ME 2002 (502 20002) Engineering Mathematics (II)
Spring, 2024
Topic 1. Orthogonal Functions and Fourier Series
Instructor: Kuo-Long Pan
Contents
1. Orthogonal Functions
2. Fourier Series
3. Fourier Cosine and Sine Series
4. Complex Fourier Series
5. Sturm-Liouville Problems
6. Bessel and Legendre Series (optional)
Reading
D. G. Zill, Advanced Engineering Mathematics, 7th Ed., Chap. 12, Sec. 12.1 − 12.6.
1. Orthogonal Functions
Definition 12.1.1 Inner Product of Functions
The inner product of two functions f1 and f2 on an interval [a, b] is the number
b
( f1 , f 2 )   f1 ( x) f 2 ( x)dx
a
Definition 12.1.2 Orthogonal Functions
Two functions f1 and f2 are said to be orthogonal on an interval [a, b] if
b
( f1 , f 2 )   f1 ( x) f 2 ( x) dx  0
a
Example: 1
Definition 12.1.3 Orthogonal Set
A set of real-valued functions {0(x), 1(x), 2(x), …} is said to be orthogonal on an interval [a,
b] if
b
(m , n )   m ( x) n ( x) dx  0, m  n
a
1
 Orthonormal Sets
The expression (u, u) = ||u||2 is called the square norm. Thus we can define the square norm
of a function as
b
n ( x)   n2 dx and n ( x) 
2

a
b
a
n2 ( x)dx
If {n(x)} is an orthogonal set on [a, b] with the property that ||n(x)|| = 1 for all n, then {n(x)}
is called an orthonormal set on [a, b].
Examples: 2, 3
 Vector Analogy
Recalling from the vectors in 3-space that
u  c1v1  c2 v 2  c3 v 3
we have
u
(u, v1 )
(u, v 2 )
(u, v 3 )
v 
v2 
v3
2 1
2
|| v1 ||
|| v 2 ||
|| v 3 ||2
3
(u, v n )
v
2 n
n 1 || v n ||

Thus we can make an analogy between vectors and functions.
Orthogonal Series Expansion
Suppose {n(x)} is an orthogonal set on [a, b]. If f(x) is defined on [a, b], we first write as
f ( x)  c00 ( x)  c11 ( x) 
 cnn ( x) 
?
Then

b
a
b
b
a
a
b
f ( x)m ( x) dx  c0  0 ( x)m ( x) dx  c1  1 ( x)m ( x) dx 
 c0 (0 , m )  c1 (1 , m ) 
 cn  n ( x)m ( x) dx 
a
 cn (n , m ) 
Since {n(x)} is an orthogonal set on [a, b], each term on the right-hand side is 0 except m = n.
In this case we have

b
a
b
f ( x)n ( x)dx  cn (n , n )  cn  n 2 ( x)dx
cn 
a

b
a
f ( x)n ( x)dx
b

a
n
2
, n  0, 1, 2,...
( x)dx
2
In other words,

f ( x)   cnn ( x), cn


b
a
n 0
f ( x)n ( x) dx
|| n ( x) ||2
Definition 12.1.4 Orthogonal Set/Weight Function
A set of real-valued functions {0(x), 1(x), 2(x), …} is said to be orthogonal with respect to a
weight function w(x) on [a, b], if

b
a
w( x)m ( x)n ( x)dx  0, m  n
Under the condition of the above definition, we have
cn


b
f ( x) w( x)n ( x) dx
a
|| n ( x) ||2
b
, || n ( x) ||2   w( x)n2 ( x) dx
a
 Complete Sets
An orthogonal set is complete if the only continuous function orthogonal to each member of
the set is the zero function.
2. Trigonometric Series
We can show that the set

x 
 2 x 
 3 x 
, cos 
, cos 
1, cos 


,
p
p
p







x
 2 x 
 3 x 
, sin 
, sin 
, sin 


,
p
p
p






is orthogonal on [−p, p]. Thus a function f defined on [−p, p] can be written as
f ( x) 
 n x 
 n x  
a0  
   an cos 
  bn sin 

