dC0+k2,)=0 dr2 (5) d4+k2ur)=0(6) dr2 其解为 u(r)=Asin(k'r+δ) r≤a (7) uo(r)=Bsin(ka+δo) r>a (8) r≤a (9) 于是 了R)=mkr+ B (10) Ro(r)=-sin(kr+6o) r>a 因R,)=,在r=0处有限,必须有u,(0=0 所以δ)=0( ) 0 ( ) 0 2 2 0 2 + k u r = dr d u r 0 0 u r A k r r a ( ) sin( ) = + 0 0 u r B kr r a ( ) sin( ) = + (6) 其解为 (7) (8) k r r a r A R0 (r) = sin( + 0 ) k r r a r B R0 (r) = sin( + 0 ) 于是 (9) (10) 因 在 处有限,必须有 r u r R r ( ) ( ) 0 0 = r = 0 (0) 0 u0 = 所以 0 = 0 ( ) 0 ( ) 0 2 2 0 2 + k u r = dr d u r (5)