数学物理方法(2)基本数学物理方程

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.

5 基础理论知识

5.1 波动方程

5.1.1 弦振动问题

\[ \left(\left.\frac{\partial u}{\partial x}\right|_{x+\Delta x}-\left.\frac{\partial u}{\partial x}\right|_{x}\right)=\Delta\left(\frac{\partial u}{\partial x}\right)\approx\frac{\partial^2u}{\partial x^2}\Delta x \]

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\[ \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2} \]

其中 \(a=\sqrt{T/\rho}\) 为振动传播的速度，即波速；\(T\) 为张力；\(\rho=m/\Delta x\)。

5.1.2 强迫振动与阻尼振动

• 强迫振动：单位长度附加力 \(F(x,t)\)，则微元附加力 \(F(x,t)\Delta x\)，方向向上。

\[ \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2}+f(x,t) \]

其中 \(f(x,t)=\frac{F(x,t)}{\rho}\)。如果附加力是重力，则

\[ \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2}-g \]

外界驱动力称为“力项”(force term)或驱动项。

• 阻尼振动：

\[ \begin{gather} F(x,t)=-k\frac{\partial u}{\partial t} \\ f(x,t)=\frac{F(x,t)}{\rho}=-\frac{k}{\rho}\frac{\partial u}{\partial t}=-2b\frac{\partial u}{\partial t} \end{gather} \]

得

\[ \frac{\partial^2u}{\partial t^2}-a^2\frac{\partial^2u}{\partial x^2}+2b\frac{\partial u}{\partial t}=0 \]

• 推广：

\[ \frac{\partial^2u}{\partial t^2}=a^2\nabla^2u+f(x,y,\ldots,t) \]

5.1.3 高频传输线问题

\[ \begin{gather} \frac{\partial^2i}{\partial x^2}=LC\frac{\partial^2i}{\partial t^2}+(RC+GL)\frac{\partial i}{\partial t}+GRi \\ \frac{\partial^2v}{\partial x^2}=LC\frac{\partial^2v}{\partial t^2}+(RC+GL)\frac{\partial v}{\partial t}+GRv \end{gather} \]

形如

\[ \frac{\partial^2u}{\partial t^2}-a^2\frac{\partial^2u}{\partial x^2}+2b\frac{\partial u}{\partial t}+cu=0 \]

的方程称为 电报方程，其中 \(a,b,c>0\)。

忽略电阻与电漏的影响，约化为

\[ \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2} \]

5.2 热传导方程

\[ \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2} \]

其中 \(a=\sqrt{k/c\rho}\) 为热扩散系数；\(k\) 为热导率；\(u\) 为温度。

• 有源热传导方程

\[ \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2}+f(x,t) \]

其中 \(f(x,t)\) 表示一个时空依赖的外热源。

• 导线内的热传导方程

\[ \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2}+\frac{j^2r}{c\rho} \]

5.4 二阶偏微分方程

5.4.1 分类与标准形式

二阶线性双变量，假定二阶混合偏导相等：

\[ A\frac{\partial^2u}{\partial x^2}+2B\frac{\partial^2u}{\partial x\partial y}+C\frac{\partial^2u}{\partial y^2}+D\frac{\partial u}{\partial x}+E\frac{\partial u}{\partial y}+Fu=G \]

引入变量代换 \(\xi=\xi(x,y),\eta=\eta(x,y)\)，为保证逆变换存在，满足

\[ \begin{vmatrix} \frac{\partial\xi}{\partial x} & \frac{\partial\xi}{\partial y} \\ \frac{\partial\eta}{\partial x} & \frac{\partial\eta}{\partial y} \\ \end{vmatrix} \neq0 \]

