\chapter{多变量函数的微分学} \section{方向导数和偏导数} \begin{definition}[方向导数] 设开集$D \subset \ndreal$,$f: D \to \realnum$,$\bvec{u} \in \realnum^n$且$\norm{\bvec{u}} = 1$,此时称$\bvec{u}$为一个方向,$\bvec{x}_0 \in D$。如果极限 \[\tolim{t}{0} \frac{f(\bvec{x}_0 + t\bvec{u}) - f(\bvec{x})}{t}\] 存在且有限,那么称这个极限是函数$f$在点$\bvec{x}_0$处沿方向$\bvec{u}$方向的导数,记为$\dfrac{\partial f}{\partial \bvec{u}} (\bvec{x}_0)$。 \end{definition} \begin{remark} 记$\phi(t) = f(\bvec{a} + t \tilde{\bvec{u}})$,则显然$\deriv{\phi}(0) = \dfrac{\partial f}{\partial \bvec{u}} (\bvec{a})$。 \end{remark} \begin{definition}[偏导数] 讨论下列单位坐标向量 \begin{align*} \bvec{e}_1 & = (1, 0, 0, \dots, 0)\\ \bvec{e}_2 & = (0, 1, 0, \dots, 0)\\ & \quad \dots\\ \bvec{e}_n & = (0, 0, \dots, 0, 1) \end{align*} 称函数$f$在点$\bvec{x}_0$处沿方向$\bvec{e}_i$的方向导数为$f$在$\bvec{x}_0$处的第$i$个一阶偏导数,记作 \[\frac{\partial f}{\partial x_i}(\bvec{x}_0)\] 或 \[D_i f(\bvec{x}_0)\] 并称$D_i = \dfrac{\partial}{\partial x_i}$为第$i$个偏微分算子,$i = 1, 2, \dots, n$。 \end{definition} \section{多变量函数的微分} 我们希望与一维函数时类似,用一个切平面来线性近似一个曲面在某一点附近的值,即如果我们已知某空间曲面$S$的函数表示为$z = f(x, y)$,那么给定$S$上一点$P = (x_0, y_0, z_0)$,考察曲面上该点上的切平面的方程。首先其方程过$P$,因此应为 \[z = z_0 + a(x - x_0) + b(y - y_0)\] 其次作为切平面应该有$z_0 = f(x_0, y_0)$,同时 \[f(x, y) - z_0 - a(x - x_0) - b(y - y_0) = o\left(\sqrt{(x - x_0)^2 + (y - y_0)^2}\right)\] 即 \[f(x, y) - f(x_0, y_0) = a(x - x_0) + b(y - y_0) + o \left(\sqrt{(x - x_0)^2 + (y - y_0)^2}\right)\] 再进一步,我们希望线性地近似一个多元函数。假设我们一直函数$u = f(x, y, z)$。那么给定一点$P = (x_0, y_0, z_0)$,考察函数在该点附近的线性近似 \[u = u_0 + a(x - x_0) + b(y - y_0) + c(z - z_0)\] 如果它是已知函数在$P$的线性近似,那么$u_0 = f(x_0, y_0, z_0)$且 \[f(x, y, z) - f(x_0, y_0, z_0) = a\Delta x + b \Delta y + c \Delta z + o\left(\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}\right)\] 其中 \[\Delta x = x - x_0, \Delta y = y - y_0, \Delta z = z - z_0\] \begin{definition}[函数的微分] 设$D \subset \ndreal$,$f: D \to \realnum$。取定一点$\bvec{x}_0 \in D\interior$。如果存在$n$维向量$\bvec{A} = (\lambda_1, \lambda_2, \dots, \lambda_n)$,满足 \[f(\bvec{x}_0 + \Delta \bvec{x}) - f(\bvec{x}_0) = \brak{\bvec{A}, \Delta \bvec{x}} + o(\norm{\Delta \bvec{x}})\] 那么称函数$f$在点$\bvec{x}_0$处可微,并称$\brak{\bvec{A}, \Delta \bvec{x}}$为$f$在$\bvec{x}_0$处的微分,记作 \[\dif f(\bvec{x}_0) = \brak{\bvec{A}, \Delta \bvec{x}}\] 其中$\bvec{A}$称为微分系数。 \end{definition} 设$f$在$\bvec{a}$点可微,$\dif f(\bvec{a}) = \brak{A, \Delta \bvec{x}}$,$A = (\lambda_1, \lambda_2, \dots, \lambda_3)$。