\gdef\red#1{{\color{cb8680}{#1}}} \gdef\green#1{{\color{4f8d63}{#1}}} \gdef\gray#1{{\color{gray}{#1}}} \gdef\purple#1{{\color{B189C6}{#1}}} \gdef\orange#1{{\color{dfa04b}{#1}}} \gdef\white#1{{\color{white}{#1}}}
Lu(r,… )=f(r,… )L u(\mathbf{r}, \dots) = f(\mathbf{r}, \dots) Lu(r,…)=f(r,…)
L波动= ∂2∂t2−a2∇2L输运= ∂∂t−a2∇2L拉普拉斯= −∇2LHelmholtz= −∇2−k2 \begin{align*} L_\text{波动} = & \ \frac{\partial^2}{\partial t^2}- a^2 \nabla^2\\ L_\text{输运} = & \ \frac{\partial}{\partial t}- a^2 \nabla^2\\ L_\text{拉普拉斯} = & \ - \nabla^2\\ L_\text{Helmholtz} = & \ - \nabla^2 - k^2 \end{align*} L波动=L输运=L拉普拉斯=LHelmholtz= ∂t2∂2−a2∇2 ∂t∂−a2∇2 −∇2 −∇2−k2
L(u1+u2)=Lu1+Lu2,L(λu1)=λLu1L(u_1 + u_2) = Lu_1 + Lu_2, \qquad L(\lambda u_1) = \lambda L u_1 L(u1+u2)=Lu1+Lu2,L(λu1)=λLu1
A(v1+v2)=Av1+Av2,A(λv1)=λAv1A(v_1 + v_2) = A v_1 + A v_2, \qquad A(\lambda v_1) = \lambda A v_1 A(v1+v2)=Av1+Av2,A(λv1)=λAv1
L:F→FL: \mathscr{F} \to \mathscr{F} L:F→F
给 F\mathscr{F}F 提供一套完备基底：
基底的构造
直角坐标、正交曲线坐标的分离变量法本质就是通过本征问题构造基底
大一线性代数：AAA 是方阵，怎么求解 Au=fAu = fAu=f?
逆矩阵 A−1A^{-1}A−1
A−1Au=A−1f⇒u=A−1f\green{A^{-1}} A u = \green{A^{-1}} f \quad \Rightarrow \quad u = \green{A^{-1}} f A−1Au=A−1f⇒u=A−1f
逆矩阵的存在性
假设 A−1A^{-1}A−1 存在；A−1A^{-1}A−1 并不总是存在
什么时候 A−1A^{-1}A−1 存在？ 行列式非零
逆矩阵的核心性质：A−1A=AA−1=IA^{-1} A = A A^{-1} = IA−1A=AA−1=I
把 LLL 看成线性映射
什么是 L−1L^{-1}L−1？
定义：微分算符 LLL 的逆称为 LLL-格林函数 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0)，满足 格林函数方程
LrG(r,r0)=δ(r−r0)L_\green{\mathbf{r}}G(\mathbf{r}, \mathbf{r}_0) = \delta(\mathbf{r} - \mathbf{r}_0) LrG(r,r0)=δ(r−r0)
格林函数方程的注意事项
下标 r\mathbf{r}r 强调求导操作是对 r\mathbf{r}r 进行的；r0\mathbf{r}_0r0 是一个固定点
方程右边 δ\deltaδ 关于 r,r0\mathbf{r}, \mathbf{r}_0r,r0 对称，但 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0) 不一定 关于 r,r0\mathbf{r}, \mathbf{r}_0r,r0 对称
格林函数的唯一性
在没有指定边界条件的情况下，格林函数 不唯一。 若 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0) 是 LLL 的格林函数，考虑任意 LLL-调和函数 Φ\PhiΦ 使得 LΦ(r)=0L\Phi(\mathbf{r}) = 0LΦ(r)=0，则
G(r,r0)+Φ(r)G(\mathbf{r}, \mathbf{r}_0) + \Phi(\mathbf{r}) G(r,r0)+Φ(r)
也是 LLL 的格林函数。
要确定一个格林函数，需要边界条件。
Delta 函数与单位矩阵的类比
δ(r−r0)\delta(\mathbf{r} - \mathbf{r}_0)δ(r−r0) 相当于无穷维单位矩阵。它对任意函数的「矩阵乘法」为
∫drδ(r−r0)u(r)=u(r0) ∼类比 ∑rδr0,rur=ur0 ∼类比 δu=u \begin{align*} \int d\mathbf{r} \delta(\mathbf{r} - \mathbf{r}_0) u(\mathbf{r}) = u(\mathbf{r}_0) \ \stackrel{\purple{\text{类比}}}{\sim} \ \sum_{\mathbf{r}} \delta_{\mathbf{r}_0, \mathbf{r}} u_\mathbf{r} = u_{\mathbf{r}_0} \ \stackrel{\purple{\text{类比}}}{\sim} \ \delta u = u \end{align*} ∫drδ(r−r0)u(r)=u(r0) ∼类比 r∑δr0,rur=ur0 ∼类比 δu=u
r\mathbf{r}r 相当于 u(r)u(\mathbf{r})u(r) 的行指标
r0,r\mathbf{r}_0, \mathbf{r}r0,r 相当于方阵 δ\deltaδ 的行、列指标: δ(r−r0)=δ(r0−r)\delta(\mathbf{r} - \mathbf{r}_0) = \delta(\mathbf{r}_0 - \mathbf{r})δ(r−r0)=δ(r0−r)，δr0,r=δr,r0\delta_{\mathbf{r}_0, \mathbf{r}} = \delta_{\mathbf{r}, \mathbf{r}_0}δr0,r=δr,r0 是对称矩阵
对 r\mathbf{r}r 积分相当于对行列指标求和
积分测度
在无界空间上求得的格林函数 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0) 也称为 基本解 (fundamental solution)
基本解的选择
基本解 不唯一，通常选一个最简单、对称性最高的作为基本解
格林函数和基本解这两个单词经常换着用
设 GGG 是 LLL 在无界空间上的基本解
定理：设 uuu 满足方程 Lu=fLu = fLu=f，则 uuu 可以用 fff 与 GGG 构造出来，
u(r)=∫dr0G(r,r0)f(r0)u(\mathbf{r}) = \int d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) f(\mathbf{r}_0) u(r)=∫dr0G(r,r0)f(r0)
从左方作用 LLL，进入积分内部，
Lu=∫dr0LG(r,r0)f(r0)=∫dr0δ(r−r0)f(r0)=f(r)Lu = \int d\purple{\mathbf{r}_0} \orange{L G(\mathbf{r}, \purple{\mathbf{r}_0})} f(\purple{\mathbf{r}_0}) = \int d\purple{\mathbf{r}_0} \orange{\delta(\mathbf{r} - \purple{\mathbf{r}_0})} f(\purple{\mathbf{r}_0}) = f(\mathbf{r}) Lu=∫dr0LG(r,r0)f(r0)=∫dr0δ(r−r0)f(r0)=f(r)
积分变量的注意事项
积分是对 r0\mathbf{r}_0r0 做的，r\mathbf{r}r 要留作自由变量被求导
三维空间中的电势满足泊松方程：uuu 是电势，ρ\rhoρ 是电荷密度
−∇2u(r)=ρ(r) - \nabla^2 u(\mathbf{r}) = \rho(\mathbf{r}) −∇2u(r)=ρ(r)
对应的格林函数方程
−∇2G(r,r0)=δ(r−r0) - \nabla^2 G(\mathbf{r}, \mathbf{r}_0) = \delta(\mathbf{r} - \mathbf{r}_0) −∇2G(r,r0)=δ(r−r0)
右边 ρ(r)=δ(r−r0)\rho(\mathbf{r}) = \delta(\mathbf{r} - \mathbf{r}_0)ρ(r)=δ(r−r0) 描述 r0\mathbf{r}_0r0 处一个单位点电荷的电荷密度
基本解
G(r,r0)=14π∣r−r0∣ G(\mathbf{r}, \mathbf{r}_0) = \frac{1}{4\pi |\mathbf{r} - \mathbf{r}_0|} G(r,r0)=4π∣r−r0∣1
就是 r0\mathbf{r}_0r0 处单位点电荷在 r\mathbf{r}r 处产生的电势。
格林函数又称为点源影响函数
泊松方程的解
u(r)=∫dr0G(r,r0)ρ(r0)=∫dr014π∣r−r0∣ρ(r0)u(\mathbf{r}) = \int d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) \rho(\mathbf{r}_0) = \int d\mathbf{r}_0 \frac{1}{4\pi |\mathbf{r} - \mathbf{r}_0|} \rho(\mathbf{r}_0) u(r)=∫dr0G(r,r0)ρ(r0)=∫dr04π∣r−r0∣1ρ(r0)
电势叠加原理
G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0) 第二个自变量控制“源”位置，第一个自变量控制“势能测量点”位置。
对源点位置 r0\mathbf{r}_0r0 求积分：把所有点电荷的贡献叠加起来
考虑无界空间中的泊松方程 (注意 符号)
−∇2u(r)=f(r) - \nabla^2 u(\mathbf{r}) = f(\mathbf{r}) −∇2u(r)=f(r)
解的非唯一性
考虑无界空间中 (−∇2)(-\nabla^2)(−∇2)-格林函数，满足如下微分方程
∇r2G(r,r0)=−δ(r−r0) \nabla^2_\mathbf{r} G(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0) ∇r2G(r,r0)=−δ(r−r0)
∇rG(r,r0)≠∇r0G(r,r0)≠∇rG(r0,r) \nabla_\mathbf{r}G(\mathbf{r}, \mathbf{r}_0) \ne \nabla_{\mathbf{r}_0}G(\mathbf{r}, \mathbf{r}_0) \ne \nabla_\mathbf{r}G(\mathbf{r}_0, \mathbf{r}) ∇rG(r,r0)=∇r0G(r,r0)=∇rG(r0,r)
∇r02G(r,r0)≠−δ(r−r0)\nabla_{\mathbf{r}_0}^2G(\mathbf{r}, \mathbf{r}_0) \ne - \delta(\mathbf{r} - \mathbf{r}_0) ∇r02G(r,r0)=−δ(r−r0)
三维基本解的物理意义
三维基本解的证明
∇2G=∇2(14π∣r−r0∣)\nabla^2 G = \nabla^2 \left({\frac{1}{4\pi |\mathbf{r} - \mathbf{r}_0|}}\right) ∇2G=∇2(4π∣r−r0∣1)
∇r2G=∇R2(14πR)=1R2ddR(R2ddR(14πR))+0\nabla^2_\mathbf{r} G = \nabla^2_\mathbf{R} \left({\frac{1}{4\pi \green{R}}}\right) = \frac{1}{R^2} \frac{d}{d R} \left(R^2 \frac{d}{d R} \left({\frac{1}{4\pi R}}\right)\right) + 0 ∇r2G=∇R2(4πR1)=R21dRd(R2dRd(4πR1))+0
