坐标系变换下的二阶偏导数求解
已知：u=f(x,y)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}∂x2∂2u−∂y2∂2u在新的坐标系下对应的表达式。{s=x+yt=x−y\left\{\begin{aligned}s & = & x...
已知：
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u=f(x,y)
u=f(x,y)有二阶连续偏导数，计算
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\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}
∂x2∂2u−∂y2∂2u在新的坐标系下对应的表达式。
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\left\{ \begin{aligned} s & = & x+y \\ t & = & x-y \\ \end{aligned} \right.
{st==x+yx−y
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：
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\begin{aligned} \frac{\partial s}{\partial x}=\frac{\partial t}{\partial x} &=1 , \end{aligned}
∂x∂s=∂x∂t=1,
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\begin{aligned} \frac{\partial s}{\partial y}&=1 ,\\ \frac{\partial t}{\partial y}&=-1. \end{aligned}
∂y∂s∂y∂t=1,=−1.
∂ 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} ∂x∂u=∂s∂u∂x∂s+∂t∂u∂x∂t=∂s∂u+∂t∂u.
∂ 2 u ∂ x 2 = ∂ ∂ 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 ∂ s 2 + 2 ∂ 2 u ∂ s ∂ t + ∂ u ∂ t 2 . \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} ∂x2∂2u=∂x∂(∂x∂u)=∂x∂(∂s∂u+∂t∂u)=∂x∂∂s∂u+∂x∂∂t∂u=∂s∂∂s∂u∂x∂s+∂t∂∂s∂u∂x∂t+∂s∂∂t∂u∂x∂s+∂t∂∂t∂u∂x∂t=∂s2∂u+2∂s∂t∂2u+∂t2∂u.
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} ∂y∂u=∂s∂u∂y∂s+∂t∂u∂y∂t=∂s∂u−∂t∂u.
∂ 2 u ∂ y 2 = ∂ ∂ 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 ∂ s 2 − 2 ∂ 2 u ∂ s ∂ t + ∂ u ∂ t 2 . \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} ∂y2∂2u=∂y∂(∂y∂u)=∂y∂(∂s∂u−∂t∂u)=∂y∂∂s∂u−∂y∂∂t∂u=∂s∂∂s∂u∂y∂s+∂t∂∂s∂u∂y∂t−∂s∂∂t∂u∂y∂s−∂t∂∂t∂u∂y∂t=∂s2∂u−2∂s∂t∂2u+∂t2∂u.
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\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}
∂y2∂2u=∂s2∂u−2∂s∂t∂2u+∂t2∂u.
则在变换坐标系下有：
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\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.
⎩⎪⎪⎨⎪⎪⎧∂x2∂2u∂y2∂2u=∂s2∂u+2∂s∂t∂2u+∂t2∂u,=∂s2∂u−2∂s∂t∂2u+∂t2∂u.
综合上述计算结果可得：
∂ 2 u ∂ x 2 − ∂ 2 u ∂ y 2 = 4 ∂ 2 u ∂ 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} ∂x2∂2u−∂y2∂2u=4∂s∂t∂2u.
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