19
MATH F211 : MATHEMATICS–III
BITS-PILANI HYDERABAD CAMPUS
Presented by
Dr. M.S. Radhakrishnan
Email: msr@bits-hyderabad.ac.in
Lecture 19
Legendre Polynomials
Ch. 8 Sections 44-45
George F. Simmons,
Differential Equations with
Applications and Historical notes,
Tata McGraw-Hill, 2nd Ed, 2003
(Twelfth reprint, 2008)
Some Special Functions of
Mathematical Physics –
Legendre Polynomials
There are many special functions that are
being used by Physicists in the solution of
physical problems, like Legendre
Polynomials, Hermite Polynomials, Bessel’s
Functions, etc. to name a few. In this lecture
we define Legendre Polynomials and study
their properties. Much of the material
covered here is taken from the book: An
Introduction to Linear Analysis - By Kreider,
Kuller, Ostberg & Perkins (Addison Wesley)
Chapter 11.
We define the Legendre polynomials,
determine the differential equations satisfied
by them, show the orthogonality of the
Legendre polynomials, and find the
recurrence relation satisfied by them and the
zeros of these polynomials.
Definition
The Legendre polynomials, Pn(x), n=0,1,2,…
are defined by the Rodrigues’ formula
n
1 d
2
n
Pn ( x)  n
( x  1)
n
2 n! dx
etc.
Thus
P0 ( x)  1
1d 2
P1 ( x) 
( x  1)
2 dx
2
x
2
1 d 2 2 1d 4
2
P2 ( x)  2
( x  1) 
( x  2 x  1)
2
2
2 2! dx
8 dx
3 2 1
 x  etc.
2
2
It is clear that
 Pn(x) is a polynomial of degree ‘n’.
 If n is even, Pn (x) contains only even
powers of x and if n is odd, Pn(x)
contains only odd powers of x.
 The coefficient of xn in Pn(x)
1 2n
1 (2n)!
 n
Pn  n
2
2 n!
2 (n !)
 The coefficient of xn-2 in Pn(x)
1
2 n2
 n  n  Pn
2 n!
1 (2n  2)!
 n
2 (n  1)!(n  2)!
Leibniz Rule for the nth derivative of
product of two functions
Let u, v be two functions of x. Then
(u v)( n )
 n  ( n1) (1)  n  ( n2) (2)
( n)
 u v   u v   u
v  ...  u v
1
 2
(n)
where superscripts denote the order of the
derivative.
The differential equation satisfied by
Pn(x):
Let
w  ( x  1)
Hence
2
n 1

w  2nx( x  1)
2
n
w  ( x 2  1) n
Multiplying both sides by (x2-1), we get
( x  1) w  2nxw
2
Differentiating both sides (n+1) times w.r.t x
(using Leibniz formula), we get
( n2)
( x 1)w
2

( n1)
( n1)
C1 2x w
( n 1)
 2n[ xw
i.e.
( x  1) w
2
( n 2)


( n)
C2 2 w
( n 1)
( n 1)
 2 xw
( n1)
(n)
C1w ]
 n(n  1) w  0
(n)
Dividing throughout by 2nn!, we get
( x  1) y  2 xy  n(n  1) y  0 where
2
n
1
1 d
(n)
2
n
y n w  n
(
x

1)
2 n!
2 n ! dx n
Thus y = Pn(x) satisfies the so-called
Legendre’s d.e. of “order” n , viz.
(1  x ) y  2 xy  n(n  1) y  0
2
Theorem: Pn(1) = 1, Pn(-1) = (-1)n
Proof:
Let w = (x2-1)n = (x+1)n(x-1)n
Differentiating both sides n times using
Leibniz rule, we get
w( n )
n 1
d
n
n
n d
 n!( x  1)  C1 n1 ( x  1)
( x  1) n
dx
dx
n2
2
d
n d
 C2 n2 ( x  1) 2 ( x  1)n  ...
dx
dx
n
n 1
d
d
n
n
n
n
 Cn1 ( x  1) n1 ( x  1)  ( x  1) n!
dx
dx
We note that in RHS, except for the first and
last term, every other term contains a factor
of (x+1) and a factor of (x-1).
Noting that w(n)=2nn!Pn(x), putting x = 1, we get
2 n! Pn (1)  2 n! Pn (1)  1
n
n
2 n! Pn (1)  n!(2)  n!(1) 2  Pn (1)  (1)
n
n
n
n
We also note for future reference, for k < n,
k
d w
 k   0 at x  1
 dx 
(as every term in w(k), ( k < n ), contains a
factor of (x+1) and a factor of (x-1)).
n
Orthogonality of the Legendre Polynomials:
We show 1
mn
0
1 Pn ( x)Pm ( x)dx  2 /(2n  1) m  n
Proof: For any function f (x), consider
1
n
d
2
n
I   f ( x) n ( x  1) dx
dx
1
On integrating by parts, we get
n 1
1
1
n 1
d
d
2
n
2
n

I  f ( x) n1 ( x  1) ]   f ( x) n1 ( x  1) dx
dx
dx
1 1
Now by the remark made earlier
d n 1 2
n
( x  1)  0
n 1
dx
n 1
1
Hence
at x = 1
d
2
n

I    f ( x) n1 ( x  1) dx
dx
1
n2
d
2
2
n
 (1)  f ( x) n2 ( x  1) dx
dx
1
1
1
 (1)n  f ( n ) ( x)( x 2  1) n dx
1
…
We note that if f (x) is a polynomial of degree
< n, f(n)(x) = 0.
Hence if m < n,
1
 P ( x)P ( x)dx 
m
1
n
1
1
dn 2
n
Pm ( x) n ( x  1) dx
n

2 n! 1
dx
1
1
n
(n)
2
n
 n (1)  Pm ( x)( x  1) dx
2 n!
1
= 0, as m < n and so Pm(n)(x)=0
1
1
P
n
2
 P ( x)P ( x)dx
( x)dx 
n
n
1
1
1
1
(n)
2
n
 (1) n  Pn ( x)( x  1) dx
2 n! 1
n
Since Pn(x) is a polynomial of degree n,
Pn(n) (x) = n![leading coefficient of Pn(x) ]
2n!
(2n)!
 n! n
 n
2
2 (n!)
2 n!
Thus
1
(2n)!
 Pn ( x)dx 
2
 2 n!
n
1
2
1
 (1  x ) dx
2 n
1
{ as (-1)n(x2-1)n = (1-x2)n }
1
2(2n)!
2 n
 2n
(1

x
) dx
2 
2 (n !) 0
1
Now look at
 (1  x
2 n
) dx
Put
x  sin 
0
We get
1
 2
0
0
2 n
(1

x
) dx 


cos 2 n 1  d
2n
2n  2
2


 .... 
2n  1 2n  1
3
2n  (2n)  (2n  2)  (2n  2)  ...  2  2

(2n  1)(2n!)
2n
2
2 (n!)

(2n  1)(2n!)
Thus
1
2
1 Pn ( x)dx  2n  1
2
In the next lecture, we shall find the
recurrence relation satisfied by the
Legendre Polynomials and the zeros of
these polynomials.
***