但是在 R=0\mathbf{R} = 0R=0，即 r=r0\mathbf{r} = \mathbf{r}_0r=r0 的地方，上面的计算有问题，因为有奇性。
把 GGG 看成点电荷产生的电势，则 ∇u∼E\nabla u \sim \mathbf{E}∇u∼E，∇2u∼∇⋅E\nabla^2 u \sim \nabla \cdot \mathbf{E}∇2u∼∇⋅E。
∫Bϵ3∇r2G(r,r0)dr= ∫Bϵ3∇R⋅(∇R14πR)dR=∫Sϵ2∇R(14πR)⋅dS= −∫Sϵ2eR4πR2⋅dS=−1\begin{align*} \int_{B_\epsilon^3}\nabla^2_\mathbf{r} \orange{G(\mathbf{r}, \mathbf{r}_0)}d\mathbf{r} = & \ \int_{B_\epsilon^3} \nabla_\mathbf{R} \cdot \left(\nabla_\mathbf{R} \orange{\frac{1}{4\pi R}}\right) d\mathbf{R} = \int_{S_\epsilon^2} \nabla_\mathbf{R} \left({\frac{1}{4\pi R}}\right) \cdot d\mathbf{S}\\ = & \ - \int_{S_\epsilon^2} \frac{\mathbf{e}_R}{4\pi R^2} \cdot d\mathbf{S} = -1 \end{align*} ∫Bϵ3∇r2G(r,r0)dr== ∫Bϵ3∇R⋅(∇R4πR1)dR=∫Sϵ2∇R(4πR1)⋅dS −∫Sϵ24πR2eR⋅dS=−1
定理：二维无界空间中 (−∇2)(-\nabla^2)(−∇2) 、满足边界 G(r,r0)∣∣r−r0∣=1=0G(\mathbf{r}, \mathbf{r}_0)\Big|_{|\mathbf{r} - \mathbf{r}_0| = 1} = 0G(r,r0)∣r−r0∣=1=0 的基本解为
G(r,r0)=−12πln∣r−r0∣ G(\mathbf{r}, \mathbf{r}_0) = - \frac{1}{2\pi} \ln |\mathbf{r} - \mathbf{r}_0| G(r,r0)=−2π1ln∣r−r0∣
二维基本解的性质
什么静电场系统具有这样的电势？ 无限长均匀带电直线
∇2G=∇2(−12πlnR)=−12π1RddR(RddRlnR)=0\nabla^2 G = \nabla^2 \left({- \frac{1}{2\pi} \ln R}\right) = - \frac{1}{2\pi} \frac{1}{R} \frac{d}{dR} \left({R \frac{d}{dR} \ln R}\right) = 0 ∇2G=∇2(−2π1lnR)=−2π1R1dRd(RdRdlnR)=0
∫Bϵ2∇2Gdr= ∫Bϵ2∇R⋅∇R(−12πlnR)dR=∫Sϵ1∇R(−12πlnR)⋅dl= −∫Sϵ1eR2πR⋅dl=−1\begin{align*} \int_{B_\epsilon^2} \nabla^2 \orange{G} d\mathbf{r} = & \ \int_{B_\epsilon^2} \nabla_\mathbf{R} \cdot \nabla_\mathbf{R} \left(\orange{\frac{-1}{2\pi}\ln R}\right) d\mathbf{R} = \int_{S_\epsilon^1} \nabla_\mathbf{R} \left({\frac{-1}{2\pi}\ln R}\right) \cdot d\mathbf{l}\\ = & \ - \int_{S_\epsilon^1} \frac{\mathbf{e}_R}{2\pi R} \cdot d\mathbf{l} = -1 \end{align*} ∫Bϵ2∇2Gdr== ∫Bϵ2∇R⋅∇R(2π−1lnR)dR=∫Sϵ1∇R(2π−1lnR)⋅dl −∫Sϵ12πReR⋅dl=−1
定理 一维无界空间中 (−∇2)(-\nabla^2)(−∇2) 的格林函数为
G(x,x0)=−(x−x0)θ(x−x0)−C(x−x0)−C′ G(x, x_0) = - (x - x_0)\theta(x - x_0) - C (x - x_0) - C' G(x,x0)=−(x−x0)θ(x−x0)−C(x−x0)−C′
一维格林函数的性质
一维空间中的格林函数是一个分段线性函数，其中 C,C′C, C'C,C′ 是常数
常用的基本解选择 C=−12C = - \frac{1}{2}C=−21
G(x,x0)=−12∣x−x0∣+C′ G(x, x_0) = - \frac{1}{2} |x - x_0| + C' G(x,x0)=−21∣x−x0∣+C′
关于 x0x_0x0 对称
求一维 ∇2\nabla^2∇2-格林函数。
d2dx2G(x,x0)=−δ(x−x0)\frac{d^2}{dx^2} G(x, x_0) = - \delta(x - x_0) dx2d2G(x,x0)=−δ(x−x0)
−ddxG(x,x0)=θ(x−x0)+C- \frac{d}{dx}G(x, x_0) = \theta(x - x_0) + C −dxdG(x,x0)=θ(x−x0)+C
证明
−G(x,x0)=C(x−x0)+∫θ(x−x0)dx+C′- G(x, x_0) = C(x - x_0) + \int \theta(x - x_0) dx + C' −G(x,x0)=C(x−x0)+∫θ(x−x0)dx+C′
∫θ(x)dx=xθ(x)\int \theta(x) dx = x \theta(x) ∫θ(x)dx=xθ(x)
−G(x−x0)=(x−x0)θ(x−x0)+C(x−x0)+C′- G(x - x_0) = (x - x_0)\theta(x - x_0) + C (x - x_0) + C' −G(x−x0)=(x−x0)θ(x−x0)+C(x−x0)+C′
G(x−x0)=−12∣x−x0∣+C′G(x - x_0) = - \frac{1}{2} |x - x_0| + C' G(x−x0)=−21∣x−x0∣+C′
∫D(∂Q∂x−∂P∂y)dr=∮∂D(Pdx+Qdy) \int_D \bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg) d\mathbf{r} = \oint_{\partial D} (Pdx + Q dy) ∫D(∂x∂Q−∂y∂P)dr=∮∂D(Pdx+Qdy)
∫Ddr(u∇2v−v∇2u)=∮∂DdS⋅(u∇v−v∇u)=∮∂DdS(u∂v∂n−v∂u∂n) \int_D d\mathbf{r} \left({u \nabla^2 v - v \nabla^2 u}\right) = \oint_{\partial D} \orange{d \mathbf{S}} \cdot \left({u \nabla v - v \nabla u}\right) = \oint_{\partial D} dS \left({u\frac{\partial v}{\partial \mathbf{n}}- v\frac{\partial u}{\partial \mathbf{n}} }\right) ∫Ddr(u∇2v−v∇2u)=∮∂DdS⋅(u∇v−v∇u)=∮∂DdS(u∂n∂v−v∂n∂u)
∂f∂n≔∇f⋅n\green{\frac{\partial f}{\partial \mathbf{n}} \coloneqq \nabla f \cdot \mathbf{n}} ∂n∂f:=∇f⋅n
证明第二格林公式
∇⋅(u∇v)=∇u⋅∇v+u∇2v,∇⋅(v∇u)=∇v⋅∇u+v∇2u\nabla \cdot (u \nabla v) = \red{\nabla u \cdot \nabla v} + \orange{u \nabla^2 v}, \qquad \nabla \cdot (v \nabla u) = \red{\nabla v \cdot \nabla u} + \orange{v \nabla^2 u} ∇⋅(u∇v)=∇u⋅∇v+u∇2v,∇⋅(v∇u)=∇v⋅∇u+v∇2u
u∇2v−v∇2u=∇⋅(u∇v−v∇u) \orange{u \nabla^2 v} - \orange{v \nabla^2 u} = \nabla \cdot (\purple{u \nabla v - v \nabla u}) u∇2v−v∇2u=∇⋅(u∇v−v∇u)
∫D(u∇2v−v∇2u)dr=∮S(u∇v−v∇u)⋅ndS=∮S(u∂v∂n−v∂u∂n)dS\int_D (u \nabla^2 v - v \nabla^2 u) d\mathbf{r} = \oint_{S} (\purple{u \nabla v - v \nabla u}) \cdot \mathbf{n} dS = \oint_{S} \left({u \frac{\partial v}{\partial \mathbf{n}} - v \frac{\partial u}{\partial \mathbf{n}}}\right) dS ∫D(u∇2v−v∇2u)dr=∮S(u∇v−v∇u)⋅ndS=∮S(u∂n∂v−v∂n∂u)dS
第二格林公式相当于针对整个微分算符 ∇2\nabla^2∇2 的分部积分
∫Ddru(∇2v)=∫Ddr(∇2u)v+∮∂DdS⋅(u∇v−(∇u)v)\int_D d \mathbf{r} u (\nabla^2 v) = \int_D d \mathbf{r} (\nabla^2 u)v + \oint_{\partial D} d \mathbf{S} \cdot (u \nabla v - (\nabla u) v) ∫Ddru(∇2v)=∫Ddr(∇2u)v+∮∂DdS⋅(u∇v−(∇u)v)
与一维分部积分的对比
对比
∫x1x2udvdxdx=∫x1x2dudxvdx+(uv)∣x1x2 \int_{x_1}^{x_2} u \frac{dv}{dx} dx = \int_{x_1}^{x_2} \frac{du}{dx} v dx + (u v)\bigg|_{x_1}^{x_2} ∫x1x2udxdvdx=∫x1x2dxduvdx+(uv)x1x2
−∇2u(r)=f(r),r∈D,u(r)∣r∈S=φ(r) .- \nabla^2 u(\mathbf{r}) =f(\mathbf{r}), \quad \mathbf{r} \in D, \qquad u(\mathbf{r})\big|_{\mathbf{r} \in S} = \varphi(\mathbf{r}) \ . −∇2u(r)=f(r),r∈D,u(r)r∈S=φ(r) .
−∇2u1(r)=f(r),r∈D,u1(r)∣S=0- \nabla^2 u_1(\mathbf{r}) = f(\mathbf{r}), \quad \mathbf{r} \in D, \qquad u_1(\mathbf{r})\big|_{S} = 0 −∇2u1(r)=f(r),r∈D,u1(r)S=0
−∇2u2(r)=0,r∈D,u2(r)∣r∈S=φ(r)- \nabla^2 u_2(\mathbf{r}) = 0, \quad \mathbf{r} \in D, \qquad u_2(\mathbf{r})\big|_{\mathbf{r} \in S} = \varphi(\mathbf{r}) −∇2u2(r)=0,r∈D,u2(r)r∈S=φ(r)
求解非齐次方程 + 齐次边界
−∇2u1(r)=f(r), r∈D,u1(r)∣S=0- \nabla^2 u_1(\mathbf{r}) = f(\mathbf{r}), \ \mathbf{r} \in D, \qquad u_1(\mathbf{r})\big|_{S} = 0 −∇2u1(r)=f(r), r∈D,u1(r)S=0
考虑满足第一类 齐次 边界条件的格林函数 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0)
∇2G(r,r0)=−δ(r−r0),r,r0∈D,G(r,r0)∣r∈S=0 .\nabla^2 G(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0), \quad \mathbf{r}, \red{\mathbf{r}_0} \in D, \qquad G(\mathbf{r}, \mathbf{r}_0)\big|_{\mathbf{r} \in S} = 0 \ . ∇2G(r,r0)=−δ(r−r0),r,r0∈D,G(r,r0)r∈S=0 .
