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 ! =

. By utilizing the Euler formula

iθ

= cos(θ) + i sin(θ);

we can equivalently re present the trigonometric Fourier series in its complex form

as follows:

f(x) =

n=¡1

; (7.1)

where the coeﬃcients F

are given by:

¡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:=

¡L

nπ

¡i

mπ

dx =

(

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=

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) =

2π

¡L

f(z)

n=¡1

i(x¡z)n!

dz;

where !

. As L approaches inﬁnity, and !

approaches zero, we interpret t he

summation a s an integral:

lim

L!1

n=¡1

i(x¡z)n!

= lim

L!1

¡L

i(x¡z)!

d!;

resulting in:

f(x) =

2π

lim

L!1

¡L



¡L

f(z) e

¡i!z



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=

¡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) =

2π

¡1

(!) 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.,

¡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

(!)j< 1:

Proof. The existence of f

is veriﬁed by the inequality:



¡1

f(x) e

¡i!x



≤

¡1

jf (x)jdx < 1;

ensuring that f

(!) converges for any !. To demonstrate continuity, let !

be ﬁxed,

and as ! approaches !

lim

!!!

¡1

f(x) e

¡i!x

dx =

¡1

lim

!!!

f(x) e

¡i!x

dx =

¡1

f(x) e

¡i!

dx:

The passage of the limit inside the integral is allowed by the dominant convergence

theorem (see the appendix of this book ). Hence,

lim

!!!

(!) = f

The ﬁnal claim is proved as follows:

sup

¡1

f(x) e

¡i!x

dx ≤sup



¡1

f(x) e

¡i!x



≤sup

¡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

¡1

deﬁned as:

¡1

2π

¡1

(!) 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

(x) being a piecewise cont inuous function. Then,

the inverse Fourier transform, denoted as F

¡1

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:

¡1

f(x

) + f(x

)

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) =

jxj

¡1 ≤x ≤1

0 otherwise

This function is integrable:

¡1

jxj

dx = 4 < 1:

By the formula, we have

(

jxj

)

= lim

L!1

¡L

jxj

¡i!x

dx = lim

L!1

(

¡L

cos(!x)

jxj

dx +

¡L

i sin(!x)

jxj

)

Since the function

jxj

is even, the s econd integral a t th e right-hand side is zero,

and we can write:

(

jxj

)

= lim

L!1

¡L

cos(!x)

jxj

dx:

The ﬁgure below depicts the Fourier transform o f the given function in ! 2(¡30; 30)

-30 -20 -10 0 10 20 30

Even though, the function is not piecewise continuous, its inverse Fourier trans-

form F

¡1

g conve rges outside of the singular point x = 0 as shown below:

-1 -0.5 0 0.5 1

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

in!

, where !

, and the magnitude F

is given by:

¡L

f(x) e

¡in!

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(!

x). This function is

periodic with a period T

2π

and the angular frequency ! = !

. Thus, the signal

has a single periodic frequency component, which is !

. We expect tha t Ffcos(!

x)g

will exhibit a spike at !

Ffcos(!

x)g= lim

L!1

¡L

cos(!

x) e

¡i!x

dx:

Using Euler's f o rmula co s(!

x) =

+ e

¡i!

, we obtain:

Ffcos(!

x)g= lim

L!1



sin[(! ¡!

)L]

! ¡!

sin[(! + !

)L]

! + !0



The ﬁgure below depicts the graph of this Fourier transform for !

= 1 for L = 20

and L = 40.

-4 -2 0 2 4

-5

-4 -2 0 2 4

-10

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

-4 -2 0 2 4

-5

-4 -2 0 2 4

-10

Now, consider an electrical or mechanical signal composed of various sub-signals

with diﬀerent angular frequencies:

f(x) = c

cos(x) + c

co(2x) + ···+ c

cos(nx):

The function Fff(x)g has spikes at physical freque nc ies ! = 1; 2; :::; n weighted

by the values c

; c

; :::; c

. The ﬁgure below depicts the function f(x) = cos(x) +

2 co s(2x) ¡ 2 cos(3x), and its Fourier transform. It is observed that Fff(x)g has

spikes at ! = 1; 2; 3 with weights 1; 2; ¡2:

-5 0 5

-4

-3

-2

-1

-4 -2 0 2 4

-150

-100

-50

100

In general, a f unction f(x) can have a continuous distribution of angular fre-

quencies !, which is commonly represented as its frequency spectrum f

(!) using

the Fourier transform F. The inverse Fourier transform allows u s to retrieve f(x)

from its frequency spectrum through the formula

f(x) =

2π

¡1

(!) e

i!x

d!:

In physics and engineering contexts, it is often convenient to use the notation !=2πf

and express the inverse Fourier transform as:

f(x) =

¡1

(!) e

i2πfx

df:

However, this notation may l ead to confusion, as we typically denote our function

by f(x).

