Chapter 7
Fourier Transform Method
In this chapter, we delve into the Fourier transform and its application in solving
linear second-order partial differential equations on unboun ded domains. Up until
now, we have focused on studying these equations solely on boun de d domains with
specific boundary conditions. Howe ver, with the Fourier transform, we are able to
extend our problem-solving capabilities to unbounded domains. The Fourier trans-
form provides us with a remarkable tool to analyze functions defined on unbounded
domains, offering insights into their frequency components and facilitating the solu-
tion of differential equations, particularly partial differential equations.
This powerful technique is closely associated with the contributions of Joseph
Fourier, a renowned French physicist and mathematician whose pioneering work
paved the way for tackling linear PDEs in new and innovative ways.
7. 1 Introduction
We will delve into the development of the Fourier transform from its roots i n Fourier
series and its application in s o lv ing partial differential equations using the eigen-
function expansion method.
7.1.1 From Fourier series to Fourier transform
In the previous sections, we explored the Fourie r series representation of a piecewise
continuously differentiable function f(x) defined on x2[¡L; L]. We demonstrated
that such a function can be elegantly expressed using trigonometric functions, namely
fcos(n!x); sin(n!x)g, where ! =
π
L
. By utilizing the Euler formula
e
= cos(θ) + i sin(θ);
we can equivalently re present the trigonometric Fourier series in its complex form
as follows:
f(x) =
X
n=¡1
1
F
n
e
i!
n
x
; (7.1)
1
where the coefficients F
n
are given by:
F
n
=
1
2L
Z
¡L
L
f(x) e
¡in!x
dx: (7.2)
These formulas are derived from the orthogonality property of the complex expo-
nential functions
hhe
in!x
; e
im! x
ii:=
Z
¡L
L
e
i
L
x
e
¡i
L
x
dx =
(
1
2L
n = m
0 n =/ m
:
In this context, we employ the co mplex version of the i nner product hh;ii, defined as
hhf ; gii=
Z
a
b
f(x) g¯(x) dx;
where g¯ (x) re presents the complex conjugate of g(x), and the integration is over
the interval [a; b].
The concept of the Fourier series representation can be extended to functions
defined on an unbounded domain by introducing the Fourier transform. Starting
from the complex Fourier series representation of a function f (x) in [¡L; L], and
substituting (7.2) into (7.1), we can write
f(x) =
1
2π
Z
¡L
L
f(z)
X
n=¡1
1
e
i(x¡z)n!
L
!
L
!
dz;
where !
L
=
π
L
. As L approaches infinity, and !
L
approaches zero, we interpret t he
summation a s an integral:
lim
L!1
X
n=¡1
1
e
i(x¡z)n!
L
!
L
= lim
L!1
Z
¡L
L
e
i(x¡z)!
d!;
resulting in:
f(x) =
1
2π
lim
L!1
Z
¡L
L
Z
¡L
L
f(z) e
¡i!z
dz
e
i!x
d!:
The inner integral in this expression is recognized as the Fourier transform of the
function f, denoted as Fff g, and defined as:
Fff g=
Z
¡1
1
f(x) e
¡i!x
dx:
Note that the integral is taken over x, resulting in a function of !. For convenience,
we denote the transformed function as f
^
(!). Thus, we obtain the final result:
f(x) =
1
2π
Z
¡1
1
f
^
(!) e
i!x
d!:
To ensure convergence of the integral for Fff g, we establish the integrability con-
dition for the function f(x).
2 Fourier Transform Method
Theorem 7.1. Assume that a function f(x) is integrable over (¡1; 1), i.e.,
Z
¡1
1
jf (x)jdx < 1:
Then, f
^
(!) exists, and it is continuous with respect to !. Furthermore, the supremum
of jf
^
(!)j is finite, i.e.,
sup
!
jf
^
(!)j< 1:
Proof. The existence of f
^
is verified by the inequality:
Z
¡1
1
f(x) e
¡i!x
dx
Z
¡1
1
jf (x)jdx < 1;
ensuring that f
^
(!) converges for any !. To demonstrate continuity, let !
0
be xed,
and as ! approaches !
0
:
lim
!!!
0
Z
¡1
1
f(x) e
¡i!x
dx =
Z
¡1
1
lim
!!!
0
f(x) e
¡i!x
dx =
Z
¡1
1
f(x) e
¡i!
0
x
dx:
The passage of the limit inside the integral is allowed by the dominant convergence
theorem (see the appendix of this book ). Hence,
lim
!!!
0
f
^
(!) = f
^
(!
0
):
The final claim is proved as follows:
sup
!
