Power Series Solutions to the Legendre Equation
Transcript of Power Series Solutions to the Legendre Equation
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)