第十章格林函数法 <script id="MathJax-Element-1" type="math/tex">\color{blue}{第十章 格林函数法}</script>

⎧ ⎩ ⎨ ⎪ ⎪ 行波法:无界空间波动问题,有局限性分离变量法:各种有界问题,其解为无穷级数积分变换法:各种无界问题,其解为无限积分  <script id="MathJax-Element-2" type="math/tex">\left \lbrace \begin{array}{l}行波法:无界空间波动问题,有局限性 \\ 分离变量法:各种有界问题,其解为无穷级数 \\ 积分变换法:各种无界问题,其解为无限积分 \end{array} \right.</script>
1.格林函数法:其解为含有格林函数的有限积分。 <script id="MathJax-Element-3" type="math/tex">1.格林函数法:其解为含有格林函数的有限积分。</script>
由§10.2:{Δu=−h(M)u| σ =f(M) → <script id="MathJax-Element-4" type="math/tex">由\S10.2:\left \lbrace \begin{array}{l} \Delta u = - h(M) \\ u |_{\sigma} = f(M) \end{array} \right. \to</script>
u(M)=∭ τ G(M,M 0 )h(M 0 )dτ 0 −∬ σ f(M 0 )∂G∂n 0  dσ 0  <script id="MathJax-Element-5" type="math/tex">\color{blue}{u(M) = \iiint_{\tau} G(M, M_0) h(M_0) d \tau_0 - \iint_{\sigma} f(M_0) \dfrac{\partial G}{\partial n_0} d \sigma_0}</script>
G(M,M 0 )−狄氏格林函数 <script id="MathJax-Element-6" type="math/tex">\color{blue}{G(M, M_0) - 狄氏格林函数}</script>
2.格林函数:点源函数,点源产生的场和影响若外力f(x,t)只在ξ点,τ时起作用 <script id="MathJax-Element-7" type="math/tex">2.格林函数:点源函数,点源产生的场和影响若外力f(x, t) 只在 \xi 点,\tau 时起作用</script>
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u tt =a 2 u xx +f(x,t),f(x,t)={0,x≠ξ,t≠τf(ξ,τ),x=ξ,t=τ u| x=0 =0,u| x=l =0u| t=0 =0,u t | t=0 =0u(x,t)−格林函数,即G(x,t|ξ,τ);f(x,t)−点源  <script id="MathJax-Element-8" type="math/tex">\left \lbrace \begin{array}{l}u_{tt} = a^2 u_{xx} + f(x, t), f(x, t) = \left \lbrace \begin{array}{l}0, x \neq \xi, t \neq \tau \\ f(\xi, \tau), x = \xi, t = \tau \end{array} \right. \\ u|_{x=0} = 0, u|_{x=l} = 0 \\ u|_{t=0} = 0, u_t|_{t=0} = 0 \\ u(x, t) -格林函数,即G(x, t| \xi, \tau); f(x, t) -点源 \end{array} \right.</script>

3.为何引入格林函数法: <script id="MathJax-Element-9" type="math/tex">3.为何引入格林函数法:</script>
5.2→{Δu=−h(M)u| σ =f(M) → <script id="MathJax-Element-10" type="math/tex">5.2 \to \left \lbrace \begin{array}{l}\Delta u = - h(M) \\ u|_{\sigma} = f(M) \end{array} \right. \to</script>
u(M)=∭ τ G(M,M 0 )h(M 0 )dτ 0 −∬ σ f(M 0 )∂G∂n 0  dσ 0  <script id="MathJax-Element-11" type="math/tex">u(M) = \iiint_{\tau}G(M, M_0)h(M_0) d\tau_0 - \iint_{\sigma} f(M_0) \dfrac{\partial G}{\partial n_0} d \sigma_0</script>
(1)解的形式(有限积分)便于理论分析和研究 <script id="MathJax-Element-12" type="math/tex">(1)解的形式(有限积分)便于理论分析和研究</script>
(2)以统一的形式研究各类定解问题 <script id="MathJax-Element-13" type="math/tex">(2)以统一的形式研究各类定解问题</script>
(3)对于线性问题,格林函数一旦求出,就可以算出任意源的场,关键就是求点源 <script id="MathJax-Element-14" type="math/tex">(3)对于线性问题,格林函数一旦求出,就可以算出任意源的场,关键就是求点源</script>

§10.1δ函数 <script id="MathJax-Element-15" type="math/tex">\color{blue}{\S 10.1 \delta函数}</script>

10.1.1δ函数的引入 <script id="MathJax-Element-16" type="math/tex">\color{blue}{10.1.1 \delta函数的引入}</script>

1.物理背景 <script id="MathJax-Element-17" type="math/tex">1.物理背景</script>
(1)金属线:总质量m=1,集中在x=0处, <script id="MathJax-Element-18" type="math/tex">(1)金属线:总质量m = 1,集中在x = 0处,</script>
则密度:ρ(x)=lim Δx→0 ΔmΔx ={0,x≠0∞,x=0  <script id="MathJax-Element-19" type="math/tex">则密度: \rho(x) = \lim \limits _{\Delta x \to 0} \dfrac{\Delta m}{\Delta x} = \left \lbrace \begin{array}{l}0, x \neq 0 \\ \infty, x = 0 \end{array} \right.</script>
∫ ∞ −∞ ρ(x)dx=1 <script id="MathJax-Element-20" type="math/tex">\int_{-\infty}^{\infty} \rho(x) dx = 1</script>

(2)带电导线:总电量q=1,集中在x=0处 <script id="MathJax-Element-21" type="math/tex">(2)带电导线:总电量q = 1,集中在x = 0处</script>
则电荷密度:ρ(x)=lim Δx→0 ΔqΔx ={0,x≠0∞,x=0  <script id="MathJax-Element-22" type="math/tex">则电荷密度:\rho(x) = \lim \limits _{\Delta x \to 0} \dfrac{\Delta q}{\Delta x} = \left \lbrace \begin{array}{l}0, x \neq 0 \\ \infty, x = 0 \end{array} \right.</script>
∫ ∞ −∞ ρ(x)dx=1 <script id="MathJax-Element-23" type="math/tex">\int_{-\infty}^{\infty} \rho(x) dx = 1</script>

2.定义: <script id="MathJax-Element-24" type="math/tex">2.定义:</script>
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ δ(x)={0,x≠0∞,x=0 −δ函数∫ ∞ −∞ δ(x)dx=1  <script id="MathJax-Element-25" type="math/tex">\color{blue}{\left \lbrace \begin{array}{l}\delta(x) = \left \lbrace \begin{array}{l}0, x \neq 0 \\ \infty, x = 0 \end{array} \right. -\delta函数 \\ \int_{-\infty}^{\infty} \delta(x) dx = 1 \end{array} \right.}</script>

