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

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by  N. Dimakis

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
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