因此 \[\tolim{\norm{\Delta \bvec{x}}}{0} \frac{\abs{f(\bvec{a} + \Delta \bvec{x}) - f(\bvec{a}) - \brak{A, \Delta \bvec{x}}}}{\norm{\Delta \bvec{x}}} = 0\] 记 \begin{align*} \bvec{u}_1 & = (1, 0, 0, \dots, 0)\\ \bvec{u}_2 & = (0, 1, 0, \dots, 0)\\ & \quad \dots\\ \bvec{u}_n & = (0, 0, \dots, 0, 1) \end{align*} 为确定微分系数,取$\Delta \bvec{x} = t \bvec{u}_i$,那么 \[\norm{\Delta \bvec{x}} = \abs{t}, \brak{A, \Delta \bvec{x}} = \brak{A, t\bvec{u}_i} = t\lambda_i\] 带入上式 \[\tolim{t}{0} \abs{\frac{f(\bvec{a} + t \bvec{u}_i) - f(\bvec{a}) - t\lambda_i}{t}} = 0\] 因此 \[\lambda_i = \tolim{t}{0} \frac{f(\bvec{a} + t\bvec{u}_i) - f(\bvec{a})}{t} = \frac{\partial f}{\partial x_i} (\bvec{a}), i = 1, 2, \dots, n\eqper\] \begin{corollary} 设$f:D \to \realnum$,$\bvec{a} \in D\interior$。如果$f$在$\bvec{a}$点可微,则在$\bvec{a}$点的偏导数存在,且 \[\dif f(\bvec{a}) = \frac{\partial f}{\partial x_1}(\bvec{a}) \Delta x_1 + \frac{\partial f}{\partial x_2}(\bvec{a}) \Delta x_2 + \dots + \frac{\partial f}{\partial x_n}(\bvec{a}) \Delta x_n\eqper\] \end{corollary} \begin{corollary} 设$f:D \to \realnum$,$\bvec{a} \in D\interior$。如果$f$在$\bvec{a}$点可微,则$f$在$\bvec{a}$点连续。 \end{corollary} \begin{proof} \begin{align*} \abs{f(\bvec{a} + \Delta \bvec{x}) - f(\bvec{a})} & = \abs{\brak{A, \Delta \bvec{x}} + o(\norm{\Delta \bvec{x}})}\\ & \leq \norm{A} \cdot \norm{\Delta \bvec{x}} + \abs{o(\norm{\Delta \bvec{x}})} \to 0\qedhere \end{align*} \end{proof} \begin{definition} 令 \[Jf(\bvec{x}) = (D_1 f(\bvec{x}), D_2f(\bvec{x}), \dots, D_n f(\bvec{x}))\] 并称它为函数$f$在点$\bvec{x}$处的Jacobian。函数的Jacobian也常记为$\gra f$或$\nabla f$,即 \[\gra f(\bvec{x}) = J f(\bvec{x})\] 称之为数量函数$f$的梯度。 \end{definition} \begin{proposition} 如果$f$在$\bvec{a}$点可微,则对于任意方向$\bvec{u} \in \ndreal$,$\norm{\bvec{u}} = 1$,那么 \[D_{\bvec{u}} f(\bvec{a}) = \frac{\partial f}{\partial \bvec{u}} (\bvec{a}) = \brak{\gra f(\bvec{a}), \bvec{u}}\eqper\] \end{proposition} \begin{corollary} 对于任意方向$\bvec{u}$,$\abs{D_{\bvec{u}} f(\bvec{a})} \leq \norm{\gra f(\bvec{a})}$。 若$\gra f(\bvec{a}) \neq 0$,$\bvec{u} = \frac{\gra f(\bvec{a})}{\norm{\gra f(\bvec{a})}}$,则$D_{\bvec{u}} f(\bvec{a}) = \norm{\gra f(\bvec{a})}$。 这说明,$f$在$\bvec{a}$的梯度向量的方向是$f$值增加最快的方向,大小是$f$在该点所有方向导数的最大值。 \end{corollary} 下面的命题给出了一个函数可微的必要条件。 \begin{proposition} 若函数$f$在$\bvec{a}$点可微,则存在 \[\gra f(\bvec{a}) = (D_1 f(\bvec{a}), \dots, D_n f(\bvec{a}))\] 从而在该点的所有方向导数都存在。 \end{proposition} 下面的命题则给出了一个函数可微的充分条件。 \begin{proposition} 如果$f$的每个偏导数$D_i f(\bvec{x}), i = 1, 2, \dots, n$在$\bvec{x} = \bvec{a}$点都存在且连续,则$f$在$\bvec{a}$点可微。 \end{proposition} \begin{proof} 以$n = 2$为例。