一般：⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ δ(x−x 0 )={0,x≠x 0 ∞,x=x 0  −δ函数∫ ∞ −∞ δ(x−x 0 )dx=1  <script id="MathJax-Element-26" type="math/tex">\color{blue}{一般：\left \lbrace \begin{array}{l}\delta(x-x_0) = \left \lbrace \begin{array}{l}0, x \neq x_0 \\ \infty, x = x_0 \end{array} \right. -\delta函数 \\ \int_{-\infty}^{\infty} \delta(x-x_0) dx = 1 \end{array} \right.}</script>

3.注意: <script id="MathJax-Element-27" type="math/tex">3.注意:</script>
(1)δ−密度函数和点源函数 <script id="MathJax-Element-28" type="math/tex">(1)\delta-密度函数和点源函数</script>
若在x=x 0 点放有m质量,总质量m,则ρ(x)=mδ(x−x 0 ) <script id="MathJax-Element-29" type="math/tex">若在x = x_0点放有m质量,总质量m,则\rho(x) = m \delta(x - x_0)</script>
若在x=x 0 点放有电量为q的点电荷,总电量为q,则ρ(x)=qδ(x−x 0 ) <script id="MathJax-Element-30" type="math/tex">若在x=x_0点放有电量为q的点电荷,总电量为q,则\rho(x) = q \delta(x-x_0)</script>

(2)δ广义函数 <script id="MathJax-Element-31" type="math/tex">(2)\delta 广义函数</script>

10.1.2δ函数的性质 <script id="MathJax-Element-32" type="math/tex">\color{blue}{10.1.2 \delta函数的性质}</script>

设f(x)在(−∞,∞)连续,则 <script id="MathJax-Element-33" type="math/tex">设f(x)在(-\infty, \infty)连续,则</script>
1.∫ ∞ −∞ f(x)δ(x−x 0 )dx=f(x 0 )[∫ ∞ −∞ f(x)δ(x)dx=f(0)] <script id="MathJax-Element-34" type="math/tex">1.\int_{-\infty}^{\infty} f(x)\delta(x - x_0) dx = f(x_0) \quad [\int_{-\infty}^{\infty} f(x) \delta(x) dx = f(0)]</script>
注意:δ也能表示连续分布的函数 <script id="MathJax-Element-35" type="math/tex">注意:\delta也能表示连续分布的函数</script>
f(t)=∫ ∞ −∞ f(τ)δ(τ−t)dτ=∫ b a f(τ)δ(τ−t)dτ <script id="MathJax-Element-36" type="math/tex">f(t) = \int_{-\infty}^{\infty} f(\tau) \delta(\tau - t) d\tau = \int_a^b f(\tau) \delta (\tau - t) d \tau</script>

附:判断函数相等的一种方法: <script id="MathJax-Element-37" type="math/tex">附:判断函数相等的一种方法:</script>
设f(x)与g(x)都是定义在(a,b)区间上的函数,若对于定义在(a,b)区间上的任意连续函数φ(x)都有如下等式成立:∫ b a f(x)φ(x)dx=∫ b a g(x)φ(x)dx,则必有:f(x)=g(x),特别：若∫ b a φ(x)g(x)dx=0,则必有g(x)=0 <script id="MathJax-Element-38" type="math/tex">设f(x)与g(x)都是定义在(a, b)区间上的函数,若对于定义在(a, b)区间上的任意连续\\ 函数\varphi(x)都有如下等式成立:\int_a^b f(x)\varphi(x) dx = \int_a^b g(x) \varphi(x) dx, 则必有:\\ f(x) = g(x), 特别：若\int_a^b \varphi(x)g(x) dx = 0, 则必有g(x) = 0</script>

2.若定义ddx δ(x)=δ ′ (x)−δ函数的导数,则 <script id="MathJax-Element-39" type="math/tex">2.若定义\dfrac{d}{dx} \delta(x) = \delta^{\prime}(x) - \delta函数的导数,则</script>
(1)∫ ∞ −∞ f(x)δ ′ (x−x 0 )dx=−f ′ (x 0 ) <script id="MathJax-Element-40" type="math/tex">(1)\int_{-\infty}^{\infty} f(x) \delta^{\prime}(x-x_0) dx = - f^{\prime}(x_0)</script>
(2)(x−x 0 )δ ′ (x−x 0 )=−δ(x−x 0 ) <script id="MathJax-Element-41" type="math/tex">(2)(x-x_0)\delta^{\prime}(x-x_0) = - \delta(x-x_0)</script>
(3)∫ ∞ −∞ f(x)δ (n) (x−x 0 )dx=(−1) n f (n) (x 0 ) <script id="MathJax-Element-42" type="math/tex">(3)\int_{-\infty}^{\infty}f(x) \delta^{(n)}(x-x_0)dx = (-1)^nf^{(n)}(x_0)</script>

3.δ[φ(x)]=∑ i=1 n δ(x−x i )|φ ′ (x i )| ,其中φ(x i )=0 <script id="MathJax-Element-43" type="math/tex">3.\delta[\varphi(x)] = \sum \limits _{i=1}^{n}\dfrac{\delta(x-x_i)}{|\varphi^{\prime}(x_i)|},其中\varphi(x_i) = 0</script>

10.1.3高维δ函数 <script id="MathJax-Element-44" type="math/tex">\color{blue}{10.1.3 高维\delta函数}</script>

1.定义： <script id="MathJax-Element-45" type="math/tex">1.定义：</script>
(1)⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ δ(M−M 0 )={0,M≠M 0 ,∞,M=M 0 , ∭ ∞ −∞ δ(M−M 0 )dv=1,为三维函数,dv=dxdydz  <script id="MathJax-Element-46" type="math/tex">(1)\left \lbrace \begin{array}{l}\delta(M-M_0) = \left \lbrace \begin{array}{l}0, M \neq M_0, \\ \infty, M = M_0, \end{array} \right. \\ \iiint_{-\infty}^{\infty}\delta(M-M_0)dv = 1, 为三维函数,dv = dxdydz \end{array} \right.</script>
其中δ(M−M 0 )=δ(x−x 0 ,y−y 0 ,z−z 0 )=δ(x−x 0 )δ(y−y 0 )δ(z−z 0 ) <script id="MathJax-Element-47" type="math/tex">其中\delta(M-M_0)=\delta(x-x_0,y-y_0,z-z_0)= \delta(x-x_0)\delta(y-y_0)\delta(z-z_0)</script>

(2)⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ δ(M−M 0 )={0,M≠M 0 ,∞,M=M 0 , ∬ ∞ −∞ δ(M−M 0 )dxdy=1为二维函数  <script id="MathJax-Element-48" type="math/tex">(2)\left \lbrace \begin{array}{l}\delta(M-M_0) = \left \lbrace \begin{array}{l}0, M \neq M_0, \\ \infty, M = M_0, \end{array} \right. \\ \iint_{-\infty}^{\infty} \delta(M-M_0)dxdy = 1 \quad 为二维函数 \end{array} \right.</script>
其中δ(M−M 0 )=δ(x−x 0 ,y−y 0 )=δ(x−x 0 )δ(y−y 0 ) <script id="MathJax-Element-49" type="math/tex">其中\delta(M-M_0) = \delta(x-x_0, y-y_0) = \delta(x-x_0)\delta(y-y_0)</script>

