5.3 First One of Two Special Equations - Legendre’s Equation
1. Legendre’s equation and solution:
Legendre’s differential equation: Ÿ1 " x 2 y UU " 2xy U nŸn 1 y 0, where n is a real number.
Ÿ1 " x 2 y UU " 2xy U 3 y 0, n 1
2
4
Example Ÿ1 " x 2 y UU " 2xy U 6y 0, n 2;
Solve Legendre’s differential equation: Ÿ1 " x 2 y UU " 2xy U nŸn 1 y 0.
Since a 2 Ÿx 1 " x 2 0 when x 1 and x "1, the equation has singular points: x o1.
.
a. Let x 0 0 (it is a regular point) and y ! m0 c m x m . Then
U
.
y ! c m mx m"1 ,
.
and
! c m mŸm " 1 x m"2 .
y UU m1
m2
b. Substitute y U and y UU into the equation Ÿ1 " x 2 y UU " 2xy U nŸn 1 y 0, or
y UU " x 2 y UU " 2xy U nŸn 1 y 0.
We have
.
.
.
.
m2
m2
m1
m0
! c m mŸm " 1 x m"2 " ! c m mŸm " 1 x m " ! 2c m mx m ! nŸn 1 c m x m 0.
In the 1st summation, replace m " 2 by k : m k 2, k m " 2, when m 2, k 0.
In the last 3 summations, replace m by k.
.
.
.
.
! c k2 Ÿk 2 Ÿk 1 x " ! c k kŸk " 1 x " ! 2c k kx ! nŸn 1 c k x k 0.
k
k0
k
k2
Ÿ2c 2 nŸn 1 c 0 .
! k2
k
k1
k0
ŸnŸn 1 " 2 c 1 6c 3 x
c k2 Ÿk 2 Ÿk 1 " c k kŸk " 1 " 2c k k nŸn 1 c k
xk 0
c. Solving from the equation, we have a recurrence relation:
nŸn 1 c0
2
nŸn 1 " 2
Ÿn " 1 Ÿn 2 c1 "
c1
ŸnŸn 1 " 2 c 1 6c 3 0, c 3 "
3!
6
c k2 Ÿk 2 Ÿk 1 " c k kŸk " 1 " 2c k k nŸn 1 c k 0, for k 2, 3, . . .
2c 2 nŸn 1 c 0 0,
c2 "
For k 2, 3, . . .
c k2 kŸk " 1 2k " nŸn 1 k 2 k " nŸn 1 Ÿn " k Ÿn k 1 ck ck "
ck
Ÿk 2 Ÿk 1 Ÿk 2 Ÿk 1 Ÿk 2 Ÿk 1 Example Find the solution of Legendre’s equation Ÿ1 " x 2 y UU " 2xy U 6y 0.
nŸ1 n 6, n 2.
2Ÿ3 nŸn 1 c0 "
c "3c 0
c2 "
2
2 0
Ÿn " 1 Ÿn 2 Ÿ1 Ÿ4 c3 "
c1 "
c1 " 2 c1
3!
3!
3
For k 2, 3, . . . ,
1
Ÿk " 2 Ÿ3 k Ÿ2 " k Ÿ3 k "Ÿk " 2 Ÿ3 k ck "
ck ck
Ÿk 2 Ÿk 1 Ÿk 2 Ÿk 1 Ÿk 2 Ÿk 1 c k2 "
k 2 c4 "
k3
c5 Ÿ1 Ÿ6 c3 Ÿ5 Ÿ4 k5
c7 Ÿ3 Ÿ8 c5 Ÿ7 Ÿ6 k7
c9 Ÿ5 Ÿ10 c7 Ÿ9 Ÿ8 B
B
k 2m " 1
c 2m1 "
Ÿ2 " 2 Ÿ5 c 2 0 ® c 2m 0 for m 2, 3, . . .
Ÿ4 Ÿ3 Ÿ1 Ÿ2 2 3! Ÿ1 Ÿ2Ÿ2 c1 "
c1 " 1 c1
5
3!
5!
2
3
Ÿ1 Ÿ2 3! Ÿ1 Ÿ3 2 4!
c1 "
c1 " 4 c1
"
5!
35
7!
Ÿ1 Ÿ2Ÿ3 Ÿ5 Ÿ4 Ÿ3 Ÿ8 Ÿ7 Ÿ6 "
Ÿ5 Ÿ10 Ÿ9 Ÿ8 "
Ÿ1 Ÿ3 2 3 4!
c1
7!
Ÿ1 Ÿ3 Ÿ5 2 4 5!
c1 " 5 c1
63
9!
"
Ÿ1 Ÿ3 Ÿ5 C2 2m"1 Ÿ2m " 3 !
c1
Ÿ2m 1 !
y c 0 Ÿ1 " 3x 2 c 1 x " 2 x 3 " 1 x 5 " 4 x 7 " 5 x 9 " C
5
3
35
63
.