6 Fourier Transform Method

Exercise 7.2. Let's compare the frequency spectrum of two functions f

(x) = e

¡0.2jxj

cos(x)

and f

(x) = e

¡0.2jxj

cos(2x). The function f

changes twice faster than f

. Draw these two

functions in x-domain. Now draw the functions f

(!) and f

(!) and explain what you observe.

You can use the following code:

x=-10:0.01:10;

f1=@(x) exp(-abs(x)/5).*cos(x);

f2=@(x) exp(-abs(x)/5).*cos(2*x);

subplot(1,2,1)

plot(x,f1(x))

subplot(1,2,2)

plot(x,f2(x))

figure()

w=-2:0.01:2;

fw1=2*integral(@(x) f1(x).*cos(w.*x),0,40, ...

'ArrayValued',true);

subplot(1,2,1)

plot(w,fw1);

w=-3:0.01:3;

fw2=2*integral(@(x) f2(x).*cos(w.*x),0,40, ...

'ArrayValued',true);

subplot(1,2,2)

plot(w,fw2)

Explain why we used in the above code the integral

(x) cos(!x) dx;

instead of the integral

¡40

(x) e

¡i!x

dx:

Your ﬁgures should be like the following ones:

-10 -5 0 5 10

-0.5

0.5

-10 -5 0 5 10

-0.5

0.5

-2 -1 0 1 2

-4 -2 0 2 4

Exercise 7.3. The function f (x) = 1 is not integrable. Its Fourier transform is

Ff1g= lim

L!1

¡L

¡i!x

dx = lim

L!1

2sin(!L)

Plot this function for diﬀerent values of L. What do you o bserve at ! =0? The function f (x)=1

is the limiting function of cos(!

x) when !

approaches 0. Compare this result to the Fourier

transform of cos(!

x) when !

!0.

7.1 Introduction 7

Exercise 7.4. If f(x) is an odd function, show that Fff g is an o dd and pure imaginary

function in !-do main. Let f (x) = sin(!

x). This function is not integrable. Plot the imaginary

part of Ffsin(!

x)g. What do you observe at !

and ¡!

? Use the following code in Matlab

to draw Ffsin(!

x)g.

w0=1;%you can change this value

w=-4:0.01:4; %you can change this interval

xinf=50% you can change this value

fw=integral(@(x) sin(w0*x).*sin(w.*x),-xinf,xinf,'ArrayValued',true);

plot(w,fw)

7.1.3 Eigen function expansion method

The Eigenfunction expansion method for problems on unbounde d domain can be

illustrated by c o nsidering the heat equation deﬁned o n x 2(¡1; 1):



= ku

u(x; 0) = f(x)

While there are no boundary conditions speciﬁed in this problem, we assume that

the solution u(x; t) is integrable to guarantee the existence o f the Fo urier transform

u^(!; t). We then write the solution u(x; t) as the following integral:

u(x; t) =

2π

¡1

u^(!; t) e

i!x

d!; (7.3)

where u^(!; t) is an undetermined function. This representation is analogou s to how

we expressed solutions using eigenfunctions fe

in!

g or equivalently, trigonometric

functions



cos

nπ



; sin

nπ



for boundary va lue problems deﬁned on a bou nded

domain (¡L; L).

To determine u^(!; t), we substitute the integral (7.3) into the heat equation

= ku

and o btain the ordinary diﬀerential equation:

(!; t) = ¡k!

u^(!; t):

This diﬀerential equation is with respect to time t, and it can be solved as:

u^(!; t) = C(!) e

¡k!

;

where C(!) is an arbitrary function with respect to !. The speciﬁc form of C(!) will

be determined by the initial condition u(x; 0)= f(x) through the Fourier transform:

u^(!; 0) = f

(!). Thus, we obtain u^(!; t) = f

(!) e

¡k!

. Consequently, the integral

solution for u(x; t) is retrieved by the inverse Fourier transform:

u(x; t) = F

¡1

(!) e

¡k!

g: (7.4)

This solution can be represented in integral form as:

u(x; y) =

2π

¡1

(!) e

¡k!

i!x

d!: (7.5)

8 Fourier Transform Method

Now, let's compare this integral solution to the series solution of the same heat

problem deﬁned on the interva l [¡L; L]:

= ku

¡L < x < L

u(¡L; t) = u(L; t)

(¡L; t) = u

(L; t)

u(x; 0) = f (x)

The solution to this problem in complex form and in terms of the eigenfunctions

in!x

g for ! =

is given as:

u(x; t) =

n=¡1

¡kn

in!

