Legendre Differential equation

Problem-1: Define the Legendre differential equation of order n and Legendre

polynomial of first kind.

Definition: Legendre D.E.: An equation of the form

𝑑2 𝑦 𝑑𝑦
(1 − 𝑥 2 ) − 2𝑥 𝑑𝑥 + 𝑛(𝑛 + 1)𝑦 = 0, where 𝑛 ≥ 0
𝑑𝑥 2

is called Legendre differential equation of order n.

Legendre Polynomial of first kind: The series solution of the Legendre differential equation is

called Legendre polynomial and can be expressed as,

𝑛
[ ]
2
(−1)𝑠 (2𝑛 − 2𝑠)! 𝑥 𝑛−2𝑠 }
𝑃𝑛 (𝑥 ) = ∑ 𝑛
{2 𝑠! (𝑛 − 𝑠)! (𝑛 − 2𝑠)!}
𝑠=0

𝑛
𝑛
, 𝑤ℎ𝑒𝑛 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
2
Where, [ 2] = {𝑛−1
, 𝑤ℎ𝑒𝑛 𝑛 𝑖𝑠 𝑜𝑑𝑑
2

Problem 2: Find the value of the Legendre polynomials, 𝑷𝒐 (𝒙), 𝑷𝟏 (𝒙), 𝑷𝟐 (𝒙), 𝑷𝟒 (𝒙).

Solution: We know the Rodrigues formula is,

1 𝑑𝑛
𝑃𝑛 (𝑥) = (𝑥 2 − 1)𝑛 … … … (1)
2𝑛 𝑛! 𝑑𝑥 𝑛

Set 𝑛 = 0 in Eq. (1), we get,

1 𝑑0
𝑝0 (𝑥) = 0 (𝑥 2 − 1)0
2 (0)! 𝑑𝑥 0

=1

Set 𝑛 = 1 in Eq. (1) , we get,

1
= (2𝑥)
2

=𝑥

Set 𝑛 = 2 in Eq. (1), we get,

1 𝑑2
𝑃2 (𝑥) = (𝑥 2 − 1)2
22 (2)! 𝑑𝑥 2

1 𝑑2
= (𝑥 4 − 2𝑥 2 + 1) [∵ (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 ]
22 (2)! 𝑑𝑥 2

1 𝑑 𝑑
= . .[ (𝑥 4 − 2𝑥 2 + 1)]
22 (2)! 𝑑𝑥 𝑑𝑥

1 𝑑
= . [4𝑥 3 − 4𝑥]
22 (2)! 𝑑𝑥

1
= (12𝑥 2 − 4)
4×2

1
= (3𝑥 2 − 1)
2

Set 𝑛 = 3 in Eq. (1) we get,

1 𝑑3
𝑃3 (𝑥) = 3 3
(𝑥 2 − 1)3
2 (3)! 𝑑𝑥

1 𝑑2 𝑑
= 8×6 𝑑𝑥 2
. [ 𝑑𝑥 (𝑥 6 − 3. (𝑥 2 )2 . 1 + 3. 𝑥 2 . (1)2 − (1)3 )] [∵ (𝑎 − 𝑏)3 = 𝑎3 − 3𝑎2 𝑏 + 3 𝑎𝑏 2 − 𝑏 3 ]

