4. 隐函数与反函数的微分

多变量函数的微分学

张瑞
中国科学技术大学数学科学学院

隐函数与反函数的微分

隐函数存在定理

由方程或方程组确定的函数关系,称为隐函数

例 1. 几个例子:

  1. $xy+x+y-1=0$ ,可以显式解出$y=f(x)$

  2. $x-y^2+\sin y=0$ ,可以显式得到$x=x(y)$

  3. $\frac{x^2}{2x^2+\sin x+1}e^{-y}-y=0 $,没法得到$y=y(x)$,但可以知道每一个$x$对应一个$y$

  4. $x^2+y^2=1$$x$不能表达为$y$的函数,$y$也不能表达为$x$的函数,但局部上是可以的

chap6-4-ex1

$\frac{x^2}{2x^2+\sin x+1}e^{-y}-y=0 $

利用方程定义函数时,有两个问题:

  1. 未必能得到解析表达
  2. 函数可能只在局部有定义

定义 1.
$F(x,y)$在区域$D\subset\mathbb{R}^2$上有定义, $(x_0,y_0)\in D$$F(x_0,y_0)=0$。若存在$(x_0,y_0)$的邻域$I\times J\subset D$,对于任一$x\in I$都有唯一的$y\in J$,使得$F(x,y)=0$,则由此对应关系确定的$I$上的函数$y=\phi(x)$称为在$(x_0,y_0)$的邻域中由方程$F(x,y)=0$所确定的隐函数。若对于任一$y\in J$都有唯一的$x\in I$,使得$F(x,y)=0$,则由此确定的$J$上的函数$x=\psi(y)$也称为在$(x_0,y_0)$的邻域中由方程$F(x,y)=0$所确定的隐函数

例 2. 方程$F(x,y)=x^3+y^3-3xy=0$所确定的曲线,在$(0,0)$附近不存在隐函数

chap6-4-ex2-3d chap6-4-ex2

隐函数定理

定理 1. (隐函数定理)
设区域$D\subset\mathbb{R}^2$$(x_0,y_0)\in D$。若$F(x,y)$$D$上有定义,且满足:

(a) $F(x,y)\in C^1(D)$。即$F$$D$上有连续的偏导数

(b) $F(x_0,y_0)=0$

(c) $F'_y(x_0,y_0)\neq 0$

则有如下结论:

(1) 方程$F(x,y)=0$$(x_0,y_0)$附近确定了隐函数$y=\phi(x)$

(2) 隐函数$y=\phi(x)$$C^1$的,且

\[\phi'(x)=\frac{F'_x(x,y)}{F'_y(x,y)} , x\in I \]

若有$F(x,\phi(x))=0$,则

\[\dfrac{d(F(x,\phi(x)))}{dx}=0 \]

\[\frac{\partial F}{\partial x}(x,\phi(x))+\frac{\partial F}{\partial y}(x,\phi(x))\phi'(x)=0 \]

所以

\[\phi'(x)=-\frac{F'_x(x,\phi(x))}{F'_y(x,\phi(x))} \]

例 3. $\ln\sqrt{x^2+y^2}=\arctan\frac{y}x$,求$y'(x)$

例 4. $z=f(x,y)$,满足$F(x,y,f(x,y))=0$,求

\[\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \]

更一般地,记

\[\vec x=(x_1,x_2,\cdots,x_n) , \vec y=(y_1,y_2,\cdots,y_m) \]
\[\vec f=(f_1,f_2,\cdots,f_m)^T \]

\[\begin{cases} & y_1=f_1(x_1,\cdots,x_n) \\ & \cdots \\ & y_m=f_m(x_1,\cdots,x_n) \end{cases} \]

可以简单记为 $ \vec y=\vec f(\vec x) $

$m$个方程组成的方程组,

\[\begin{cases} & F_1(x_1,\cdots,x_n,y_1,\cdots,y_m)=0 \\ & \cdots \\ & F_m(x_1,\cdots,x_n,y_1,\cdots,y_m)=0 \end{cases} \]

\[\begin{aligned} \vec F=(F_1,F_2,\cdots,F_m)^T \\ \end{aligned} \]

则,可以简单记为 $\vec F(\vec x,\vec y)=0 $。 记

\[J_xF=\begin{pmatrix} \frac{\partial F_1}{\partial x_1} & \cdots & \frac{\partial F_1}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial F_m}{\partial x_1} & \cdots & \frac{\partial F_m}{\partial x_n} \end{pmatrix} , J_yF=\begin{pmatrix} \frac{\partial F_1}{\partial y_1} & \cdots & \frac{\partial F_1}{\partial y_m} \\ \vdots & & \vdots \\ \frac{\partial F_m}{\partial y_1} & \cdots & \frac{\partial F_m}{\partial y_m} \end{pmatrix} \]

隐映射定理

定理 2. (隐映射定理)
$D\subset\mathbb{R}^{n+m}$为开集,映射 $F:D\to\mathbb{R}^m$满足下列条件:

(1) $F\in C^1(D)$

(2) $F(x_0, y_0)=0$

(3) $\mbox{det} J_yF( x_0, y_0)\neq0$

其中$x_0\in\mathbb{R}^n$, $y_0\in\mathbb{R}^m$, 则有下列结论:

(1) 方程组$F(x_0,y_0)=0$$(x_0,y_0)$附近确定了隐映射$y=f(x)$$f=(f_1,\cdots,f_m)^T$;即存在$(x_0,y_0)$的邻域$U\times V\subset D$及映射$f: U\to V$,使得$y=f(x)$$V$中满足$F(x,y)=0$的唯一元素($x\in U$);

(2) 隐映射 $\vec y=\vec f(\vec x)$$C^1$的,且

\[Jf(x)=-[J_yF(x,y)]^{-1}J_xF(x,y) , x\in U \]

