数学物理方法(3)
本文最后更新于 2025年8月14日 星期四 17:24
顾樵. 数学物理方法. 科学出版社, 2012. 2022 年 12 月第 25 次印刷. 978-7-03-033064-2.
10 行波法
10.2 一维波动方程的 d'Alembert 公式
10.2.1 d'Alembert 公式的推导
\[ \begin{cases} \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2} & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x),\quad\left.\frac{\partial u}{\partial t}\right|_{t=0}=\psi(x) & (x\in\mathbb R) \end{cases} \]
\[ u(x,t)=\frac12[\phi(x+at)+\phi(x-at)]+\frac{1}{2a}\int_{x-at}^{x+at}\psi(\xi)\mathrm{d}\xi \]
10.4 一阶线性微分方程的特征线法
\[ \begin{cases} \frac{\partial u}{\partial t}=a(x,t)\frac{\partial u}{\partial x}+b(x,t)u=f(x,t) & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \end{cases} \]
\(u(x,t)=u[x(t),t]\equiv U(t)\) 满足常微分方程。
由
\[ \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t}=a(x,t) \\ x(0)=c \end{cases} \]
解得 \(x(t)\) 的表达式。又由
\[ \begin{cases} \frac{\mathrm{d}U}{\mathrm{d}t}+b[x(t),t]U=a(x,t) \\ \left.U\right|_{t=0}=\phi(c) \end{cases} \]
解得 \(U(t)\) 的表达式。
10.5 非齐次波动方程：齐次化原理
\[ \begin{cases} \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2}+f(x,t) & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x),\quad\left.\frac{\partial u}{\partial t}\right|_{t=0}=\psi(x) & (x\in\mathbb R) \end{cases} \]
\[ u(x,t)=\frac12[\phi(x+at)+\phi(x-at)]+\frac{1}{2a}\int_{x-at}^{x+at}\psi(\xi)\mathrm{d}\xi+\frac{1}{2a}\int_0^t\int_{x-a(t-\tau)}^{x+a(t-\tau)}f(\xi,\tau)\mathrm{d}\xi\mathrm{d}\tau \]
10.6 三维波动方程
\[ \begin{cases} \frac{\partial^2u}{\partial t^2}=a^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}\right) & (x,y,z\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x,y,z),\quad\left.\frac{\partial u}{\partial t}\right|_{t=0}=\psi(x,y,z) & (x,y,z\in\mathbb R) \end{cases} \]
10.6.1 三维波动方程的球对称解
\[ u(r,t)=\frac{f_1(r+at)}{r}+\frac{f_2(r-at)}{r} \]
10.6.2 三维波动方程的 Poisson 公式
\[ \begin{gather} u(x,t)=\frac{\partial}{\partial t}[t\langle\phi(x,t)\rangle]+t\langle\psi(x,t)\rangle \\ u(x,y,z,t)=\frac{1}{4\pi a}\frac{\partial}{\partial t}\iint_{S_{at}^M}\frac{\phi(M')}{at}\mathrm{d}S+\frac{1}{4\pi a}\iint_{S_{at}^M}\frac{\psi(M')}{at}\mathrm{d}S \end{gather} \]
其中 \(M'(x+r\sin\theta\cos\phi,y+r\sin\theta\sin\phi,z+r\cos\theta)\)。
10.6.3 Poisson 公式的物理意义
- 二维波动方程
\[ \begin{align} u(x,y,t) &= \frac{1}{2\pi a}\frac{\partial}{\partial t}\iint_{\Sigma_{at}^M}\frac{\phi(\xi,\eta)}{\sqrt{(at)^2-(\xi-x)^2-(\eta-y)^2}}\mathrm{d}\xi\mathrm{d}\eta \notag \\ & \quad+\frac{1}{2\pi a}\iint_{\Sigma_{at}^M}\frac{\psi(\xi,\eta)}{\sqrt{(at)^2-(\xi-x)^2-(\eta-y)^2}}\mathrm{d}\xi\mathrm{d}\eta \end{align} \]
10.7 旁轴波动方程： Green 算符法
10.7.1 旁轴波动方程的解
一列沿 \(z\) 轴正向自由传播的单色光束
\[ u(\mathbf r,z,t)=\psi(\mathbf r,z)\exp[i(kz-\omega t)] \]
得
\[ i\beta\frac{\partial}{\partial z}\psi(\mathbf r,z)=\hat H\psi(\mathbf r,z) \]