在$P = (a, b)$点附近考虑函数$f(x, y)$: \[f(a + \Delta x, b + \Delta y) - f(a, b) = f(a + \Delta x, b + \Delta y) - f(a + \Delta x, b) + f(a + \Delta x, b) - f(a, b)\] 应用一元函数中值定理,存在$\eta, \theta \in (0, 1)$满足 \begin{align*} f(a + \Delta x, b + \Delta y) - f(a + \Delta x, b) & = D_y f(a + \Delta x, b + \eta \Delta y) \Delta y\\ f(a + \Delta x, b) - f(a, b) & = D_x f(a + \theta \Delta x, b)\Delta x \end{align*} 可以将上式凑配为 \begin{align*} f(a + \Delta x, b + \Delta y) - f(a + \Delta x, b) & = D_y f(a, b) \Delta y + [D_y f(a + \Delta x, b + \eta \Delta y) - D_y f(a, b)] \Delta y\\ f(a + \Delta x, b) - f(a, b) & = D_x f(a, b) \Delta x + [D_x f(a + \theta \Delta x, b) - D_x f(a, b)] \Delta x \end{align*} 记 \begin{align*} [\alpha] & = D_y f(a + \Delta x, b + \eta \Delta y) - D_y f(a, b)\\ [\beta] & = D_x f(a + \theta \Delta x, b) - D_x f(a, b) \end{align*} 那么 \[f(a + \Delta x, b + \Delta y) - f(a, b) = D_x f(a, b) \Delta x + D_y f(a, b) \Delta y + [\alpha] \Delta x + [\beta] \Delta y\] 于是我们只需证明$[\alpha] \Delta x + [\beta] \Delta y = o\left(\sqrt{x^2 + y^2}\right)$。我们已知$D_x f(x, y), D_y f(x, y)$在$P = (a, b)$点连续,因此 \[\frac{\abs{[\alpha]\Delta x + [\beta] \Delta y}}{\sqrt{\Delta x^2 + \Delta y^2}} \leq \abs{[\alpha]} + \abs{[\beta]} \to 0\] 综上, \[f(a + \Delta x, b + \Delta y) - f(a, b) = D_x f(a, b) \Delta x + D_y f(a, b) \Delta y + o\left[\sqrt{\Delta x^2 + \Delta y^2}\right] \eqper \qedhere\] \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。 \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 在$\bvec{x}_0$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} 设$D \in \ndreal$,$\bvec{f}: D \to \realnum^m$,$\bvec{g}: \Omega \to \realnum^k$,$\bvec{f}(D) \subset \Omega \subset \realnum^m$。如果\boldf 在$\bvec{x}_0 \in D\interior$上可微,$\bvec{g}$在$\boldf(\bvec{x}_0)$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 点$(x_0, y_0) \in D$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 对$x \in (x_0 - \delta, x_0 + \delta)$,$y = f(x)$,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 点$(\bvec{x}_0, y_0) \in D$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 对$\bvec{x} \in B_\delta (\bvec{x}_0)$,$y = f(\bvec{x})$,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$J_y \bvec{F}$是$m$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 点$(\bvec{x}_0, \bvec{y}_0) \in D$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 对$\bvec{x} \in B_\delta (\bvec{x}_0)$,$\bvec{y} = \bvec{f}(\bvec{x})$,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} \section{逆映射定理} 给定$\boldf: D \to \ndreal$,$D \subset \ndreal$。考察\boldf 的反函数及其性质:$\bvec{y} = \boldf^{-1}(\bvec{x})$。 考虑应用隐函数定理。$\bvec{y} = \boldf^{-1}(\bvec{x})$意味着$\bvec{x} = \boldf(\bvec{x})$。