电磁场与波（6）：进一步分离变量

直角坐标系分离变量

求解如下边值问题：
在一沿xyz轴长度分别为abc的长方体中，只有一个面上电势不为零

$$\begin{aligned} &\nabla^2 \varphi = 0 \\ &\varphi |_{z=c} = f(x,y)\\ &\varphi_{x=0} = \varphi_{x=a} = 0\\ &\varphi_{y=0} = \varphi_{y=b} = 0\\ &\varphi_{z=0} = 0 \end{aligned}$$

解：考虑如下可分离变量的形式的解

$$\varphi(x,y,z) = X(x)Y(y)Z(z)$$

代入方程中，得到

$$\nabla^2 \varphi = \frac{d^2X}{dx^2}Y(z)Z(z) + X(x)\frac{d^2Y}{dy^2}Z(z) + X(x)Y(y)\frac{d^2Z}{dz^2} = 0\\$$

$$\frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2} + \frac{1}{Z}\frac{d^2Z}{dz^2} = 0$$

三项都只与一个变量有关，因此每一项都应该等于一个常数

$$\frac{1}{X}\frac{d^2X}{dx^2} = \text{const} = -k_x^2\\ \frac{1}{Y}\frac{d^2Y}{dy^2} = \text{const} = -k_y^2\\ \frac{1}{Z}\frac{d^2Z}{dz^2} = \text{const} = -k_z^2\\ k_x^2 + k_y^2 + k_z^2 = 0$$

边界条件

$$\begin{aligned} \varphi|_{x=0} &= 0 \Rightarrow X(0) = 0\\ \varphi|_{x=a} &= 0 \Rightarrow X(a) = 0\\ \varphi|_{y=0} &= 0 \Rightarrow Y(0) = 0\\ \varphi|_{y=b} &= 0 \Rightarrow Y(b) = 0\\ \varphi|_{z=0} &= 0 \Rightarrow Z(0) = 0\\ \varphi|_{z=c} &= f(x,y) \nRightarrow X(x)Y(y)Z(c) = f(x,y) \end{aligned}$$

$X(x)$可能的解为：

• $k_x^2 = 0,X(x) = Ax+B$
• $k_x^2 > 0,X(x) = A\sin(k_xx) + B\cos(k_xx)$
• $k_x^2 < 0,X(x) = A\exp(k_xx) + B\exp(-k_xx) = A\sh k_xx+B\ch k_xx$

双曲正弦：$\sh x = \frac{e^x - e^{-x}}{2}$
双曲余弦：$\ch x = \frac{e^x + e^{-x}}{2}$

根据边界条件$X(0) = 0,X(a) = 0$，我们可以得知只有

$$k_x^2 > 0,X(x) = A\sin(k_xx) + B\cos(k_xx)$$

成立，并且根据边界条件可以得到

$$X(x) = A\sin k_xx,k_x = \frac{m\pi}{a},m=1,2,3,\cdots$$

同理有

$$Y(y) = C\sin k_yy,k_y = \frac{n\pi}{b},n=1,2,3,\cdots$$

由于

$$k_x^2 + k_y^2 + k_z^2 = 0$$

我们得到

$$k_z^2 = -k_x^2 - k_y^2 = -\frac{m^2\pi^2}{a^2} - \frac{n^2\pi^2}{b^2} = -K_{mn}^2 < 0$$

由$Z(0) = 0$得到$F=0$
所以$Z(z) = E\sh K_{mn}z$
至此可以得到

$$\varphi_{mn}(x,y,z) = A\sin k_xx \sin k_yy \sh K_{mn}z$$

取$\varphi_{mn}(x,y,z)$的线性组合

$$\varphi(x,y,z) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{mn}\sin k_xx \sin k_yy \sh K_{mn}z$$

令$\varphi(x,y,z)|_{z=c} = f(x,y)$

$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{mn}\sin k_xx \sin k_yy \sh K_{mn}c = f(x,y)$$

得到

$$A_{mn} = \frac{4\int_0^a dx \int_0^b f(x,y) \sin \frac{m\pi}{a}x \sin \frac{n\pi}{b}y dy}{ab\sh K_{mn}c}$$

完备正交基

$[0,a]$上的完备正交基

• Fourier级数：$1,\sin \frac{2\pi}{a}x,\cos \frac{2\pi}{a}x,\sin \frac{4\pi}{a}x,\cos \frac{4\pi}{a}x,\cdots$
• Cos级数：$1,\cos \frac{2\pi}{a}x,\cos \frac{4\pi}{a}x,\cdots$
• Sin级数：$\sin \frac{2\pi}{a}x,\sin \frac{4\pi}{a}x,\cdots$

