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

iθ

= 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

nπ

L

x

e

¡i

mπ

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 ﬁxed,

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