@RogerLuo 2015-03-03T14:48:30.000000Z 字数 3109 阅读 565

波动方程的推导

数学物理方法

因为对于波动方程书上和大家已经给了利用微元法推导的过程，我换用上学期潘老师讲的连续体系拉格朗日力学来得到波动方程。(图是盗的，和写的有点不一样)

首先建立一个离散的模型，如图：

记u(x,t)为位于x位置的弹簧振子偏离平衡位置的距离，其质量为m，用劲度系数为k的弹簧相连

写 出 中 间 u (x 0 + i h, t) 一 个 振 子 的 拉 格 朗 日 量 (x = x 0 + i h) :

$写出中间u(x_0+ih,t)一个振子的拉格朗日量(x=x_0+ih):$

L i = 1 2 m (\partial u ( x 0 + i h , t ) \partial t) 2 - 1 2 k (u (x 0 + (i + 1) h, t) - u (x 0 + i h, t)) 2 + 1 2 k (u (x 0 + i h, t) - u (x 0 + (i - 1) h, t)) 2

$L_i= \frac{1}{2} m(\frac{\partial u(x_0+ih,t)}{\partial t})^{2}-\frac{1}{2}k (u(x_0+(i+1)h,t)-u(x_0+ih,t))^{2}+\frac{1}{2}k(u(x_0+ih,t)-u(x_0+(i-1)h,t))^{2}$
系统的拉格朗日量就为：

L = \sum i L i

$L=\sum_{i} L_i$
对于连续体系(如一根弦)，让h趋近于0可以得到拉格朗日密度:

k = N \cdot K, N = l h, m = M N = M \cdot h l

$k=N\cdot K,N=\frac{l}{h},m=\frac{M}{N}=\frac{M\cdot h}{l}$

L ˜ = lim h \to 0, n \to \infty L i = 1 2 M \cdot h l (\partial u \partial t) 2 - 1 2 l \cdot K h [(\partial u ( x 0 + i h , t ) \partial ( x 0 + i h )) 2 - (\partial u ( x 0 + ( i - 1 ) h , t ) \partial ( x 0 + ( i - 1 ) h )) 2]

$\widetilde L=\lim_{h\rightarrow 0,n\rightarrow \infty} L_{i}=\frac{1}{2} \frac{M\cdot h}{l} (\frac{\partial u}{\partial t})^{2}-\frac{1}{2} \frac{l\cdot K}{h}[(\frac{\partial u(x_0+ih,t)}{\partial (x_0+ih)})^2-(\frac{\partial u(x_0+(i-1)h,t)}{\partial (x_0+(i-1)h)})^2]$

= M \cdot h 2 l ((\partial u ( x , t ) \partial t) 2 - K l 2 M [ ( \partial u ( x 0 + i h , t ) \partial ( x 0 + i h ) ) 2 - ( \partial u ( x 0 + ( i - 1 ) h , t ) \partial ( x 0 + ( i - 1 ) h ) ) 2 ] h)

$=\frac{M\cdot h}{2l}((\frac{\partial u(x,t)}{\partial t})^{2}-\frac{Kl^2}{M}\frac{[(\frac{\partial u(x_0+ih,t)}{\partial (x_0+ih)})^2-(\frac{\partial u(x_0+(i-1)h,t)}{\partial (x_0+(i-1)h)})^2]}{h})$

= M \cdot h 2 l ((\partial u ( x , t ) \partial t) 2 - K l 2 M \cdot (\partial u ( x , t ) \partial t) 2)

$=\frac{M\cdot h}{2l}((\frac{\partial u(x,t)}{\partial t})^{2}-\frac{Kl^2}{M}\cdot (\frac{\partial u(x,t)}{\partial t})^2)$

= [1 2 ρ (\partial u ( x , t ) \partial t) 2 - 1 2 K l (\partial u ( x , t ) \partial x) 2] d x

$=[\frac{1}{2}\rho(\frac{\partial u(x,t)}{\partial t})^{2}-\frac{1}{2}Kl(\frac{\partial u(x,t)}{\partial x})^2]dx$
故系统的拉格朗日量为

L = \int l 0 L ˜ d x = \int l 0 [1 2 ρ (\partial u ( x , t ) \partial t) 2 - 1 2 K l (\partial u ( x , t ) \partial x) 2] d x

$L=\int_0^l \widetilde L dx=\int_0^l [\frac{1}{2}\rho(\frac{\partial u(x,t)}{\partial t})^{2}-\frac{1}{2}Kl(\frac{\partial u(x,t)}{\partial x})^2]dx$
由最小作用量原理，应有

δ \int L d t = δ \int \int l 0 L ˜ (q, \partial q \partial t, \partial q \partial x, t) d x d t

$\delta \int Ldt = \delta\int\int_0^l \widetilde L(q,\frac{\partial q}{\partial t},\frac{\partial q}{\partial x},t)dx dt$

= \int \int l 0 [\partial L ˜ \partial q δ q + \partial L ˜ \partial ( \partial q \partial t ) \partial \partial t δ q + \partial L ˜ \partial ( \partial q \partial x ) \partial \partial x δ q] d x d t

$=\int \int_0^l [\frac{\partial \widetilde L}{\partial q}\delta q+\frac{\partial \widetilde L}{\partial (\frac{\partial q}{\partial t})}\frac{\partial}{\partial t} \delta q+\frac{\partial \widetilde L}{\partial (\frac{\partial q}{\partial x})}\frac{\partial}{\partial x}\delta q]dx dt$

对时间空间分部积分:

δ \int B A L d t = \int l 0 [\partial L ˜ \partial ( \partial q \partial t ) δ q] B A d x + \int B A [\partial L ˜ \partial ( \partial q \partial x ) δ q] x 2 x 1 d t + \int B A \int l 0 [\partial L ˜ \partial q - \partial \partial t \partial L ˜ \partial ( \partial q \partial t ) - \partial \partial x \partial L ˜ \partial ( \partial q \partial x )] δ q d x d t = 0

$\delta \int_A^B Ldt=\int_0^l[\frac{\partial \widetilde L}{\partial (\frac{\partial q}{\partial t})}\delta q]_A^B dx+\int_A^B[\frac{\partial \widetilde L}{\partial (\frac{\partial q}{\partial x})}\delta q]_{x_1}^{x_2} dt+\int_A^B\int_0^l[ \frac{\partial \widetilde L}{\partial q}-\frac{\partial}{\partial t}\frac{\partial \widetilde L}{\partial (\frac{\partial q}{\partial t})}-\frac{\partial}{\partial x}\frac{\partial \widetilde L}{\partial(\frac{\partial q}{\partial x})}]\delta q dxdt=0$

\Rightarrow \partial L ˜ \partial q - \partial \partial t \partial L ˜ \partial ( \partial q \partial t ) - \partial \partial x \partial L ˜ \partial ( \partial q \partial x ) = 0

$\Rightarrow \frac{\partial \widetilde L}{\partial q}-\frac{\partial}{\partial t}\frac{\partial \widetilde L}{\partial (\frac{\partial q}{\partial t})}-\frac{\partial}{\partial x}\frac{\partial \widetilde L}{\partial(\frac{\partial q}{\partial x})}=0$
带入可得波动方程:

\partial 2 u \partial t 2 - K l 2 M \partial 2 u \partial x 2 = 0

$\frac{\partial^2 u}{\partial t^2}-\frac{Kl^2}{M}\frac{\partial^2 u}{\partial x^2}=0$

波动方程的推导

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