2.性质: <script id="MathJax-Element-50" type="math/tex">2.性质:</script>
(1)∭ ∞ −∞ f(M)δ(M−M 0 )dxdydz=f(x 0 ,y 0 ,z 0 )=f(M 0 ) <script id="MathJax-Element-51" type="math/tex">(1)\iiint_{-\infty}^{\infty}f(M)\delta(M-M_0)dxdydz = f(x_0, y_0, z_0) = f(M_0)</script>
(2)∬ ∞ −∞ f(M)δ(M−M 0 )dxdy=f(x 0 ,y 0 )=f(M 0 ) <script id="MathJax-Element-52" type="math/tex">(2)\iint_{-\infty}^{\infty}f(M)\delta(M-M_0)dxdy = f(x_0,y_0) = f(M_0)</script>

10.1.4例题 <script id="MathJax-Element-53" type="math/tex">\color{blue}{10.1.4 例题}</script>

1.∫ 2 1 sinxδ(x−12 )dx=0 <script id="MathJax-Element-54" type="math/tex">1.\int_1^2 \sin x \delta(x - \dfrac{1}{2})dx = 0</script>
2.∫ 2 1 sinxδ(x)dx=0 <script id="MathJax-Element-55" type="math/tex">2.\int_1^2 \sin x \delta(x) dx = 0</script>
3.∫ ∞ −∞ ∫ ∞ −∞ sin(x+y)δ(x+2)δ(y−1)dxdy=sin(−1) <script id="MathJax-Element-56" type="math/tex">3.\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sin(x+y) \delta(x+2) \delta(y-1)dxdy = \sin(-1)</script>
4.长为1，密度为ρ的弦两端固定,初位移为零,初始时刻在x=x 0 点受到一横向冲量I 0 .试写出弦的横震动的定解问题. <script id="MathJax-Element-57" type="math/tex">4.长为1，密度为\rho的弦两端固定,初位移为零,初始时刻在x = x_0点受到一横向\\ 冲量I_0.试写出弦的横震动的定解问题.</script>
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u tt =a 2 u xx u| x=0 =0,u| x=l =0u| t=0 =0u t | t=0 =I 0 ρ δ(x−x 0 )  <script id="MathJax-Element-58" type="math/tex">\left \lbrace \begin{array}{l}u_{tt} = a^2 u_{xx} \\ u|_{x=0} = 0, u|_{x=l} = 0 \\ u|_{t=0} = 0 \\ u_t | _{t=0} = \dfrac{I_0}{\rho}\delta(x-x_0) \end{array} \right.</script>

§10.2泊松方程的狄氏问题 <script id="MathJax-Element-59" type="math/tex">\color{blue}{\S 10.2 泊松方程的狄氏问题}</script>

{Δu=−h(M),M∈τ(1)u| σ=0 =g(M)(2)  <script id="MathJax-Element-60" type="math/tex">\left \lbrace \begin{array}{l}\Delta u = -h(M), M \in \tau \quad (1) \\ u|_{\sigma=0} = g(M) \quad (2) \end{array} \right.</script>

10.2.1格林公式 <script id="MathJax-Element-61" type="math/tex">\color{blue}{10.2.1 格林公式}</script>

1.为何引入格林公式 <script id="MathJax-Element-62" type="math/tex">1.为何引入格林公式</script>
(1)积分公式的起点是通过直接积分或分部积分将未知函数从微分号下解脱出来 <script id="MathJax-Element-63" type="math/tex">(1)积分公式的起点是通过直接积分或分部积分将未知函数从微分号下解脱出来</script>
(2)我们要求解的三类数值方程中均含有Δ,格林公式是将未知函数,从微分算法Δ下解脱出来的工具.设u(x,y,z),v(x,y,z)在τ中具有连续的二阶导数,在τ ¯ 上具有连续的一阶导数,则有如下格林公式: <script id="MathJax-Element-64" type="math/tex">(2)我们要求解的三类数值方程中均含有\Delta,格林公式是将未知函数,从微分算法\Delta\\ 下解脱出来的工具.设u(x, y, z), v(x, y, z)在\tau中具有连续的二阶导数,在\bar \tau上具\\ 有连续的一阶导数,则有如下格林公式:</script>
2.格林第一公式 <script id="MathJax-Element-65" type="math/tex">2.格林第一公式</script>
∫ τ uΔvdτ+∫ τ ∇u⋅∇vdτ=∫ σ u∂v∂n dσ(3) <script id="MathJax-Element-66" type="math/tex">\int_{\tau} u \Delta v d \tau + \int_{\tau} \nabla u \cdot \nabla v d \tau = \int_{\sigma} u \dfrac{\partial v}{\partial n} d \sigma \quad (3)</script>
∫ τ vΔudτ+∫ τ ∇u⋅∇vdτ=∫ σ v∂u∂n dσ(4) <script id="MathJax-Element-67" type="math/tex">\int_{\tau} v \Delta u d \tau + \int_{\tau} \nabla u \cdot \nabla v d \tau = \int_{\sigma} v \dfrac{\partial u}{\partial n} d \sigma \quad (4)</script>

3.格林第二公式 <script id="MathJax-Element-68" type="math/tex">3.格林第二公式</script>
∫ τ uΔvdτ−∫ τ vΔudτ=∫ σ (u∂v∂n −v∂u∂n )dσ(5) <script id="MathJax-Element-69" type="math/tex">\int_{\tau} u \Delta v d \tau - \int_{\tau} v \Delta u d \tau = \int_{\sigma}(u \dfrac{\partial v}{\partial n} - v \dfrac{\partial u}{\partial n}) d\sigma \quad (5)</script>
意义:(1)将u,v,Δu,Δv的值与u,v,∂v∂n ,∂u∂n 的边值联系起来. <script id="MathJax-Element-70" type="math/tex">意义:(1)将u,v,\Delta u, \Delta v的值与u, v, \dfrac{\partial v}{\partial n}, \dfrac{\partial u}{\partial n}的边值联系起来.</script>
(2)u,v堆成 <script id="MathJax-Element-71" type="math/tex">(2)u,v堆成</script>
(3)若已知v,Δv=0,及v| σ ,则由格林公式可能求得{Δu=−h(M)(1)u| σ =g(M)(2) 之解. <script id="MathJax-Element-72" type="math/tex">(3)若已知v, \Delta v = 0,及v|_{\sigma},则由格林公式可能求得\left \lbrace \begin{array}{l}\Delta u = -h(M) \quad(1) \\ u|_{\sigma} = g(M) \quad (2) \end{array} \right.之解.</script>

