证明 设$\mathit{a}\in {\mathbb{R}}^3$为任意常向量，由Gauss定理可得$$\begin{array}{}\text{(2)}& \begin{array}{rl}\mathit{a}\cdot {\oint }_{\partial \mathrm{\Omega }}\mathit{n}\phi \mathrm{d}S& ={\oint }_{\partial \mathrm{\Omega }}\mathit{n}\cdot (\phi \mathit{a})\mathrm{d}S={\int }_{\mathrm{\Omega }}\mathrm{\nabla }\cdot (\phi \mathit{a})\mathrm{d}V=\mathit{a}\cdot {\int }_{\mathrm{\Omega }}\mathrm{\nabla }\phi \mathrm{d}V\\ \mathit{a}\cdot {\oint }_{\partial \mathrm{\Omega }}\mathit{n}\times \mathit{f}\mathrm{d}S& ={\oint }_{\partial \mathrm{\Omega }}\mathit{a}\cdot (\mathit{n}\times \mathit{f})\mathrm{d}S={\oint }_{\partial \mathrm{\Omega }}\mathit{n}\cdot (\mathit{f}\times \mathit{a})\mathrm{d}S\\ & ={\int }_{\mathrm{\Omega }}\mathrm{\nabla }\cdot (\mathit{f}\times \mathit{a})\mathrm{d}V=\mathit{a}\cdot {\int }_{\mathrm{\Omega }}\mathrm{\nabla }\times \mathit{f}\mathrm{d}V\end{array}\end{array}$$ 由于$\mathit{a}$是任意常向量，故有$$\begin{array}{}\text{(3)}& {\oint }_{\partial \mathrm{\Omega }}\mathit{n}\phi \mathrm{d}S={\int }_{\mathrm{\Omega }}\mathrm{\nabla }\phi \mathrm{d}V,{\oint }_{\partial \mathrm{\Omega }}\mathit{n}\times \mathit{f}\mathrm{d}S={\int }_{\mathrm{\Omega }}\mathrm{\nabla }\times \mathit{f}\mathrm{d}V\end{array}$$ $◻$

证明 注意到被积函数可表示为$$\begin{array}{}\text{(17)}& \frac{\mathit{P}({\mathit{r}}^{\prime })\cdot (\mathit{r}-{\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }{\parallel }^3}=\mathit{P}({\mathit{r}}^{\prime })\cdot {\mathrm{\nabla }}^{\prime }\frac{1}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }={\mathrm{\nabla }}^{\prime }\cdot \frac{\mathit{P}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }-\frac{{\mathrm{\nabla }}^{\prime }\cdot \mathit{P}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\end{array}$$ 由Gauss定理可得$$\begin{array}{}\text{(18)}& \begin{array}{rl}\phi (\mathit{r})& =\frac{1}{4\pi {\epsilon }_0}{\int }_{\mathrm{\Omega }}{\mathrm{\nabla }}^{\prime }\cdot \frac{\mathit{P}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }-\frac{1}{4\pi {\epsilon }_0}{\int }_{\mathrm{\Omega }}\frac{{\mathrm{\nabla }}^{\prime }\cdot \mathit{P}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }\\ & =\frac{1}{4\pi {\epsilon }_0}{\oint }_{\partial \mathrm{\Omega }}\mathit{n}\cdot \frac{\mathit{P}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{S}^{\prime }+\frac{1}{4\pi {\epsilon }_0}{\int }_{\mathrm{\Omega }}\frac{-{\mathrm{\nabla }}^{\prime }\cdot \mathit{P}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }\\ & =\frac{1}{4\pi {\epsilon }_0}{\oint }_{\partial \mathrm{\Omega }}\frac{{\sigma }_{\mathrm{p}}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{S}^{\prime }+\frac{1}{4\pi {\epsilon }_0}{\int }_{\mathrm{\Omega }}\frac{{\rho }_{\mathrm{p}}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }\end{array}\end{array}$$ $◻$

