Implicit Function

隐函数

定义

设$E\subset {\mathbb{R}}^2$,函数$F:E\to \mathbb{R}$, 对于方程

$$F(x,y)=0$$

如果存在集合$I,J\in \mathbb{R}$使得对任意的$x\in I$, 有唯一确定的$y\in J$满足$(x,y)\in E$和上述方程，则该方程确定了一个$I\to J$的隐函数.

例如方程

$$xy+y-1=0$$

确定了一个定义在$(-\infty ,-1)\cup (-1,+\infty )$的隐函数$y=f(x)$，且可以写成显函数的形式

$$y=\frac{1}{1+x}$$

隐函数存在唯一性定理

若函数$F(x,y)$满足以下条件：

1. $F$在以$P_0(x_0,y_0)$为内点的某一区域$D\subset {\mathbb{R}}^2$上连续
2. $F(x_0,y_0)=0$ （通常称为初始条件）
3. $F$在$D$上存在连续的偏导数$F_y(x,y)$
4. $F_y(x_0,y_0)\ne 0$ 则
5. 存在点$P_0$的某个邻域$U(P_0)\subset D$，在$U(P_0)$上$F(x,y)=0$唯一确定了一个定义在某区间$(x_0-\alpha ,x_0+\alpha )$上的隐函数$y=f(x)$，使得当$x\in (x_0-\alpha ,x_0+\alpha )$时，$(x,f(x))\in U(P_0)$且$F(x,f(x))\equiv 0,f(x_0)=y_0$
6. $f(x)$在$(x_0-\alpha ,x_0+\alpha )$上连续

隐函数可微性定理

设$F(x,y)$满足隐函数存在唯一性定理，且在$D$上还存在连续的偏导数$F_x(x,y)$, 则由方程$F(x,y)=0$确定的隐函数$y=f(x)$在其定义域$(x_0-\alpha ,x_0+\alpha )$上有连续导函数，且

$$f^{\prime }(x)=-\frac{F_x(x,y)}{F_y(x,y)}$$

高阶导数

关键是要注意到这里$y$也是$x$的函数

$$\begin{array}{c}F(x,y)=0\\ \frac{\partial F}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{\partial F}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ \frac{{\partial }^{2}F}{\partial x^2}\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{\partial F}{\partial x}y\frac{\mathrm{d}y}{\mathrm{d}x}+\left(\frac{\partial F}{\partial y}x\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{{\partial }^{2}F}{\partial y^2}\frac{\mathrm{d}y}{\mathrm{d}x}\right)\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial F}{\partial y}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=0\\ ⟹y^{\prime \prime }=\frac{2F_x{F}_y{F}_{xy}-F_y^2{F}_{xx}-F_x^2{F}_{yy}}{F_y^3}\end{array}$$

隐函数极值问题

1. 求$y^{\prime }$为$0$的点，即方程组$\{\begin{array}{l}F(x,y)=0\\ F_x(x,y)=0\end{array}$的解$A(x,y)$, 注意需满足$F_y(x,y)\ne 0$
2. 对于每一个$A$, 计算$y^{\prime \prime }{|}_A=-\frac{F_{xx}}{F_y}{|}_A$，并由此判断是极大值还是极小值

隐函数组和隐映射

如由这样的方程组

$$\{\begin{array}{l}F(x,y,u,v)=0\\ G(x,y,u,v)=0\end{array}$$

所确定的隐函数组

$$\{\begin{array}{l}u=f(x,y)\\ v=g(x,y)\end{array}(x,y)\in D,(u,v)\in W$$

本质上可以看成由一个映射方程

$$f(x,y)=(f_1(x,y),f_2(x,y))=0x=(x_1,x_2),y=(y_1,y_2)$$

确定的隐映射

$$y=g(x),x\in D$$

隐映射定理

设$W$为${\mathbb{R}}^{n+m}$中的开集，$W$中的点用$(x,y)$来表示，其中$x=(x_1,\dots ,x_n),y=(y_1,\dots ,y_m)$. $f:W\to {\mathbb{R}}^m$为$C^k$映射:

$$f(x,y)=(f_1(x,y),f_2(x,y),\dots ,f_m(x,y))$$

设$(x^0,y^0)\in W,f(x^0,y^0)=0$且$\det J{f}_y(x^0,y^0)\ne 0$，其中

$$J{f}_y(x,y)={\left(\frac{\partial f_i}{\partial y_j}(x,y)\right)}_{m\times m}$$

