格林函数
$\quad\\$
格林函数
$\quad\\$
  泊松方程
$$\Delta u=f(\boldsymbol{r})\qquad\boldsymbol{r}\in T$$
  格林公式
$$\iint_\Sigma u\nabla v\cdot\dd\boldsymbol{S}=\iiint_T \nabla\cdot(u\nabla v)\dd V\\=\iiint_T u\Delta v\dd V+\iiint_T \nabla u\cdot\nabla v\dd V\\
\quad\\
\iint_\Sigma v\nabla u\cdot\dd\boldsymbol{S}=\iiint_T \nabla\cdot(v\nabla u)\dd V\\=\iiint_T v\Delta u\dd V+\iiint_T \nabla v\cdot\nabla u\dd V\\
\quad\\
\iint_\Sigma\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\dd S=\iiint_T (u\Delta v-v\Delta u)\dd V$$
  取$v$为单位强度点源产生的势场，即
$$\Delta v(\boldsymbol{r},\boldsymbol{r_0})=\delta(\boldsymbol{r}-\boldsymbol{r_0})$$
  在$T\backslash B(\boldsymbol{r_0},\varepsilon\to 0)$应用格林公式
$$\iint_{\Sigma-S_\varepsilon}\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\dd S=\iiint_{T-B_\varepsilon} (u\Delta v-v\Delta u)\dd V\\
\quad\\
\iint_{\Sigma}\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\dd S-\oint_{S_\varepsilon}\left(u\frac{1}{4\pi\varepsilon^2}+\frac{1}{4\pi\varepsilon}\frac{\partial u}{\partial n}\right)\varepsilon^2\dd\Omega\\=-\iiint_{T-B_\varepsilon}vf\dd V$$
$$u(\boldsymbol{r_0})=\iiint_{T}v(\boldsymbol{r},\boldsymbol{r_0})f(\boldsymbol{r})\dd V-\iint_{\Sigma}\left[v(\boldsymbol{r},\boldsymbol{r_0})\frac{\partial u(\boldsymbol{r})}{\partial n}-u(\boldsymbol{r})\frac{\partial v(\boldsymbol{r},\boldsymbol{r_0})}{\partial n}\right]\dd S\quad$$
$\quad\\$
  边界条件
$$\left[\alpha\frac{\partial u}{\partial n}+\beta u\right]_\Sigma=\varphi(M)$$
  再令$v$满足齐次边界条件
$$\left[\alpha\frac{\partial v}{\partial n}+\beta v\right]_\Sigma=0$$
  称$v$为格林函数，记为$G(\boldsymbol{r},\boldsymbol{r_0})$
  第一类边界条件$u|_\Sigma=\varphi,G|_\Sigma=0$
$$u(\boldsymbol{r_0})=\iiint_{T}G(\boldsymbol{r},\boldsymbol{r_0})f(\boldsymbol{r})\dd V+\iint_{\Sigma}\varphi(\boldsymbol{r})\frac{\partial G(\boldsymbol{r},\boldsymbol{r_0})}{\partial n}\dd S\quad$$
  第三类边界条件$\displaystyle\left[\alpha\frac{\partial u}{\partial n}+\beta u\right]_\Sigma=\varphi,\left[\alpha\frac{\partial G}{\partial n}+\beta G\right]_\Sigma=0$
$$u(\boldsymbol{r_0})=\iiint_{T}G(\boldsymbol{r},\boldsymbol{r_0})f(\boldsymbol{r})\dd V-\frac{1}{\alpha}\iint_{\Sigma}G(\boldsymbol{r},\boldsymbol{r_0})\varphi(\boldsymbol{r})\dd S\quad$$
  第二类边界条件$\displaystyle\frac{\partial u}{\partial n}=\varphi,\frac{\partial G}{\partial n}=0$
  引入推广的格林函数
$$\Delta G=\delta(\boldsymbol{r}-\boldsymbol{r_0})-\frac{1}{V_T}$$
