Identites and properties for associated Legendre functions

Transcription

1 Identites and properties for associated Legendre functions DBW This note is a persona note with a persona history; it arose out off y incapacity to find references on the internet that prove reations that exist between the associated Legendre functions. The goa is to put notes on the internet that at east contain the 14 recurrence reations and soe other identites found on Wikipedia 1, without resorting to advanced ethods, such as generating functions; in these notes erey basic anaysis and agebraic anipuations are used. Wikipedia cites any references, such as Hibert and Courant 1953, or Abraowitz and Stegun But ost of these were not at y disposa, and soe are very usefu atheatica tabes, but with very itte proofs. I start with the definition and soe basic properties of Legendre poynoias P n, then introduce associated Legendre functions P. Then foows the ain text, in which I give proofs of a nuber of reations aong the P. I then consider the nuber of zeroes of the P n and P. After that, I show proofs of soe integra reations, which are cited in the ain text. On notation: I often oit the arguent x of the functions and write d for differentiation with respect to x on any occasions. Finay I give part of a screen shot of the Wikipedia-website showing the 14 recurrence reations entioned above. I have itte hope this text wi be free of typos and ore serious errors; pease write an eai if you find one to westradennis at gai dot co! 1 Legendre Poynoias We define the Legendre poynoias P for = 0, 1, 2,... by Rodriguez forua P x = 1 d x ! dx By inspection, P has degree, and we have P x = P x as the th derivative of an even function has parity. The P for an orthogona set of poynoias on [; 1] 1 See the section recurrence reations fro poynoias. 1

2 and proofs of the reation P xp k xdx = δ k, 2 + 1, 2 can be found in any onine resources, and we present one in section 4. The sybo δ k, is the Kronecker deta, which equas 1 if k = and zero otherwise. Our first reation needs the orthogonaity reation and soe hard work, athough a proof using the generating function is faster but then, one first has to show the generating function has the right properties. The Legendre poynoias satisfy the foowing recurrence reation n + 1P n+1 x = 2n + 1xP n x np n x 3 Proof: Consider the poynoia xp n x. It has degree n + 1 and is thus in the inear span of P 0,..., P n+1. We can hence write xp n x as a inear cobination of the first n + 2 Legendre poynoias and the kth Legendre poynoia appears with coefficient a k = 2k xp n xp k xdx. We are interested in integras of xp r xp s x for genera r and s. In section 4, we show that these integras vanish uness r = s ± 1 and for this case, we can use xp r xp r xdx = 2r 2r 12r + 1. Writing xp n x = αp n+1 x + βp n x, and first integrating the product with P n+1 we find α = n+1 n 2n+1 and siiary β = 2n+1. Hence This is what we wanted to prove. 2n + 1xP n x = n + 1P n+1 x + np n x. The difference P +1 P satsifies the foowing differentia reation: 2 + 1P x = d P +1 x P x. 4 dx Proof: The second derivative of x equas d 2 x = 2 + 1x x 2 x 2 1 = x x 2 1. Acting on this resut with ! d gives the required resut. 5 2

3 2 Associated Legendre Functions We define the associated Legendre functions P for by P x = 2! 1 x 2 /2 d +x dx One iediatey sees that P 0 = P and that for 0 we have P x = 1 x 2 /2 d P x. 7 dx The functions P are poynoias of degree for even and if is odd, then P is 1 x 2 ties a poynoia of degre 1. The parity of P is +. If one tries to appy the defining forua 6 for > one finds zero. We have: 1 x 2 P x = 1 P P +1 +1, 0. 8 Proof: We act on identity 4 with 1 x 2 +1/2 d and the resut foows. Later, when we have shown 21 we wi see that the condition 0 can be reaxed. The reation 8 is the sixth on the ist of recurrence reations of Wikipedia. For negative we can use the foowing reation between P P and P :! = +! P. 9 Proof: By induction one easiy shows that for k = 0, 1, 2, 3,... and soe differentiabe functions f and g k k d k fg = d r f d k r g. r We consider f = x 1 and g = x + 1, so that and for p = 0, 1, 2, 3,... r=0 df p = pf p, d k f r = r k f r k where r k is defined to be 0 if r < k and r k = r! r k! for r k. We now show that d f g! = fg +! d+ f g. 3