2 n 1 
 p 
 p 
Now we calculate the coefficients.

p
p
Since cos⁡(
𝑛𝜋𝑥
𝑝
f ( x)dx 
) and sin⁡(
 
p
p
 n x 
 n x  
a0 p
dx

an  cos 
dx  bn  sin 



dx 

p
p
2 p
n 1 
 p 
 p  
𝑛𝜋𝑥
𝑝
)are orthogonal to 1 on this interval, then (3) becomes
3




p
p
p
a p
a
f ( x)dx  0  dx  0 x  pa0
2 p
2 p
Thus we have
1 p
f ( x)dx
p  p
a0 
In addition,

p
p

 m x 
f ( x)cos 
dx
 p 

p
p

a0 p
 m x 
 m x   n x 
 m x   n x  
cos
dx

a
cos
cos
dx

b
cos

n
n







 p  p   p 
 p  p  sin  p  dx 
2  p
 p 
n 1 
By orthogonality we have
p
 m x 
 p cos  p  dx  0, m  0,
And
 m x   n x 
cos 
 sin 
 dx  0
p
 p   p 

p
 0, m  n
 m x   n x 
cos
cos
dx

 p  p   p   p, m  n
p
Thus reduces to

p
p
So
an 
 n x 
f ( x)cos 
 dx  an p
 p 
 n x 
1 p
f ( x)cos 
 dx

p p
 p 
Likewise
 m x 
sin
 p  p  dx  0, m  0,
p
 m x   n x 
sin
 p  p  cos  p  dx  0
p
And
 0, m  n
 m x   n x 
sin
sin
dx

 p  p   p   p, m  n
p
we find that
bn 
1 p
 n x 
f ( x)sin 
 dx


p
p
 p 
Definition 12.2.1 Fourier Series
The Fourier series of a function f defined on the interval (−p, p) is given by
 n x 
 n x  
a0  
f ( x)     an cos 

b
sin
n

 p 
2 n1 
 p 


a0 
1 p
1 p
n
1 p
n
f
(
x
)
dx
,
a

f
(
x
)
cos
x
dx
,
b

f
(
x
)
sin
x dx
n
n
p  p
p  p
p
p  p
p
4
Example: 1
Theorem 12.2.1 Conditions for Convergence
Let f and f’ be piecewise continuous on the interval (−p, p); that is, let f and f’ be continuous
except at a finite number of points in the interval and have only finite discontinuities at these
points. Then the Fourier series of f on the interval converge to f(x) at a point of continuity.
At a point of discontinuity, the Fourier series converges to the average
f ( x  )  f ( x )
2
where f(x+) and f(x－) denote the limit of f at x from the right and from the left, respectively.
Example: 2
 Periodic Extension
Fig 12.2.2 is the periodic extension of the function f
in Example 1.
Thus the discontinuity at x = 0, 2, 4, …will
converge to
f (0 )  f (0) 

2
2
and at x = , 3, … will converge to
f ( )  f (  )
0
2
 Sequence of Partial Sums
Referring to (13), we write the partial
sums as (see Fig 12.2.3)
S1 
S3 

4

4
, S2 


4

2

cos x  sin x,
2
1
cos x  sin x  sin 2 x

2
12.3 Fourier Cosine and Sine Series
 Even and Odd Functions
even : f(−x) = f(x) odd : f(−x) = −f(x)
5
Theorem 12.3.1 Properties of Even/Odd Functions
(a) The product of two even functions is even.
(b) The product of two odd functions is even.
(c) The product of an even function and an odd function is odd.
(d) The sum (difference) of two even functions is even.
(e) The sum (difference) of two odd functions is odd.
(f) If f is even then

a

a
a
(g) If f is odd then
a
a
f ( x)dx  2  f ( x)dx
0
f ( x)dx  0
 Cosine and Sine Series
If f is even on (−p, p) then
1 p
2 p
f
(
x
)
dx

