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Legendre Polynomials Pn(x)
Tapas Mazumdar (2018A8B40427P)
October 2019
Introduction
The Legendre polynomials form the family of solutions for the Legendre differential equation
(1 − x^2 )y′′^ − 2 xy′^ + n(n + 1)y = 0 (1)
or d dx
[
(1 − x^2 )y′
]
for n ∈ Z+ 0.
Solution
It can be easily verified that the above differential equation has regular singular points at x = − 1 , 1. General series solution around x = 0 is
y = a 0
[
n(n + 1) 2! x^2 + (n − 2)n(n + 1)(n + 3) 4! x^4 + (n − 4)(n − 2)n(n + 1)(n + 3)(n + 5) 6! x^6 + · · ·
]
Φ 1
+a 1
[
x − (n^ −^ 1)(n^ + 2) 3! x^3 + (n^ −^ 3)(n^ −^ 1)(n^ + 2)(n^ + 4) 5! x^5 − (n^ −^ 5)(n^ −^ 3)(n^ −^ 1)(n^ + 2)(n^ + 4)(n^ + 6) 7! x^7 + · · ·
]
Φ 2
Note that for n ∈ Z+ 0 we have { Φ 1 is a polynomial, n = 2k, k ∈ Z+ 0 Φ 2 is a polynomial, n = 2k + 1, k ∈ Z+ 0
this leads us to the following theorem.
Theorem 1. If Ψ is a polynomial solution of (1), then
Ψ = cΦ 1 or Ψ = cΦ 2 Proof. Since Ψ is a solution, Ψ = c 1 Φ 1 + c 2 Φ 2 However, for any given n ∈ Z+ 0 , only one of Φ 1 or Φ 2 can be a polynomial. Since Ψ is a polynomial solu- tion, hence if Φ 1 is a polynomial then c 2 = 0 and vice-versa.
- Solution near x = 1: (1) can be written as (x − 1)(x + 1)y′′^ + 2xy′^ − n(n + 1)y = 0 and can be transformed to a hypergeometric equation about x = 1 by substitution t = (^) −x 1 −−^11 = 1 − 2 x. The equation tranforms to t(1 − t)y′′^ + (1 − 2 t)y′^ + n(n + 1)y = 0 where c = 1, a + b + 1 = 2 and ab = −n(n + 1), hence the solution is
y = F (−n, n + 1, 1 , t) = F
−n, n + 1, 1 , 1 − x 2
Theorem 2. F
−n, n + 1, 1 , 1 − 2 x
is a polynomial for n ∈ Z+ 0. Proof.
F
−n, n + 1, 1 , 1 − x 2
∑^ ∞
k=
(−n) · · · (−n + k − 1)(n + 1) · · · (n + k) (k!)^2
1 − x 2
)k
∑^ ∞
k=
(−1)k^ (n − k + 1) 2 n (k!)^2
1 − x 2
)k
∑^ n
k=
(−1)k^ (n − k + 1) 2 n (k!)^2
1 − x 2
)k
:^0
∑^ ∞
k=n+
(−1)k^ (n − k + 1) 2 n (k!)^2
1 − x 2
)k
∑^ n
k=
(−1)k^ (n − k + 1) 2 n (k!)^2
1 − x 2
)k
∑^ n k=
n k
n + k k
x − 1 2
)k
which is a polynomial.
Both of the above representations are the series representations for the Legendre Polynomials, Pn(x).
- Another representation of Pn(x) The recurrence relation obtained for a series solution about x = 0 to obtain (3) can also be made a decreasing recurrence as am− 2 = −m(m − 1) (n − m + 2)(n + m − 1) am This gives
Pn(x) = an
[
xn^ − n(n − 1) 2(2n − 1) xn−^2 + · · · + (−1)kn(n − 1)(n − 2) · · · (n − (2k − 1)) 2 · 4 · 6 · · · 2 k · (2n − 1)(2n − 3) · · · (2n − (2k − 1)) xn−^2 k^ + · · ·
]
where an = (2n!) 2 n(n!)^2
Note 1. Since all powers of x in Pn(x) are of the same parity and this parity is the same as the parity of n, we have Pn(−x) = (−1)nPn(x)
Orthogonal Property
The Legendre polynomials satisfy the orthogonality condition ∫ (^1)
− 1
Pn(x) · Pm(x) dx = 2 2 n + 1 δmn
where δmn is the Kronecker delta function which is the Boolean function δij = [i = j]. (The proof of the result is skipped here (see Appendix).)
Formulas to obtain Legendre Polynomials
- Rodrigue’s formula: Pn(x) =
2 n(n!)
dn dxn^ (x^2 − 1)n