有

\[ \begin{align} \frac{\partial u}{\partial x}&=\frac{\partial u}{\partial\xi}\frac{\partial\xi}{\partial x}+\frac{\partial u}{\partial\eta}\frac{\partial\eta}{\partial x} \\ \frac{\partial u}{\partial y}&=\frac{\partial u}{\partial\xi}\frac{\partial\xi}{\partial y}+\frac{\partial u}{\partial\eta}\frac{\partial\eta}{\partial y} \\ \frac{\partial^2u}{\partial x^2}&=\frac{\partial^2u}{\partial\xi^2}\left(\frac{\partial\xi}{\partial x}\right)^2+2\frac{\partial^2u}{\partial\xi\partial\eta}\frac{\partial\xi}{\partial x}\frac{\partial\eta}{\partial x}+\frac{\partial^2u}{\partial\eta^2}\left(\frac{\partial\eta}{\partial x}\right)^2+\frac{\partial u}{\partial\xi}\frac{\partial^2\xi}{\partial x^2}+\frac{\partial u}{\partial\eta}\frac{\partial^2\eta}{\partial x^2} \\ \frac{\partial^2u}{\partial x\partial y}&=\frac{\partial^2u}{\partial\xi^2}\frac{\partial\xi}{\partial x}\frac{\partial\xi}{\partial y}+\frac{\partial^2u}{\partial\xi\partial\eta}\left(\frac{\partial\xi}{\partial x}\frac{\partial\eta}{\partial y}+\frac{\partial\xi}{\partial y}\frac{\partial\eta}{\partial x}\right)+\frac{\partial^2u}{\partial\eta^2}\frac{\partial\eta}{\partial x}\frac{\partial\eta}{\partial y}+\frac{\partial u}{\partial\xi}\frac{\partial^2\xi}{\partial x\partial y}+\frac{\partial u}{\partial\eta}\frac{\partial^2\eta}{\partial x\partial y} \\ \frac{\partial^2u}{\partial y^2}&=\frac{\partial^2u}{\partial\xi^2}\left(\frac{\partial\xi}{\partial y}\right)^2+2\frac{\partial^2u}{\partial\xi\partial\eta}\frac{\partial\xi}{\partial y}\frac{\partial\eta}{\partial y}+\frac{\partial^2u}{\partial\eta^2}\left(\frac{\partial\eta}{\partial x}\right)^2+\frac{\partial u}{\partial\xi}\frac{\partial^2\xi}{\partial y^2}+\frac{\partial u}{\partial\eta}\frac{\partial^2\eta}{\partial y^2} \end{align} \]

得

\[ a\frac{\partial^2u}{\partial \xi^2}+2b\frac{\partial^2u}{\partial \xi\partial \eta}+c\frac{\partial^2u}{\partial \eta^2}+d\frac{\partial u}{\partial \xi}+e\frac{\partial u}{\partial \eta}+fu=g \]

其中

\[ \begin{align} a & = A\left(\frac{\partial\xi}{\partial x}\right)^2+2B\frac{\partial\xi}{\partial x}\frac{\partial\xi}{\partial y}+C\left(\frac{\partial\xi}{\partial y}\right)^2 \\ b & = A\frac{\partial\xi}{\partial x}\frac{\partial\eta}{\partial x}+B\left(\frac{\partial\xi}{\partial x}\frac{\partial\eta}{\partial y}+\frac{\partial\xi}{\partial y}\frac{\partial\eta}{\partial x}\right)+C\frac{\partial\xi}{\partial y}\frac{\partial\eta}{\partial y} \\ c & = A\left(\frac{\partial\eta}{\partial x}\right)^2+2B\frac{\partial\eta}{\partial x}\frac{\partial\eta}{\partial y}+C\left(\frac{\partial\eta}{\partial y}\right)^2 \\ d & = A\frac{\partial^2\xi}{\partial x^2}+2B\frac{\partial^2\xi}{\partial x\partial y}+C\frac{\partial^2\xi}{\partial y^2}+D\frac{\partial\xi}{\partial x}+E\frac{\partial\xi}{\partial y} \\ e & = A\frac{\partial^2\eta}{\partial x^2}+2B\frac{\partial^2\eta}{\partial x\partial y}+C\frac{\partial^2\eta}{\partial y^2}+D\frac{\partial\eta}{\partial x}+E\frac{\partial\eta}{\partial y} \\ f & = F \\ g & = G \end{align} \]

特征方程

\[ A\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2-2B\frac{\mathrm{d}y}{\mathrm{d}y}+C=0 \]

曲线 \(\xi(x,y)=\gamma_1\) 和 \(\eta(x,y)=\gamma_2\) 称为特征线。特征方程解为

\[ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{B\pm\sqrt{B^2-AC}}{A} \]

记 \(\Delta=B^2-AC\)，讨论：

1. 当 \(\Delta>0\) 时，\(a=c=0\)，双曲型方程

\[ \frac{\partial^2u}{\partial\xi\partial\eta}=-\frac{1}{2b}\left(d\frac{\partial u}{\partial\xi}+e\frac{\partial u}{\partial\eta}+fu-g\right) \]