f ( x)dx
p  p
p 0
1 p
n
2 p
n
an   f ( x)cos x dx   f ( x)cos xdx
p p
p
p 0
p
a0 
even
bn 
1
n
f ( x)sin x dx  0


p
p
p
p
odd
Similarly, if f is odd on (−p, p) then
an  0, n  0, 1, 2,...
bn 
2 p
n
f ( x)sin xdx

p 0
p
Definition 12.3.1 Fourier Cosine and Sine Series
i)
The Fourier series of an even function f on the interval (−p, p) is the cosine series
6
a0 
n
  an cos
x
2 n1
p
2 p
2 p
n
a0   f ( x) dx, an   f ( x)cos
x dx
0
0
p
p
p
The Fourier series of an odd function f on the interval (−p, p) is the sine series
f ( x) 
ii)

f ( x)   bn sin
i 1
bn 
n
x
p
2 p
n
f ( x)sin
x dx

0
p
p
Examples: 1, 2
 Gibbs Phenomenon
Fig 12.3.6 shows the partial sums of (7).
We can see there are pronounced spikes near the discontinuities.
This overshooting of SN does not smooth out but remains fairly constant even when N is
large. This is so-called Gibbs phenomenon
 Half-Range Expansions
If a function f is defined only on 0 < x < L, we can make arbitrary definition of the function on
−L < x < 0.
7
If y = f(x) is defined on 0 < x < L,
(i)
reflect the graph about the y-axis onto −L < x < 0; the function is now even (Fig 12.3.7).
(ii)
reflect the graph through the origin onto −L < x < 0; the function is now odd (Fig 12.3.8).
(iii)
define f on −L < x < 0 by f(x) = f(x + L). See Fig 12.3.9.
Example: 3
 Periodic Driving Force
Consider the following physical system

d 2x
n
m 2  k x  f (t ), x p (t )   Bn sin
t
dt
p
n 1
is a half-range sine expansion.
Example: 4
12.4 Complex Fourier Series
 Euler’s formula
eix  cos( x)  i sin( x)
e ix  cos( x)  i sin( x)
 Complex Fourier Series
From Euler’s formula, we have
cos x 
Using this relation to replace⁡cos⁡(
𝑛𝜋𝑥
𝑝
eix  e  ix
eix  e  ix
, sin x 
2
2i
) and sin⁡(
𝑛𝜋𝑥
𝑝
8
), then
a0   ein x /p  e in x /p
ein x /p  e in x /p 
   an
 bn

2 n1 
2
2i


a0   1
1
   (an  ibn )ein x /p  (an  ibn )e in x /p 
2 n1  2
2



n 1
n 1
 c0   cnein x /p   c ne  in x /p
When the function f is real, cn and c-n are complex conjugates. We have
1 1 p
c0    f ( x) dx
2 p p
1
(an  ibn )
2
 n x 
 n x  
1 1 p
1 p
   f ( x) cos 
dx  i  f ( x) sin 

 dx 
2  p p
p p
 p 
 p  
cn 


 n x 
 n x  
1 p
f
(
x
)
cos

i
sin




  dx
2 p  p
 p 
 p 


1 p
f ( x)e  in x / p dx


p
2p
1
(an  ibn )
2
 n x 
 n x  
1 1 p
1 p
   f ( x) cos 
dx  i  f ( x) sin 

 dx 
2  p p
p p
 p 
 p  
c n 


 n x 
 n x  
1 p
f ( x) cos 
 i sin 

  dx

2 p p
 p 
 p 


1 p
f ( x)ein x / p dx


p
2p
Definition 12.4.1 Complex Fourier Series
The Complex Fourier Series of function f defined on an interval (p, p) is given by

