ebook img

Orthogonality of the Ferrers' Associated Legendre Functions of the Second Kind with Imaginary Argument PDF

0.12 MB·
Save to my drive
Quick download
Download
Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.

Preview Orthogonality of the Ferrers' Associated Legendre Functions of the Second Kind with Imaginary Argument

Orthogonality of the Associated Legendre Function of the Second Kind with Imaginary Argument N. Dimakis ∗ Instituto de Ciencias F´ısicas y Matem´aticas Universidad Austral de Chile, Valdivia, Chile 7 1 0 2 Abstract n a J In this work study the associated Legendre functions of the second kind with 9 a purely imaginary argument Qk( x). We derive the conditions under which they provide a set of square integrablℓeifunctions when x R and we prove the relevant ] ∈ A orthogonality relation that they satisfy. C . h 1 Introduction t a m The importance of the associated Legendre equation is well known in applied mathe- [ matics. Its significance is also evident in mathematical physics as it appears in various 1 different areas of study; from geodesy [1] to quantum mechanics [2]. In the latter it usu- v 9 ally emerges through a separation of values process over the eigenvalue equation for the 1 Casimir invariant of the so(3) algebra of the angular momentum operators [2], that leads 2 2 to 0 d dY(z) k2 . (1 z2) + ℓ(ℓ+1) Y(z) = 0. (1.1) 1 dz − dz − 1 z2 0 (cid:20) (cid:21) (cid:20) − (cid:21) 7 The solution of (1.1) is expressed as a linear combination of the two associate Legendre 1 functions of the first Pk(z) and second kind Qk(z), : ℓ ℓ v i X Y(z) = c Pk(z)+c Qk(z), (1.2) 1 ℓ 2 ℓ r a which can be defined by means of a more general function: (1+z)k/2 1 z Pk(z) = F˜ ( ℓ,ℓ+1;1 k; − ) (1.3a) ℓ (1 z)k/22 1 − − 2 − π (1+z)k/2 1 z Qk(z) = cos(kπ) F˜ ( ℓ,ℓ+1;1 k; − ) ℓ 2sin(kπ) (1 z)k/2 2 1 − − 2 " − (1.3b) Γ(k +ℓ+1) (1 z)k/2 1 z − F˜ ( ℓ,ℓ+1;k +1; − ) − Γ( k +ℓ+1)(1+z)k/22 1 − 2 # − ∗ [email protected] 1 where Γ(α) represents the Gamma function, while +∞ (a) (b) F˜ (a,b;c;z) = s s zs, z < 1 (1.4) 2 1 Γ(c+s)s! | | s=0 X is the regularized Gauss hypergeometric function, in which also appears the Pochhammer symbol (x) = Γ(x+n)/Γ(x). n There exists an extensive bibliography regarding the properties of (1.3) [3–6]. Most of those that are well known refer to the associate Legendre function of the first kind Pk(z), ℓ with the most recognizable being of course the orthogonality relation 1 2(k +m)!δ Pk(x)Pk(x)dx = mn , k,m,n N. (1.5) m n (2m+1)(m k)! ∈ Z−1 − Other orthogonality relations regarding Pk(z) in terms of Dirac’s delta function when ℓ the order or the degree is complex are also known, see for example [7–10]. In this work however, we concentrate onQk(z) andto the orthogonalityrelation to which it leads when ℓ the argument is purely imaginary. In particular, we prove that when z = x, x R, the i ∈ following relation holds (−1)ℓπ3/2A0(ℓ)2δℓℓ′ k (ℓ+i)(ℓ i+1), k > ℓ+1 +∞ Γ(12−ℓ)ℓ! i=ℓ+2 − Qk( x)Qk( x)dx=  (1.6) Z−∞ ℓ i ℓ′ i (−1)Γℓ(π13−/2ℓ)Aℓ0!