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
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
