柱函数例 1.计算∫(xdk [xz,(x)=x"Z-1( xIxJ3 (x)]dx=xd[xj2(x) X J2(x) 2x 计算W[J(x)J(x),其中Wyy] W[(x),,(x) W U,(x),J- (x)=U U-v),,(v),1 ()()+,(J,)=-(J)().-J,()=J(,)-J、(
柱函数例 1. 计算 J (x)dx 3 x Z x x Z x dx d ( ) ] ( ) [ +1 = − [ ( )] ( ) 1 x Z x x Z x dx d = − x [x J (x)]dx 3 2 2 − = [ ( )] 2 2 2 x d x J x − = − − = −J x − x J (x)dx 2 1 ( ) 2 1 2 x J x J x 2 ( ) ( ) 1 = − 2 − 2. 计算 W[J (x), J (x)] − ，其中 x x y y y y W y y ' ' [ , '] = x x J J J J ( ) ( ) − − W[J (x), J− (x)] = x x J (J ) J (J ) = − − − W [J (x), J (x)] x − x x x [J (J ) J (J ) ] = − − − x x xx x x xx (J ) (J ) J (J ) (J ) (J ) J (J ) = − + − − − − − xx xx J (J ) J (J ) = − − − 1
原方程:d2R,1dB dx- x dx +(1--2)R=0 )3+(1-2)]+J-[(J)2+(1--2)J WIJ { xWIJ(x),J,(x)}=x:HL,(x),,(x)+W1J(x),J-(x)=0 x WI,(x),J(x)]=C W, (x),J_ (x) X→ co(x-Wz-x/4)()≈-1=3c0x-W7-x/4)+1sn(x-z-/4 cos(x+z-x/4)(,)2≈--cox+-7/4)+1-sm(x+z-/4
( ) (1 ) ] 1 ( ) (1 ) ] [ 1 [ 2 2 2 2 J x J x J J x J x J = − − x + − − + − x + − (1 ) 0 1 2 2 2 2 + + − R = x m dx dR dx x d R xx xx J (J ) J (J ) = − − − [ ( ) ( ) ] 1 x x J J J J x = − − − − [ , ] 1 = − − W J J x x {x W[J (x), J (x)]} − x W [J (x), J (x)] W[J (x), J (x)] x − + − = = 0 x C x W[J (x), J− (x)] = C W[J (x), J − (x)] = cos( / 4) 2 x − − x J sin( / 4)] 2 cos( / 4) 2 2 1 ( ) [ 3 − − − + x − − x x x J x cos( / 4) 2 − x + − x J sin( / 4)] 2 cos( / 4) 2 2 1 ( ) [ 3 − − + − + x + − x x x J x x → 原方程：
WU,(x),J_(x) cos(x-W-/4)-1-3cos(x+-/4 sin(x+VI-/4) coS(x+VIT-T/4 zr 3 cos(x-vII-T/4) sin(x-VT- /41) 2 --[cos(x-Vxz-丌/4)sn(x+vz-m/4)-cos(x+W-m/4)sn(x-w-7/4) 与x的值无关: WI,(x),,(x)=--sin 2v WIm(x),m(x)=0
sin( / 4)]} 2 cos( / 4) 2 2 1 cos( / 4)[ 2 sin( / 4)] 2 cos( / 4) 2 2 1 cos( / 4)[ 2 { [ ( ), ( )] 3 3 − + − − − − − − − = − − − + − − + − − = x x x x x x x x x x x x W J x J x [cos( / 4)sin( / 4) cos( / 4)sin( / 4)] 2 = − x − − x + − − x + − x − − x sin 2 2 [ ( ), ( )] x W J x J − x = − x → W[Jm (x), J−m (x)] = 0 与 x 的值无关：