(ℓ)2δℓℓ′Q, k = ℓ+1 2  whenever k Z , ℓ N, with k > ℓ. The A (ℓ) appearing in (1.6) is given by relation + 0 ∈ ∈ A (ℓ) = (1 + x2)(ℓ+1)/2Qℓ+1( x) and it is (a real finite) constant with respect to x. Its 0 ℓ i value depends only on ℓ and it is calculated explicitly in the appendix. The definite integral of the left hand side denotes the improper Riemann integral defined as +∞ a F(x)dx = lim F(x)dx. (1.7) a→+∞ Z−∞ Z−a The consideration of an imaginary argument for the independent variable z = x, i turns (1.1) into d dY(x) k2 (1+x2) ℓ(ℓ+1) Y(x) = 0 (1.8) dx dx − − 1+x2 (cid:20) (cid:21) (cid:20) (cid:21) with the solution being expressed of course as any linear combination of Pk( x) and ℓ i Qk( x). This equation is our starting point in the consecutive analysis. The structure ℓ i of the paper is the following: In section 2 we briefly summarize some of the properties that are to be used. In particular, we are interested in the behaviour of Qk( x) at the ℓ i boundary x and for which values the function vanishes. In section 3 we start → ±∞ from (1.8) and exhibit under which conditions Qk( x) is orthogonal in x ( ,+ ) ℓ i ∈ −∞ ∞ for different values of ℓ, while in the final section we prove that it is square integrable and derive expression (1.6). 2 2 Some relevant properties of Qk(z) ℓ First of all, let us investigate how Qk(z), with z = x, behaves as a function in relation ℓ i to its order and degree. In what follows we assume that the order of the function is a positive integer i.e. k Z and we distinguish the following two cases: + ∈ (1) ℓ N ∈ Then, Qk(z) can be expressed with the help of the subsequent relations ℓ dk Qk(z) = ( 1)k(1 z2)k/2 Q (z) (2.1a) ℓ − − dzk ℓ ℓ 1 1+z P (z)P (z) j−1 ℓ−j Q (z) = P (z)ln (2.1b) ℓ ℓ 2 1 z − j (cid:18) − (cid:19) j=1 X 1 dℓ P (z) = z2 1 ℓ , (2.1c) ℓ 2ℓℓ!dzℓ − h i (cid:0) (cid:1) where P , Q are the Legendre functions of the first and second kind respectively. By ℓ ℓ virtue of (2.1) it can be shown that Qk(z) assumes the general form [11] ℓ (z2−1)−k/2 1A(1)(z)ln z+1 +A(2)(z) , ℓ k (ℓ−k)! 2 kℓ z−1 kℓ ≥ Qk(z) =  h i (2.2) ℓ (cid:0) (cid:1)   (−1)k A(2)(z), ℓ < k (z2−1)k/2 kℓ   (i)  where A (z), i = 1,2, are polynomials in z, with the following properties: kℓ is of order ℓ+k, when ℓ k A(1)(z) ≥ (= 0, when ℓ < k (2.3) is of order ℓ+k 1, when ℓ k A(2)(z) − ≥ (is of order k ℓ 1, when ℓ < k − − and additionally The coefficients of the powers of z in A(i)(z) are all integers. • If the order of the polynomial is even (odd) then all the powers of z are even or • odd. Hence, for k > ℓ, it follows that when k ℓ 1 is even, Qk( x)is a real function, else − − ℓ i it is purely imaginary. What is more, as can