拉普拉斯变换解微分方程

使用拉普拉斯变换解一阶、二阶微分方程。

一、核心内容

拉氏变换对

时域	s域
eat\displaystyle e^{at}eat	1s−a\displaystyle \frac{1}{s-a}s−a1
1\displaystyle 11	1s\displaystyle \frac{1}{s}s1
t\displaystyle tt	1s2\displaystyle \frac{1}{s^2}s21
t2\displaystyle t^2t2	2s3\displaystyle \frac{2}{s^3}s32

导数的拉氏变换
f′(x)=sF(x)−f(0)f′′(x)=s2F(x)−sf(0)−f′(0)f'(x) = sF(x) - f(0) \\ f''(x) = s^2F(x) - sf(0) - f'(0) f′(x)=sF(x)−f(0)f′′(x)=s2F(x)−sf(0)−f′(0)
留数法分解因式

二、例子

例1

y′=1,y(0)=1y' = 1, \qquad y(0) = 1 y′=1,y(0)=1

解：两边进行拉氏变换

sY(s)−y(0)=1s⇒sY(s)−1=1s⇒Y(s)=1s+1s2\begin{aligned} & sY(s) - y(0) = \frac{1}{s} \\ \Rightarrow &sY(s) - 1 = \frac{1}{s} \\ \Rightarrow &Y(s) = \frac{1}{s} + \frac{1}{s^2} \end{aligned}⇒⇒sY(s)−y(0)=s1sY(s)−1=s1Y(s)=s1+s21

两边进行反拉氏变换

y(t)=1+ty(t) = 1+t y(t)=1+t

这个例子简单，杀鸡用了牛刀，但很有利于熟悉公式及验证算法正确性。下面看个二阶的。

例2
y′′−y=e−t,y(0)=1,y′(0)=0y'' - y = e^{-t}, \qquad y(0) = 1, \quad y'(0)=0 y′′−y=e−t,y(0)=1,y′(0)=0

解：两边进行拉氏变换

s2Y−sy(0)−y′(0)−Y=11+s⇒s2Y−s−Y=1s+1⇒Y=s2+s+1(s+1)2(s−1)=As+1+B(s+1)2+Cs−1\begin{aligned} & s^2Y-sy(0)-y'(0) - Y= \frac{1}{1+s} \\ \Rightarrow & s^2Y - s - Y = \frac{1}{s+1} \\ \Rightarrow & Y = \frac{s^2 +s + 1}{(s+1)^2 (s-1)} \\ &\quad = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{s-1} \end{aligned}⇒⇒s2Y−sy(0)−y′(0)−Y=1+s1s2Y−s−Y=s+11Y=(s+1)2(s−1)s2+s+1=s+1A+(s+1)2B+s−1C

留数法分解因式

A=lim⁡s→−1dds(s+1)2s2+s+1(s+1)2(s−1)=lim⁡s→−1ddss2+s+1s−1=lim⁡s→−1(2s+1)(s−1)−(s2+s+1)⋅1(s−1)2=14\begin{aligned} A &= \lim_{s\to -1}\frac{d}{ds} (s+1)^2 \frac{s^2 +s + 1}{(s+1)^2 (s-1)} \\ &= \lim_{s\to -1} \frac{d}{ds} \frac{s^2 +s + 1}{s-1} \\ &= \lim_{s\to -1} \frac{(2s + 1)(s-1) - (s^2+s+1) \cdot 1}{(s-1)^2} \\ &= \frac{1}{4} \end{aligned} A=s→−1limdsd(s+1)2(s+1)2(s−1)s2+s+1=s→−1limdsds−1s2+s+1=s→−1lim(s−1)2(2s+1)(s−1)−(s2+s+1)⋅1=41

B=lim⁡s→−1(s+1)2s2+s+1(s+1)2(s−1)=lim⁡s→−1s2+s+1s−1=−12\begin{aligned} B &= \lim_{s\to -1} (s+1)^2 \frac{s^2 +s + 1}{(s+1)^2 (s-1)} \\ &= \lim_{s\to -1} \frac{s^2 +s + 1}{s-1} \\ &= -\frac{1}{2} \end{aligned}B=s→−1lim(s+1)2(s+1)2(s−1)s2+s+1=s→−1lims−1s2+s+1=−21

C=lim⁡s→1(s−1)s2+s+1(s+1)2(s−1)=lim⁡s→1s2+s+1(s+1)2=34\begin{aligned} C &= \lim_{s\to 1} (s-1) \frac{s^2 +s + 1}{(s+1)^2 (s-1)} \\ &= \lim_{s\to 1} \frac{s^2 +s + 1}{(s+1)^2} \\ &= \frac{3}{4} \end{aligned}C=s→1lim(s−1)(s+1)2(s−1)s2+s+1=s→1lim(s+1)2s2+s+1=43

因此

Y=141s+1−121(s+1)2+341s−1Y =\frac{1}{4} \frac{1}{s+1} - \frac{1}{2} \frac{1}{(s+1)^2} + \frac{3}{4} \frac{1}{s-1} Y=41s+11−21(s+1)21+43s−11

拉氏反变换得

y(t)=14e−t−12te−t+34ety(t) = \frac{1}{4} e^{-t} - \frac{1}{2} te^{-t} +\frac{3}{4} e^{t} y(t)=41e−t−21te−t+43et

中间这一项的反变换不太懂，下次再说。