证明 注意到被积函数可表示为$$\begin{array}{}\text{(23)}& \frac{\mathit{M}({\mathit{r}}^{\prime })\times (\mathit{r}-{\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }{\parallel }^3}=\mathit{M}({\mathit{r}}^{\prime })\times {\mathrm{\nabla }}^{\prime }\frac{1}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }=-{\mathrm{\nabla }}^{\prime }\times \frac{\mathit{M}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }+\frac{{\mathrm{\nabla }}^{\prime }\times \mathit{M}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\end{array}$$ 由推广的Gauss定理可得$$\begin{array}{}\text{(24)}& \begin{array}{rl}\mathit{A}(\mathit{r})& =-{\int }_{\mathrm{\Omega }}{\mathrm{\nabla }}^{\prime }\times \frac{\mathit{M}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }+{\int }_{\mathrm{\Omega }}\frac{{\mathrm{\nabla }}^{\prime }\times \mathit{M}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }\\ & ={\oint }_{\partial \mathrm{\Omega }}\frac{\mathit{M}({\mathit{r}}^{\prime })\times \mathit{n}}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{S}^{\prime }+{\int }_{\mathrm{\Omega }}\frac{{\mathrm{\nabla }}^{\prime }\times \mathit{M}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }\\ & ={\oint }_{\partial \mathrm{\Omega }}\frac{{\mathit{i}}_{\mathrm{m}}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{S}^{\prime }+{\int }_{\mathrm{\Omega }}\frac{{\mathit{j}}_{\mathrm{m}}({\mathit{r}}^{\prime })}{\parallel \mathit{r}-{\mathit{r}}^{\prime }\parallel }\mathrm{d}{V}^{\prime }\end{array}\end{array}$$ $◻$

证明 注意到$$\begin{array}{}\text{(26)}& \mathrm{\nabla }\cdot \frac{\mathit{r}}{\parallel \mathit{r}{\parallel }^3}=0,\mathrm{\forall }\mathit{r}\ne \mathbf{0}\end{array}$$

(1) 若$P_0\notin \mathrm{\Omega }$，则$\mathit{r}\ne \mathbf{0}$，由Gauss定理可得$$\begin{array}{}\text{(27)}& {\oint }_{\partial \mathrm{\Omega }}\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}S={\int }_{\mathrm{\Omega }}\mathrm{\nabla }\cdot \frac{\mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}V=0\end{array}$$

(2) 若$P_0\in \mathrm{\Omega }$，则$P_0$为$\mathrm{\Omega }$的内点。选择合适的$\epsilon >0$，令$\tilde{\mathrm{\Omega }}=\mathrm{\Omega }\setminus B(P_0,\epsilon )$，由Gauss定理可得$$\begin{array}{}\text{(28)}& {\oint }_{\partial \mathrm{\Omega }}\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}S-{\oint }_{\partial B(P_0,\epsilon )}\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}S={\oint }_{\partial \tilde{\mathrm{\Omega }}}\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}S={\int }_{\tilde{\mathrm{\Omega }}}\mathrm{\nabla }\cdot \frac{\mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}V=0\end{array}$$ 因此$$\begin{array}{}\text{(29)}& {\oint }_{\partial \mathrm{\Omega }}\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}S={\oint }_{\partial B(P_0,\epsilon )}\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}\mathrm{d}S={\oint }_{\partial B(P_0,\epsilon )}\frac{1}{{\epsilon }^2}\mathrm{d}S=4\pi \end{array}$$

综上所述，得证。 $◻$

证明 (1) 由Gauss定理可得$$\begin{array}{}\text{(39)}& {\oint }_{\partial \mathrm{\Omega }}v\frac{\partial u}{\partial \mathit{n}}\mathrm{d}S={\int }_{\mathrm{\Omega }}\mathrm{\nabla }\cdot (v\mathrm{\nabla }u)\mathrm{d}V={\int }_{\mathrm{\Omega }}v\mathrm{\Delta }u\mathrm{d}V+{\int }_{\mathrm{\Omega }}\mathrm{\nabla }v\cdot \mathrm{\nabla }u\mathrm{d}V\end{array}$$