4 This wi prove the cai about the associated Legendre poynoias. To show the interediate stateent, we first rewrite + + d + f g = d r f d + r g = r r=0 + d r f d + r g r since the oitted ters in the su vanish; d r f s = 0 if r > s. Then we can write put in the expressions for d r f and for d + r g, and change the suation variabe to s = r to obtain d + f g = s=0 Second, we rewrite the other ter d f g out as d f g = r=0 r=! 2 +! s!s!s +! s! f s g s. 10!! r! r! r! f r! r! g r which by inspection is proportiona to fg ties 10, and the constant of proportionaity can be read off to be! +!. This proves that what we set out to show. Equation 9 akes it cear why P = 0 if >. The foowing reation is the first of the 14 recurrence reations isted by Wikipedia: 2 + 1xP = + 1P P. 11 Proof: First we reark that it suffices to proof 11 for nonnegative : If 11 hods for positive, then one can insert the reation P = +!! P and one shows that it foows that 2 + 1xP = + + 1P +1 + P, which is precisey 11 subject to the substitution. Differentiating reation 3 ties and utipying with 1 x 2 /2 we find for nonnegative the reation + 1P+1 x = 2 + 1xP x x 2 1/2 P x P x. 12 Cobining eqns.12 and 8 one obtains the resut. The foowing two resuts are rather cheap. 4

5 If we differentiate the defining reation 6 we obtain: dp x = x 1 x 2 P x P +1 x x 2 Now it is a sipe step for the foowing. Mutipying 13 with 1 x 2 one obtains nr. 13 of the ist of recurrence reations fro Wikipedia: x 2 1dP x = xp x + 1 x 2 P The foowing reation gives an easy way to et go one up: P +1 x = xp x + 1 x 2 P x. 15 Proof: We consider d ++1 x = d xx 2 1, and use the binoia reation for derivatives d r fg = r k d k fd r k g to rewrite the righthand side as Mutipying with d + x xd + x ! 1 x2 /2 one obtains the resut. The next is the second in the ist on Wikipedia: 2xP x = 1 x 2 P +1 x P x. 16 Proof: We consider the foowing identity d ++1 x = d ++1 x 2 1 x 2 1 and use the binoia identity for derivatives rewrite the right-hand side as x 2 1d ++1 x xd + x d + x 2 1. Mutipying through by 2! 1 x 2 /2 and regrouping soe ters we find 2 + 1P+1 2x + + 1P = 1 x 2 P P. 17

6 If we insert reation 15 to eiinate P+1 we obtain the 16. This identity is the third on the ist of Wikipedia: 1 P 1 x 2 x = 1 2 P +1 x P x. 18 Proof: We start with 16 and substitute 2xP which foows fro 15. Then it foows that 2P+1 = 1 x 2 P +1 = 2P x 2 P, P fro which 18 foows iediatey if we change to 1. The foowing identity is not on Wikipedia s ist, but is definitey usefu to reate soe identities. Its proof is rather cubersoe and tedious. The foowing is rather surprising: P P = P P , Reark: Athough at first gance the right-hand side sees to be of degree + 1, the highest-order ters cance, and so is aso of degree 1. Proof: If we pug in the definitions of the P, we see that we are to prove that the two ters L = x 2 d d + 2 x 2 1 and R = 1 x 2 d d + x are equa. We distinguish between two cases: I + = 2n, II + = 2n + 1. We first consider case I, and eiinate a occurrences of. We expand x 2 1 ±1 by the binoia theore, appy d A x B = B! B A! xb A and find that where L k = 2 +n+k+1 1 n + k 1 n L = 4 + 1L k x 2k, k=0 2k + 2n 2! 2k! 2n+2 n k+k2k n2n. 6

7 We aso expand R = n+1 k=0 R k x 2k and find + 1 2k + 2n! R k = 2 +n+k 2n+2k+1+1 n k+k2k 2 2n+1 n+1. n + k1 2k! In particuar, R n+1 = 0. We can write 4 + 1L k = 4 + 1!2k + 2n 2! n + k! + 1 n k!2k! +n++1 2n n ka k with and siiary a k = 2n n k + k2k 1 n2n 1 R k = 4 + 1!2k + 2n 2! n + k! + 1 n k!2k! +n++1 n + k2k + 2n 1b k with b k = 2 2n + 1 n + 1 2n + 2k n k k2k 1. By inspection, it suffices to show 2a k + 1 n k = 2k + 2n 1b k, which is seen rather easiy if we rewrite a k and b k as a k = 2n2n + 2k 1, b k = 2 2n + 1 n k. This shows identity 19 for + is even. For the case, where + is odd, we take + = 2n + 1, and it bois down to proving that the ters L = 1 x 2 d 2n+1 + 2n2n + 1d 2n x 2 1 and R = 1 x 2 d 2n n2 2n + 1d 2n+1 x are reated by 4+1L = R. Again we write L = n k=1 x2k L k and R = n+1 k=1 x 2k R k. We find with soe agebra that L k = 2 +n+k+1 1!2n + 2k 2! n + k 1! n k!2k 1! a k where a k = 2n 12n + 2k 1. This eads to +n+k+1 + 1!2n + 2k!2n + 2k 1! 4 + 1L k = 4 2n 1. n + k! n k!2k 1! 7

