第八周。

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\end{proof} \end{proof}
总结起来,偏导数在$\bvec{a}$点都连续可以推出函数在$\bvec{a}$点可微,进而可以推出函数在$\bvec{a}$点连续,也可以推出函数在$\bvec{a}$点所有方向导数都存在。 总结起来,偏导数在$\bvec{a}$点都连续可以推出函数在$\bvec{a}$点可微,进而可以推出函数在$\bvec{a}$点连续,也可以推出函数在$\bvec{a}$点所有方向导数都存在。
\section{向量值函数的微分}
\begin{definition}
如果映射\boldf 满足存在Jacobian $J \boldf (\bvec{x}_0)$且满足
\[\boldf (\bvec{x}_0 + \Delta \bvec{x}) - \boldf (\bvec{x}_0) = J \boldf (\bvec{x}_0) \Delta \bvec{x} + o\left(\norm{\Delta \bvec{x}}\right)\]
其中
\[J\boldf (\bvec{x}_0) =
\begin{bmatrix}
D_1 f_1(\bvec{x}_0) & \cdots & D_n f_1(\bvec{x}_0)\\
\vdots & \ddots & \vdots\\
D_1 f_m(\bvec{x}_0) & \cdots & D_n f_m(\bvec{x}_0)
\end{bmatrix}
=
\begin{bmatrix}
\gra f_1 (\bvec{x}_0)\\
\vdots\\
\gra f_m (\bvec{x}_0)
\end{bmatrix}\]
此时\boldf$\bvec{x}_0$点的微分记为
\[\dif \boldf (\bvec{x}_0) = J \boldf (\bvec{x}_0) \Delta \bvec{x} \eqper\]
\end{definition}
\begin{theorem}
若映射\boldf 在开集$D$上存在Jacobian $J \boldf$,且$J \boldf$的各元素在点$\bvec{x}_0$处都连续,则映射\boldf 在点$\bvec{x}_0$处可微。
\end{theorem}
\section{复合求导}
\begin{theorem}
$D \in \ndreal$$\bvec{f}: D \to \realnum^m$$\bvec{g}: \Omega \to \realnum^k$$\bvec{f}(D) \subset \Omega \subset \realnum^m$。如果\boldf$\bvec{x}_0 \in D\interior$上可微,$\bvec{g}$$\boldf(\bvec{x}_0)$上可微,那么复合映射$\bvec{g} \circ \boldf$在点$\bvec{x}_0$处可微,且
\[J(\bvec{g} \circ \bvec{f}) = J \bvec{g}(\boldf (\bvec{x}_0)) J \boldf(\bvec{x}_0)\eqper\]
\end{theorem}
如果我们记$\bvec{u} = \bvec{g}(\bvec{y}), \bvec{y} = \bvec{f}(\bvec{x})$,那么$\bvec{g} \circ \bvec{f}$的Jacobin可以写为
\[\begin{bmatrix}
\dfrac{\partial u_1}{\partial x_1} & \dfrac{\partial u_1}{\partial x_2} & \cdots & \dfrac{\partial u_1}{\partial x_n}\\[1em]
\dfrac{\partial u_2}{\partial x_1} & \dfrac{\partial u_2}{\partial x_2} & \cdots & \dfrac{\partial u_2}{\partial x_n}\\[1ex]
\vdots & \vdots & \ddots & \vdots\\
\dfrac{\partial u_k}{\partial x_1} & \dfrac{\partial u_k}{\partial x_2} & \cdots & \dfrac{\partial u_k}{\partial x_n}
\end{bmatrix}
=
\begin{bmatrix}
\dfrac{\partial u_1}{\partial y_1} & \dfrac{\partial u_1}{\partial y_2} & \cdots & \dfrac{\partial u_1}{\partial y_m}\\[1em]
\dfrac{\partial u_2}{\partial y_1} & \dfrac{\partial u_2}{\partial y_2} & \cdots & \dfrac{\partial u_2}{\partial y_m}\\[1ex]
\vdots & \vdots & \ddots & \vdots\\