be seen from (2.2) and (2.3), Qk( x) and all its derivatives are finite for every x R. ℓ i ∈ Due to these properties it is clear that when the inequality k > ℓ holds, Qk( x) is a ℓ i solution of (1.8) which vanishes at infinity and at the same time does not posses any singularities for x R. Although this is neither a sufficient nor an adequate condition ∈ for a function to be square integrable, it is a first indication of a behaviour that may lead to such a result. 3 (2) ℓ R ∈ In this case let us use the following expression that is valid for values of z outside the unit circle 2−(ℓ+2)e kπ√π(1 z2)k/2 Qk(z) = i − ℓ cos(ℓπ)zk+ℓ+1 +∞ k−ℓ k−ℓ−1 22ℓ+1Γ(k ℓ)sin((k ℓ)π)z2ℓ+1 2 s 2 s (2.4) ×"i − − s=0 Γ(cid:0)(s−(cid:1)ℓ(cid:0)+ 21)s!z(cid:1)2s X +∞ k+ℓ+1 k+ℓ+2 + cos((k +ℓ)π)+e (ℓ−k)π Γ(k +ℓ+1) 2 s 2 s . i Γ(s+ℓ+ 3)s!z2s s=0 (cid:0) (cid:1) (cid:0) 2 (cid:1) # (cid:0) (cid:1) X With the help of (2.4) let us examine what happens to Qk( x) as x and in ℓ i → ±∞ particular possible values of ℓ for which the function vanishes at the boundary. Given the fact that only the s = 0 terms are important in the sums of (2.4) as z tends to infinity, we can distinguish the following two cases: (a) Firstly, ℓ 0, theninordertohave azero atinfinitywe must demandk ℓ Z , + ≥ − ∈ which is again what we got in in the previous investigation. (b) On the other hand, when ℓ < 0, we must necessarily enforce ℓ > 1. − Asaresult we canstatethat Qk( x)vanishes at theboundary either ifk > ℓ N ℓ i ±∞ ∈ or when ℓ ( 1,0). ∈ − 3 Orthogonality We start from equation (1.8) and consider the two following relations that hold identically for Qk( x) and Qm( x) ℓ i n i d dQk k2 (1+x2) ℓ ℓ(ℓ+1) Qk = 0 (3.1a) dx dx − − 1+x2 ℓ (cid:20) (cid:21) (cid:20) (cid:21) d dQm m2 (1+x2) n n(n+1) Qm = 0. (3.1b) dx dx − − 1+x2 n (cid:20) (cid:21) (cid:20) (cid:21) IfwemultiplythefirstandthesecondwithQm( x),Qk( x)respectively andsucceedingly n i ℓ i subtract the one from the other we arrive at k2 m2 d dQk dQm (ℓ n)(ℓ+n+1) − QkQm = (1+x2) Qm ℓ Qk n , − − 1+x2 ℓ n dx n dx − ℓ dx (cid:20) (cid:21) (cid:20) (cid:18) (cid:19)(cid:21) which, for k = m and ℓ = n becomes 6 +∞ 1+x2 dQk dQm +∞ (ℓ+n+1) Qk( x)Qm( x)dx= Qm ℓ Qk n . (3.2) ℓ i n i ℓ n n dx − ℓ dx Z−∞ (cid:20) − (cid:18) (cid:19)(cid:21)−∞ If we want an orthogonality condition to be satisfied when ℓ = n the boundary term that 6 appears on the right hand of (3.2) must be zero. Until now we have deduced the two subsequent possibilities from the previous section: 4 1. k Z, ℓ ( 1,0) ∈ ∈ − 2. k Z , ℓ N and k > ℓ. + ∈ ∈ For the first case let us check the behaviour of the product x2Qk dQmn for the leading terms ℓ dx as x , which can be deduced with the help of (2.4): → ±∞ dQm α α β α x2Qk n x2 1 + 2 1 + 2 ℓ dx ∼ x|ℓ| x1−|ℓ| x1+|n| x2−|n| (cid:18) (cid:19) (cid:16)γ γ (cid:17) γ γ 1 2 3 1 + + + , ∼ x|n|+|ℓ|−1 