$$u(\boldsymbol{r_0})=\iiint_{T}G(\boldsymbol{r},\boldsymbol{r_0})f(\boldsymbol{r})\dd V-\iint_{\Sigma}G(\boldsymbol{r},\boldsymbol{r_0})\frac{\partial\varphi(\boldsymbol{r})}{\partial n}\dd S\quad$$
  下证格林函数具有对称性，以$u=G(\boldsymbol{r},\boldsymbol{r_1})$，$v=G(\boldsymbol{r},\boldsymbol{r_2})$代入格林公式
$$\iint_{\Sigma-S_{\varepsilon 1}-S_{\varepsilon 2}}\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\dd S=\iiint_{T-B_{\varepsilon 1}-B_{\varepsilon 2}} (u\Delta v-v\Delta u)\dd V\\
\quad\\
\iint_{S_{\varepsilon 1}}\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\dd S+\iint_{S_{\varepsilon 2}}\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\dd S=0\\
\quad\\
v(\boldsymbol{r_1})=u(\boldsymbol{r_2})\Leftrightarrow G(\boldsymbol{r_1},\boldsymbol{r_2})=G(\boldsymbol{r_2},\boldsymbol{r_1})$$
  三类边界条件下解的积分表达式分别成为
$$u(\boldsymbol{r})=\iiint_{T_0}G(\boldsymbol{r},\boldsymbol{r_0})f(\boldsymbol{r_0})\dd V_0+\iint_{\Sigma_0}\varphi(\boldsymbol{r_0})\frac{\partial G(\boldsymbol{r},\boldsymbol{r_0})}{\partial n_0}\dd S_0\\
\quad\\
u(\boldsymbol{r})=\iiint_{T_0}G(\boldsymbol{r},\boldsymbol{r_0})f(\boldsymbol{r_0})\dd V_0-\iint_{\Sigma_0}G(\boldsymbol{r},\boldsymbol{r_0})\frac{\partial\varphi(\boldsymbol{r_0})}{\partial n_0}\dd S_0\\
\quad\\
u(\boldsymbol{r})=\iiint_{T_0}G(\boldsymbol{r},\boldsymbol{r_0})f(\boldsymbol{r_0})\dd V_0-\frac{1}{\alpha}\iint_{\Sigma_0}G(\boldsymbol{r},\boldsymbol{r_0})\varphi(\boldsymbol{r_0})\dd S_0\\$$
$\quad\\$
电像法
$\quad\\$
  称无界区域格林函数解为基本解，记为$G_0$，也就是带$-\varepsilon$电量的点电荷产生的电势，一般解$G=G_0+G_1$，二维与三维基本解分别为
$$G_0(\boldsymbol{r},\boldsymbol{r_0})=-\frac{1}{2\pi}\ln\frac{C}{|\boldsymbol{r}-\boldsymbol{r_0}|}\\
\quad\\
G_0(\boldsymbol{r},\boldsymbol{r_0})=-\frac{1}{4\pi}\frac{1}{|\boldsymbol{r}-\boldsymbol{r_0}|}$$
  得到$G_0$后，$G_1$满足拉普拉斯方程$\Delta G_1=0$，可由分离变数等方法求得级数解，当区域边界比较规则时，$G_1$可以等效为点电荷产生的电势，这些点电荷称为产生$G_0$的$\boldsymbol{r_0}$处点电荷的电像，以下列出第一边界条件$u|_\Sigma=f$下几种边界的电像，格林函数以及拉普拉斯方程$\Delta u=0$的积分形式的解
  $r=R$的圆环电像
$$\boldsymbol{r_像}=\frac{R^2}{r_0^2}\boldsymbol{r_0}\qquad q_像=\varepsilon\\
\quad\\
G=-\frac{1}{2\pi}\ln\frac{1}{|\boldsymbol{r}-\boldsymbol{r_0}|}+\frac{1}{2\pi}\ln\frac{\displaystyle\frac{R}{r_0}}{\left|\boldsymbol{r}-\displaystyle\frac{R^2}{r_0^2}\boldsymbol{r_0}\right|}\\
\quad\\
u(r,\varphi)=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{R^2-r^2}{R^2-2 R r\cos(\varphi-\varphi_0)+r^2}f(Re^{i\varphi_0})\dd\varphi_0=f(re^{i\varphi})$$