获得同时满足方程和边界条件的解
u1(r)=∫Ddr0G(r,r0)f(r0)u_1(\mathbf{r}) = \int_D d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) f(\mathbf{r}_0) u1(r)=∫Ddr0G(r,r0)f(r0)
积分变量
注意，积分是对 r0\mathbf{r}_0r0 做的。r0\mathbf{r}_0r0 既出现在 GGG 中，也出现在 fff 中。
∇2u2(r)=0, r∈D,u2(r)∣S=φ(r)\nabla^2 u_2(\mathbf{r}) = 0, \ \mathbf{r} \in D, \qquad u_2(\mathbf{r})\big|_{S} = \red{\varphi(\mathbf{r})} ∇2u2(r)=0, r∈D,u2(r)S=φ(r)
u2′(r)=u2(r)−v(r),v(r)∣∂D=φ(r)⇒−∇2u2′(r)=∇2v(r), r∈D,u2′(r)∣∂D=0\begin{align*} u'_2(\mathbf{r}) = u_2(\mathbf{r}) - v(\mathbf{r}), \qquad \green{v(\mathbf{r})|_{\partial D} = \varphi(\mathbf{r})} \\ \\ \Rightarrow\qquad - \nabla^2 u'_2(\mathbf{r}) = \purple{\nabla^2 v(\mathbf{r})}, \ \mathbf{r}\in D ,\qquad \orange{u'_2(\mathbf{r})|_{\partial D} = 0} \end{align*} u2′(r)=u2(r)−v(r),v(r)∣∂D=φ(r)⇒−∇2u2′(r)=∇2v(r), r∈D,u2′(r)∣∂D=0
u2′(r)=∫Ddr0G(r,r0)∇r02v(r0),u2(r)=u2′(r)+v(r)u'_2(\mathbf{r}) = \int_D d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) \purple{\nabla_{\mathbf{r}_0}^2 v(\mathbf{r}_0)}, \qquad u_2(\mathbf{r}) = u_2'(\mathbf{r}) + v(\mathbf{r}) u2′(r)=∫Ddr0G(r,r0)∇r02v(r0),u2(r)=u2′(r)+v(r)
利用第二格林公式 将 ∇r02\nabla_{\mathbf{r}_0}^2∇r02 移动到 GGG 上，得到三项
u2′(r)= ∫Ddr0v(r0)∇r02G(r,r0) +∮∂DdS0(G(r,r0)∇r0v(r0)−v(r0)∇r0G(r,r0))⋅n\begin{align*} u_2'(\mathbf{r}) = & \ \int_D d\mathbf{r}_0 v(\mathbf{r}_0)\nabla_{\mathbf{r}_0}^2 G(\mathbf{r}, \mathbf{r}_0)\\ & \ + \oint_{\partial D} dS_0 \Big({G(\mathbf{r}, \mathbf{r}_0) \nabla_{\mathbf{r}_0} v(\mathbf{r}_0) - \purple{v(\mathbf{r}_0)} \nabla_{\mathbf{r}_0} G(\mathbf{r}, \mathbf{r}_0)}\Big) \cdot \mathbf{n} \end{align*} u2′(r)= ∫Ddr0v(r0)∇r02G(r,r0) +∮∂DdS0(G(r,r0)∇r0v(r0)−v(r0)∇r0G(r,r0))⋅n
体积分里，是对 GGG 的第二个变量 r0\mathbf{r}_0r0 求导
好像更复杂了
转机：本情况下，格林函数 GGG 具有对称性
G(r,r0)=G(r0,r)G(\mathbf{r}, \mathbf{r}_0) = G(\mathbf{r}_0, \mathbf{r}) G(r,r0)=G(r0,r)
稍后证明
∫Ddr0v(r0)∇r02G(r0,r)=−∫Ddr0v(r0)δ(r0−r)=−v(r)\int_D d\mathbf{r}_0 v(\mathbf{r}_0)\nabla_{\mathbf{r}_0}^2 \purple{G(\mathbf{r}_0, \mathbf{r})} = - \int_D d\mathbf{r}_0 v(\mathbf{r}_0) \delta(\mathbf{r}_0 - \mathbf{r}) = \red{- v(\mathbf{r})} ∫Ddr0v(r0)∇r02G(r0,r)=−∫Ddr0v(r0)δ(r0−r)=−v(r)
G(r0,r)∣r0∈S=0⇒∮∂DdS0G(r,r0)∇r0v(r0)⋅n=0G(\mathbf{r}_0, \mathbf{r})|_{\mathbf{r}_0 \in S} = 0 \quad\Rightarrow\quad \oint_{\partial D} dS_0 G(\mathbf{r}, \mathbf{r}_0) \nabla_{\mathbf{r}_0} v(\mathbf{r}_0) \cdot \mathbf{n} = 0 G(r0,r)∣r0∈S=0⇒∮∂DdS0G(r,r0)∇r0v(r0)⋅n=0
u2′(r)=−v(r)−∮dS0φ(r0)∇r0G(r,r0)u'_2(\mathbf{r}) = - v(\mathbf{r}) - \oint dS_0 \varphi(\mathbf{r}_0) \nabla_{\mathbf{r}_0} G(\mathbf{r}, \mathbf{r}_0) u2′(r)=−v(r)−∮dS0φ(r0)∇r0G(r,r0)
总结：齐次方程 + 非齐次边界条件的解为
u2(r)=u2′(r)+v(r)= −v(r)−∮∂Dv(r0)∇r0G(r,r0)⋅ndS0+v(r)= −∮φ(r0)∂G(r,r0)∂n0dS0\begin{align*} u_2(\mathbf{r}) = u'_2(\mathbf{r}) + \green{v(\mathbf{r})} = & \ \red{- v(\mathbf{r})} - \oint_{\partial D} \purple{v(\mathbf{r}_0)} \orange{\nabla_{\mathbf{r}_0}G(\mathbf{r}, \mathbf{r}_0) \cdot \mathbf{n} dS_0} + \green{v(\mathbf{r})}\\ = & \ - \oint \purple{\varphi(\mathbf{r}_0)} \orange{\frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}_0}dS_0} \end{align*} u2(r)=u2′(r)+v(r)== −v(r)−∮∂Dv(r0)∇r0G(r,r0)⋅ndS0+v(r) −∮φ(r0)∂n0∂G(r,r0)dS0
物理解释
边界势 φ\varphiφ 在边界 SSS 诱发 表面电偶极矩 D=φD = \varphiD=φ，并在 r\mathbf{r}r 处诱发的响应函数 ∂G(r,r0)/∂n0\partial G(\mathbf{r}, \mathbf{r}_0)/\partial \mathbf{n}_0∂G(r,r0)/∂n0 的叠加
对称性定理：−∇2-\nabla^2−∇2 的齐次第一类边界格林函数具有对称性
齐次边界：u1(r)= ∫Ddr0G(r,r0)f(r0)非齐次边界：u2(r)= −∮∂Dφ(r0)∂G(r,r0)∂n0dS0\begin{align*} \small{\text{齐次边界：}}u_1(\mathbf{r}) = & \ \int_D d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) f(\mathbf{r}_0)\\ \small{\text{非齐次边界：}}u_2(\mathbf{r}) = & \ - \oint_{\partial D} \varphi(\mathbf{r}_0) \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}_0}dS_0 \end{align*} 齐次边界：u1(r)=非齐次边界：u2(r)= ∫Ddr0G(r,r0)f(r0) −∮∂Dφ(r0)∂n0∂G(r,r0)dS0
u(r)=u1(r)+u2(r)=∫Ddr0G(r,r0)f(r0)−∮∂Dφ(r0)∂G(r,r0)∂n0dS0\begin{align*} u(\mathbf{r}) = u_1(\mathbf{r}) + u_2(\mathbf{r}) = \int_D d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) f(\mathbf{r}_0) - \oint_{\partial D} \varphi(\mathbf{r}_0) \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}_0}dS_0 \end{align*} u(r)=u1(r)+u2(r)=∫Ddr0G(r,r0)f(r0)−∮∂Dφ(r0)∂n0∂G(r,r0)dS0
证明对称性
G(r,r0)−G(r0,r)= ∫Ddr1G(r1,r0)δ(r1−r)−∫Ddr1G(r1,r)δ(r1−r0)\begin{align*} & \ G(\purple{\mathbf{r}}, \mathbf{r}_0) - G(\purple{\mathbf{r}_0}, \mathbf{r}) \\ = & \ \int_D d\mathbf{r}_1 G(\red{\mathbf{r}_1}, \mathbf{r}_0) \delta(\red{\mathbf{r}_1} - \purple{\mathbf{r}}) - \int_D d\mathbf{r}_1 G(\red{\mathbf{r}_1}, \mathbf{r}) \delta(\red{\mathbf{r}_1} - \purple{\mathbf{r}_0})\\ \end{align*} = G(r,r0)−G(r0,r) ∫Ddr1G(r1,r0)δ(r1−r)−∫Ddr1G(r1,r)δ(r1−r0)
第一项 r↔r0\mathbf{r} \leftrightarrow \mathbf{r}_0r↔r0 就得到第二项
=−[∫Ddr1G(r1,r0)∇r12G(r1,r)−∫Ddr1G(r1,r)∇r12G(r1,r0)] = - \left[{\int_D d\mathbf{r}_1 G(\mathbf{r}_1, \mathbf{r}_0) \nabla_{\red{\mathbf{r}_1}}^2G(\mathbf{r}_1, \mathbf{r}) - \int_D d\mathbf{r}_1 G(\mathbf{r}_1, \mathbf{r}) \nabla_{\red{\mathbf{r}_1}}^2 G(\mathbf{r}_1, \mathbf{r}_0)}\right] =−[∫Ddr1G(r1,r0)∇r12G(r1,r)−∫Ddr1G(r1,r)∇r12G(r1,r0)]
=−∫∂DdS1(G(r1,r0)∇r1G(r1,r)−G(r1,r)∇r1G(r1,r0))⋅n= - \int_{\partial D} \purple{dS_1} \Big({\red{G(\mathbf{r}_1, \mathbf{r}_0)} \nabla_{\mathbf{r}_1} G(\mathbf{r}_1, \mathbf{r}) - \red{G(\mathbf{r}_1, \mathbf{r})} \nabla_{\mathbf{r}_1} {G(\mathbf{r}_1, \mathbf{r}_0)}}\Big) \cdot \mathbf{n} =−∫∂DdS1(G(r1,r0)∇r1G(r1,r)−G(r1,r)∇r1G(r1,r0))⋅n
G(r,r0)−G(r0,r)=0G(\mathbf{r}, \mathbf{r}_0) - G(\mathbf{r}_0, \mathbf{r}) = 0 G(r,r0)−G(r0,r)=0
讲义上提供泊松方程第一边值问题另一种解法
第二解法
∇r2G(r,r0)=−δ(r−r0), r∈D,G(r,r0)∣r∈S=0\orange{\nabla_\mathbf{r}^2 G(\mathbf{r}, \mathbf{r}_0)} = - \delta(\mathbf{r} - \mathbf{r}_0), \ \mathbf{r} \in D, \qquad G(\mathbf{r}, \mathbf{r}_0)\bigg|_{\mathbf{r} \in S} = 0 ∇r2G(r,r0)=−δ(r−r0), r∈D,G(r,r0)r∈S=0
G(r,r0)∇r2u(r)−u(r)∇r2G(r,r0)=−G(r,r0)f(r)+u(r)δ(r−r0)\purple{G(\mathbf{r}, \mathbf{r}_0)\nabla^2_\mathbf{r}u(\mathbf{r})} - \orange{u(\mathbf{r})\nabla_\mathbf{r}^2 G(\mathbf{r}, \mathbf{r}_0)} = \purple{- G(\mathbf{r}, \mathbf{r}_0) f(\mathbf{r})} + \orange{u(\mathbf{r}) \delta(\mathbf{r} - \mathbf{r}_0) } G(r,r0)∇r2u(r)−u(r)∇r2G(r,r0)=−G(r,r0)f(r)+u(r)δ(r−r0)
∫D[G(r,r0)∇r2u(r)−u(r)∇r2G(r,r0)]dr=−∫DG(r,r0)f(r)dr+u(r0)\int_D \left[{\purple{G(\mathbf{r}, \mathbf{r}_0)\nabla^2_\mathbf{r}u(\mathbf{r})} - \orange{u(\mathbf{r})\nabla_\mathbf{r}^2 G(\mathbf{r}, \mathbf{r}_0)}}\right]d\mathbf{r} = - \int_D G(\mathbf{r},\mathbf{r}_0) f(\mathbf{r}) d\mathbf{r} + u(\mathbf{r}_0) ∫D[G(r,r0)∇r2u(r)−u(r)∇r2G(r,r0)]dr=−∫DG(r,r0)f(r)dr+u(r0)
∮∂DdS⋅[G(r,r0)∇ru(r)−u(r)∇rG(r,r0)]+∫DG(r,r0)f(r)dr=u(r0)\oint_{\partial D} d\mathbf{S} \cdot \Big[{G(\mathbf{r}, \mathbf{r}_0) \nabla_\mathbf{r} u(\mathbf{r}) - u(\mathbf{r}) \nabla_\mathbf{r} G(\mathbf{r}, \mathbf{r}_0)}\Big] + \int_D G(\mathbf{r}, \mathbf{r}_0) f(\mathbf{r})d\mathbf{r} = u(\mathbf{r}_0) ∮∂DdS⋅[G(r,r0)∇ru(r)−u(r)∇rG(r,r0)]+∫DG(r,r0)f(r)dr=u(r0)
u(r0)=∫DG(r,r0)f(r)dr+∮∂DdS[G(r,r0)∂u(r)∂n−u(r)∂G(r,r0)∂n]u(\mathbf{r}_0) = \int_D \orange{G(\mathbf{r}, \mathbf{r}_0)} f(\mathbf{r})d\mathbf{r} + \oint_{\partial D} dS \Big[ \red{G(\mathbf{r}, \mathbf{r}_0)} \frac{\partial u(\mathbf{r})}{\partial \mathbf{n}} - \purple{u(\mathbf{r})} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}}\Big] u(r0)=∫DG(r,r0)f(r)dr+∮∂DdS[G(r,r0)∂n∂u(r)−u(r)∂n∂G(r,r0)]
u(r0)=∫DG(r,r0)f(r)dr+∮∂DdS[−φ(r)∂G(r,r0)∂n]u(\mathbf{r}_0) = \int_D \orange{G(\mathbf{r}, \mathbf{r}_0)} f(\mathbf{r})d\mathbf{r} + \oint_{\partial D} dS \Big[ - \purple{\varphi(\mathbf{r})} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}} \Big] u(r0)=∫DG(r,r0)f(r)dr+∮∂DdS[−φ(r)∂n∂G(r,r0)]
u(r0)=∫DG(r0,r)f(r)dr−∮∂Dφ(r)∂G(r0,r)∂ndSu(\mathbf{r}_0) = \int_D \orange{G(\mathbf{r}_0, \mathbf{r})} f(\mathbf{r})d\mathbf{r} - \oint_{\partial D} \varphi(\mathbf{r}) \frac{\partial \orange{G(\mathbf{r}_0, \mathbf{r})}}{\partial \mathbf{n}}dS u(r0)=∫DG(r0,r)f(r)dr−∮∂Dφ(r)∂n∂G(r0,r)dS
u(r)=∫DG(r,r0)f(r0)dr0−∮∂Dφ(r0)∂G(r,r0)∂n0dS0u(\mathbf{r}) = \int_D G(\mathbf{r}, \mathbf{r}_0) f(\mathbf{r}_0)d\mathbf{r}_0 - \oint_{\partial D} \varphi(\mathbf{r}_0) \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}_0}dS_0 u(r)=∫DG(r,r0)f(r0)dr0−∮∂Dφ(r0)∂n0∂G(r,r0)dS0
∇2u(r)=0,r<a,u(r)∣r=a=f(θ,φ)\nabla^2 u(\mathbf{r}) = 0, \quad r < a, \qquad u(\mathbf{r})\big|_{r = a} = f(\theta, \varphi) ∇2u(r)=0,r<a,u(r)r=a=f(θ,φ)
∇2G(r,r0)=−δ(r−r0),r,r0<a,G(r,r0)∣r=a=0\nabla^2 G(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0), \quad r, r_0 < a, \qquad G(\mathbf{r}, \mathbf{r}_0)\big|_{r = a} = 0 ∇2G(r,r0)=−δ(r−r0),r,r0<a,G(r,r0)r=a=0
太难了，不知道怎么直接求
知之为知之，不知为不知。是知也。
——孔子《论语·为政》
通过镜像法构建 等效感应电荷
在球外放置一个或多个假想的点电荷，与 r0\mathbf{r}_0r0 物理电荷共同构建满足接地边界条件的电势
平替 镜像电荷是诱导电荷的平替，它在球内产生的电势与诱导电荷一模一样：用了镜像电荷就 不用再考虑 诱导电荷
为什么要在球外放置假想电荷，而不是球内？
尝试单一像电荷 qqq：与物理点电荷位于球心共线同侧，位置为 r0′\mathbf{r}'_0r0′
像电荷位置的选择
对称性可以帮助推测共线，同侧是为了满足球面第一类齐次边界条件
但是假想电荷大小 qqq 和具体位置 ∣r0′∣|\mathbf{r}_0'|∣r0′∣ 还需要通过边界条件来确定
真实电荷和假想电荷在球内 r\mathbf{r}r 处的电势
g(r,r0)=14π∣r−r0∣+q4π∣r−r0′∣,∣r∣≤a,∣r0′∣>ag(\mathbf{r}, \mathbf{r}_0) = \purple{\frac{1}{4\pi|\mathbf{r} - \mathbf{r}_0|}} + \frac{q}{4\pi|\mathbf{r} - \mathbf{r}'_0|}, \qquad |\mathbf{r}| \le a, \quad |\mathbf{r}_0'| > a g(r,r0)=4π∣r−r0∣1+4π∣r−r0′∣q,∣r∣≤a,∣r0′∣>a
第一项是 基本解，第二项是对基本解的修正：为了满足边界条件
该电势函数显然满足球内格林函数方程
∇r2g(r,r0)=−δ(r−r0),∣r∣<a\nabla^2_\mathbf{r} g(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0), \qquad |\mathbf{r}|< a ∇r2g(r,r0)=−δ(r−r0),∣r∣<a
why?
