基础知识

对于 $z=f (u (x, y), v (x, y))$ 偏导数为:

$$ \begin{array}{l} \frac{\partial z}{\partial x}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}=f_{1}^{\prime} \cdot \frac{\partial u}{\partial x}+f_{2}^{\prime} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y}=f_{1}^{\prime} \cdot \frac{\partial u}{\partial y}+f_{2}^{\prime} \cdot \frac{\partial v}{\partial y} \end{array} $$

复合函数偏导数求导,就是分析从外函数到自变量要走过的路,把每条路的求导加起来就是了。一般来说我们使用广度优先的策略,一层一层展开求导。

【要点】

  1. 层层展开求导。
  2. 导数的复合路径图不变。

题型

复合函数偏导数

【例子】 设 $w = f (x+y+z, xyz)$ 求 $\frac {\partial w}{\partial x}$,$ \frac {\partial ^2w}{\partial x \partial z} $

【分析与解答】

首先分析复合路径:

f
    x+y+z
        x
        y
        z
    xyz
        x
        y
        z
$$ \begin{align} \frac{\partial w}{\partial x} &= f_1' (x+y+z)' + f_2' (xyz)'\\ &= f_1' + f_2' yz \end{align} $$
$$ \begin{align} \frac{\partial ^2w}{\partial x \partial z} &=(f_{11}^{''}1 + f_{12}^{''} xy) + [(f_{21}^{''} 1 + f_{22}^{11})yz + f_2' y]\\ &= f_11^{''} + (xy+yz)f_{12}^{''} + xy^2z f_{22}^{''}+yf_2' \end{align} $$

【例子】 设 $u=f (x,y,z), y = g (x,t), t = h (x,z)$ ,都有一阶连续偏导数,求 $\frac {\partial u}{\partial x}, \frac {\partial u}{\partial z}$

【分析与解答】

首先分析复合路径

u = f
    x
    y = g
        x
        t = h
            x
            z
    t = h
        x
        z

到 $x$ 一共有 4 条路径。一层一层展开

$$ \begin{align} \frac{\partial u}{\partial x} &= f_1' \cdot (x)' + f_2' \cdot (y)' + f_3' \cdot (z)'\\ &= f_1' + f_2'(g_1' + g_2' \cdot t') + 0 \\ &= f_1' + f_2' (g_1' + g_2' \cdot (h_1' )) \\ &= f_1' + f_2'g_1' + f_2'g_2'h_1' \end{align} $$

【例子】 $z = f (u, x, y)$,$u = xe^y$ ,求 $\frac {\partial ^2 z}{\partial x \partial y} $

【分析与解答】

分析复合路径

z = f
    u
        x
        y
    x
    y

先求 $\frac {\partial z}{\partial x} $

$z\to y$ 的分支就不考虑了,一定是 $0$。

$$ \begin{align} \frac{\partial z}{\partial x} &= f_1'\cdot u' + f_2'\cdot x'\\ &= f_1'e^y + f_2' \end{align} $$

然后求 $\frac {\partial ^2 z}{\partial x \partial y} $

$$ \begin{align} \frac{\partial ^2 z}{\partial x \partial y} &= [(f_{11}^{''}u' + f_{13}^{''}y')e^y + f_1' e^y] + (f_{21}^{''}u'+f_{23}^{''}y')\\ &= [(f_{11}^{''}xe^y + f_{13}^{''}y')e^y + f_1' e^y] + (f_{21}^{''}xe^y+f_{23}^{''}y')\\ &= xe^{2y}f_{11}^{''} + e^y f_{13}^{''} + e^y f_1' + xe^yf_{21}^{''} + f_{23}^{''} \end{align} $$

【例子】 用变换 $\left\{\begin {array}{l} u=x-2 y \\v=x+a y\end {array}\right.$ 可把 方程 $6 \frac {\partial^{2} z}{\partial x^{2}}+\frac {\partial^{2} z}{\partial x \partial y}-\frac {\partial^{2} z}{\partial y^{2}}=0$ 化简为 $\frac {\partial^{2} z}{\partial u \partial v}=0$,求 $a$;$z$ 有二阶连续偏导数 。

【分析与解答】

这道题看似恐怖,其实只是像写 CURD 一样的体力活。下面来计算吧:

$6 \frac {\partial^{2} z}{\partial x^{2}}+\frac {\partial^{2} z}{\partial x \partial y}-\frac {\partial^{2} z}{\partial y^{2}}=0$ 是 $z$ 关于 $x,y$ 的方程,替换后变成了 $z$ 关于 $u,v$ 的方程。

分析复合路径:

  • z
    • u
      • x
      • y
    • v
      • x
      • y
$$ \begin{align} \frac{\partial z}{\partial x} &= {\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1} \end{align} $$
$$ \begin{align} \frac{\partial ^2 z}{\partial x^2} &= {\color{red}[\frac{\partial z}{\partial u}(\frac{\partial z}{\partial u} \cdot 1 + \frac{\partial z}{\partial v} \cdot 1) ]} + {\color{blue}[\frac{\partial z}{\partial v} (\frac{\partial z}{\partial u} \cdot 1 + \frac{\partial z}{\partial v} \cdot 1)]}\\ &= \frac{\partial ^2z }{\partial u^2} + 2 \frac{\partial ^2 z}{\partial uv} + \frac{\partial ^2 z}{\partial v^2} \tag{1} \end{align} $$

PS:二阶偏导连续所以可以合并混合导

$$ \begin{align} \frac{\partial z}{\partial y} &= \frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a \end{align} $$

套娃开始(

$$ \begin{align} \frac{\partial^2 z}{\partial x \partial y} &= -2 \frac{\partial z}{\partial u} [{\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1}] + a \frac{\partial z}{\partial v} [{\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1}]\\ &= -2 \frac{\partial ^2 z}{\partial u^2} + (a -2) \frac{\partial ^2 z}{\partial u \partial v} + a \frac{\partial ^2 z}{\partial v^2}\tag{2} \end{align} $$
$$ \begin{align} \frac{\partial^2 z}{\partial^2 y} &= -2 \frac{\partial z}{\partial u} [\frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a] + a \frac{\partial z}{\partial v} [\frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a]\\ &= 4 \frac{\partial ^2 z}{\partial u^2} -4a \frac{\partial ^2 z }{\partial u \partial v} + a^2 \frac{\partial ^2 z}{\partial v ^2} \tag{3} \end{align} $$

将 $1,2,3$ 式代入,得到:

$$ (10+5a) \frac{\partial ^2 z}{\partial u \partial v} + (6 + a - a^2) \frac{\partial ^2 z}{\partial v^2} $$

想要 $\frac {\partial ^2 z}{\partial u \partial v} = 0 $ ,只需 $6 + a - a^2 = 0 \wedge 10 + 5a \ne 0$。方程解为

$$ a = 3 $$