§1.3球函数 缔合 Legendre函数 1.缔合 Legendre方程的有限解(本征值)问题 (1-x)y.2xyPr2=0() y→有限 x=0—常点,当然可用常微分方程的级数解法 求解。现用下述方法求解,以使之与研究详 尽,较简单的P(x)联系
§1.3球函数 一、缔合Legendre函数 1.缔合Legendre方程的有限解(本征值)问题 ( ) ( ) ( ) 2 2 2 1- - 2 1 - 0 1 1- m x y xy l l y x é ù ¢¢ ¢ + + = ê ú ë û x 1 y =± →有限, x=0一常点，当然可用常微分方程的级数解法 求解。现用下述方法求解，以使之与研究详 尽,较简单的 P x l ( )联系
令()=-x)=()(2) 代入(得 (1-x2)y”(x)-2(m+1)xw/(x) +[(+1)-m(m+)y=0(3) 又1-x2)Px)-2xP(x)+1(+1)P(x)=0(4) dx ()(x)2()-2xm( (+)m(m+p()0(4)
( ) ( ) ( ) ( ) 2 2 1- 2 m 令y x = x x n 代入(1)得 ( ) ( ) ( ) ( ) ( ) ( ) 2 (1 ) - 2 1 1 - 1 0 3 x x m x x l l m m n n n - + ¢¢ ¢ + é ù + + = ë û ( ) ( ) ( ) ( ) ( ) 2 (1- ) - 2 1 0 4 l l l 又 x P¢¢ ¢ x xP x + l l + = P x ( ) ( ) ( ) 1 - 1 ( ) 0 4( ) m l l l m m P x ¢ + é ù + + = ë û ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 : 1- - 2 1 m m m m l d x P x x m P x dx l ² ¢ é ù + é ù ë û ë û
对比(3)(4)得v(x)=P(x) 代入(2) y(x)=(-x)P()=P(x) 缔合 Legendre函数
对比 ( )( ) ( ) ( ) 3 4 ( ) m l 得n x = P x ( ) ( ) ( ) ( ) ( ) 2 1- 2 m m l l m y x = x P x = = P x 记 称 代入(2): 缔合Legendre函数
故本征值问题(1)的本征值为:+1,1=0,2, 本征函数为: 由(5)→y(x)=P0(x)=(1-x) d P(x),m=0.,1 d x 7(5) P()=p() P(x)=(-+2()=(1-x2)2,oP(s)=sme (x)=x d x 822 1()=0-x)22()=3-x) or p'lcos0)==sin 20 2(x)=3(1-x),or2(cs)=3si2
故本征值问题(1)的本征值为：l(l+1),l=0,1,2,… 本征函数为： ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1- , 0,1,... 5 m m m l l m d y x P x x P x m l dx 由(5) ® = = = ( ) ( ) 0 Pl l x = P x ( ) ( ) ( ) ( ) 1 1 1 2 2 2 2 1 1 1 (1 ) (1 ) , cos sin P x x d P x x P x x orP dx q q = ¢ ¢ = - = - = ( ) ( ) ( ) ( ) 1 2 3 -1 2 1 1 2 2 2 2 2 2 (1 ) 3 1- , x d P x x P x x dx ¢ = - = ( ) ( ) 2 3 cos sin 2 6 2 or P¢ q q = ( ) ( ) ( ) 2 2 2 2 2 2 P x = = 3 1- x ,orP cosq q 3sin
2.Pm(x)的微分式 1 d P(r) 211 dx (2-) d 27! dx 1+m 在(1)中,m换-m形式不变。 故有Pm(x) d 2ll dx/-m (x2 3P"(x)的积分式
在(1)中，m 换-m形式不变。 2. ( ) m P x l 的微分式： ( ) ( ) 1 2 -1 2 ! l l l l l d P x x l dx Q = ( ) ( ) 2 2 2 1- ( 1) (7) 2 ! m l m m l l l l m x d P x x l dx + + = - ( ) ( ) ( ) - 2 2 - - 2 - 1- -1 2 ! m l m l m l l l m x d P x x l dx 故有 = . ( ) m 3 P x l 的积分式：
()=2n f(5) 2 n+1 兀l 故由(7)立即可证: .x)2(2+m)( ()=-2n2 f、d2(8) 丌l 二、缔合 Legendre多项式的性质 1.递推公式: (1+1-m)Pm(x)-(21+1)xP"(x) (1+m)(x)=09)
