Computational Physics - Department of Physics

5.3 Gaussian Quadrature 119

∫

f(x)dx≈

∫

P 2 N− 1 (x)dx=

N− 1

∑

i= 0

P 2 N− 1 (xi)ωi,

The reason why we can represent a functionf(x)with a polynomial of degree 2 N− 1 is due

to the fact that we have 2 Nequations,Nfor the mesh points andNfor the weights.

The mesh points are the zeros of the chosen orthogonal polynomialof orderN, and the
weights are determined from the inverse of a matrix. An orthogonal polynomials of degreeN
defined in an interval[a,b]has preciselyNdistinct zeros on the open interval(a,b).
Before we detail how to obtain mesh points and weights with orthogonal polynomials, let
us revisit some features of orthogonal polynomials by specializing to Legendre polynomials.
In the text below, we reserve hereafter the labellingLNfor a Legendre polynomial of orderN,
whilePNis an arbitrary polynomial of orderN. These polynomials form then the basis for the
Gauss-Legendre method.

5.3.1 Orthogonal polynomials, Legendre

The Legendre polynomials are the solutions of an important differential equation in Science,
namely

C( 1 −x^2 )P−m^2 lP+ ( 1 −x^2 )d

dx

(

( 1 −x^2 )dP

dx

)

= 0.

HereCis a constant. Forml= 0 we obtain the Legendre polynomials as solutions, whereas
ml 6 = 0 yields the so-called associated Legendre polynomials. This differential equation arises
in for example the solution of the angular dependence of Schrödinger’s equation with spher-
ically symmetric potentials such as the Coulomb potential.
The corresponding polynomialsPare

Lk(x) =

1

2 kk!

dk

dxk

(x^2 − 1 )k k= 0 , 1 , 2 ,...,

which, up to a factor, are the Legendre polynomialsLk. The latter fulfil the orthogonality
relation ∫ 1

− 1

Li(x)Lj(x)dx=

2

2 i+ 1

δi j, (5.10)

and the recursion relation

(j+ 1 )Lj+ 1 (x)+jLj− 1 (x)−( 2 j+ 1 )xLj(x) = 0. (5.11)

It is common to choose the normalization condition

LN( 1 ) = 1.

With these equations we can determine a Legendre polynomialof arbitrary order with input
polynomials of orderN− 1 andN− 2.
As an example, consider the determination ofL 0 ,L 1 andL 2. We have that

L 0 (x) =c,

withca constant. Using the normalization equationL 0 ( 1 ) = 1 we get that

L 0 (x) = 1.