3.计算rax 2 sin 2. WU,(x),(x)] x(x) ()-J,()x]x x(x) 2sin 2VT DX 丌 2sin 2VIJ,(x) (x) 丌 2sin 2VITJJ,(x)3J+(x 丌 2sin 2VIT J,(x)
3. 计算 ( ) 2 xJ x dx x W J x J x 1 [ ( ), ( )] 2sin 2 − − = − − − = − ( ) [ ( ) ( ) ] ( ) 2sin 2 2 2 J x J J J J dx x J x dx x x ] ( ) ( ) [ 2sin 2 2 − − = − − J x J dJ J x dJ ] ( ) ( ) [ 2sin 2 2 − − = − − J x J dJ J x dJ ] ( ) 2 ( ) ( ) [ 2sin 2 2 − − = − − J x J dJ J x J x
求方程 1-2a L"+ n+(2=x-)2+a=yyp=0的通解 ∠ 设 u== v(z) =C v(2)+2 dz dz 公=a(a-1)2“-2v()+2ah,d2 +2 dz a(a-1)z v(Bz)+2az -1 dy +2 dz 1-2c [cav()+z]+【(B 厶 22+a-小+(B1)+-2 =v=0 [(Bx)2+ lv=0
4. 求方程 ' [( ) ] 0 1 2 '' 2 2 2 2 1 2 = − + + − + − u z u z z u 的通解。 u z v ( z ) 设 = dz dv z v z z dz du = + − ( ) 1 2 2 2 1 2 2 ( 1) ( ) 2 dz d v z dz dv z v z z dz d u = − + + − − [ ( ) ] [( ) ] 0 1 2 ( 1) ( ) 2 2 2 2 2 1 1 2 22 2 1 = − + + + − + − + + + − − − − z v z z dz dv z v z z z dz d v z dz dv z v z z [( ) ] 0 22 2 1 1 2 2 2 = − + + + − − z v z z dz dv z dz d v z [( ) ] 0 22 2 1 2 2 2 = − + + + − v z z zdz dv dz d v
(B2)22(B)2=2+p2 lv=0 (B) dd(z d(z dz dz Bre2-d= d (z (Br dz (Bdz (Br)dz (Br)zr-dz Iv=0 d1() (By)2=2d(B)2=2- (B) d(2)(22+-1 dv Iv=0 (B) l=0 d(Bz. B2'Brds*l7 (Br )2v+ [1 ]v=0 d(Brr) Bz d(Bz. (Br
dz d z dz d d z d ( ) / ( ) = z dz d −1 = ) ( ) ( ( ) ) ( ) ( 1 1 2 z dz d z dz d d z d − − = z dz d z dz d 1 2 2 2 2 1 2 ( ) ( 1) ( ) − − − = − ] 0 ( ) [1 ( ) ( ) 2 2 1 2 2 2 2 1 2 + + − = − − v z dz z dv z dz d v ] 0 ( ) [1 ( ) ( ) ( 1) ) ( ) ( 2 2 2 2 1 2 2 1 2 + + − = − + − − v z dz z dv z dz dv v d z d ] 0 ( ) [1 ( ) ( ) ( 2 2 2 2 1 2 + + − = − v z dz z dv v d z d ] 0 ( ) [1 1 ) ( ) ( 2 2 1 2 + + − = − v z dz z dv z v d z d ] 0 ( ) [1 ( ) 1 ) ( ) ( 2 2 2 + + − v = d z z dv z v d z d
这是v一阶贝塞耳函数。通解为 v(=)=C1J()+C2N(B) l()=z(=) 5.求下列方程的通解 a.l"+au=0; b.z2l"-2zl4(=4-1)=0 cu'-3u+zu=0 1-2c Z"+ l+(B) =0 b=2(y-1) b u+laz+ a=0 +1a=(6y) 4z24 ? 2 B b+2 2b+2(2)=C1VEJ(2a ×)、b+2)+C 2√ab+2 b b+2
