1 o, 1=vo)sin aot +(o )cos aot 八)F F|(0)c0s0+ sin aot 由上次例题 FFAp(x+at)+p(x-ar) +F-F lara y(s)dE 2p(x+a)+0(x-a)+ x+at a ar y (r dx
( ) ( ) a t ( ) a t a u t y w w j w w w w cos ~ sin 1 ~ , ~ \ = + u(x t) F [u ( ,t)] ~ , 1 w - = [ ( ) ] ( ) ú û ù ê ë é = + - - a t a F a t F w w y w j w w sin ~ cos 1 ~ 1 由上次例题 [ ( ) ( )] ú û ù ê ë é = + + - - F F j x at j x at 2 1 1 [ ( ) ( )] ( ) ò + - = + + - + x at x at x dx a j x at j x at y 2 1 2 1 ( ) ú û ù ê ë é + ò + - - x at x at d a F F y x x 2 1 1
由此我们看出傅氏变换的解题三大步骤: 对定解问题各项实行傅氏变换→>常微 分方程(选择恰当的变量) 二、解常微分方程 三、对解进行傅氏逆变换
由此我们看出傅氏变换的解题三大步骤: 一、对定解问题各项实行傅氏变换 常微 分方程(选择恰当的变量) 二、解常微分方程 三、对解进行傅氏逆变换 ®
二输运问题 f(x,1)() n(x,0)=g(x) (8) 记x=(,) f(r, t e o dx=flo, t) 0(x)d=o() +a2o2=f(o,t)(9) i(o,0)=() 0)
二.输运问题 1. u(x t)e dx u ( t) i x , ~ , w w ò ¥ -¥ - 记 = ( ) ( ) ò ¥ -¥ - f x t e dx = f t i x , ~ , w w 则 ( ) ( ) ò ¥ -¥ - j = j w w ~ x e dx i x ( ) ( ) (10 ) ~ ,0 ~u w = j w ( , ) ( ) 9 ~ ~ ~ 2 2 a u f t dt d u + w = w u (x , 0 ) = j (x ) (8 ) ( , ) (7 ) 2 u a u f x t t - xx =
对于y(x)+pxy=(x) 有小()=c(j(mk+c ∴l( 小=c/x 又由(0c=00) n(o,1)=20o()+.(o,x) d t
Q对于y ¢(x)+ p(x)y = Q(x) ( ) ( ) ú û ù ê ë é + ò ò = ò - y x e Q x e dx c pdx pdx 有 ( ) ( ) úû ù êë é \ = + ò - t a t a u t e f e d c 0 2 2 2 2 , ~ , ~ w w t t w w t j(w) ~ c = ( ) ( ) ( ) ( ) ò - - - \ = + t a t a t u t e f e d 0 2 2 2 2 , ~ ~ , ~ w j w w t t w w t 又由(10):
ux, t=F-li(o, t) Fl(0)0 +∫FV(o:0 而dk=F()+F-」 f(o, t e a2o2(-r)=F x,)*F F a ot dx 2兀 cos a tdx 丌
u(x t) F [u ( ,t)] ~ , 1 w - = [ ( ) ] a t F e 2 2 1 ~ w j w - - = ( ) ( ) [ ] ò - - - + t a t F f e d 0 1 2 2 , ~ w t t w t [ ] ò ¥ - ¥ - - - F e = e e dx a w t a w t iw t p 2 2 2 2 2 1 1 ( ) [ ( ) [ ]] a t a t e F x F e 2 2 2 2 ~ w 1 w j w j - - - 而 = * ( ) ( ) ( ) ( ) [ [ ]] w t w t w t t - - - - - = * a t a t f e F f x F e 2 2 2 2 1 , , ~ ò ¥ - = 0 cos 1 2 2 e tdx a t w p w