1 p
f ( x)   cnein x / p , cn 
f ( x)e  in x / p dx, n  0,  1,  2,


p
2p
n 
If f satisfies the hypotheses of Theorem 12.2.1, a complex Fourier series converges to f(x)
at a point of continuity and to the average
9
at a point of discontinuity.
f ( x  )  f ( x )
2
Example: 1
 Fundamental Frequency
The fundamental period is T = 2p and then p = T/2.
The Fourier series becomes

a0 
  (an cos n x  bn sin n x) and  cnein x
2 n1
n 
where  = 2/T is called the fundamental angular frequency.
 Frequency Spectrum
If f is periodic and has fundamental period T, the plot of the points (n, |cn|) is called the
frequency spectrum of f.
Examples: 2, 3 (correction!)
12.5 Sturm-Liouville Problem
 Eigenvalue and Eigenfunctions
Recall the ODE with boundary condition
y   y  0,
y (0)  0,
y ( L)  0
This equation possesses nontrivial solutions only when  takes on the values
n 2 2
n  2 , n  1, 2,3... called eigenvalues. The corresponding nontrivial solutions
L
 n x 
 n x 
y  c2 sin 
 are called the eigenfunctions.
 or simply y  sin 
 L 
 L 
Example: 1
 Regular Sturm-Liouville Problem
Let p, q, r and r be real-valued functions continuous on [a, b], and let r(x) > 0 and p(x) > 0 for
every x in the interval.
Then solve
10
d
[r ( x) y]  (q ( x)   p ( x)) y  0
dx
(3)
Subject to
 A1 y (a)  B1 y(a)  0

 A2 y (b)  B2 y(b)  0
is said to be a regular Sturm-Liouville problem. The coefficients are assumed to be real and
independent of .
Theorem 12.5.1 Properties of the Regular Sturm–Liouville Problem
(a) There exist an infinite number of real eigenvalues that can be arranged in increasing order
1 < 2 < 3 < … < n < … such that n →  as n → .
(b) For each eigenvalue there is only one eigenfunction (except for nonzero constant
multiples).
(c) Eigenfunctions corresponding to different eigenvalues are linearly independent.
(d) The set of eigenfunctions corresponding to the set of eigenvalues is orthogonal with
respect to the weight function p(x) on the interval [a, b].
Proof of (d):
Let ym and yn be eigenfunctions corresponding to eigenvalues m and n. Then
d
[r ( x) ym ]  ( q( x)  m p( x)) ym  0
dx
d
[r ( x) yn ]  (q ( x)  n p ( x)) yn  0
dx
we have
(m  n ) p ( x) ym yn  ym
d
d
 r ( x) yn   yn  r ( x) ym 
dx
dx
Integrating the above equation from a to b, then
b
(m  n )  p ( x) ym yn dx
a
 r (b)[ ym (b) yn (b)  yn (b) ym (b)]
 r (a )[ ym (a ) yn (a )  yn (a ) ym (a )]
Since all solutions must satisfy the boundary condition, we have
11
(8)
A1 ym (a )  B1 ym (a )  0
A1 yn (a )  B1 yn (a )  0
For this system to be satisfied by A1 and B1 not both zero, the determinant must be zero
ym (a ) yn (a )  yn (a ) ym (a )  0
ym (b) yn (b)  yn (b) ym (b)  0
Similarly, we have
Hence we have the orthogonality relation

b
a
p( x) ym ( x) yn ( x) dx  0, m  n
(9)
Example: 2
 Singular Sturm-Liouville problem
There are several important conditions for the boundary condition of Sturm-Liouville problem
i) r(a) = 0 and a boundary condition of the type is specified at x = b. (Singular BVP)
ii) r(b) = 0 and a boundary condition of the type is specified at x = a. (Singular BVP)
iii) r(a) = r(b) = 0 and no boundary condition is specified at either x = a or at x = b;
iv) r(a) = r(b) and boundary conditions y(a) = y(b), y′(a) = y′(b). (Periodic BVP)
 By assuming the solutions of (3) are bounded on [a, b], from (8) we have

If r(a) = 0, then the orthogonality relation (9) holds with no boundary condition at x = a;

If r(b) = 0, then the orthogonality relation (9) holds with no boundary condition at x = b;

If r(a) = r(b) = 0, then the orthogonality relation (9) holds with no boundary conditions
specified at either x = a or x = b;

