下册 7.2 多元函数的可微性 第3题
📝 题目
3．求解或证明下列各题．
（1）设 $\phi(t), \psi(t)$ 有二阶连续导数，$\displaystyle u=\phi\left(\frac{y}{x}\right)+x \psi\left(\frac{y}{x}\right)$ ，求 $\displaystyle x^{2} \frac{\partial^{2} u}{\partial x^{2}}+2 x y \frac{\partial^{2} u}{\partial x \partial y}+y^{2} \frac{\partial^{2} u}{\partial y^{2}}$ ．
（2）设 $\displaystyle z=x f\left(\frac{x}{y}\right)+2 y f\left(\frac{y}{x}\right)$ ，其中 $f(x)$ 在 $(-\infty,+\infty)$ 上有连续的二阶导数，求 $z_{x}, z_{y}, z_{x y}$ ．
（3）设 $\displaystyle u=y f\left(\frac{x}{y}\right)+x g\left(\frac{y}{x}\right)$ ，其中 $f, g$ 具有二阶连续偏导数，求 $\displaystyle x \frac{\partial^{2} u}{\partial x^{2}}+y \frac{\partial^{2} u}{\partial x \partial y}$ 。
（4）设 $\displaystyle z=x^{2} f\left(\frac{y}{x}\right)+\frac{1}{y^{2}} g\left(\frac{y}{x}\right)$ ，且 $f, g$ 是任意的二阶可导函数，则
$$
x^{2} \frac{\partial^{2} z}{\partial x^{2}}+2 x y \frac{\partial^{2} z}{\partial x \partial y}+y^{2} \frac{\partial^{2} z}{\partial y^{2}}+x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=4 z \text {. }
$$
（5）设 $z=y f\left(x^{2}-y^{2}\right), f$ 为任意可微分函数，求 $\displaystyle y^{2} \frac{\partial z}{\partial x}+x y \frac{\partial z}{\partial y}$ 。
💡 答案解析
\section*{解题过程：}
（1）$\displaystyle u_{x}=-\frac{y}{x^{2}} \phi^{\prime}+\psi-\frac{y}{x} \psi^{\prime} ; u_{y}=\frac{1}{x} \phi^{\prime}+\psi^{\prime}$ ．
$$
\begin{aligned}
& u_{x x}=\frac{2 y}{x^{3}} \phi^{\prime}+\frac{y^{2}}{x^{4}} \phi^{\prime \prime}-\frac{y}{x^{2}} \psi^{\prime}+\frac{y}{x^{2}} \psi^{\prime}+\frac{y^{2}}{x^{3}} \psi^{\prime \prime}=\frac{2 y}{x^{3}} \phi^{\prime}+\frac{y^{2}}{x^{4}} \phi^{\prime \prime}+\frac{y^{2}}{x^{3}} \psi^{\prime \prime} \\
& u_{x y}=-\frac{1}{x^{2}} \phi^{\prime}-\frac{y}{x^{3}} \phi^{\prime \prime}+\frac{1}{x} \psi^{\prime}-\frac{1}{x} \psi^{\prime}-\frac{y}{x^{2}} \psi^{\prime \prime}=-\frac{1}{x^{2}} \phi^{\prime}-\frac{y}{x^{3}} \phi^{\prime \prime}-\frac{y}{x^{2}} \psi^{\prime \prime} \\
& u_{y y}=\frac{1}{x^{2}} \phi^{\prime \prime}+\frac{1}{x} \psi^{\prime \prime}
\end{aligned}
$$
故 $\displaystyle x^{2} \frac{\partial^{2} u}{\partial x^{2}}+2 x y \frac{\partial^{2} u}{\partial x \partial y}+y^{2} \frac{\partial^{2} u}{\partial y^{2}}=0$ ．
（2）$\displaystyle z_{x}=f\left(\frac{x}{y}\right)+\frac{x}{y} f^{\prime}\left(\frac{x}{y}\right)-2 \frac{y^{2}}{x^{2}} f^{\prime}\left(\frac{y}{x}\right), z_{y}=-\frac{x^{2}}{y^{2}} f^{\prime}\left(\frac{x}{y}\right)+2 f\left(\frac{y}{x}\right)+2 \frac{y}{x} f^{\prime}\left(\frac{y}{x}\right)$ ．
$$
\begin{aligned}
z_{x y} & =-\frac{x}{y^{2}} f^{\prime}\left(\frac{x}{y}\right)-\frac{x^{2}}{y^{3}} f^{\prime \prime}\left(\frac{x}{y}\right)-\frac{x}{y^{2}} f^{\prime}\left(\frac{x}{y}\right)-\frac{4 y}{x^{2}} f^{\prime}\left(\frac{y}{x}\right)-\frac{2 y^{2}}{x^{3}} f^{\prime \prime}\left(\frac{y}{x}\right) \\
