Laplace方程与复变函数方法

1. Laplace 方程定义

二维情形： $$ \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 $$

2. 坐标变换（复变函数方法）

设： $$ \xi = x + iy,\quad \eta = x - iy $$

则： $$ \frac{\partial}{\partial x} = \frac{\partial \xi}{\partial x} \frac{\partial}{\partial \xi} + \frac{\partial \eta}{\partial x} \frac{\partial}{\partial \eta} = \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} $$

\[ \frac{\partial}{\partial y} = \frac{\partial \xi}{\partial y} \frac{\partial}{\partial \xi} + \frac{\partial \eta}{\partial y} \frac{\partial}{\partial \eta} = i\left(\frac{\partial}{\partial \xi} - \frac{\partial}{\partial \eta} \right) \]

3. 二阶偏导的展开

计算： $$ \frac{\partial^2 u}{\partial x^2} = \left( \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} \right)^2 u = \frac{\partial^2 u}{\partial \xi^2} + 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2} $$

\[ \frac{\partial^2 u}{\partial y^2} = \left( i\left(\frac{\partial}{\partial \xi} - \frac{\partial}{\partial \eta} \right) \right)^2 u = -\left( \frac{\partial^2 u}{\partial \xi^2} - 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2} \right) \]

两式相加： $$ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 4\frac{\partial^2 u}{\partial \xi \partial \eta} = 0 $$

即： $$ \frac{\partial^2 u}{\partial \xi \partial \eta} = 0 $$

4. 解的形式

由于混合偏导为 0，因此： $$ u(\xi, \eta) = f(\xi) + g(\eta) $$

代入原变量得： $$ u(x, y) = f(x + iy) + g(x - iy) $$

此为 Laplace 方程的通解，表达为任意解析函数之和。

波动方程与D'Alembert解法

1. 波动方程定义（无限区间）

设 \( u = u(x, t) \)，满足： $$ \frac{\partial^2 u}{\partial t^2} - a^2 \frac{\partial^2 u}{\partial x^2} = 0,\quad -\infty < x < \infty,\, t > 0 $$

初始条件： $$ u(x, 0) = \varphi(x),\quad \frac{\partial u}{\partial t}(x, 0) = \psi(x) $$

2. 坐标变换法（引入特征变量）

令： $$ \xi = x + at,\quad \eta = x - at $$

偏导数变换为： $$ \frac{\partial}{\partial x} = \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta},\quad \frac{\partial}{\partial t} = a\left( \frac{\partial}{\partial \xi} - \frac{\partial}{\partial \eta} \right) $$

代入波动方程可得： $$ \frac{\partial^2 u}{\partial \xi \partial \eta} = 0 $$

故通解为： $$ u(x, t) = f(x - at) + g(x + at) $$

3. 利用初值条件求特解

由初始位移： $$ u(x, 0) = f(x) + g(x) = \varphi(x) \quad \text{(1)} $$

由初始速度： $$ \frac{\partial u}{\partial t}(x, 0) = -a f'(x) + a g'(x) = \psi(x) \quad \text{(2)} $$

即： $$ f'(x) - g'(x) = -\frac{1}{a} \psi(x) $$

对 (2) 积分： $$ f(x) - g(x) = -\frac{1}{a} \int_0^x \psi(\xi) d\xi + C $$

联立 (1) 与上式，解得： $$ f(x) = \frac{1}{2} \varphi(x) - \frac{1}{2a} \int_0^x \psi(\xi) d\xi + \frac{C}{2} $$

\[ g(x) = \frac{1}{2} \varphi(x) + \frac{1}{2a} \int_0^x \psi(\xi) d\xi - \frac{C}{2} \]

4. D'Alembert公式（标准形式）

令 \( C = 0 \)，最终解表示为： $$ u(x, t) = \frac{1}{2}[\varphi(x - at) + \varphi(x + at)] + \frac{1}{2a} \int_{x - at}^{x + at} \psi(\xi) d\xi $$

此即一维波动方程在无限区间上的通解。