《数学物理方程》课程PPT教学课件（讲稿）第八章（8-1）习题1

9.1习题2 1.细杆的一端弹性固定,初始时刻在另一端受到一纵向 冲量作用,但初始位移为零,试求杆的纵振动。 解 0 u(x, t)+hu, (x,D)x-l=0 △P 1=0=0/ 只出现于ε 小段 0(E<x≤ 0区 n1=0-1/(x≤E) mu= pau

9.1 习题2 1. 细杆的一端弹性固定，初始时刻在另一端受到一纵向 冲量作用，但初始位移为零，试求杆的纵振动。 k 0 l I 解： 0 2 utt − a uxx = u t=0 = 0 . / ( ) 0 ( ) 0       = =    I x x l ut t ux (x,t) x=0 = 0 u(x,t) + hux (x,t) x=l = 0 0  I = P 只出现于ε 一小段。 = mut = ut

分离变量+a2T=0 X"+λX=0(0):=0k(+hx()=0 A>0 X(x)=C, cos ax+ C2 sin vax X(x)=√AC2cos√Ax-Clsn√4x AX(0)=0國2= xX(O)+hN=0國→ X()+bX(D)=C[cos√l-hλsn√=0 cgl=h√

'' 0; 2 T +a T = X ''+X = 0; X '(0) = 0 X (l) + hX '(l) = 0.   0 X (x) C cos  x C sin  x = 1 + 2 X '(0) = 0  C2 = 0 '( ) [ cos sin ] 2 1 X x =  C  x −C  x X (l) + hX '(l) = 0.  X (l) + hX '(l) = C1 [cos l − h  sin l] = 0 分离变量 ctg l = h 

以0=此式确定本征值园,为交点 AX(x)=C2COS√λx T+nT a2 naat T=0 T(t)=Acos+Bsin nzA、B是积分常数 n=12.3… 42(x)=(A,cos√2+ B. sin√2a)cosy1x 1(x:)∑(4syxm+Bsm√a)coyx

cot   l h = 此式确定本征值 n ，为交点。 2 4 6 8 10 12 -30 -20 -10 10 20 30 X x C x n ( ) cos = 2 '' 0; 2 2 2 2 + T = l n a T  ( ) cos sin , l n at B l n at T t A   = + A、B 是积分常数。 u (x,t) (A cos at B sin at)cos x. n = n n + n n n n =1,2,3 ( , ) ( cos sin ) cos . 1 u x t A at B at x n n n n n n =   +    =

1+cos2、λ,x [cos a,x]'dx dx=2+cos2√nxx 0 sn2√,x sIn 2 2sin√ 2.l cos√Al sIn lcot√ X vAn cot2√λ,l+122x,h2+1 cot√x h cot√=h√ 1-cos2√n sn√nxax= 0 2 224h2+1

2 1 1 cot 1 2 cot 2 1 2 sin cot 2 1 2 2sin cos 4 1 2 sin 2 4 1 2 sin 2 4 1 2 cos 2 2 1 2 2 1 cos 2 [cos ] 2 2 2 0 0 0 0 2 2 + = + + = + = + = + = + = + = + + = =    h l h l l l l l l l l l l l x l xdx l dx x N x dx n n n n n n n n n n n n l n n l n l n l n n                  2 1 1 2 2 1 cos 2 [sin ] 2 0 0 2 + = − − =   h l h dx x x dx n l n l n    cot   l h =

cOS X cOS ∫cos√)x+cox+√m) sin( sin( 2-√m n SIn( 22+ siny2,lcos√nl-cos√λ iI sin√nl,sin√2lcos√nl+cos√λ il sin a, sin a, I cos cosa, sin v am/ √si2 sin ascot√ λ.cot Sin V: m ,sn√ √ n V/hsin√ x sin am x-√m√hsi xSin VAmx=0

0 0 0 0 1 cos cos [cos( ) cos( ) ] 2 1 sin( ) sin( ) { } 2 1 sin( ) sin( ) [ ] 2 1 sin cos cos sin sin cos cos sin [ ] 2 s l l n m n m n m n m n m l l n m n m n m n m n m n m n m n m n m n m n m n m n x xdx x x dx x x l l l l l l l l l l                                    = − + + − + = + − + − + = + − + − + = + − + =   in cos cos sin sin sin cot cot sin sin sin sin sin sin 0 n m m n m n n m m m n n m n m n m nm n n m l l l l l l l l l l h x x h x x                      − = − = − =

t=0 ∑ A, cos a,x=0 0(E<x≤D) n aB co 0(E<x≤D) t|t=0 l/ps(x≤) n=1 /pE(x≤E) sm、E B coS,xdx= N2√A,apE N n 1/N2 2 h+1(2h2+1) B a, ap h+l(a,h+ 2 (x,t)= sIn at cos aDa√nh+l(xh2+1)