& =-\frac{2 x}{y^{2}} f^{\prime}\left(\frac{x}{y}\right)-\frac{4 y}{x^{2}} f^{\prime}\left(\frac{y}{x}\right)-\frac{x^{2}}{y^{3}} f^{\prime \prime}\left(\frac{x}{y}\right)-\frac{2 y^{2}}{x^{3}} f^{\prime \prime}\left(\frac{y}{x}\right)
\end{aligned}
$$
（3）$\displaystyle u_{x}=f^{\prime}\left(\frac{x}{y}\right)+g\left(\frac{y}{x}\right)-\frac{y}{x} g\left(\frac{y}{x}\right)$ ．
$$
u_{x x}=\frac{1}{y} f^{\prime \prime}\left(\frac{x}{y}\right)-\frac{y}{x^{2}} g^{\prime}\left(\frac{y}{x}\right)+\frac{y}{x^{2}} g\left(\frac{y}{x}\right)+\frac{y^{2}}{x^{3}} g^{\prime}\left(\frac{y}{x}\right)
$$
$$
u_{x y}=-\frac{x}{y^{2}} f^{\prime \prime}\left(\frac{x}{y}\right)+\frac{1}{x} g^{\prime}\left(\frac{y}{x}\right)-\frac{1}{x} g\left(\frac{y}{x}\right)-\frac{y}{x^{2}} g^{\prime}\left(\frac{y}{x}\right)
$$
于是 $\displaystyle x \frac{\partial^{2} u}{\partial x^{2}}+y \frac{\partial^{2} u}{\partial x \partial y}=0$ ．
（4）$\displaystyle z_{x}=2 x f\left(\frac{y}{x}\right)-y f^{\prime}\left(\frac{y}{x}\right)-\frac{1}{y x^{2}} g^{\prime}\left(\frac{y}{x}\right), z_{y}=x f^{\prime}\left(\frac{y}{x}\right)-\frac{2}{y^{3}} g\left(\frac{y}{x}\right)+\frac{1}{x y^{2}} g^{\prime}\left(\frac{y}{x}\right)$ ．
$$
\begin{aligned}
z_{x x} & =2 f\left(\frac{y}{x}\right)-\frac{2 y}{x} f^{\prime}\left(\frac{y}{x}\right)-\frac{y^{2}}{x^{2}} f^{\prime \prime}\left(\frac{y}{x}\right)+\frac{2}{y x^{3}} g^{\prime}\left(\frac{y}{x}\right)+\frac{1}{x^{4}} g^{\prime \prime}\left(\frac{y}{x}\right) . \\
z_{y y} & =f^{\prime \prime}\left(\frac{y}{x}\right)+\frac{6}{y^{4}} g\left(\frac{y}{x}\right)-\frac{2}{x y^{3}} g^{\prime}\left(\frac{y}{x}\right)-\frac{2}{x y^{3}} g^{\prime}\left(\frac{y}{x}\right)+\frac{1}{x^{2} y^{2}} g^{\prime \prime}\left(\frac{y}{x}\right) \\
& =f^{\prime \prime}\left(\frac{y}{x}\right)+\frac{6}{y^{4}} g\left(\frac{y}{x}\right)-\frac{4}{x y^{3}} g^{\prime}\left(\frac{y}{x}\right)+\frac{1}{x^{2} y^{2}} g^{\prime \prime}\left(\frac{y}{x}\right) . \\
z_{x y} & =\frac{\partial}{\partial y}\left(2 x f\left(\frac{y}{x}\right)-y f^{\prime}\left(\frac{y}{x}\right)-\frac{1}{y x^{2}} g^{\prime}\left(\frac{y}{x}\right)\right) \\
& =2 f^{\prime}\left(\frac{y}{x}\right)-f^{\prime}\left(\frac{y}{x}\right)-\frac{y}{x} f^{\prime \prime}\left(\frac{y}{x}\right)+\frac{1}{x^{2} y^{2}} g^{\prime}\left(\frac{y}{x}\right)-\frac{1}{x^{3} y} g^{\prime \prime}\left(\frac{y}{x}\right) \\
& =f^{\prime}\left(\frac{y}{x}\right)-\frac{y}{x} f^{\prime \prime}\left(\frac{y}{x}\right)+\frac{1}{x^{2} y^{2}} g^{\prime}\left(\frac{y}{x}\right)-\frac{1}{x^{3} y} g^{\prime \prime}\left(\frac{y}{x}\right) .
