Power Series Solutions to the Legendre Equation
Power Series Solutions to the Legendre Equation
Department of MathematicsIIT Guwahati
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
The Legendre equation
The equation
(1− x2)y ′′ − 2xy ′ + α(α + 1)y = 0, (1)
where α is any real constant, is called Legendre’s equation.
When α ∈ Z+, the equation has polynomial solutions calledLegendre polynomials. In fact, these are the same polynomialthat encountered earlier in connection with the Gram-Schmidtprocess.
The Eqn. (1) can be rewritten as
[(x2 − 1)y ′]′ = α(α + 1)y ,
which has the form T (y) = λy , where T (f ) = (pf ′)′, withp(x) = x2 − 1 and λ = α(α + 1).
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Note that the nonzero solutions of (1) are eigenfunctions of Tcorresponding to the eigenvalue α(α + 1).
Since p(1) = p(−1) = 0, T is symmetric with respect to theinner product
(f , g) =
∫ 1
−1f (x)g(x)dx .
Thus, eigenfunctions belonging to distinct eigenvalues areorthogonal.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Power series solution for the Legendre equation
The Legendre equation can be put in the form
y ′′ + p(x)y ′ + q(x)y = 0,
where
p(x) = − 2x
1− x2and q(x) =
α(α + 1)
1− x2, if x2 6= 1.
Since 1(1−x2) =
∑∞n=0 x
2n for |x | < 1, both p(x) and q(x) have
power series expansions in the open interval (−1, 1).
Thus, seek a power series solution of the form
y(x) =∞∑n=0
anxn, x ∈ (−1, 1).
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Differentiating term by term, we obtain
y ′(x) =∞∑n=1
nanxn−1 and y ′′ =
∞∑n=2
n(n − 1)anxn−2.
Thus,
2xy ′ =∞∑n=1
2nanxn =
∞∑n=0
2nanxn,
and
(1− x2)y ′′ =∞∑n=2
n(n − 1)anxn−2 −
∞∑n=2
n(n − 1)anxn
=∞∑n=0
(n + 2)(n + 1)an+2xn −
∞∑n=0
n(n − 1)anxn
=∞∑n=0
[(n + 2)(n + 1)an+2 − n(n − 1)an]xn.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Substituting in (1), we obtain
(n+2)(n+1)an+2−n(n−1)an−2nan+α(α+1)an = 0, n ≥ 0,
which leads to a recurrence relation
an+2 = −(α− n)(α + n + 1)
(n + 1)(n + 2)an.
Thus, we obtain
a2 = −α(α+ 1)
1 · 2a0,
a4 = − (α− 2)(α+ 3)
3 · 4a2 = (−1)2α(α− 2)(α+ 1)(α+ 3)
4!a0,
...
a2n = (−1)nα(α− 2) · · · (α− 2n + 2) · (α+ 1)(α+ 3) · · · (α+ 2n − 1)
(2n)!a0.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Similarly, we can compute a3, a5, a7, . . . , in terms of a1 and obtain
a3 = − (α− 1)(α+ 2)
2 · 3a1
a5 = − (α− 3)(α+ 4)
4 · 5a3 = (−1)2 (α− 1)(α− 3)(α+ 2)(α+ 4)
5!a1
...
a2n+1 = (−1)n (α− 1)(α− 3) · · · (α− 2n + 1)(α+ 2)(α+ 4) · · · (α+ 2n)
(2n + 1)!a1
Therefore, the series for y(x) can be written as
y(x) = a0y1(x) + a1y2(x), where
y1(x) = 1 +∑∞
n=1(−1)nα(α−2)···(α−2n+2)·(α+1)(α+3)···(α+2n−1)
(2n)! x2n, and
y2(x) = x +∑∞
n=1(−1)n(α−1)(α−3)···(α−2n+1)·(α+2)(α+4)···(α+2n)
(2n+1)! x2n+1.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Note: The ratio test shows that y1(x) and y2(x) converges for|x | < 1. These solutions y1(x) and y2(x) satisfy the initialconditions
y1(0) = 1, y ′1(0) = 0, y2(0) = 0, y ′2(0) = 1.
