}t}\right\},\end{array}$$$$\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}=\frac{1}{{\left[{\phi }^{\prime }\left(t\right)\right]}^5}\{{\left[{\phi }^{\prime }\left(t\right)\right]}^2\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{t}^3}-3{\phi }^{\prime }\left(t\right){\phi }^{\prime \prime }\left(t\right)\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}$$$$\begin{array}{}\text{(6.59c)}& +\left[3{\left[{\phi }^{\prime \prime }\left(t\right)\right]}^2-{\phi }^{\prime }\left(t\right){\phi }^{\prime \prime \prime }\left(t\right)\right]\frac{\mathrm{d}y}{\mathrm{d}x}\},\cdots .\end{array}$$

情况 $1\mathrm{b}$ 若 $x,t$ 之间的关系不是以显形式给出,而是满足隐函数方程

$$\begin{array}{}\text{(6.60)}& \mathrm{\Phi }\left(x,t\right)=0,\end{array}$$

则可按同样的公式计算导数 $\frac{\mathrm{d}y}{\mathrm{d}x},\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2},\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}$ ,不过导数 ${\phi }^{\prime }\left(t\right),{\phi }^{\prime \prime }\left(t\right),{\phi }^{\prime \prime \prime }\left(t\right)$ 必须根据隐函数法则来计算. 在这种情况下可能 (6.58b) 中包含着变量 $x$ ,为了消去 $x$ ,可利用 (6.60).

情况 2 若用函数 $u\left(x\right)$ 代替函数 $y$ ,它们间的关系为

$$\begin{array}{}\text{(6.61a)}& y=\phi \left(u\right),\end{array}$$

则可利用下面的公式计算导数:

$$\begin{array}{}\text{(6.61b)}& \frac{\mathrm{d}y}{\mathrm{d}x}={\phi }^{\prime }\left(u\right)\frac{\mathrm{d}u}{\mathrm{d}x},\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}={\phi }^{\prime }\left(u\right)\frac{{\mathrm{d}}^{2}u}{\mathrm{d}{x}^2}+{\phi }^{\prime \prime }\left(u\right){\left(\frac{\mathrm{d}u}{\mathrm{d}x}\right)}^2,\end{array}$$$$\begin{array}{}\text{(6.61c)}& \frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}={\phi }^{\prime }\left(u\right)\frac{{\mathrm{d}}^{3}u}{\mathrm{d}{x}^3}+3{\phi }^{\prime \prime }\left(u\right)\frac{\mathrm{d}u}{\mathrm{d}x}\frac{{\mathrm{d}}^{2}u}{\mathrm{d}{x}^2}+{\phi }^{\prime \prime \prime }\left(u\right){\left(\frac{\mathrm{d}u}{\mathrm{d}x}\right)}^3,\cdots .\end{array}$$

情况 3 用新的变量 $t,u$ 代替 $x,y$ ,它们间的关系为

$$\begin{array}{}\text{(6.62a)}& x=\phi \left(t,u\right),y=\psi \left(t,u\right),\end{array}$$

则可利用下面的公式计算导数:

$$\begin{array}{}\text{(6.62b)}& \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\partial \psi }{\partial t}+\frac{\partial \psi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}}{\frac{\partial \phi }{\partial t}+\frac{\partial \phi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}}\end{array}$$$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{\frac{\partial \psi }{\partial t}+\frac{\partial \psi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}}{\frac{\partial \phi }{\partial t}+\frac{\partial \phi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}}\right]=\frac{1}{\frac{\partial \phi }{\partial t}+\frac{\partial \phi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}}\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\frac{\partial \psi }{\partial t}+\frac{\partial \psi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}}{\frac{\partial \phi }{\partial t}+\frac{\partial \phi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}}\right],$$

(6.62c)

$$\begin{array}{}\text{(6.62d)}& \frac{1}{B}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{A}{B}\right)=\frac{1}{B^3}\left(B\frac{\mathrm{d}A}{\mathrm{d}t}-A\frac{\mathrm{d}B}{\mathrm{d}t}\right),\end{array}$$

其中

$$\begin{array}{}\text{(6.62e)}& A=\frac{\partial \psi }{\partial t}+\frac{\partial \psi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}\end{array}$$$$\begin{array}{}\text{(6.62f)}& B=\frac{\partial \phi }{\partial t}+\frac{\partial \phi }{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}.\end{array}$$

类似地,可确定三阶导数 $\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}$ .

口利用

$$\begin{array}{}\text{(6.63a)}& x=\rho \cos \phi ,y=\rho \sin \phi \end{array}$$

可把笛卡儿坐标变成极坐标. 由以下公式可计算一阶和二阶导数:

$$\begin{array}{}\text{(6.63b)}& \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{{\rho }^{\prime }\sin \phi +\rho \cos \phi }{{\rho }^{\prime }\cos \phi -\rho \sin \phi },\end{array}$$$$\begin{array}{}\text{(6.63c)}& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{{\rho }^2+2{\rho }^{\prime 2}-\rho {\rho }^{\prime \prime }}{{({\rho }^{\prime }\cos \phi -\rho \sin \phi )}^3}.\end{array}$$

