8.1习题 a2u=0 边界条件:v(0.1)=0u(1,)=0 初始条件: l(x,0)= bx(l-x) 解: u(x,)=∑ccp- alsin n7 0)=∑C b (-x); 5(-=5=2dmn7(y-y 86 l12 3(2k+1)3 86 u(x, t= (2k *13exp (2k+1)Tat,:(2k+1)r
8.1 习题 2. 0 2 ut − a uxx = 边界条件： u(0,t) = 0 u(l,t) = 0 初始条件： 2 ( ) ( ,0) l bx l x u x − = ( , ) exp[ ]sin ; 2 2 2 2 1 l n x l n a t u x t C n n = − = 解： ( ,0) sin ( ); 2 1 x l x l b l n x u x C n = k = − = sin ( ); 2 2 0 = − l l b l n d l C l k 2 sin ( ); 2 1 0 = b dy n y y − y 3 3 (2 1) 8 + = k b ; (2 1) ]sin (2 1) exp[ (2 1) 8 1 ( , ) 2 2 2 2 0 3 3 l k x l k a t k b u x t k + + − + = =
3 l2-a2l1=0a(x,)=0 (x,1) Xx 0 0(x<x-6 V2(x0-8,x+O 0(x>x-) (x,1)=∑(Acos naat nat n7 +B, sin -s =00x,)=∑B naat nZA B ds 2/v 5 27v )nr(x0-6) nx0-8 (n)2a (n)2(osnr(x+δ) 4/h sIn (n丌)2a 47v (x,1)=∑ nto. nnat. na
3. 0 2 utt − a uxx = u(x,t) x=0 = 0 u(x,t) x=l = 0 0 u t=0 = − − + − = = 0 ( ) ,( , ) 0 ( ) 0 0 0 0 0 0 x x v x x x x ut t (1) ( , ) ( cos sin )sin . 1 l n x l n at B l n at u x t An n n = + = 0 u t=0 = sin . 2 0 0 0 d l n n a v B x x n + − = ( , ) sin sin . 1 l n x l n at u x t Bn n = = + = − − 0 0 cos ( ) 2 2 0 x x l n n a lv ] ( ) cos ( ) [cos ( ) 2 0 0 2 0 l n x l n x n a lv − − + = − sin sin . ) sin sin 4 1 ( , ) 0 2 1 2 0 l n x l n at l n l n x a n lv u x t n = = l n l n x n a lv sin sin ( ) 4 0 2 0 =
X< ,(x0-6,x+o) 26 (x>x-) n(x,1)=∑Bsin naat. nTA sIn 2p0+ n 2v n丌 B COS 5 (cos -COS+sin sin )S ds 26 220 226 2 n COS COS ds +sin 5n丌 sIn d2} n 26 2 2 x0- COS Isin( )5+sn( )2]d5 2 126 [cos( nt,I nZ (x0-6)-cos( xo +ol+ cos(:-o8)-cos nT 1 (x0+) 2( l26 126 n n 2/"+1m +)6+ -{sn( 2( 126 26
(2) − − + − − = = 0 ( ) ,( , ) 2 cos 0 ( ) 0 0 0 0 0 0 0 x x x x x x v x x ut t ( , ) sin sin . 1 l n x l n at u x t Bn n = = sin . 2 cos 2 0 0 0 0 d l x n n a v B x x n + − − = )sin . 2 sin 2 sin 2 cos 2 (cos 2 0 0 0 0 0 d l x x n n a v x x + − = + sin }. 2 sin 2 sin sin 2 cos 2 {cos 2 0 0 0 0 0 0 0 d l x n d l x n n a v x x x x + − + − = + d l n l n d l n x x x x ) ] 2 1 ) sin( 2 1 [sin( 2 1 sin 2 cos 0 0 0 0 = + + − + − + − )( )] 2 1 )( ) cos( 2 1 [cos( ) 2 1 2( 1 )( )] 2 1 )( ) cos( 2 1 [cos( ) 2 1 2( 1 0 0 0 0 − − − − + − + − − + + + + = x l n x l n l n x l n x l n l n ) 2 1 ) sin( 2 1 [sin( ) 2 1 2( 1 ) ] 2 1 ) sin( 2 1 [sin( ) 2 1 2( 1 0 − 0 − − + + + + = l n x l n l l n n x l n l n
l(x)8v61 nZA nTo. naat. nzx COS u-a u 0 u (xt) tt l,(x,)x==0 (x,1)=4+B+∑(,cos naat naat noX + B, sin-cos n=0(x1)=4+∑40sm(eon =1 2 28 Ao d5, 5d5=-, 2J5c05 4 nZ nIel 48 d nz 1 nz dssin nrs n E 5 COS (n丌) (n) (2k+1)2丌 naat nZA l(x,1)=-E+ COS (2k+
sin cos sin sin . ) 2 1 ( 8 1 1 ( , ) 0 1 2 3 0 l n x l n at l n l n x l a n n v u x t n = − = 4. 0 2 utt − a uxx = ux (x,t) x=0 = 0 ux (x,t) x=l = 0 x ut t=0 = 0 l u t 2 =0 = − ( , ) ( cos sin ) cos . 1 0 0 l n x l n at B l n at u x t A B t An n n = + + + = ut t=0 = 0 ( , ) cos cos . 1 0 l n x l n at u x t A An n = = + , 2 0 0 2 d l A l = − , 2 0 2 = − = − d l l cos ; 4 0 2 d l n l A l n = − sin ; 4 0 l n d n l l = − sin ; 4 sin 4 0 0 l n d l n l n n l l l = − + l l n n 2 0 cos ( ) 4 = − [1 ( ) ] ( ) 4 2 n n = − − . (2 1) 8 2 2 + = k cos cos . (2 1) 8 1 ( , ) 1 2 2 l n x l n at k u x t n = + = − +
l2-a2l1=0a(x,)a=0u(x,1 Xx △lF 4n= =0 yS (2k+1)t uK (, t)=(Ak cos 2k+d)tat +b, sin Si)(2k+1)z 2 27 27 0u(x,1)=∑Acos (2k+1)mt(2k+1) SIn 27 Fo 2F n元 4F x→4 Ed cos 2k+1)5 YS lYS (2k +DrYS 27 4F (2k+1 (2k+1)n2S pzg-」 (2k+1)x5 cOs 21 8Fl (2k+1)r5 8F01( (2k+1)x2S (2k+1)22YS 8Fl )(2k+1)m(2k+1)m u(x,t= COS sin T YS k=o (2k+1)
5. 0 2 utt − a uxx = u(x,t) x=0 = 0 ux (x,t) x=l = 0 x ut t=0 = 0 YS F u t 0 =0 = YS F l l x u 0 = = ut t=0 = 0 . 2 (2 1) )sin 2 (2 1) sin 2 (2 1) ( , ) ( cos l k x l k at B l k at uk x t Ak k + + + + = . 2 (2 1) sin 2 (2 1) ( , ) cos 0 l k x l k at u x t Ak k + + = = x YS F u t 0 =0 = sin . 2 0 0 d l n lYS F A l k = l k d k YS F l 2 (2 1) cos (2 1) 4 0 0 + + = − } 2 (2 1) cos 2 (2 1) { cos (2 1) 4 0 0 0 d l k l k k YS F l l + − + + = − l l k k YS F l 2 2 0 0 2 (2 1) sin (2 1) 8 + + = k YS F l k 2 2 0 (2 1) 8 ( ) + − = . 2 (2 1) sin 2 (2 1) cos (2 1) 8 ( ) ( , ) 0 2 2 0 l k x l k at YS k F l u x t k k + + + − = =