问题及解答

设 $z=z(x,y)$ 可微, 且满足 $x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=z^2$. 作变换

\[
\begin{cases}
u&=x,\\
v&=\frac{1}{y}-\frac{1}{x},
\end{cases}\quad\text{及}\quad w=\frac{1}{z}-\frac{1}{x},
\]

证明: $\frac{\partial w}{\partial u}=0$.

注: 题目来自于 https://www.bilibili.com/video/BV1kv4y1E7kK/

根据变换 $w=\frac{1}{z}-\frac{1}{x}=\frac{1}{z}-\frac{1}{u}$.

\[
\frac{\partial w}{\partial u}=-\frac{1}{z^2}\cdot\frac{\partial z}{\partial u}+\frac{1}{u^2}.
\]

由 $v=\frac{1}{y}-\frac{1}{x}$ 及 $u=x$ 可以反解出 $x=u$, $y=\frac{u}{1+uv}$. 于是

\[
z(x,y)=z(u,\frac{u}{1+uv})
\]

因此

\[
\begin{split}
\frac{\partial z}{\partial u}&=\frac{\partial z}{\partial x}\cdot 1+\frac{\partial z}{\partial y}\cdot(\frac{u}{1+uv})'_u\\
&=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot\frac{1\cdot(1+uv)-u\cdot v}{(1+uv)^2}\\
&=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot\frac{1}{(1+uv)^2}
\end{split}
\]

因此,

\[
\begin{split}
\frac{\partial w}{\partial u}&=-\frac{1}{z^2}\cdot\biggl(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot\frac{1}{(1+uv)^2}\biggr)+\frac{1}{u^2}\\
&=\frac{1}{u^2}-\frac{z'_x+\frac{1}{(1+uv)^2}z'_y}{x^2 z'_x+y^2 z'_y}\\
&=\frac{x^2 z'_x+y^2 z'_y-u^2 z'_x-\frac{u^2}{(1+uv)^2}z'_y}{u^2(x^2 z'_x+y^2 z'_y)}\\
&=\frac{(y^2-\frac{u^2}{(1+uv)^2}) z'_y}{u^2(x^2 z'_x+y^2 z'_y)}\\
&=0.
\end{split}
\]