【高等数学】坐标系变换下的二阶偏导数求解

Mr_zwX
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已知：u=f(x,y)u=f(x,y)有二阶连续偏导数，计算∂2u∂x2−∂2u∂y2\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。

{s=x+yt=x−y\left\{\begin{aligned}s & = & x+y \\t & = & x-y \\\end{aligned}\right.

这道偏导数的题实质上是利用中间变量求导，在书写的过程中容易出错，写一个解答过程，也算是一次markdown语法编辑的练习了

对于变换坐标系：

∂s∂x=∂t∂x=1,\begin{aligned} \frac{\partial s}{\partial x}=\frac{\partial t}{\partial x} &=1 , \end{aligned} ∂s∂y=1,∂t∂y=−1.\begin{aligned} \frac{\partial s}{\partial y}&=1 ,\\ \frac{\partial t}{\partial y}&=-1. \end{aligned}

∂u∂x=∂u∂s∂s∂x+∂u∂t∂t∂x=∂u∂s+∂u∂t.\begin{aligned} \frac{\partial u}{\partial x} &=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x} \\ &=\frac{\partial u}{\partial s}+\frac{\partial u}{\partial t}. \end{aligned}

∂2u∂x2=∂∂x(∂u∂x)=∂∂x(∂u∂s+∂u∂t)=∂∂x∂u∂s+∂∂x∂u∂t=∂∂s∂u∂s∂s∂x+∂∂t∂u∂s∂t∂x+∂∂s∂u∂t∂s∂x+∂∂t∂u∂t∂t∂x=∂u∂s2+2∂2u∂s∂t+∂u∂t2.\begin{aligned} \frac{\partial ^2u}{\partial x^2}&=\frac{\partial }{\partial x}(\frac{\partial u}{\partial x}) \\ &=\frac{\partial }{\partial x}(\frac{\partial u}{\partial s}+\frac{\partial u}{\partial t})\\ &=\frac{\partial }{\partial x}\frac{\partial u}{\partial s}+\frac{\partial }{\partial x}\frac{\partial u}{\partial t}\\ &=\frac{\partial }{\partial s}\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial }{\partial t}\frac{\partial u}{\partial s}\frac{\partial t}{\partial x}+\frac{\partial }{\partial s}\frac{\partial u}{\partial t}\frac{\partial s}{\partial x}+\frac{\partial }{\partial t }\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}\\ &=\frac{\partial u}{\partial s^2 }+2\frac{\partial ^2u}{\partial s \partial t}+\frac{\partial u}{\partial t^2 }.\\ \end{aligned}

由于代码编辑量较大...所以同理可得 算了，还是老老实实写完整过程，因为中间出现的负号容易出错...

∂u∂y=∂u∂s∂s∂y+∂u∂t∂t∂y=∂u∂s−∂u∂t.\begin{aligned} \frac{\partial u}{\partial y} &=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial y} \\ &=\frac{\partial u}{\partial s}-\frac{\partial u}{\partial t}. \end{aligned}

∂2u∂y2=∂∂y(∂u∂y)=∂∂y(∂u∂s−∂u∂t)=∂∂y∂u∂s−∂∂y∂u∂t=∂∂s∂u∂s∂s∂y+∂∂t∂u∂s∂t∂y−∂∂s∂u∂t∂s∂y−∂∂t∂u∂t∂t∂y=∂u∂s2−2∂2u∂s∂t+∂u∂t2.\begin{aligned} \frac{\partial ^2u}{\partial y^2}&=\frac{\partial }{\partial y}(\frac{\partial u}{\partial y}) \\ &=\frac{\partial }{\partial y}(\frac{\partial u}{\partial s}-\frac{\partial u}{\partial t})\\ &=\frac{\partial }{\partial y}\frac{\partial u}{\partial s}-\frac{\partial }{\partial y}\frac{\partial u}{\partial t}\\ &=\frac{\partial }{\partial s}\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial }{\partial t}\frac{\partial u}{\partial s}\frac{\partial t}{\partial y}-\frac{\partial }{\partial s}\frac{\partial u}{\partial t}\frac{\partial s}{\partial y}-\frac{\partial }{\partial t }\frac{\partial u}{\partial t}\frac{\partial t}{\partial y}\\ &=\frac{\partial u}{\partial s^2 }-2\frac{\partial ^2u}{\partial s \partial t}+\frac{\partial u}{\partial t^2 }.\\ \end{aligned}

∂2u∂y2=∂u∂s2−2∂2u∂s∂t+∂u∂t2.\begin{aligned} \frac{\partial ^2u}{\partial y^2}&=\frac{\partial u}{\partial s^2 }-2\frac{\partial ^2u}{\partial s \partial t}+\frac{\partial u}{\partial t^2 }. \end{aligned}

则在变换坐标系下有： {∂2u∂x2=∂u∂s2+2∂2u∂s∂t+∂u∂t2,∂2u∂y2=∂u∂s2−2∂2u∂s∂t+∂u∂t2.\left\{\begin{aligned}\frac{\partial ^2u}{\partial x^2} & = \frac{\partial u}{\partial s^2 }+2\frac{\partial ^2u}{\partial s \partial t}+\frac{\partial u}{\partial t^2}, \\\frac{\partial ^2u}{\partial y^2} & = \frac{\partial u}{\partial s^2}-2\frac{\partial^2u}{\partial s \partial t}+\frac{\partial u}{\partial t^2 }. \\\end{aligned}\right.

综合上述计算结果可得：

∂2u∂x2−∂2u∂y2=4∂2u∂s∂t.\begin{aligned}\frac{\partial ^2u}{\partial x^2}-\frac{\partial ^2u}{\partial y^2}&=4\frac{\partial ^2u}{\partial s \partial t} .\end{aligned}

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