跳转至

隐函数定理及其运用

1. 隐函数

  • 隐函数存在唯一性、连续性定理:若 \(F(x, y)\) 满足

    • \(F\) 在以 \(P_0(x_0, y_0)\) 为内点的某一区域 \(D\) 上连续
    • \(F(x_0, y_0) = 0\)
    • \(F\)\(D\) 上有连续偏导数 \(F_2'(x, y)\)
    • \(F_2'(x_0, y_0) \ne 0\) 则在 \(P_0\) 的某邻域内,方程 \(F(x, y) = 0\) 唯一确定了一个定义在某区间 \((x_0 - \alpha, x_0 + \alpha)\) 内的函数 \(y = f(x)\),使
    • \(f(x_0) = y_0\)\(\forall x \in (x_0 - \alpha, x_0 + \alpha), (x, f(x)) \in U(P_0), F(x, f(x)) \equiv 0\)
    • \(f(x)\)\((x_0 - \alpha, x_0 + \alpha)\) 连续
  • 隐函数可导性定理:在上述基础上,若 \(F_1'(x, y)\)\(D\) 上连续,则 \(f(x)\)\((x_0 - \alpha, x_0 + \alpha)\) 可导,且

\[\frac{dy}{dx} = - \frac{F_1'(x, y)}{F_2'(x, y)}\]

2. 隐函数组

  • 雅可比 (Jacobi) 行列式\(\frac{\partial(F, G)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial F}{\partial u} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v} \end{vmatrix}\)

  • 隐函数组定理:若 \(F(x, y, u, v), G(x, y, u, v)\) 满足

    • \(F, G\) 在以 \(P_0(x_0, y_0, u_0, v_0)\) 为内点的某一区域 \(D\) 上连续
    • \(F(x_0, y_0, u_0, v_0) = G(x_0, y_0, u_0, v_0) = 0\)
    • \(F, G\)\(D\) 上有一阶连续偏导数
    • \(J = \frac{\partial(F, G)}{\partial(u, v)} \ne 0\) 则在 \(P_0\) 的某邻域内,方程 \(F(x, y) = 0\) 唯一确定了两个定义在 \(Q(x_0, y_0)\) 的某领域 \(U(Q)\) 内的函数 \(u = f(x, y), v = g(x, y)\),使
    • \(u_0 = f(x_0, y_0), v = g(x_0, y_0)\)\(\forall(x, y) \in U(Q), (x, y, f(x, y), g(x, y)) \in U(Q), F(x, y, f(x, y), g(x, y)) = \equiv 0, G(x, y, f(x, y), g(x, y)) = \equiv 0\)
    • \(f(x, y), g(x, y)\)\(U(Q)\) 连续
    • \(f(x, y), g(x, y)\) 存在连续偏导数
    \[\frac{\partial u}{\partial x} = - \frac{1}{J} \frac{\partial(F, G)}{\partial(x, v)}\]
    \[\frac{\partial u}{\partial y} = - \frac{1}{J} \frac{\partial(F, G)}{\partial(y, v)}\]
    \[\frac{\partial v}{\partial x} = - \frac{1}{J} \frac{\partial(F, G)}{\partial(u, x)}\]
    \[\frac{\partial v}{\partial y} = - \frac{1}{J} \frac{\partial(F, G)}{\partial(u, y)}\]
  • 反函数定理:设 \(f : \mathbb{R}^n \to \mathbb{R}^n\) 一阶导数连续,\(a \in \mathbb{R}^n, J_f(a) \ne 0\),则存在 \(a\) 的邻域 \(U(a)\)\(f(a)\) 的邻域 \(V\),使 \(f|_U: U \to V\) 为双射,且逆映射 \(f^{-1} : V \to U\) 导函数也连续,其导数为

\[D(f^{-1})(y) \times Df(f^{-1}(y))= 1\]

