legendrepolynomials2.pdf

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Legendre Polynomials

Vladimir ZakharovDepartment of MathematicsUniversity of Arizona

Lectures in Mathematics

The following lecture introduces the Legendre polynomials.It includes their derivation, and the topics of orthogonality,

normalization, and recursion.

I. General Formula

We start with a solution to the Laplace equation in 3 - dimensional space:

U = 0 (1)

A solution is:

U = 1r r 0 r 0 = 0 (2)

Let us introduce the spherical coordinate system. The z-axis is along r 0.In this system the axial symmetric Laplace equation looks as follows:

1r 2 r r 2 U r + 1r 2 1sin sin( ) U (3)


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Solution (2) is:

U = 11 rx z2

(4)

x = cos

Equation (3) can be rewritten as follows:1r 2

r r 2 U

r +1r 2 x

(1- x2) U x 0 (5)

For (r


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a k 2 = -n k n k 1

k 2 k 1 a k (10)

Condition (10) is satisfied if P n(x) is presented by the series

P n(x) =12n k 0

1 k 2 n 2 k k n k n 2 k x

n 2 k (11)

The series (11) is terminated as far as one of the factorials in thedenominator becomes infinite

Remember that

n! = 0 t n 1 e t t

Hence, n! = , if n is a negative integer

Relation (10), proving satisfaction of equation (8) can be checkedby the use of (11) immediately.To be sure that P(1) = 1, we formulate the following:

TheoremP n(x) can be presented by the Rodriges formula:

P n =1

2n n

n

xn x2 1

n(12)

Indeed x2 1n= k 0

nk n k 1

k x2 n k

n

xn x2 n k = 2 n 2 k n 2 k x

2 n k

By comparison with (11) we obtain the desired result.Then,

P n(x) = 12n n2

xn x 1 n x 1 n (13)Let x 1, In this limit we should differentiate only x 1 n,otherwise we get zero.

P n(1) =1

2n n n! 1 1n 1

Legendre Polynomials.nb 3


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II. Orthogonality

Let us consider equations for P n , P m(14)

x x 2 1

P n

x n n 1 P n 0 P n

x x 2 1

P m

x m m 1 P m 0 P m

14

From (14) we get

P m x x2 1 P n x - P n x x

2 1P m

xn n 1 m m 1 P n P m

1. Orthogonality Let P n (x), P m |y) - two polynomials, satisfy the following equations:

dxx 2 1

P n

xn n 1 P n 0 (1)

x x2 1 P m x m m 1 P m 0 (2)

Let us multiply (1) by P m and (2) by P n and subtract, we get:

P m x x2 1 P n x - P n x x

2 1 P m x + [n(n + 1) - m(m+1)] P n P m = 0 (3)

Equation (3) can be presented as follows:

dxx 2 1 (P m

P n

x- P n

P m x ) +[n(n + 1) - m(m+1)] P n P m = 0 (4)

After integration of (4) by x we get:

1 x 2 (P mP n

x- P n

P m x )

1

1+ [n(n + 1) - m(m+1)] 1

1 P n P m x = 0

Hence, 11 P n P m x 0



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2. Normalization

U(x,r) = 11 2 xr r 2

= n 0 r nP n(x)

U 2(x,r) = 1

1 2 xrr 2

= n 0 m 0r n mP n(x)P m(x)

After integration by x we get:

11 U 2(x,r)dx = m 0r

2 n 11 P n2(x)dx

11 U 2dx = 1

1 x1 2 xr r 2

12 r 1

1 x x 12 r

12

= 12 r ln( x12 r

12 )

1

1= - 12 ln

1 r 1 r

2

= 1r (ln(1+r) - ln(1-r)) =2r (r +

r 3

3r 5

5 + ...) = 2 n 0r 2 n

2 n 1

Hence,

11 P n2 x x 22 n 1

3. Recursion Relations

U = 11 2 xr r 2

1

2

U

dr

x r

1 2 xr r 2U

1 2 xr r 2 ) Udr = (x-r)UU = r 0 r

nP n(x)U

dr= n 0nr

n 1P n(x) = n 0 n 1 r nP n 1(x)

r Udr

= nrnP n(x)

r 2 Udr

= n 0nrn 1P n(x) = n 1 n 1 r

nP n 1(x)

rU = n 0r n 1P n(x) = n 1r

nP n 1(x)

Collecting all together we get:n 1 n 1 P n 1 - x((2n+1) P n+nP n 1 x r

n = 0

Finally, we get the recursion relation

n 1 P n 1 x 2 n 1 P n n P n 1 x 0



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Then,U

dx= r

1 2 xr r 2U

U

dr

x r

1 2 xr r 2 U(x-r) U

dx= r U

dr

r Udr

= nrnP n(x)U

dx= r nP n '(x)

x Udx

= r nxPn '(x)

-r Udr

= r n 1P n '(x) = r nP n 1 '(x)

Thus, we getP n ' xPn 1 ' nPn 1


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