Machine-translated from Chinese.
看论文的时候用到了一个结论, 想了一会发现确实是对的.
震撼于这么简单而有用的命题为什么之前没学过, 在此记录一下.
命题 1. 调和函数的任意$k$次Taylor展开为调和多项式.
证: 取调和函数$u,$ 记其$k$次Taylor展开式为$h_k.$ $k=0,1$时平凡.
$k=2$时, $x_ix_j$项不对调和项产生影响. 对于$x_i^2$项,
求$\Delta$后为常数, 因此只需说明$\Delta h_k$后在一点处取零即可.
我们在原点考察, 函数本身在原点有$\Delta u=0,$ 同时余项$u-h_k$为高阶项,
求二阶导后在原点也取零, 这就迫使$\Delta h_k$在原点也取零, 进而处处为零.
注意对余项的考察用到了调和函数的解析性.
对于$k\ge 3,$ 归纳地, 假设$k-1$次展开式都是调和多项式.
$\partial_iu$的$k-1$次Taylor展开式为$\partial_ih_k,$
由于调和函数的导数仍是调和函数,
$\partial_i\Delta h_k=\Delta \partial_ih_k=0.$
这样多项式$\Delta h_k$各阶导数为零, $\Delta h_k$为常数.
而$h_k-h_{k-1}$为$k$次齐次多项式,
$\Delta (h_k-h_{k-1})=\Delta h_k$为$k-2$次齐次多项式, 又是常数,
只能取零.
文章最后更新于 2022-03-30 16:45:54
When I was reading the paper, I came up with a conclusion, and after thinking about it for a while, I found that it was indeed correct.
I am shocked that I have never learned such a simple and useful proposition before, so I will record it here.
Proposition 1. Any Taylor expansion of degree $k$ of a harmonic function is a harmonic polynomial.
Certificate: Take the harmonic function $u,$ and record its $k$ degree Taylor expansion as $h_k.$ $k=0,1$, which is trivial.
When $k=2$ is used, the $x_ix_j$ item does not affect the harmonic term. For the $x_i^2$ item,
After finding $\Delta$, it is a constant, so it only needs to be stated that after $\Delta h_k$, it is zero at one point.
We examine it at the origin. The function itself has $\Delta u=0,$ at the origin and the remainder $u-h_k$ is a higher-order term.
After finding the second derivative, it also takes zero at the origin, which forces $\Delta h_k$ to also take zero at the origin, and then it becomes zero everywhere.
Note that the analysis of the remainder uses the analytic properties of the harmonic function.
For $k\ge 3,$, inductively, it is assumed that the expansions of $k-1$ times are all harmonic polynomials.
The $k-1$ degree Taylor expansion of $\partial_iu$ is $\partial_ih_k,$
Since the derivative of the harmonic function is still a harmonic function,
$\partial_i\Delta h_k=\Delta \partial_ih_k=0.$
In this way, the derivatives of each order of polynomial $\Delta h_k$ are zero, and $\Delta h_k$ is a constant.
And $h_k-h_{k-1}$ is a homogeneous polynomial of degree $k$,
$\Delta (h_k-h_{k-1})=\Delta h_k$ is a homogeneous polynomial of degree $k-2$, which is also a constant.
It can only take zero.
The article was last updated on 2022-03-30 16:45:54