4.球面平均值公式 <script id="MathJax-Element-73" type="math/tex">4.球面平均值公式</script>
(1)定义: <script id="MathJax-Element-74" type="math/tex">(1)定义:</script>
u ¯ (r,t)=14πr 2  ∬ S M 0  r  u(M,t)ds <script id="MathJax-Element-75" type="math/tex">\bar u(r, t) = \dfrac{1}{4 \pi r^2}\iint_{S_r^{M_0}} u(M, t) ds</script>
=14π ∬ S M 0  r  u(M,t)dΩdΩ=dsr 2  =sinθdθdφ <script id="MathJax-Element-76" type="math/tex">= \dfrac{1}{4 \pi}\iint_{S_r^{M_0}}u(M, t) d \Omega \quad d\Omega = \dfrac{ds}{r^2} = \sin \theta d \theta d \varphi</script>
−u(M,t)在以M 0 为中心，r为半径的球面S M 0  r 上的平均值. <script id="MathJax-Element-77" type="math/tex">-u(M, t)在以M_0为中心，r为半径的球面S_r^{M_0}上的平均值.</script>
(2)显然u(M 0 ,t 0 )=lim r→0 u ¯ (r,t 0 ) <script id="MathJax-Element-78" type="math/tex">(2)显然 u(M_0, t_0) = \lim \limits_{r \to 0} \bar u(r, t_0)</script>

10.2.2积分公式−格林函数法 <script id="MathJax-Element-79" type="math/tex">\color{blue}{10.2.2 积分公式-格林函数法}</script>

1.(三维)狄氏积分公式:M,M 0 ∈τ <script id="MathJax-Element-80" type="math/tex">1.(三维)狄氏积分公式:M,M_0 \in \tau</script>
{Δu=−h(M)(1)u| σ =f(M)(2) {ΔG=−δ(M−M 0 )(3)G| σ =0(4)  <script id="MathJax-Element-81" type="math/tex">\left \lbrace \begin{array}{l}\Delta u = -h(M) \quad (1) \\ u|_{\sigma} = f(M) \quad (2) \end{array} \right. \quad \left \lbrace \begin{array}{l}\Delta G = -\delta(M-M_0) \quad (3) \\ G|_{\sigma} = 0 \quad (4) \end{array} \right.</script>
则(3)→G=14πr (10.3.4),r=(x−x 0 ) 2 +(y−y 0 ) 2 +(z−z 0 ) 2  − − − − − − − − − − − − − − − − − − − − − − − − − −  √  <script id="MathJax-Element-82" type="math/tex">则(3) \to G = \dfrac{1}{4 \pi r} (10.3.4), r= \sqrt{(x-x_0)^2 +(y-y_0)^2 + (z-z_0)^2}</script>
[(1)⋅G−(3)⋅u]在[τ−τ ε ]中积分有: <script id="MathJax-Element-83" type="math/tex">[(1)\cdot G - (3) \cdot u]在[\tau - \tau_{\varepsilon}]中积分有:</script>
∫ τ−τ ε  [GΔu−uΔGdτ=∫ τ−τ ε  uδ(M−M 0 )dτ−∫ τ−τ ε  Gh(M)dτ <script id="MathJax-Element-84" type="math/tex">\int_{\tau - \tau_{\varepsilon}}[G\Delta u - u \Delta G d \tau = \int_{\tau - \tau_{\varepsilon}} u \delta(M-M_0) d \tau - \int_{\tau - \tau_{\varepsilon}} Gh(M) d \tau</script>
对左边用格林第二公式有: <script id="MathJax-Element-85" type="math/tex">对左边用格林第二公式有:</script>
∫ σ+σ ε  (G(∂u∂n −u∂G∂n )dσ=∫ τ−τ ε  uδ(M−M 0 )dτ−∫ τ−τ ε  Gh(M)dτ <script id="MathJax-Element-86" type="math/tex">\int_{\sigma + \sigma_{\varepsilon}}(G(\dfrac{\partial u}{\partial n} - u \dfrac{\partial G}{\partial n}) d \sigma = \int_{\tau - \tau_{\varepsilon}} u \delta(M-M_0) d \tau - \int_{\tau - \tau_{\varepsilon}} Gh(M) d \tau</script>
∫ σ (G∂u∂n −u∂G∂n )dσ−u(M 0 )=−∫ τ G(M,M 0 )h(M)dτ <script id="MathJax-Element-87" type="math/tex">\int_{\sigma}(G\dfrac{\partial u}{\partial n} - u\dfrac{\partial G}{\partial n}) d \sigma - u(M_0) = -\int_{\tau} G(M, M_0)h(M) d \tau</script>
将边界条件(2)(4)代入上式有: <script id="MathJax-Element-88" type="math/tex">将边界条件(2)(4)代入上式有:</script>
u(M 0 )=∫ τ G(M,M 0 )h(M)dτ−∫ σ f(M)∂G∂n dσ <script id="MathJax-Element-89" type="math/tex">u(M_0) = \int_{\tau}G(M, M_0)h(M)d \tau - \int_{\sigma}f(M) \dfrac{\partial G}{\partial n} d \sigma</script>
u(M)=∫ τ G(M,M 0 )h(M 0 )dτ 0 −∫ σ f(M 0 )∂G∂n 0  dσ 0  <script id="MathJax-Element-90" type="math/tex">\color{blue}{u(M) = \int_{\tau}G(M, M_0)h(M_0)d \tau_0 - \int_{\sigma}f(M_0)\dfrac{\partial G}{\partial n_0} d \sigma_0}</script>
∵G(M,M 0 )=G(M 0 ,M) <script id="MathJax-Element-91" type="math/tex">\because G(M, M_0) = G(M_0, M)</script>

2.狄氏积分公式的物理意义: <script id="MathJax-Element-92" type="math/tex">2.狄氏积分公式的物理意义:</script>
第一项:体内源产生的场的和。 <script id="MathJax-Element-93" type="math/tex">第一项:体内源产生的场的和。</script>
第二项:边界上源产生的场的和。 <script id="MathJax-Element-94" type="math/tex">第二项:边界上源产生的场的和。</script>

3.(二维)狄氏积分公式 <script id="MathJax-Element-95" type="math/tex">3.(二维)狄氏积分公式</script>
{Δu(M)=−h(M)M∈σu| l =f(M)  <script id="MathJax-Element-96" type="math/tex">\left \lbrace \begin{array}{l}\Delta u(M) = -h(M) \quad M \in \sigma \\ u|_l = f(M) \end{array} \right.</script>
u(M)=∬ σ G(M,M 0 )h(M 0 )dσ 0 −∫ l f(M 0 )∂G∂n 0  dl 0  <script id="MathJax-Element-97" type="math/tex">u(M) = \iint_{\sigma}G(M, M_0)h(M_0)d \sigma_0 - \int_l f(M_0) \dfrac{\partial G}{\partial n_0} dl_0</script>

10.2.3小结 <script id="MathJax-Element-98" type="math/tex">\color{blue}{10.2.3 小结}</script>