It turns out that when L!1, this solution converges to the integral solution (7.5).

Problem 7.1. Consider the following heat problem



= u

+ h(x)

u(x; 0) = 0

Assume that h is integrable. Find an integral solution to this problem.

7. 2 Fou rier tra nsform of important functions

In this section, we will review the Fourier transform of some important functions

that will be useful in solving heat and wave equations on unbounded domains in the

future examples. We wi ll examine the Fourier transform of commonly encountered

functions, which will aid us in ﬁnding inte g ral solutions and understanding the

behavior of these fun ctions unde r Fourier transformation.

7.2.1 Pulse or rectangle function

Consider the following function:

f(x) =



1 ¡a < x < a

0 otherw ise

;

where a > 0 is a constant. By deﬁnition of the Fourier transform, we have

Fff g=

¡a

¡i!x

dx =

2 sin(a!)

In particular, when a = 1, t he transform function is f

(!) = 2 sinc(!) for the sinc

function: sinc(z) =

sin(z)

7.2 Fourier transform of important functions 9

-20 -10 0 10 20

-0.5

0.5

1.5

Now, let's investigate the inverse Fourier transform of

2 sin(a!)

. The expression

for the inverse transform is given as:

¡1



2sin(a!)



¡1

sin(a!)

cos(!x) d!:

The ﬁgure below depicts the inverse transform with integration over interval s ! 2

(¡01; 10) and ! 2(¡50; 50):

-4 -2 0 2 4

-0.2

0.2

0.4

0.6

0.8

-4 -2 0 2 4

-0.2

0.2

0.4

0.6

0.8

If a approaches 1, the function f(x) a pproaches the co nstant function f ≡ 1.

Even though the constant function f ≡1 is not integrable, we can write

Ff1

g= lim

a!1

2 sin(a!)

It can be shown that the limiting function at the right hand side is a the Dirac delta

function.

Problem 7.2. Show ﬁrst the equality

¡1

sin(az)

dz = π:

Now, let f be a continuous function and integrable. Show th e relation

lim

a!1

¡1

sin(az)

πz

f(z) dz = f (0):

Now, with these results, we can symbolically write:

Ff1

g= 2πδ(!):

10 Fourier Transform Method

Note also that

¡1

f2πδ(!)g=

2π

¡1

2πδ(!) e

i!x

d! = 1

Using the inverse transform formula, we can ﬁnd the transform of the function

f(x) =

sin(ax)

. We have

¡1

sin(a!)

i!x

d! =



π ¡a < x < a

0 otherwise

Changing ! !x and x !¡!, we obtain

¡1

sin(ax)

¡i!x

dx =



π ¡a < ! < a

0 otherwise

;

and thus



sin(ax)





π ¡a < ! < a

0 otherwise

The function at the right-hand side represents a low-pass ﬁlter in the frequency

domain. This ﬁlter allows low frequencies to pass through and ﬁlters out high fre-

quencies in a given function. Such ﬁlters are commonly used in signal processing

and communication systems to eliminate unwanted noise and interference while

preserving essential information.

Exercise 7.5. Consider the function

(x) =

(

¡a < x < a

0 otherwise

;

for a > 0. Let a approaches 0, ﬁnd Fff

(x)g. What is limiting function f

(x) when a !0?

7.2.2 Exponential functions

Let f be given by: f

(x) = e

¡ajxj

; a > 0. This function is integrable and continuously

diﬀerentiable eve rywhere except x = 0. By deﬁnition of the Fourier transform, we

have

Fff

(x)g=

¡1

¡ajxj

¡i!x

dx =

¡1

(a¡ i!) x

¡(a+i!)x

+ !

The ﬁgure below illustrates the reconstruction of f

(x) for a = 1 by its Fourier

transform:

-2 -1 0 1 2

0.2

0.4

0.6

0.8

7.2 Fourier transform of important functions 11

When a approaches zero, f

(x) approaches the constant function f ≡ 1. Then,

we expect that Fff

(x)g approaches the Dirac delta function 2πδ(!).

Problem 7.3. Show ﬁrst that

¡1

+ x

dx = π;

for a ny a > 0. Let h(x) be a conti nuous function which is zero outside an interval (¡R; R) for

some R > 0. Show the following relation

lim

a!0

¡1

ah(x)

+ x

dx = πh(0):

Exercise 7.6. Use the inverse Fourier transform and show the following relation

cos(ax)

1 + x

dx =

¡jaj

Exercise 7.7. Use the inverse Fourier transform and show the relation



+ x



¡aj! j

Exercise 7.8. The function given by

f(x) =



1 x > 0

¡1 x < 0

is not integrable. The function f (x) is the limiting function of f

(x)=

(

¡"x

x > 0

¡e

x < 0

. Find Fff

and let " approaches zero.

