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

abababa

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。

战巡

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
\[\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=(\frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2})((\frac{\partial u}{\partial y})^2+(\frac{\partial v}{\partial y})^2)=0\]

abababa

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。

青青子衿

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