其中 \(\beta=-2k\)，\(\hat H=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\)。
\[ i\beta\frac{\partial}{\partial z}|\psi(z)\rangle=\hat H|\psi(z)\rangle \]
积分
\[ \int_{|\psi(z_0)\rangle}^{|\psi(z)\rangle}\frac{\mathrm{d}|\psi(z)\rangle}{|\psi(z)\rangle}=-\frac{i}{\beta}\hat H\int_{z_0}^z\mathrm{d}z \\ \]
得到态矢量
\[ |\psi(z)\rangle=\hat G_L|\psi(z_0)\rangle \]
Green 算符
\[ \begin{gather} \hat G_L=\exp\left(-i\frac{L}{\beta}\hat H\right)=\exp\left(i\frac{L}{2k}\hat H\right) \\ L=z-z_0 \end{gather} \]
坐标表象的波函数
\[ \langle\mathbf r|\psi(z)\rangle=\psi(\mathbf r,z)=\int_{-\infty}^\infty G_L(\mathbf r,\mathbf r_0)\psi(\mathbf r_0,z_0)\mathrm{d}\mathbf r_0 \]
其中 矩阵元 \(G_L(\mathbf r,\mathbf r_0)=\langle\mathbf r|\hat G_L|\mathbf r_0\rangle\) 称为 Green 函数。
11 积分变换法
11.1 Fourier 变换法
11.1.1 热传导问题与 Gauss 核
- Gauss 核(标准差为 \(\sigma=a\sqrt{2t}\) 的 Gauss 分布)
\[ \begin{gather} K(x,t)= \begin{cases} \frac{1}{2a\sqrt{\pi t}}\exp\left(-\frac{x^2}{4a^2t}\right) & (t>0) \\ 0 & (t\leq0) \end{cases} \\ \mathscr F\{K(x,t)\}=\mathrm{e}^{-\omega^2a^2t} \end{gather} \]
- Poisson 核
\[ \begin{gather} P(x,y)=\frac{1}{\pi}\frac{y}{x^2+y^2} \\ \mathscr F\{P(x,y)\}=\mathrm{e}^{-|\omega|y} \end{gather} \]
11.2 Laplace 变换法
- 余误差函数
\[ \mathrm{erfc}(x)=\frac{2}{\sqrt{\pi}}\int_x^\infty \mathrm{e}^{-x^2}\mathrm{d}x \]
11.3 联合变换法
11.3.3 有源热传导问题
- 误差函数
\[ \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x \mathrm{e}^{-x^2}\mathrm{d}x \]
12 Green 函数法
12.4 Green 公式
12.4.3 Green 公式
第一 Green 公式
\[ \iiint_\Omega(u\nabla^2v)\mathrm{d}V=\iint_\Gamma u\frac{\partial v}{\partial n}\mathrm{d}S-\iiint_\Omega(\nabla u\cdot\nabla v)\mathrm{d}V \]
第二 Green 公式(简称 Green 公式)
\[ \iiint_\Omega(u\nabla^2v-v\nabla^2u)\mathrm{d}V=\iint_\Gamma\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\mathrm{d}S \]
二维 Green 公式
\[ \iint_D(u\nabla^2v-v\nabla^2u)\mathrm{d}\sigma=\int_C\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)\mathrm{d}l \]
12.5 Laplace 方程和 Poisson 方程
12.5.1 Laplace 方程的基本解
- 三维 Laplace 方程(球坐标系)
\[ \begin{gather} u(r)=c_1\frac1r+c_2 \\ u(r)=\frac{1}{4\pi r} \end{gather} \]
- 二维 Laplace 方程(极坐标系)
\[ \begin{gather} u(r)=c_1\ln r+c_2 \\ u(r)=\frac{1}{2\pi}\ln\frac1r \end{gather} \]
12.5.2 Poisson 方程的基本积分公式
\[ \begin{gather} \nabla^2u(\mathbf r)=-f(\mathbf r) \\ \nabla^2G(\mathbf r,\mathbf r_0)=-\delta(\mathbf r-\mathbf r_0) \\ u(\mathbf r)=\iiint_\Omega G(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}V_0+\iint_\Gamma\left[G(\mathbf r,\mathbf r_0)\frac{\partial u(\mathbf r_0)}{\partial n_0}-u(\mathbf r_0)\frac{\partial G(\mathbf r,\mathbf r_0)}{\partial n_0}\right]\mathrm{d}S_0 \\ u(\mathbf r)=\iint_DG(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}\sigma_0+\int_C\left[G(\mathbf r,\mathbf r_0)\frac{\partial u(\mathbf r_0)}{\partial n_0}-u(\mathbf r_0)\frac{\partial G(\mathbf r,\mathbf r_0)}{\partial n_0}\right]\mathrm{d}l_0 \end{gather} \]
