\\begin{defn}
\n Suppose $\\Omega\\subset\\R^n$ is a domain, and $u\\in C^2$ is a function on $\\Omega$. Define \\iemph{Laplace operator} $\\Delta$ as
\n \\[
\n \\Delta u\\eqdef\\sum_{i=1}^n\\D_{ii} u=\\sum_{i=1}^n\\frac{\\pt^2 u}{(\\pt x_i)^2}.
\n \\]
\n Then the equation
\n \\[
\n \\Delta u=0,
\n \\]
\n is called the \\iemph{Laplace equation}, the solution of Laplace equation is called \\iemph{harmonic functions}.
\n\\end{defn}
\n
\nThe 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
\n\\[
\n\\bar u(\\bar x)=u(x), \\quad x=A\\bar x,
\n\\]
\nthen $\\Delta\\bar u=0$. In fact,
\n\\[\\begin{split}
\n\\frac{\\pt\\bar u}{\\pt\\bar x_i}&
\n =\\frac{\\pt u}{\\pt x_j}\\frac{\\pt x_j}{\\pt\\bar x_i}
\n =\\frac{\\pt u}{\\pt x_j}\\frac{\\pt (a_{jk}\\bar x_k)}{\\pt\\bar x_i}
\n =\\frac{\\pt u}{\\pt x_j}a_{ji},\\\\
\n\\frac{\\pt^2\\bar u}{\\pt (\\bar x_i)^2}&
\n =\\frac{\\pt \\left(\\frac{\\pt u}{\\pt x_j}a_{ji}\\right)}{\\pt\\bar x_i}
\n =\\frac{\\pt^2 u}{\\pt x_j\\pt x_k}\\frac{\\pt x_k}{\\pt\\bar x_i}a_{ji}\\\\
\n &=\\frac{\\pt^2 u}{\\pt x_j\\pt x_k}a_{ki}a_{ji}
\n =\\frac{\\pt^2 u}{\\pt x_j\\pt x_k}\\delta_{jk}\\\\
\n &=\\frac{\\pt^2 u}{(\\pt x_i)^2}.
\n\\end{split}\\]
\nThus,
\n\\[
\n\\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.
\n\\]
\nAnother 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
\n\\begin{prop}
\n 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
\n \\[
\n \\bar u(\\bar x)=u(x),\\quad x=A\\bar x+b
\n \\]
\n satisfy $\\Delta\\bar u=0$.
\n\\end{prop}
\nBy 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
\n\\[\\begin{split}
\n 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)
\n =\\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}\\\\
\n &=\\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\\}\\\\
\n &=u_{rr}+\\frac{n-1}{r}u_r.
\n\\end{split}\\]
\nThus, $u_r=cr^{1-n}$ and if $r>0$, then
\n\\[
\nu=\\begin{cases}
\n c\\ln r,&n=2\\\\
\n \\frac{c}{2-n}r^{2-n},&n\\geq3.
\n\\end{cases}
\n\\]
\nNow, fix a point $y\\in\\R^n$ and \\iemph{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 \\iemph{fundamental solution} $\\Gamma(x-y)$ or $\\Gamma(|x-y|)$ of Laplace equation:
\n\\[
\n\\Gamma(x-y)=\\begin{cases}
\n \\frac{1}{2\\pi}\\ln |x-y|,&n=2\\\\
\n \\frac{1}{n(2-n)\\omega_n}|x-y|^{2-n},&n\\geq3.
\n\\end{cases}
\n\\]
\n\\begin{rem}
\n\\begin{itemize}
\n\\item The fundamental solution $\\Gamma(x-y)$ of Laplace equation is defined on $\\R^n\\setminus\\set{x}$.
\n\\item The normalization makes us to have
\n\\[
\n\\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.
\n\\]
\n\\end{itemize}
\n\\end{rem}
\n\\begin{prob}
\nGive a detailed proof that if we do have a solution for Laplace equation, then from \\ref{prop:2}, we have spherical symmetrical solution and vice versa.
\n\\end{prob}
\n\\begin{excs}
\nTry to prove
\n\\[
\nD_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},
\n\\]
\nand
\n\\[
\n D_{ij}\\Gamma\\eqdef\\frac{\\pt^2\\Gamma}{\\pt x_i\\pt x_j}
\n =\\frac{1}{n\\omega_n}\\left(
\n \\frac{\\delta_{ij}}{|x-y|^n}-\\frac{n (x_i-y_i)(x_j-y_j)}{|x-y|^{n+2}}
\n \\right).
\n\\]
\n\\end{excs}
\n\\begin{excs}
\nProve the following estimate
\n\\[\\begin{split}
\n|D_i\\Gamma(x-y)|&\\leq\\frac{1}{n\\omega_n}|x-y|^{1-n};\\\\
\n|D_{ij}\\Gamma(x-y)|&\\leq\\frac{1}{\\omega_n}|x-y|^{-n};\\\\
\n|D^\\beta\\Gamma(x-y)|&\\leq C|x-y|^{2-n-|\\beta|}, \\quad C=C(n,|\\beta|).
\n\\end{split}\\]
\n\\end{excs} <\/p>\n","protected":false},"excerpt":{"rendered":"
\\begin{defn} Suppose $\\Omega\\subset\\R^n$ …<\/p>\n