GENERATING FUNCTION FOR LEGENDRE POLYNOMIALS

5. GENERATING FUNCTION FOR LEGENDRE POLYNOMIALS

The expression

Φ(x, h) = (1 − 2xh + h 2 ) −1/2 ,

|h| < 1,

is called the generating function for Legendre polynomials. We shall show that

Φ(x, h) = P (x) + hP (x) + h 2 P

h 0 l 1 2 (x) + · · · = P l (x),

l=0

where the functions P l (x) are the Legendre polynomials. (For discussion of conver- gence of the series, see Chapter 14, Problem 2.43.) Let us first verify a few terms

of (5.2). For simplicity put 2xh − h 2 = y into (5.1), expand (1 − y) −1/2 in powers of y, then substitute back y = 2xh − h 2 and collect powers of h to get

1 · 3 Φ = (1 − y) −1/2 =1+ 1 2 y+ 2 2 y 2 +···

2! =1+ 1 2 2 2 (2xh − h )+ 3 8 2 (2xh − h ) +··· (5.3)

1 h 2 + 3 (4x 2 h 2 = 1 + xh − 3

8 − 4xh +h 2 4 )+···

= 1 + xh + h 2 ( 3 x 2 2 1 − 2 )+··· =P 0 (x) + hP 1 (x) + h 2 P 2 (x) + · · · .

570 Series Solutions of Differential Equations Chapter 12

This is not a proof that the functions called P l (x) in (5.2) are really Legendre polynomials, but merely a verification of the first few terms. To prove in general that the polynomials called P l (x) in (5.2) are Legendre polynomials we must show that they satisfy Legendre’s equation and that they have the property P l (1) = 1. The latter is easy to prove; putting x = 1 in (5.1) and (5.2), we get

=1+h+h Φ(1, h) = (1 − 2h + h 2 +···

≡P 0 (1) + P 1 (1)h + P 2 (1)h 2 +···.

Since this is an identity in h, the functions P l (x) in (5.2) have the property P l (1) = 1. To show that they satisfy Legendre’s equation, we shall use the following identity which can be verified from (5.1) by straightforward differentiation and some algebra (Problem 2):

Substituting the series (5.2) for Φ into (5.5), we get

This is an identity in h, so the coefficient of each power of h must be zero. Setting the coefficient of h l equal to zero, we get

(5.7) (1 − x 2 )P l ′′ (x) − 2xP l ′ (x) + l(l + 1)P l (x) = 0. This is Legendre’s equation, so we have proved that the functions P l (x) in (5.2)

satisfy it as claimed. Recursion Relations The generating function is useful in deriving the recursion

relations (also called recurrence relations) for Legendre polynomials. These recur- sion relations are identities in x and are used (as trigonometric identities are) to simplify work and to help in proofs and derivations. Some examples of recursion relations are:

(a) lP l (x) = (2l − 1)xP l−1 (x) − (l − 1)P l−2 (x), (b) xP l ′ (x) − P ′ l−1 (x) = lP l (x), (c) P ′ l (x) − xP ′ l−1 (x) = lP l−1 (x),

(d) (1 − x 2 )P l ′ (x) = lP l−1 (x) − lxP l (x),

(e) (2l + 1)P l (x) = P l+1 ′ (x) − P ′ l−1 (x), (f)

(1 − x 2 )P ′ l−1 (x) = lxP l−1 (x) − lP l (x). We shall now derive (5.8a); the problems outline derivations of the other equations.

From (5.1) we get ∂Φ

2 (1 − 2xh + h ) −3/2 (−2x + 2h); (5.9)

= (x − h)Φ.

∂h

Section 5 Generating Function for Legendre Polynomials 571

Substituting the series (5.2) and its derivative with respect to h into (5.9), we get

h P l (x).

l=1

l=0

This is an identity in h so we equate coefficients of h l−1 . Carefully adjusting indices so that we select the term in h l−1 each time, we find

(5.10) lP l (x) − 2x(l − 1)P l−1 (x) + (l − 2)P l−2 (x) = xP l−1 (x) − P l−2 (x) which simplifies to (5.8a). The recursion relation (5.8a) gives the simplest way

of finding any Legendre polynomial when we know the Legendre polynomials for smaller l (Problem 3).

