(1)

Assume the series solution as y = a0 + a1 x + a2 x 2 + a3 x 3 + ⋅ ⋅ ⋅ + an x n + ⋅ ⋅ ⋅ ∞ ∑ = an x n , n=0

We would like to ﬁnd a value for constant a.

(2)

Series Solutions of Diﬀerential Equations: Cont’d

We diﬀerentiate the series once y ′ = a1 + 2a2 x + 3a3 x 2 + ⋅ ⋅ ⋅ + nan x n−1 + ⋅ ⋅ ⋅ ∞ ∑ = nan x n−1 , n=1

(3)

Series Solutions of Diﬀerential Equations: Cont’d Substitute Eqn. (2) and (3) into Eqn. (1) a1 +2a2 x+3a3 x 2 +4a4 x 3 +⋅ ⋅ ⋅ = 2a0 x+2a1 x 2 +2a2 x 3 +2a3 x 4 +2a4 x 5 +⋅ ⋅ ⋅ (4) By collecting x term, we get a1 = 0

a2 = a0

2 a3 = a1 3

1 a4 = a2 2

(5)

or in general: { nan = 2an−2 ,

an =

0, odd n 2 an−2 , even n n

(6)

Series Solutions of Diﬀerential Equations: Cont’d

Putting n = 2m, Since only even term exists, we get

a2 m =

2 1 1 1 a2m−2 = a2m−2 = a2m−4 = ⋅ ⋅ ⋅ = a0 . 2m m m m!

(7)

Substituting Eqn. (7) into Eqn. (2), we get the assumed solution,

y = a0 + a0 x 2 +

∞ ∑ 1 x 2m 1 a0 x 4 ⋅ ⋅ ⋅ + a0 x 2m + ⋅ ⋅ ⋅ = a0 . (8) 2! m! m! m=0

Legendre’s Equation

The Legendre diﬀerential equation is ′′

′

(1 − x 2 )y − 2xy + l(l + 1)y = 0, Arises in 1. Solution of Partial Diﬀerential Equations in spherical coordinates 2. Problems in Mechanics, Quantum Mechanics, Electromagnetic Theory, Heat with spherical symmetry.

(9)

Legendre’s Equation: Cont’d

Assume the series solution as y = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + ⋅ ⋅ ⋅ + an x n + ⋅ ⋅ ⋅ y ′ = a1 + 2a2 x + 3a3 x 2 + 4a4 x 3 + ⋅ ⋅ ⋅ + nan x n−1 + ⋅ ⋅ ⋅ y ′′ = 2a2 + 6a3 x + 12a4 x 2 + 20a5 x 3 + ⋅ ⋅ ⋅ + n(n − 1)an x n−2 + ⋅ ⋅ ⋅ (10)

Legendre’s Equation: Cont’d

Inserting Eqn. (10) into Eqn. (9), we get 2a2 + 6a3 x + 12a4 x 2 + 20a5 x 3 + ⋅ ⋅ ⋅ + (n + 2)(n + 1)an+2 x n + ⋅ ⋅ ⋅ − (2a2 x 2 + 6a3 x 3 + 12a4 x 4 + 20a5 x 5 + ⋅ ⋅ ⋅ + n(n − 1)an x n + ⋅ ⋅ ⋅ ) − (2a1 x + 4a2 x 2 + 6a3 x 3 + 8a4 x 4 + ⋅ ⋅ ⋅ + 2nan x n + ⋅ ⋅ ⋅ + l(l + 1)(a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + ⋅ ⋅ ⋅ + an x n + ⋅ ⋅ ⋅ ) =0

Legendre’s Equation: Cont’d

Collecting x term 2a2 + l(l + 1)a0 = 0 or 6a3 + (l 2 + l − 2)a1 = 0 or 12a4 + (l 2 + l − 6)a2 = 0 or

l(l + 1) a0 ; 2 (l − 1)(l + 2) a1 ; a3 = − 6 (l − 2)(l + 3) a4 = − a2 12 l(l + 1)(l − 2)(l + 3) = a0 ; 4!

a2 = −

Legendre’s Equation: Cont’d

For x n , we get (n + 2)(n + 1)an+2 + (l 2 + l − n2 − n)an = 0. So an+2 is an+2 = −

(l − n)(l + n + 1) an (n + 2)(n + 1)

Legendre’s Equation: Cont’d

The general solution for Eqn. (9) is ] [ l(l + 1) 2 l(l + 1)(l − 2)(l + 3) 4 x + x − ⋅⋅⋅ y = a0 1 − 2! 4! [ ] (l − 1)(l + 2) 3 (l − 1)(l + 2)(l − 3)(l + 4) 5 + a1 x − x + x − ⋅⋅⋅ 3! 5! (11)

Legendre’s Equation: Cont’d

If l = 0 then Eqn. (11) will become y = a0 [1 − 0 + 0 − ⋅ ⋅ ⋅ ] ] [ 1 3 1 5 + a1 x + x + x − ⋅ ⋅ ⋅ 3 5 = a0 [ ] 1 2 1 4 + a1 x 1 + x + x − ⋅ ⋅ ⋅ 3 5 When x 2 = 1, a1 will diverge. So we are left with y = 1, if we set a0 = 1.

Legendre’s Equation: Cont’d

If l = 1 then Eqn. (11) will become [ ] 1 4 2 y = a0 1 − x − x − ⋅ ⋅ ⋅ 3 + a1 [x − 0 + 0 − ⋅ ⋅ ⋅ ] [ ] 1 4 2 = a0 1 − x − x − ⋅ ⋅ ⋅ 3 + a1 x When x 2 = 1, a0 will diverge. So we are left with y = x, if we set a1 = 1.

Legendre’s Equation: Cont’d

Taking a0 = 1 and a1 = 1 l = 2,

y = 1 − 3x 2

l = 4,

y = 1 − 10x 2 +

l =3 l =5

35 4 x 3

5 y = x − x3 3 14 21 y = x − x3 + x5 3 5

Legendre Polynomials Multiplying the solution y with a constant C is also a solution. The multiplication C ⋅ y = 1 when x = 1, for example when l = 2, Pl (x) = C ⋅ y = C ⋅ (1 − 3) = 1 1 Thus C = − . This polynomial is called Legendre Polynomials. 2 P0 (x) = 1, 1 P2 (x) = (3x 2 − 1) 2 1 P4 (x) = (35x 4 − 30x 2 + 3) 8

P1 (x) = x 1 P3 (x) = (5x 3 − 3x) 2 1 P5 (x) = (63x 5 − 70x 3 + 15x) 8

Legendre Polynomials: Cont’d

Legendre polynomials in general form Pl (x) =

N ∑ n=0

(−1)n

(2l − 2n)! x l−2n 2l n!(l − n)!(l − 2n)!