1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

340 Chapter 5 Higher Dimensions and Other Coordinates

Theorem. If f(x)is sectionally smooth on the interval− 1 <x< 1 ,thenatevery
point of that interval the Legendre series of f is convergent, and

∑∞

n= 0

bnPn(x)=f(x+)+f(x−)

2

.

From Eq. (10) for the coefficient of a Legendre series and from the fact that
the Legendre polynomials are odd or even, we see that an odd function will
have only odd-indexed coefficients that are nonzero, and an even function will
have only even-indexed coefficients that are nonzero. Furthermore, if a func-
tionfisgivenontheinterval0<x<1, then its odd and even extensions have
odd and even Legendre series, andfis represented by either in that interval:

f(x)=

∑

neven

bnPn(x), 0 <x< 1 ,

bn=( 2 n+ 1 )

∫ 1

0

f(x)Pn(x)dx (neven), (11)

f(x)=

∑

nodd

bnPn(x), 0 <x< 1 ,

bn=( 2 n+ 1 )

∫ 1

0

f(x)Pn(x)dx (nodd). (12)

Because thePn(x)are polynomials, the integral equation (10) for any spe-
cific coefficient can be done in closed form for a variety of functionsf(x).
However, gettinganas a function ofnis not so easy. Fortunately, some ele-
mentary integrals can be done using the differential equation
(
( 1 −x^2 )Pn′

)′

+n(n+ 1 )Pn= 0.

(1) First, separate the two terms of the differential equation and integrate:

n(n+ 1 )

∫

Pn(x)dx=

∫

−

(

( 1 −x^2 )Pn′

)′

dx

=−

(

1 −x^2

)

Pn′(x).

This equation may be solved for the integral ifn=0.
(2) Now multiply through the differential equation byx, separate terms, and
integrate:

n(n+ 1 )

∫

xPn(x)dx=

∫

−x

(

( 1 −x^2 )Pn′

)′

dx

=−x

(

1 −x^2

)

Pn′+

∫ (

1 −x^2

)

Pn′dx

=−x

(

1 −x^2

)

Pn′+

(

1 −x^2

)

Pn−

∫

(− 2 x)Pndx.