2022考研数二第20题解析：多元复合函数的偏导数、一阶偏微分方程

一、题目

二、解析

第（1）问

首先，由 $g\left(x, y\right) = f\left(x, y – x\right)$ 和多元复合函数的链式求导法则，可知：

$$
\begin{aligned}
\frac{\partial g\left(x, y\right)}{\partial x} & = f^{\prime}_{u} – f^{\prime}_{v} \\ \\
& = \frac{\partial f \left( u, v \right)}{\partial u} – \frac{\partial f \left( u, v \right)}{\partial v} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\frac{\partial f\left(u, v\right)}{\partial u} – \frac{\partial f\left(u, v\right)}{\partial v} = 2 \left(u – v\right) \mathrm{e}^{-\left(u + v\right)}} \\ \\
& = 2 \left(u – v\right) \mathrm{e}^{-\left(u + v\right)} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{u = x, \ v = y – x} \\ \\
& = 2 \left(x – y + x\right) \mathrm{e}^{-y} \\ \\
& = \textcolor{lightgreen}{ 2 \left(2 x – y\right) \mathrm{e}^{-y} }
\end{aligned}
$$

第（2）问

由第（1）问可知：

$$
\frac{\partial g\left(x, y\right)}{\partial x} = 2 \left(2 x – y\right) \mathrm{e}^{-y}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 所以：

$$
\begin{aligned}
g \left(x, y\right) & = \int \frac{\partial g\left(x, y\right)}{\partial x} \mathrm{~d} x \\ \\
& = \int 2 \left(2 x – y\right) \mathrm{e}^{-y} \mathrm{~d}x \\ \\
& = 2 \mathrm{e}^{-y} \int \left(2 x – y\right) \mathrm{~d}x \\ \\
& = 2 \mathrm{e}^{-y} \int 2 x \mathrm{~d}x – 2 \mathrm{e}^{-y} \int y \mathrm{~d}x \\ \\
& = 2 \mathrm{e}^{-y} \textcolor{pink}{ \int 2 x \mathrm{~d}x } – 2 y \mathrm{e}^{-y} \textcolor{lightblue}{ \int 1 \mathrm{~d}x } \\ \\
& = 2 \textcolor{pink}{ x^{2} } \mathrm{e}^{-y} – 2 x \textcolor{lightblue}{ y } \mathrm{e}^{-y} + \varphi\left(y\right) \\ \\
& = \textcolor{lightgreen}{ 2 x \left(x – y\right) \mathrm{e}^{-y} + \varphi\left(y\right) }
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 又因为 $g\left(x, y\right) = f\left(x, y – x\right)$, 所以：

$$
\begin{aligned}
f\left(x, y – x\right) & = 2 x \left(x – y\right) \mathrm{e}^{-y} + \varphi\left(y\right) \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{u = x, \ v = y – x} \\ \\
& = \textcolor{lightgreen}{ -2 u v \mathrm{e}^{-\left(u + v\right)} + \varphi\left(u + v\right) = f \left( u, v \right) } \\ \\
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 又因为：

$$
f\left(u, 0\right) = u^{2} \mathrm{e}^{-u}
$$

所以，将 $v = 0$ 代入 $f \left( u,v \right) = -2 u v \mathrm{e}^{-\left(u + v\right)} + \varphi\left(u + v\right)$, 得：

$$
\varphi\left(u\right) = u^{2} \mathrm{e}^{-u}
$$

于是：

$$
\begin{aligned}
f\left(u, v\right) & = -2 u v \mathrm{e}^{-\left(u + v\right)} + \left(u + v\right)^{2} \mathrm{e}^{-\left(u + v\right)} \\ \\
& = \textcolor{lightgreen}{ \left(u^{2} + v^{2}\right) \mathrm{e}^{-\left(u + v\right)} }
\end{aligned}
$$

在上面的计算过程中，我们是在计算出 $f\left(x, y – x\right)$ 含有 $\varphi\left(y\right)$ 的表达式后立即进行了 $u = x, \ v = y – x$ 的代换，我们也可以在计算出 $f\left(x, y – x\right)$ 不含有 $\varphi\left(y\right)$ 的表达式后再进行这个代换，所得的结果是一样的：

首先：

$$
f\left(x, y – x\right) = 2 x \left(x – y\right) \mathrm{e}^{-y} + \varphi\left(y\right)
$$

又因为：

$$
\begin{aligned}
f\left(u, 0\right) & = u^{2} \mathrm{e}^{-u} \\ \\
g\left(x, y\right) & = f\left(x, y – x\right)
\end{aligned}
$$

所以：

$$
f \left( x,0 \right) = g \left( x,x \right) = \varphi (x) = x^{2} \mathrm{e}^{-x}
$$

于是：

$$
g \left( x,y \right) = 2 x \left(x – y\right) \mathrm{e}^{-y} + x^{2} \mathrm{e}^{-x} = f \left( x, y-x \right) = g \left( x, y \right)
$$