写成三角函数的组合
g(r,r0)=14πr2+r02−2rr0cosψ+q4πr2+(r0′)2−2rr0′cosψg(\mathbf{r}, \mathbf{r}_0) = \frac{1}{4\pi\sqrt{r^2 + r_0^2 - 2 \green{r r_0 \cos \psi}}} + \frac{q}{4\pi\sqrt{r^2 + (r'_0)^2 - 2 r r_0' \cos \psi}} g(r,r0)=4πr2+r02−2rr0cosψ1+4πr2+(r0′)2−2rr0′cosψq
距离可以写成矢量差的平方，
∣r−r0∣=(r−r0)⋅(r−r0)=r2+r02−2r⋅r0 |\mathbf{r} - \mathbf{r}_0| = \sqrt{(\mathbf{r} - \mathbf{r}_0)\cdot (\mathbf{r} - \mathbf{r}_0)} = \sqrt{\mathbf{r}^2 + \mathbf{r}_0^2 - 2 \mathbf{r} \cdot \mathbf{r}_0} ∣r−r0∣=(r−r0)⋅(r−r0)=r2+r02−2r⋅r0
结合 r⋅r0=rr0cosψ\green{\mathbf{r} \cdot \mathbf{r}_0 = r r_0 \cos \psi}r⋅r0=rr0cosψ，以及 r⋅r0′=rr0′cosψ\mathbf{r} \cdot \mathbf{r}_0' = r r_0' \cos \psir⋅r0′=rr0′cosψ
g(r,r0)∣r=a=14πa2+r02−2ar0cosψ+q4πa2+(r0′)2−2ar0′cosψ g(\mathbf{r}, \mathbf{r}_0)\big|_{r = a} = \frac{1}{4\pi\sqrt{a^2 + r_0^2 - 2 a r_0 \cos \psi}} + \frac{q}{4\pi\sqrt{a^2 + (r'_0)^2 - 2 a r_0' \cos \psi}}\\ g(r,r0)r=a=4πa2+r02−2ar0cosψ1+4πa2+(r0′)2−2ar0′cosψq
=14πr01+a2/r02−2(a/r0)cosψ+q4πa1+(r0′)2/a2−2(r0′/a)cosψ= \frac{1}{4\pi \red{r_0} \sqrt{\orange{1} + \green{a^2/r_0^2} - 2 (\green{a/r_0}) \orange{\cos \psi}}} + \frac{q}{4\pi \red{a} \sqrt{\orange{1} + \green{(r'_0)^2/a^2} - 2 (\green{r_0'/a}) \orange{\cos \psi}}} =4πr01+a2/r02−2(a/r0)cosψ1+4πa1+(r0′)2/a2−2(r0′/a)cosψq
qa=−1r0,ar0=r0′a⇒g(r,r0)∣r=a=0\frac{q}{\red{a}} = - \frac{1}{\red{r_0}}, \quad \green{\frac{a}{r_0}} = \green{\frac{r_0'}{a}}\quad \Rightarrow \quad g(\mathbf{r}, \mathbf{r}_0)\Big|_{r = a} = 0 aq=−r01,r0a=ar0′⇒g(r,r0)r=a=0
像电荷的性质
∣q∣=a/r0>1|q| = a/r_0 > 1∣q∣=a/r0>1，像电荷的绝对值大于单位物理电荷，才能在球面远端与物理电荷产生的电势抵消
r0′\mathbf{r}_0'r0′ 称为 r0\mathbf{r}_0r0 的对称点或者反演点
该格林函数只适用于球内：静电屏蔽，球外物理电势为零
球内满足齐次第一类边界的格林函数 G(r,r0)=g(r,r0)G(\mathbf{r}, \mathbf{r}_0) = g(\mathbf{r}, \mathbf{r}_0)G(r,r0)=g(r,r0)
G(r,r0)= 14πr2+r02−2rr0cosψ−a4πr0r2+(a2/r0)2−2ra2/r0cosψ= 14πr2+r02−2rr0cosψ−a4πr2r02+a4−2rr0a2cosψ\begin{align*} G(\mathbf{r}, \mathbf{r}_0) = & \ \frac{1}{4\pi \sqrt{r^2 + r_0^2 - 2 r r_0 \cos \psi}} - \frac{a}{4\pi r_0 \sqrt{r^2 + (a^2/r_0)^2 - 2 r a^2/r_0 \cos \psi}}\\ = & \ \frac{1}{4\pi \sqrt{r^2 + r_0^2 - 2 r r_0 \cos \psi}} - \frac{a}{4\pi \sqrt{r^2r_0^2 + a^4 - 2 r r_0 a^2 \cos \psi}}\\ \end{align*} G(r,r0)== 4πr2+r02−2rr0cosψ1−4πr0r2+(a2/r0)2−2ra2/r0cosψa 4πr2+r02−2rr0cosψ1−4πr2r02+a4−2rr0a2cosψa
验证 G(r,r0)=G(r0,r)G(\mathbf{r}, \mathbf{r}_0) = G(\mathbf{r}_0, \mathbf{r})G(r,r0)=G(r0,r)
感应电荷分布
σ(r∈S)=−∂g(r,r0)∂n朝内∣r=a= +∂g(r,r0)∂r∣r=a= r02−a24πa(a2+r02−2ar0cosψ)3/2\begin{align*} \sigma(\mathbf{r} \in S) = - \frac{\partial g(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}_\text{朝内}}\bigg|_{r = a} = & \ + \frac{\partial g(\mathbf{r}, \mathbf{r}_0)}{\partial r}\bigg|_{r = a}\\ = & \ \frac{r_0^2 - a^2}{4\pi a(a^2 + r_0^2 - 2ar_0 \cos \psi)^{3/2}} \end{align*} σ(r∈S)=−∂n朝内∂g(r,r0)r=a== +∂r∂g(r,r0)r=a 4πa(a2+r02−2ar0cosψ)3/2r02−a2
∫dSσ=∫σa2sinψdψdφ=−1 \int dS \sigma = \int \sigma a^2 \sin \psi d\psi d\varphi = \red{-1} ∫dSσ=∫σa2sinψdψdφ=−1
u(r)= (−)a2∮sinθ0dθ0dϕ0∂G(r,r0)∂n0f(θ0,ϕ0)= a(a2−r2)4π∮dθ0dϕ0sinθ0f(θ0,ϕ0)(a2+r2−2arcosψ)3/2\begin{align*} u(\mathbf{r}) = & \ (-) a^2 \oint \sin\theta_0 d\theta_0 d\phi_0 \red{\frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}_0}} f(\theta_0, \phi_0)\\ = & \ \frac{a(a^2 - r^2)}{4\pi} \oint d\theta_0 d\phi_0 \sin\theta_0 \frac{f(\theta_0, \phi_0) }{\purple{(a^2 + r^2 - 2 a r \cos \psi)^{3/2}}} \end{align*} u(r)== (−)a2∮sinθ0dθ0dϕ0∂n0∂G(r,r0)f(θ0,ϕ0) 4πa(a2−r2)∮dθ0dϕ0sinθ0(a2+r2−2arcosψ)3/2f(θ0,ϕ0)
∂G(r,r0)∂n0=∂G(r,r0)∂r0∣r0=a=r2−a24πa(a2+r2−2arcosψ)3/2\red{\frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial \mathbf{n}_0}} = \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial r_0}\bigg|_{r_0 = a} = \frac{r^2 - a^2}{4\pi a\purple{ (a^2 + r^2 - 2 a r \cos \psi)^{3/2}}} ∂n0∂G(r,r0)=∂r0∂G(r,r0)r0=a=4πa(a2+r2−2arcosψ)3/2r2−a2
对称性
根据对称性 G(r,r0)=G(r0,r)G(\mathbf{r}, \mathbf{r}_0) = G(\mathbf{r}_0, \mathbf{r})G(r,r0)=G(r0,r)，法向导数
∂G(r,r0)∂n0∣r0=a↔r↔r0∂G(r,r0)∂n∣r=a\frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0}\bigg|_{r_0 = a} \xleftrightarrow{\mathbf{r} \leftrightarrow \mathbf{r}_0} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n}\bigg|_{r = a} ∂n0∂G(r,r0)r0=ar↔r0∂n∂G(r,r0)r=a
r2−a24πa(a2+r2−2arcosψ)3/2→r↔r0r02−a24πa(a2+r02−2ar0cosψ)3/2 \frac{r^2 - a^2}{4\pi a\purple{ (a^2 + r^2 - 2 a r \cos \psi)^{3/2}}} \xrightarrow{\mathbf{r} \leftrightarrow \mathbf{r}_0} \frac{r_0^2 - a^2}{4\pi a(a^2 + r_0^2 - 2ar_0 \cos \psi)^{3/2}} 4πa(a2+r2−2arcosψ)3/2r2−a2r↔r04πa(a2+r02−2ar0cosψ)3/2r02−a2
夹角的说明
ψ\psiψ 是 r\mathbf{r}r 与 r0\mathbf{r}_0r0 的夹角：依赖积分变量 θ0,ϕ0\theta_0, \phi_0θ0,ϕ0 以及外部变量 r,θ,ϕr, \theta, \phir,θ,ϕ (r\mathbf{r}r 的位置)
−∇2u(ρ)=0,−∞<x<∞,0<y<∞ u(ρ)∣y=0=φ(x)\begin{align*} & \ - \nabla^2 u(\boldsymbol{\rho}) = 0, \quad -\infty < x < \infty, \quad 0 < y < \infty\\ & \ u(\boldsymbol{\rho})\big|_{y = 0} = \varphi(x) \end{align*} −∇2u(ρ)=0,−∞<x<∞,0<y<∞ u(ρ)y=0=φ(x)
格林函数方程与第一类齐次边界条件
−∇2G(ρ,ρ0) =δ(ρ−ρ0),−∞<x,x0<∞,0<y,y0<∞G(ρ,ρ0)∣y=0 =0\begin{align*} - \nabla^2 G(\boldsymbol{\rho}, \boldsymbol{\rho}_0) & \ = \delta(\boldsymbol{\rho} - \boldsymbol{\rho}_0), \quad -\infty < x, x_0 < \infty, \quad 0 < y, y_0 < \infty\\ G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)\big|_{y = 0}& \ = 0 \end{align*} −∇2G(ρ,ρ0)G(ρ,ρ0)y=0 =δ(ρ−ρ0),−∞<x,x0<∞,0<y,y0<∞ =0
物理背景
相当于在 ρ0\boldsymbol{\rho}_0ρ0 处放置一个电量为 Q=1Q = 1Q=1 的点电荷，xxx 轴接地
绿色区域是求解区域
镜像法：在下半平面放置一个等效点电荷，电量为 −q′-q'−q′，与物理电荷共同构建满足边界条件的电势
像电荷电量
很快可以证明 q′=1q' = 1q′=1
物理电荷在 ρ0=(x0,y0)\boldsymbol{\rho}_0 = (x_0, y_0)ρ0=(x0,y0)：像电荷在 ρ0′=(x0′,y0′)\boldsymbol{\rho}'_0= (x'_0, y'_0)ρ0′=(x0′,y0′)
像电荷位置
很快可以证明 (x0′,y0′)=(x0,−y0)(x'_0, y'_0) = (x_0, - y_0)(x0′,y0′)=(x0,−y0)