If r(a) = r(b), then the orthogonality relation (9) holds with periodic boundary conditions
y(a) = y(b), y′(a) = y′(b).
 Self-Adjoint Form
r ( x) y  r ( x) y  (q( x)   p( x)) y  0
Thus we can write the Legendre’s differential equation (1  x 2 ) y " 2 xy ' n(n  1) y  0 as
d
[(1  x 2 ) y]  n(n  1) y  0
dx
12
Here we find that the coefficient of y is the derivative of the coefficient of y.
In addition, if the coefficients are continuous and a(x)  0 for all x in some interval, then any
second-order differential equation
a( x) y  b( x) y  (c( x)   d ( x)) y  0
can be recast into the so-called self-adjoint form (3).
To understand the above fact, we start from a1(x)y + a0(x)y = 0
Let P = a0(x)/a1(x), then y + Py = 0, y + Py = 0, μ′ = Pμ,  = e P(x)dx d(y)/dx = 0.
Now, let Y = y, and the integrating factor be e [b(x)/a(x)] dx, then
e
e
Y 
b ( x )/ a ( x ) dx
( b / a ) dx
y 
b ( x )/ a ( x ) dx 
b( x)  b ( x )/ a ( x ) dx
d
e
Y  ...  e 
Y  ...

a( x)
dx 
b( x)  ( b / a ) dx
d ( x)  ( b / a ) dx 
 c( x)  ( b / a ) dx
e
y  
e

e
y 0
a( x)
a( x)
 a( x)

d   ( b / a ) dx   c( x)  (b / a ) dx
d ( x)  (b / a ) dx 
e
y  
e

e
y 0
  a ( x)
dx 
a ( x)

r ( x)  e 
( b / a ) dx
, q( x) 
c( x)  (b / a ) dx
d ( x)  (b / a ) dx
e
, p( x) 
e
a( x)
a( x)
Examples: 3 (correction!), 4
12.6 Bessel and Legendre Series (for your own reading and optional exploration)
 Fourier-Bessel Series
We have shown that {Jn(ix)}, i = 1, 2, 3, … is orthogonal w.s.t. p(x) = x on [0, b] when the i
are defined by
A2 J n ( b)  B2 J n ( b)  0
This orthogonal series expansion or generalized Fourier series of a function f defined on (0, b) in
b
terms of this orthogonal set is

 xJ n ( i x) f ( x) dx
f ( x)   ci J n ( i x), ci  0
2
J n ( i x)
i 1
The square norm of the function Jn(ix) is defined by
b
J n ( i x)   xJ n2 ( i x)dx
2
0
f(x) is called a Fourier-Bessel series.
13
 Differential Recurrence Relations
d n
d n
[ x J n ( x)]  x n J n1 ( x),
[ x J n ( x)]   x  n J n1 ( x)
dx
dx
 Square Norm
b
J n ( i x)   xJ n2 ( i x)dx is dependent on i   i 2
2
0
If y = Jn(x) we have
 2 n2 
d

 xy     x   y  0
dx
x

After we multiply by 2xy’, then
d
d
 xy2   2 x 2  n 2  [ y ]2  0
dx
dx
Integrating by parts on [0, b], then