\end{aligned}
$$
于是 $\displaystyle \quad x^{2} \frac{\partial^{2} z}{\partial x^{2}}+2 x y \frac{\partial^{2} z}{\partial x \partial y}+y^{2} \frac{\partial^{2} z}{\partial y^{2}}+x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=4 x^{2} f\left(\frac{y}{x}\right)+\frac{4}{y^{2}} g\left(\frac{y}{x}\right)=4 z$ ．
（5）$z_{x}=2 x y f^{\prime}\left(x^{2}-y^{2}\right), z_{y}=f\left(x^{2}-y^{2}\right)-2 y^{2} f^{\prime}\left(x^{2}-y^{2}\right)$ 。于是 $\displaystyle y^{2} \frac{\partial z}{\partial x}+x y \frac{\partial z}{\partial y}=x z$ 。
📋 详细解题步骤
步骤 1/7
目标：引入中间变量简化表达式
令 $v = \frac{y}{x}$，则 $u = \phi(v) + x \psi(v)$。
提示：注意 $v$ 是 $x$ 和 $y$ 的函数，求导时要使用链式法则。
步骤 2/7
目标：计算一阶偏导数
对 $x$ 求偏导：$u_x = \phi'(v) \cdot (-\frac{y}{x^2}) + \psi(v) + x \psi'(v) \cdot (-\frac{y}{x^2}) = -\frac{y}{x^2} \phi' + \psi - \frac{y}{x} \psi'$。
对 $y$ 求偏导：$u_y = \phi'(v) \cdot \frac{1}{x} + x \psi'(v) \cdot \frac{1}{x} = \frac{1}{x} \phi' + \psi'$。
公式：链式法则：$\frac{\partial}{\partial x} f(v) = f'(v) \frac{\partial v}{\partial x}$
提示：注意 $\frac{\partial v}{\partial x} = -\frac{y}{x^2}$，$\frac{\partial v}{\partial y} = \frac{1}{x}$。
步骤 3/7
目标：计算二阶偏导数 $u_{xx}$
对 $u_x$ 再对 $x$ 求偏导：
$u_{xx} = \frac{\partial}{\partial x} \left( -\frac{y}{x^2} \phi' + \psi - \frac{y}{x} \psi' \right)$。
逐项求导：
第一项：$\frac{\partial}{\partial x} \left( -\frac{y}{x^2} \phi' \right) = \frac{2y}{x^3} \phi' - \frac{y}{x^2} \phi'' \cdot (-\frac{y}{x^2}) = \frac{2y}{x^3} \phi' + \frac{y^2}{x^4} \phi''$。
第二项：$\frac{\partial}{\partial x} \psi = \psi' \cdot (-\frac{y}{x^2})$。
第三项：$\frac{\partial}{\partial x} \left( -\frac{y}{x} \psi' \right) = \frac{y}{x^2} \psi' - \frac{y}{x} \psi'' \cdot (-\frac{y}{x^2}) = \frac{y}{x^2} \psi' + \frac{y^2}{x^3} \psi''$。
合并：$u_{xx} = \frac{2y}{x^3} \phi' + \frac{y^2}{x^4} \phi'' + \frac{y^2}{x^3} \psi''$。
提示：注意 $\psi$ 对 $x$ 求导时也要用链式法则，且 $\psi$ 本身是 $v$ 的函数。
步骤 4/7
目标：计算二阶偏导数 $u_{xy}$
对 $u_x$ 再对 $y$ 求偏导：
$u_{xy} = \frac{\partial}{\partial y} \left( -\frac{y}{x^2} \phi' + \psi - \frac{y}{x} \psi' \right)$。