0 u t=0 = cos 0. 1  =  = A x n n n  . / ( ) 0 ( ) 0       = =    I x x l ut t .. / ( ) 0 ( ) cos 1        =  =      I x x l aB x n n n n An = 0              N a I N a I xdx I N a B n n n n n n n n n n 2 0 2 0 2 sin cos 1 → =  = → ( 1) 2 1/ 2 2 + + = h l h N n n  ( 1) 2 2 + + = a h l h I B n n n    sin cos . ( 1) 2 1 1 ( , ) 1 2 at x a h l h I u x t n n n n n        = + + =

2.均匀弦两端固定,初始位移和初始速度为零。 求在重力下的弦振动。 解 L.- 8x) x=0 u(xt t=0 0 (x1)=∑T(sin n7 带入泛定方程∑o+(mr()m g|Z(o)=0T'(O)=0 =0 Tn"()+(",)2Tn(1)=2|(-g)sn n兀 ds 8An75|1 cOS 4 g k=0,1, (2k+1) k"((2k+1)m -)27(t) g (2k+1)

2. 均匀弦两端固定，初始位移和初始速度为零。 求在重力下的弦振动。 解： utt − a uxx = −g 2 0 u t=0 = 0. ut t=0 = u(x,t) x=0 = 0 u(x,t) x=l = 0   = = 0 ( , ) ( )sin n n l n x u x t T t  带入泛定方程 g l n x T t l n a T t n  n + n = −  =0 2 [ ''( ) ( ) ( )]sin       d l n g l T t l n a T t l n n ( )sin 2 ''( ) ( ) ( ) 0 2  + = − l l n n g 0 cos 2    = [( ) 1] 2 = − − n n g  0,1, (2 1) 4 = + = k k g    (2 1) 4 ) ( ) (2 1) ''( ) ( 2 + = + + k g T t l k a T t k k Tn (0) = 0 Tn '(0) = 0

(2k+1)m )27(t) g (2k+1) 7(O)=0T(O)=0 两边在拉普拉斯变换: 4 p27(p)+( 2k+1)m g P p(2k+1) 4 (p)= g [p2+( (2k+1)ma )2]p(2k+1)丌 作拉普拉斯反演(如6.4应用题): 1=sin at L[-]= p +@@ 4gl 7(0) (2k+1)m SIn (t-rdt (2k+1)2r2an

  (2 1) 4 ) ( ) (2 1) ''( ) ( 2 + = + + k g T t l k a T t k k Tn (0) = 0 Tn '(0) = 0 两边在拉普拉斯变换：   (2 1) 4 ( ) ~ ) (2 1) ( ) ( 2 ~ 2 + = + + p k g T p l k a p Tk p k   ) ] (2 1) (2 1) [ ( 4 ( ) ~ 2 2 + + + = p k l k a p g Tk p 作拉普拉斯反演（如6.4应用题）： [ ] sin , 2 2 1 t p    = + − L ] 1, 1 [ 1 = − p L     t d l k a k a gl T t t k ( ) (2 1) sin (2 1) 4 ( ) 0 2 2 − + + = 

4gl () (2k+1)m sIn 2k+1)2z2a (t-r dr g4 Cos(2k+1)ria (2k+1)2a 4 g (2k+1)mt COS (2k+1)3ra 得 丌a/2k+[1-、(2k+1)mt 4g12 nIA l(x,) ∑ Isin

    t d l k a k a gl T t t k ( ) (2 1) sin (2 1) 4 ( ) 0 2 2 − + + =  t t l k a k a gl 3 3 2 0 2 ( ) (2 1) cos (2 1) 4    − + + = ] (2 1) [1 cos (2 1) 4 3 3 2 2 l k at k a gl   + − + =   = + − + = 0 3 2 3 2 ]sin (2 1) [1 cos (2 1) 4 1 ( , ) k l n x l k at a k gl u x t    得

高维波动方程例 均匀正方形薄膜固定于边沿,初始形状为A(-xX-y), A是常数。初始速度为零。求其自由振动规律 解:泛定方程 u-af(ux+uw)=0 边界条件1(x,y2)-0=00xy: u(x,), D)y-o=0 u(x, oI 初始条件p=4A(-x1-y 0

高维波动方程例： 均匀正方形薄膜固定于边沿，初始形状为 ， A是常数。初始速度为零。求其自由振动规律。 Axy(l − x)(l − y) ( ) 0 2 utt −a uxx +uyy = ( )( ) 0 u Axy l x l y t= = − − 0. ut t=0 = u(x, y,t) x=0 = 0 u(x, y,t) x=l = 0 解： 泛定方程 边界条件 ( , , ) 0 u(x, y,t) y=l = 0 u x y t y=0 = 初始条件