g(ρ,ρ0)= −12πln∣ρ−ρ0∣+q′2πln∣ρ−ρ0′∣+c= −14πln(∣ρ−ρ0∣2∣ρ−ρ0′∣2q′)+c=−14πln((x−x0)2+(y−y0)2((x−x0′)2+(y−y0′)2)q′)+c\begin{align*} g(\boldsymbol{\rho}, \boldsymbol{\rho}_0) = & \ - \frac{1}{2\pi} \ln |\boldsymbol{\rho} - \boldsymbol{\rho}_0| + \frac{q'}{2\pi} \ln |\boldsymbol{\rho} - \boldsymbol{\rho}'_0| + c\\ = & \ \frac{- 1}{4\pi} \ln \left({\frac{|\boldsymbol{\rho} - \boldsymbol{\rho}_0|^2}{|\boldsymbol{\rho} - \boldsymbol{\rho}'_0|^{2q'}}}\right) + c = \frac{-1}{4\pi} \ln \left({\frac{(x - x_0)^2 + (y - y_0)^2}{((x - x'_0)^2 + (y - y'_0)^2)^{q'}}}\right) + c \end{align*} g(ρ,ρ0)== −2π1ln∣ρ−ρ0∣+2πq′ln∣ρ−ρ0′∣+c 4π−1ln(∣ρ−ρ0′∣2q′∣ρ−ρ0∣2)+c=4π−1ln(((x−x0′)2+(y−y0′)2)q′(x−x0)2+(y−y0)2)+c
g((x,0),(x0′,y0′))=−14πln((x−x0)2+y02[(x−x0′)2+(y0′)2]q′)+c=0g((x, 0), (x_0', y_0')) = - \frac{1}{4\pi} \ln \left({\frac{(x - x_0)^2 + y_0^2}{[(x - x'_0)^{2} + (y'_0)^2]^{q'}}}\right) + c = 0 g((x,0),(x0′,y0′))=−4π1ln([(x−x0′)2+(y0′)2]q′(x−x0)2+y02)+c=0
dndxn∣x=0g((x,0),(x0′,y0′))=0,n=0,1,2,3,⋯\frac{d^n}{dx^n} \purple{\bigg|_{x = 0}}g((x, 0), (x_0', y_0')) = 0, \qquad n = 0 , 1, 2, 3, \cdots dxndnx=0g((x,0),(x0′,y0′))=0,n=0,1,2,3,⋯
−2x0x02+y02+2qx0′x0′2+y0′2=0,2q(x0′−y0′)(x0′+y0′)(x0′2+y0′2)2−2(x0−y0)(x0+y0)(x02+y02)2= 04qx0′(x0′2−3y0′2)(x0′2+y0′2)3−4x0(x02−3y02)(x02+y02)3= 0\begin{align*} - \frac{2x_0}{x_0^2 + y_0^2} + \frac{2q x'_0}{x_0'^2 + y_0'^2} = 0, \quad \frac{2 q (x_0'-y_0') (x_0'+y_0')}{\left(x_0'^2+y_0'^2\right)^2}-\frac{2 (x_0-y_0) (x_0+y_0)}{\left(x_0^2+y_0^2\right)^2} = & \ 0\\ \frac{4 q x_0' \left(x_0'^2-3 y_0'^2\right)}{\left(x_0'^2+y_0'^2\right)^3}-\frac{4 x_0 \left(x_0^2-3 y_0^2\right)}{\left(x_0^2+y_0^2\right)^3} = & \ 0 \end{align*} −x02+y022x0+x0′2+y0′22qx0′=0,(x0′2+y0′2)22q(x0′−y0′)(x0′+y0′)−(x02+y02)22(x0−y0)(x0+y0)=(x0′2+y0′2)34qx0′(x0′2−3y0′2)−(x02+y02)34x0(x02−3y02)= 0 0
x0′=x0,y0′=−y0,q′=1x_0' = x_0, \qquad y_0' = - y_0, \qquad q' = 1 x0′=x0,y0′=−y0,q′=1
G(ρ,ρ0)=g(ρ,ρ0)=14πln((x−x0)2+(y−y0)2(x−x0)2+(y+y0)2)G(\boldsymbol{\rho}, \boldsymbol{\rho}_0) = g(\boldsymbol{\rho}, \boldsymbol{\rho}_0) = \frac{1}{4\pi} \ln \left({\frac{(x - x_0)^2 + (y - y_0)^2}{(x - x_0)^2 + (y + y_0)^2}}\right) G(ρ,ρ0)=g(ρ,ρ0)=4π1ln((x−x0)2+(y+y0)2(x−x0)2+(y−y0)2)
σ(x)=−∂G(ρ,ρ0)∂y∣y=0=−y0π1(x−x0)2+y02\sigma(x) = - \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial y}\bigg|_{y = 0} = - \frac{y_0}{\pi} \frac{1}{(x - x_0)^2 + y_0^2} σ(x)=−∂y∂G(ρ,ρ0)y=0=−πy0(x−x0)2+y021
感应电荷密度函数
求格林函数的方向导数
边界法向说明
边界法向 n⃗\vec nn 朝着 −y-y−y 方向
∂G(ρ,ρ0)∂n0∣y0=0=−∂G(ρ,ρ0)∂y0∣y0=0=−yπ1(x−x0)2+y2\frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial n_0}\bigg|_{y_0 = 0} = - \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial y_0}\bigg|_{y_0 = 0} = - \frac{y}{\pi} \frac{1}{(x - x_0)^2 + y^2} ∂n0∂G(ρ,ρ0)y0=0=−∂y0∂G(ρ,ρ0)y0=0=−πy(x−x0)2+y21
上半平面的解，ρ=(x,y)\boldsymbol{\rho} = (x, y)ρ=(x,y)
u(ρ)=−∫∂Dφ(ρ)∂G(ρ,ρ0)∂n0dS0=∫−∞+∞dx0yπ1(x−x0)2+y2φ(x0)u(\boldsymbol{\rho}) = - \int_{\partial D} \varphi(\rho) \frac{\partial G(\rho, \rho_0)}{\partial n_0}dS_0 = \int_{-\infty}^{+\infty} dx_0 \frac{y}{\pi} \frac{1}{(x - x_0)^2 + y^2} \varphi(x_0) u(ρ)=−∫∂Dφ(ρ)∂n0∂G(ρ,ρ0)dS0=∫−∞+∞dx0πy(x−x0)2+y21φ(x0)
边界条件的验证
利用
limy→0yπ((x−x0)2+y2)=δ(x−x0)\lim_{y \to 0} \frac{y}{\pi((x - x_0)^2 + y^2)} = \delta(x - x_0) y→0limπ((x−x0)2+y2)y=δ(x−x0)
可以验证 u(x,0)=φ(x)u(x, 0) = \varphi(x)u(x,0)=φ(x)
平面上圆内 Laplace 方程第一边值问题
∇2u(ρ)=0,ρ=∣ρ∣<a,u(ρ)∣ρ=a=f(φ)\begin{align*} & \nabla^2 u(\boldsymbol{\rho}) = 0, \qquad \rho = |\boldsymbol{\rho}| < a, \qquad & u(\rho)|_{\rho = a} = f(\varphi) \end{align*} ∇2u(ρ)=0,ρ=∣ρ∣<a,u(ρ)∣ρ=a=f(φ)
格林函数方程与边界条件
∇2G(ρ,ρ0)=−δ(ρ−ρ0),ρ,ρ0<a,G(ρ,ρ0)∣ρ=a=0\begin{align*} & \nabla^2 G(\boldsymbol{\rho}, \boldsymbol{\rho}_0) = - \delta(\boldsymbol{\rho} - \boldsymbol{\rho}_0), \qquad \rho, \rho_0 < a,\\ & G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)|_{\rho = a} = 0 \end{align*} ∇2G(ρ,ρ0)=−δ(ρ−ρ0),ρ,ρ0<a,G(ρ,ρ0)∣ρ=a=0
圆内 (ρ0,φ0)(\rho_0, \varphi_0)(ρ0,φ0) 放置一个电量为 Q=1Q = 1Q=1 的点电荷，圆导体圈接地
在 ρ0′\boldsymbol{\rho}'_0ρ0′ 放置电量为 −q′- q'−q′ 的假想镜像电荷
在 ρ0′\boldsymbol{\rho}'_0ρ0′ 放置电量为 −q′\orange{- q'}−q′ 的点电荷，则圆内电势叠加
g(ρ)= −12πln∣ρ−ρ0∣−−q′2πln∣ρ−ρ0′∣+c=−14πln∣ρ−ρ0∣2∣ρ−ρ0′∣2q+c= −14πln(ρ2+ρ02−2ρρ0cosψ(ρ2+ρ0′2−2ρρ0′cosψ)q′)+c\begin{align*} g(\boldsymbol{\rho}) = & \ - \frac{1}{2\pi} \ln |\boldsymbol{\rho} - \boldsymbol{\rho}_0| - \frac{\orange{ - q'}}{2\pi} \ln |\boldsymbol{\rho} - \boldsymbol{\rho}'_0| + c = - \frac{1}{4\pi} \ln \frac{|\boldsymbol{\rho} - \boldsymbol{\rho}_0|^2}{|\boldsymbol{\rho} - \boldsymbol{\rho}'_0|^{2q}} + \purple{c}\\ = & \ - \frac{1}{4\pi} \ln \left({\frac{\rho^2 + \rho_0^2 - 2 \rho \rho_0 \cos \psi}{(\rho^2 + \rho_0'^2 - 2 \rho \rho_0' \cos \psi)^{q'}}}\right) + c \end{align*} g(ρ)== −2π1ln∣ρ−ρ0∣−2π−q′ln∣ρ−ρ0′∣+c=−4π1ln∣ρ−ρ0′∣2q∣ρ−ρ0∣2+c −4π1ln((ρ2+ρ0′2−2ρρ0′cosψ)q′ρ2+ρ02−2ρρ0cosψ)+c
常数项和对称性假设
ccc 是待定常数。根据问题对称性，已经假设 ρ0∥ρ0′\boldsymbol{\rho}_0 \parallel \boldsymbol{\rho}'_0ρ0∥ρ0′ 且同侧
g(ρ,ρ0)∣ρ=a=14πln(a2+ρ02−2aρ0cosψ(a2+ρ0′2−2aρ0′cosψ)q′)+c=0g(\boldsymbol{\rho}, \boldsymbol{\rho}_0)\big|_{\rho = a} = \frac{1}{4\pi} \ln \left({\frac{a^2 + \rho_0^2 - 2 a \rho_0 \cos \psi}{(a^2 + \rho_0'^2 - 2 a \rho_0' \cos \psi)^{q'}}}\right) + c = 0 g(ρ,ρ0)ρ=a=4π1ln((a2+ρ0′2−2aρ0′cosψ)q′a2+ρ02−2aρ0cosψ)+c=0
左边对数表达式应当与 ψ\psiψ 无关。
dndφn∣φ=0ln(a2+ρ02−2aρ0cosψ(a2+ρ0′2−2aρ0′cosψ)q′)=0,n=1,2,⋯ ,∞\frac{d^n}{d \varphi^n}\bigg|_{\varphi = 0} \ln \left({\frac{a^2 + \rho_0^2 - 2 a \rho_0 \cos \psi}{(a^2 + \rho_0'^2 - 2 a \rho_0' \cos \psi)^{q'}}}\right) = 0, \qquad n = 1, 2, \cdots, \infty dφndnφ=0ln((a2+ρ0′2−2aρ0′cosψ)q′a2+ρ02−2aρ0cosψ)=0,n=1,2,⋯,∞
这无穷多个等式可以确定 ρ0′\rho_0'ρ0′ 与 ρ0\rho_0ρ0 的关系
ρ0′=a2ρ0,q′=1,⇒ρ0′=a2ρ02ρ0\rho_0' = \frac{a^2}{\rho_0}, \qquad q' = 1, \qquad \Rightarrow \qquad \purple{\boldsymbol{\rho}_0' = \frac{a^2}{\rho_0^2} \boldsymbol{\rho}_0} ρ0′=ρ0a2,q′=1,⇒ρ0′=ρ02a2ρ0
像电荷的位置和电量
在圆外反演点放置一个等效点电荷，电量为 −1-1−1
−14πln(a2+ρ02−2aρ0cosψa2+(a2/ρ0)2−2a(a2/ρ0)cosψ)+c=12πlnaρ0+c=0- \frac{1}{4\pi}\ln \left({\frac{a^2 + \rho_0^2 - 2 a \rho_0 \cos \psi}{a^2 + (a^2/\rho_0)^2 - 2 a (a^2/\rho_0) \cos \psi}}\right) + c = \frac{1}{2\pi} \ln \frac{a}{\rho_0} + c = 0 −4π1ln(a2+(a2/ρ0)2−2a(a2/ρ0)cosψa2+ρ02−2aρ0cosψ)+c=2π1lnρ0a+c=0
