Lecture Notes for Math110A on Aug 25, 2014
Last Time:
Key observation:
!
Here,
l
sin
0
X1 (x) = sin
nπx
l
nπx
mπx
sin
dx =
l
l
and
"
0,
if n != m
l/2 if n = m
X2 (x) = sin
X !! (x) = λX(x),
mπx
l
are both eigenfunctions of:
0<x<l
The eigenfunctions are “orthogonal” to each other.
§ 5.3 Orthogonality and General Fourier Series
inner product: If f and g are functions defined on (a, b), then their inner product
is defined below:
!
b
(f, g) =
f (x)g(x)dx
a
f and g are orthogonal if
(f, g) = 0
X !! (x) = λX(x),
0<x<l
Both X1 and X2 are eigenvectors. Why are they orthogonal?
Key: Green’s second identity:
!
b
a
#b
#
(−X1 X2 + X2 X1 )dx = (−X1 X2 + X2 X1 )##
!!
!!
!
!
a
1
How about the r.h.s.?
#b
#
(−X1 X2 + X2 X1 )## = 0?
!
!
(∗)
a
There are cases where (∗) is indeed zero:
• Dirichlet:
X(a) = X(b) = 0
• Neumann:
X ! (a) = X ! (b) = 0
• Periodic:
In general,
X ! (a) = X ! (b)
X(a) = X(b),
#b
#
(−X1 X2 + X2 X1 )## != 0
!
!
a
However, if
#b
#
(−X1 X2 + X2 X1 )## = 0
!
!
a
we say that the corresponding boundary conditions are symmetric.
Theorem 1. If you have symmetric boundary conditions, then any two eigenfunctions with different eigenvalues are orthogonal.
2
What if we have complex-valued eigenvalues and eigenfunctions?
Define inner product:
! b
(f, g) =
f (x)g(x)dx
a
Green’s second identity:
!
b
a
#b
#
(−X1 X2 + X2 X1 )dx = (−X1 X2 + X2 X1 )##
!!
!!
!
!
a
Boundary condition is symmetric or hermitian if
#b
#
!
!
(−X1 X2 + X2 X1 )## = 0
a
Theorem 2. Under symmetric boundary conditions, all eigenvalues are real
numbers.
Theorem 3. Under symmetric boundary conditions, if
#b
#
!
f (x)f (x)## ≤ 0
a
for all f (x) satisfying the boundary conditions, then there is no negative eigenvalues.
3
§ 5.4, 5.5. Completeness and Gibbs Phenomenon (Not Required)
Sin Coeff
150
Cos Coeff
2.5
1
2
100
1.5
0.5
1
50
0.5
0
0
0
−0.5
−50
−1
−0.5
−1.5
−100
−2
−150
−1
−0.5
0
0.5
−2.5
−1
1
−0.5
0
0.5
1
−1
−1
−0.5
Sin Coeff
3.5
0
0.5
1
Cos Coeff
0.8
1.4
0.6
1.2
3
1
0.4
0.8
2.5
0.2
0.6
2
0
0.4
−0.2
1.5
0.2
−0.4
0
1
−0.6
0.5
−1
−0.5
0
0.5
1
−0.8
−1
−0.2
−0.5
0
0.5
1
−0.4
−1
−0.5
0
0.5
1
Three notions of convergence:
$N
Pointwise Convergence:
n=1 fn (x) converges to f (x) pointwisely in (a, b),
if for any a < x < b we have
#
#
N
#
#
%
#
#
f
(x)
−
f
(x)
as N → ∞
#
#→0
n
#
#
n=1
$
Uniform Convergence: N
n=1 fn (x) converges to f (x) uniformly in [a, b], if
#
#
N
#
#
%
#
#
max #f (x) −
fn (x)# → 0
as N → ∞
a≤x≤b #
#
n=1
Sin Coeff
Cos Coeff
1.5
1.2
1.2
0.5
1
1.1
1
0.8
1
0.6
0.9
0.5
0
0.8
0.4
0.7
0.2
0
0.6
0
0.5
0.4
−0.5
0
0.5
1
−0.2
0
0.5
1
4
−0.5
0
0.5
1
0
0.5
1
$N
Mean-Square (or L2 ) Convergence:
n=1 fn (x) converges in the meansquare (or L2 ) sense to f (x) in (a, b),
#2
! b ##
N
#
%
#
#
as N → ∞
fn (x)# dx → 0
#=
#f (x) −
#
a #
n=1
3.5
3.5
3.5
3
3
3
3.5
3
2.5
2.5
2.5
2.5
2
2
2
2
1.5
1.5
1.5
1.5
1
1
1
1
0.5
−1
0
1
0.5
−1
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1
0
1
−1
−1
N =7
0
1
0.5
−1
0
0.5
0.5
−1
0
1
0
1
0.4
0.2
0
0
−0.2
0
1
−0.5
−1
0
N = 70
# = 0.0079
1
1
−0.4
−1
N = 350
# = 0.0027
N = 700
# = 0.0012
# = 0.00068
Convergence Theorems:
Theorem (Pointwise Convergence):
#%
#
#
#
An Xn (x) − f (x)# → 0
#
for all a < x < b
if f (x), f ! (x) are continuous on a ≤ x ≤ b.
Theorem (Uniform Convergence):
#%
#
#
#
max #
An Xn (x) − f (x)# → 0
a≤x≤b
"
if
(i) f (x), f ! (x), f !! (x) exist, and are continuous on a ≤ x ≤ b.
(ii) f (x) satisfies the given boundary conditions.
Theorem (L2 Convergence):
! b
If
|f (x)|2 dx < ∞,
then
a
5
!
b
a
|
%
An Xn (x) − f (x)|2 dx
L2 space contains all the functions f (x), with
!
b
a
|f (x)|2 dx < ∞
L2 norm
'f ' =
&
(f, f ) =
'
!
b
a
|f (x)|2 dx
L2 metric
&
'f − g' = (f − g, f − g) =
'
!
b
a
|f (x) − g(x)|2 dx
To prove Theorem (L2 Convergence), we need:
• L2 space is complete. (Hilbert space/Banach space)
• The eigenfunctions (trignometric functions) are dense in L2 .
• Least-Square Approximation.
Parseval’s Equality
%
2
2
|An | 'Xn ' =
!
b
a
|f (x)|2 dx
L2 Convergence of the Fourier series is equivalent with the Parseval’s Equality
6
To prove Theorem (Pointwise Convergence)
Key obeservation
SN (x) =
=
!
!
π
−π
π
−π
with
(
1+2
N
%
)
(cos ny cos nx + sin ny sin nx) f (x)
n=1
KN (x − y)f (y)
dy
2π
KN (θ) = 1 + 2
N
%
cos nθ =
n=1
sin(N + 12 )θ
sin 12 θ
Dirichlet Kernel
50
40
30
20
10
0
−10
−4
−3
−2
−1
0
1
7
2
3
4
dy
2π
To prove Theorem (Uniform Convergence)
Key obervation
If An Bn are Fourier coefficients of f (x); and A! B ! are Fourier coefficients of
f (x), then
1
1
An = − Bn! ,
and Bn = An
n
n
!
8