12.5.3 Poisson 方程的边值问题
第一类边值问题
\[ \begin{gather} \begin{cases} \nabla^2u(\mathbf r)=-f(\mathbf r) \\ u(\mathbf r)|_\Gamma=g(\mathbf r) \end{cases} \\ \begin{cases} \nabla^2G(\mathbf r,\mathbf r_0)=-\delta(\mathbf r-\mathbf r_0) \\ G|_\Gamma=0 \end{cases} \\ u(\mathbf r)=\iiint_\Omega G(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}V_0-\iint_\Gamma g(\mathbf r_0)\frac{\partial G(\mathbf r,\mathbf r_0)}{\partial n_0}\mathrm{d}S_0 \\ u(\mathbf r)=\iint_DG(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}\sigma_0-\int_C g(\mathbf r_0)\frac{\partial G(\mathbf r,\mathbf r_0)}{\partial n_0}\mathrm{d}l_0 \end{gather} \]
第二类边值问题
\[ \begin{gather} \begin{cases} \nabla^2u(\mathbf r)=-f(\mathbf r) \\ \left.\frac{\partial u}{\partial n}\right|_\Gamma=h(\mathbf r) \end{cases} \\ \begin{cases} \nabla^2G(\mathbf r,\mathbf r_0)=-\delta(\mathbf r-\mathbf r_0) \\ \left.\frac{\partial G}{\partial n}\right|_\Gamma=0 \end{cases} \\ u(\mathbf r)=\iiint_\Omega G(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}V_0+\iint_\Gamma G(\mathbf r,\mathbf r_0)h(\mathbf r_0)\mathrm{d}S_0 \\ u(\mathbf r)=\iint_DG(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}\sigma_0+\int_C G(\mathbf r,\mathbf r_0)h(\mathbf r_0)\mathrm{d}l_0 \end{gather} \]
第三类边值问题
\[ \begin{gather} \begin{cases} \nabla^2u(\mathbf r)=-f(\mathbf r) \\ \left.\left[u+\alpha\frac{\partial u}{\partial n}\right]\right|_\Gamma=z(\mathbf r) \end{cases} \\ \begin{cases} \nabla^2G(\mathbf r,\mathbf r_0)=-\delta(\mathbf r-\mathbf r_0) \\ \left.\left[G+\alpha\frac{\partial G}{\partial n}\right]\right|_\Gamma=0 \end{cases} \\ u(\mathbf r)=\iiint_\Omega G(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}V_0+\frac{1}{\alpha}\iint_\Gamma G(\mathbf r,\mathbf r_0)z(\mathbf r_0)\mathrm{d}S_0 \\ u(\mathbf r)=\iint_DG(\mathbf r,\mathbf r_0)f(\mathbf r_0)\mathrm{d}\sigma_0+\frac{1}{\alpha}\int_C G(\mathbf r,\mathbf r_0)z(\mathbf r_0)\mathrm{d}l_0 \end{gather} \]
12.6 Green 函数法的应用：电像法
- 上半空间的第一类边值问题
\[ \begin{align} G(M,M_0)&= \frac1{4\pi}\frac1{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}} \\ & \quad-\frac1{4\pi}\frac1{\sqrt{(x-x_0)^2+(y-y_0)^2+(z+z_0)^2}} \\ \frac{\partial G}{\partial z_0} &= \frac1{4\pi}\frac{z-z_0}{[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]^{3/2}} \\ & \quad+\frac1{4\pi}\frac{z+z_0}{[(x-x_0)^2+(y-y_0)^2+(z+z_0)^2]^{3/2}} \\ \left.\frac{\partial G}{\partial n_0}\right|_\Gamma &=-\left.\frac{\partial G}{\partial z_0}\right|_{z_0=0}=-\frac{z}{2\pi}\frac{1}{[(x-x_0)^2+(y-y_0)^2+z^2]^{3/2}} \end{align} \]