应用举例

依旧在之前的长方体中，问题如下

$$\begin{aligned} &\nabla^2 \varphi = 0 \\ &\varphi |_{z=c} = f(x,y)\\ &\varphi_{y=b} = g(x,z)\\ &\varphi_{\text{其余}} = 0 \end{aligned}$$

令$\varphi = \varphi_1 + \varphi_2$
其中$\varphi_1$满足

$$\begin{aligned} &\nabla^2 \varphi_1 = 0 \\ &\varphi_1 |_{z=c} = f(x,y)\\ &\varphi_{\text{其余}} = 0 \end{aligned}$$

$\varphi_2$满足

$$\begin{aligned} &\nabla^2 \varphi_2 = 0 \\ &\varphi_2 |_{y=b} = g(x,z)\\ &\varphi_{\text{其余}} = 0 \end{aligned}$$

后面进行相似的求解即可

分离变量一般步骤

1. 写出可分离变量的形式解
2. 根据齐次边界条件，确定函数系$\varphi_{mn}(x,y,z)$
3. 取通解：$\varphi = \sum_{m,n} A_{mn}\varphi_{mn}(x,y,z)$
4. 代入非齐次边界条件，得到$A_{mn}$的表达式

球坐标分离变量

$$\varphi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$$

$$\nabla^2 \varphi = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial \varphi}{\partial r})\Theta\Phi + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta \frac{\partial \varphi}{\partial \theta}) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \varphi}{\partial \phi^2} = 0$$

也即

$$\begin{aligned} &\frac{d^2\Phi}{d\phi^2} + m^2\Phi = 0\\ &\frac{d}{dr}(r^2\frac{dR}{dr}) -n(n+1) R = 0\\ &\frac{1}{\sin\theta}\frac{d}{d\theta}(\sin\theta \frac{d\Theta}{d\theta}) + [n(n+1) - \frac{m^2}{\sin^2\theta}]\Theta = 0 \end{aligned}$$

最后一个方程被称为关联Legendre方程
对于$\Phi(\phi)$：只考虑轴对称的情况，$\Phi$与$\phi$无关，所以$\Phi = \text{const}$
对于$R(r)$：

$$\begin{aligned} &R(r) = Ar^n + Br^{-n-1},n^2\geq0\\ &R(r) \text{不存在},n^2<0\\ \end{aligned}$$

对于$\Theta(\theta)$：

$$\Theta(\theta) = AP_n(\cos\theta) + BQ_n(\cos\theta),n=0,1,2,\cdots$$

其中$P_n(\theta)$为第一类Legendre函数，$Q_n(\theta)$为第二类Legendre函数
Recall：根据上学期PDE的内容，第二类Legendre函数通常会被消掉

Legendre函数性质

• 在极轴附近$\theta \rightarrow 0$，$\cos(\theta) \rightarrow \pm 1$时

$$|P_n(\pm 1)| = 1,|Q_n(\pm 1)| \rightarrow \infty$$

• 正交性

$$\int_{-1}^1 P_n(x)P_m(x)dx = \frac{2}{2n+1}\delta_{mn}$$

• 完备性
  $[-1,1]$上的任意函数$f(x)$都可以展开为

$$f(x) = \sum_{n=0}^{\infty} C_n P_n(x)$$

其中

$$C_n = \frac{2n+1}{2}\int_{-1}^1 f(x)P_n(x)dx$$

*求解区域包含极轴，并且解具有轴对称性质的情况

此时通解可以写作

$$\varphi(r,\theta) = \sum_{n=0}^{\infty} (A_n r^n + B_n r^{-n-1})P_n(\cos\theta)\\$$

常见的Legendre函数有

$$\begin{aligned} &P_0(\cos\theta) = 1\\ &P_1(\cos\theta) = \cos\theta\\ &P_2(\cos\theta) = \frac{1}{2}(3\cos^2\theta - 1)\\ \end{aligned}$$

例题

无限大的均匀外场$\vec{E_0}$中放置一个不带电的导体球，球半径为$a$，求空间电势

解：以球心为原点，$\vec{E_0}$方向为极轴，建立球坐标系
求解区域为球外空间，包含极轴，且有轴对称性
则通解可以写作

$$\varphi = \sum_{n=0}^{\infty} (A_n r^n + B_n r^{-n-1})P_n(\cos\theta)\\$$

• 由$r\rightarrow \infty$时，$\varphi \rightarrow -{E_0} {r}\cos \theta$，得到

$$\sum_{n=0}^{\infty} A_n r^n P_n(\cos\theta) = -E_0 r \cos\theta$$

逐项比较得到

$$A_0 = 0,A_1 = -E_0,A_n = 0,n\geq2$$

• 由$\varphi|_{r=a} = \text{const}$，得到

$$-E_0 a \cos\theta + \sum_{n=0}^{\infty} B_n a^{-n-1} P_n(\cos\theta) = \text{const}$$