1.∫ τ uΔvdτ−∫ τ vΔudτ=∫ σ (u∂v∂n −v∂u∂n )dσ(5) <script id="MathJax-Element-99" type="math/tex">1.\int_{\tau}u\Delta v d\tau - \int_{\tau}v \Delta u d \tau = \int_{\sigma}(u\dfrac{\partial v}{\partial n} - v \dfrac{\partial u}{\partial n}) d \sigma \quad (5)</script>
2.{Δu=−h(M),M∈τ(1)u| σ =f(M)(2)  <script id="MathJax-Element-100" type="math/tex">2.\left \lbrace \begin{array}{l}\Delta u = -h(M), M \in \tau \quad (1) \\ u|_{\sigma} = f(M) \quad (2) \end{array} \right.</script>
u(M)=∭ τ G(M,M 0 )h(M 0 )dτ 0 −∬ σ f(M 0 )∂G∂n 0  dσ 0 (6) <script id="MathJax-Element-101" type="math/tex">u(M) = \iiint_{\tau} G(M, M_0)h(M_0) d\tau_0 - \iint_{\sigma} f(M_0)\dfrac{\partial G}{\partial n_0} d \sigma_0 \quad (6)</script>
{ΔG=−δ(M−M 0 )(3)G| σ =0(4) G(M,M 0 )=G(M 0 ,M) <script id="MathJax-Element-102" type="math/tex">\left \lbrace \begin{array}{l}\Delta G = -\delta(M-M_0) \quad (3) \\ G|_{\sigma} = 0 \quad (4) \end{array} \right. G(M, M_0) = G(M_0, M)</script>

§10.3格林函数 <script id="MathJax-Element-103" type="math/tex">\color{blue}{\S 10.3 格林函数}</script>

10.3.1泊松方程的格林函数 <script id="MathJax-Element-104" type="math/tex">\color{blue}{10.3.1 泊松方程的格林函数}</script>

1.三维:ΔG=−δ(M−M 0 ) <script id="MathJax-Element-105" type="math/tex">1.三维:\Delta G = -\delta(M-M_0)</script>
ΔG=1r 2  ∂∂r (r 2 ∂G∂r =−δ(r) <script id="MathJax-Element-106" type="math/tex">\Delta G = \dfrac{1}{r^2}\dfrac{\partial}{\partial r}(r^2 \dfrac{\partial G}{\partial r} = -\delta(r)</script>
(1)若r≠0:G=−C 1 1r  <script id="MathJax-Element-107" type="math/tex">(1)若r \neq 0: G = -C_1 \dfrac{1}{r}</script>
(2)若r=0:考虑∭ τ ε  ΔGdv=−∭ τ ε  δ(r)dv=−1 <script id="MathJax-Element-108" type="math/tex">(2)若r = 0: 考虑\iiint_{\tau_{\varepsilon}}\Delta G dv = -\iiint_{\tau_{\varepsilon}}\delta(r) dv = -1</script>
又∭ τ ε  ΔGdv=∭ τ ε  ∇⋅∇Gdv=∬ σ ε  ∇G⋅dσ ⃗ =C 1 4π <script id="MathJax-Element-109" type="math/tex">又\iiint_{\tau_{\varepsilon}}\Delta G dv = \iiint_{\tau_{\varepsilon}} \nabla \cdot \nabla G dv = \iint_{\sigma_{\varepsilon}} \nabla G \cdot d \vec \sigma = C_1 4 \pi </script>
→C 1 =−14π , <script id="MathJax-Element-110" type="math/tex">\to C_1 = -\dfrac{1}{4 \pi},</script>
(1),(2)→G=14πr −泊松方程格林函数 <script id="MathJax-Element-111" type="math/tex">(1),(2) \to \color{blue}{G = \dfrac{1}{4 \pi r} - 泊松方程格林函数}</script>

2.二维:ΔG=−δ(M−M 0 ) <script id="MathJax-Element-112" type="math/tex">2.二维:\Delta G = - \delta(M-M_0)</script>
ΔG=1r ∂∂r (r∂G∂r )=−δ(r),r=(x−x 0 ) 2 +(y−y 0 ) 2  − − − − − − − − − − − − − − − − −  √  <script id="MathJax-Element-113" type="math/tex">\Delta G = \dfrac{1}{r}\dfrac{\partial}{\partial r}(r\dfrac{\partial G}{\partial r}) = - \delta(r), r = \sqrt{(x-x_0)^2 + (y-y_0)^2}</script>
∬ σ ∇⋅∇udσ=∫ l ∇u⋅dl ⃗  <script id="MathJax-Element-114" type="math/tex">\iint_{\sigma}\nabla \cdot \nabla u d \sigma = \int_l \nabla u \cdot d \vec l</script>
G(M,M 0 )=12π ln1r  <script id="MathJax-Element-115" type="math/tex">\color{blue}{G(M, M_0) = \dfrac{1}{2 \pi} \ln \dfrac{1}{r}}</script>

10.3.2狄氏格林函数 <script id="MathJax-Element-116" type="math/tex">\color{blue}{10.3.2 狄氏格林函数}</script>

1.三维： <script id="MathJax-Element-117" type="math/tex">1.三维：</script>
{ΔG=−δ(x−x 0 ,y−y 0 ,z−z 0 ),M∈τG| σ =0  <script id="MathJax-Element-118" type="math/tex">\left \lbrace \begin{array}{l}\Delta G = - \delta(x-x_0, y-y_0, z-z_0), M \in \tau \\ G|_{\sigma} = 0 \end{array} \right.</script>
令G(M,M 0 )=F(M,M 0 )+g(M,M 0 ) <script id="MathJax-Element-119" type="math/tex">令G(M, M_0) = F(M, M_0) + g(M, M_0)</script>
使ΔF(M,M 0 )=−δ(M−M 0 )M∈τ,则 <script id="MathJax-Element-120" type="math/tex">使\Delta F(M, M_0) = -\delta(M - M_0) \quad M \in \tau, 则</script>
G(M,M 0 )=14πr +g−狄氏格林函数 <script id="MathJax-Element-121" type="math/tex">\color{blue}{G(M, M_0) = \dfrac{1}{4 \pi r} + g -狄氏格林函数}</script>
⎧ ⎩ ⎨ Δg=0,M∈τg| σ =−14πr | σ   <script id="MathJax-Element-122" type="math/tex">\color{blue}{\left \lbrace \begin{array}{l}\Delta g = 0, M \in \tau \\ g|_{\sigma} = -\dfrac{1}{4 \pi r}|_{\sigma} \end{array} \right.}</script>

2.二维: <script id="MathJax-Element-123" type="math/tex">2.二维:</script>
{ΔG=−δ(x−x 0 ,y−y 0 )G| l =0  <script id="MathJax-Element-124" type="math/tex">\left \lbrace \begin{array}{l}\Delta G = -\delta(x-x_0, y-y_0) \\ G|_l = 0 \end{array} \right.</script>
G=12π ln1r +g−狄氏格林函数 <script id="MathJax-Element-125" type="math/tex">\color{blue}{G = \dfrac{1}{2 \pi} \ln \dfrac{1}{r} + g -狄氏格林函数}</script>
⎧ ⎩ ⎨ g=0,M∈σg| l =−12π ln1r | l   <script id="MathJax-Element-126" type="math/tex">\color{blue}{\left \lbrace \begin{array}\Delta g = 0, M \in \sigma \\ g|_l = -\dfrac{1}{2 \pi} \ln \dfrac{1}{r} |_l \end{array} \right.}</script>