8 Siiar and siiary tedious agebra eads to with R k = 2 +n+k+1 + 1!2n + 2k! n + k! + 1 n k!2k 1! b k b k = n2 2n + 1 2k 1k 1 2n + 2k n k, which can be brought into b k = 2 2n n k. Substituting this expression into R k shows that 4 + 1L k = R k and the proof is done indeed, aso R n+1 = 0. Nr. 4 on Wikipedia s ist foows directy fro 18 and x 2 P = P P Nr. 5 on Wikipedia s with 0 ist foows directy fro 8 and 19 1 x 2 P = P P. 21 Let us now show: Equations 8 and 21 aso hod for 0. By virtue of 9 we substitute P = +!! P, P +1 = +1 +! +2! P and P = +!! P +1 in 21, which eads directy to 1 x 2 P P +1 +1, which is precisey = P 8 for negative. The sae substitutions in 8 show that 21 aso hods for negative. As nr. 6 was aready proven, we now proceed to nuber 7 on Wikipedia s ist. The foowing identity hods 1 x 2 P +1 = x P + P. 22 Proof: We use 16 to write 1 x 2 P +1 = x 2 P 2xP. 8

9 Then we insert 8 to eiinate 1 x 2 P and find 1 x 2 P +1 = P Inserting now + 1P+1 = 2 + 1xP agebra then finishes the proof P xP. + P, a.k.a. 11, and doing the The foowing identity is nr. 8 on Wikipedia s ist: 1 x 2 P +1 = + 1P xP. 23 Proof: We rewrite the right-hand side of 22 as x P + P = 2 + 1xP + + 1xP + P. The proof is finished by appying 11 to eiinate the ter 2 + 1xP. We now turn to soe differentia properties of the associated Legendre functions. We present nr. 9: 1 x 2 dp x = 1 2 P +1 x P 2 x. 24 We can rewrite equation 13 as 1 x 2 dp x = x 1 x 2 P x P +1 x. Using equation 16 and eiinating xp x eads to the resut. The reaining equations fro the ist of identities on Wikipedia are then ainy just sipe agebra and cobining a the other resuts: Nuber 10: 1 x 2 dp = P + 1P Proof: Using eqn.24 we write the eft-hand side of equation to be proven as x 2 2 P 1 x 2 P +1. Then we use equation 8 for 1 x 2 P and equation 21 for 1 x 2 P +1. This finishes the proof. 9

10 Nuber 11: x 2 1dP = xp + P. 26 Proof: If cobine 11 and 25 this identity foows iediatey. Nuber 12: x 2 1dP = + 1xP P Proof: The proof foows if we use = and write by virtue of 26 x 2 1dP and then use reation 11. = 2 + 1xP + 1xP + P As we aready had nuber 13 before, we go on to the ast on the ist: Nuber 14: x 2 1dP First we write xp = 2xP 1 x 2 P +1, which shows = x 2 P xp. 28 xp, then we insert 16 and subsequenty we add xp + 1 x 2 P +1 = 1 x P xp, which is the right-hand side of the equation to be proven by virtue of On the zeros of associated Legendre poynoias In this section we prove the foowing cais: 1 The Legendre poynoias P x have distinct sipe zeros on the interva ; +1; 2 The associated Legendre functions P x have distinct sipe zeros on the interva ; +1. The case = 0 is trivia, since P 0 x = 1. Cai 1 is cai 2 restricted to = 0. We first consider this case. To ake the notation easier, we introduce the short-hand f, g = fxgxdx.