(2) 由(1)即得证。

(3) $P_0$为$\mathrm{\Omega }$的内点，令$v=\frac{1}{\parallel \mathit{r}\parallel }$，则$$\begin{array}{}\text{(40)}& \frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}=-\mathrm{\nabla }v\cdot \mathit{n}=-\frac{\partial v}{\partial \mathit{n}},\mathrm{\Delta }v=0,\mathrm{\forall }\mathit{r}\ne \mathbf{0}\end{array}$$ 选择合适的$\epsilon >0$，令$\tilde{\mathrm{\Omega }}=\mathrm{\Omega }\setminus B(P_0,\epsilon )$，由Gauss定理和积分中值定理可得$$\begin{array}{}\text{(41)}& \begin{array}{rl}\mathrm{RHS}& =\frac{1}{4\pi }({\oint }_{\partial \mathrm{\Omega }}-{\oint }_{\partial B(P_0,\epsilon )}+{\oint }_{\partial B(P_0,\epsilon )})(v\frac{\partial u}{\partial \mathit{n}}-u\frac{\partial v}{\partial \mathit{n}})\mathrm{d}S\\ & =\frac{1}{4\pi }{\int }_{\tilde{\mathrm{\Omega }}}(v\mathrm{\Delta }u-u\mathrm{\Delta }v)\mathrm{d}V+\frac{1}{4\pi }{\oint }_{\partial B(P_0,\epsilon )}(u\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}+\frac{1}{\parallel \mathit{r}\parallel }\frac{\partial u}{\partial \mathit{n}})\mathrm{d}S\\ & =\frac{1}{4\pi }{\oint }_{\partial B(P_0,\epsilon )}\frac{u(P)}{{\epsilon }^2}\mathrm{d}S+\frac{1}{4\pi \epsilon }{\int }_{B(P_0,\epsilon )}\mathrm{\Delta }u\mathrm{d}V=u(P),P\in B(P_0,\epsilon )\end{array}\end{array}$$ 令$\epsilon \to 0^{+}$可得$$\begin{array}{}\text{(42)}& u(P_0)=\frac{1}{4\pi }{\oint }_{\partial \mathrm{\Omega }}(u\frac{\mathit{n}\cdot \mathit{r}}{\parallel \mathit{r}{\parallel }^3}+\frac{1}{\parallel \mathit{r}\parallel }\frac{\partial u}{\partial \mathit{n}})\mathrm{d}S\end{array}$$

(4) 令$\mathrm{\Omega }=B(P_0,R)$，由(3)可得$$\begin{array}{}\text{(43)}& u(P_0)=\frac{1}{4\pi R^2}{\oint }_{\partial B(P_0,R)}u(P)\mathrm{d}S+\frac{1}{4\pi R}{\int }_{B(P_0,R)}\mathrm{\Delta }u\mathrm{d}V=\frac{1}{4\pi R^2}{\oint }_{\partial B(P_0,R)}u(P)\mathrm{d}S\end{array}$$ $◻$

证明 (1)(2)(3)(4) 与上例相似，利用散度形式的Green公式即可证明，故略去。

(5) 设$f$在$z_0=x_0+\mathrm{i}{y}_0$处可导，则²$$\begin{array}{}\text{(50)}& \mathrm{\Delta }f=\mathrm{\Delta }u+\mathrm{i}\mathrm{\Delta }v=A\mathrm{\Delta }z+\rho (\mathrm{\Delta }z)\mathrm{\Delta }z,\rho (\mathrm{\Delta }z)=o(1)\end{array}$$ 记$A=\alpha +\mathrm{i}\beta$、$\rho (\mathrm{\Delta }z)={\rho }_1(\mathrm{\Delta }z)+\mathrm{i}{\rho }_2(\mathrm{\Delta }z)$，则$$\begin{array}{}\text{(51)}& \mathrm{\Delta }u+\mathrm{i}\mathrm{\Delta }v=(\alpha \mathrm{\Delta }x-\beta \mathrm{\Delta }y+{\rho }_1\mathrm{\Delta }x-{\rho }_2\mathrm{\Delta }y)+\mathrm{i}(\beta \mathrm{\Delta }x+\alpha \mathrm{\Delta }y+{\rho }_2\mathrm{\Delta }x+{\rho }_1\mathrm{\Delta }y)\end{array}$$ 故有$$\begin{array}{}\text{(52)}& \begin{array}{rl}\mathrm{\Delta }u& =\alpha \mathrm{\Delta }x-\beta \mathrm{\Delta }y+o\left(\sqrt{\mathrm{\Delta }{x}^2+\mathrm{\Delta }{y}^2}\right)\\ \mathrm{\Delta }v& =\beta \mathrm{\Delta }x+\alpha \mathrm{\Delta }y+o\left(\sqrt{\mathrm{\Delta }{x}^2+\mathrm{\Delta }{y}^2}\right)\end{array}\end{array}$$ 故$u(x,y),v(x,y)$在$(x_0,y_0)$处均可微，且成立Cauchy-Riemann方程$$\begin{array}{}\text{(53)}& \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=\alpha ,\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}=\beta \end{array}$$ 因此$$\begin{array}{}\text{(54)}& \frac{{\partial }^{2}u}{\partial x^2}=\frac{\partial }{\partial x}\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}\frac{\partial v}{\partial x}=-\frac{{\partial }^{2}u}{\partial y^2}\Rightarrow u_{xx}+u_{yy}=0\end{array}$$ 故$u$为$D$上的调和函数。同理，$v$也是$D$上的调和函数。