Ÿ1 Ÿ3 Ÿ5 C2 2m"1 Ÿ2m " 3 ! 2m1
x
c 0 Ÿ1 " 3x 2 c 1 x " 2 x 3 " !
3
Ÿ2m 1 !
m2
Example Find the solution of Legendre’s equation Ÿ1 " x 2 y UU " 2xy U 3 y 0.
4
nŸ1 n 1 3 3 , n 1 .
2
4
2 2
1 3
nŸn 1 Ÿ1 Ÿ3 c2 "
c 0 " 2 2 c 0 Ÿ"1 c0 " 3 c0
2
2
8
2!Ÿ4 Ÿn " 1 Ÿn 2 c3 "
c1 "
3!
"1
2
3!
5
2
c 1 Ÿ"1 2
Ÿ1 Ÿ5 c1 5 c1
24
3!Ÿ4 For k 2, 3, . . . ,
c k2
Ÿ 1 " k Ÿ 3 k Ÿk " 1 Ÿ3 k Ÿn " k Ÿn k 1 Ÿ1 " k Ÿ3 k 2
ck " 2
ck "
ck ck
"
4Ÿk 2 Ÿk 1 4Ÿk 2 Ÿk 1 Ÿk 2 Ÿk 1 Ÿk 2 Ÿk 1 k2
c4 k4
c6 B
k 2Ÿm " 1 2
Ÿ1 Ÿ5 c2 4Ÿ4 Ÿ3 Ÿ3 Ÿ7 c4 4Ÿ6 Ÿ5 " 7 c0
1024
Ÿ1 Ÿ5 4Ÿ4 Ÿ3 Ÿ3 Ÿ7 4Ÿ6 Ÿ5 Ÿ"1 "
Ÿ1 Ÿ3 c0
2!Ÿ4 "
Ÿ1 2 Ÿ3 Ÿ5 c0
4!Ÿ4 2 B
c 2m "
Ÿ1 2 Ÿ3 Ÿ5 CŸ2m " 3 2 Ÿ2m " 1 Ÿ2m 1 c0
Ÿ2m ! Ÿ4 m Ÿ1 2 Ÿ3 Ÿ5 c0 " 5 c0
128
4!Ÿ4 2 "
Ÿ1 2 Ÿ3 2 Ÿ5 Ÿ7 c0
6!Ÿ4 3 k3
c5 k5
Ÿ2 Ÿ6 c3 4Ÿ5 Ÿ4 Ÿ2Ÿ1 Ÿ2Ÿ3 4Ÿ5 Ÿ4 Ÿ1 Ÿ5 c1
3!Ÿ4 2 2 Ÿ1 2 Ÿ3 Ÿ5 c1 1 c1
32
5!Ÿ4 2 2 2 Ÿ1 2 Ÿ3 Ÿ5 Ÿ2Ÿ2 Ÿ2Ÿ4 Ÿ4 Ÿ8 c5 c1
4Ÿ7 Ÿ6 4Ÿ7 Ÿ6 5!Ÿ4 2 1 c1
168
2 3 Ÿ2! Ÿ4! Ÿ5 Ÿ2Ÿ3 Ÿ2Ÿ5 Ÿ6 Ÿ10 c9 c7 c1
4Ÿ9 Ÿ8 4Ÿ9 Ÿ8 7!4 3
5 c1
4032
B
2 2m"3 Ÿ2m " 5 !Ÿ2m " 3 !Ÿ5 c 2m1 c1
Ÿ2m 1 ! Ÿ4 m c7 k7
B
k 2m " 1
y c 0 1 " 3 x 2 " 5 x 4 " " 7 x 6 ". . .
8
128
1024
2 3 Ÿ2! Ÿ4! Ÿ5 c1
7!4 3
2 5 Ÿ3! Ÿ5! Ÿ5 c1
9!4 4
c 1 x 1 x 3 1 x 5 5 x 7 . . .
32
628
4032
.
c0
Ÿ1 2 Ÿ3 Ÿ5 CŸ2m " 3 2 Ÿ2m " 1 Ÿ2m 1 2m
1 " 3 x2 " !
x
8
Ÿ2m ! Ÿ4 m m2
.
2 2m"3 Ÿ2m " 5 !Ÿ2m " 3 !Ÿ5 2m1
x
c1 x 1 x3 !
32
Ÿ2m 1 ! Ÿ4 m m2
Remarks:
a. The general solution of a Legendre equation
For any n,
nŸn 1 Ÿn " 1 Ÿn 2 c0,
c3 "
c1
c2 "
2
3!
Ÿn " k Ÿn k 1 c k2 "
c k , for k 2, 3, . . .
Ÿk 2 Ÿk 1 k2
k3
nŸn 1 Ÿn " 2 Ÿn 3 Ÿn " 2 Ÿn 3 c2 "
c0
"
2
Ÿ4 Ÿ3 Ÿ4 Ÿ3 Ÿn " 2 Ÿn 3 nŸn 1 c0
Ÿ"1 2
4!
Ÿn " 1 Ÿn 2 Ÿn " 3 Ÿn 4 Ÿn " 3 Ÿn 4 c5 "
c3 "
c1
"
3!
