Chapter 3 - Orthogonal Functions and Sturm-Liouville Eigenvalue Problems

Question 1

The Boundary Value Problem

Answer 1

y’’ + λy = 0
y(0) = 0
y(1) = 0

Question 2

The Boundary Value Problem

Case 1 : λ= -k²<0

Answer 2

-here the general solution is:
y(x) = c1*e^(kx) + c2*e^(-kx)
-boundary conditions imply:
y(0) = c1 + c2 = 0
y(1)  = c1e^k + c2e^(-k)
=> c1 = c2 = 0
-since e^(2k) has no real solutions for real k≠0
-therefore the only solution is the trivial solution:
y(x) = 0

Question 3

The Boundary Value Problem

Case 2: λ = 0

Answer 3

-here the general solution is:
y(x)  = c1 + c2*x
-boundary conditions imply:
y(0) = c1 = 0
y(1) = c1 + c2 = 0
=> c1 = c2 = 0
-the only solution is the trivial solution:
y(x) = 0

Question 4

The Boundary Value Problem

Case 3 : λ = ω²>0

Answer 4

-here the general solution is:
y(x) = c1*cos(ωx) + c2*sin(ωx)
-boundary conditions imply:
y(0) = c1 = 0
so:
y(1) = c2*sin(ω) = 0
-to avoid the trivial solution (c2=0), we must CHOOSE ω to satisfy sinω=0
-this means:
ω = nπ, for some integer n
-so λn = ω² = n²π²

Question 5

Eigenvalues and Eigenfunctions Lemma

Answer 5

-the boundary value problem has nontrivial solutions only for an infinite sequence of values λ labelled λn
-for each eigenvalue λn, there is a unique eigenfunction yn given by:
λn = n²π² and yn = sin(nπx)
-for n ≥1

Question 6

The Lagrange Identity

Answer 6

LHS = d/dx(yn*ym' - ym*yn') 
-use the product rule:
= yn*ym'' - ym*yn''
-sub in using the BVP:
= (λn-λm) yn ym = RHS
-integrate both sides over interval [0,1] and use the boundary conditions of the BVP:
(λn-λm) ∫ yn ym dx
= [ynym' - ymyn']|0,1 = 0
-therefore λn≠λm =>
 ∫ yn ym dx = 0 for m≠n

Question 7

Inner Product Space

Answer 7

-we interpret the orthogonality relation and lagrange identity as an inner product on the linear space of functions L, which are smooth and satisfy:
∫ (f(x))² dx < ∞
-where the integral is taken from 0 to 1

Question 8

Inner Product

Answer 8

-on the inner product space we define:
⟨f,g⟩ = ∫ f(x) g(x) dx
-we see that with respect to this inner product:
⟨ym,yn⟩ = 0 for m≠n

Question 9

Orthogonality and Inner Product Formula

Answer 9

⟨ym,yn⟩ = 1/2 * 𝛿m,n
-where 𝛿m,n is the Kroneker delta:
𝛿mn = {0, for m≠n 1, for m=n

Question 10

Sturm-Liouville Form

Starting Form

Answer 10

A(x)y’’ + B(x)y’ + (C(x)+λD(x))y = 0

• where A, B, C and D are some functions of x
• and λ is some parameter
• any equation in this form can be put into Sturm-Liouville form

Question 11

Sturm-Liouville Form

Form

Answer 11

d/dx (p(x)*dy/dx) + q(x)y + λσ(x)y = 0

Question 12

Putting an Equation into Sturm-Liouville Form

Answer 12

-any equation of the form:
A(x)y'' + B(x)y' + (C(x)+λD(x))y = 0
-can be put into Sturm-Liouville form
-multiply by an integrating factor R(x)
-use the product rule to rewrite the first term R(x)A(x)y'' as (RAy')'-(RA)'y'
-choose R(x) in order to eliminate the coefficient of y'(x):
R(x)B(x) - d/dx[R(x)A(x)] = 0
=>
R'/R = B/A - A'/A
R(x) = 1/A(x) * exp(∫B/A dx)
-giving Sturm-Liouville form:
d/dx (p(x)*dy/dx) + q(x)y + λσ(x)y = 0
-where:
p(x) = R(x)A(x)
q(x) = R(x)C(x)
σ(x) = R(x)D(x)

Question 13

Sturm-Liouville Eigenvalue Problem

Standard Form

Answer 13

-we want to consider the eigenvalue problem (BVP) associated with the general equation of Sturm-Liouville type:
d/dx (p*dy/dx) + qy + λσy = 0
a0 for all x in the range a≤x≤b and λ interpreted as the eigenvalue

Question 14

Sturm-Liouville Eigenvalue Problem

Equation

Answer 14

-let ym and yn be the be eigenfunctions of:
d/dx (pdy/dx) + qy + λσy = 0
-then using the Lagrange Identity;
(λn-λm) ∫σymyn dx
= [p * (ynym’ - ym*yn’)]|a,b
-where the integral is evaluated between a and b
-there are two conditions which would make the RHS vanish:
i) α1yn(a) + α2yn’(a) = 0
AND
β1yn(b) + β2yn’(b) = 0

OR
ii) p(x) satisfies p(a)=p(b)=0
-if one of these conditions holds then we have:
λm≠λn 
=> ∫σ*ym*yn dx = 0