x|n|−|ℓ| x|ℓ|−|n| x1−|n|−|ℓ| where the α , β and γ ’s are constants. It is clear that the expression diverges at the i i i boundary for all n,ℓ ( 1,0). ∈ − On the contrary, when we turn to the last case and remember (2.2) together with properties (2.3) we see that Qk( x) decays as x−ℓ−1, thus ℓ i dQm 1 1 1 x2Qk n x2 = ℓ dx ∼ xℓ+1xn+2 xℓ+n+1 which vanishes on the boundary, since now ℓ,n 0. Hence we can see that, functions ≥ Qk( x), for k Z , ℓ N, with k > ℓ form an orthogonal set in the region x ℓ i ∈ + ∈ ∈ ( ,+ ). −∞ ∞ 4 Square integrability By having proven that +∞ Qk( x)Qk( x)= 0, ℓ = n (4.1) ℓ i n i 6 Z−∞ for the given range of values of k,ℓ and n, we need only investigate what happens when ℓ = n, case where equation (3.2) is not applicable. We work in a similar way to the derivation of the normalization of the Legendre polynomials of the first kind Pk(x) when ℓ x [ 1,1] [12]. At first we use definition (2.1a) to write ∈ − (1+x2)k dk dk Qk( x)Qk( x)= Q ( x) Q ( x). ℓ i ℓ i 2k dxk ℓ i dxk ℓ i i If we integrate the previous relation, we get +∞ 1 dk dk−1 +∞ Qk( x)Qk( x)dx= (1+x2)k Q ( x) Q ( x) ℓ i ℓ i 2k dxk ℓ i dxk−1 ℓ i Z−∞ i (cid:20) (cid:21)−∞ (4.2) 1 +∞ d dkQ ( x) dk−1Q ( x) (1+x2)k ℓ i ℓ i dx, − 2k dx dxk dxk−1 i Z−∞ (cid:18) (cid:19) where the surface term of the right hand side of (4.2) is zero, since the expression in the brackets decayes at infinity like x−2ℓ−1. Let us now see what happens with the second term. At first we need to calculate d dk dk+1 dk (1+x2)k Q ( x) = (1+x2)k Q ( x)+2kx(1+x2)k−1 Q ( x), (4.3) dx dxk ℓ i dxk+1 ℓ i dxk ℓ i (cid:18) (cid:19) 5 then we need to remember that the Legendre function of the second kind Q ( x)satisfies ℓ i the differential equation d2 d (1+x2) Q ( x)+2x Q ( x) ℓ(ℓ+1)Q ( x)= 0. (4.4) dx2 ℓ i dx ℓ i − ℓ i By differentiating (k 1) with respect to x and with the help of Leibnitz’s formula − dk k k! dk−m dm [A(x)B(x)] = A(x) B(x), (4.5) dxk m!(k m)!dxk−m dxm m=0 − X we are led to the following relation dk+1 dk dk−1 (1+x2) Q ( x)+2kx Q ( x)= (ℓ+k)(ℓ k+1)(1+x2)k−1 Q ( x). (4.6) dxk+1 ℓ i dxk ℓ i − dxk−1 ℓ i The left hand side of (4.6) is equal to the right of (4.3), hence we can deduce that +∞ (ℓ+k)(ℓ k +1) +∞ dk−1Q ( x)dk−1Q ( x) Qk( x)Qk( x)dx= − (1+x2)k−1 ℓ i ℓ i dx ℓ i ℓ i 2k dxk−1 dxk−1 Z−∞ −i Z−∞ +∞ =(ℓ+k)(ℓ k +1) Qk−1( x)Qk−1( x)dx. − ℓ i ℓ i Z−∞ By repeating the same procedure k ℓ+1 times we get − +∞ k +∞ Qk( x)Qk( x)dx= (ℓ+i)(ℓ i+1) Qℓ+1( x)Qℓ+1( x)dx. (4.7) ℓ i ℓ i − ℓ i ℓ i Z−∞ i=ℓ+2 Z−∞ Y At this point we need only prove that the integral on the right hand side is bounded. Let (2) us recall properties (2.3); for k = ℓ+1, polynomial A (z) is of zero-th order, hence (2.2) kℓ implies A (ℓ) Qℓ+1( x)= 0 , (4.8) ℓ i (1+x2)(ℓ+1)/2 where A (ℓ) is defined by (4.8) and its value with respect of ℓ given by (A.11). Addition- 0 ally, it is