Ÿ5 Ÿ4 Ÿ5 Ÿ4 Ÿn " 1 Ÿn 2 Ÿn " 3 Ÿn 4 Ÿn " 3 Ÿn 4 c3 "
c1
"
"
3!
Ÿ5 Ÿ4 Ÿ5 Ÿ4 c4 "
nŸn 1 2
Ÿn " 2 Ÿn 3 nŸn 1 4
x Ÿ"1 2
x . . .
2
4!
Ÿn " 1 Ÿn 2 3
Ÿn " 3 Ÿn 4 Ÿn " 1 Ÿn 2 5
x Ÿ"1 2
x . . .
c1 x "
6
5!
c0 y1 c1 y2
y c0 1 "
where c 0 and c 1 are arbitrary constants. Observe that y 1 is even and y 2 is odd. Hence, y 1 and y 2 are
linearly independent, and
y c0 y1 c1 y2
3
is the general solution of Legendre’s equation.
b. Radius of convergence
y is called a Legendre function for x in the interval of convergence. The radius R of convergence of
the power series for a Legendre function is 1 which is the distance to the nearest singular point x 1 or
x "1. That is, y is a Legendre function for "1 x 1.
2. Legendre Polynomials:
Observe that
nŸn 1 Ÿn " 1 Ÿn 2 c0,
c3 "
c1
2
3!
Ÿn " k Ÿn k 1 ck,
k 2, 3, . . . .
"
Ÿk 2 Ÿk 1 c2 "
c k2
When n is a nonnegative integer and n " k 0,
c k2 0 for k u n.
In this case, y 1 Ÿif k is even) or y 2 Ÿif k is odd) is a polynomial. For example,
when n 0, c 2 0 and then c 4 0, c 6 0, . . . . So, y 1 c 0 , a polynomial of degree 0;
when n 1, c 3 0 and then c 5 0, c 7 0, . . . . So, y 2 c 1 x, a polynomial of degree 1;
when n 2, c 4 0 and then c 6 0, c 8 0, . . . . So, y 1 c 0 c 2 x 2 , a polynomial of degree 2;
when n 3, c 5 0 and then c 7 0, c 9 0, . . . . So, y 2 c 1 x c 3 x 3 , a polynomial of degree 3.
The polynomials derived from the Legendre’s function are called Legendre polynomials. With choosing
Ÿ1 Ÿ3 Ÿ5 . . . Ÿn " 1 Ÿ1 Ÿ3 Ÿ5 . . . Ÿn c 0 Ÿ"1 n/2
,
and c 1 Ÿ"1 Ÿn"1 /2
,
Ÿ2 Ÿ4 . . . Ÿn Ÿ2 Ÿ4 . . . Ÿn " 1 P 0 Ÿx 1
P 2 Ÿx P 4 Ÿx 1
2
1
8
P 1 Ÿx x
Ÿ3x 2 " 1 P 3 Ÿx Ÿ35x 4 " 30x 2 3 P 5 Ÿx B
1
2
1
8
Ÿ5x 3 " 3x Ÿ63x 5 " 70x 3 15x B
1
0.8
0.6
y
0.4
0.2
-1
-0.8
-0.6
-0.4
-0.2
0
-0.2
0.2
0.4 x 0.6
0.8
-0.4
-0.6
-0.8
-1
y P0, P1, P2, P3, P4, P5
Example Show that P 5 Ÿx 4
1
8
Ÿ63x 5 " 70x 3 15x satisfies Legendre equation.
1
Since n 5, consider the differential equation: Ÿ1 " x 2 y UU " 2xy U 30y 0.
U
P 5 Ÿx 315 x 4 " 105 x 2 15 , P UU5 Ÿx 315 x 3 " 105 x
4
2
8
8
2
105
105
15
315
315
1
x3 "
x " 2x
x4 "
x2 30
Ÿ1 " x 2 Ÿ63x 5 " 70x 3 15x 2
4
8
2
8
8
0
3. Properties of Legendre’s Polynomials:
a. P 2m ’s are even and P 2m1 ’s are odd.
b. P m ’s are orthogonal on "1, 1 .
Recall: Functions f and g are orthogonal on a, b if
b
; a fŸx gŸx dx 0.
Recall: If f is odd, then
a
; "a fŸx dx 0.
1
1
; "1 P 2 Ÿx dx ; "1
1 Ÿ3x 2 " 1 dx 0
2
1
1
; "1 P 4 Ÿx dx ; "1 18 Ÿ35x 4 " 30x 2 3 dx 0
1
1
; "1 P 2 Ÿx P 4 Ÿx dx ; "1 12 Ÿ3x 2 " 1 18 Ÿ35x 4 " 30x 2 3 dx 0
We know that £1, x, x 2 , . . . , x m ¤ forms a basis for polynomials of degree m. Legendre’s polynomials
P 0 , P 1 , . . . , P m also forms an orthogonal basis for polynomials of degree m.
5