Question 15

Sturm-Liouville Eigenvalue Problem

Inner Product Space

Answer 15

-we have chosen a and b so that RHS=0, then
λm≠λn =>  ∫σ*ym*yn dx = 0
-we again interpret this as an inner product on the linear space of functions L which are smooth and satisfy:
∫σ*ym*yn dx < ∞
-on space we define the inner product:
⟨f,g⟩ = ∫σ*f(x)*g(x) dx
-then:
⟨ym,yn⟩ = ∫σ*ym*yn dx = 0 
for m≠n

Question 16

Legendre’s Equation

Sturm-Liouville Form

Answer 16

d/dx [(1-x²)y’] + λy = 0, -1 0 for all x

-and (a.b) = (-1,1) to avoid the singular points x=±1

Question 17

What is a generating function?

Answer 17

• a generating function is a way of encoding an infinite sequence of numbers an, by treating them as coefficients of a power series
• the sum of this infinite series is the generating function

Question 18

The Legendre Polynomial

Generating Function

Answer 18

-the Legendre polynomial Pn is the coefficient of t^n in the following expansion:
G(x,t) = (1-2tx+t²)^(-1/2)
= ΣPn(x)*t^n
-using a binomial expansion, the first few terms are easily computed:
G(x,t) = 1 + xt + 1/2(3x²-1)t² + 1/2(5x³-3x)t³ + ….
-giving the first few Legendre polynomials as:
Po = 1
P1 = x
P2 = 1/2(3x²-1)
P3 = 1/2(5x³-3x)

Question 19

2nd Order Recurrence Relation

Formula

Answer 19

(n+1) P = (2n+1)xP-n*P

-where <> indicates subscript

Question 20

2nd Order Recurrence Relation

Derivation

Answer 20

• start with the generating formula and differentiate with respect to t
• write G’ in terms of G
• then sub in G=ΣP(x)*t^n
• sub in for coefficients of t^n, this is the recurrence relation

Question 21

2nd Order Recurrence Relation

Initial Conditions

Answer 21

-since this is a 2nd order relation we need two initial conditions, Po and P1
-from the binomial expansion of the generating function we already have Po=1
-sub this in to the recurrence relation for n=0:
P1 = x*Po = x
-continue subbing in for P2, P3, … etc.

Question 22

Inner Product of Legendre’s Polynomials

Answer 22

-from general Sturm-Liouville theory, we have:
⟨f,g⟩ = ∫f(x)g(x) dx
-where the integral is taken between -1 and 1
Pn = Eigenfunction
λn = n(n+1) = eigenvalue
λn≠λm if n≠m
-but what about n=m?
-we can show that in general:
⟨Pn,Pn⟩ = 2/(2n+1)
-by induction

Question 23

Legendre Polynomials

Rodrigues’ Formula

Answer 23

-the Legendre Polynomials can also be generated by the formula:
Pn = 1/(n!*2^n) d’n/dx^n (x²-1)^n
-known as Rodrigues’ Formula
-this is straight forward when n is small, just sub in n=1, 2, 3 etc.

Question 24

Expansions in Terms of Legendre Polynomials

Answer 24

-given a function f(x), its Taylor Expansion about x=0 is:
f(x) = Σ f’n (0) * x^n/n!
-the formulae for the Legendre Polynomials
Po=1, P1=x, P2=1/2(3x²-1), …
-these formulae can be inverted:
1 = Po , x=P1, x² = 1/3(2P2+Po), …
-this is a change of basis so the Taylor Expansion can be replaced by:
f(x) = an*Pn(x)
-to find coeffiecients an, ise the orthogonality relations:
⟨Pn,Pm⟩ = 2/(2n+1) * 𝛿n,m
-giving:
am = (2m+1)/2 * ⟨f,Pn⟩
= (2m+1)/2 * ∫ f(x) * Pm(x) dx
-where the integral is taken from -1 to +1

Question 25

Hermite Polynomials

Sturm-Liouville Form

Answer 25

-Hermite's equation has Sturm-Liouville form:
d/dx (e^(-x²/2) dy/dx) 
\+ λ*e^(-x²/2) * y
-so σ(x) = e^(-x²/2)
-and the range (a,b) is chosen as (-∞,∞)

Question 26

Hermite Polynomials

Inner Product

Answer 26

⟨f,g⟩ = ∫ f(x)g(x)e^(-x²/2) dx

-where the integral is taken between -∞ and ∞

Question 27

Hermite Polynomials

Generating Function

Answer 27

G(x,t) = e^(xt-t²/2)

= Σ Hn(x) * t^n/n!

Question 28

Hermite Polynomials

Rodrigues’ Formula

Answer 28

-the Hermite polynomials can also be generated using Rodrigues’ formula:
Hn = (-1)^n * e^(x²/2) * d^n/dx^n[e^(-x²/2)]

Question 29

Hermite Polynomials

Formula for ⟨Hn,Hn⟩

Answer 29

⟨Hn,Hn⟩ = n! *⟨Ho,Ho⟩
-where:
⟨Ho,Ho⟩ = ∫ e^(x²/2) dx = √(2π)

Question 30

The Boundary Value Problem

Eigenvalues and Eigenfunctions

Answer 30

λn = n²π²
yn = sin(nπx)