# Harmonic Functions and Fundamental Solution of Laplace Equation

Vanabel/ 2月 10, 2012/ Elliptic Equations/ 0 comments

$\Delta u\eqdef\sum_{i=1}^n\D_{ii} u=\sum_{i=1}^n\frac{\pt^2 u}{(\pt x_i)^2}.$
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

$\bar u(\bar x)=u(x),\quad x=A\bar x+b$
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:
$\Gamma(x-y)=\begin{cases} \frac{1}{2\pi}\ln |x-y|,&n=2\\ \frac{1}{n(2-n)\omega_n}|x-y|^{2-n},&n\geq3. \end{cases}$

• The fundamental solution $\Gamma(x-y)$ of Laplace equation is defined on $\R^n\setminus\set{x}$.
• The normalization makes us to have
$\int_{\pt B_r}\frac{\pt\Gamma}{\pt\n}\rd S=\int_{\pt B_r} \langle u_r\frac{x-y}{r},\frac{x-y}{r}\rangle\rd S=1.$

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

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