11 Then the integra of P n over the interva [; +1] equas zero as this integra equas P, 1 = P, P 0 = 0. Thus P, being continuous, ust change sign soewhere on ; +1 and thus has at east one zero of odd utipicity. Let ξ 1,..., ξ k be a the different zeros of odd utipicity of P, so that 1 k and the continuous function ZP with Z = x ξ 1 x ξ 2 x ξ k does not change on ; +1. Thus Z, P 0. If k < then Z is a poynoia of degree saer than and thus can be written as a inear cobination of P 0, P 1,..., P. But then Z, P vanishes, which cannot be, and thus k =. Thus P has distinct zeros. For the second cai, we need a preiinary resut, which is P, P which is proved in section 4. = +1 P xp dx = !! δ, 29 Since we ay assue > 0 and > 0, we first consider =. Then P is proportiona to 1 x 2 /2 and thus has zero zeros on ; +1. Thus in this case the cai is true. For the reaining we fix > > 0. For each k 0 the functions Pk is of the for 1 x 2 p k x, where p k is a poynoia of degree k. Thus for given N > the poynoias 1 x 2 /2 P x, 1 x 2 /2 P+1 x, 1 x2 /2 PN x span the space of poynoias with degree at ost N. Since P, P = 0 as > and P does not change sign, so does P have to change sign. Thus there are k 1 zeros ξ 1,..., ξ k of P of odd utipicity on ; +1. We introduce Z = 1 x 2 /2 x ξ 1 x ξ k, so that ZP does not change sign on ; +1 and therefore Z, P 0. If < k then Z can be written as a inear su of ess then associated Legendre functions P,... Pk, so that in this case we shoud have Z, P = 0, which we aready argued cannot be. Hence = k and the cai is proved. 4 Proof of integra reations We consider the function Ba, b defined by A partia integration shows 0 t a 1 t b dt = b a + 1 Ba, b = 0 0 t a 1 t b dt. t a+1 1 t b dt = 11 b Ba + 1, b 1. a + 1

12 Since Ba, 0 = 1 a+1, it then foows rather easiy Ba, b = Using the substitution x = 2t 1 we deduce a!b! a + b + 1! x 2 dx = B, = 2 2+1!! 2 + 1!. 31 Let us now show that P k xp xdx = Pugging in the definition, we have 1 1 P k xp xdx = 2 k+ d k x 2 1 k d x 2 1 dx.!k! Since d a x 2 1 b vanishes at x = ±1 if a < b, partia integration eads to P k xp xdx = 2 k+!k! d k+ x 2 1 k x 2 1 dx. If k we ay assue > k and then d k+ x 2 1 k = 0. Hence the integra vanishes uness k =. In this case we use d 2 x 2 1 = d 2 x 2 = 2! and x 2 1 = 1 x 2 to find 2! 1 P k xp xdx = δ k, 2 2 x 2 1 dx,!! which directy eads to the resut using 31. Let us now show that 2r xp s xp r xdx = δ r,s+1 2r 12r δ 2s s,r+1 2s + 12s First we consider the case that r = s. In this case the integra is easiy found to be zero P r has parity r, and thus xp r 2 has parity. Therefore the integra over the syetric interva [; 1] vanishes. Now we consider the case r > s. We find, using Rodriguez forua 1, partia integration and xf n = xf n +nf n, that xp rxp s xdx equas up to soe nuerica factors the expression x 2 1 r[ d r+sx d r+sx x 2 1 s + r 2 1 s] dx. dx dx 12

13 We see that the first ter vanishes since r > s; the second ter ony survives if r s + 1. Hence we need ony consider the case s = r 1. Writing c r = 1 2 r r! we find, again using Rodriguez forua, partia integration and xfn = xf n + nf n but aso that the 2r 2th derivative of x 2 1 r equas 2r 2!, that xp r xp r xdx = r c r c r 2r 2! Putting in the nubers, we find xp rxp r xdx = to the sae expression, but s and r interchanged. Finay, we show +1 Pugging in the definition, we have +1 P k xp xdx = Pk xp 2 xdx = δ k, k+ k!! x 2 1 r dx. 2r 2r2r+1. The case s > r eads +!! x 2 d k+ x 2 1 k d + x 2 1 dx. The degree of gx = 1 x 2 d k+ x 2 k is k+. The degree of hx = d + x 2 is. If we first assue k, we ay as we take > k. The first derivatives vanish at x = ±1 and the first 1 derivatives of x 2 1 vanish at we at x = ±. Therefire, if we perfor partia integrations, oving the derivatives fro h to g, a boundary ters vanish. Since d + gx = 0 if > k, the integra 34 vanishes if k. If k =, we find +1 Pk xp xdx = + 2 k+!! x 2 1 d + 1 x 2 d + x 2 1 dx. Since the ter in brackets has degree +, we ony need to consider the highest order ter; d + 1 x 2 d + x 2 1 = d + x 2 d + x 2 2! +! =.! Cobining this resut and the previous integras one easiy finds the stated resut. 13

14 5 The ist of recurrence reations fro Wikipedia 14