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 |
|---|---|
| e a t \displaystyle e^{at} eat | 1 s − a \displaystyle \frac{1}{s-a} s−a1 |
| 1 \displaystyle 1 1 | 1 s \displaystyle \frac{1}{s} s1 |
| t \displaystyle t t | 1 s 2 \displaystyle \frac{1}{s^2} s21 |
| t 2 \displaystyle t^2 t2 | 2 s 3 \displaystyle \frac{2}{s^3} s32 |
- 导数的拉氏变换
f ′ ( x ) = s F ( x ) − f ( 0 ) f ′ ′ ( x ) = s 2 F ( x ) − s f ( 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 ) = 1 y’ = 1, \qquad y(0) = 1 y′=1,y(0)=1
解：两边进行拉氏变换
s Y ( s ) − y ( 0 ) = 1 s ⇒ s Y ( s ) − 1 = 1 s ⇒ Y ( s ) = 1 s + 1 s 2 \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 + t y(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 ) = 0 y” – 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
s 2 Y − s y ( 0 ) − y ′ ( 0 ) − Y = 1 1 + s ⇒ s 2 Y − s − Y = 1 s + 1 ⇒ Y = s 2 + s + 1 ( s + 1 ) 2 ( s − 1 ) = A s + 1 + B ( s + 1 ) 2 + C s − 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 → − 1 d d s ( s + 1 ) 2 s 2 + s + 1 ( s + 1 ) 2 ( s − 1 ) = lim s → − 1 d d s s 2 + s + 1 s − 1 = lim s → − 1 ( 2 s + 1 ) ( s − 1 ) − ( s 2 + s + 1 ) ⋅ 1 ( s − 1 ) 2 = 1 4 \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 ) 2 s 2 + s + 1 ( s + 1 ) 2 ( s − 1 ) = lim s → − 1 s 2 + s + 1 s − 1 = − 1 2 \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 ) s 2 + s + 1 ( s + 1 ) 2 ( s − 1 ) = lim s → 1 s 2 + s + 1 ( s + 1 ) 2 = 3 4 \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 = 1 4 1 s + 1 − 1 2 1 ( s + 1 ) 2 + 3 4 1 s − 1 Y =\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 ) = 1 4 e − t − 1 2 t e − t + 3 4 e t y(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。
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