又因为 $u = x, \ v = y – x$, 所以：

$$
\begin{aligned}
f \left( u, v \right) & = – 2 u v \mathrm{e}^{- \left( u+v \right)} + u^{2} \mathrm{e}^{- \left( u+v \right)} \\ \\
& = \textcolor{lightgreen}{ \left( u^{2} + v^{2} \right) \mathrm{e}^{- \left( u+v \right)} }
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 接着，根据《二元函数非条件极值的判定和计算方法》可知，首先令：

$$
\begin{aligned}
\frac{\partial f}{\partial u} = f^{\prime}_{u} & = 2 u \mathrm{e}^{-\left(u + v\right)} – \left(u^{2} + v^{2}\right) \mathrm{e}^{-\left(u + v\right)} \\
& = \textcolor{lightgreen}{ – \mathrm{e}^{- \left( u+v \right)} \left( 2 u – u^{2} – v^{2} \right)^{2} = 0 } \\ \\
\frac{\partial f}{\partial v} = f^{\prime}_{v} & = 2 v \mathrm{e}^{-\left(u + v\right)} – \left(u^{2} + v^{2}\right) \mathrm{e}^{-\left(u + v\right)} \\
& = \textcolor{lightgreen}{ – \mathrm{e}^{- \left( u+v \right)} \left( 2 v – u^{2} – v^{2} \right) = 0 }
\end{aligned}
$$

即：

$$
\begin{cases}
2 u – u^{2} – v^{2} = 0 \\
2 v – u^{2} – v^{2} = 0
\end{cases} \textcolor{lightgreen}{ \leadsto } u = v
$$

将 $u = v$ 代回原式 $\begin{cases} 2 u – u^{2} – v^{2} = 0 \\ 2 v – u^{2} – v^{2} = 0 \end{cases}$, 得：

$$
\begin{cases}
2 u – u^{2} – v^{2} = 0 \\
2 v – u^{2} – v^{2} = 0
\end{cases} \textcolor{lightgreen}{ \leadsto } \begin{cases}
u – u^{2} = 0 \\
v – v^{2} = 0
\end{cases} \textcolor{lightgreen}{ \leadsto } \textcolor{lightgreen}{ \begin{cases}
u = v = 0 \\
u = v = 1
\end{cases} }
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 又：

$$
\begin{aligned}
\boldsymbol{A} = f_{u u}^{\prime\prime} = f_{1 1}^{\prime\prime} & = 2 \mathrm{e}^{-\left(u + v\right)} – 2 u \mathrm{e}^{-\left(u + v\right)} – 2 u \mathrm{e}^{-\left(u + v\right)} + \left(u^{2} + v^{2}\right) \mathrm{e}^{-\left(u + v\right)} \\
& = \left(2 – 4 u + u^{2} + v^{2}\right) \mathrm{e}^{-\left(u + v\right)} \\ \\
\boldsymbol{B} = f ^{\prime \prime} _{u, v} = f_{1 2}^{\prime\prime} & = -2 u \mathrm{e}^{-\left(u + v\right)} – 2 v \mathrm{e}^{-\left(u + v\right)} + \left(u^{2} + v^{2}\right) \mathrm{e}^{-\left(u + v\right)} \\
& = \left(u^{2} + v^{2} – 2 u – 2 v\right) \mathrm{e}^{-\left(u + v\right)} \\ \\
\boldsymbol{C} = f_{v v}^{\prime\prime} = f_{2 2}^{\prime\prime} & = \left(2 – 4 v + v^{2} + u^{2}\right) \mathrm{e}^{-\left(u + v\right)}
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 将 $u = v = 0$ 代入上面的 $A$, $B$, $C$ 式，得：

$$
\begin{aligned}
\boldsymbol{A}\left(0, 0\right) & = 2 \\ \\
\boldsymbol{B}\left(0, 0\right) & = 0 \\ \\
\boldsymbol{C}\left(0, 0\right) & = 2
\end{aligned}
$$

$\textcolor{lightgreen}{\blacktriangleright}$ 将 $u = v = 1$ 代入上面的 $A$, $B$, $C$ 式，得：

$$
\begin{aligned}
\boldsymbol{A}\left(1, 1\right) & = 0 \\ \\
\boldsymbol{B}\left(1, 1\right) & = -2 \mathrm{e}^{-2} \\ \\
\boldsymbol{C}\left(1, 1\right) & = 0
\end{aligned}
$$

综上：

$\textcolor{lightgreen}{\blacktriangleright}$ 在 $\left(0, 0\right)$ 点处，有 $\boldsymbol{A} \boldsymbol{C} – \boldsymbol{B}^{2} = 4 > 0, \boldsymbol{A} > 0$, 所以，函数 $f \left( u,v \right)$ 在这一点取得极小值 $0$;

$\textcolor{lightgreen}{\blacktriangleright}$ 在 $\left(1, 1\right)$ 点处，有 $\boldsymbol{A} \boldsymbol{C} – \boldsymbol{B}^{2} < 0$, 所以，函数 $f \left( u, v \right)$ 在该点处没有极值.

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