  $y=0$的直线电像
$$(x_像,y_像)=(x_0,-y_0)\qquad q_像=\varepsilon\\
\quad\\
G=-\frac{1}{2\pi}\ln\frac{1}{|\boldsymbol{r}-(x_0,y_0)|}+\ln\frac{1}{2\pi}\frac{1}{\left|\boldsymbol{r}-(x_0,-y_0)\right|}\\
\quad\\
u(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(x-x_0)^2+y^2}f(x_0)\dd x_0$$
  $r=R$的球面电像
$$\boldsymbol{r_像}=\frac{R^2}{r_0^2}\boldsymbol{r_0}\qquad q_像=\frac{R}{r_0}\varepsilon\\
\quad\\
G=-\frac{1}{4\pi}\frac{1}{|\boldsymbol{r}-\boldsymbol{r_0}|}+\frac{R}{r_0}\frac{1}{4\pi}\frac{1}{\left|\boldsymbol{r}-\displaystyle\frac{R^2}{r_0^2}\boldsymbol{r_0}\right|}\\
\quad\\
u(r,\theta,\varphi)=\frac{R}{4\pi}\int_0^\pi\int_0^{2\pi}\frac{R^2-r^2}{(R^2-2Rr\cos{\Theta}+r^2)^{\frac{3}{2}}}f(\theta_0,\varphi_0)\sin{\theta_0}\dd\theta_0\dd\varphi_0\\
\quad\\
\cos\Theta=\cos\theta\cos\theta_0+\sin\theta\sin\theta_0\cos(\varphi-\varphi_0)$$
  $z=0$的平面电像
$$(x_像,y_像,z_像)=(x_0,y_0,-z_0)\qquad q_像=\varepsilon\\
\quad\\
G=-\frac{1}{4\pi}\frac{1}{|\boldsymbol{r}-(x_0,y_0,z_0)|}+\frac{1}{4\pi}\frac{1}{\left|\boldsymbol{r}-(x_0,y_0,-z_0)\right|}\\
\quad\\
u(x,y,z)=\frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{z}{[(x-x_0)^2+(y-y_0)^2+z^2]^{\frac{3}{2}}}f(x_0,y_0)\dd x_0\dd y_0$$
$\quad\\$
含时格林函数
$\quad\\$
  振动定解问题
$$u_{tt}-a^2\Delta u=f(\boldsymbol{r},t)\\
\quad\\
\left.\left(\alpha\frac{\partial u}{\partial n}+\beta u\right)\right|_\Sigma=\theta(M,t)\\
\quad\\
u|_{t=0}=\varphi(\boldsymbol{r})\qquad u_t|_{t=0}=\psi(\boldsymbol{r})$$
  分解作用力
$$f(\boldsymbol{r},t)=\iiint_T\int_t f(\boldsymbol{r_0},\tau)\delta(\boldsymbol{r}-\boldsymbol{r_0})\delta(t-\tau)\dd V_0\dd \tau$$
  引入格林函数，记$G(\boldsymbol{r},t;\boldsymbol{r_0},t_0)$为点位脉冲点力引起的振动，要求其满足齐次边界条件
$$G_{tt}-a^2\Delta G=\delta(\boldsymbol{r}-\boldsymbol{r_0})\delta(t-t_0)\\
\quad\\
\left.\left(\alpha\frac{\partial G}{\partial n}+\beta G\right)\right|_\Sigma=0\\
\quad\\
G|_{t=0}=0\qquad G_t|_{t=0}=0$$
  下证格林函数具有对称性$G(\boldsymbol{r_1},t_1;\boldsymbol{r_2},t_2)=G(\boldsymbol{r_2},-t_2;\boldsymbol{r_1},-t_1)$，
分别记$G(\boldsymbol{r},t;\boldsymbol{r_1},t_1)=G(\boldsymbol{r},-t;\boldsymbol{r_2},-t_2)$为$u,v$，由格林公式
$$\iiint_T\int_{-\infty}^{t’}[(u_{tt}-a^2\Delta u)v-(v_{tt}-a^2\Delta v)u]\dd V\dd t\\
=\iiint_T(u_tv-v_tu)|_{-\infty}^{t’}\dd V+a^2\int_{-\infty}^{t’}\left[\iint_\Sigma\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\dd S\right]\dd t\\
\quad\\
v(\boldsymbol{r_1},t_1)-u(\boldsymbol{r_2},t_2)=0$$