3.狄氏格林函数的物理意义: <script id="MathJax-Element-127" type="math/tex">3.狄氏格林函数的物理意义:</script>
{ΔG=−δ(M−M 0 ),M∈τG| σ =0  <script id="MathJax-Element-128" type="math/tex">\left \lbrace \begin{array}{l}\Delta G = - \delta(M - M_0), M \in \tau \\ G | _{\sigma} = 0 \end{array} \right.</script>
G(M,M 0 )=14πr +g⎧ ⎩ ⎨ Δg=0,M∈τg| σ =−14πr | σ   <script id="MathJax-Element-129" type="math/tex">G(M, M_0) = \dfrac{1}{4 \pi r} + g \quad \left \lbrace \begin{array}{l} \Delta g = 0, M \in \tau \\ g |_{\sigma} = -\dfrac{1}{4 \pi r} | _{\sigma} \end{array} \right.</script>
G−M点点位⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ε 0 提供:14πε 0  ε 0 r =14πr 感应电荷提供v:⎧ ⎩ ⎨ Δv=0,M∈τ∵v=gv| σ =−14πr | σ    <script id="MathJax-Element-130" type="math/tex">G-M点点位\left \lbrace \begin{array}{l}\varepsilon_0提供:\dfrac{1}{4 \pi \varepsilon_0} \dfrac{\varepsilon_0}{r} = \dfrac{1}{4 \pi r} \\ 感应电荷提供v:\left \lbrace \begin{array}{l}\Delta v = 0, M \in \tau \because v = g \\ v|_{\sigma} = -\dfrac{1}{4 \pi r} | _{\sigma} \end{array} \right. \end{array} \right.</script>
求G→求M点点位→求感应电荷产生的点位 <script id="MathJax-Element-131" type="math/tex">求G \to 求M点点位 \to 求感应电荷产生的点位</script>

对于三维:即求:⎧ ⎩ ⎨ Δg=0,M∈τg| σ =−14πr | σ   <script id="MathJax-Element-132" type="math/tex">对于三维:即求:\left \lbrace \begin{array}{l}\Delta g = 0, M \in \tau \\ g|_{\sigma} = -\dfrac{1}{4 \pi r} | _{\sigma} \end{array} \right.</script>
对于二维:即求:⎧ ⎩ ⎨ Δg=0,M∈σg| l =−12π ln1r | l   <script id="MathJax-Element-133" type="math/tex">对于二维:即求:\left \lbrace \begin{array}{l}\Delta g = 0, M \in \sigma \\ g|_l = - \dfrac{1}{2 \pi} \ln \dfrac{1}{r} | _l \end{array} \right.</script>

10.3.3用电像法求狄氏格林函数 <script id="MathJax-Element-134" type="math/tex">\color{blue}{10.3.3 用电像法求狄氏格林函数}</script>

1.问题引入: <script id="MathJax-Element-135" type="math/tex">1.问题引入:</script>
求解球内狄氏问题:{Δu=0,ρ<au| ρ=a =f(M)  <script id="MathJax-Element-136" type="math/tex">求解球内狄氏问题:\left \lbrace \begin{array}{l}\Delta u = 0, \rho < a \\ u|_{\rho = a} = f(M) \end{array} \right.</script>
解:u(M)=−∬ σ f(M 0 )∂G∂n 0  dσ 0  <script id="MathJax-Element-137" type="math/tex">解:u(M) = -\iint_{\sigma} f(M_0) \dfrac{\partial G}{\partial n_0} d \sigma_0</script>
G(M,M 0 )=14πr +g;⎧ ⎩ ⎨ Δg=0,M∈ρ<ag| ρ=a =−14πr | ρ=a   <script id="MathJax-Element-138" type="math/tex">G(M, M_0) = \dfrac{1}{4 \pi r} + g; \left \lbrace \begin{array}{l}\Delta g = 0, M \in \rho < a \\ g|_{\rho = a} = - \dfrac{1}{4 \pi r} | _{\rho = a} \end{array} \right.</script>
求u→求G→求M点电位→求感应电荷产生的点位g <script id="MathJax-Element-139" type="math/tex">求u \to 求G \to 求M点电位 \to 求感应电荷产生的点位g</script>

2.用电像法求g: <script id="MathJax-Element-140" type="math/tex">2.用电像法求g:</script>
(1)分析:若能在σ外的某点M 1 放一适当的负q,则 <script id="MathJax-Element-141" type="math/tex">(1)分析:若能在\sigma外的某点M_1放一适当的负q,则</script>
Δ(−q4πε 0 r 1  )=0,M∈ρ<a <script id="MathJax-Element-142" type="math/tex">\Delta(\dfrac{-q}{4 \pi \varepsilon_0 r_1}) = 0, M \in \rho < a</script>
使:−q4πε 0 r 1  | ρ=a =−14πr | ρ=a  <script id="MathJax-Element-143" type="math/tex">使:-\dfrac{q}{4 \pi \varepsilon_0 r_1} |_{\rho = a} = -\dfrac{1}{4 \pi r} | _{\rho = a}</script>
则g=−qrπε 0 r 1   <script id="MathJax-Element-144" type="math/tex">则g = -\dfrac{q}{r \pi \varepsilon_0 r_1}</script>
∴求g→a)确定M 1 的位置;b)确定q大小问题 <script id="MathJax-Element-145" type="math/tex">\therefore 求g \to a)确定M_1的位置; b)确定q大小问题</script>

(2)求球域的G <script id="MathJax-Element-146" type="math/tex">(2)求球域的G</script>
a)r 1 =?记|OM 0 |=ρ 0 ,|OM 1 |=ρ 1 ,使ρ 0 ⋅ρ 1 =a 2 ,即ρ 0 a =aρ 1   <script id="MathJax-Element-147" type="math/tex">a) r_1 = ?记|OM_0| = \rho_0, |OM_1| = \rho_1,使\rho_0 \cdot \rho_1 = a^2, 即\dfrac{\rho_0}{a} = \dfrac{a}{\rho_1}</script>
则称M 1 为M 0 关于球面ρ=a的像 <script id="MathJax-Element-148" type="math/tex">则称M_1为M_0关于球面\rho = a的像</script>
b)q=?1r | σ =? <script id="MathJax-Element-149" type="math/tex">b)q = ? \dfrac{1}{r}|_{\sigma} = ?</script>
∵ΔOM 0 M∼ΔOM 1 M <script id="MathJax-Element-150" type="math/tex">\because \Delta OM_0M \sim \Delta OM_1M</script>
∴ρ 0 a =aρ 1  =rr 1  ,即1r | ρ=a =a/ρ 0 r 1  | ρ=a  <script id="MathJax-Element-151" type="math/tex">\therefore \dfrac{\rho_0}{a} = \dfrac{a}{\rho_1} = \dfrac{r}{r_1}, 即\dfrac{1}{r}|_{\rho = a} = \dfrac{a/\rho_0}{r_1} | _{\rho = a}</script>
g=−ε 0 a/ρ 0 4πε 0 r 1  =−a/ρ 0 4πr 1  ,q=ε 0 aρ 0   <script id="MathJax-Element-152" type="math/tex">g = \dfrac{-\varepsilon_0 a/ \rho_0}{4 \pi \varepsilon_0 r_1} = \dfrac{-a/\rho_0}{4 \pi r_1}, \quad q = \dfrac{\varepsilon_0 a}{\rho_0}</script>
G=14πr −a/ρ 0 4πr 1  ,−q=−ε 0 aρ 0  是ε 0 的电像 <script id="MathJax-Element-153" type="math/tex">G = \dfrac{1}{4 \pi r} - \dfrac{a/\rho_0}{4 \pi r_1}, -q = -\dfrac{\varepsilon_0 a}{\rho_0}是\varepsilon_0的电像</script>