则存在$x^0$的开邻域$V\subset {\mathbb{R}}^n$以及唯一的$C^k$映射$g:V\to {\mathbb{R}}^m$使得

1. $y^0=g(x^0),f(x,g(x))=0,\forall x\in V$
2. $Jg(x)=-{\left[J{f}_y(x,g(x))\right]}^{-1}J{f}_x(x,g(x))$ 特别地，当$n=m=2$时，利用伴随矩阵可以直接解得

$$\begin{array}{rl}& \\ (y_1,y_2)& =g(x_1,x_2)\\ Jg=-(J{f}_y{)}^{-1}J{f}_x& =-\frac{1}{\det (J{f}_y)}(J{f}_y{)}^{\ast }J{f}_x\\ ⟹\left(\begin{array}{cc}\frac{\partial y_1}{\partial x_1}& \frac{\partial y_1}{\partial x_2}\\ \frac{\partial y_2}{\partial x_1}& \frac{\partial y_2}{\partial x_2}\end{array}\right)& =-{\left|\begin{array}{cc}\frac{\partial f_1}{\partial y_1}& \frac{\partial f_1}{\partial y_2}\\ \frac{\partial f_2}{\partial y_1}& \frac{\partial f_2}{\partial y_2}\end{array}\right|}^{-1}\left(\begin{array}{cc}\frac{\partial f_2}{\partial y_2}& -\frac{\partial f_1}{\partial y_2}\\ -\frac{\partial f_2}{\partial y_1}& \frac{\partial f_1}{\partial y_1}\end{array}\right)\left(\begin{array}{cc}\frac{\partial f_1}{\partial x_1}& \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1}& \frac{\partial f_2}{\partial x_2}\end{array}\right)\end{array}$$

符号说明

若需要明确指出$f;x,y$的每一个分量的具体的符号，也可以把上面的$J{f}_x,J{f}_y$写成$\frac{D(f_1,\dots ,f_m)}{D(x_1,\dots ,x_n)},\frac{D(f_1,\dots ,f_m)}{D(y_1,\dots ,y_m)}$ 并定义$\frac{\partial (f_1,\dots ,f_m)}{\partial (x_1,\dots ,x_m)}=\left|\frac{D(f_1,\dots ,f_m)}{D(x_1,\dots ,x_m)}\right|,\frac{\partial (f_1,\dots ,f_m)}{\partial (y_1,\dots ,y_m)}=\left|\frac{D(f_1,\dots ,f_m)}{D(y_1,\dots ,y_m)}\right|$ 注意在需要求行列式的时候上面得取$m=n$，也就是Jacobian矩阵应当为一方阵

逆映射和变量代换

若上面的确定的隐映射

$$y=g(x)$$

是双射，那么就有逆映射

$$x=g^{-1}(y)$$

它存在当且仅当（隐映射定理）

$$\det J{y}_x=\left|\frac{D(y_1,\dots ,y_m)}{D(x_1,\dots ,x_m)}\right|\ne 0$$

特别地，我们有

$$\det J{y}_x\cdot \det J{x}_y=\left|\frac{D(y_1,\dots ,y_m)}{D(x_1,\dots ,x_m)}\right|\left|\frac{D(x_1,\dots ,x_m)}{D(y_1,\dots ,y_m)}\right|=1$$

这个工具可以用于分析变量代换，如直角坐标和球坐标之间的变换

$$\{\begin{array}{l}x=r\sin \phi \cos \theta \\ y=r\sin \phi \sin \theta \\ z=r\cos \theta \end{array}$$

