题号:914    题型:解答题    来源:1996年全国硕士研究生招生考试试题
设变换 $\left\{\begin{array}{l}u=x-2 y, \\ v=x+a y\end{array}\right.$ 可把方程 $6 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}-\frac{\partial^{2} z}{\partial y^{2}}=0$ 简化为 $\frac{\partial^{2} z}{\partial u \partial v}=0$, 求常数 $a$. (这里应假设 $z$ 有二阶连续偏导数. )
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答案:
【解析】由多元复合函数求导法则, 得
$$
\begin{aligned}
&\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}=\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v} \\
&\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}=-2 \frac{\partial z}{\partial u}+a \frac{\partial z}{\partial v}
\end{aligned}
$$

所以
$$
\begin{aligned}
\frac{\partial^{2} z}{\partial x^{2}} &=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial u}\right)+\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial v}\right)=\frac{\partial^{2} z}{\partial u^{2}} \cdot \frac{\partial u}{\partial x}+\frac{\partial^{2} z}{\partial u \partial v} \cdot \frac{\partial v}{\partial x}+\frac{\partial^{2} z}{\partial v^{2}} \cdot \frac{\partial v}{\partial x}+\frac{\partial^{2} z}{\partial v \partial u} \frac{\partial u}{\partial x} \\
&=\frac{\partial^{2} z}{\partial u^{2}}+2 \frac{\partial^{2} z}{\partial u \partial v}+\frac{\partial^{2} z}{\partial v^{2}} \\
\frac{\partial^{2} z}{\partial x \partial y} &=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial u}\right)+\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial v}\right)=\frac{\partial^{2} z}{\partial u^{2}} \cdot \frac{\partial u}{\partial y}+\frac{\partial^{2} z}{\partial u \partial v} \cdot \frac{\partial v}{\partial y}+\frac{\partial^{2} z}{\partial v^{2}} \cdot \frac{\partial v}{\partial y}+\frac{\partial^{2} z}{\partial v \partial u} \frac{\partial u}{\partial y} \\
&=-2 \frac{\partial^{2} z}{\partial u^{2}}+(a-2) \frac{\partial^{2} z}{\partial u \partial v}+a \frac{\partial^{2} z}{\partial v^{2}}, \\
\frac{\partial^{2} z}{\partial y^{2}} &=-2 \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial u}\right)+a \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial v}\right) \\
&=-2\left(\frac{\partial^{2} z}{\partial u^{2}} \cdot \frac{\partial u}{\partial y}+\frac{\partial^{2} z}{\partial u \partial v} \cdot \frac{\partial v}{\partial y}\right)+a\left(\frac{\partial^{2} z}{\partial v^{2}} \cdot \frac{\partial v}{\partial y}+\frac{\partial^{2} z}{\partial v \partial u} \cdot \frac{\partial u}{\partial y}\right) \\
&=4 \frac{\partial^{2} z}{\partial u^{2}}-4 a \frac{\partial^{2} z}{\partial u \partial v}+a^{2} \frac{\partial^{2} z}{\partial v^{2}}
\end{aligned}
$$
代入 $6 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}-\frac{\partial^{2} z}{\partial y^{2}}=0$, 并整理得
$$
6 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}-\frac{\partial^{2} z}{\partial y^{2}}=(10+5 a) \frac{\partial^{2} z}{\partial u \partial v}+\left(6+a-a^{2}\right) \frac{\partial^{2} z}{\partial v^{2}}=0
$$

于是, 令 $6+a-a^{2}=0$ 得 $a=3$ 或 $a=-2$.
$a=-2$ 时, $10+5 a=0$, 故舍去, $a=3$ 时, $10+5 a \neq 0$, 因此仅当 $a=3$ 时化简为 $\frac{\partial^{2} z}{\partial u \partial v}=0$.
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