(3)电像法:这种在像点放一虚构的点电荷，来等效代替边界面上的感应电荷所产生的点位的方法称之为电像法。 <script id="MathJax-Element-154" type="math/tex">(3)电像法:这种在像点放一虚构的点电荷，来等效代替边界面上的感应电荷所产生的\\ 点位的方法称之为电像法。</script>

3.求u(M) <script id="MathJax-Element-155" type="math/tex">3.求u(M)</script>
∂G∂n =14π [∂∂ρ (1r )−aρ 0  ∂∂ρ (1r 1  )] <script id="MathJax-Element-156" type="math/tex">\dfrac{\partial G}{\partial n} = \dfrac{1}{4 \pi}[\dfrac{\partial}{\partial \rho}(\dfrac{1}{r}) - \dfrac{a}{\rho_0}\dfrac{\partial}{\partial \rho}(\dfrac{1}{r_1})]</script>
r=ρ 2 +ρ 2 0 −2ρρ 0 cosγ − − − − − − − − − − − − − − − −  √ r 1 =ρ 2 +ρ 2 1 −2ρρ 1 cosγ − − − − − − − − − − − − − − − −  √  <script id="MathJax-Element-157" type="math/tex">r = \sqrt{\rho^2 + \rho_0^2 - 2 \rho \rho_0 \cos \gamma} \quad r_1 = \sqrt{\rho^2 + \rho_1^2 - 2 \rho \rho_1 \cos \gamma}</script>
∂∂ρ 1r =ρ 2 0 −ρ 2 −r 2 2ρr 3  →∂∂ρ (1r ) ρ=a =ρ 2 0 −a 2 −r 2 2ar 3   <script id="MathJax-Element-158" type="math/tex">\dfrac{\partial}{\partial \rho}\dfrac{1}{r} = \dfrac{\rho_0^2 - \rho^2 - r^2}{2 \rho r^3} \to \dfrac{\partial}{\partial \rho}(\dfrac{1}{r})_{\rho=a} = \dfrac{\rho_0^2 - a^2 - r^2}{2a r^3}</script>
类似有:aρ 0  ∂∂ρ (1r 1  ) ρ=a =aρ 0  ρ 2 1 −a 2 −r 2 1 2ar 3 1  =a 2 −r 2 −ρ 2 0 2ar 3   <script id="MathJax-Element-159" type="math/tex">类似有:\dfrac{a}{\rho_0}\dfrac{\partial}{\partial \rho}(\dfrac{1}{r_1})_{\rho=a} = \dfrac{a}{\rho_0} \dfrac{\rho_1^2 - a^2 - r_1^2}{2ar_1^3} = \dfrac{a^2 - r^2 - \rho_0^2}{2ar^3}</script>
∂G∂n | ρ=a =14πa ρ 2 0 −a 2 r 3   <script id="MathJax-Element-160" type="math/tex">\dfrac{\partial G}{\partial n}|_{\rho = a} = \dfrac{1}{4 \pi a} \dfrac{\rho_0^2 - a^2}{r^3}</script>

u(M)=−∬ σ f(M 0 )∂G∂n 0  dσ 0  <script id="MathJax-Element-161" type="math/tex">u(M) = -\iint_{\sigma} f(M_0) \dfrac{\partial G}{\partial n_0} d \sigma_0</script>
=14πa ∫ 2π 0 ∫ π 0 f(θ 0 ,φ 0 )a 2 −ρ 2 (a 2 +ρ 2 −2aρcosγ) 32   a 2 sinθ 0 dθ 0 dφ 0  <script id="MathJax-Element-162" type="math/tex">= \dfrac{1}{4 \pi a} \int_0^{2 \pi} \int_0^{\pi}f(\theta_0, \varphi_0) \dfrac{a^2 - \rho^2}{(a^2 + \rho^2 - 2a \rho \cos \gamma)^{\frac{3}{2}}}a^2 \sin \theta_0 d \theta_0 d \varphi_0</script>

u(M)=a4π ∫ 2π 0 ∫ π 0 f(θ 0 ,φ 0 )a 2 −ρ 2 (a 2 +ρ 2 −2aρcosγ) 32   sinθ 0 dθ 0 dφ 0 −球的泊松积分公式 <script id="MathJax-Element-163" type="math/tex">\color{blue}{u(M) = \dfrac{a}{4 \pi}\int_0^{2 \pi} \int_0^{\pi} f(\theta_0, \varphi_0) \dfrac{a^2 - \rho^2}{(a^2 + \rho^2 - 2a \rho \cos \gamma)^{\frac{3}{2}}} \sin \theta_0 d \theta_0 d \varphi_0 -球的泊松积分公式}</script>
其中，cosγ=sinθsinθ 0 cos(φ−φ 0 )+cosθcosθ 0  <script id="MathJax-Element-164" type="math/tex">其中，\cos \gamma = \sin \theta \sin \theta_0 \cos(\varphi - \varphi_0) + \cos \theta \cos \theta_0</script>

10.3.4注释 <script id="MathJax-Element-165" type="math/tex">\color{blue}{10.3.4 注释}</script>

