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Introduction to Acoustics

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equation, the two differential equations being (with η<br />

replacing kr)<br />

�<br />

1 d<br />

sin θ<br />

sin θ dθ<br />

dPℓ<br />

�<br />

+ λℓ Pℓ = 0 , (3.361)<br />

dθ<br />

� �<br />

d 2 dÊℓ<br />

η − λℓÊℓ + η<br />

dη dη<br />

2 Êℓ = 0 . (3.362)<br />

Here λℓ is a constant, termed the separation constant.<br />

Equivalently, with Pℓ regarded as a function of ξ = cos θ,<br />

the first of these two differential equations can be written<br />

�<br />

d<br />

�<br />

1 − ξ<br />

dξ<br />

2� �<br />

dPℓ<br />

+ λℓ Pℓ = 0 . (3.363)<br />

dξ<br />

Legendre Polynomials<br />

Usually, one desires solutions that are finite at both θ = 0<br />

and θ = π,oratξ = 1andξ =−1, but the solutions for<br />

the θ-dependent fac<strong>to</strong>r are usually singular at one of<br />

the other of these two end points. However, for special<br />

values (eigenvalues) of the separation constant λℓ,<br />

there exist particular solutions (eigenfunctions) that are<br />

finite at both points. To determine these functions, one<br />

postulates a series solution of the form<br />

∞�<br />

Pℓ(ξ) = aℓ,nξ n , (3.364)<br />

n=0<br />

and derives the recursion relation<br />

n(n − 1)aℓ,n = [(n − 1)(n − 2) − λℓ] aℓ,n−2 .<br />

(3.365)<br />

The series diverges as ξ →±1 unless it only has a finite<br />

number of terms, and such may be so if for some n the<br />

quantity in brackets on the right side is zero. The general<br />

choice of the separation constant that allows this is<br />

λℓ = ℓ(ℓ + 1) , (3.366)<br />

where ℓ is an integer, so that the recursion relation<br />

becomes<br />

n(n − 1)aℓ,n = [(n − ℓ − 2)(n + ℓ − 1)] aℓ,n−2 .<br />

(3.367)<br />

However, aℓ,0 and aℓ,1 can be chosen independently and<br />

the recursion relation can only terminate one of the two<br />

possible infinite series. Consequently, one must choose<br />

aℓ,1 = 0 if ℓ even ;<br />

aℓ,0 = 0 if ℓ odd . (3.368)<br />

If ℓ is even the terms correspond <strong>to</strong> n = 0, n = 2, n = 4,<br />

up <strong>to</strong> n = ℓ, while if ℓ is odd the terms correspond <strong>to</strong><br />

Basic Linear <strong>Acoustics</strong> 3.11 Spherical Waves 69<br />

n = 1, n = 3, up <strong>to</strong> n = ℓ. The cus<strong>to</strong>mary normalization<br />

is that Pℓ(1) = 1, and the polynomials that are derived are<br />

termed the Legendre polynomials. A general expression<br />

that results from examination of the recursion relation<br />

for the coefficients is<br />

�<br />

Pℓ(ξ) = aℓ,ℓ ξ ℓ ℓ(ℓ − 1)<br />

−<br />

2(2ℓ − 1) ξℓ−2<br />

+ ℓ(ℓ − 2)(ℓ − 1)(ℓ − 3)<br />

(2)(4)(2ℓ − 1)(2ℓ − 3) ξℓ−4 �<br />

+ ... ,<br />

(3.369)<br />

where the last term has ξ raised <strong>to</strong> either the power of 1 or<br />

0, depending on whether ℓ is odd or even. Equivalently,<br />

if one sets,<br />

aℓ,ℓ = Kℓ<br />

(2ℓ)!<br />

2ℓ , (3.370)<br />

(ℓ!) 2<br />

where Kℓ is <strong>to</strong> be selected, the series has the relatively<br />

simple form<br />

M(ℓ) �<br />

Pℓ(ξ) = Kℓ (−1) m (2ℓ − 2m)!<br />

2ℓm!(ℓ − m)!(ℓ − 2m)! ξℓ−2m<br />

m=0<br />

M(ℓ) �<br />

= Kℓ bℓ,mξ ℓ−2m . (3.371)<br />

m=0<br />

Here M(ℓ) = ℓ/2 ifℓ is even, and M(ℓ) = (ℓ − 1)/2 if<br />

ℓ is odd, so M(0) = 0, M(1) = 0, M(2) = 1, M(3) = 1,<br />

M(4) = 2, etc.<br />

The coefficients bℓ,m as defined here satisfy the<br />

relation<br />

(ℓ + 1)bℓ+1,m = (2ℓ + 1)bℓ,m − ℓbℓ−1,m−1 ,<br />

(3.372)<br />

as can be verified by algebraic manipulation. A consequence<br />

of this relation is<br />

(ℓ + 1) Pℓ+1(ξ)<br />

Kℓ+1<br />

= (2ℓ + 1)ξ Pℓ(ξ)<br />

Kℓ<br />

− ℓ Pℓ−1(ξ)<br />

Kℓ−1<br />

(3.373)<br />

when ℓ ≥ 1.<br />

The cus<strong>to</strong>mary normalization is <strong>to</strong> take Pℓ(1) = 1.<br />

The series for ℓ = 0andℓ = 1 are each of only one<br />

term, and the normalization requirement leads <strong>to</strong> K0 = 1<br />

and K1 = 1. The relation (3.373) indicates that the normalization<br />

requirement will result, via induction, for all<br />

successive ℓ if one takes Kℓ = 1forallℓ. With this definition,<br />

the relation (3.373) yields the recursion relation<br />

among polynomials of different orders<br />

(ℓ + 1)Pℓ+1(ξ) = (2ℓ + 1)ξ Pℓ(ξ) − ℓPℓ−1(ξ) .<br />

(3.374)<br />

Part A 3.11

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