2 2  xy 2 dx   xy  ( 2 x 2  n 2 ) y 2
b
0
2

b
0
Since y = Jn(x), the lower limit is 0 for n > 0, because Jn(0) = 0.
For n = 0, at x = 0. Thus
b
2 2  xJ n2 ( x) dx   2b 2 [ J n ( b)]2  ( 2b 2  n 2 )[ J n  b)]2
0
where y '   J n ( x).
Now we consider three cases for the condition
A2 J n ( b)  B2 J n ( b)  0
Case I: If we choose A2 = 1 and B2 = 0, then
J n ( b)  0
There are an infinite number of positive roots xi = ib of (8) (see Fig 5.3.1), which defines i =
xi/b. The eigenvalues are positive and then i = i2 = xi2/b2. No new eigenvalues result from the
negative roots of (8) since Jn(−x) = (−1)nJn(x).
The number 0 is not an eigenvalue for any n since Jn(0) = 0, n= 1, 2, 3, … and J0(0) = 1.
b2 2
|| J n i ( x) ||  J n1 ( ib)
2
2
Case II: If we choose A2 = h  0 and B2 = b, then
hJ n ( b)   bJ n ( b)  0
14
There are an infinite number of positive roots xi = ib for n = 1, 2, 3, …. As before i = i2 =
xi2/b2.  = 0 is not an eigenvalue for n = 1, 2, 3, ….
|| J n ( i x) ||2 
 i2b 2  n 2  h 2 2
J n ( ib)
2 i2
Case III: If h = 0 and n = 0 in
hJ n ( b)   bJ n ( b)  0
i are defined from the roots of
J 0 ( b)  0
Though this is a special case, it is the only situation for which  = 0 is an eigenvalue.
For n = 0, the result implies J0(b) = 0 is equivalent to J1(b) = 0.
Since x1 = ib = 0 is a root of the last equation and because J0(0) = 1 is nontrivial,
we conclude that from 1 = 12 = x12/b2 that 1 = 0 is an eigenvalue.
But we cannot use
 i2b 2  n 2  h 2 2
|| J n ( i x) || 
J n ( ib)
2 i2
2
when 1 = 0, h = 0, n = 0, and n = 0. However from
b
J n ( i x)   xJ n2 ( i x)dx
2
0
we have
b
||1||2   x dx 
0
b2
2
For i > 0 we can use
|| J n ( i x) ||2 
with h = 0 and n = 0:
 i2b 2  n 2  h 2 2
J n ( ib)
2 i2
b2 2
|| J 0 ( i x) ||  J 0 ( ib)
2
2
Definition 12.6.1 Fourier–Bessel Series
The Fourier-Bessel series of a function f defined on the interval (0, b) is given by
i)

f ( x)   ci J n ( i x), ci 
i 1
b
2
xJ n ( i x) f ( x) dx
b 2 J n21 ( ib) 0
where the i are defined by Jn(b) = 0.
ii)

f ( x)   ci J n ( i x), ci 
i 1
b
2 i2
xJ n ( i x) f ( x) dx
2 2
2
2
2

( i b  n  h ) J n ( ib) 0
15
where the i are defined by hJn(b) + bJ’n(b) = 0.
iii)

b
2 b
2
x f ( x ) dx, ci  2 2
xJ 0 ( i x ) f ( x ) dx
2 0

b
b J 0 ( ib) 0
i 2
where the i are defined by J′0(b) = 0.
f ( x)  c1   ci J 0 ( i x), c1 
Theorem 12.6.1 Conditions for Convergence
If f and f ′ are piecewise continuous in the open interval (0, b), then a Fourier-Bessel expansion
of f converges to f(x) at any point where f is continuous and to the average
f ( x  )  f ( x )
2
at a point where f is discontinuous.
Examples: 1, 2
Definition 12.6.2 Fourier–Legendre Series
The Fourier-Legendre series of a function f defined on the interval (–1, 1) is given by

f ( x)   cn Pn ( x), cn 
n 0
2n  1 1
f ( x) Pn ( x) dx
2 1
Theorem 12.6.2 Conditions for Convergence
If f and f ′ are piecewise continuous in the open interval (–1, 1), then a Fourier-Legendre series
converges to f(x) at a point of continuity ant to the average
f ( x  )  f ( x )
2
at a point of discontinuity.
Example: 3
 Alternative Form of Series
If we let x = cos , the x = 1 implies  = 0, x = −1 implies  = . Since dx = −sin  d, FourierLegendre series become, respectively,
16

F ( )   cn Pn (cos ), cn 
n 0
2n  1 
F ( ) Pn (cos )sin  d
2 0
where f(cos ) has been replaced by F().
Exercises
․
․
․
․
․
․
12.1: 16, 19, 22, 25, 27, 28, 30
12.2: 21, 22, 24, 25
12.3: 37, 39, 40, 42, 43, 46, 47, 49, 50 (try 53, 54, 55)
12.4: for self-practice
12.5: 3, 4, 9, 10, 11
12.6: (optional) for your own study as much as you wish.
17