Harmonic Functions and Fundamental Solution of Laplace Equation

定义 1. Suppose $\Omega\subset\R^n$ is a domain, and $u\in C^2$ is a function on $\Omega$. Define Laplace operator $\Delta$ as
\[
\Delta u\eqdef\sum_{i=1}^n\D_{ii} u=\sum_{i=1}^n\frac{\pt^2 u}{(\pt x_i)^2}.
\]
Then the equation
\[
\Delta u=0,
\]
is called the Laplace equation, the solution of Laplace equation is called harmonic functions.


The most important property of Laplace equation is that its solution has spherical symmetry, i.e., if $u=u(x)$, $x\in \Omega$ satisfy $\Delta u=0$, and for any rotation $A$, define a new function $\bar u(\bar x)$, $\bar x\in\bar \Omega=A^{-1}(\Omega)\subset\R^n$ by
\[
\bar u(\bar x)=u(x), \quad x=A\bar x,
\]
then $\Delta\bar u=0$. In fact,
\[\begin{split}
\frac{\pt\bar u}{\pt\bar x_i}&
=\frac{\pt u}{\pt x_j}\frac{\pt x_j}{\pt\bar x_i}
=\frac{\pt u}{\pt x_j}\frac{\pt (a_{jk}\bar x_k)}{\pt\bar x_i}
=\frac{\pt u}{\pt x_j}a_{ji},\\
\frac{\pt^2\bar u}{\pt (\bar x_i)^2}&
=\frac{\pt \left(\frac{\pt u}{\pt x_j}a_{ji}\right)}{\pt\bar x_i}
=\frac{\pt^2 u}{\pt x_j\pt x_k}\frac{\pt x_k}{\pt\bar x_i}a_{ji}\\
&=\frac{\pt^2 u}{\pt x_j\pt x_k}a_{ki}a_{ji}
=\frac{\pt^2 u}{\pt x_j\pt x_k}\delta_{jk}\\
&=\frac{\pt^2 u}{(\pt x_i)^2}.
\end{split}\]
Thus,
\[
\Delta\bar u=\sum_{i=1}^n\frac{\pt^2\bar u}{\pt (\bar x_i)^2}=\frac{\pt^2 u}{(\pt x_i)^2}=0.
\]
Another more apparent thing is that if we define $\bar u(\bar x)=u(x)$, and $x=\bar x+b$, for some translation $b\in\R^n$, then we also have $\Delta\bar u=0$. Consequently, we have proved
命题 2. If $u(x)$, $x\in \Omega\subset\R^n$, is a solution of $\Delta u=0$, then for any rigid transformation $G:x\to Ax+b$, the function on $G^{-1}(\Omega)$ by
\[
\bar u(\bar x)=u(x),\quad x=A\bar x+b
\]
satisfy $\Delta\bar u=0$.

By the above proposition, we want to find spherical symmetrical solution of Laplace equation $\Delta u=0$. Fix $x\in\Omega$ and let $r=r(y)=|x-y|=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}$, then
\[\begin{split}
0&=\Delta u(r)=\sum_{i=1}^n\frac{\pt}{\pt y_i}\left(\frac{\pt u}{\pt r}\frac{\pt r}{\pt y_i}\right)
=\sum_{i=1}^n\frac{\pt^2 u}{\pt r^2}\left(\frac{\pt r}{\pt y_i}\right)^2+\sum_{i=1}^n\frac{\pt u}{\pt r}\frac{\pt^2 r}{(\pt y_i)^2}\\
&=\sum_{i=1}^n \left\{u_{rr} \left(-\frac{x_i-y_i}{r}\right)^2+u_r\left(\frac{1}{r}-\frac{(x_i-y_i)^2}{r^3}\right)\right\}\\
&=u_{rr}+\frac{n-1}{r}u_r.
\end{split}\]
Thus, $u_r=cr^{1-n}$ and if $r>0$, then
\[
u=\begin{cases}
c\ln r,&n=2\\
\frac{c}{2-n}r^{2-n},&n\geq3.
\end{cases}
\]
Now, fix a point $y\in\R^n$ and normalize the solution by taking $c=\frac{1}{n\omega_n}$, where $\omega_n=\frac{2\pi^{n/2}}{n\Gamma(n/2)}$ is the volume of unit sphere in $\R^n$, we obtain the fundamental solution $\Gamma(x-y)$ or $\Gamma(|x-y|)$ of Laplace equation:
\[
\Gamma(x-y)=\begin{cases}
\frac{1}{2\pi}\ln |x-y|,&n=2\\
\frac{1}{n(2-n)\omega_n}|x-y|^{2-n},&n\geq3.
\end{cases}
\]
注记 1.
  • The fundamental solution $\Gamma(x-y)$ of Laplace equation is defined on $\R^n\setminus\set{x}$.
  • The normalization makes us to have
    \[
    \int_{\pt B_r}\frac{\pt\Gamma}{\pt\n}\rd S=\int_{\pt B_r} \langle u_r\frac{x-y}{r},\frac{x-y}{r}\rangle\rd S=1.
    \]

问题 1. Give a detailed proof that if we do have a solution for Laplace equation, then from prop:2, we have spherical symmetrical solution and vice versa.

练习 1. Try to prove
\[
D_i\Gamma(x-y)\eqdef\frac{\pt\Gamma(x-y)}{\pt x_i}=\frac{1}{n\omega_n}\frac{x_i-y_i}{|x-y|^n},
\]
and
\[
D_{ij}\Gamma\eqdef\frac{\pt^2\Gamma}{\pt x_i\pt x_j}
=\frac{1}{n\omega_n}\left(
\frac{\delta_{ij}}{|x-y|^n}-\frac{n (x_i-y_i)(x_j-y_j)}{|x-y|^{n+2}}
\right).
\]

练习 2. Prove the following estimate
\[\begin{split}
|D_i\Gamma(x-y)|&\leq\frac{1}{n\omega_n}|x-y|^{1-n};\\
|D_{ij}\Gamma(x-y)|&\leq\frac{1}{\omega_n}|x-y|^{-n};\\
|D^\beta\Gamma(x-y)|&\leq C|x-y|^{2-n-|\beta|}, \quad C=C(n,|\beta|).
\end{split}\]

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