因此定义$\bvec{F}:\tilde{D} \to \ndreal$,满足 \[\bvec{F}(\bvec{x}, \bvec{y}) = \bvec{x} - \boldf(\bvec{y}), (\bvec{x}, \bvec{y}) \in \ndreal \times D = \tilde{D} \subset \realnum^{2n}\] 再取$\bvec{y}_0 \in D\interior$,$\bvec{x}_0 = \boldf(\bvec{y}_0)$,那么$(\bvec{x}_0, \bvec{y}_0) \in \tilde{D}\interior$,$\bvec{F}(\bvec{x}_0, \bvec{y}_0) = \bvec{0}$。 如果我们假设$\bvec{f} \in C^1$,那么 \[J_{\bvec{y}} \bvec{F}(\bvec{x}, \bvec{y}) = -J \bvec{f}(\bvec{y})\] 且 \[J_{\bvec{x}} \bvec{F}(\bvec{x}, \bvec{y}) = J\bvec{x} = \bvec{I}_n \eqper\] \begin{theorem}[逆映射定理(局部)] 设$\bvec{f} \in C^1(D, \ndreal), D \subset \ndreal, \bvec{y}_0 \in D\interior$满足$\det(J\bvec{f}(\bvec{y}_0)) \neq 0$,那么存在$\delta, \eta > 0$以及函数$\bvec{g}: B_\delta (\bvec{x}_0) \to B_\eta(\bvec{y}_0)$,其中$\bvec{x}_0 = \bvec{f}(\bvec{y}_0)$满足以下性质: \begin{enumerate} \item 对任意的$\bvec{x}$满足$\norm{\bvec{x} - \bvec{x}_0} < \delta$,$\bvec{g}(\bvec{x}_0) = \bvec{y}_0$,$\bvec{f}(\bvec{g}(\bvec{x})) = \bvec{x}$; \item $\bvec{g} \in C^1(B_\delta(\bvec{x}_0), \ndreal)$; \item $J\bvec{g}(\bvec{x}) = [J\bvec{f}(\bvec{y})]^{-1}$,其中$\bvec{y} = \bvec{g}(\bvec{x})$。 \end{enumerate} \end{theorem} \begin{theorem}[逆映射定理] 设$\bvec{f} \in C^1(D, \ndreal), D \subset\ndreal$为开集,且$\bvec{f}:D \to \ndreal$为单射,对任意的$\bvec{y}$,$\det(J\bvec{f}(\bvec{y})) \neq 0$。则记$\Omega = \bvec{f}(D)$,存在$\bvec{f}$的反函数$\bvec{f}^{-1} \in C^1 (\Omega, \ndreal)$且对任意的$\bvec{x} \in \Omega$, \[J\bvec{f}^{-1}(\bvec{x}) = (J\bvec{f}(\bvec{y}))^{-1}, \bvec{y} = \bvec{f}^{-1}(\bvec{x})\eqper\] \end{theorem} \section{高阶偏导数} \begin{definition} 设在开集$D$上的每一点,函数$f$存在偏导数 \[D_i f(\bvec{x}) = \frac{\partial f}{\partial x_i}(\bvec{x}), i = 1, 2, \dots, n\] 称他们为$f$的一阶偏导数,如果对这些偏导函数又可取偏导数,得出的就是$f$的二阶偏导函数,仿照此可以定义三阶偏导函数乃至更高阶的偏导数。我们将以一阶偏导数$\dfrac{\partial f}{\partial x_j}$再对$x_i$求偏导数时,把$\dfrac{\partial}{\partial x_i}\left(\dfrac{\partial f}{\partial x_j}\right)$记作$\dfrac{\partial^2 f}{\partial x_i \partial x_j}$,如果$i = j$,那么把$\dfrac{\partial^2 f}{\partial x_i \partial x_i}$记作$\dfrac{\partial^2 f}{\partial x_i^2}$。 \end{definition} \begin{theorem}[Clairaut定理] 设$f:D \to \realnum$,$D \subset \realnum^2$是开集,$P = (x_0, y_0) \in D$。若$\dfrac{\partial^2 f}{\partial x \partial y}$和$\dfrac{\partial^2 f}{\partial y \partial x}$在$D$内存在且在$P$点连续,那么二者在该点相等。 \end{theorem} \begin{corollary} 设$f: D \to \realnum$,$D \subset \ndreal$是开集。若$f$在$D$内所有$k$阶偏导数都存在且连续,则$k$阶偏导数的值与关于自变量的求导次序无关。 \end{corollary} \section{拟微分平均值定理} 这个下面这个定理主要阐述的是多元数值函数的中值定理。 \begin{theorem} 设定义在凸区域$D \subset \ndreal$上的函数$f$可微,则对任何两点$\bvec{a}, \bvec{b} \in D$,在由$\bvec{a}$与$\bvec{b}$确定的直线段上有一点$\bvec{\xi}$使得 \[f(\bvec{b}) - f(\bvec{a}) = Jf(\bvec{\xi})(\bvec{b} - \bvec{a})\eqper\] \end{theorem} 对于向量值函数,中值定理不一定成立,而有下面的拟微分平均值定理: \begin{theorem}[拟微分平均值定理] 设凸区域$D \subset \ndreal$且$\boldf: D \to \realnum^m$,映射\boldf 在$D$上可微,则对于任何$\bvec{a}, \bvec{b} \in D$,在由$\bvec{a}, \bvec{b}$确定的线段上必有一点$\bvec{\xi}$使得 \[\norm{\boldf(\bvec{b}) - \boldf(\bvec{a})} \leq \norm{J\boldf(\bvec{\xi})} \norm{\bvec{b} - \bvec{a}}\eqper\] \end{theorem} \begin{corollary} 设区域$D \subset \ndreal$,$\bvec{f}:D \to \realnum^m$,如果$J\bvec{f} = \bvec{0}$在$D$上成立,则\boldf 在$D$上为一常向量。 \end{corollary} \section{Taylor公式} 考虑用多项式近似一个多元函数,即对在$\bvec{a}$点有$m + 1$阶连续偏导数的$n$元函数$f(x)$,是否有$m$次多项式$P_m(\bvec{x})$,使得 \[f(\bvec{a} + \Delta \bvec{x}) = P_m (\Delta \bvec{x}) + o(\norm{\Delta \bvec{x}}^m)\] 成立? 我们设$f \in C^{m + 1}(B_r(\bvec{a})), \bvec{a} \in \ndreal, r > 0$,那么对任意满足$\norm{\Delta \bvec{x}} < r$的$\Delta \bvec{x}$,定义 \[\varphi(t) = f(\bvec{a} + t\Delta \bvec{x}) \in C^{m + 1}[0, 1]\] 应用一元函数的Taylor公式:对任意的$t \in [0, 1]$,都存在$\theta \in (0, 1)$满足 \[\varphi(t) = \sum_{k = 1}^m \frac{\varphi^{(k)}(0)}{k!}t^k + \frac{\varphi^{(m + 1)}(\theta t)}{(m + 1)!} t^{m + 1}\] 特别取$t = 1$得到 \[\varphi(1) = \sum_{k = 1}^m \frac{\varphi^{(k)}(0)}{k!} + \frac{\varphi^{(m + 1)}(\theta)}{(m + 1)!}\] 将引入的一元函数的表达式带入(特别注意求导项的带入): 计算 \begin{align*} & \deriv{\varphi} (t) = \sum_{i = 1}^n \frac{\partial f(\bvec{a} + t\Delta \bvec{x})}{\partial x_i} \Delta x_i, \deriv{\varphi}(0) = \sum_{i = 1}^n \frac{\partial f(\bvec{a})}{\partial x_i}\Delta x_i\\ & \varphi^{\prime \prime} (t) = \sum_{i, j = 1}^n \frac{\partial^2 f(\bvec{a} + t\Delta \bvec{x})}{\partial x_i \partial x_j} \Delta x_i \Delta x_j, \varphi^{\prime \prime} (0) = \sum_{i, j = 1}^n \frac{\partial ^2 f(\bvec{a})}{\partial x_i \partial x_j}\Delta x_i \Delta x_j\\ & \dots \dots \end{align*} 我们引入记号$\alpha = (\alpha_1, \dots, \alpha_n)$,其中每个$\alpha_i$都是非负整数,记 \[\abs{\alpha} = \alpha_1 + \dots + \alpha_n, \alpha! = \alpha_1! \dots \alpha_n!\] 如果$\bvec{x} = (x_1, \dots, x_n)$,那么记$x^\alpha = x_1^{\alpha_1} \dots x_n^{\alpha_n}$。 对于多重指标$\alpha = (\alpha_1, \dots, \alpha_n)$,我们还引进记号 \[D^\alpha f(\bvec{a}) = \frac{\partial^{\alpha_1 + \dots + \alpha_n}f}{\partial x_1^{\alpha_1} \dots \partial x_n^{\alpha_n}} (\bvec{a})\eqper\] 于是我们可以叙述多元函数Taylor公式: \begin{theorem}[多元函数Taylor公式]\label{多元函数Taylor公式} 设$D \subset \ndreal$是一个凸区域,$f \in C^{m + 1}(D)$。$\bvec{a} = (a_1, \dots, a_n)$和$\bvec{a} + \bvec{h} = (a_1 + h_1, \dots, a_n + h_n)$是$D$中两点,则必存在$\theta \in (0, 1)$使得 \[f(\bvec{a} + \bvec{h}) = \sum_{k = 0}^m \sum_{\abs{\alpha} = k} \frac{D^\alpha f(\bvec{a})}{\alpha!} h^\alpha + R_m\] 其中 \[R_m = \sum_{\abs{\alpha} = m + 1} \frac{D^\alpha f(\bvec{a} + \theta \bvec{h})}{\alpha!} h^\alpha\] 称为Lagrange余项。 \end{theorem} 这个定理中的和式的意思是,对每个$\alpha$,都求对$f$求$\alpha$次偏导的导函数在$\bvec{a}$处的值,并且对$x_i$求了$\alpha_i$次偏导就要在后面乘上$\dfrac{x_i^{\alpha_i}}{\alpha_i!}$。 我们再引入一个高阶微分的记号。对$\bvec{h} = (h_1, h_2, \dots, h_3)$,我们定义 \begin{align*} & \qquad\left(h_1 \frac{\partial}{\partial x_1} + \dots + h_n \frac{\partial}{\partial x_n}\right)^k f(\bvec{a})\\ & = \sum_{\abs{\alpha} = k} \frac{k!}{\alpha!} \frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\dots \frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}}f(\bvec{a}) \bvec{h}^\alpha\\ & = \sum_{\abs{\alpha} = k} \frac{k!}{\alpha!}D^\alpha f(\bvec{a}) h^\alpha \end{align*} 特别考虑二元函数的情况。如果设$x, y$的改变量为$h, k$,那么 \begin{align*} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y}\right) f(x, y) & = \frac{\partial f(x, y)}{\partial x}h + \frac{\partial f(x, y)}{\partial y}k\\ \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y}\right)^2 f(x, y) & = \frac{\partial^2 f(x, y)}{\partial x^2}h^2 + 2 \frac{\partial^2 f(x, y)}{\partial x \partial y}hk + \frac{\partial^2 f(x, y)}{\partial y^2}k^2\\ \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y}\right)^m f(x, y) & = \sum_{i = 0}^m \binom{m}{i} \frac{\partial^m f(x, y)}{\partial x^ \partial y^{m - i}} h^i k^{m - i}, m = 1, \dots, n + 1 \end{align*} 在一般的应用中,特别重要的是Taylor公式的前三项。把他们具体写出来: \[f(\bvec{a} + \bvec{h}) = f(\bvec{a}) + \sum_{i = 1}^n \frac{\partial f}{\partial x_i}(\bvec{a})h_i + \frac{1}{2} \sum_{i, j = 1}^n \frac{\partial^2 f}{\partial x_i \partial x_j} (\bvec{a}) h_i h_j + \cdots\] 如果记 \[Hf(\bvec{a}) = \begin{bmatrix} \dfrac{\partial^2 f}{\partial x_1^2}(\bvec{a}) & \cdots & \dfrac{\partial^2 f}{\partial x_1 \partial x_n}(\bvec{a})\\[1em] \vdots & \ddots & \vdots\\[1ex] \dfrac{\partial^2 f}{\partial x_n \partial x_1}(\bvec{a}) & \cdots & \dfrac{\partial^2 f}{\partial x_n^2}(\bvec{a}) \end{bmatrix}\] 进一步简记为 \[Hf(\bvec{a}) = \begin{bmatrix} D_{11} f(\bvec{x}) & \cdots & D_{1n} f(\bvec{x})\\ \vdots & \ddots & \vdots\\ D_{n1} f(\bvec{x}) & \cdots & D_{nn} f(\bvec{x}) \end{bmatrix}\eqper\] 这$Hf$称为$f$的Hessian,它是一个$n$阶对称方阵。 那么前面的Taylor公式可以写成 \[f(\bvec{a} + \bvec{h}) = f(\bvec{a}) + Jf(\bvec{a})\bvec{h} + \frac{1}{2} \bvec{h}^{\mathrm{T}} Hf(\bvec{a}) \bvec{h} + \cdots\] \begin{theorem} 在定理\ref{多元函数Taylor公式}的条件下, \[R_m = O(\norm{\bvec{h}}^{m + 1})\eqper\] 于是我们可以把Taylor公式写成Peano余项的形式: \[f(\bvec{a} + \bvec{h}) = f(\bvec{a}) + Jf(\bvec{a}) \bvec{h} + \frac{1}{2} \bvec{h}^{\mathrm{T}} Hf(\bvec{a}) \bvec{h} + o(\norm{\bvec{h}}^2)\] \end{theorem} \section{极值} \begin{definition} 设$D \subset \ndreal$,函数$f:D \to \realnum$,点$\bvec{p}_0 \in D\interior$,如果存在一个球$B_r(\bvec{p}_0) \subset D\interior$,使得$f(\bvec{p}) \geq f(\bvec{p}_0)$($f(\bvec{p}) > f(\bvec{p}_0)$)对一切$\bvec{p} \in B_r(\hat{\bvec{p}}_0)$成立,那么$\bvec{p}_0$称为$f$的一个(严格)极小值点,而$f(\bvec{p}_0)$称为函数$f$的一个(严格)极小值。 