\dfrac{\partial u_k}{\partial y_1} & \dfrac{\partial u_k}{\partial y_2} & \cdots & \dfrac{\partial u_k}{\partial y_m}
\end{bmatrix}
\begin{bmatrix}
\dfrac{\partial y_1}{\partial x_1} & \dfrac{\partial y_1}{\partial x_2} & \cdots & \dfrac{\partial y_1}{\partial x_n}\\[1em]
\dfrac{\partial y_2}{\partial x_1} & \dfrac{\partial y_2}{\partial x_2} & \cdots & \dfrac{\partial y_2}{\partial x_n}\\[1ex]
\vdots & \vdots & \ddots & \vdots\\
\dfrac{\partial y_m}{\partial x_1} & \dfrac{\partial y_m}{\partial x_2} & \cdots & \dfrac{\partial y_m}{\partial x_n}
\end{bmatrix}\eqper\]
\section{隐函数定理}
\begin{theorem}[隐函数定理]
设开集$D \subset \realnum^2$,函数$F: D \to \realnum$满足条件:
\begin{enumerate}[label=(\roman{*})]
\item $F \in C^1(D)$
\item$(x_0, y_0) \in D$使得$F(x_0, y_0) = 0$
\item $\dfrac{\partial F(x_0, y_0)}{\partial y} \neq 0$
\end{enumerate}
则存在$\delta, \eta > 0$以及唯一的函数$f: (x_0 - \delta, x_0 + \delta) \to (y_0 - \eta, y_0 + \eta)$具有性质
\begin{enumerate}
\item 对任意的$\abs{x - x_0} < \delta$$f(x_0) = y_0$,有$F(x, f(x)) = 0$
\item $f \in C^1(x_0 - \delta, x_0 + \delta)$
\item$x \in (x_0 - \delta, x_0 + \delta)$$y = f(x)$,有
\[\deriv{f}(x) = -\frac{\dfrac{\partial F}{\partial x}(x, y)}{\dfrac{\partial F}{\partial y}(x, y)}\eqper\]
\end{enumerate}
\end{theorem}
\begin{theorem}
设开集$D \subset \realnum^{n + 1}$$F: D \to \realnum$,满足条件:
\begin{enumerate}[label=(\roman{*})]
\item $F \in C^{(1)}(D)$
\item$(\bvec{x}_0, y_0) \in D$使得$F(\bvec{x}_0, y_0) = 0$
\item $\dfrac{\partial F(\bvec{x}_0, y_0)}{\partial y} \neq 0$
\end{enumerate}
则存在$\delta, \eta > 0$以及唯一的函数$f: B_\delta (\bvec{x}_0) \to (y_0 - \eta, y_0 + \eta)$具有性质
\begin{enumerate}
\item 对任意的$\norm{\bvec{x} - \bvec{x}_0} < \delta$$f(\bvec{x}_0) = y_0$,有$F(\bvec{x}, f(\bvec{x})) = 0$
\item $f \in C^1 (B_\delta (\bvec{x}_0))$
\item$\bvec{x} \in B_\delta (\bvec{x}_0)$$y = f(\bvec{x})$,有
\[D_i f(x) = -\frac{\dfrac{\partial F}{\partial x_i}(\bvec{x}, y)}{\dfrac{\partial F}{\partial y}(\bvec{x}, y)}, i = 1, 2, \dots, n\eqper\]
\end{enumerate}
\end{theorem}
\section{隐映射定理}
我们先引入几个记号。设想有$m$个方程形成的方程组
\[\begin{cases}
F_1(x_1, \dots, x_n, y_1, \dots, y_m) = 0,\\
\qquad \dots\dots\\
F_m(x_1, \dots, x_n, y_1, \dots, y_m) = 0
\end{cases}\label{隐映射定理1}\tag{1}\]
如果这个方程组是一个合适的约束,那么我们可以期望从中解出$y_1, \dots, y_m$,使得其中的每一个都是$x_1, \dots, x_n$的函数,即
\[\begin{cases}
y_1 = f_1(x_1, \dots, x_n)\\