(6) 假定³$u,v\in {\mathcal{C}}^1$，由Green公式可得$$\begin{array}{}\text{(55)}& \begin{array}{rl}{\oint }_{\partial \mathrm{\Omega }}f(z)\mathrm{d}z& ={\oint }_{\partial \mathrm{\Omega }}(u\mathrm{d}x-v\mathrm{d}y)+\mathrm{i}{\int }_{\partial \mathrm{\Omega }}(v\mathrm{d}x+u\mathrm{d}y)\\ & ={\int }_{\mathrm{\Omega }}(-\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})\mathrm{d}x\mathrm{d}y+\mathrm{i}{\int }_{\mathrm{\Omega }}(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y})\mathrm{d}x\mathrm{d}y=0\end{array}\end{array}$$ 选择合适的$\epsilon >0$，令$\tilde{\mathrm{\Omega }}=\mathrm{\Omega }\setminus B(P_0,\epsilon )$，由Green公式可得$$\begin{array}{}\text{(56)}& \mathrm{RHS}=\frac{1}{2\pi \mathrm{i}}({\oint }_{\partial \mathrm{\Omega }}-{\oint }_{\partial B(z_0,\epsilon )}+{\oint }_{\partial B(z_0,\epsilon )})\frac{f(z)}{z-z_0}\mathrm{d}z=\frac{1}{2\pi \mathrm{i}}({\oint }_{\partial \tilde{\mathrm{\Omega }}}+{\oint }_{\partial B(z_0,\epsilon )})\frac{f(z)}{z-z_0}\mathrm{d}z\end{array}$$ 由于$g:z\mapsto \frac{f(z)}{z-z_0}$在$\tilde{\mathrm{\Omega }}$上解析，由Cauchy-Goursat定理可得$$\begin{array}{}\text{(57)}& {\oint }_{\partial \tilde{\mathrm{\Omega }}}\frac{f(z)}{z-z_0}\mathrm{d}z=0\end{array}$$ 由积分中值定理可得$$\begin{array}{}\text{(58)}& \begin{array}{rl}\mathrm{RHS}& =\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial B(z_0,\epsilon )}\frac{f(z)}{z-z_0}\mathrm{d}z\stackrel{z=z_0+\epsilon {\mathrm{e}}^{\mathrm{i}\theta }}{=}\frac{1}{2\pi \mathrm{i}}{\int }_{-\pi }^{\pi }\frac{f(z_0+\epsilon {\mathrm{e}}^{\mathrm{i}\theta })}{\epsilon {\mathrm{e}}^{\mathrm{i}\theta }}\epsilon \mathrm{i}{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{d}\theta \\ & =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(z_0+\epsilon {\mathrm{e}}^{\mathrm{i}\theta })\mathrm{d}\theta =f(z_0+\epsilon {\mathrm{e}}^{\mathrm{i}\tilde{\theta }})\end{array}\end{array}$$ 令$\epsilon \to 0^{+}$可得$$\begin{array}{}\text{(59)}& f(z_0)=\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial \mathrm{\Omega }}\frac{f(z)}{z-z_0}\mathrm{d}z\end{array}$$ $◻$