easy to verify that 1 1 3 dx = x F ( ,ℓ+1; ; x2)+const. (4.9) (1+x2)ℓ+1 2 1 2 2 − Z With the use of sin((b a)π) ( z)−a 1 ˜ ˜ − F (a,b;c;z) = − F (a,a c+1;a b+1; ) 2 1 2 1 π Γ(b)Γ(c a) − − z − (4.10) ( z)−b 1 ˜ − F (b,b c+1;b a+1; ) 2 1 −Γ(a)Γ(c b) − − z − which holds outside the unit circle z = 1 [5], definition (1.4) and the relation that con- nectstheregularized F˜ withtheord|in| aryGausshypergeometricfunction,i.e. F (a,b;c;z) = 2 1 2 1 Γ(z) F˜ (a,b;c;z), we deduce that 2 1 1 3 ( 1)ℓπ π1/2 ( x2)−(ℓ+1) 1 F ( ,ℓ+1; ; x2) = − − +O[ ]. (4.11) 2 1 2 2 − Γ(1 ℓ) 2Γ(ℓ+1) x − 2Γ(3 +ℓ) x2 2 − (cid:20) | | 2 (cid:21) 6 Given that the smallest value of ℓ is zero - so that the second term in the previous bracket can be neglected as x - from relations (4.8)-(4.11) we get → ±∞ +∞ ( 1)ℓπ3/2 +∞ Qℓ+1( x) 2dx = A (ℓ)2 x − ℓ i 0 2Γ(1 ℓ)Γ(ℓ+1) x Z−∞ (cid:20) 2 − | |(cid:21)−∞ (4.12) (cid:0) (cid:1) ( 1)ℓπ3/2A (ℓ)2 0 = − Γ(1 ℓ)Γ(ℓ+1) 2 − which leads (4.7) to be written as +∞ ( 1)ℓπ3/2A2 k Qk( x)Qk( x)dx= − 0 (ℓ+i)(ℓ i+1), (4.13) ℓ i ℓ i Γ(1 ℓ)ℓ! − Z−∞ 2 − i=ℓ+2 Y which as we see is a bounded expression. As a result we can state that, the Qk( x)’s for k Z , ℓ N with k > ℓ form an orthogonal set of square integrable functℓioins in + ∈ ∈ the region x ( ,+ ). What is more, by combining (4.1), (4.12) and (4.13) we have ∈ −∞ ∞ proven that (1.6) holds. A Calculation of A (ℓ) 0 Let us proceed with the calculation of A (ℓ) by using definitions (2.1). From the latter 0 we can derive the relation ( 1)ℓ+1 dℓ+1 1+z Qℓ+1(z) = − (1 z2)(ℓ+1)/2 P(z) ln , (A.1) ℓ 2 − dzℓ+1 ℓ 1 z (cid:20) (cid:18) − (cid:19)(cid:21) since the second term of (2.1b) is a polynomial of order ℓ 1 and its ℓ+1 order derivative − is zero. By application of Leibnitz’s formula (4.5) we can write dℓ+1 1+z ℓ+1 (ℓ+1)! dℓ+1−m dm 1+z P(z) ln = P (z) ln , (A.2) dzℓ+1 ℓ 1 z m!(ℓ+1 m)!dzℓ+1−m ℓ dzm 1 z (cid:20) (cid:18) − (cid:19)(cid:21) m=1 − (cid:18) − (cid:19) X where the summation starts from m = 1 and not m = 0 because P (z) is a polynomial of ℓ order ℓ and its ℓ+1-th derivative is zero. By mathematical induction it is easy to show that dm 1+z ( 1)m−1 1 ln = (m 1)! − + dzm 1 z − (1+z)m (1 z)m (cid:18) − (cid:19) (cid:20) − (cid:21) (A.3) m (m 1)! m = − ( 1)m−1+s +1 zs , (1 z2)m s − − s=0 (cid:20)(cid:18) (cid:19) (cid:21) X (cid:0) (cid:1) where for the last equality the binomial theorem n n (α+β)n = αn−sβs (A.4) s s=0 (cid:18) (cid:19) X 7 has been employed. Another useful relation that can be found in the bibliography [1] is the following: 1 r ( 1)p(2ℓ 2p)! P (z) = − − zℓ−2p (A.5) ℓ 2ℓ p!(ℓ p)!(ℓ 2p)! p=0 − − X where r is the greatest integer that is smaller than ℓ, i.e. it is r = ℓ if ℓ is even, else it is 2 2 r = ℓ−1. Since, 2 dm m (n+1 m) zn = zn−m (n j +1) = − (n+2 m) zn−m (A.6) dzm − n+1 − m j=1 Y where in the last equality we generalize the result with the help of the Pochhammer symbol (x) = Γ(x+n)/Γ(x). As a result, we can now write n dℓ+1−m 1 r ( 1)p(2ℓ 2p)! (m 2p) P (z) = − − − (m+1 2p) zm−2p−1. (A.7) dzℓ+1−m ℓ 2ℓ p!