Expansion of a Potential The generating function is useful in problems involv- ing the potential associated with any inverse square force. Recall that the gravi- tational force between two point masses separated by a distance d is proportional

to 1/d 2 and the associated potential energy is proportional to 1/d. Similarly, the electrostatic force between two electric charges a distance d apart is proportional to 1/d 2 and the associated electrostatic potential energy is proportional to 1/d.

Example 1. In either case we can write the potential as

V=

where K is an appropriate constant. In Figure 5.1, let the two masses (or charges)

be at the heads of vectors r and R. Then, by the law of cosines, the distance between them is

Figure 5.1 and the gravitational or electric potential is

K 2r r 2 −1/2

For |r| < |R|, we make the change of variables

(Note: x is not a coordinate but just a new variable standing for cos θ.) Then in terms of the generating function Φ in (5.1) we have

(1 − 2hx + h 2 ) −1/2 =

572 Series Solutions of Differential Equations Chapter 12

Using (5.2), we can write the potential V as an infinite series

or in terms of r and θ [using (5.14)]

r l P l (cos θ) (5.17)

K ∞ r l P (cos θ)

In many applications the distance |R| is much larger than |r|. Then the terms of the series (5.17) decrease rapidly in magnitude because of the factor (r/R) l , and the potential can be approximated by using only a few terms in the series.

We can make (5.17) more general and useful by considering the following prob- lem. (We shall discuss the electrical case for definiteness—the gravitational case could be discussed in parallel fashion.)

Example 2. Suppose there are a large number of charges q i at points r i . The electrostatic potential V i at the point R due to the charge q i at r i means the electrostatic potential energy of a pair of charges, namely, a unit charge at R and the charge q i at r i ; this is given by (5.11) and (5.12), or by (5.17), with r = r i ,θ=θ i , and K=q i ·1·K ′ , where K ′ is a numerical constant depending on the choice of units:

r i P l (cos θ i )

The total potential V at R due to all the charges q i is then a sum over i of all the series (5.18), namely

i q i r i P l (cos θ i (5.19) ) V= V

r i P l (cos θ i )

i =K

=K

R Example 3. If, instead of a set of discrete charges, we have a continuous charge distribu-

tion, then the sum over i becomes an integral, namely (5.20)

r l P l (cos θ)ρ dτ, where ρ is the charge density, and the integral is over the space occupied by the

r l P l (cos θ) dq

or

charge distribution. Then (5.19) becomes

V=K ′

r l+1 l P l (cos θ)ρ dτ.

The terms of the series (5.21) can be interpreted physically. The l = 0 term is

ρ dτ =

· (total charge).

Thus if R is large enough compared to all the r i or all the values of r at points of the charge distribution, we can approximate the potential of the distribution as

Section 5 Generating Function for Legendre Polynomials 573

that of a single charge at the origin of magnitude equal to the total charge of the distribution. The l = 1 term of the series (5.21) is

r cos θ ρ dτ.

To interpret this recall that the electric dipole moment of a pair of charges +q and −q a distance d apart (as in Figure 5.2) is defined as the vector qd, where d is the vector from −q to +q.

Since the vector qd is equal to q(r 1 −r 2 ) = qr 1 − qr 2 , we often call qr 1 and −qr 2 the dipole moments of +q and −q about O;

Figure 5.2 then the total dipole moment due to the two charges is just the

sum of the two moments. Suppose we calculate the dipole moment about O of all the charges q i ; this is the vector sum i q i r i cos θ i , since θ i is the angle between R and r i . In the case of a continuous charge distribution this sum becomes

r cos θ ρ dτ.

Thus we see from (5.23) and (5.24) that the second term of the series (5.21) is 1/R 2 times the component in the R direction of the dipole moment of the charge distribution. If you consider the fact that the first term of (5.21) involves the total charge (a scalar, that is, a tensor of rank zero) and the second term involves the dipole moment (a vector, that is, a tensor of rank one), it may not surprise you to learn that the third term involves a 2 nd -rank tensor known as the quadrupole moment of the charge distribution, the fourth term involves a 3 rd -rank tensor known as the octopole moment, etc. (See Problem 15 for more detail.)

