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题：设$\,\Gamma=\partial G\,$为光滑闭曲线,在区域$\,G\,$上$\,\Delta u=0.$,VZBnd
$\qquad$则$\,u{\small(a,b)}=\frac{1}{2\pi}{\small\displaystyle\int}_{\Gamma}\big(u\frac{\partial\ln r}{\partial n}-\frac{\partial u}{\partial n}\ln r)ds\;\small((a,b)\in G)$3&ZKE
$\qquad$其中$\small\,r=|(x-a,y-b)|,\;(x,y)\in\Gamma,\;\scriptsize\dfrac{\partial}{\partial n}\,$是方向导数.4kwf`+
证：取$\,\rho(\,>0)$使$\,\bar{N}_{\rho}(a,b)\subset G^{\circ},\;\Gamma_{\rho}=\partial N_{\rho}(a,b).$lI
$\qquad$在$\,G-\bar{N}_{\rho}(a,b)\,$上$\,\Delta\ln r=\Delta u=0\,\big(\frac{\partial^2(\ln r)}{\partial x^2}=$s]H9E
$\qquad\frac{(y-b)^2-(x-a)^2}{r^4}\big).\qquad\qquad$现作下列分析：EC
 u
$\qquad$如图,从$\,G\,$的内部挖去$\bar{N}_{\rho}(a,b),\,$所剩部分被分A{
$\qquad$为二单连通区域$\small\,G_1,\,G_2.\,$对它们应用Green第二DG91
$\qquad$型公式立即得$\,\small\displaystyle\int_{\Gamma+\Gamma_{\rho}^{-}}=0:$L5D
$\qquad\small\displaystyle\int_{\Gamma}\big(u\frac{\partial\ln r}{\partial n}-\frac{\partial u}{\partial n}\ln r\big)ds=\int_{\Gamma_{\rho}}\big(u\frac{\partial\ln r}{\partial n}-\frac{\partial u}{\partial n}\ln r\big)ds$1"GE
$\qquad$在$\,\Gamma_{\rho}\,$上$\displaystyle\,{\small\frac{\partial u}{\partial n}}ds=(\mathbf{n}\cdot\nabla u) ds = {\small\frac{\partial u}{\partial x}}dy-{\small\frac{\partial u}{\partial y}}dx$*Q#x
$\qquad{\small\displaystyle\int}_{\Gamma_{\rho}}\big({\small\dfrac{\partial u}{\partial n}}\ln r\big)ds=\ln\rho{\small\displaystyle\iint}_{|(x,y)|\le\rho}(\Delta u)dxdy=0$uar
$\qquad$最后$,\,\frac{1}{\large 2\pi}{\small\displaystyle\int}_{\Gamma_{\rho}}\big(u{\small\dfrac{\partial\ln r}{\partial n}}\big)ds=$2/
$\qquad\frac{1}{\large 2\pi}{\small\displaystyle\int}_0^{2\pi}u(a+\rho\cos\theta,b+\rho\sin\theta)d\theta\overset{\rho\to 0+}{\to }u(a,b)${2
$\quad\;\;\boxed{u{\small(a,b)}={\small\frac{1}{2\pi}}\int_{\Gamma}\big(u{\small\frac{\partial\ln r}{\partial n}}-{\small\frac{\partial u}{\partial n}}\ln r\big)ds\;\small((a,b)\in G)}$j%3,%:
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2018/08/16 06:14am 
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