《数学物理方程》课程PPT教学课件（讲稿）第八章（8-1）习题2

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(xx-) (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 =

Xx-) 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    + + + − =   =

l2-a2l1=0a(x,)a=0u(x,1 Xx 0 (2k+1)t uK (, t)=(Ak cos 2k+d)tat +b, sin Si)(2k+1)z 2 27 27 =0=0 n()=Bsn(②2k+1)m(2 k+D) 27 B 1t=0 ds nal n75 n7 2v.l niS! 16vl(-) cos (n)2a (nr)a (2k+1)n3a u(xt= *13 Sin(2k+1)ai -sin (2k+1) (2k

7. 0 2 utt − a uxx = u(x,t) x=0 = 0 ux (x,t) x=l = 0 0 u t=0 = x l v 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  +  +  + + = u t=0 = 0  . 2 (2 1) sin 2 (2 1) ( , ) sin l k x l k at uk x t Bk +  +  = x l v ut t 0 =0 =  sin . 2 0 0      d l n n al v B l n  = [ cos cos ]. ( ) 2 0 2 0 0        d l n l n n a v l l  = − − 2 1 n = k + l l n n a v l 3 0 0 sin ( ) 2   = k a v l k 3 3 0 (2 1) 16 ( ) +  − = . 2 (2 1) sin 2 (2 1) sin (2 1) 16 ( ) ( , ) 3 0 3 0 l k x l k at a k v l u x t k k    + + + − =   =

8 at 22= Bu u(x, t)=X(x)T( XT-a2'X"T=BYT XT XT axt rT-B/a T-Br a'x T X T"+(a2-B)T=0 X+X"=0 又 X(0)=0和X()=0 n X(x)=C,sin n丌 n=1,2,3 T(t=Cexp[(B-2)t B 1=a/B

8. u x u a t u =    −   2 2 2 u(x,t) = X (x)T(t) XT'−a X''T = XT 2 2 2 / ' '' a XT X T a XT XT − =    = = − − X a X T T' T '' 2 ' ( ) 0 2 T + a −  T = X + X'' = 0 又 X (0) = 0 和 X (l) = 0. l n x X x C  ( ) = 2 sin 2 2 2 l n   = n =1,2,3 ( ) exp[( ) ]; 2 2 2 2 t l n a T t C  =  − 2 2 ( ) a l n c   = n =1 l c = a / 

9 au (x,1)x=0=N 0 (x,O)=0 (x,)x==N 非齐次边界条件,设:u(x,)=v(x)+N avaI v(X 0 v(x, t at v(x,0)=-N (x,1)=∑B,eNl nnat nZ n B=3 n丌 2N sin 5 o dcos ni5 2No nsu COS n丌 n (2k+1) (x,1)=N、4N0 (2k+1)(x)2t1(2k+1)m sIn 2k+1) 大t: (x0)=、4-7令 (m)2t 丌

9. N0 N0 0 2 2 2 =   −   x u a t u 0 0 u(x,t) x= = N 0 u(x,t) x=l = N u(x,0) = 0 非齐次边界条件，设： 0 u(x,t) = v(x,t) + N 0 2 2 2 =   −   x v a t v v(x,t) x=0 = 0 v(x,t) x=l = 0 v(x,0) = −N ]sin . ( ) ( , ) exp[ 2 2 1 l n x l n a t u x t Bn n   = −  = sin . 2 0 0    d l n l N B l n  = − cos . 2 0 0 l n d n N l    = cos . 2 0 0 l l n n N    = (2 1) 4 0 + = k N . (2 1) ]sin (2 1) ( ) exp[ (2 1) 4 ( , ) 2 2 2 1 0 0 l k x l k a t k N u x t N n    + + − + = −  = 大t: ]sin . ( ) exp[ 4 ( , ) 2 2 0 0 l x l N a t u x t N    = − −

10 u(x,t 0 0 u(x,0)=0 u, (x, t X(x)=C,cos n/D

10. 0 2 2 2 =   −   x u a t u ux (x,t) x=0 = 0 ux (x,t) x=l = 0 u(x,0) = 0 0 0 l n x X x C  ( ) cos = 1 x

Ll.=0 边界条件 0 0=4(b-y) 0 b Bsin 卩+A(b-y)a-x) 0 v +v B =0 J=aBsin Ab(a-x) X(x)=C2sin n=1.2 v(x,y)=∑{ ny+B, ex n Y()=Expl nyu+bex nz =BSi Ab(a-x) Bsin" -Ab(a-x)=2(A,+B,sin 0 b ∑{4ep["]+B,ext- }

11. x y b 0 a 0 A(b − y ) 0 ax B  sin + = 0 xx yy u u 边界条件： ( ) 0 u A b y x = = − u x = a = 0 ax u y B  = 0 = sin = 0 . u y = b u = v + A ( b − y)( a − x ) + = 0 xx yy v v v x = 0 = 0 v x = a = 0 sin ( ) 0 Ab a x ax v y = = B − −  = 0 . y = b v l n x X x C  ( ) = 2 sin n = 1 , 2 ,  ( ) exp[ ] exp[ ] an y B an y Y y A   = + − a n x a n y B a n y v x y A n n n    ( , ) { exp[ ] exp[ ]}sin 1 =  + − = sin ( ) 0 Ab a x ax v y = = B − −  a n x Ab a x A B ax B n n n   sin ( ) { }sin 1 − − =  + = =0 . y = b v a n x a n b B a n b A n n n    0 { exp[ ] exp[ ]}sin 1 =  + − =