100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes $7.99
Add to cart

Class notes

Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes

 3 views  0 purchase
  • Course
  • Institution

Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes

Preview 2 out of 7  pages

  • December 12, 2022
  • 7
  • 2019/2020
  • Class notes
  • Justin ko
  • All classes
avatar-seller
July 6, 2020 APM346 – Week 7 Justin Ko


1 Eigenvalue Problems
We want to study the wave and heat equation on the finite interval. Before we do that, we review
some self contained background material that will be used when we do separation of variables.
We begin introducing a class of Sturm–Liouville eigenvalue problems. Consider the second order
ODE on [a, b] subject to some boundary conditions

00
−X (x) = λX(x)
 a<x<b
0 0 (1)
a1 X(a) + b1 X(b) + c1 X (a) + d1 X (b) = 0
0 0

a2 X(a) + b2 X(b) + c2 X (a) + d2 X (b) = 0


where a1 , . . . , d2 ∈ R. A non-trivial solution X to (1) is called an eigenfunction, and the corresponding
value of λ is called an eigenvalue.
Remark 1. This terminology should remind you of a concept from linear algebra. Recall that the
eigenvalues λ and eigenvectors v 6= 0 of a matrix A ∈ Rn×n are solutions to

Av = λv.
2
d
Since we are in finite dimensions, there are at most n eigenvalues. We can think of L = − dx as a
linear operator on X. In this context, solutions to the ODE in (1) satisfy

LX = λX.

In this “infinite” dimensional case, there are infinitely many eigenvalues.

1.1 Common Eigenvalue Problems
We summarize the eigenfunctions and eigenvalues of several common eigenvalue problems.

1. Dirichlet Boundary Conditions:
(
−X 00 (x) = λX(x) 0<x<L
(2)
X(0) = X(L) = 0

Eigenvalues: λn = ( nπ 2
L ) for n ≥ 1
Eigenfunctions: Xn = sin( nπx
L ) for n ≥ 1

2. Neumann Boundary Conditions:
(
−X 00 (x) = λX(x) 0<x<L
(3)
X 0 (0) = X 0 (L) = 0

Eigenvalues: λn = ( nπ 2
L ) for n ≥ 0
Eigenfunctions: Xn = cos( nπx
L ) for n ≥ 1 and X0 =
1
2 for n = 0.
3. Periodic Boundary Conditions:
(
−X 00 (x) = λX(x) −L < x < L
(4)
X(−L) − X(L) = X 0 (−L) − X 0 (L) = 0

Eigenvalues: λn = ( nπ 2
L ) for n ≥ 0
Eigenfunctions: Xn = cos( nπx nπx
L ) and Yn = sin( L ) for n ≥ 1 and X0 =
1
2 for n = 0.


Page 1 of 7

, July 6, 2020 APM346 – Week 7 Justin Ko


4. Dirichlet–Neumann Mixed Boundary Conditions:
(
−X 00 (x) = λX(x) 0<x<L
(5)
X(0) = X 0 (L) = 0

Eigenvalues: λn = ( (2n+1)π
2L )2 for n ≥ 0
Eigenfunctions: Xn = sin( (2n+1)πx
2L ) for n ≥ 0.
5. Dirichlet–Neumann Mixed Boundary Conditions:
(
−X 00 (x) = λX(x) 0<x<L
(6)
X 0 (0) = X(L) = 0

Eigenvalues: λn = ( (2n+1)π
2L )2 for n ≥ 0
Eigenfunctions: Xn = cos( (2n+1)πx
2L ) for n ≥ 0.
Remark 2. Notice that if X is an eigenfunction of (1), then cX is also an eigenfunction for any
number c 6= 0. This means that the eigenfunctions in the table are unique up to a scaling factor.

1.2 Orthogonality of Eigenfunctions
Definition 1. Consider continuous functions f, g defined on [a, b]. The L2 -inner product of these
functions are given by
Z b
hf, gi = f (x)g(x) dx.
a
We say that the functions f and g are orthogonal if
hf, gi = 0.
Definition 2. The boundary conditions of (1) are symmetric if
 x=b
f 0 (x)g(x) − f (x)g 0 (x) = f 0 (b)g(b) − f (b)g 0 (b) − f 0 (a)g(a) + f (a)g 0 (a) = 0. (7)
x=a
for functions f and g that solve (1). All the standard eigenvalue problems we encounter in this course
will have symmetric boundary conditions.
Theorem 1 (Orthogonality of Eigenfunctions)
If the eigenvalue problem (1) has symmetric boundary conditions, then the eigenfunctions corre-
sponding to distinct eigenvalues are orthogonal.

Proof. Let X1 and X2 be distinct solutions to (1), that is for λ1 6= λ2 ,
−X100 = λ1 X1 and − X200 = λ2 X2 .
We can check orthogonality directly,
Z b Z b
(λ2 − λ1 )hX1 , X2 i = (λ2 − λ1 ) X1 (x)X2 (x) dx = X100 (x)X2 (x) − X1 (x)X200 (x) dx.
a a
Integrating by parts implies
Z b x=b
X100 (x)X2 (x) − X1 (x)X200 (x) dx = X10 (x)X2 (x) − X1 (x)X20 (x)

=0
a x=a

because X1 and X2 satisfy the symmetric boundary condition (7). Since λ1 − λ2 6= 0, hX1 , X2 i = 0 so
X1 and X2 are orthogonal.
Remark 3. We can have distinct eigenfunctions for repeated eigenvalue. They might not be orthog-
onal, but we can use the Gram–Schmidt process extract a orthogonal set.


Page 2 of 7

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller 9kfhgia89h1. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $7.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

52355 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$7.99
  • (0)
Add to cart
Added