Power Series Solutions to the Legendre EquationPower Series Solutions to the Legendre Equation Note...

Power Series Solutions to the Legendre Equation


Department of MathematicsIIT Guwahati

RA/RKS MA-102 (2016)


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)


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 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)


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)


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)


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)


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)


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)


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)


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)


Figure : Legendre polynomial over the interval [−1, 1]

RA/RKS MA-102 (2016)


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)


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)


• 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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