武汉大学：《数学物理方法》课程教学资源（课件讲稿）第四章（4.4）拉普拉斯变换法

§4.4*拉普拉斯变换法 n几 1.P178 T 7()=f() 325){7(0)=0 327)7()=0 记(=7(L[(=F(p) 则p27(p)p70)-7(0) 7(p)= Fp SIn n p

( ) T ( )t f ( )t l n a T t ÷ = ø ö ç è æ ¢¢ + 2 p T ¢(0) = 0 T (0) = 0 L[T(t)] T ( p) L[f (t)] F(p) ~ , ~ 记 = = P178 (3.2.5) (3.2.7) 1. ( ) ( ) ( ) T( )t F( ) p l n a p T p pT T ~ 0 0 ~ 2 2 ÷ = ø ö ç è æ - - ¢ + p 则 ( ) ( ) ( ) ú û ù ê ë é = * ÷ ø ö ç è æ + \ = t l n a n a l L f t l n a p F p T p p p p sin ~ 2 2 § 4.4* 拉普拉斯变换法

7()2()im (t-rdt n几 2.(傅氏变换主要用于解无界问题),拉氏适合解混 合问题 L.=a2l00 n(0)=f()im(x)=0(≥0 x→00 n(:0)=0.,(x0)=0

( ) ( ) ( ) ò \ = - t t d l n a f n a l T t 0 sin t t p t p 2.(傅氏变换主要用于解无界问题),拉氏适合解混 合问题 ,0 , 0 2 utt = a uxx (0, ) = ( ), lim ( , ) = 0 ( ³ 0) ®¥ u t f t u x t t x u(x,0) = 0, u (x,0) = 0 t

仍选t作变换∵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 项=(cp)()=f() 则 p2u(x, p)-pu(x,0)u, (x,)=au(x,p) 10p)=F0) limu(x, p)=0 x→00

的边界条件未给出 函数的在 和一阶导数的值 而关于 仍选 作变换 二阶微商的变换要涉及 x 0, , t t = Q L[u(t)] u(x p) L[f (t)] F(p) ~ , , ~ = = ( ) ( ) ( ) u (x p) x p u x p pu x u x a t , ~ , ,0 ,0 ~ 2 2 2 2 ¶ ¶ - - = ( ) ( ) ( , ) 0 ~ , lim ~ 0, ~ = = ®¥ u p F p u x p x 则

2l-2=0 即 n(0)=F limu=o u =c,ple a tc plea 由0)=F设:c(p)+c()=F()

0 ~ ~ 2 2 2 2 - u = a p u dx d u ( ) F ~ 0 ~ = 0 ~ lim = ®¥ u x 即 ( ) ( ) x a p x a p u c p e c p e 1 2 ~ = + - u ( ) F c ( p) c (p) F(p) ~ : ~ 0 ~ 由 = 设 1 + 2 =

由i(x)=0c(=0c2=0 c1(p)=F(p) (x, p)=f(pe a n(c)=|(p)e n(:)=L2∠

( ) 0 ( ) 0 0 ~u ¥ = c2 p e = \ c2 = x a p 由 c ( p ) F ( p ) ~ 1 = ( ) ( ) x a p u x p F p e - = ~ , ~ ( ) ( ) ú û ù ê ë é = × - - x a p u x t L F p e ~ , 1 ( ) ú û ù ê ë é ÷ ø ö ç è æ = - - a x u x t L L f t 1 , ÷ ø ö ç è æ = - a x f t

3求解 l1=a2a200 n,(.)=0.,(.1)=1 (x0) pux,p-ur,0=g? 02 对{2,(0,)=0 (,p)=u1/p

3.求解 ,0 , 0 2 ut = a uxx ( ) ( ) 1 u x 0, t = 0, u l, t = u ( ,0) , 0 u x = u 对t ( ) ( ) u (x p) x pu x p u x a , ~ , ,0 ~ 2 2 2 ¶ ¶ - = (0, ) 0, ~ ux p = u (l, p) u / p ~ = 1

d u+ 0 d x 即么2 p)=4/p 通解:n(xp)=24+c( shIPs+c(khyx

即 0 2 ~ 0 2 2 2~ - + = a u u a p dx d u (0, ) 0, ~ ux p = u(l, p) u / p ~ = 1 ( ) ( ) ( ) x a p x c p a p c p p u : u x, p 1 sh 2 ch 0 \通解 = + +

由边界条件 ch vx 通解:l(x,p) aL PP chVPI Lp 4(y2 a22(2k-1)2 2k-1)x S 兀2k-121

由边界条件: ( ) 1 1 , ch ch : , 0 1 0 1 =ú û ù ê ë - é \ = + - p L l a p x a p p u u p u 通解 u x p ( ) ( ) ( ) ( ) å ¥ = - - - - - = + 1 4 2 1 1 2 2 2 2 2 2 1 cos 2 1 4 1 , k l k a k e l k x k u x t u p p p

6. -a2u=0.00 n(x0)=0 4(x:0)=v(x) 对x:令F(x:)=(oF(x)=(0 Fv(x)=vlo)

0,0 , 0 2 utt - a uxx = u(x,0) = j(x) u (x ) (x) t ,0 =y 6. [ ( )] ( ) [ ( )] ( ), ~ , , ~ 对x :令F u x,t = u w t F j x = j w [y ( )] y (w) ~ F x =

d +a2oi(o,t)=0t>0 (00)=0(0) ,(0,0)=v(0 以o为参量对进行拉氏变换

( , ) 0 0 ~ ~ 2 2 2 + a u t = t > dt d u w w ( ) ( ), ~ ,0 ~u w = j w (w ) y (w) ~ ,0 ~ut = 以w为参量对t进行拉氏变换