文档介绍：§ * 拉普拉斯变换法
2
ænπaö
1. P178 T¢¢()t+ç ÷T()t =f()t
è l ø
() T(0)=0
() T¢(0)=0
~ ~
记L[T(t)]=T(p), L[f(t)]=F(p)
2
~ æ nπaö ~
则p2T ()p  pT()0 T¢()0 + ç ÷ T()t= F()p
è lø
~ F(p) é l nπa ù
\T p= =L ft* sin t
() 2 ê () ú
ænπaö ë nπa l û
p2 +ç ÷
è l ø
l t nπa
\T()t= f()τsin (tτ)dτ
nπaò0 l
2.(傅氏变换主要用于解无界问题),拉氏适合解混
合问题
2
utt = a u xx ,0 < x < ¥,t > 0
u(0,t)= f (t), lim u(x,t)= 0(t ³ 0)
x®¥
u(x,0)=0, ut(x,0)=0
仍选t作变换Q二阶微商的变换要涉及
函数的在t = 0,和一阶导数的值,而关于
x的边界条件未给出
~
L[u(t)] = u~(x, p), L[ f (t)] = F(p)
¶ 2
p 2u~(x, p)  pu(x,0)  u (x,0) = a 2 u~(x, p)
则 t 2
~ ¶x
u~(0, p) = F(p), limu~(x, p) = 0
x®¥
d 2 p 2
u~  u~ = 0
即 dx 2 a 2
~
u~(0) = F
limu~ = 0
x®¥
p p
x x
~ a a
u=c1()pe +c2()pe
~ ~ ~
由u(0) = F 设: c1(p)+ c2 (p) = F(p)
p
x
~ a
由u(¥) =0 c2()pe =0 \c2=0
~
c1(p)=F(p)
p
~ x
u~(x,p)=F()pea
p
x
1é~ ù
u(x,t)=LêF()p×eaú
ë û
1éæ xöù
u(x,t)=LLêfçt ÷ú
ëè aøû
æ xö
=fçt ÷
è aø

2
ut = a u xx ,0 < x < ¥,t > 0
u x (0, t ) = 0, u(l, t ) = u 1
u(x,0) = u0 ,
¶ 2
pu~(x, p) u(x,0) = a 2 u~(x, p)
¶x 2
对t ~
u x (0, p) = 0,
~
u (l, p) = u1 / p
2~
d u p ~ u
− u + 0 = 0
dx2 a2 a2
即~
ux(0, p) =0,
~
u(l, p) =u1 / p
u p p
\通解:u(x, p) = 0 + c()psh x+ c()pch x
p 1 a 2 a
由边界条件:
p
ch x
u0 u1  u0 a 1 é 1 ù
\通解: u(x, p) = + , L ê ú =1
p p p ë pû
ch l
a
k a2π 2 (2k1)2
¥ 
4 (1) (2k 1)πx 4l 2
u(x,t) = u1 + å cos e
π k=1 2k 1 2l
u a 2u 0,0 x ,t 0
6. tt  xx = < < ¥ >
u(x,0) = ϕ(x)
ut (x,0) =ψ(x)
对x :令F[u(x,t)]= u~(ω,t), F[ϕ(x)]= ϕ~(ω),
F[ψ(x)] =ψ~(ω)
d 2u~
+ a 2ωu~(ω,t) = 0 t > 0
dt 2
u~(ω,0) = ϕ~(ω),
~ ~
ut (ω,0) =ψ(ω)
以ω为参量对t进行拉氏变换