例 5. $z=z(x,y)$是由方程$ax^2+by^2+cz^2=1$所确定的隐函数,求

\[\frac{\partial ^2z}{\partial x^2},\frac{\partial ^2z}{\partial y^2}, \frac{\partial ^2z}{\partial x\partial y}, \frac{\partial z}{\partial x} \]

例 6. 已知

\[\begin{cases} & xu-yv=0 \\ & yu+xv=1 \end{cases} \]

$\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$

例 7. 已知 $x=\phi(u,v), y=\psi(u,v), z=f(u,v)$,求

\[\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \]

例 5

例 6

例 7

反函数存在定理与微分

在一维中,对于反函数的导数,有如下公式

\[(f^{-1}(y))'=\frac1{f'(x)} \]

其中,$y=f(x)$, $x=f^{-1}(y)$


多元映射也是类似的。

定理 3. (逆映射定理)
$D\subset\mathbb{R}^n$为开集,$f:D\to\mathbb{R}^n$$C^1$映射,$x_0\in D$$\mbox{det}Jf(x_0)\neq 0$,则存在$x_0$的邻域$U$,使得$V=f(U)$为一一映射,从而有逆映射$f^{-1}:V\to U$,且对$\forall y\in V$,有

\[Jf^{-1}(y)=[Jf(x)]^{-1} \]

其中$x\in U$$f(x)=y$

一维中,由

\[f^{-1}(y)=f^{-1}(f(x))=x \]

两边对$x$求导,由链式法则,有

\[(f^{-1}(y))'f'(x)=1 \]

则有

\[(f^{-1}(y))'=\frac1{f'(x)} \]

多变量也是类似。如$x=(x_1,x_2)$, $y=(y_1,y_2)$$f=(f_1,f_2)$,且有

\[\begin{cases} & y_1=f_1(x_1,x_2) \\ & y_2=f_2(x_1,x_2) \end{cases} \]

\[\begin{cases} & x_1=f_1^{-1}(y_1,y_2) \\ & x_2=f_2^{-1}(y_1,y_2) \end{cases} \]

同样有

\[\begin{cases} & x_1=f_1^{-1}(f_1(x_1,x_2),f_2(x_1,x_2)) \\ & x_2=f_2^{-1}(f_1(x_1,x_2),f_2(x_1,x_2)) \end{cases} \]

第一式对$x_1$$x_2$求偏导,可以得到

\[\begin{aligned} \frac{\partial f_1^{-1}}{\partial y_1}\frac{\partial y_1}{\partial x_1}+\frac{\partial f_1^{-1}}{\partial y_2}\frac{\partial y_2}{\partial x_1}=1 \\ \frac{\partial f_1^{-1}}{\partial y_1}\frac{\partial y_1}{\partial x_2}+\frac{\partial f_1^{-1}}{\partial y_2}\frac{\partial y_2}{\partial x_2}=0 \end{aligned} \]

写成矩阵形式为

\[\begin{pmatrix} \frac{\partial f_1^{-1}}{\partial y_1} & \frac{\partial f_1^{-1}}{\partial y_2} \end{pmatrix} \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{pmatrix} =\begin{pmatrix} 1 & 0 \end{pmatrix} \]

第二式对$x_1$$x_2$求偏导,可以得到

\[\begin{aligned} \frac{\partial f_2^{-1}}{\partial y_1}\frac{\partial y_1}{\partial x_1}+\frac{\partial f_2^{-1}}{\partial y_2}\frac{\partial y_2}{\partial x_1}=0 \\ \frac{\partial f_2^{-1}}{\partial y_1}\frac{\partial y_1}{\partial x_2}+\frac{\partial f_2^{-1}}{\partial y_2}\frac{\partial y_2}{\partial x_2}=1 \end{aligned} \]

写成矩阵形式为

\[\begin{pmatrix} \frac{\partial f_2^{-1}}{\partial y_1} & \frac{\partial f_2^{-1}}{\partial y_2} \end{pmatrix} \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{pmatrix} =\begin{pmatrix} 0 & 1 \end{pmatrix} \]

组合后,即有

\[\begin{pmatrix} \frac{\partial f_1^{-1}}{\partial y_1} & \frac{\partial f_1^{-1}}{\partial y_2} \\ \frac{\partial f_2^{-1}}{\partial y_1} & \frac{\partial f_2^{-1}}{\partial y_2} \end{pmatrix} \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{pmatrix} =\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

\[Jf^{-1}(y)=[Jf(x)]^{-1} \]

例 8. (例6.4.5) 求极坐标变换的反变换的偏导数。即由

\[\begin{cases} & x=r\cos\theta \\ & y=r\sin\theta \end{cases} \]

\[\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y} \]

由定理知

\[Jf^{-1}=[Jf]^{-1} \]

则有

\[\begin{pmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} \end{pmatrix} =\begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{pmatrix}^{-1} \]
\[=\begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}^{-1} =\begin{pmatrix} \cos\theta & \sin\theta \\ -\frac1r \sin\theta & \frac1r\cos\theta \end{pmatrix} \]

或者,则$x=r\cos\theta$,两边对$x$求导,有

\[1=\frac{\partial r}{\partial x}\cos\theta+r(-\sin\theta)\frac{\partial \theta}{\partial x} \]

$y=r\sin\theta$,两边对$x$求导,有

\[0=\frac{\partial r}{\partial x}\sin\theta+r(\cos\theta)\frac{\partial \theta}{\partial x} \]

两式联立,可解出

\[\begin{cases} & \frac{\partial r}{\partial x}=\cos\theta \\ & \frac{\partial \theta}{\partial x}=-\frac1r\sin\theta \end{cases} \]

目录

谢谢

例 9. 本节读完

9.