Example 4. Given a charge or mass distribution, the moments of various ranks and the terms in (5.21) can be computed. The opposite process is often of great interest in applied problems. Consider a satellite circling the earth; it is moving in the gravitational field of the earth’s mass. If the mass distribution of the earth were spherically symmetric, then only the first term would appear in the series for the gravitational potential [this series would be (5.21) with ρ a mass density instead of a charge density]. But since the earth is not a perfect sphere (equatorial bulge, etc.), other terms are present in (5.21) and the corresponding forces affect the motion of satellites. From accurate measurements of the satellite orbits, it is now possible to calculate many terms of the series (5.21). Similarly, in the electrical case, experimental measurements give us information about the distribution of electric charge inside atoms and nuclei; our discussion here and equation (5.21) provide the basis for the interpretation of such measurements, and the terminology used in discussing them.

PROBLEMS, SECTION 5

1. Find P 3 (x) by getting one more term in the generating function expansion (5.3). 2. Verify (5.5) using (5.1). 3. Use the recursion relation (5.8a) and the values of P 0 (x) and P 1 (x) to find P 2 (x),

P 3 (x), P 4 (x), P 5 (x), and P 6 (x). [After you have found P 3 (x), use it to find P 4 (x), and so on for the higher order polynomials.]

574 Series Solutions of Differential Equations Chapter 12

4. Show from (5.1) that

Substitute the series (5.2) for Φ, and so prove the recursion relation (5.8b). 5. Differentiate the recursion relation (5.8a) and use the recursion relation (5.8b) with

l replaced by l − 1 to prove the recursion relation (5.8c). 6. From (5.8b) and (5.8c), obtain (5.8d) and (5.8f). Then differentiate (5.8d) with

respect to x and eliminate P ′ l−1 (x) using (5.8b). Your result should be the Legendre equation. The derivation of Problems 4 to 6 constitutes an alternative proof [to that of equations (5.5) to (5.7)] that the functions P l (x) in (5.2) are Legendre polynomials.

7. Write (5.8c) with l replaced by l + 1 and use it to eliminate the xP l ′ (x) term in (5.8b). You should get (5.8e).

Express each of the following polynomials as linear combinations of Legendre polynomials. Hint: Start with the highest power of x and work down in finding the correct combination.

10. x 4 11. x−x 3

8. 5 − 2x

9. 3x 2 +x−1

12. 7x 4 − 3x + 1 5 13. x 14. Show that any polynomial of degree n can be written as a linear combination of

Legendre polynomials with l ≤ n. 15. Expand the potential V = K/d in (5.11) in the following way in order to see how

the terms depend on the tensors mentioned above. In Figure 5.1 let R have the coordinates X, Y , Z and r have coordinates x, y, z. [Note: The coordinate x here is not the x in (5.14).] Then

V= K = K[(X − x) 2 + (Y − y) 2 + (Z − z) 2 ] −1/2 .

Consider X, Y , Z as constants and expand V (x, y, z) in a three variable power series about the origin. (See Chapter 4, Section 2, for discussion of two-variable power series and generalize the method.) You should find

and similar terms in y, z, y 2 , xz, and so on. Now letting r = r i and K = K ′ q i for a charge distribution as in (5.18), and summing (or integrating) over the charge distribution, show that: the first term is just (K ′ /R) · total charge; the next group of terms (in x, y, z) involve the three components of the electric dipole moment; the

sum of these terms is (K ′ /R 2 )·component of the dipole moment in the R direction; the next group (quadratic terms) involve six quantities of the form

ZZZ

and similar y, z integrals, ZZZ

x 2 ρ dτ

and similar xz, yz integrals. If we split the 2xy term into xy and yx (and similarly for the 2xz and 2yz terms), we

2xyρ dτ

have the nine components of a 2 nd -rank tensor called the quadrupole moment. Use the “direct product” method of Chapter 10, Section 2 to show that it is a 2 nd -rank

Section 6 Complete Sets of Orthogonal Functions 575

tensor. (Remember from Chapter 10 that, by definition, x, y, z are the components of a vector, that is, a 1 st -rank tensor.) Just as two charges +q and −q form an + s s electric dipole, so four charges like this - s s + - form an electric quadrupole and the quadratic terms in the V series give the potential of such a charge configuration. Again using Chapter 10, Section 2, show that the third-order terms in x, y, z form

a3 rd -rank tensor; this is known as the octopole moment. It can be represented physically by two quadrupoles side by side just as the quadrupole above was formed by two dipoles side by side.