6.2.4.2 二元函数

已知一函数 $w\left(x,y\right)$ 以及一个含自变量、函数 $w\left(x,y\right)$ 及其偏导数的微分表达式 $F$ :

$$\begin{array}{}\text{(6.64a)}& \omega =f\left(x,y\right),\end{array}$$$$\begin{array}{}\text{(6.64b)}& F=F\left(x,y,\omega ,\frac{\partial \omega }{\partial x},\frac{\partial \omega }{\partial y},\frac{{\partial }^2\omega }{\partial x^2},\frac{{\partial }^2\omega }{\partial x\partial y},\frac{{\partial }^2\omega }{\partial y^2},\cdots \right).\end{array}$$

若用新的变量 $u,v$ 代替 $x,y$

$$\begin{array}{}\text{(6.65a)}& x=\phi \left(u,v\right),y=\psi \left(u,v\right),\end{array}$$

则一阶偏导数可由方程组

$$\begin{array}{}\text{(6.65b)}& \frac{\partial \omega }{\partial u}=\frac{\partial \omega }{\partial x}\frac{\partial \phi }{\partial u}+\frac{\partial \omega }{\partial y}\frac{\partial \psi }{\partial u},\frac{\partial \omega }{\partial v}=\frac{\partial \omega }{\partial x}\frac{\partial \phi }{\partial v}+\frac{\partial \omega }{\partial y}\frac{\partial \psi }{\partial v}\end{array}$$

给出,设 $A,B,C,D$ 为新变量 $u,v$ 的新函数,则

$$\begin{array}{}\text{(6.65c)}& \frac{\partial \omega }{\partial x}=A\frac{\partial \omega }{\partial u}+B\frac{\partial \omega }{\partial v},\frac{\partial \omega }{\partial y}=C\frac{\partial \omega }{\partial u}+D\frac{\partial \omega }{\partial v}.\end{array}$$

用同样的公式可计算二阶偏导数,只不过不再用 $w$ ,而是用偏导数 $\frac{\partial w}{\partial x},\frac{\partial w}{\partial y}$ ,例如

$$\frac{{\partial }^2\omega }{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial \omega }{\partial x}\right)=\frac{\partial }{\partial x}\left(A\frac{\partial \omega }{\partial u}+B\frac{\partial \omega }{\partial v}\right)$$$$=A\left(A\frac{{\partial }^2\omega }{\partial u^2}+B\frac{{\partial }^2\omega }{\partial u\partial v}+\frac{\partial A}{\partial u}\frac{\partial \omega }{\partial u}+\frac{\partial B}{\partial u}\frac{\partial \omega }{\partial v}\right)$$$$\begin{array}{}\text{(6.66)}& +B\left(A\frac{{\partial }^2\omega }{\partial u\partial v}+B\frac{{\partial }^2\omega }{\partial v^2}+\frac{\partial A}{\partial v}\frac{\partial \omega }{\partial u}+\frac{\partial B}{\partial v}\frac{\partial \omega }{\partial v}\right).\end{array}$$

用同样的方法可以计算高阶偏导数.

$◼$ 把拉普拉斯算子 (参见第 934 页 13.2.6.5) 用极坐标 (参见第 255 页 3.5.2.1, 2.) 表示:

$$\begin{array}{}\text{(6.67a)}& \mathrm{\Delta }\omega =\frac{{\partial }^2\omega }{\partial x^2}+\frac{{\partial }^2\omega }{\partial y^2},\end{array}$$$$\begin{array}{}\text{(6.67b)}& x=\rho \cos \phi ,y=\rho \sin \phi .\end{array}$$

计算如下:

$$\frac{\partial \omega }{\partial \rho }=\frac{\partial \omega }{\partial x}\cos \phi +\frac{\partial \omega }{\partial y}\sin \phi ,\frac{\partial \omega }{\partial \phi }=-\frac{\partial \omega }{\partial x}\rho \sin \phi +\frac{\partial \omega }{\partial y}\rho \cos \phi ,$$$$\frac{\partial \omega }{\partial x}=\cos \phi \frac{\partial \omega }{\partial \rho }-\frac{\sin \phi }{\rho }\frac{\partial \omega }{\partial \phi },\frac{\partial \omega }{\partial y}=\sin \phi \frac{\partial \omega }{\partial \rho }+\frac{\cos \phi }{\rho }\frac{\partial \omega }{\partial \phi },$$$$\frac{{\partial }^2\omega }{\partial x^2}=\cos \phi \frac{\partial }{\partial \rho }\left(\cos \phi \frac{\partial \omega }{\partial \rho }-\frac{\sin \phi }{\rho }\frac{\partial \omega }{\partial \phi }\right)-\frac{\sin \phi }{\rho }\frac{\partial }{\partial \phi }\left(\cos \phi \frac{\partial \omega }{\partial \rho }-\frac{\sin \phi }{\rho }\frac{\partial \omega }{\partial \phi }\right).$$

类似地,可以计算 $\frac{{\partial }^{2}w}{\partial y^2}$ ,最终有

$$\begin{array}{}\text{(6.67c)}& \mathrm{\Delta }\omega =\frac{{\partial }^2\omega }{\partial {\rho }^2}+\frac{1}{{\rho }^2}\frac{{\partial }^2\omega }{\partial {\phi }^2}+\frac{1}{\rho }\frac{\partial \omega }{\partial \rho }.\end{array}$$

注 若含有两个以上变量的函数被代换, 可得到类似的代换公式.