\[ \begin{align} u(\mathbf r) &= \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{z_0=0}^\infty G(M,M_0)f(x_0,y_0,z_0)\mathrm{d}x_0\mathrm{d}y_0\mathrm{d}z_0 \\ & \quad+\frac{z}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{g(x_0,y_0)}{[(x-x_0)^2+(y-y_0)^2+z^2]^{3/2}}\mathrm{d}x_0\mathrm{d}y_0 \end{align} \]
\[ \begin{align} G(M,M_0)&= \frac1{4\pi}\ln\frac{(x-x_0)^2+(y+y_0)^2}{(x-x_0)^2+(y-y_0)^2} \\ \left.\frac{\partial G}{\partial n_0}\right|_C &=-\left.\frac{\partial G}{\partial y_0}\right|_{y_0=0}= -\frac{1}{\pi}\frac{y}{(x-x_0)^2+y^2} \end{align} \]
\[ \begin{align} u(\mathbf r) &= \frac{1}{4\pi}\int_{-\infty}^\infty\int_{y_0=0}^\infty f(x_0,y_0)\ln\frac{(x-x_0)^2+(y+y_0)^2}{(x-x_0)^2+(y-y_0)^2}\mathrm{d}x_0\mathrm{d}y_0 \\ & \quad+\frac{y}{\pi}\int_{-\infty}^\infty\frac{g(x_0)}{(x-x_0)^2+y^2}\mathrm{d}x_0 \end{align} \]
- 球域内第一类边值问题
\[ \begin{align} G(M,M_0) &= \frac1{4\pi}\frac1{\sqrt{r_0^2+r^2-2r_0r\cos\gamma}} \\ & \quad-\frac1{4\pi}\frac{R}{\sqrt{r_0^2r^2+R^4-2R^2r_0r\cos\gamma}} \\ \frac{\partial G}{\partial r_0} &= -\frac{1}{4\pi}\frac{r_0-r\cos\gamma}{(r_0^2+r^2-2r_0r\cos\gamma)^{3/2}} \\ & \quad+\frac1{4\pi}\frac{R(r_0r^2-R^2r\cos\gamma)}{(r_0^2r^2+R^4-2R^2r_0r\cos\gamma)^{3/2}} \\ \left.\frac{\partial G}{\partial n_0}\right|_\Gamma &=\left.\frac{\partial G}{\partial r_0}\right|_{r_0=R}=-\frac{1}{4\pi R}\frac{R^2-r^2}{(R^2+r^2-2rR\cos\gamma)^{3/2}} \end{align} \]
\[ \begin{align} u(r,\theta,\phi) &= \int_{0}^R\int_{0}^\pi\int_{0}^{2\pi}G(M,M_0)f(r_0,\theta_0,\phi_0)r_0^2\sin\theta_0\mathrm{d}r_0\mathrm{d}\theta_0\mathrm{d}\phi_0 \\ & \quad+\frac{R(R^2-r^2)}{4\pi}\int_{0}^{2\pi}\int_{0}^\pi\frac{g(R,\theta_0,\phi_0)}{(R^2+r^2-2rR\cos\gamma)^{3/2}}\sin\theta_0\mathrm{d}\theta_0\mathrm{d}\phi_0 \end{align} \]
\[ \cos\gamma=\sin\theta\sin\theta_0\cos(\phi-\phi_0)+\cos\theta\cos\theta_0 \]
- 圆域内第一类边值问题
\[ \begin{gather} G(M,M_0)=\frac1{4\pi}\ln\frac{r_0^2r^2+R^4-2r_0rR^2\cos(\theta-\theta_0)}{R^2[r^2+r_0^2-2r_0r\cos(\theta-\theta_0)]} \\ \frac{\partial G}{\partial r_0}=-\frac{1}{2\pi}\frac{(R^2-r^2)\left[r_0(r^2+R^2)-r(r_0^2+R^2)\cos(\theta-\theta_0)\right]}{(r_0^2r^2+R^4-2r_0rR^2\cos(\theta-\theta_0))(r_0^2+r^2-2r_0r\cos(\theta-\theta_0))} \\ \left.\frac{\partial G}{\partial n_0}\right|_C=\left.\frac{\partial G}{\partial r_0}\right|_{r_0=R}=-\frac{1}{2\pi R}\frac{R^2-r^2}{R^2+r^2-2rR\cos(\theta-\theta_0)} \end{gather} \]
\[ \begin{align} u(r,\theta) &= \int_{0}^R\int_{0}^{2\pi}G(M,M_0)f(r_0,\theta_0)r_0\mathrm{d}r_0\mathrm{d}\theta_0 \\ & \quad+\frac{R^2-r^2}{2\pi}\int_{0}^{2\pi}\frac{g(\theta_0)}{R^2+r^2-2rR\cos(\theta-\theta_0)}\mathrm{d}\theta_0 \end{align} \]
- 圆域的 Poisson 核
\[ P(r,\theta)=\frac{R^2-r^2}{R^2+r^2-2rR\cos(\theta-\theta_0)}\quad(0\leq r<R,0\leq\theta\leq2\pi) \]