复习上次课 1建立了定义 U()小=/(xk==G(o) FGo)2∫Go-o=/() 2.(常用)性质 F[/(x)=(o)[/(x F‖f(x 小=1F[(x F[=F小*F/
复习上次课 1.建立了定义 [ ( )] ( ) ( ) ò ¥ -¥ - = = w w F f x f x e dx G i x î í ì [ ( )] ( ) ( ) ò ¥ -¥ - F G = G e d = f x i x w w w 1 w 2.(常用)性质 ( ) F [f (x)] (i ) F[f (x)] n n = w [ ( )] F [f ( ) x ] i F f x x x w 1 0 = ò [ ] [ ] [ ] 1 2 1 2 F f × f = F f * F f
§42傅里叶变换法 用傅氏变换解数理方程 一、波动问题 a2u=0.-∞0 (1) (2) Lo=y() (3) 曾由行波法求得 )=2l 四{x+a)+(x-c)+ ara v(a do
§4.2 傅里叶变换法 用傅氏变换解数理方程 一、波动问题 u tt - a 2 u xx = 0,-¥ 0 （1） u t = 0 = j (x ) （2） u t t = 0 = y (x ) （3） 曾由行波法求得 ( ) [ ( ) ( )] ( ) ò + - = + + - + x at x at d a u x t j x at j x at y a a 2 1 2 1
现用傅氏变换法求解 为此对定解问题各项施行傅氏变换 ∫n(x,t 02u 10X dx-a e1o dx=0 ax ∫(.0kd=」.o(x)kd !n(x0)k“h=(x)
现用傅氏变换法求解 为此对定解问题各项施行傅氏变换 ( , ) 0 2 2 2 2 2 = ¶ ¶ - ¶ ¶ ò ò ¥ - ¥ - ¥ - ¥ - e dx x u u x t e dx a t iw x iw x ( ) ( ) ò ò ¥ - ¥ ¥ - ¥ - - u x e dx = x e dx iw x iw x ,0 j ( ) ( ) ò ò ¥ - ¥ - ¥ - ¥ - = ¶ ¶ u x e dx x e dx t iw x iw x ,0 y
(r, t e -io dx=io, t) 记 p(x=0(0 y(x e-ioxdx=o) 则 tabula, )=0(4) i(o,t)=0(o) (5) i,(o,0)=V()
记 u (x t )e dx u ( t ) i x , ~ , w w ò = ¥ - ¥ - j ( ) j (w ) w ~ ò = ¥ - ¥ - x e dx i x y ( ) y (w ) w ~ ò = ¥ - ¥ - x e dx i x 则 ( ) ( , ) 0 ( ) 4 , ~ ~ 2 2 2 + a u t = dt d u t w w w ( ) ( ) (5 ) ~ , ~u w t = j w ( ) ( ) (6 ) ~ ,0 ~u t w = y w
解(4得 i0,)=40)cd0+80ao() (5)代入(7) A(o)=0(0) (,1)=0o)0sa0+B(o) sin aot(7) (6)代入(7): Blo ao=vla ∴B()=-v(o)
解(4)得 ( , ) ( )cos ( )sin (7) ~ u w t = A w awt + B w awt (5)代入(7): (w) j(w) ~ A = ( ) ( )cos ( )sin (7 ) ~ , ~ \u w t = j w awt + B w awt ¢ (6)代入(7): (w) w y(w) ~ B a = ( ) y( ) w w w 1 ~ a \B =
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