G(ρ,ρ0)=−14πln(∣ρ−ρ0∣2∣ρ−ρ0′∣2a2ρ02),ρ0′=a2ρ02ρ0G(\boldsymbol{\rho}, \boldsymbol{\rho}_0) = - \frac{1}{4\pi} \ln \left({\frac{|\boldsymbol{\rho} - \boldsymbol{\rho}_0|^2}{|\boldsymbol{\rho} - \purple{\boldsymbol{\rho}'_0}|^2} \orange{\frac{a^2}{\rho_0^2}}}\right), \qquad \purple{\boldsymbol{\rho}_0' = \frac{a^2}{\rho_0^2} \boldsymbol{\rho}_0} G(ρ,ρ0)=−4π1ln(∣ρ−ρ0′∣2∣ρ−ρ0∣2ρ02a2),ρ0′=ρ02a2ρ0
等势线分布
G(ρ,ρ0)=14πln(a2+ρ02−2aρ0cosψa2+a4ρ02−2aa2ρ0cosψ)−12πlnaρ0=14πln(a2(ρ2+ρ02−2ρρ0cosψ)a4+ρ2ρ02−2a2ρρ0cosψ)G(\boldsymbol{\rho}, \boldsymbol{\rho}_0) = \frac{1}{4\pi} \ln \left({\frac{a^2 + \rho_0^2 - 2 a \rho_0 \cos \psi}{a^2 + \frac{a^4}{\rho_0^2} - 2 a \frac{a^2}{\rho_0} \cos \psi}}\right) - \frac{1}{2\pi}\ln \frac{a}{\rho_0}\\ = \frac{1}{4\pi} \ln \bigg( \frac{a^2(\rho^2 + \rho_0^2 - 2\rho \rho_0\cos \psi)}{a^4 + \rho^2 \rho_0^2 - 2 a^2 \rho \rho_0 \cos \psi} \bigg) G(ρ,ρ0)=4π1ln(a2+ρ02a4−2aρ0a2cosψa2+ρ02−2aρ0cosψ)−2π1lnρ0a=4π1ln(a4+ρ2ρ02−2a2ρρ0cosψa2(ρ2+ρ02−2ρρ0cosψ))
∂G(ρ,ρ0)∂n0∣ρ0=a=∂G(ρ,ρ0)∂ρ0∣ρ0=a=ρ2−a22πa(a2+ρ2−2aρcosψ)\frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial n_0}\bigg|_{\rho_0 = a} = \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial \rho_0}\bigg|_{\rho_0 = a} = \frac{\red{\rho^2 - a^2}}{2\pi \red{a} (a^2 + \rho^2 - 2 a \rho \cos \psi)} ∂n0∂G(ρ,ρ0)ρ0=a=∂ρ0∂G(ρ,ρ0)ρ0=a=2πa(a2+ρ2−2aρcosψ)ρ2−a2
−∂G(ρ,ρ0)∂n∣ρ=a=−∂G(ρ,ρ0)∂ρ∣ρ=a=−ρ02−a22πa(a2+ρ02−2aρ0cosψ)- \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial n}\bigg|_{\rho = a} = - \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial \rho}\bigg|_{\rho = a} = - \frac{\red{\rho_0^2 - a^2}}{2\pi \red{a} (a^2 + \rho_0^2 - 2 a \rho_0 \cos \psi)} −∂n∂G(ρ,ρ0)ρ=a=−∂ρ∂G(ρ,ρ0)ρ=a=−2πa(a2+ρ02−2aρ0cosψ)ρ02−a2
ψ=φ−φ0\psi = \varphi - \varphi_0ψ=φ−φ0
圆内解为 (ρ=(ρ,φ)\boldsymbol{\rho} = (\rho, \varphi)ρ=(ρ,φ))
u(ρ)= −∫∂Df(ρ0)∂G(ρ,ρ0)∂n0dS0= ∫02πdφ0a2−ρ22πf(φ0)a2+ρ2−2aρcos(φ−φ0)\begin{align*} u(\boldsymbol{\rho}) = & \ \red{\boldsymbol{-}} \int_{\partial D} f(\boldsymbol{\rho}_0) \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial n_0} dS_0\\ = & \ \int_0^{2\pi} d\varphi_0 \frac{\red{a^2 - \rho^2}}{2\pi}\frac{f(\varphi_0)}{a^2 + \rho^2 - 2 a \rho \cos(\green{\varphi - \varphi_0})} \end{align*} u(ρ)== −∫∂Df(ρ0)∂n0∂G(ρ,ρ0)dS0 ∫02πdφ02πa2−ρ2a2+ρ2−2aρcos(φ−φ0)f(φ0)
积分变量说明
其中 ψ=φ−φ0\psi = \green{\varphi - \varphi_0}ψ=φ−φ0，dS0=adφ0dS_0 = \red{a} d\varphi_0dS0=adφ0，aaa 跟 GGG 导数分母抵消
∇r2u(r)=f(r),r∈D,u(r)∣∂D=φ(r)\nabla^2_\mathbf{r} u(\mathbf{r}) = \green{f(\mathbf{r})}, \quad \mathbf{r} \in D, \qquad u(\mathbf{r})|_{\partial D} = \red{\varphi(\mathbf{r})} ∇r2u(r)=f(r),r∈D,u(r)∣∂D=φ(r)
∇r2G(r,r0)=−δ(r−r0),r,r0∈D,G(r,r0)∣∂D=0\nabla^2_\mathbf{r} G(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0), \quad \mathbf{r}, \mathbf{r}_0 \in D, \qquad G(\mathbf{r}, \mathbf{r}_0)|_{\partial D} = 0 ∇r2G(r,r0)=−δ(r−r0),r,r0∈D,G(r,r0)∣∂D=0
泊松方程 + 第一边值问题最终解
u(r)=∫Ddr0G(r,r0)f(r0)−∮∂DdS0φ(r0)∂G(r,r0)∂n0u(\mathbf{r}) = \int_D d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) \green{f(\mathbf{r}_0)} - \oint_{\partial D} dS_0 \red{\varphi(\mathbf{r}_0)} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0} u(r)=∫Ddr0G(r,r0)f(r0)−∮∂DdS0φ(r0)∂n0∂G(r,r0)
两个积分分别负责一种非齐次性
系统可以拆分成两个子系统：总电势是子系统电势的直接叠加
子系统 1 可以进一步拆分成无穷多个点电荷的电势叠加
每一个点电荷产生的电势分布恰好为 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0)
u1(r)=∫dr0f(r0)G(r,r0)u_1(\mathbf{r}) = \int d\mathbf{r}_0 f(\mathbf{r}_0) G(\mathbf{r}, \mathbf{r}_0) u1(r)=∫dr0f(r0)G(r,r0)
子系统 2：边界上设有表面电势 φ(r)\varphi(\mathbf{r})φ(r)
依然可以用第一类齐次边界条件的格林函数 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0) 来求解
u2(r)=−∮∂DdS0φ(r0)∂G(r,r0)∂n0u_2(\mathbf{r}) = - \oint_{\partial D} dS_0 \varphi(\mathbf{r}_0) \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0} u2(r)=−∮∂DdS0φ(r0)∂n0∂G(r,r0)
物理图像复杂：涉及电偶极矩薄层 (dipole layer)，位置由 r0∈∂D\mathbf{r}_0 \in \partial Dr0∈∂D 标记
推论：泊松方程 + 第一类边界条件问题的解存在且唯一
就由那两个积分给出。
此事在第 7 章亦有记载。
推论：拉普拉斯方程 + 第一类齐次边界条件的解存在且 平凡
∇2u=0,u∣∂D=0⇒u=0\nabla^2 u = 0, \quad u|_{\partial D} = 0 \qquad \Rightarrow \qquad u = 0 ∇2u=0,u∣∂D=0⇒u=0
因为 f=0f = 0f=0，φ=0\varphi = 0φ=0，
u(r)=∫Ddr0G(r,r0)f(r0)−∮∂DdS0φ(r0)∂G(r,r0)∂n0=0u(\mathbf{r}) = \int_D d\mathbf{r}_0 G(\mathbf{r}, \mathbf{r}_0) \green{f(\mathbf{r}_0)} - \oint_{\partial D} dS_0 \red{\varphi(\mathbf{r}_0)} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0} = 0 u(r)=∫Ddr0G(r,r0)f(r0)−∮∂DdS0φ(r0)∂n0∂G(r,r0)=0
推论：拉普拉斯方程 + 第一类非齐次常数边界条件的解存在且 几乎平凡
∇2u=0,u∣∂D=u0⇒u=u0\nabla^2 u = 0, \quad u|_{\partial D} = u_0 \qquad \Rightarrow \qquad u = u_0 ∇2u=0,u∣∂D=u0⇒u=u0
可以直接验证。电势整体平移不影响电场分布。
∮∂D∂G(r,r0)∂ndS=−1\oint_{\partial D} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n} dS = -1 ∮∂D∂n∂G(r,r0)dS=−1
u(r)=−∮∂DdS01∂G(r,r0)∂n0=1u(\mathbf{r}) = - \oint_{\partial D} dS_0 \red{1} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0} = 1 u(r)=−∮∂DdS01∂n0∂G(r,r0)=1
说明
φ(r∈∂D)=1⇒u(r∈D)=1\varphi(\mathbf{r}\in \partial D) = 1 \quad \Rightarrow \quad u(\mathbf{r} \in D) = 1 φ(r∈∂D)=1⇒u(r∈D)=1
⇒u(r)=−∫∂Ddr0(1×)∂G(r,r0)∂n0dS0=1\Rightarrow u(\mathbf{r}) = - \int_{\partial D} d\mathbf{r}_0 (1 \times) \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0} dS_0 = 1 ⇒u(r)=−∫∂Ddr0(1×)∂n0∂G(r,r0)dS0=1
∫∂D∂G(r,r0)∂n0dS0=∫∂D∂G(r0,r)∂n0dS0=−1\int_{\partial D} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0}dS_0 = \int_{\partial D} \frac{\partial G(\mathbf{r}_0, \mathbf{r})}{\partial n_0}dS_0 = -1 ∫∂D∂n0∂G(r,r0)dS0=∫∂D∂n0∂G(r0,r)dS0=−1
∫∂D∂G(r,r0)∂ndS=−1\int_{\partial D} \frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n}dS = -1 ∫∂D∂n∂G(r,r0)dS=−1
∂G(r,r0)∂n0=∇r0G(r,r0)⋅n\frac{\partial G(\mathbf{r}, \mathbf{r}_0)}{\partial n_0} = \nabla_{\mathbf{r}_0}G(\mathbf{r}, \mathbf{r}_0) \cdot \mathbf{n} ∂n0∂G(r,r0)=∇r0G(r,r0)⋅n