同样地可以定义(严格)极大值点和(严格)极大值。极小值和极大值统称极值。 \end{definition} 类似于Fermat引理,我们可以得到极值点的必要条件 \begin{theorem} 设$n$元函数$f$在$\bvec{p}_0$取得极值,且$Jf(\bvec{p}_0)$存在,那么必须有$Jf(\bvec{p}_0) = \bvec{0}$。 设$\bvec{u}$是任意方向向量,则$D_{\bvec{u}} f(\bvec{a}) = 0$。 \end{theorem} \begin{definition} $D$中使得$Jf(\bvec{p}) = \bvec{0}$的一切内点称为函数函数$f$的驻点。极值点一定是煮点,而驻点未必是极值点。 \end{definition} \begin{definition} 设 \(A = \begin{bmatrix} a_{ij} \end{bmatrix}\) 是一个$n$阶对称方阵。设 \[\bvec{x} = \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix}\] 称 \[Q(\bvec{x}) = \bvec{x}^\mathrm{T} A \bvec{x} = \sum_{i, j = 1}^n a_{ij}x_i x_j\] 为$x_1, x_2, \dots, x_n$的一个二次型,方阵$A$称为二次型$Q$的系数方阵。 如果对任意$\bvec{x} \neq \bvec{0}$都有$Q(\bvec{x}) \geq 0$($\leq 0$),则称二次型$Q$是正(负)定的,其系数方阵$A$相应地称为正(负)定方阵。 如果对任意$\bvec{x} \neq \bvec{0}$都有$Q(\bvec{x}) > 0$($< 0$),则称二次型$Q$是严格正(负)定的,其系数方阵$A$相应地称为严格正(负)定方阵。 如果总存在$\bvec{p}, \bvec{q} \in \ndreal$,使得$Q(\bvec{p}) < 0 < Q(\bvec{q})$,旧称二次型$Q$是不定的,其系数方阵$A$相应地称为不定方阵。 \end{definition} \begin{theorem} 设 \(A = \begin{bmatrix} a_{ij} \end{bmatrix}\) 是一个$n$阶方阵。方阵$A$为严格正定的一个必要充分条件是它的各级顺序主子式均大于零。 \end{theorem} 欲证明一个方阵$A$负定,只需证明$-A$是正定的即可。 \begin{theorem} 设二阶对称方阵 \[A = \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\] $A$为严格正(负)定的一个必要充分条件是 \[a_{11} > 0(a_{11} < 0),\] \[\begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix} > 0\] $A$为不定矩阵的一个充分必要条件是 \[\begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix} < 0\eqper\] \end{theorem} \begin{theorem} 设$\bvec{x}_0$是函数$f$的一个驻点,函数$f$在$\bvec{x}_0$的某一临域内有连续的二阶偏导数。 \begin{enumerate} \item 如果Hessian $Hf(\bvec{x}_0)$是严格正定(负)方阵,那么$\bvec{x}_0$是$f$的一个严格极小(大)值点。 \item 如果Hessian $Hf(\bvec{x}_0)$是不定方阵,那么$\bvec{x}_0$不是$f$的极值点。 \end{enumerate} \end{theorem} \section{条件极值} 首先我们引入一个问题。设$f: D \to \realnum$,$\Phi: D \to \realnum^m$,$D \subset \realnum^{n + m}$是开集,$(\bvec{x}, \bvec{y}) \in \realnum^{n + m}$。求满足$\Phi(\bvec{x}, \bvec{y}) = \bvec{0}$的条件下$f(\bvec{x}, \bvec{y})$的最大/最小值,记为 \[\begin{cases} \max f(\bvec{x}, \bvec{y})\\ \Phi(\bvec{x}, \bvec{y}) = 0 \end{cases} \text{或} \begin{cases} \min f(\bvec{x}, \bvec{y})\\ \Phi(\bvec{x}, \bvec{y}) = 0 \end{cases}\] 其中$f(\bvec{x}, \bvec{y})$称为目标函数,$\Phi(\bvec{x}, \bvec{y}) = \bvec{0}$称为约束条件。 设在约束$\varphi(\bvec{x}, y) = 0$下$f(\bvec{x}, y)$在$P = (\bvec{x}_0, y_0) \in D$点达到极值,且$\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0) \neq 0$,应用隐函数定理:在$P$点附近,方程$\varphi(\bvec{x}, y) = \bvec{0}$确定了隐函数$y = y(\bvec{x})$。