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f对u,v满足拉普拉斯方程,求教何时对x,y也满足此方程

已知$f(u,v)$满足方程$\frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2}=0$,现在$u,v$又都是$x,y$的函数,求教当什么情况下能满足$\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=0$?
我最近做了几个题,发现都有这种规律,查了一下书,感觉是$u(x,y),v(x,y)$满足柯西黎曼方程时,$f$对$x,y$也满足拉普拉斯方程。但我证明不了。
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回复 1# abababa


\[\frac{\partial^2f}{\partial x^2}=\frac{\partial^2f}{\partial v^2}·(\frac{\partial v}{\partial x})^2+2\frac{\partial u}{\partial x}·\frac{\partial v}{\partial x}·\frac{\partial^2f}{\partial u\partial v}+(\frac{\partial u}{\partial x})^2\frac{\partial^2f}{\partial u^2}+\frac{\partial f}{\partial u}·\frac{\partial^2 u}{\partial x^2}+\frac{\partial f}{\partial v}·\frac{\partial^2 v}{\partial x^2}\]
对$y$同理,那么有
\[\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\frac{\partial^2f}{\partial v^2}((\frac{\partial v}{\partial x})^2+(\frac{\partial v}{\partial y})^2)+\frac{\partial^2f}{\partial u^2}((\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2)+2\frac{\partial^2f}{\partial u\partial v}(\frac{\partial u}{\partial x}·\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}·\frac{\partial v}{\partial y})+\frac{\partial f}{\partial u}(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2})+\frac{\partial f}{\partial v}(\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2})=0\]
所以当柯西-黎曼方程满足时
\[(\frac{\partial v}{\partial x})^2+(\frac{\partial v}{\partial y})^2=(\frac{\partial u}{\partial y})^2+(\frac{\partial v}{\partial y})^2=(\frac{\partial u}{\partial y})^2+(\frac{\partial u}{\partial x})^2\]
\[\frac{\partial u}{\partial x}·\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}·\frac{\partial v}{\partial y}=-\frac{\partial u}{\partial x}·\frac{\partial u}{\partial y}+\frac{\partial u}{\partial y}·\frac{\partial u}{\partial x}=0\]
\[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial \frac{\partial v}{\partial y}}{\partial x}-\frac{\partial \frac{\partial v}{\partial x}}{\partial y}=\frac{\partial^2v}{\partial x\partial y}-\frac{\partial^2v}{\partial x\partial y}=0\]
同理
\[\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}=0\]
因此原式变成
\[\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=(\frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2})((\frac{\partial u}{\partial y})^2+(\frac{\partial v}{\partial y})^2)=0\]

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本帖最后由 abababa 于 2016-12-25 11:11 编辑

回复 2# 战巡
谢谢,这个方法后来我也想到了,就是硬算偏导数,最后拿出来等于0的化简。但不知有没有利用一些定理,能让过程变得简单一点的,不做这么多的计算。比如解析函数的实、虚部都是调和函数,调和函数也必能和它的共轭组成解析函数,调和函数满足拉普拉斯方程,等等,像这类的构造。
另外发现计算过程中需要用到偏导数交换顺序
\[(\frac{\partial^2f}{\partial v\partial u}+\frac{\partial^2f}{\partial u\partial v})(\frac{\partial u}{\partial x}\cdot\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\cdot\frac{\partial v}{\partial y})=0\]
上面这部分不需要能交换顺序,但下面这部分需要偏导数能交换顺序
\[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0\]
看来我在主楼感觉到的这个规律只在混合偏导数连续的情况下才有,连续就能成为充要条件了。可能因为我做的几个题都是连续的,没发现特殊情况。

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本帖最后由 青青子衿 于 2020-10-22 21:58 编辑

回复 3# abababa
《复变函数论习题集》
(苏)沃尔科维斯基(L.Volkovysky)等著;宋国栋 等译
463215128.png
2020-10-22 21:58

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