13.07.2015 Views

Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

18.2 ASSOCIATED LEGENDRE FUNCTIONSSince d 2l (1 − x 2 ) l /dx 2l =(−1) l (2l)!, <strong>and</strong> noting that (−1) 2l+2m = 1, we have1 (2l)!(l + m)!I lm =2 2l (l!) 2 (l − m)!We have already shown in section 18.1.2 that∫ 1∫ 1−1(1 − x 2 ) l dx.K l ≡ (1 − x 2 ) l dx = 22l+1 (l!) 2−1(2l +1)! ,<strong>and</strong> so we obtain the final resultI lm = 2 (l + m)!2l +1(l − m)! . ◭The orthogonality <strong>and</strong> normalisation conditions, (18.36) <strong>and</strong> (18.37) respectively,mean that the associated Legendre functions Pl m (x), with m fixed, may beused in a similar way to the Legendre polynomials to exp<strong>and</strong> any reasonablefunction f(x) on the interval |x| < 1 in a series of the <strong>for</strong>mf(x) =∞∑a m+k Pm+k(x), m (18.38)k=0where, in this case, the coefficients are given by2l +1(l − m)!a l = f(x)Pl m (x) dx.2 (l + m)! −1We note that the series takes the <strong>for</strong>m (18.38) because Pl m (x) =0<strong>for</strong>m>l.Finally, it is worth noting that the associated Legendre functions Pl m (x) mustalso obey a second orthogonality relationship. This has to be so because one mayequally well write the associated Legendre equation (18.28) in Sturm–Liouville<strong>for</strong>m (py) ′ +qy+λρy = 0, with p =1−x 2 , q = l(l+1), λ = −m 2 <strong>and</strong> ρ =(1−x 2 ) −1 ;once again the natural interval is [−1, 1]. Since the associated Legendre functionsPl m (x) are regular at the end-points x = ±1, they must there<strong>for</strong>e be mutuallyorthogonal with respect to the weight function (1 − x 2 ) −1 over this interval <strong>for</strong> afixed value of l, i.e.∫ 1−1∫ 1P m l (x)P k l (x)(1 − x 2 ) −1 dx =0 if|m| ≠ |k|. (18.39)One may also show straight<strong>for</strong>wardly that the corresponding normalisation conditionwhen m = k is given by∫ 1Pl m (x)Pl m (x)(1 − x 2 ) −1 (l + m)!dx =−1m(l − m)! .In solving physical problems, however, the orthogonality condition (18.39) is notof any practical use.591

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!