令 \(\xi=\mu+\nu,\eta=\mu-\nu\)，则

\[ \frac{\partial^2u}{\partial\mu^2}-\frac{\partial^2u}{\partial\nu^2}=-\frac{1}{b}\left[(d+e)\frac{\partial u}{\partial\mu}+(d-e)\frac{\partial u}{\partial\nu}+2fu-2g\right] \]

2. 当 \(\Delta=0\) 时，\(a=b=0\)，抛物线型方程

\[ \frac{\partial^2u}{\partial\eta^2}=-\frac{1}{c}\left(d\frac{\partial u}{\partial\xi}+e\frac{\partial u}{\partial\eta}+fu-g\right) \]

3. 当 \(\Delta<0\) 时，\(a=c,b=0\)，椭圆型方程

\[ \frac{\partial^2u}{\partial\xi^2}+\frac{\partial^2u}{\partial\eta^2}=-\frac{1}{a}\left(d\frac{\partial u}{\partial\xi}+e\frac{\partial u}{\partial\eta}+fu-g\right) \]

5.4.2 常系数方程

\[ y-\frac{B+\sqrt{B^2-AC}}{A}x=\gamma_1,\quad y-\frac{B-\sqrt{B^2-AC}}{A}x=\gamma_2 \]

1. 当 \(\Delta>0\) 时，\(\xi=y-\frac{B+\sqrt{B^2-AC}}{A}x\)，\(\eta=y-\frac{B-\sqrt{B^2-AC}}{A}x\) \[ \frac{\partial^2u}{\partial\xi\partial\eta}=d_1\frac{\partial u}{\partial\xi}+e_1\frac{\partial u}{\partial\eta}+f_1u+g_1(\xi,\eta) \]
2. 当 \(\Delta=0\) 时，\(\xi=y-\frac{B}{A}x\)，\(\eta=y\)(任意选择) \[ \frac{\partial^2u}{\partial\eta^2}=d_2\frac{\partial u}{\partial\xi}+e_2\frac{\partial u}{\partial\eta}+f_2u+g_2(\xi,\eta) \]
3. 当 \(\Delta<0\) 时，\(\xi=y-\frac{B}{A}x\)，\(\eta=\frac{\sqrt{AC-B^2}}{A}x\) \[ \frac{\partial^2u}{\partial\xi^2}+\frac{\partial^2u}{\partial\eta^2}=d_3\frac{\partial u}{\partial\xi}+e_3\frac{\partial u}{\partial\eta}+f_3u+g_3(\xi,\eta) \]

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• 电报方程 \[ \frac{\partial^2u}{\partial x^2}-\alpha^2\frac{\partial^2u}{\partial y^2}+2\beta\frac{\partial u}{\partial x}+\delta u=0 \] 由 \(\xi=y-\alpha x,\eta=y+\alpha x\)，得标准形式 \[ \frac{\partial^2u}{\partial\xi\partial\eta}=-\frac{\beta}{2\alpha}\left(\frac{\partial u}{\partial\xi}-\frac{\partial u}{\partial\eta}-\frac{\delta}{2\alpha\beta}u\right) \]

5.5 定解问题

5.5.3 初始条件与边界条件

1. 第一类边界条件：\(\left.u\right|_{x=L}=0\)
2. 第二类边界条件：\(\left.\frac{\partial u}{\partial x}\right|_{x=L}=0\)
3. 第三类边界条件：\(\left[u+\beta\frac{\partial u}{\partial x}\right]_{x=L}=0\)，\(\beta=T/k\)，\(k\) 为倔强系数

6 分离变量法

6.1 弦振动问题

6.1.1 弦振动问题的求解

\[ \begin{cases} \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2} & (0<x<L,t>0) \\ \left.u\right|_{t=0}=\phi(x),\quad\left.\frac{\partial u}{\partial t}\right|_{t=0}=\psi(x) & (0\leq x\leq L) \\ \left.u\right|_{x=0}=0,\quad\left.u\right|_{x=L}=0 & (t>0) \end{cases} \]

设变量分离的形式解

\[ u(x,t)=X(x)T(t) \]

满足第三行边界条件，即

\[ X(0)T(0)=0,\quad X(L)T(t)=0\Rightarrow X(0)=0,\quad X(L)=0 \]

• 空间函数 \[ \begin{cases} X''(x)+\lambda X(x)=0 \\ X(0)=0,\quad X(L)=0 \end{cases} \]

