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5.2 Legendre’s Differential Equation
Legendre’s equation arises when solving partial differential equations involving the Laplacian in
spherical coordinate. The ODE in the r direction of spherical coordinate usually takes the form
dy
d2y
(1 - x ) 2 - 2x
+ y = 0,
dx
dx
2
-1<x<1
(5.2-1)
 2x

, q(x) =
, and r(x) = 0 are analytic at x = 0, the equation can be
2
1 x
1  x2
represented by a power series solution of the form
Since p(x) =

y=
a
m 0
m
xm
(5.2-2)
Differentiating the series solution (5.2-2) yields

y’ =
 mam x m1 , y” =
m 1

 m(m  1)a
m
x m 2
m 2
Substituting the function and its derivatives into Eq. (5.2-1) yields

(1 - x2)  m( m  1)a m x m 2 - 2x
m 2

 m(m  1)am x m2 -
m 2


m 1
m 0
 mam x m1 +   a m x m = 0



m 2
m 1
m 0
 m(m  1)am x m - 2  mam x m +   a m x m = 0
(5.2-3)
The terms with xm-2 can be changed to xm by replacing m with m + 2

 m(m  1)am x m2 =
m 2

 (m  2)( m  1)a
m2
xm
m  22
Equation (5.2-3) becomes

 (m  2)( m  1)am2 x m -
m 0

 (m  2)( m  1)a
m2

 m(m  1)am x m - 2x
m 2


m 1
m 0
 mam x m1 +   a m x m = 0
 [m( m  1)  2m   ]a m x m + 2a2 + a0 + (6a3 - 2a1 + a1)x = 0
m 2
a2 = 

a0 ,
2
a3 =
110
2
a1
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am+2 =
m( m  1)  
am
( m  2)( m  1)
( n  m)( n  m  1)
m(m  1)  n 2  n
am = am
( m  2)( m  1)
(m  2)( m  1)
For  = n ( n  1)  am+2 =
( n  1)( n  2)
a1
3!
a2 = 
n( n  1)
a0
2!
a4 = 
( n  2)( n  3)
( n  2)n ( n  1)( n  3)
a2 =
a0
3 4
4!
a5 = 
( n  3)( n  4)
( n  3)( n  1)( n  2)( n  4)
a3 =
a1
45
5!
a3 = 
The general solution is then
y (x ) = a0y1 (x ) + a1y2 (x )
where
y1 (x ) = 1 
n( n  1) 2 ( n  2)n ( n  1)( n  3) 2
x +
x  ...
2!
4!
y2 (x ) = x 
( n  1)( n  2) 3 ( n  3)( n  1)( n  2)( n  4) 5
x +
x + ...
3!
5!
If n is an even integer then (when m = n) an+2 = -
( n  n )( n  n  1)
an = 0 = an+4 = an+6 = ...
( n  2)( n  1)
2( 3) 2
x = 1  3x2
2!
If n = 2,
y1 (x ) = 1 
If n = 4,
y1 (x ) = 1  10x2 +
35 4
x
3
If n is an even integer then y1 (x ) is a polynomial of degree n, and y2 (x ) has the form of an
infinite series. Similarly, if n is an odd integer then y2 (x ) is a polynomial of degree n, and
y1 (x ) is an infinite series.
If n = 1,
y2 (x ) = x
If n = 3,
y2 (x ) = x 
5 3
x
3
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If n = 5,
y2 (x ) = x 
14 3 21 5
x +
x
3
5
For the polynomial solutions, the non-vanishing coefficients can be expressed in terms of the
coefficients an of the highest power of x of the polynomial. We need a backward recurrence
relation that takes us from am to am-2. From the recurrence relation
am+2 = -
( n  m)( n  m  1)
am
( m  2)( m  1)
Replacing m by m – 2 yields
am = -
( n  m  2)( n  m  1)
am-2
m( m  1)
The backward recurrence relation becomes
am-2 = -
m( m  1)
am
( n  m  2)( n  m  1)
an-2 = -
n( n  1)
an
2( 2n  1)
For m = n,
( 2n )!
so that all the
2 n ( n! ) 2
polynomials will have the value unity at x = 1. It we do that, the polynomials are called Legendre
polynomials of degree n: Pn(x)
The coefficient an is still arbitrary. It is standard to choose an =
Pn(x) =
where
1
2n
M
 (1)
m 0
m
(2n  2m)!
xn-2m
m! (n  m)! (n  2m)!
M = n/2 if n is even, and M = (n – 1)/2 if n is odd.
The first few Legendre polynomials are
P0(x) = 1
P1(x) = x
P2(x) =
1
(3x2  1)
2
P3(x) =
1
(5x3  3x)
2
P4(x) =
1
(35x4  30x2 + 3)
8
P5(x) =
1
(63x5  70x3 + 15x)
8
In summary: For n = 0, 1, 2, ... the Legrendre equation of order n
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dy
d2y
(1 - x ) 2 - 2x
+ n(n + 1)y = 0,
dx
dx
2
-1<x<1
has two linearly independent solutions y1 (x ) and y2 (x )
y1 (x ) = 1 
n( n  1) 2 ( n  2)n ( n  1)( n  3) 2
x +
x  ...
2!
4!
y2 (x ) = x 
( n  1)( n  2) 3 ( n  3)( n  1)( n  2)( n  4) 5
x +
x + ...
3!
5!
When n is an even integer y1 is a polynomial Pn(x) of degree n and y2 has the form of an infinite
series. Legendre’s function of the second kind is defined as
Qn(x) = y1 (x ) y2 (x )
n even
When n is an odd integer y2 is a polynomial Pn(x) of degree n and y1 has the form of an infinite
series. Legendre’s function of the second kind for this case is defined as
Qn(x) = - y1 (x ) y2 (x )
n odd
The general solution of Legrendre equation is then
y (x ) = C1 Pn(x) + C2 Qn(x)
Legendre function of the second kind converges on the interval - 1 < x < 1 and diverges at the
end points. The first few Legendre functions of the second kind are
Q0(x) =
1 1 x
ln
2 1 x
Q1(x) = x Q0(x)  1
Q2(x) = P2(x)Q0(x) 
3
x
2
Q3(x) = P3(x)Q0(x) 
5 2 2
x +
2
3
Q4(x) = P4(x)Q0(x) 
35 3 55
x +
x
24
8
Q5(x) = P5(x)Q0(x) 
63 4 49 2 8
x +
x 
8
15
8
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