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

  1. 拉氏变换对
时域 s域

e

a

t

\displaystyle e^{at}

eat

1

s

a

\displaystyle \frac{1}{s-a}

sa1

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
  1. 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)
  2. 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

yy=et,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}

s2Ysy(0)y(0)Y=1+s1s2YsY=s+11Y=(s+1)2(s1)s2+s+1=s+1A+(s+1)2B+s1C

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=s1limdsd(s+1)2(s+1)2(s1)s2+s+1=s1limdsds1s2+s+1=s1lim(s1)2(2s+1)(s1)(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=s1lim(s+1)2(s+1)2(s1)s2+s+1=s1lims1s2+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=s1lim(s1)(s+1)2(s1)s2+s+1=s1lim(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+1121(s+1)21+43s11

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)=41et21tet+43et

中间这一项的反变换不太懂,下次再说。


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