3. 几何运用

  • 曲线切线与法平面

    • 切线方程
      • \(\frac{x-x_0}{x'(t_0)} = \frac{y-y_0}{y'(t_0)} = \frac{z-z_0}{z'(t_0)}\)
      • \(\frac{x-x_0}{\frac{\partial(F, G)}{\partial(y, z)}} = \frac{y-y_0}{\frac{\partial(F, G)}{\partial(z, x)}} = \frac{z-z_0}{\frac{\partial(F, G)}{\partial(x, y)}}\)
    • 法平面方程\(\frac{\partial(F, G)}{\partial(y, z)}(x-x_0) + \frac{\partial(F, G)}{\partial(z, x)}(y-y_0) + \frac{\partial(F, G)}{\partial(x, y)}(z-z_0) = 0\)
  • 曲面法线与切平面

    • 法线方程\(\frac{x-x_0}{F_1'(x_0, y_0, z_0)} = \frac{y-y_0}{F_2'(x_0, y_0, z_0)} = \frac{z-z_0}{F_3'(x_0, y_0, z_0)}\)
    • 切平面方程\(F_1'(x_0, y_0, z_0)(x - x_0) + F_2'(x_0, y_0, z_0)(y - y_0) + F_3'(x_0, y_0, z_0)(z - z_0) = 0\)

4. 条件极值

  • 拉格朗日乘数法: 设 \(\varphi_k(x_1,x_2,\cdots,x_n)=0,k=1,2,\cdots,m\quad(m<n)\), 求 \(y=f(x_1,x_2,\cdots,x_n)\) 的极值, 其中 \(f\)\(\varphi_k (k=1, 2, \cdots, m)\) 在区域 \(D\) 上有连续的一阶偏导数. 若 \(D\) 的内点 \(P_0 (x_1^{(0)}, \cdots, x_n^{(0)})\) 是上述问题的极值点, 且雅可比矩阵
\[\begin{pmatrix} \frac{\partial \varphi_1}{\partial x_1} & \cdots & \frac{\partial \varphi_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial \varphi_m}{\partial x_1} & \cdots & \frac{\partial \varphi_m}{\partial x_n} \end{pmatrix}_{P_0}\]

的秩为 \(m\), 则存在 \(m\) 个常数 \(\lambda_1^{(0)}, \cdots, \lambda_m^{(0)}\), 使得 \((x_1^{(0)}, \cdots, x_n^{(0)}, \lambda_1^{(0)}, \cdots, \lambda_m^{(0)})\)拉格朗日函数

\[\begin{aligned} & L(x_{1},x_{2},\cdots,x_{n},\lambda_{1},\lambda_{2},\cdots,\lambda_{m}) \\ & =f(x_{1},x_{2},\cdots,x_{n})+\sum_{k=1}^{m}\lambda_{k}\varphi_{k}(x_{1},x_{2},\cdots,x_{n}), \end{aligned}\]

的稳定点, 即 \((x_1^{(0)}, \cdots, x_n^{(0)}, \lambda_1^{(0)}, \cdots, \lambda_m^{(0)})\)\(n+m\) 个方程

\[\begin{cases} L_{x_1} = \frac{\partial f}{\partial x_1} + \sum_{k=1}^m \lambda_k \frac{\partial \varphi_k}{\partial x_1} = 0, \\ \cdots \\ L_{x_n} = \frac{\partial f}{\partial x_n} + \sum_{k=1}^m \lambda_k \frac{\partial \varphi_k}{\partial x_n} = 0, \\ L_{\lambda_1} = \varphi_1 (x_1, \cdots, x_n) = 0, \\ \cdots \\ L_{\lambda_m} = \varphi_m (x_1, \cdots, x_n) = 0 \end{cases}\]

的解.

Note

最值一定存在且极值怀疑点唯一,那么其为最值。若最值两端均存在且有两个极值极值怀疑点,则其分别对应最大值与最小值

评论