第三章 勒让德多项式习题

1、 u r =R = Al R l Pl ( x = f ( x l =0 0 2l + 1 1 2l + 1 1 ( ( ( + = f x P x dx u P x dx l 1 u 0 Pl (x dx 0 0 l 2 R l 1 2R l 0 (2l + 1u 0 1 = Pl ( x dx Pl ( x dx l 1 0 2R Al = 当 l = 2n 时 A2 n = 0 (4n + 1u 0 1 P2 n ( x dx = P2 n ( x dx P2 n ( x d ( x 2 n 0 1 1 0 2R 2R 0 (4n + 1u 0 1 = (4n + 1u 0 1 P (

2、x dx 1 P ( x d ( x ( ( ( P x dx P x d x = + 2 2 n n 2n 1 0 2n 0 0 2 n 2R 2R 2n =0 2n (4n + 1u 0 P ( x dx 1 2n 0 当 l = 2n + 1 时 A2 n +1 = 0 = (4n + 3u 0 1 P ( x dx 0 P ( x d ( x ( ( P x dx P x dx n + n + 2 1 2 1 1 1 2n+1 0 0 2 n +1 2 R 2 n +1 2 R 2 n +1 1 1 0 1 (4n + 3u 0 (4n + 3u 0 P2 n +1 ( x dx P2

3、 n +1 ( x d (x = P ( x dx + P2 n +1 ( x d ( x = 2 n +1 0 0 1 0 2 n +1 2R 2 R 2 n +1 1 (4n + 3u 0 P2 n +1 ( x dx = R 2 n +1 0 (4n + 3u 0 1 2 n +1 d 2 n +1 2 1 x 1 dx = 2 n +1 2 n +1 2 n +1 0 R 2 (2n + 1! dx 1 (4n + 3u 0 ( 2 n +1 d 2n 2 1 x 1 = 2 n +1 2 n +1 2n R 2 (2n + 1! dx 上限为零，只剩下限 (4n + 3u 0 ( 1

4、 0 = = (4n + 3u 0 R 2 n +1 d 2 n 2 n +1 1 ( 1k (2n + 1! x 4 n+ 2k 2 n +1 2 n +1 2n k!(2n + 1 k ! R 2 (2n + 1! dx k =0 x =0 只有2n = 4n + 2 k , 即k = n + 1时，上式不为零 所以 (4n + 3u 0 2 n +1 d 2n 2 1 x 1 2 n +1 2n 2 (2n + 1! dx ( x =0 R n (4n + 3u 0 = ( 1 R 2 n +1 n (4n + 3u 0 = ( 1 R 2 n +1 n (4n + 3u 0 = ( 1

5、 R 2 n +1 A2 n +1 = (4n + 3u 0 2 n +1 (2n ! (n + 1!n! 2 (2n ! (2n + 2!(2n ! (2n 1! (2n + 2! 1 2 n +1 2 2 n +1 1 ( 1n+1 (2n + 1! (2n ! (n + 1!n! (2n + 1! u (r , = ( 1 n =0 n u 0 (4n + 3(2n 1! r (2n + 2! R 2 n +1 P2 n +1 (cos 0 2 4、在电场强度为 E 0 的均匀电场中放置一个接地导体球，球半径为 a ，求球外任意一点 电势？ 解：选取沿着电场强度方向作为球坐标系的极轴方向

6、，则定解问题变为轴对称问题 球外总电势 u (r , 由匀强电场的电势 u1 (r , 和感应电荷的电势 u 2 (r , 叠加而成： u (r , = u1 (r , + u 2 (r , 设坐标原点电势为 u 0 ，则任意一点匀强电场电势： u1 (r , = u 0 E 0 z = u 0 E 0 r cos 感应电荷电势 u 2 (r , 满足定解问题： 2 u = 0 u 2 r = a = u1 r = a = E 0 a cos u 0 (接地，总电势为零 u 2 r 0 通解为 B u 2 (r , = Al r l + l +l1 Pl ( x r l =0 u2 r 0 ，

7、得到 Al = 0 u 2 (r , = l =0 Bl Pl ( x r l +1 u2 r =a = l =0 Bl Pl ( x = E 0 a cos u 0 a l +1 B1 = E 0 a 3 Bl = 0(l 2 B0 = u 0 a u 2 (r , = u 0 a E0 a 3 + 2 cos r r u 0 a E0 a 3 + 2 cos u (r , = u1 (r , + u 2 (r , = u 0 E 0 r cos r r 3 a a = u 0 1 E 0 r 1 r 3 cos r 5、真空中存在一电场强度为 E 0 的匀强电场，将一个半径为 a 、相对介

8、电常数为 的介 质球放置匀强电场中，求球内外电势分布？ 1 球内电势所满足定解方程 2 u内 = 0 u内 r =0 = 有限 (r a (r a B u内 (r , = Al r l + l +l1 Pl (cos r l =0 由 u内 r =0 = 有限 ，得到 Bl = 0 u内 (r , = Al r l Pl (cos l =0 (r a u 外 = 0 u 外 r = u 0 E 0 r cos D u 外 (r , = C l r l + l +l1 Pl (cos r l =0 (r a 由 u外 r = C l r l Pl (x = u 0 E 0 r cos ， l =

9、0 得到 C 0 = u 0 , C1 = E 0 , C l = 0(l 0,1 u 外 (r , = u 0 E 0 rP1 (cos + l =0 Dl Pl (cos r l +1 (r a 3 边界连接条件 u 内 r = a = u 外 r = a u 内 u 外 0 r =a = 0 r r r =a Dl l ( ( A a P cos u E aP cos P (cos = + 0 0 1 l l l +1 l l =0 l =0 a Dl (l + 1 A la l 1 P (cos = E P (cos Pl (cos 0 1 l l a l +2 l =0 l =0 比

10、较两边系数，得 D A0 = u 0 + 0 a 0 = D0 a2 A0 = u 0 D0 = 0 1 A1 a = E 0 a + D1 2 a A = E 2 D1 1 0 a3 3 A1 = +2 D = 1 a 3 E 1 0 +2 Al = 0 Dl = 0 1 Al a l = Dl l +1 a (l 0,1 lA a l 1 = (l + 1 Dl l a l +1 (l 0,1 u内 (r , = u 0 3 (r a 6、 （1） 1 k +1 k xPk ( x Pl ( x dx = Pk +1 ( x + Pk 1 (x Pl ( x dx 1 1 2k + 1 2

11、k + 1 1 1 k +1 k Pk +1 (x Pl (x dx + Pk 1 ( x Pl ( x dx = 1 2k + 1 2k + 1 1 k +1 2 k 2 k +1,l + k 1,l = 2k + 1 2l + 1 2k + 1 2l + 1 2(k + 1 (2k + 1(2k + 3 , l = k + 1 2k , l = k 1 = (2k + 1(2k 1 0, l k + 1, k 1 1 （2） 1 l +1 l ( ( xP x P x dx = Pl +1 ( x + Pl 1 ( x Pl +1 (x dx 1 + l l 1 1 2l + 1 2l + 1 l +1 2 2(l + 1 = = 2l + 1 2(l + 1 + 1 (2l + 1(2l + 3 1 （3） 1 l +1 l l+2 l + 3 x 2 Pl