u(z) z v(z) = ( ) ( ) ( ) 1 2 v z = C J z +C N z 这是 − 阶贝塞耳函数。通解为 5. 求下列方程的通解 . '' 3 ' 0 . '' 2 ' 4( 1) 0; . '' 0; 2 4 − + = − + − = + = c zu u zu b z u zu z u a u az u b a. ' [( ) ] 0 1 2 '' 2 2 2 2 1 2 = − + + − + − u z u z z u 2 1 = ] 0 4 1 4 1 '' [ 2 2 + + − u = z z u az b b = 2( −1) 1 2 = + b 2 a = () 2 2 + = b a 2 1 = 2 1 + = b ) 2 2 ) ( 2 2 ( ) ( 2 1 2 2 1 1 + + + + + = b b z b a z C zN b a u z C z J
P346,习题7. 均匀膜的微小振动,方程(P142(7.1.16)) 解 △L=0 tt 1.分离变量 XT-a TAX=0 T+kfat=o △X+k2X=0 在极坐标中 02X1aX102X 0p2Ppd∞2+k2X=0 再分离变量 X=R(p)O(o) rag Rd+-r+ P∞+k2R=0 R+R+ R R
P346,习题7. 均匀膜的微小振动，方程（P142(7.1.16) ） 2 0 2 解 utt − a u = 1. 分离变量 u =TX '' 2 0 2 XT −a T X = '' 0 2 2 T +k a T = 0 2 2 X + k X = 在极坐标中 0 1 1 2 2 2 2 2 2 + = + + k X X X X 再分离变量 X = R()() '' ' 0 2 2 2 2 + = + + k R R R R 2 2 2 2 2 2 1 '' R' k m R R R = + + = −
opp p2R"+pR+(k2p2-m2)R=0 令 x2R"+xR+(x2-m2)R=0 为m阶贝塞耳方程 边界条件 p20,1)=0 初始条件:(p=0)=(1-a(9t=0)=0 2通解 与0无关。m=0。由边界条件:(p,9,1)=0 P, t)=2(C+Dp)o(krp)[A, cos(akf 1)+B, sin( akot) 其中J(k0)p)=0决定k0
'' ' ( ) 0 2 2 2 2 R +R + k − m R = 0 2 2 2 + = m 令 k = x '' ' ( ) 0 2 2 2 x R +x R+ x − m R = 为m阶贝塞耳方程 边界条件： u(0 ,,t) = 0 初始条件： 2 0 0 2 u( , ,t 0) (1 )u = = − ut (,,t = 0) = 0 2.通解 与 无关。m=0。 ( , , ) ( ) ( )[ cos( ) sin( )] (0) (0) (0) 0 1 u t C D J k A ak t B ak t n n n n n n = + + = 其中 ( 0 ) 0 (0) J0 kn = 决定 (0) n k 由边界条件： u(0 ,,t) = 0
3定解 t=0)=∑(+D)(kp)42=(-21) C=1,D=0 u, (p,p, t=0)=2 Jo(hi p)akn b,=0 Bn,=0 p-)0(k0)p)p Pa-P2Vo(ki p)adp= Jo(k p)odp-aS Jo(ko)p)p'dp o J(x)xdx x0) (0)2 (0)4 lo do(rdx P334结果 x(0) (0)2 J1(x)x on h(04p2LrJ1-2x2J,1=go 210
3.定解 2 0 0 2 (0) 0 1 u( , ,t 0) (C D )J (kn )An (1 )u n = = + = − = C =1, D = 0 ( , , 0) ( ) 0 (0) (0) 0 1 = = = = n n n n ut t J k ak B Bn = 0 J k d J k u A n n n (1 ) ( ) ( ) 2 (0) 0 0 2 0 2 2 0 (0) 1 2 0 0 0 = − J kn d J kn d J kn d (0) 3 0 0 2 0 (0) 0 0 (0) 0 0 2 0 2 ( ) 1 (1 ) ( ) ( ) 0 0 0 − = − J x x dx k J x xdx k n n x n x n 3 0 0 2 0 0 (0)4 0 (0)2 ( ) 1 ( ) 1 (0) (0) = − (0) (0) 2 0 2 1 3 2 0 (0)2 1 0 (0)4 [ 2 ] 1 ( ) 1 n n x n x n x J x J k J x x k = − − P334结果