LEGENDRE POLYNOMIALS AND APPLICATIONS Legendre equation ...

LEGENDRE POLYNOMIALS AND APPLICATIONS

We construct Legendre polynomials and apply them to solve Dirichlet problems in spherical coordinates.

1. Legendre equation: series solutions

The Legendre equation is the second order differential equation

(1)

(1 - x2)y - 2xy + y = 0

which can also be written in self-adjoint form as

(2)

[ (1

-

x2)y]

+

y

=

0

.

This equation has regular singular points at x = -1 and at x = 1 while x = 0 is an

ordinary point. We can find power series solutions centered at x = 0 (the radius of

convergence will be at least 1). Now we construct such series solutions.

Assume that

y = cj xj

j=0

be a solution of (1). By substituting such a series in (1), we get

(1 - x2) j(j - 1)cjxj-2 - 2x jcjxj-1 + cjxj = 0

j=2

j=1

j=0

j(j - 1)cjxj-2 - j(j - 1)cjxj - 2jcjxj + cjxj = 0

j=2

j=2

j=1

j=0

After re-indexing the first series and grouping the other series, we get

(j + 2)(j + 1)cj+2xj - (j2 + j - )cjxj = 0

j=0

j=0

and then

[ (j

+

2)(j

+

1)cj+2

-

(j2

+

j

-

)cj] xj

=

0

.

j=0

By equating each coefficient to 0, we obtain the recurrence relations

(j + 1)j - cj+2 = (j + 2)(j + 1) cj ,

j = 0, 1, 2, ? ? ?

We can obtain two independent solutions as follows. For the first solution we

make c0 = 0 and c1 = 0. In this case the recurrence relation gives

2-

12 -

c3 = 6 c1 = 0, c5 = 20 c3 = 0, ? ? ? , codd = 0 .

The coefficients with even index can be written in terms of c0:

-

(2 ? 3 - ) (2 ? 3 - )(-)

c2 = 2 c0, c4 = 4 ? 3 c2 =

4!

c0

Date: April 19, 2016. 1

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LEGENDRE POLYNOMIALS AND APPLICATIONS

we prove by induction that

[(2k - 1)(2k - 2) - ][(2k - 3)(2k - 4) - ] ? ? ? [3 ? 2 - ][-]

c2k =

(2k)!

c0

We can write this in compact form as

(

)

c2k

=

c0 (2k)!

k-1 [(2i + 1)(2i) - ]

i=1

This give a solution

(

)

k-1

x2k

y1(x) = c0

[(2i + 1)(2i) - ] (2k)!

k=0 i=0

A second series solution (independent from the first) can be obtained by making

c0 = 0 and c1 = 0. In this case ceven = 0 and

(

)

c2k+1

=

c1 (2k +

1)!

k [2i(2i - 1) - ]

i=1

The corresponding solution is

(

)

k

x2k+1

y2(x) = c1

[(2i)(2i - 1) - ] (2k + 1)!

k=0 i=1

Remark 1. It can be proved by using the ratio test that the series defining y1 and y2 converge on the interval (-1, 1) (check this as an exercise). 2. It is also proved that for every either y1 or y2 is unbounded on (-1, 1). That is, as x 1 or as x -1, one of the following holds, either |y1(x)| or |y2(x)| . 3. The only case in which Legendre equation has a bounded solution on [-1, 1] is when the parameter has the form = n(n + 1) with n = 0 or n Z+. In this case either y1 or y2 is a polynomial (the series terminates). This case is considered below.

2. Legendre polynomials

Consider the following problem

Problem. Find the parameters R so that the Legendre equation

(3)

[ (1

-

x2)y]

+

y

=

0,

-1 x 1 .

has a bounded solution. This is a singular Sturm-Liouville problem. It is singular because the function

(1 - x2) equals 0 when x = ?1. For such a problem, we don't need boundary conditions. The boundary conditions are replaced by the boundedness of the solution. As was pointed out in the above remark, the only values of for which we have bounded solutions are = n(n + 1) with n = 0, 1, 2, ? ? ? . These values of are the eigenvalues of the SL-problem.

To understand why this is so, we go back to the construction of the series solutions and look again at the recurrence relations giving the coefficients

(j + 1)j - cj+2 = (j + 2)(j + 1) cj ,

j = 0, 1, 2, ? ? ?

LEGENDRE POLYNOMIALS AND APPLICATIONS

3

If = n(n + 1), then

(n + 1)n - cn+2 = (n + 2)(n + 1) cn = 0.

By repeating the argument, we get cn+4 = 0 and in general cn+2k = 0 for k 1. This means

? if n = 2p (even), the series for y1 terminates at c2p and y1 is a polynomial of degree 2p. The series for y2 is infinite and has radius of convergence equal to 1 and y2 is unbounded.

? If n = 2p + 1 (odd), then the series for y2 terminates at c2p+1 and y2 is a polynomial of degree 2p + 1 while the solution y1 is unbounded.

For = n(n+1), we can rewrite the recurrence relation for a polynomial solution in terms of cn. We have,

(j + 2)(j + 1)

(j + 2)(j + 1)

cj

=

j(j

+ 1) -

n(n

+ 1) cj+2

=

- (n

- j)(n + j

+ 1) cj+2,

for j = n - 2, n - 4, ? ? ? 1 or 0. Equivalently,

(n - 2k + 2)(n - 2k + 1) cn-2k = - (2k)(2n - 2k + 1) cn-2k+2, k = 1, 2, ? ? ? , [n/2].

I will leave it as an exercise to verify the following formula for cn-2k in terms of cn: (-1)k n(n - 1) ? ? ? (n - 2k + 1)

cn-2k = 2k k! (2n - 1)(2n - 3) ? ? ? (2n - 2k + 1) cn .