逐项比较得到

$$B_1 = E_0a^3,B_n = 0,n\geq2$$

（目前还没有确定$B_0$的值）

• 导体球总电荷为零

$$\oint_{r=a} \frac{\partial \varphi}{\partial r} dS = 0\\ \oint_{r=a} (-E_0 \cos \theta - \frac{B_0}{a^2} - 2\frac{E_0a^3}{a^3} \cos\theta) dS = 0 \Rightarrow B_0 = 0$$

综上

$$\varphi = -E_0 r \cos\theta + \frac{E_0 a^3}{r^2} \cos\theta$$

其中第一项为匀强外场的电势，第二项为感应电势

考虑原点处沿极轴方向的电偶极子$\vec{P}$
电势：

$$\varphi = \frac{1}{4\pi \varepsilon_0} \frac{\vec{P} \cdot \vec{r}}{r^3} = \frac{1}{4\pi \varepsilon_0} \frac{P\cos\theta}{r^2}$$

对比可知道

$$P = 4\pi \varepsilon_0 E_0 a^3$$

球面电荷密度

$$\rho_S = -\varepsilon_0\frac{\partial \varphi}{\partial r} |_{r=a} = 3\varepsilon_0 E_0\cos\theta$$

*Green函数法

待求解问题

$$\begin{aligned} &\nabla^2 \varphi = -\frac{\rho}{\epsilon} \\ &\text{边界条件}\varphi|_{S} \text{或} \frac{\partial \varphi}{\partial n}|_{S} \end{aligned}$$

Green函数性质

$$\nabla^2 G(\vec{x},\vec{x'}) = -\frac{\delta(\vec{x}-\vec{x'})}{\varepsilon}$$

方程求解

使用Green第二恒等式

$$\int_V (\varphi \nabla^2 \psi - \psi \nabla^2 \varphi) dV = \oint_S (\varphi \nabla \psi - \psi \nabla \varphi) \cdot d\vec{S}$$

取$\psi = G(\vec{x},\vec{x'})$

$$\int_V (\varphi \nabla^2 G - G \nabla^2 \varphi) dV = \oint_S (\varphi \nabla G - G \nabla \varphi) \cdot d\vec{S}\\ \int_V (-\varphi \frac{\delta(\vec{x}-\vec{x'})}{\varepsilon} - G \nabla^2 \frac{\rho(\vec{x}')}{\varepsilon}) dV = \oint_S (\varphi \nabla G - G \nabla \varphi) \cdot d\vec{S}$$

有形式解

$$\phi(\vec{x'}) = \int_V G(\vec{x},\vec{x'}) \rho(\vec{x'}) dV + \varepsilon \oint_S G(\vec{x},\vec{x'}) \frac{\partial \varphi(\vec{x})}{\partial n}ds - \varepsilon \oint_S \varphi(\vec{x}) \frac{\partial G(\vec{x},\vec{x'})}{\partial n}ds$$

将$\vec{x},\vec{x}'$对换

$$\phi(\vec{x}) = \int_V G(\vec{x},\vec{x'}) \rho(\vec{x'}) dV' + \varepsilon \oint_S G(\vec{x},\vec{x'}) \frac{\partial \varphi(\vec{x'})}{\partial n}ds' - \varepsilon \oint_S \varphi(\vec{x'}) \frac{\partial G(\vec{x},\vec{x'})}{\partial n}ds'$$

• 第一类边值问题，$\varphi|_S$已知
  取$G|_{S} = 0$，则

$$\phi(\vec{x}) = \int_V G(\vec{x},\vec{x'}) \rho(\vec{x'}) dV' - \varepsilon \oint_S \varphi(\vec{x'}) \frac{\partial G(\vec{x},\vec{x'})}{\partial n}ds'$$

• 第二类边值问题，$\frac{\partial \varphi}{\partial n}|_S$已知
  取$\frac{\partial G}{\partial n}|_S = \text{const} = -\frac{1}{\varepsilon S}$，则

$$\varphi = \int_V G(\vec{x},\vec{x'}) \rho(\vec{x'}) dV' + \varepsilon \oint_S G(\vec{x},\vec{x'}) \frac{\partial \varphi(\vec{x'})}{\partial n}ds' + \langle\varphi \rangle_S$$

其中

$$\langle\varphi \rangle_S = \frac{1}{S} \oint_S \varphi(\vec{x'}) dS'$$

第一类边值问题

$$\begin{aligned} &\nabla^2 G = -\frac{\delta(\vec{x}-\vec{x'})}{\varepsilon}\\ &G|_S = 0 \end{aligned}$$

课本中可能出现的差异

《电磁场》

$$\nabla^2 G = -\delta(\vec{x}-\vec{x'})\\$$

《电动力学》和《电磁场》中$\hat{n}$方向相反（前者为外法线，后者为内法线）

静电场部分结束

静磁场

虚晃一枪