( ) ( ) ( ) ( ) 1 ! 2 - n l n n f f z d i z x x p x + Q = ò 故由(7)立即可证： ( ) ( ) ( ) ( ) ( ) 2 2 2 * 1 1- ! -1 (8) 2 ! 2 - m l m l l l l m x l m P x d l i x x x p x + + + = ò 二、缔合Legendre多项式的性质 1.递推公式： ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 -1 1- - 2 1 0 9 m m l l m l l m P x l xP x l m P x + + + + + =
(1+1)P1(x)-(21+1)xP(x) +P1(x)=0(B) ()(+2(x)(21+)yP(x) m(21+1)m(x)+m(x)=0( 又(21+1)P(x)=P1(x)P1(x)
( ) ( ) ( ) ( ) ( ) ( ) 1 -1 1 - 2 1 0 l l l l P x l xP x lP x B + + + + = Q ( ) ( ) ( ) ( ) ( ) ( ) : 1 1 - 2 1 ( ) m m m m l l d B l P x l xP x dx + + + ( ) ( ) ( ) ( ) ( ) ( ) -1 -1 - 2 1 0 10 m m m l l l + P x + = lP x ( ) ( ) ( ) ( ) 1 -1 2 1 - l l l l P x P x P x + 又 + = ¢ ¢
故有-m(21+1)0(x) cmpifm ()+mp- (x)(11) )代入 2.正交性 5 pm() pm(x dx +m)!2 2/+1
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) -1 1 -1 - 2 1 - 11 m l m m l l m l P x mP x mP x + + = + 故有 ( ) ( ) ( ) 2 2 11 10 1- m é ù× Þ x ë û 代入 (9) 2.正交性 ( ) ( ) ( ) ( ) 1 -1 ! 2 - ! 2 1 m m l k kl l m P x P x dx l m l d + ò = +
证明:记=P(x)P(x)tx I (l-x2)p(m (x). pom)(x)cr -x)y0(x)p( 出(小2(xp( ((-x)()女
证明： ( ) ( ) 1 , -1 m m m l k l k 记I = ò P x P x dx ( ) ( ) ( ) ( ) ( ) 1 2 -1 -1 1- m m m l k d x P x P x dx dx = ò ( ) ( ) ( ) ( ) ( ) 1 -1 2 -1 - 1- m m m k l d P x x P x dx dx é ù ò ê ú ë û ( ) ( ) ( ) ( ) ( ) 1 2 -1 -1 1- m m m l k = x P x P x ( ) ( ) ( ) ( ) ( ) 1 -1 2 -1 - 1- m m m k d P x x P x dx dx l é ù = ò ê ú ë û
而由(4)1(-x) -x2)yPm3(2(m)1-x2)Pm(x) +[(+1)m(m+(1-x)P0()=0 那(-)p- [(+1)-m(m+1)(1-x)P(x) 故=[1(+1)m(m+)(1-x2)“P0(x)Pm(x) [(+1)-m(m+1)]/m (1+m)(1-m+1)7m
( ) 4 而由 ¢ ( ) 2 1- : m × x ( ) ( ) ( )( ) ( ) ( ) 1 2 ( 2) 2 1 1- - 2 1 1- m m m l l x P x x m x P x + + + + ( ) ( ) ( ) ( ) ( ) 2 1 - 1 1- 0 m m l + é ù l l + m m + = x P x ë û ( ) ( ) ( ) 1 2 1 1- m m l d x P x dx + é ù + ê ú ë û 即 ( ) ( ) ( ) ( ) ( ) 2 - 1 - 1 1- m m l = é ù l l + + m m x P x ë û ( ) ( ) ( ) ( ) ( ) ( ) ( ) -1 1 2 -1 -1 , -1 1 - 1 1- m m m m l k l k I = é ù l l + + m m ò x P x P x dx 故 ë û ( ) ( ) -1 , 1 - 1 m l k = é ù l l + + m m I ë û ( )( ) -1 , - 1 m l k = l + + m l m I
