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高阶偏导数
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2022-12-31 08:13
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设函数 $z=f(x, y)$ 在区域 $D$ 内具有偏导数 $$ \frac{\partial z}{\partial x}=f_x(x, y), \quad \frac{\partial z}{\partial y}=f_y(x, y), $$ 则在 $D$ 内 $f_x(x, y)$ 和 $f_y(x, y)$ 都是 $x, y$ 的函数. 如果这两个函数的偏导数存在, 则称它们是函数 $z=f(x, y)$ 的二阶偏导数. 按照对变量求导次序的不同,共有下 列四个二阶偏导数: $$ \begin{aligned} & \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial^2 z}{\partial x^2}=f_{x x}(x, y), \quad \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial^2 z}{\partial x \partial y}=f_{x y}(x, y) \\ & \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial^2 z}{\partial y \partial x}=f_{y x}(x, y), \quad \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial^2 z}{\partial y^2}=f_{y y}(x, y) \end{aligned} $$ 其中第二、第三两个偏导称为混合偏导数. 类似地,可以定义三阶、四阶、拟及 $n$ 阶偏导数. 我们把二阶及二阶以上的偏导数统称为高阶偏导数. 例 7 设 $z=4 x^3+3 x^2 y-3 x y^2-x+y$ ,求 $$ \frac{\partial^2 z}{\partial x^2}, \frac{\partial^2 z}{\partial y \partial x}, \frac{\partial^2 z}{\partial x \partial y}, \frac{\partial^2 z}{\partial y^2}, \frac{\partial^3 z}{\partial x^3} $$ 解 $\frac{\partial z}{\partial x}=12 x^2+6 x y-3 y^2-1, \quad \frac{\partial z}{\partial y}=3 x^2-6 x y+1$; $$ \begin{aligned} & \frac{\partial^2 z}{\partial x^2}=24 x+6 y, \frac{\partial^3 z}{\partial x^3}=24 \quad \frac{\partial^2 z}{\partial y^2}=-6 x, \\ & \frac{\partial^2 z}{\partial x \partial y}=6 x-6 y, \frac{\partial^2 z}{\partial y \partial x}=6 x-6 y . \end{aligned} $$ 例 8 求 $z=x \ln (x+y)$ 的二阶偏导数. $$ \text { 解 } \begin{aligned} & \frac{\partial z}{\partial x}=\ln (x+y)+\frac{x}{x+y} , \frac{\partial z}{\partial y}=\frac{x}{x+y}, \\ & \frac{\partial^2 z}{\partial x^2}=\frac{1}{x+y}+\frac{x+y-x}{(x+y)^2}=\frac{x+2 y}{(x+y)^2} , \frac{\partial^2 z}{\partial y^2}=\frac{-x}{(x+y)^2} , \\ & \frac{\partial^2 z}{\partial x \partial y}=\frac{1}{x+y}+\frac{-x}{(x+y)^2}=\frac{y}{(x+y)^2} , \\ & \frac{\partial^2 z}{\partial y \partial x}=\frac{(x+y)-x}{(x+y)^2}=\frac{y}{(x+y)^2} . \end{aligned} $$ 例 9 求函数 $z=x^y$ 的二阶偏导数. 解 $$ \begin{aligned} & \frac{\partial z}{\partial x}=y x^{y-1} , \frac{\partial z}{\partial y}=x^y \ln x \\ & \frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=y(y-1) x^{y-2}, \frac{\partial^2 z}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=x^y \ln ^2 x , \\ & \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=x^{y-1}+y x^{y-1} \ln x=x^{y-1}(1+y \ln x) , \\ & \frac{\partial^2 z}{\partial y \partial x}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=y x^{y-1} \ln x+x^y \cdot \frac{1}{x}=x^{y-1}(1+y \ln x) \end{aligned} $$ 例 10 验证函数 $u(x, y)=\ln \sqrt{x^2+y^2}$ 满足方程 $$ \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 \text {. } $$ 证 因为 $\ln \sqrt{x^2+y^2}=\frac{1}{2} \ln \left(x^2+y^2\right)$ , 所以 $$ \frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}, \frac{\partial u}{\partial y}=\frac{y}{x^2+y^2}, $$ $$ \begin{aligned} & \frac{\partial^2 u}{\partial x^2}=\frac{\left(x^2+y^2\right)-x \cdot 2 x}{\left(x^2+y^2\right)^2}=\frac{y^2-x^2}{\left(x^2+y^2\right)^2}, \quad \frac{\partial^2 u}{\partial y^2}=\frac{\left(x^2+y^2\right)-y \cdot 2 y}{\left(x^2+y^2\right)^2}=\frac{x^2-y^2}{\left(x^2+y^2\right)^2} \\ & \text { 从而 } \quad \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{y^2-x^2}{\left(x^2+y^2\right)^2}+\frac{x^2-y^2}{\left(x^2+y^2\right)^2}=0 . \end{aligned} $$ 例 10 验证函数 $u(x, y)=\ln \sqrt{x^2+y^2}$ 满足方程 $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$. 注 方程 $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ 称为 Laplace (拉普拉斯) 方程,它是数学物理方程中 的一种很重要的方程,若引入记号(算子) $\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$ ,则拉普拉斯方程可写 成 $\Delta u=0$. 上述算子也称为拉普拉斯算子. 我们在例 7 ,例 8,例 9 中都看到 $\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial^2 z}{\partial y \partial x}$ ,这不是偶然的. 事实上,有 如下的定理. 定理 1 如果函数 $z=f(x, y)$ 的两个二阶混合偏导数 $\frac{\partial^2 z}{\partial y \partial x}$ 及 $\frac{\partial^2 z}{\partial x \partial y}$ 在区域 $D$ 内 连续,则在该区域内有 $\frac{\partial^2 z}{\partial y \partial x}=\frac{\partial^2 z}{\partial x \partial y}$. 证明从略. 这就是说,连续的二阶混合偏导数与求导次序无关.
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