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/

令 $f(x,y)=y\sin\frac{1}{x}$, 由于 $\sin()$ 函数有界, 故

\[
\lim_{(x,y)\rightarrow(0,0)}y\sin\frac{1}{x}=0,
\]

\[
\lim_{x\rightarrow 0}\lim_{y\rightarrow 0}y\sin\frac{1}{x}=\lim_{x\rightarrow 0}0=0,
\]

但是

\[
\lim_{x\rightarrow 0}y\sin\frac{1}{x}
\]

不存在.

以下内容来自于楼红卫 著《数学分析  要点、难点、拓展》P.111.  这里稍作修改.

如果 $f(x,y)$ 在 $(x_0,y_0)$ 点的两个混合偏导数都存在, 且其中之一在 $(x_0,y_0)$ 点连续, 则 $f_{xy}(x_0,y_0)=f_{yx}(x_0,y_0)$.

证明.  不妨设 $(x_0,y_0)=(0,0)$, 且 $f_{xy}$ 在 $(0,0)$ 点连续, 则由微分中值定理,

\[
\begin{split}
&\frac{f(x,y)-f(x,0)-f(0,y)+f(0,0)}{xy}\\
=&\frac{1}{xy}\Bigl[\bigl(f(x,y)-f(x,0)\bigr)-\bigl(f(0,y)-f(0,0)\bigr)\Bigr]\\
=&\frac{1}{x}\Bigl(f_y(x,\theta y)-f_y(0,\theta_1 y)\Bigr)\\
=&\frac{1}{x}\Bigl(f_y(x,\theta y)-f_y(0,\theta y)+f_y(0,\theta y)-f_y(0,\theta_1 y)\Bigr)\\
=&f_{xy}(\sigma x,\theta y)+\frac{f_y(0,\theta y)-f_y(0,\theta_1 y)}{x}
\end{split}
\]

于是

\[
\begin{split}
&\lim_{x\rightarrow 0}\lim_{y\rightarrow 0}\frac{f(x,y)-f(x,0)-f(0,y)+f(0,0)}{xy}\\
=&\lim_{x\rightarrow 0}\varlimsup_{y\rightarrow 0}f_{xy}(\sigma x,\theta y)+\lim_{x\rightarrow 0}\lim_{y\rightarrow 0}\frac{f_y(0,\theta y)-f_y(0,\theta_1 y)}{x}\\
=&f_{xy}(0,0)+\lim_{x\rightarrow 0}\frac{f_y(0,0)-f_y(0,0)}{x}\\
=&f_{xy}(0,0)
\end{split}
\]

这里倒数第二个等号是因为 $f_{xy}$ 在 $(0,0)$ 点处连续, 故 $f_y$ 在 $(0,0)$ 处连续.

类似的,

\[
\begin{split}
&\frac{f(x,y)-f(x,0)-f(0,y)+f(0,0)}{yx}\\
=&\frac{1}{yx}\Bigl[\bigl(f(x,y)-f(0,y)\bigr)-\bigl(f(x,0)-f(0,0)\bigr)\Bigr]\\
=&\frac{1}{y}\Bigl(f_x(\sigma x,y)-f_x(\sigma_1 x,0)\Bigr)\\
=&\frac{1}{y}\Bigl(f_x(\sigma x,y)-f_x(\sigma x,0)+f_x(\sigma x,0)-f_x(\sigma_1 x,0)\Bigr)\\
=&f_{yx}(\sigma x,\theta y)+\frac{f_x(\sigma x,0)-f_x(\sigma_1 x,0)}{y}
\end{split}
\]

于是

\[
\begin{split}
&\lim_{y\rightarrow 0}\lim_{x\rightarrow 0}\frac{f(x,y)-f(x,0)-f(0,y)+f(0,0)}{xy}\\
=&\lim_{y\rightarrow 0}\varlimsup_{x\rightarrow 0}f_{yx}(\sigma x,\theta y)+\lim_{y\rightarrow 0}\lim_{x\rightarrow 0}\frac{f_x(\sigma x, 0)-f_x(\sigma_1 x,0)}{y}\\
=&f_{yx}(0,0)+\lim_{y\rightarrow 0}\frac{f_x(0,0)-f_x(0,0)}{y}\\
=&f_{yx}(0,0).
\end{split}
\]

由 $f_{xy}$ 在 $(0,0)$ 点连续, 可得

\[
\lim_{(x,y)\rightarrow (0,0)}\frac{f(x,y)-f(x,0)-f(0,y)+f(0,0)}{xy}=f_{xy}(0,0).
\]

这样, 利用定理 16.1 可得

\[
\begin{split}
&\lim_{y\rightarrow 0}\lim_{x\rightarrow 0}\frac{f(x,y)-f(x,0)-f(0,y)+f(0,0)}{xy}\\
=&\lim_{x\rightarrow 0}\lim_{y\rightarrow 0}\frac{f(x,y)-f(x,0)-f(0,y)+f(0,0)}{xy}.\\
\end{split}
\]

即 $f_{xy}(0,0)=f_{yx}(0,0)$.

注: 这个问题也可参见 [1] P. 416, 习题 10.

References:

[1]  梅加强  著  《数学分析》

设 $2\sin(x+2y-3z)=x+2y-3z$, 证明: $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=1$.

设 $w=f(x+y+z,xyz)$, $f$ 具有二阶连续偏导数, 求 $\frac{\partial^2 w}{\partial x\partial z}$.

设 $z=xy+yf(\frac{x}{y})$, 其中 $f(u)$ 为可导函数, 证明: $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=xy+z$.

计算下列数的近似值.

(1)  $(1.007)^{2.98}$

(4)  $z=\frac{2x-y}{x+2y}$

(5)  $u=x(x+y^2+z^3)$

设 $r=\sqrt{x^2+y^2+z^2}$, 证明: $\frac{\partial^2(\ln r)}{\partial x^2}+\frac{\partial^2(\ln r)}{\partial y^2}+\frac{\partial^2(\ln r)}{\partial z^2}=\frac{1}{r^2}$. 即 $\Delta(\ln r)=\frac{1}{r^2}$.

1.

$\lim\limits_{(x,y)\rightarrow(0,0)}\frac{x+y}{x-y}$