1.cosγ=? <script id="MathJax-Element-166" type="math/tex">1.\cos \gamma = ?</script>
设I ⃗ 为OM → 方向单位向量,I 0  → 为OM 0  → 方向单位向量,则 <script id="MathJax-Element-167" type="math/tex">设\vec I为\vec{OM}方向单位向量,\vec{I_0}为\vec{OM_0}方向单位向量, 则</script>
I ⃗ =xi ⃗ +yj ⃗ +zk ⃗ =sinθcosφi ⃗ +sinθsinφj ⃗ +cosθk ⃗  <script id="MathJax-Element-168" type="math/tex">\vec I = x \vec i + y \vec j + z \vec k = \sin \theta \cos \varphi \vec i + \sin \theta \sin \varphi \vec j + \cos \theta \vec k</script>
I 0  → =x 0 i ⃗ +y 0 j ⃗ +z 0 k ⃗ =sinθ 0 cosφ 0 i ⃗ +sinθ 0 sinφ 0 j ⃗ +cosθ 0 k ⃗  <script id="MathJax-Element-169" type="math/tex">\vec{I_0} = x_0 \vec i + y_0 \vec j + z_0 \vec k = \sin \theta_0 \cos \varphi_0 \vec i + \sin \theta_0 \sin \varphi_0 \vec j + \cos \theta_0 \vec k</script>
∴I ⃗ ⋅I ⃗  0 =|I ⃗ ⋅I ⃗  0 |cosγ=cosγ <script id="MathJax-Element-170" type="math/tex">\therefore \vec I \cdot \vec I_0 = |\vec I \cdot \vec I_0| \cos \gamma = cos \gamma</script>
=sinθcosφsinθ 0 cosφ 0 +sinθsinφsinθ 0 sinφ 0 +cosθcosθ 0  <script id="MathJax-Element-171" type="math/tex">=\sin \theta \cos \varphi \sin \theta_0 \cos \varphi_0 + \sin \theta \sin \varphi \sin \theta_0 \sin \varphi_0 + \cos \theta \cos \theta_0</script>

2.对于球外的狄氏问题: <script id="MathJax-Element-172" type="math/tex">2.对于球外的狄氏问题:</script>
{Δu 1 =0,ρ>au 1 | ρ=a =f(M)  <script id="MathJax-Element-173" type="math/tex">\left \lbrace \begin{array}{l}\Delta u_1 = 0, \rho > a \\ u_1|_{\rho = a} = f(M) \end{array} \right.</script>
u 1 (M)=−u(M) <script id="MathJax-Element-174" type="math/tex">u_1(M) = -u(M)</script>
u 1 (M)=−a4π ∫ 2π 0 ∫ π 0 f(θ 0 ,φ 0 )a 2 −ρ 2 (a 2 +ρ 2 −2aρcosγ) 32   sinθ 0 dθ 0 dφ 0  <script id="MathJax-Element-175" type="math/tex">\color{blue}{u_1(M) = \dfrac{-a}{4 \pi}\int_0^{2 \pi} \int_0^{\pi} f(\theta_0, \varphi_0) \dfrac{a^2 - \rho^2}{(a^2 + \rho^2 - 2a \rho \cos \gamma)^{\frac{3}{2}}} \sin \theta_0 d \theta_0 d \varphi_0}</script>

3.对于圆内的狄氏问题: <script id="MathJax-Element-176" type="math/tex">3.对于圆内的狄氏问题:</script>
G=12π ln1r +g <script id="MathJax-Element-177" type="math/tex">G = \dfrac{1}{2 \pi} \ln \dfrac{1}{r} +g</script>
{Δu=0,ρ<au| ρ=a =f(φ) ⎧ ⎩ ⎨ Δg=0,M∈σg| ρ=a =−12π ln1r | ρ=a   <script id="MathJax-Element-178" type="math/tex">\left \lbrace \begin{array}{l}\Delta u = 0, \rho < a \\ u|_{\rho=a} = f(\varphi) \end{array} \right. \quad \left \lbrace \begin{array}{l}\Delta g = 0, M \in \sigma \\ g|_{\rho = a} = -\dfrac{1}{2 \pi} \ln \dfrac{1}{r} |_{\rho = a} \end{array} \right.</script>
u(M)=12π ∫ 2π 0 f(φ 0 )a 2 −ρ 2 a 2 +ρ 2 −2aρcos(φ−φ 0 ) dφ 0  <script id="MathJax-Element-179" type="math/tex">\color{blue}{u(M) = \dfrac{1}{2 \pi}\int_0^{2\pi} f(\varphi_0) \dfrac{a^2 - \rho^2}{a^2 + \rho^2 - 2a \rho \cos(\varphi - \varphi_0)} d \varphi_0}</script>

4.对于圆外的狄氏问题: <script id="MathJax-Element-180" type="math/tex">4.对于圆外的狄氏问题:</script>
{Δu 1 =0,ρ>au 1 | ρ=a =f(φ) u 1 (M)=−u(M) <script id="MathJax-Element-181" type="math/tex">\left \lbrace \begin{array}{l}\Delta u_1 = 0, \rho > a \\ u_1 |_{\rho = a} = f(\varphi) \end{array} \right. \quad u_1(M) = -u(M)</script>
u(M)=12π ∫ 2π 0 ∫ 2π 0 f(φ 0 )ρ 2 −a 2 a 2 +ρ 2 −2aρcos(φ−φ 0 ) dφ 0  <script id="MathJax-Element-182" type="math/tex">\color{blue}{u(M) = \dfrac{1}{2 \pi}\int_0^{2\pi}\int_0^{2 \pi} f(\varphi_0)\dfrac{\rho^2 - a^2}{a^2 + \rho^2 - 2a\rho \cos(\varphi - \varphi_0)} d \varphi_0}</script>

10.3.5小结 <script id="MathJax-Element-183" type="math/tex">\color{blue}{10.3.5 小结}</script>

ΔG=−δ(M−M 0 )→G=14πr  <script id="MathJax-Element-184" type="math/tex">\Delta G = -\delta(M - M_0) \to G = \dfrac{1}{4 \pi r}</script>
{ΔG=−δ(M−M 0 ),M∈τG| σ =0 →G(M,M 0 )=14πr +g <script id="MathJax-Element-185" type="math/tex">\left \lbrace \begin{array}{l} \Delta G = -\delta(M-M_0), M \in \tau \\ G|_{\sigma} = 0 \end{array} \right. \to G(M, M_0) = \dfrac{1}{4 \pi r} + g</script>
ρ<a:G=14πr −a/ρ 0 4πr 1   <script id="MathJax-Element-186" type="math/tex">\rho < a: G = \dfrac{1}{4 \pi r} - \dfrac{a/\rho_0}{4 \pi r_1}</script>
⎧ ⎩ ⎨ Δg=0g| σ =−14πr | σ   <script id="MathJax-Element-187" type="math/tex">\left \lbrace \begin{array}{l}\Delta g = 0 \\ g|_{\sigma} = -\dfrac{1}{4 \pi r}|_{\sigma} \end{array} \right.</script>
{Δu=0,ρ<au| ρ=a =f(M)  <script id="MathJax-Element-188" type="math/tex">\left \lbrace \begin{array}{l}\Delta u = 0, \rho < a \\ u|_{\rho = a} =f(M) \end{array} \right.</script>
u(M)=a4π ∫ 2π 0 ∫ π 0 f(θ 0 ,φ 0 )a 2 −ρ 2 (a 2 +ρ 2 −2aρcosγ) 32   sinθ 0 dθ 0 dφ 0  <script id="MathJax-Element-189" type="math/tex">u(M) = \dfrac{a}{4 \pi}\int_0^{2\pi}\int_0^{\pi}f(\theta_0, \varphi_0)\dfrac{a^2 - \rho^2}{(a^2 + \rho^2 - 2a \rho \cos \gamma)^{\frac{3}{2}}}\sin \theta_0 d \theta_0 d \varphi_0</script>