二阶偏导数在坐标系变换中的求解
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简介：
本文探讨了二阶偏导数在不同坐标系之间转换时的计算方法，分析其变化规律，并提供具体实例来阐述求解技巧和应用。
已知函数 $u=f(x,y)$ 有二阶连续偏导数，要求计算 $\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}$ 在新的坐标系下对应的表达式。设新坐标系为：
$$
s = x + y \\
t = x - y
$$
这个问题实质上是利用中间变量求导，在书写过程中容易出错，下面是一个解答过程。
首先对 $u$ 关于 $x$ 和 $y$ 的偏导数进行链式法则变换。由于新坐标系下的关系为：
$$
s = x + y \\
t = x - y
$$
可以解得原变量 $x,y$ 与中间变量 $s,t$ 的关系：
$$
x = \frac{s+t}{2},\quad y=\frac{s-t}{2}
$$
接下来对函数 $u=f(x,y)$ 求偏导数。首先求一阶偏导：
1. 对于 $\frac{\partial u}{\partial x}$ 和 $\frac{\partial u}{\partial y}$：
$$
\begin{aligned}
&\frac{\partial u}{\partial x} = \frac{\partial f(x,y)}{\partial s}\cdot \frac{\partial s}{\partial x} + \frac{\partial f(x,y)}{\partial t}\cdot \frac{\partial t}{\partial x}\\
&= \left(\frac{\partial u}{\partial s}\right)\cdot 1 + \left(\frac{\partial u}{\partial t}\right) \cdot 1 = \frac{\partial u}{\partial s} + \frac{\partial u}{\partial t}
\\
&\frac{\partial u}{\partial y} = \frac{\partial f(x,y)}{\partial s}\cdot \frac{\partial s}{\partial y} + \frac{\partial f(x,y)}{\partial t}\cdot \frac{\partial t}{\partial y}\\
&= \left(\frac{\partial u}{\partial s}\right)\cdot 1 - \left(\frac{\partial u}{\partial t}\right) \cdot 1 = \frac{\partial u}{\partial s} - \frac{\partial u}{\partial t}
\\
\end{aligned}
$$
2. 接下来求二阶偏导：
$$
\begin{aligned}
&\frac{\partial^2 u}{\partial x^2} = \frac{\partial }{\partial x}\left(\frac{\partial u}{\partial s} + \frac{\partial u}{\partial t}\right)\\
&= \left( \frac{\partial^2 u}{\partial s^2}+\frac{\partial^2 u}{\partial t^2}\right)\cdot1+2\left(\frac{\partial^2 u}{\partial s \, \partial t}\right)
\\
&\frac{\partial^2 u}{\partial y^2} = \frac{\partial }{\partial y}\left( \frac{\partial u}{\partial s}-\frac{\partial u}{\partial t} \right)\\
&= \left(\frac{\partial^2u}{\partial s^2}+\frac{ {\partial ^2u}}{\partial t^{2}}\right)\cdot1-2\left( \frac{\partial ^{2}u }{{\partial s}\, {\partial t }} \right)
\\
& \therefore \frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}=4\cdot \left( \frac{\partial ^{2}u }{{\partial s}\, {\partial t }} \right)
\\
\end{aligned}
$$
因此，$\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}$ 在新的坐标系下对应的表达式为 $4 \cdot \left( \frac{\partial ^{2}u }{{\partial s}\, {\partial t }} \right)$.
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