Since y1(x) and y2(x) are independent, the general solution ofthe Legendre equation over (−1, 1) is
y(x) = a0y1(x) + a1y2(x)
with arbitrary constants a0 and a1.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
ObservationsCase I. When α = 0 or α = 2m, we note that
α(α− 2) · · · (α− 2n + 2) = 2m(2m − 2) · · · (2m − 2n + 2) =2nm!
(m − n)!
and
(α+ 1)(α+ 3) · · · (α+ 2n − 1) = (2m + 1)(2m + 3) · · · (2m + 2n − 1)
=(2m + 2n)!m!
2n(2m)!(m + n)!.
Then, in this case, y1(x) becomes
y1(x) = 1 +(m!)2
(2m)!
m∑k=1
(−1)k (2m + 2k)!
(m − k)!(m + k)!(2k)!x2k ,
which is a polynomial of degree 2m. In particular, forα = 0, 2, 4(m = 0, 1, 2), the corresponding polynomials are
y1(x) = 1, 1− 3x2, 1− 10x2 +35
3x4.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Note that the series y2(x) is not a polynomial when α is evenbecause the coefficients of x2n+1 is never zero.
Case II. When α = 2m + 1, y2(x) becomes a polynomial andy1(x) is not a polynomial.
In this case,
y2(x) = x +(m!)2
(2m + 1)!
m∑k=1
(−1)k (2m + 2k + 1)!
(m − k)!(m + k)!(2k + 1)!x2k+1.
For example, when α = 1, 3, 5 (m = 0, 1, 2), the correspondingpolynomials are
y2(x) = x , x − 5
3x3, x − 14
3x3 +
21
5x5.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
The Legendre polynomialLet
Pn(x) =1
2n
[n/2]∑r=0
(−1)r (2n − 2r)!
r !(n − r)!(n − 2r)!xn−2r ,
where [n/2] denotes the greatest integer ≤ n/2.
• When n is even, it is a constant multiple of the polynomialy1(x).
• When n is odd, it is a constant multiple of the polynomialy2(x).
The first five Legendre polynomials are
P0(x) = 1, P1(x) = x , P2(x) =1
2(3x2 − 1)
P4(x) =1
8(35x4 − 30x2 + 3), P5(x) =
1
8(63x5 − 70x3 + 15x).
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Figure : Legendre polynomial over the interval [−1, 1]
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Rodrigues’s formula for the Legendre polynomialsNote that
(2n − 2r)!
(n − 2r)!xn−2r =
dn
dxnx2n−2r and
1
r !(n − r)!=
1
n!
(nr
).
Thus, Pn(x) in (2) can be expressed as
Pn(x) =1
2nn!
dn
dxn
[n/2]∑r=0
(−1)r(
nr
)x2n−2r .
When [n/2] < r ≤ n, the term x2n−2r has degree less than n,so its nth derivative is zero. This gives
Pn(x) =1
2nn!
dn
dxn
n∑r=0
(−1)r(
nr
)x2n−2r =
1
2n n!
dn
dxn(x2−1)n,
which is known as Rodrigues’ formula.RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
Properties of the Legendre polynomials Pn(x)
• For each n ≥ 0, Pn(1) = 1. Moreover, Pn(x) is the onlypolynomial which satisfies the Legendre equation
(1− x2)y ′′ − 2xy ′ + n(n + 1)y = 0
and Pn(1) = 1.
• For each n ≥ 0, Pn(−x) = (−1)nPn(x).
• ∫ 1
−1Pn(x)Pm(x)dx =
{0 if m 6= n,
22n+1
if m = n.
RA/RKS MA-102 (2016)
Power Series Solutions to the Legendre Equation
• If f (x) is a polynomial of degree n, we have
f (x) =n∑
k=0
ckPk(x), where
ck =2k + 1
2
∫ 1
−1f (x)Pk(x)dx .
• It follows from the orthogonality relation that∫ 1
−1g(x)Pn(x)dx = 0
for every polynomial g(x) with deg(g(x)) < n.
*** End ***
RA/RKS MA-102 (2016)
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