∇=ex∂∂x+ey∂∂y+ez∂∂z\nabla = \mathbf{e}_x \frac{\partial}{\partial x} + \mathbf{e}_y \frac{\partial}{\partial y} + \mathbf{e}_z \frac{\partial}{\partial z} ∇=ex∂x∂+ey∂y∂+ez∂z∂
∇=eρ∂∂ρ+eφ1ρ∂∂φ+ez∂∂z\nabla = \mathbf{e}_\rho\frac{\partial }{\partial \rho} + \mathbf{e}_\varphi\frac{1}{\rho}\frac{\partial }{\partial \varphi} + \mathbf{e}_z\frac{\partial }{\partial z} ∇=eρ∂ρ∂+eφρ1∂φ∂+ez∂z∂
∇=er∂∂r+eθ1r∂∂θ+eφ1rsinθ∂∂φ\nabla = \mathbf{e}_r\frac{\partial}{\partial r} + \mathbf{e}_\theta \frac{1}{r}\frac{\partial}{\partial \theta} + \mathbf{e}_\varphi \frac{1}{r\sin\theta}\frac{\partial}{\partial \varphi} ∇=er∂r∂+eθr1∂θ∂+eφrsinθ1∂φ∂
∇ρ0G(ρ,ρ0)=eρ0∂G(ρ,ρ0)∂ρ0+eφ01ρ0∂G(ρ,ρ0)∂φ0\nabla_{\boldsymbol{\rho}_0} G(\boldsymbol{\rho}, \boldsymbol{\rho}_0) = \mathbf{e}_{\rho_0}\frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial \rho_0} + \mathbf{e}_{\varphi_0}\frac{1}{\rho_0} \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial \varphi_0} ∇ρ0G(ρ,ρ0)=eρ0∂ρ0∂G(ρ,ρ0)+eφ0ρ01∂φ0∂G(ρ,ρ0)
⇒∂G(ρ,ρ0)∂n0=∇ρ0G(ρ,ρ0)⋅n0=∂G(ρ,ρ0)∂ρ0eρ0⋅eρ0=∂G(ρ,ρ0)∂r0\Rightarrow \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial n_0} = \nabla_{\boldsymbol{\rho}_0} G(\boldsymbol{\rho}, \boldsymbol{\rho}_0) \cdot \mathbf{n}_0 = \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial \rho_0} \mathbf{e}_{\rho_0} \cdot \mathbf{e}_{\rho_0} = \frac{\partial G(\boldsymbol{\rho}, \boldsymbol{\rho}_0)}{\partial r_0} ⇒∂n0∂G(ρ,ρ0)=∇ρ0G(ρ,ρ0)⋅n0=∂ρ0∂G(ρ,ρ0)eρ0⋅eρ0=∂r0∂G(ρ,ρ0)
难度的流动
泊松方程 + 第一类边界条件问题 →\to→ 第一类齐次边界条件 的格林函数求解
镜像法
镜像法具有读博猜测的成分
可以根据对称性、直觉以及边界条件的基本要求来提高命中精度
像电荷必须位于求解区域 以外
如果区域边界复杂，像电荷还会有自己的像电荷
可能会因此产生无穷多像电荷
当代拉马努金：直接看出格林函数形式
考虑 nnn 维无界欧式空间中的泊松方程
∇2u(r)=−f(r),r=(x1,⋯ ,xn)∈Rn\nabla^2 u(\mathbf{r}) = - f(\mathbf{r}), \quad \mathbf{r} = (x_1, \cdots, x_n) \in \mathbb{R}^n ∇2u(r)=−f(r),r=(x1,⋯,xn)∈Rn
其基本解满足格林函数方程
∇r2G(r,r0)=−δ(r−r0)\nabla^2_\mathbf{r} G(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0) ∇r2G(r,r0)=−δ(r−r0)
可以暂时不添加边界条件，等需要做决定再添加
未知函数 G(r,r0)G(\mathbf{r}, \mathbf{r}_0)G(r,r0) 关于 r\mathbf{r}r 的傅里叶 (逆) 变换/傅里叶展开
G(r,r0)= ∫dk1⋯dkn2πnG~r0(k1,k2,⋯ ,kn)ei(k1x1+k2x2+⋯+knxn)= ∫dnk(2π)n/2G~r0(k)eik⋅r\begin{align*} G(\mathbf{r}, \mathbf{r}_0) = & \ \int \frac{d k_1 \cdots d k_n}{\sqrt{2\pi}^n} \widetilde{G}_{\mathbf{r}_0}( k_1, k_2, \cdots, k_n) e^{i ( k_1 {x}_1 + k_2 {x}_2 + \cdots + k_n {x}_n)} \\ = & \ \int \frac{d^n \mathbf{k}}{(2\pi)^{n/2}} \widetilde{G}_{\mathbf{r}_0}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} \end{align*} G(r,r0)== ∫2πndk1⋯dknGr0(k1,k2,⋯,kn)ei(k1x1+k2x2+⋯+knxn) ∫(2π)n/2dnkGr0(k)eik⋅r
这里 r0\mathbf{r}_0r0 看成一个固定的参数。用第 8 章的语言：
G(r,r0)↔G~r0(k)G(\mathbf{r}, \mathbf{r}_0) \leftrightarrow \widetilde{G}_{\mathbf{r}_0}(\mathbf{k}) G(r,r0)↔Gr0(k)
δ\deltaδ-函数的傅里叶 (逆) 变换/傅里叶展开
δ(r)=∫dk1⋯dkn2πn12πnei(k1x1+k2x2+⋯+knxn)=∫dnk(2π)n/212πneik⋅r\delta(\mathbf{r}) = \int \frac{d k_1 \cdots d k_n}{\sqrt{2\pi}^n} \frac{1}{\sqrt{2\pi}^n} e^{i ( k_1 {x}_1 + k_2 {x}_2 + \cdots + k_n {x}_n)} = \int \frac{d^n \mathbf{k}}{(2\pi)^{n/2}} \frac{1}{\sqrt{2\pi}^n} e^{i \mathbf{k} \cdot \mathbf{r}} δ(r)=∫2πndk1⋯dkn2πn1ei(k1x1+k2x2+⋯+knxn)=∫(2π)n/2dnk2πn1eik⋅r
用笛卡尔坐标书写 δ(r)=δ(x1)δ(x2)⋯δ(xn)\delta(\mathbf{r}) = \delta(x_1)\delta(x_2) \cdots \delta(x_n)δ(r)=δ(x1)δ(x2)⋯δ(xn)
用第 8 章的符号
δ(r)↔Δ(k)=1(2π)n/2\delta(\mathbf{r}) \leftrightarrow \Delta(\mathbf k) = \frac{1}{(2\pi)^{n/2}} δ(r)↔Δ(k)=(2π)n/21
坐标平移
δ(r−r0)=∫dnk(2π)n/212πneik⋅(r−r0)\delta(\mathbf{r} - \mathbf{r}_0) = \int \frac{d^n \mathbf{k}}{(2\pi)^{n/2}} \frac{1}{\sqrt{2\pi}^n} e^{i \mathbf{k} \cdot (\mathbf{r} - \mathbf{r}_0)} δ(r−r0)=∫(2π)n/2dnk2πn1eik⋅(r−r0)
δ(r−r0)↔Δ(k)e−ik⋅r0\delta(\mathbf{r} - \mathbf{r}_0) \leftrightarrow \Delta(\mathbf{k})e^{ - i \mathbf{k} \cdot \mathbf{r}_0} δ(r−r0)↔Δ(k)e−ik⋅r0
傅里叶变换将微分运算转化为代数运算
∇r2G(r,r0)=∫dk1⋯dkn2πnG~r0(k)(−k2)eik⋅r\nabla^2_\mathbf{r}G(\mathbf{r}, \mathbf{r}_0) = \int \frac{d k_1 \cdots d k_n}{\sqrt{2\pi}^n} \widetilde{G}_{\mathbf{r}_0}(\mathbf{k}) (- \mathbf k^2) e^{i \mathbf{k} \cdot \mathbf{r}} ∇r2G(r,r0)=∫2πndk1⋯dknGr0(k)(−k2)eik⋅r
因为
∂xie−ikixi=−ikie−ikixi⇒∂xi2e−ikixi=−ki2e−ikixi\partial_{x_i} e^{- i k_i x_i} = - i k_i e^{- i k_i x_i} \quad \Rightarrow \quad \partial_{x_i}^2 e^{- i k_i x_i} = - k_i^2 e^{- i k_i x_i} ∂xie−ikixi=−ikie−ikixi⇒∂xi2e−ikixi=−ki2e−ikixi
从而
∇r2e−ik⋅r=−∑i=1nki2e−ik⋅r=−k2e−ik⋅r\nabla^2_\mathbf{r} e^{- i \mathbf{k}\cdot \mathbf{r}} = - \sum_{i=1}^{n} k_i^2 e^{- i \mathbf{k}\cdot \mathbf{r}} = - \mathbf{k}^2 e^{- i \mathbf{k}\cdot \mathbf{r}} ∇r2e−ik⋅r=−i=1∑nki2e−ik⋅r=−k2e−ik⋅r
∇r2G(r,r0)=−δ(r−r0)\nabla^2_\mathbf{r}G(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0) ∇r2G(r,r0)=−δ(r−r0)
⇒−k2G~r0(k)=−Δ(k)e−ik⋅r0=−1(2π)n/2e−ik⋅r0\Rightarrow \quad - \mathbf{k}^2 \widetilde{G}_{\mathbf{r}_0}(\mathbf{k}) = - \Delta(\mathbf{k}) e^{- i \mathbf{k} \cdot \mathbf{r}_0} = - \frac{1}{(2\pi)^{n/2}} e^{- i \mathbf{k} \cdot \mathbf{r}_0} ⇒−k2Gr0(k)=−Δ(k)e−ik⋅r0=−(2π)n/21e−ik⋅r0
G~r0(k)=1k21(2π)n/2e−ik⋅r0\widetilde{G}_{\mathbf{r}_0}(\mathbf{k}) = \frac{1}{\mathbf{k}^2 } \frac{1}{(2\pi)^{n/2}} e^{ - i \mathbf{k} \cdot \mathbf{r}_0} Gr0(k)=k21(2π)n/21e−ik⋅r0
G(r,r0)=∫dnk(2π)n1k2eik⋅(r−r0)G(\mathbf{r}, \mathbf{r}_0) = \int \frac{d^n \mathbf{k}}{(2\pi)^{n}} \frac{1}{\mathbf{k}^2} e^{i \mathbf{k} \cdot (\mathbf{r} - \mathbf{r}_0)} G(r,r0)=∫(2π)ndnkk21eik⋅(r−r0)
G(x,x0)=∫−∞+∞dk2π1k2eik(x−x0)G(x, x_0) = \int_{-\infty}^{+\infty} \frac{dk}{2\pi} \frac{1}{k^2} e^{i k (x - x_0)} G(x,x0)=∫−∞+∞2πdkk21eik(x−x0)
能否用之前介绍的留数定理方法计算？ 不能，因为在实轴上有二阶极点
这个积分的主值确实是 发散 的
称为红外发散，因为来自低能区 k→0k \to 0k→0 的贡献
Gm(x,x0)=∫−∞+∞dk2π1k2+m2eik(x−x0)G_m(x, x_0) = \int_{-\infty}^{+\infty} \frac{dk}{2\pi} \frac{1}{k^2 + m^2} e^{i k (x - x_0)} Gm(x,x0)=∫−∞+∞2πdkk2+m21eik(x−x0)
∫−∞+∞dk2π1k2+m2eik(x−x0)=2πi2πResk=im(eik(x−x0)k2+m2)=12me−m(x−x0)\int_{-\infty}^{+\infty} \frac{dk}{2\pi} \frac{1}{k^2 + m^2} e^{i k (x - x_0)} = \frac{2\pi i}{2\pi} \operatorname{Res}_{k = im} \left({\frac{e^{i k (x - x_0)}}{k^2 + m^2}}\right) = \frac{1}{2m} e^{- m (x - x_0 )} ∫−∞+∞2πdkk2+m21eik(x−x0)=2π2πiResk=im(k2+m2eik(x−x0))=2m1e−m(x−x0)