由于

$$\begin{array}{rl}\frac{\partial \left(x,y,z\right)}{\partial \left(r,\phi ,\theta \right)}& =\left|\begin{array}{ccc}\sin \phi \cos \theta & r\cos \phi \cos \theta & -r\sin \phi \sin \theta \\ \sin \phi \sin \theta & r\cos \phi \sin \theta & r\sin \phi \cos \theta \\ \cos \phi & -r\sin \phi & 0\end{array}\right|\\ & =r^2\sin \phi ,\end{array}$$

所以在$r^2\sin \phi \ne 0$即除去$z$轴上的一切点，都可以给出$r,\phi ,\theta$为$x,y,z$的函数

如果我们要求（比如说需要在一个关于$u=u(x,y,z)$微分方程中变换变量为$r,\phi ,\theta$）给出$u_r,u_{\phi },u_{\theta }$，则由链式法则

$$\begin{array}{rl}\left(\begin{array}{c}{u}_x\\ u_y\\ u_z\end{array}\right)& =(\begin{array}{ccc}{r}_x& {\theta }_x& {\phi }_x\\ r_y& {\theta }_y& {\phi }_y\\ r_z& {\theta }_z& {\phi }_z\end{array})\left(\begin{array}{c}{u}_r\\ u_{\theta }\\ u_{\phi }\end{array}\right)={\left(\begin{array}{ccc}{x}_r& y_r& z_r\\ x_{\theta }& y_{\theta }& z_{\theta }\\ x_{\phi }& y_{\phi }& z_{\phi }\end{array}\right)}^{-1}\left(\begin{array}{c}{u}_r\\ u_{\theta }\\ u_{\phi }\end{array}\right)\\ & =\frac{1}{J}(\begin{array}{ccc}{r}^2{\sin }^2\theta \cos \phi & r^2{\sin }^2\theta \sin \phi & r^2\sin \theta \cos \theta \\ r\sin \theta \cos \theta \cos \phi & r\sin \theta \cos \theta \sin \phi & -r{\sin }^2\theta \\ -r\sin \phi & r\cos \phi & 0\end{array})\left(\begin{array}{c}{u}_r\\ u_{\theta }\\ u_{\phi }\end{array}\right).\end{array}$$

其中

$$J=\frac{\partial (x,y,z)}{\partial (r,\phi ,\theta )}=r^2\sin \phi$$

这里用到了

$$(\begin{array}{ccc}{r}_x& {\theta }_x& {\phi }_x\\ r_y& {\theta }_y& {\phi }_y\\ r_z& {\theta }_z& {\phi }_z\end{array})\left(\begin{array}{ccc}{x}_r& y_r& z_r\\ x_{\theta }& y_{\theta }& z_{\theta }\\ x_{\phi }& y_{\phi }& z_{\phi }\end{array}\right)=E$$

因为有

$$\begin{array}{c}\frac{\partial r}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial r}{\partial y}\frac{\partial y}{\partial r}+\frac{\partial r}{\partial z}\frac{\partial z}{\partial r}=\frac{\partial r}{\partial r}=1\\ \cdots \end{array}$$

对矩阵积的对角线元素成立，和

$$\begin{array}{c}\\ \frac{\partial r}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial r}{\partial y}\frac{\partial y}{\partial \theta }+\frac{\partial r}{\partial z}\frac{\partial z}{\partial \theta }=\frac{\partial r}{\partial \theta }=0\\ \cdots \end{array}$$