讨论：

1. \(\lambda=0\)，通解 \(X(x)=A+Bx\)，\(A=B=0\) 导致平庸解 \(u(x,t)\equiv0\)
2. \(\lambda<0\)，通解 \(X(x)=A\cosh kx+B\sinh kx\)，\(k=\sqrt{-\lambda}\)，\(A=B=0\) 导致平庸解 \(u(x,t)\equiv0\)
3. \(\lambda>0\)，通解 \(X(x)=A\cos kx+B\sin kx\)，\(k=\sqrt{\lambda}\)

本征值

\[ \lambda=\lambda_n=\left(\frac{n\pi}{L}\right)^2 \]

本征函数

\[ X(x)=X_n(x)=B_n\sin\frac{n\pi}{L}x \]

• 时间函数 \[ T''(t)+\left(\frac{an\pi}{L}\right)^2T(t)=0 \] 通解 \[ T(t)=T_n(t)=c_n\cos\frac{an\pi}{L}t+d_n\sin\frac{an\pi}{L}t \]

\[ u_n(x,t)=\left(C_n\cos\frac{an\pi}{L}t+D_n\sin\frac{an\pi}{L}t\right)\sin\frac{n\pi}{L}x,\quad n=1,2,\cdots \]

其中 \(C_n=B_nc_n\)，\(D_n=B_nd_n\)。

\[ \begin{gather} u(x,t)=\sum_{n=1}^\infty u_n(x,t)=\sum_{n=1}^\infty\left(C_n\cos\frac{an\pi}{L}t+D_n\sin\frac{an\pi}{L}t\right)\sin\frac{n\pi}{L}x \\ \frac{\partial u(x,t)}{\partial t}=\sum_{n=1}^\infty\frac{an\pi}{L}\left(-C_n\sin\frac{an\pi}{L}t+D_n\cos\frac{an\pi}{L}t\right)\sin\frac{n\pi}{L}x \\ \phi(x)=\sum_{n=1}^\infty C_n\sin\frac{n\pi}{L}x \\ \psi(x)=\sum_{n=1}^\infty D_n\frac{an\pi}{L}\sin\frac{n\pi}{L}x \\ C_n=\frac2{L}\int_0^L\phi(x)\sin\frac{n\pi}{L}x\mathrm{d}x \\ D_n=\frac2{an\pi}\int_0^L\psi(x)\sin\frac{n\pi}{L}x\mathrm{d}x \end{gather} \]

6.1.2 解的物理意义：驻波条件

本征解可以改写成本征波

\[ u_n(x,t)=N_n\sin(\omega_n t+\theta_n)\sin\frac{n\pi}{L}x \]

其中 \(N_n=\sqrt{C_n^2+D_n^2}\)，\(\omega_n=\frac{an\pi}{L}\)，\(\tan\theta_n=\frac{C_n}{D_n}\)。

• 不同系统的驻波条件 \(L=\frac\lambda 2n\)

系统	波	\(L\)	学科
弦振动	机械波	弦的长度	数学物理方法
单模激光	电磁波	谐振腔长度	激光物理学
势阱中的粒子	几率波	势阱宽度	量子力学

6.2 基本定解问题

本征方程 \(X''(x)+\lambda X(x)=0\) 的本征值和本征函数

\(\left[u\right]_{x=0}=0\)	\(\left[u\right]_{x=L}=0\)	\(\lambda=\left(\frac{n\pi}{L}\right)^2\)	\(X_n(x)=B_n\sin\frac{n\pi}{L}x,n=1,2,\cdots\)
\(\left[\frac{\partial u}{\partial x}\right]_{x=0}=0\)	\(\left[\frac{\partial u}{\partial x}\right]_{x=L}=0\)	\(\lambda=\left(\frac{n\pi}{L}\right)^2\)	\(X_n(x)=A_n\cos\frac{n\pi}{L}x,n=0,1,\cdots\)
\(\left[u\right]_{x=0}=0\)	\(\left[\frac{\partial u}{\partial x}\right]_{x=L}=0\)	\(\lambda=\left[\frac{(2n+1)\pi}{2L}\right]^2\)	\(X_n(x)=B_n\sin\frac{(2n+1)\pi}{2L}x,n=0,1,\cdots\)
\(\left[\frac{\partial u}{\partial x}\right]_{x=0}=0\)	\(\left[u\right]_{x=L}=0\)	\(\lambda=\left[\frac{(2n+1)\pi}{2L}\right]^2\)	\(X_n(x)=A_n\cos\frac{(2n+1)\pi}{2L}x,n=0,1,\cdots\)