∫dk2π1k2+m2eik(x−x0)=∫dk2π1(−k)2+m2e+i(−k)(x−x0)=12me−m(x0−x)\int \frac{dk}{2\pi} \frac{1}{k^2 + m^2}e^{i k (x - x_0)} = \int \frac{dk}{2\pi} \frac{1}{(-k)^2 + m^2} e^{+ i (-k) (x - x_0)} = \frac{1}{2m} e^{- m (x_0 - x )} ∫2πdkk2+m21eik(x−x0)=∫2πdk(−k)2+m21e+i(−k)(x−x0)=2m1e−m(x0−x)
Gm(x,x0)=12me−m∣x−x0∣G_m(x, x_0) = \frac{1}{2m}e^{-m |x - x_0 |} Gm(x,x0)=2m1e−m∣x−x0∣
Gm(x,x0)=12m−m2m∣x−x0∣+O(m)G_m(x, x_0) = \frac{1}{2m} - \frac{m}{2m}|x - x_0| + O(m) Gm(x,x0)=2m1−2mm∣x−x0∣+O(m)
G(x,x0)=Regularm→0Gm(x,x0)=−12∣x−x0∣G(x, x_0) = \operatorname{Regular}_{m \to 0} G_m(x, x_0) = - \frac{1}{2} |x - x_0| G(x,x0)=Regularm→0Gm(x,x0)=−21∣x−x0∣
其它解哪里去了？
格林函数方程是二阶线性常微分方程：有两个积分常数
k2G~x0(k)=12πe−ikx0⇏G~x0(k)=12πk2e−ikx0k^2 \widetilde{G}_{x_0}(k) = \frac{1}{\sqrt{2\pi}} e^{ - ikx_0} \qquad \red{\not \Rightarrow \qquad \widetilde{G}_{x_0}(k) = \frac{1}{\sqrt{2\pi} k^2}e^{ - ikx_0}} k2Gx0(k)=2π1e−ikx0⇒Gx0(k)=2πk21e−ikx0
中小学生：？？？
k2δ(k)=0,k2δ′(k)=0k^2 \delta(k) = 0, \qquad k^2 \delta'(k) = 0 k2δ(k)=0,k2δ′(k)=0
n=2n = 2n=2
G(r,r0)=∫d2k(2π)21k2eik⋅(r−r0)G(\mathbf{r}, \mathbf{r}_0) = \int \frac{d^2 \mathbf{k}}{(2\pi)^{2}} \frac{1}{\mathbf{k}^2} e^{i \mathbf{k} \cdot (\mathbf{r} - \mathbf{r}_0)} G(r,r0)=∫(2π)2d2kk21eik⋅(r−r0)
改用极坐标计算
d2k=kdkdφ\purple{d^2\mathbf{k} = kdkd\varphi}d2k=kdkdφ，k⋅(r−r0)=k∣r−r0∣cosφ\mathbf{k} \cdot (\mathbf{r} - \mathbf{r}_0) = k |\mathbf{r} - \mathbf{r}_0| \cos \varphik⋅(r−r0)=k∣r−r0∣cosφ
G(r,r0)=∫0+∞dk∫02πdφk(2π)21k2eik∣r−r0∣cosφG(\mathbf{r}, \mathbf{r}_0) = \int_0^{+\infty} \purple{dk} \int_0^{2\pi} \purple{d\varphi} \frac{\purple{k}}{(2\pi)^2} \frac{1}{k^2} e^{i k |\mathbf{r} - \mathbf{r}_0| \cos \varphi} G(r,r0)=∫0+∞dk∫02πdφ(2π)2kk21eik∣r−r0∣cosφ
r−r0\green{\mathbf{r} - \mathbf{r}_0}r−r0
k\red{\mathbf{k}}k
φ\purple{\varphi}φ
利用 Bessel 函数的积分表示 (下一章节的内容)
J0(x)=12π∫02πdφeixcosφ\green{J_0}(x) = \frac{1}{2\pi} \int_0^{2\pi} d\varphi e^{i x \cos \varphi} J0(x)=2π1∫02πdφeixcosφ
得到
G(r,r0)=∫0+∞dk12πJ0(k∣r−r0∣)G(\mathbf{r}, \mathbf{r}_0) = \int_0^{+\infty} dk \frac{1}{2\pi} \green{J_0}(k |\mathbf{r} - \mathbf{r}_0|) G(r,r0)=∫0+∞dk2π1J0(k∣r−r0∣)
红外发散：J0(x→0)=1J_0(x \to 0) = 1J0(x→0)=1，呈现对数型发散
Gm(r,r0)=∫0+∞dk12πkk2+m2J0(k∣r−r0∣)=−12πK0(m∣r−r0∣)G_m(\mathbf{r}, \mathbf{r}_0) = \int_0^{+\infty} dk \frac{1}{2\pi} \frac{k}{k^2 + m^2} \green{J_0}(k |\mathbf{r} - \mathbf{r}_0|) = - \frac{1}{2\pi} K_0(m |\mathbf{r} - \mathbf{r}_0|) Gm(r,r0)=∫0+∞dk2π1k2+m2kJ0(k∣r−r0∣)=−2π1K0(m∣r−r0∣)
K0(z→0)∼−lnz2−γ+O(z2)K_0(z \to 0) \sim - \ln \frac{z}{2} - \gamma + O(z^2) K0(z→0)∼−ln2z−γ+O(z2)
代入 z=m∣r−r0∣z = m |\mathbf{r} - \mathbf{r}_0|z=m∣r−r0∣
Gm(r,r0)∼+12πlnm∣r−r0∣2+O(1)∼12πln∣r−r0∣+lnm2π+O(1)G_m(\mathbf{r}, \mathbf{r}_0) \sim + \frac{1}{2\pi} \ln \frac{m|\mathbf{r} - \mathbf{r}_0|}{2} + O(1) \sim \frac{1}{2\pi} \ln |\mathbf{r} - \mathbf{r}_0| + \frac{\ln m}{2\pi} + O(1) Gm(r,r0)∼+2π1ln2m∣r−r0∣+O(1)∼2π1ln∣r−r0∣+2πlnm+O(1)
去掉发散的 lnm\ln mlnm 以及调整常数：得到之前的结果
G(r,r0)=∫d3k(2π)31k2eik⋅(r−r0)G(\mathbf{r}, \mathbf{r}_0) = \int \frac{d^3\mathbf{k}}{(2\pi)^3} \frac{1}{\mathbf{k^2}} e^{i \mathbf{k} \cdot (\mathbf{r} - \mathbf{r}_0)} G(r,r0)=∫(2π)3d3kk21eik⋅(r−r0)
∫0+∞dk∫0πsinθdθ∫02πdφk2(2π)31k2eik∣r−r0∣cosθ\int_0^{+\infty} {\purple{dk} \int_0^{\pi} \purple{\sin \theta d\theta} \int_0^{2\pi} \purple{d\varphi} \frac{\purple{k^2}}{(2\pi)^3} \frac{1}{k^2} e^{i k |\mathbf{r} - \mathbf{r}_0| \cos \theta}} ∫0+∞dk∫0πsinθdθ∫02πdφ(2π)3k2k21eik∣r−r0∣cosθ
∫0πeik∣r−r0∣cosθsinθdθ=−∫0πeik∣r−r0∣cosθd(cosθ)=2sin(k∣r−r0∣)k∣r−r0∣\int_0^{\pi} e^{i k |\mathbf{r} - \mathbf{r}_0| \cos \theta}\sin \theta d\theta = - \int_0^\pi e^{i k |\mathbf{r} - \mathbf{r}_0| \cos \theta}d(\cos \theta) = \frac{2 \sin(k |\mathbf{r} - \mathbf{r}_0|)}{k |\mathbf{r} - \mathbf{r}_0|} ∫0πeik∣r−r0∣cosθsinθdθ=−∫0πeik∣r−r0∣cosθd(cosθ)=k∣r−r0∣2sin(k∣r−r0∣)
G(r,r0)=2×2π(2π)3∫0+∞dkk2k2sin(k∣r−r0∣)k∣r−r0∣G(\mathbf{r}, \mathbf{r}_0) = \frac{2 \times 2\pi}{(2\pi)^3} \int _0^{+\infty} dk \frac{k^2}{k^2} \frac{\sin(k |\mathbf{r} - \mathbf{r}_0|)}{k|\mathbf{r} - \mathbf{r}_0|} G(r,r0)=(2π)32×2π∫0+∞dkk2k2k∣r−r0∣sin(k∣r−r0∣)
G(r,r0)=12π21∣r−r0∣∫0+∞dxsinxxG(\mathbf{r}, \mathbf{r}_0) = \frac{1}{2\pi^2} \frac{1}{|\mathbf{r} - \mathbf{r}_0|} \int_0^{+\infty} dx \frac{\sin x}{x} G(r,r0)=2π21∣r−r0∣1∫0+∞dxxsinx
可不可以用留数定理计算？
可以！实轴上是单极点
∫0+∞sinxxdx=12∫−∞+∞sinxxdx\int_0^{+\infty} \frac{\sin x}{x} dx = \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\sin x}{x} dx ∫0+∞xsinxdx=21∫−∞+∞xsinxdx
=12i∫−∞+∞cosx+isinxxdx,or,=12Im∫−∞+∞eixxdx= \frac{1}{2 i} \int_{-\infty}^{+\infty} \frac{\cos x + i \sin x}{x} dx, \qquad \text{or}, \qquad = \frac{1}{2} \operatorname{Im} \int_{-\infty}^{+\infty} \frac{e^{i x}}{x} dx =2i1∫−∞+∞xcosx+isinxdx,or,=21Im∫−∞+∞xeixdx
=2πi2Im[12Resz=0(eizz)]=12Im[2πi2×1]=π2= \frac{2\pi i}{2} \operatorname{Im} \left[{\frac{1}{2}\operatorname{Res}_{z = 0} \left({\frac{e^{i z}}{z}}\right)}\right] = \frac{1}{2} \operatorname{Im} \left[{\frac{2\pi i}{2} \times 1}\right] = \frac{\pi}{2} =22πiIm[21Resz=0(zeiz)]=21Im[22πi×1]=2π
G(r,r0)=12π21∣r−r0∣×π2=14π1∣r−r0∣G(\mathbf{r}, \mathbf{r}_0) = \frac{1}{2\pi^2} \frac{1}{|\mathbf{r} - \mathbf{r}_0|} \times \frac{\pi}{2} = \frac{1}{4\pi} \frac{1}{|\mathbf{r} - \mathbf{r}_0|} G(r,r0)=2π21∣r−r0∣1×2π=4π1∣r−r0∣1
(∇2+m2)G(r,r0)=−δ(r−r0)(\nabla^2 + m^2) G(\mathbf{r}, \mathbf{r}_0) = - \delta(\mathbf{r} - \mathbf{r}_0) (∇2+m2)G(r,r0)=−δ(r−r0)
G(r,r0)=∫dnk(2π)n1k2−m2eik⋅(r−r0)G(\mathbf{r}, \mathbf{r}_0) = \int \frac{d^n \mathbf{k}}{(2\pi)^n} \frac{1}{\mathbf{k}^2 - m^2} e^{i \mathbf{k} \cdot (\mathbf{r} - \mathbf{r}_0)} G(r,r0)=∫(2π)ndnkk2−m21eik⋅(r−r0)
∫dk2π1k2−m2eik(x−x0)=−sinm∣x−x0∣2m\int \frac{dk}{2\pi} \frac{1}{k^2 - m^2}e^{i k (x - x_0)} = - \frac{\sin m |x - x_0|}{2m} ∫2πdkk2−m21eik(x−x0)=−2msinm∣x−x0∣
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