1 𝑑2 𝑑
= . 2 .[ (𝑥 6 − 3𝑥 4 + 3𝑥 2 − 1)]
48 𝑑𝑥 𝑑𝑥

1 𝑑2
= . 2 [6𝑥 5 − 12𝑥 3 + 6𝑥]
48 𝑑𝑥
 1 𝑑 𝑑
= [ (6𝑥 5 − 12𝑥 3 + 6𝑥)]
48 𝑑𝑥 𝑑𝑥

1 𝑑
= ( 30𝑥 4 − 36 𝑥 2 + 6)
48 𝑑𝑥

1
= (120𝑥 3 − 72𝑥)
48

1
= × 24 (5𝑥 3 − 3𝑥)
48

1
= (5 𝑥 3 − 3𝑥)
2

Set 𝑛 = 4 in Eq. (1) we get,

1 𝑑4
𝑃4 (𝑥) = 4 (𝑥 2 − 1)4
2 (4)! 𝑑𝑥 4

1 𝑑3 𝑑
= 3
[ {(𝑥 2 )4 − 4 (𝑥 2 )3 . (1) + 6. (𝑥 2 )2 . (1)2 − 4. (𝑥 2 )1 . (1)3 + (1)4 }]
16 × 24 𝑑𝑥 𝑑𝑥

1 𝑑3 𝑑 8
= [ (𝑥 − 4 𝑥 6 + 6 𝑥 4 − 4𝑥 2 + 1)]
16 × 24 𝑑𝑥 3 𝑑𝑥

1 𝑑3
= (8 𝑥 7 − 24 𝑥 5 + 24 𝑥 3 − 8 𝑥)
16 × 24 𝑑𝑥 3

1 𝑑2 𝑑
= ×8× 2× (𝑥 7 − 3𝑥 5 + 3𝑥 3 − 𝑥)
16 × 24 𝑑𝑥 𝑑𝑥

1 𝑑2
= × 2 × (7𝑥 6 − 15 𝑥 4 + 9𝑥 2 − 1)
16 × 3 𝑑𝑥

1 𝑑 𝑑
= × × (7𝑥 6 − 15 𝑥 4 + 9𝑥 2 − 1)
16 × 3 𝑑𝑥 𝑑𝑥

1 𝑑
= × × (42 𝑥 5 − 60 𝑥 3 + 18 𝑥)
16 × 3 𝑑𝑥

1 𝑑
= ×6× (7𝑥 5 − 10 𝑥 3 + 3𝑥)
16 × 3 𝑑𝑥
 1
= (35 𝑥 4 − 30 𝑥 2 + 3)
8

1 1
Therefore, 𝑃0 (𝑥) = 1, 𝑃1 (𝑥) = 𝑥, 𝑃2 (𝑥) = 2 (3𝑥 2 − 1), 𝑃3 (𝑥) = 2 (5 𝑥 3 − 3𝑥),

1
𝑃4 (𝑥) = 8 (35 𝑥 4 − 30 𝑥 2 + 3). (Answer)

Problem 3: Express the function in terms of Legendre polynomial

𝒇(𝒙) = 𝒂𝒙𝟑 + 𝒃𝒙𝟐 + 𝒄𝒙 + 𝒅

Solution: Given function,

𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 … … … (1)

Now, we know, 𝑃0 (𝑥) = 1, 𝑃1 (𝑥) = 𝑥,

1
𝑃2 (𝑥) = 2 (3𝑥 2 − 1)

⇒ 2 𝑃2 (𝑥) = 3𝑥 2 − 1

⇒ 2 𝑃2 (𝑥) + 1 = 3𝑥 2

1
⇒ 𝑥 2 = 3 (2𝑃2 (𝑥) + 1)

1
Again, 𝑃3 (𝑥) = 2 (5 𝑥 3 − 3𝑥)

⇒ 2𝑃3 (𝑥) = 5𝑥 3 − 3 𝑃1

⇒ 2𝑃3 (𝑥) + 3𝑃1 (𝑥) = 5 𝑥 3

1
⇒ 𝑥 3 = 5 {2𝑃3 (𝑥) + 3𝑃1 (𝑥)}

Now, Substituting the values of 𝑥, 𝑥 2 , 𝑥 3 in Eq. (1), we get,

1 1
𝑓(𝑃) = 𝑎 × {2𝑃3 (𝑥) + 3𝑃1 (𝑥)} + 𝑏 × (2𝑃2 (𝑥) + 1) + 𝑐 × 𝑃1 (𝑥) + 𝑑 × 𝑃0 (𝑥)
5 3

Which is the required polynomial.