逐项求导：
第一项：$\frac{\partial}{\partial y} \left( -\frac{y}{x^2} \phi' \right) = -\frac{1}{x^2} \phi' - \frac{y}{x^2} \phi'' \cdot \frac{1}{x} = -\frac{1}{x^2} \phi' - \frac{y}{x^3} \phi''$。
第二项：$\frac{\partial}{\partial y} \psi = \psi' \cdot \frac{1}{x}$。
第三项：$\frac{\partial}{\partial y} \left( -\frac{y}{x} \psi' \right) = -\frac{1}{x} \psi' - \frac{y}{x} \psi'' \cdot \frac{1}{x} = -\frac{1}{x} \psi' - \frac{y}{x^2} \psi''$。
合并：$u_{xy} = -\frac{1}{x^2} \phi' - \frac{y}{x^3} \phi'' - \frac{y}{x^2} \psi''$。
提示：注意 $\frac{\partial v}{\partial y} = \frac{1}{x}$。
步骤 5/7
目标：计算二阶偏导数 $u_{yy}$
对 $u_y$ 再对 $y$ 求偏导：
$u_{yy} = \frac{\partial}{\partial y} \left( \frac{1}{x} \phi' + \psi' \right) = \frac{1}{x} \phi'' \cdot \frac{1}{x} + \psi'' \cdot \frac{1}{x} = \frac{1}{x^2} \phi'' + \frac{1}{x} \psi''$。
提示：注意 $\phi'$ 和 $\psi'$ 都是 $v$ 的函数。
步骤 6/7
目标：代入表达式并化简
计算 $x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy}$：
$x^2 u_{xx} = x^2 \left( \frac{2y}{x^3} \phi' + \frac{y^2}{x^4} \phi'' + \frac{y^2}{x^3} \psi'' \right) = \frac{2y}{x} \phi' + \frac{y^2}{x^2} \phi'' + \frac{y^2}{x} \psi''$。
$2xy u_{xy} = 2xy \left( -\frac{1}{x^2} \phi' - \frac{y}{x^3} \phi'' - \frac{y}{x^2} \psi'' \right) = -\frac{2y}{x} \phi' - \frac{2y^2}{x^2} \phi'' - \frac{2y^2}{x} \psi''$。
$y^2 u_{yy} = y^2 \left( \frac{1}{x^2} \phi'' + \frac{1}{x} \psi'' \right) = \frac{y^2}{x^2} \phi'' + \frac{y^2}{x} \psi''$。
相加：$\left( \frac{2y}{x} \phi' - \frac{2y}{x} \phi' \right) + \left( \frac{y^2}{x^2} \phi'' - \frac{2y^2}{x^2} \phi'' + \frac{y^2}{x^2} \phi'' \right) + \left( \frac{y^2}{x} \psi'' - \frac{2y^2}{x} \psi'' + \frac{y^2}{x} \psi'' \right) = 0$。
提示：注意合并同类项时系数要准确。
步骤 7/7
目标：得出最终结果
因此，$x^{2} \frac{\partial^{2} u}{\partial x^{2}}+2 x y \frac{\partial^{2} u}{\partial x \partial y}+y^{2} \frac{\partial^{2} u}{\partial y^{2}} = 0$。
提示：结果为零，说明该表达式是某种齐次线性偏微分方程的形式。
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