(ℓ p)!(ℓ 2p)!ℓ 2p+1 − (ℓ−m+1) p=0 − − − X Due to the properties of Qk(z) we know that ℓ A (ℓ) = (1 z2)(ℓ+1)/2Qℓ+1(z), ℓ N (A.8) 0 − ℓ ∈ is constant in respect to z [11]. By substitution of (A.1)-(A.3) and (A.7) in (A.8) we see that we have successive products of polynomials in z that result into a constant value, hence for the calculation of this value we need only consider the constant terms of each (e.g. the s = 0 of (A.3) and p = m−1 of (A.7)). A straightforward calculation yields 2 ( 1)ℓ+1 ℓ+1 ( 1)(m−1)/2(ℓ+1)!(2ℓ m+1)!(m 1)! A (ℓ) = − − − − ( 1)m−1 +1 , (A.9) 0 2ℓ+1 m!(ℓ+1 m)!Γ(1+m)Γ(ℓ+ 3−m) − " # m=1 − 2 2 X (cid:0) (cid:1) where we have also used the equality (2) (1+ℓ−m) = (ℓ+1 m)!. (A.10) ℓ+2 m − − Because of the (( 1)m−1 +1) multiplication term, it is clear that only the odd values of − m contribute in the summation in (A.9). Hence, the Γ functions appearing in it can be substituted by factorials due to the fact that no half integer values are produced in the arguments. Furthermore, by setting m = 2n+1 we can significantly simplify (A.9) and obtain ( 1)ℓ+1(ℓ+1)! r ( 1)n(2ℓ 2n)!(2n)! A (ℓ) = − − − , (A.11) 0 2ℓ (2n+1)!(ℓ 2n)!(ℓ n)!n! n=0 − − X where now r = ℓ/2 if ℓ is even, else r = ℓ−1. 2 Acknowledgements TheauthoracknowledgesfinancialsupportbyFONDECYTpostdoctoralgrantNo. 3150016. 8 References [1] W. A. Heiskanen and H. Moritz, Physical Geodesy, Springer, Wien, New York (2005) [2] A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, Princeton, New Jersey (1957) [3] W. Magnus and F. Oberhettinger, Formulas and Theorems for the Functions of Mathematical Physics, Chealsea Publishing Company, New York (1954) [4] M. Abramowitz and I. A. Stegun Handbook of Mathematical Functions, National Bureau of Standards Applied Mathematics Series, U.S. Government Printing Office, Washington, D.C. (1964) [5] F. W. J. Olver, Asymptotics and Special Functions, A.K.Peters Wellesley, Mas- sachusets (1997) [6] N. Virchenko and I. Fedotova, Generalized Associated Legendre Functions and their Applications, World Scientific, Singapore, New Jersey, London, Hong Kong (2001) [7] R. G. Van Nostrand, J. Math. and Phys., 33 (1954), 276-282 [8] F. Go¨tze, Indag. Math., 27 (1965), 396-404 [9] S. Yu. Kalmykov and G. Shvets, Phys. Plasmas 11 (2004), 46864694 [10] S. Bielski, Integral Transforms and Special Functions, 24 (2012), 331-337 [11] D.Suschowk, Mathematical Tablesand Other Aids to Computation,13No.68(1959), 303-305 [12] K. T. Tang, Mathematical Methods for Engineers and Scientists Vol. 3, Springer- Verlag, Berlin, Heidelberg (2007) 9

See more

The list of books you might like

Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.