对矩阵积的非对角线元素成立

几何应用

平面曲线的切线和法线

对于平面曲线

$$F(x,y)=0$$

如果在$(x_0,y_0)$附近满足隐函数定理的条件，那么由

$$f^{\prime }(x)=-\frac{F_x}{F_y}$$

得到切线

$$F_x(x_0,y_0)(x-x_0)+F_y(x_0,y_0)(y-y_0)=0$$

和法线

$$F_x(x_0,y_0)(x-x_0)-F_y(x_0,y_0)(y-y_0)=0$$

空间曲线的切线和法平面

对于由参数方程确定的空间曲线

$$L:x=x(t),y=y(t),z=z(t),\alpha \le t\le \beta$$

由对称式方程，可以得到$L$在$(x_0,y_0,z_0)$的切线方程为

$$\frac{x-x_0}{x^{\prime }(t_0)}=\frac{y-y_0}{y^{\prime }(t_0)}=\frac{z-z_0}{z^{\prime }(t_0)}$$

而法平面$\Pi$满足

$$(x^{\prime }(t_0),y^{\prime }(t_0),z^{\prime }(t_0))\cdot (x-x_0,y-y_0,z-z_0)=0$$

即

$$x^{\prime }(t_0)(x-x_0)+y^{\prime }(t_0)(y-y_0)+z^{\prime }(t_0)(z-z_0)=0$$

如果能够给出确定$L$的隐映射

$$L:f(x,y,z)=(f_1(x,y,z),f_2(x,y,z))=(0,0)$$

且在$(x_0,y_0,z_0)$的某邻域满足隐映射定理 要求$\frac{\mathrm{d}x}{\mathrm{d}z}$和$\frac{\mathrm{d}y}{\mathrm{d}z}$，则可取自变量$X=(z)$，因变量$Y=(x,y)$，满足$(x,y)=g(z)$有

$$\begin{array}{rl}Jg(z)& =-(J{f}_Y(x,y){)}^{-1}(J{f}_X(x,y))\\ & =-{\left|\begin{array}{cc}\frac{\partial f_1}{\partial x}& \frac{\partial f_1}{\partial y}\\ \frac{\partial f_2}{\partial x}& \frac{\partial f_2}{\partial y}\end{array}\right|}^{-1}\left(\begin{array}{cc}\frac{\partial f_2}{\partial y}& -\frac{\partial f_1}{\partial y}\\ -\frac{\partial f_2}{\partial x}& \frac{\partial f_1}{\partial x}\end{array}\right)\left(\begin{array}{c}\frac{\partial f_1}{\partial z}\\ \frac{\partial f_2}{\partial z}\end{array}\right)\end{array}$$

即可算出

$$\frac{\mathrm{d}x}{\mathrm{d}z}=\frac{\partial x}{\partial z}=\frac{\partial g_1}{\partial z};\frac{\mathrm{d}y}{\mathrm{d}z}=\frac{\partial y}{\partial z}=\frac{\partial g_2}{\partial z}$$

于是有法平面方程

$$\frac{x-x_0}{\frac{\mathrm{d}x}{\mathrm{d}z}}+\frac{y-y_0}{\frac{\mathrm{d}y}{\mathrm{d}z}}+(z-z_0)=0$$

和切线方程

$$\frac{x-x_0}{\frac{\mathrm{d}x}{\mathrm{d}z}}=\frac{y-y_0}{\frac{\mathrm{d}y}{\mathrm{d}z}}=\frac{z-z_0}{1}$$

曲面的切平面与法线

设曲面方程

$$F(x,y,z)=0$$

在$(x_0,y_0,z_0)$的某邻域满足隐映射定理 结合可微性的几何意义，要先确定隐函数$z=f(x,y)$才能写出法向量

$$\mathbf{n}=(f_x(x_0,y_0),f_y(x_0,y_0),-1)$$

由于

$$\left(\begin{array}{cc}\frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}\end{array}\right)=-{\left(\begin{array}{c}\frac{\partial F}{\partial z}\end{array}\right)}^{-1}\left(\begin{array}{cc}\frac{\partial F}{\partial x}& \frac{\partial F}{\partial y}\end{array}\right)$$

故有法线方程

$$\frac{x-x_0}{F_x}=\frac{y-y_0}{F_y}=\frac{z-z_0}{F_z}$$

和切平面方程

$$F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0$$