Z
¡1
1
f(x) e
¡i!x
dx sup
!
Z
¡1
1
f(x) e
¡i!x
dx
sup
!
Z
¡1
1
jf (x)jdx < 1;
confirming that the supremum of jf
^
(!)j is finite, and this completes the proof.
The following th eorem, known a s the Fourier theorem states the convergence of
F
¡1
defined as:
F
¡1
ff
^
g=
1
2π
Z
¡1
1
f
^
(!) e
¡i!x
d!:
Theorem 7.2. Let f (x) be an integrable function defined on (¡1; 1) and piecewise
cont inuously differentiable, with f
0
(x) being a piecewise cont inuous function. Then,
the inverse Fourier transform, denoted as F
¡1
ff
^
g, is equal to f(x) at continuity
point s of f. However, at discontinuity points of f, the inverse Fourier transform is
given by:
F
¡1
ff
^
g=
f(x
+
) + f(x
¡
)
2
where f (x
+
) and f(x
¡
) represent the right and left limits of f at x, respectively.
7.1 Introduction 3
Example 7.1. Consider the function f(x) given by:
f(x) =
8
<
:
1
jxj
p
¡1 x 1
0 otherwise
:
This function is integrable:
Z
¡1
1
1
jxj
p
dx = 4 < 1:
By the formula, we have
F
(
1
jxj
p
)
= lim
L!1
Z
¡L
L
1
jxj
p
e
¡i!x
dx = lim
L!1
(
Z
¡L
L
cos(!x)
jxj
p
dx +
Z
¡L
L
i sin(!x)
jxj
p
dx
)
:
Since the function
1
jxj
p
is even, the s econd integral a t th e right-hand side is zero,
and we can write:
F
(
1
jxj
p
)
= lim
L!1
Z
¡L
L
cos(!x)
jxj
p
dx:
The figure below depicts the Fourier transform o f the given function in ! 2(¡30; 30)
-30 -20 -10 0 10 20 30
0
1
2
3
4
Even though, the function is not piecewise continuous, its inverse Fourier trans-
form F
¡1
ff
^
g conve rges outside of the singular point x = 0 as shown below:
-1 -0.5 0 0.5 1
0
5
10
15
4 Fourier Transform Method
Exercise 7.1. Show that if f(x) is an even function, Fff g is an even function in ! and real.
If f (x) is an odd function, Fff g is an odd function in ! and pure imaginary.
7.1.2 Fourier transform as frequency distribution
When we view a function f (x) as an electrical signal or a wave in time or space, its
Fourier t ransform f
^
(!) = Fff(x)g reveals the frequency components embedded in
the s ignal. For a function f(x) defined on x2[¡L; L], the frequency distribution is
discrete, characterized by the terms F
n
e
in!
L
x
, where !
L
=
π
L
, and the magnitude F
n
is given by:
F
n
=
1
2L
Z
¡L
L
f(x) e
¡in!
L
x
dx:
As we move to functions defined on the entire real line (¡1; 1), this f requency
spectrum evolves into a continuous di stribution over the !-domain.
To illustrate this, let's consider the function f(x) = cos(!
0
x). This function is
periodic with a period T
0
=
2π
!
0
and the angular frequency ! = !
0
. Thus, the signal
has a single periodic frequency component, which is !
0
. We expect tha t Ffcos(!
0
x)g
will exhibit a spike at !
0
:
Ffcos(!
0
x)g= lim
L!1
Z
¡L
L
cos(!
0
x) e
¡i!x
dx:
Using Euler's f o rmula co s(!
0
x) =
e
i!
0
x
+ e
¡i!
0
x
2
, we obtain:
Ffcos(!
0
x)g= lim
L!1
sin[(! ¡!
0
)L]
! ¡!
0
+
sin[(! + !
0
)L]
! + !0
:
The figure below depicts the graph of this Fourier transform for !
0
= 1 for L = 20
and L = 40.
-4 -2 0 2 4
-5
0
5
10
15
20
25
-4 -2 0 2 4
-10
0
10
20
30
40
As obs erved, the spectrum or frequency distribution of the fun ction cos(x)
exhibits a spike at the physical frequency ! = 1 an d a non-physical (or purely math-
ematical) frequency at ! = ¡1, which becomes more pronounced as L approaches
infinity. Therefore, f
^
(!) = 0 for ! =/ ± 1.
Now, let's compare this with the function f(x) = cos(2x). This function rotates
twice faster than cos(x). The function Ffcos(2x)g exhibits spikes at ! = 2, and
! = ¡2.
7.1 Introduction 5