函数$F(\bvec{x}) = f(\bvec{x}, y(\bvec{x}))$在$\bvec{x}_0$达到极值,因此$J F(\bvec{x}_0) = \bvec{0}$。 因此 \[\frac{\partial F}{\partial x_i} (\bvec{x}_0) = \frac{\partial f}{\partial x_i}(\bvec{x}_0, y_0) + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x_i}(\bvec{x}_0) = 0, i = 1, 2, \cdots, n\] 而根据隐函数定理, \[\frac{\partial y}{\partial x_i}(\bvec{x}_0) = -\frac{\dfrac{\partial \varphi}{\partial x_i}(\bvec{x}_0, y_0)}{\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0)}\] 将它带入上式, \[\frac{\partial f}{\partial x_i}(\bvec{x}_0, y_0) - \frac{\dfrac{\partial f}{\partial y}{(\bvec{x}_0, y_0)} \dfrac{\partial \varphi}{\partial x_i}(\bvec{x}_0, y_0)}{\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0)} = 0\] 引入参数 \[\lambda = -\frac{\dfrac{\partial f}{\partial y}(\bvec{x}_0, y_0)}{\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0)}\] 那么 \[\frac{\partial f}{\partial x_i}(\bvec{x}_0, y_0) + \lambda \frac{\partial \varphi}{\partial x_i}(\bvec{x}_0, y_0) = 0, i = 1, 2, \cdots, n\] 因此我们得到 \[Jf(\bvec{x}_0, y_0) + \lambda J\varphi(\bvec{x}_0, y_0) = \bvec{0}\] 是条件极值的一个必要条件。 \begin{theorem}[条件极值的必要条件] 设$f \in C^1(D)$,$\Phi \in C^1 (D, \realnum^m)$,$D \subset \realnum^{n + m}$是开集。记$P = (\bvec{x}_0, \bvec{y}_0) \in D$,又设$\det(J_{\bvec{y}}\Phi(\bvec{x}_0, \bvec{y}_0)) \neq 0$。如果$f(\bvec{x}, \bvec{y})$在约束$\Phi(\bvec{x}, \bvec{y}) = \bvec{0}$下在$P$点达到极值,则存在$\bvec{\Lambda} = (\lambda_1, \dots, \lambda_m)$使得 \[Jf(\bvec{x}_0, \bvec{y}_0) + \bvec{\Lambda} J \Phi(\bvec{x}_0, \bvec{y}_0) = \bvec{0}\eqper\] 这也就是说$(\bvec{x}_0, \bvec{y}_0)$满足方程组 \begin{align*} & \frac{\partial f}{\partial x_k}(\bvec{x}_0, \bvec{y}_0) + \sum_{i = 1}^m \lambda_i \frac{\partial \Phi_i}{\partial x_k}(\bvec{x}_0, \bvec{y}_0) = 0, k = 1, \cdots, n\\ & \frac{\partial f}{\partial y_j}(\bvec{x}_0, \bvec{y}_0) + \sum_{i = 1}^m \lambda_i \frac{\partial \Phi_i}{\partial y_j}(\bvec{x}_0, \bvec{y}_0) = 0, j = 1, \cdots, m\\ \end{align*} \end{theorem} 根据这个定理,我们可以引入Lagrange乘数法。定义函数$L: D \times \realnum^m \to \realnum$, \[L(\bvec{z}, \bvec{\Lambda}) = f(\bvec{z}) + \bvec{\lambda} \Phi(\bvec{z}), (\bvec{z}, \bvec{\Lambda}) \in D \times \realnum^m\] $L$称为条件极值问题的Lagrange函数,$\bvec{\Lambda}$称为Lagrange乘数/乘子。根据条件极值的表要条件,在条件极值点$\bvec{z}_0 \in D$,存在$\bvec{\Lambda} \in \realnum^m$满足 \[J_{\bvec{z}} L (\bvec{z}_0, \bvec{\Lambda}) = J_{\bvec{z}} f(\bvec{z}_0) + \bvec{\Lambda} J_{\bvec{z}}\Phi(\bvec{z}_0) = \bvec{0}\] 此外 \[J_{\bvec{\Lambda}} L(\bvec{z}_0, \bvec{\Lambda}) = \Phi(\bvec{z}_0) = \bvec{0}\] 因此 \[JL(\bvec{z}, \bvec{\Lambda}) = (J_{\bvec{z}}, J_{\bvec{\Lambda}}) = \bvec{0} \eqper\]