Mathematical Tools for Physics - Department of Physics - University

4—Differential Equations 89

and the general solution is a sum of these.
This procedure gives both solutions to the differential equation, one with even powers and one
with odd powers. Both are infinite series and are called Legendre Functions. An important point about

both of them is that they blow up asx→ ± 1. This fact shouldn’t be too surprising, because the

differential equation (4.55) has a singular point there.

y′′−

2 x

(1 +x)(1−x)

y′+

C

(1 +x)(1−x)

y= 0 (4.59)

It’s a regular singular point, but it is still singular. A detailed calculation in the next section shows that

these solutions behave asln(1−x)nearx= 1.

There is an exception! If the constantC is for exampleC= 6, then withs= 0the equations

(4.57) are

a 2 =a 0

− 6

2

, a 4 =a 2

6 − 6

12

= 0, a 6 =a 8 =...= 0

The infinite series terminates in a polynomial

a 0 +a 2 x^2 =a 0 [1− 3 x^2 ]

This (after a conventional rearrangement) is a Legendre Polynomial,

P 2 (x) =

3

2

x^2 −

1

2

The numerator in Eq. (4.56) foran+2is[(n+s)(n+s+ 1)−C]. If this happen to equal zero

for some value ofn=N, thenaN+2= 0and so then all the rest ofaN+4...are zero too. The series

is a polynomial. This will happen only for special values ofC, such as the valueC= 6above. The

values ofCthat have this special property are

C=(+ 1), for `= 0, 1 , 2 , ... (4.60)

This may be easier to see in the explicit representation, Eq. (4.58). When a numerator equals zero,

all the rest that follow are zero too. WhenC=(+ 1)for even`, the first series terminates in a

polynomial. Similarly for odd`the second series is a polynomial. These are the Legendre polynomials,

denotedP`(x), and the conventional normalization is to require that their value atx= 1is one.

P 0 (x) = 1 P 1 (x) =x P 2 (x) =^32 x^2 −^12

P 3 (x) =^52 x^3 −^32 x P 4 (x) =^358 x^4 −^308 x^2 +^38

(4.61)

The special case for which the series terminates in a polynomial is by far the most commonly used
solution to Legendre’s equation. You seldom encounter the general solutions as in Eq. (4.58).

A few properties of theP`are

(a)

∫ 1

− 1

dxPn(x)Pm(x) =

2

2 n+ 1

δnm where δnm=

{

1 ifn=m

0 ifn 6 =m

(b) (n+ 1)Pn+1(x) = (2n+ 1)xPn(x)−nPn− 1 (x)

(c) Pn(x) =

(−1)n

2 nn!

dn

dxn

(1−x^2 )n

(d) Pn(1) = 1 Pn(−x) = (−1)nPn(x)

(e)

(

1 − 2 tx+t^2

)− 1 / 2

=

∑∞

n=0

tnPn(x)

(4.62)