The polynomial solution is therefore

y(x)

=

[ n/2]

(-1)k 2k k!

(2n

n(n - 1) ? - 1)(2n -

? ? (n - 2k + 1) 3) ? ? ? (2n - 2k

+

1) cnxn-2k

k=0

where cn is an arbitrary constant. The n-th Legendre polynomial Pn(x) is the above polynomial of degree n for the particular value of cn

(2n)! cn = 2n(n!)2 .

This particular value of cn is chosen to make Pn(1) = 1. We have then (after simplification)

Pn(x)

=

1 2n

[ n/2]

(-1)k(2n - 2k)! xn-2k k!(n - k)!(n - 2k)!

.

k=0

Note that if n is even (resp. odd), then the only powers of x involved in Pn are even (resp. odd) and so Pn is an even (resp. odd).

The first six Legendre polynomials are.

P0(x) = 1

P2(x)

=

1 (3x2 2

-

1)

P4(x)

=

1 (35x4 8

-

30x2

+

3)

We have the following proposition.

P1(x) = x

P3(x)

=

1 (5x3 2

- 3x)

P5(x)

=

1 (63x5 8

- 70x3

+ 15x)

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LEGENDRE POLYNOMIALS AND APPLICATIONS

P0 P

2

P1 P3

P4 P6

P5 P7

Proposition. If y(x) is a bounded solution on the interval (-1, 1) of the Legendre equation (1) with = n(n + 1), then there exists a constant K such that

y(x) = KPn(x)

where Pn is the n-th Legendre polynomial.

Remark. When = n(n + 1) a second solution of the Legendre equation, inde-

pendent from Pn, can be found in the form

1

1+x

Qn(x) = 2 Pn(x) ln 1 - x + Rn(x)

where Rn is a polynomial of degree n - 1. The construction of Qn can be achieved by the method of reduction of order. Note that |Qn(x)| as x ?1. The general solution of the Legendre equation is then

y(x) = APn(x) + BnQn(x)

and such a function is bounded on the interval (-1, 1) if and only if B = 0.

3. Rodrigues' formula

The Legendre polynomials can be expressed in a more compact form.

Theorem 1. (Rodrigues' Formula) The n-th Legendre polynomial Pn is given by the following

(4)

Pn(x)

=

1 2n n!

dn dxn

[(x2

-

1)n]

(thus expression (4) gives a solution of (3) with = n(n + 1)).

Proof. Let y = (x2 - 1)n. We have following

Claim. The k-th derivative y(k)(x) of y satisfies the following:

(5)

(1

-

x2

)

d2y(k) dx2

+

2(n

-

k

-

dy(k) 1)x

dx

+

(2n

-

k)(k

+

1)y(k)

=

0.

Proof of the Claim. By induction. For k = 0, y = y(0). We have

y = 2nx(x2 - 1)n-1 (1 - x2)y + 2nxy = 0

and after differentiation, we get (1 - x2)y + 2(n - 1)xy + 2ny = 0

LEGENDRE POLYNOMIALS AND APPLICATIONS

5

So formula (5) holds when k = 0. By induction suppose the (5) holds up to order k - 1. We can rewrite (5) for k - 1 as

(1 - x2) dy(k) + 2(n - k)xy(k) + (2n - k + 1)ky(k-1) = 0. dx

We differentiate to obtain

(1

-

x2)

d2y(k) dx2

+

[2(n

-

k)

-

dy(k) 2]x

dx

+

[2(n

-

k)

+

k(2n

-

k

+

1)]y(k)

=

0.

which is precisely (5).

Now if we let k = n in (5), we obtain

(1

-

x2

)

d2y(n) dx2

-

dy(n) 2x

dx

+

n(n

+

1)y(k)

=

0.

Hence y(n) solves the Legendre equation with = n(n + 1). Since y(n) is a polyno-

mial of degree 2n, then by Proposition 1, it is a multiple of Pn. There is a constant K such that Pn(x) = Ky(n)(x). To complete the proof, we need to find K. For this notice that the coefficient of xn in Pn is (2n)!/(2n (n!)2). The coefficient of xn

in y(n) is that of

dn(x2n) dxn

=

(2n)(2n

-

1) ? ? ? (2n

-

n

+

1)xn

=

(2n)! xn n!

Hence

(2n)! (2n)!

1

K n!

= 2n (n!)2 K = 2n n! .

This completes the proof of the Rodrigues' formula.

A consequence of this formula is the following property between three consecutive Legendre polynomials.

Proposition. The Legendre polynomials satisfy the following

(6)

(2n + 1)Pn(x) = Pn +1(x) - Pn -1(x)

Proof. From Rodrigues' formula we have

Pk (x)

=

d dx

( 1

2k k!

dk dxk

[(x2

) - 1)k]

=

2k 2k k!

dk dxk

[x(x2

-

1)k-1]

=

1 2k-1 (k

- 1)!

dk-1 dxk-1

[ ((2k

- 1)x2

- 1)(x2

- 1)k-2]

For k = n + 1, we get

Pn +1(x)

=

1 2n n!

dn dxn

[ ((2n

+

1)x2

-

1)(x2

-

1)n-1]

From Rodrigues' formula at n - 1, we get

Pn -1(x)

=

d dx

( 1

2n-1(n

-

1)!

dn-1 dxn-1

[(x2

-

) 1)n-1]

=

2n 2n n!

dn dxn

[(x2

-

1)n-1]

As a consequence, we have

Pn +1(x)

-

Pn -1(x)

=

2n + 1 2n n!

dn dxn

[(x2

-

1)n]

=

(2n

+

1)Pn(x)

.

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