\qquad \dots\dots\\
y_m = f_m(x_1, \dots, x_n)
\end{cases}\label{隐映射定理2}\tag{2}\]
为了缩短记号,可令
\[\bvec{F} = \begin{bmatrix}
F_1\\ \vdots\\ F_m
\end{bmatrix},
\boldf = \begin{bmatrix}
f_1\\ \vdots\\ f_m
\end{bmatrix}\]
那么\eqref{隐映射定理1}式可以写为
\[\bvec{F}(\bvec{x}, \bvec{y}) = \bvec{0}\]
\eqref{隐映射定理2}式可以写为
\[\bvec{y} = \boldf (\bvec{x})\eqper\]
我们设$\bvec{F}$定义在开集$D \subset \realnum^{m + n}$,那么在$m \times (n + m)$矩阵
\[J \bvec{F} = \begin{bmatrix}
\dfrac{\partial F_1}{\partial x_1} & \cdots & \dfrac{\partial F_1}{\partial x_n} & \dfrac{\partial F_1}{\partial y_1} & \cdots & \dfrac{\partial F_1}{y_m}\\[1ex]
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\
\dfrac{\partial F_m}{\partial x_1} & \cdots & \dfrac{\partial F_m}{\partial x_n} & \dfrac{\partial F_m}{\partial y_1} & \cdots & \dfrac{\partial F_m}{y_m}
\end{bmatrix}\]
中作分块$J\bvec{F} = \begin{bmatrix}
J_x \bvec{F} & J_y \bvec{F}
\end{bmatrix}$
其中
\[J_x \bvec{F} = \begin{bmatrix}
\dfrac{\partial F_1}{\partial x_1} & \cdots & \dfrac{\partial F_1}{\partial x_n}\\
\vdots & \ddots & \vdots\\
\dfrac{\partial F_m}{\partial x_1} & \cdots & \dfrac{\partial F_m}{\partial x_n}
\end{bmatrix},
J_y \bvec{F} = \begin{bmatrix}
\dfrac{\partial F_1}{\partial y_1} & \cdots & \dfrac{\partial F_1}{\partial y_m}\\
\vdots & \ddots & \vdots\\
\dfrac{\partial F_m}{\partial y_1} & \cdots & \dfrac{\partial F_m}{\partial y_m}
\end{bmatrix}\]
其中$J_y \bvec{F}$$m$阶方阵。
\begin{theorem}[隐映射定理]
设开集$D \subset \realnum^{n + m}$,映射$\bvec{F}: D \to \realnum^m$,满足下列条件:
\begin{enumerate}[label=(\roman{*})]
\item $\bvec{F} \in C^1(D)$
\item$(\bvec{x}_0, \bvec{y}_0) \in D$使得$\bvec{F}(\bvec{x}_0, \bvec{y}_0) = \bvec{0}$
\item $\det[J_y \bvec{F}(\bvec{x}_0, \bvec{y}_0)] \neq 0$
\end{enumerate}
则存在$\delta, \eta > 0$以及唯一的函数$\boldf: B_\delta (\bvec{x}_0) \to B_\eta (\bvec{y}_0)$具有性质
\begin{enumerate}
\item 对任意的$\norm{\bvec{x} - \bvec{x}_0} < \delta$$\bvec{f}(\bvec{x}_0) = \bvec{y}_0$,有$\bvec{F}(\bvec{x}, f(\bvec{x})) = \bvec{0}$
\item $\bvec{f} \in C^1 (B_\delta (\bvec{x}_0), \realnum^m)$
\item$\bvec{x} \in B_\delta (\bvec{x}_0)$$\bvec{y} = \bvec{f}(\bvec{x})$,有
\[J\bvec{f}(\bvec{x}) = -(J_y \bvec{F}(\bvec{x}, \bvec{y}))^{-1} J_x \bvec{F}(\bvec{x}, \bvec{y})\eqper\]
\end{enumerate}
\end{theorem}

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