Problem 7.4. Let g

(x) =

(x). Show that g

(x) is a Dirac delta sequence function, i.e.,

lim

a!1

¡a

(x) h(x) dx = h(0);

for any continuous function h(x) which vanishes outside an interval (¡R; R) for some R > 0.

7.2.3 Gaussian function

The Gaussian function shows itself in several ﬁel ds of pure and applied sciences. One

of its beauty is that the Fourier transform of a Gaussian function is a Gaussian. Let

f(x) be given by: f(x) = e

¡ax

. Its Fourier tran sform i s:

Ffe

¡ax

¡1

¡ax

¡i!x

dx = e

¡1

¡a



x¡



dx:

The last integral in the above equality is

¡1

¡a



x¡



dx =

;

and thus

Ffe

¡ax

In particular, if a =

, we obtain

= 2π

12 Fourier Transform Method

Exercise 7.9. Show the relation

¡1

¡t!

4πt

;

for t > 0. Show also the equality

¡1

4πt

dx = 1:

Furthermore, show that the function f

(x) =

4πt

is a Dirac delta sequence function for

t ! 0, i.e., for any continuous function h(x) vanishing outside an interval (¡R; R) for so me

R > 0, it satisﬁes the relation

lim

t!0

¡1

(x) h(x) dx = h(0):

7.2.4 Dirac delta function

Recall that the Dirac delta function δ(x) is deﬁned by the relation:

¡1

h(x) δ(x) dx = h(0);

for any continuous and bounded function h. The Dirac delta function plays a fun-

damental role in the theory of Fourier transform and is essential for solving linear

partial diﬀerential equations using Fourier transform. It acts a s a distribution and

has many important properties, including the shifting prop erty:

¡1

h(x) δ(x ¡x

) dx = h(x

Additionally, we can deﬁne the Dirac delta function sequence a s follows: A sequence

of functions δ

(x) is called a Dirac delta sequence if for any continuous and bounded

function h, we have:

lim

n!1

¡a

h(x) δ

(x) dx = h(0):

The Dirac delta f unction sequence is used to approximate the Dirac delta function

δ(x) for a continuous function h. By considering the limit of the Dirac delta s equence

as n approaches inﬁnity, we c a n de ﬁne the Dirac d elta fun ction as:

δ(x)= lim

n!1

(x):

Exercise 7.10. Prove that the following functions are Dirac delta sequences:

a) The sequence

(x) =

(

< x <

0 otherwise

b) The sequence

(x) =

x + n ¡

< x < 0

¡n

x + n 0 < x <

0 otherwise

7.2 Fourier transform of important functions 13

The Fourier transform of the function f(x) = δ(x ¡x

) is deﬁned as

Ffδ(x ¡x

)g=

¡1

δ(x ¡x

) e

¡i!x

dx = e

¡i!x

¡1

δ(x ¡x

) dx = e

¡i!x

In particular, if x

= 0, then Ffδ(x)g= 1

. On the other hand, by the deﬁnition of

the inverse Fourier transform, we have

¡1

2π

lim

L!1

¡L

i!x

d! =

lim

L!1

sin(xL)

The sequence

sin(nx)

is a Dirac delta sequence for L !1, and thus:

¡1

g= δ(x):

Exercise 7.11. Show that the sequence of functions

sin(nx)

is a Dirac delta sequence.

Problem 7.5. A function f (x) is called self-dual if f

(!) = f (!). For th is reason, and to make

a balance between F and F

¡1

, some textbooks change the deﬁnitions as follows:

Fff(x)g=

2π

¡1

f(x) e

¡i!x

dx;

and

¡1

(!)g=

2π

¡1

(!) e

i!x

d!:

It is simply seen that f (x) = 2π

is a self-dual function with the above deﬁnitions as:

2π

= 2π

= f (!):

Consider the function deﬁned as:

f(x) = 2π

n=¡1

δ(x ¡2nπ):

Show that its Fourier transform with respect to the balanced form of F is:

(!) = lim

n!1

sin[(2n + 1)π!]

sin(π!)

Draw this functio n for some values of n and explain what you observe at n 2Z. We can show

that the function at these point behave like a Dirac delta function. Fix n = n

and assume that

h is function continuous in the interval ( n

¡"; n

+ ") for small " > 0 and zero outside of this

interval. Show the relation

lim

n!1

¡"

sin[(2n + 1)π!]

sin(π!)

h(x) = h(n

This implies that f

(!) behaves like a Dirac delta sequence function at ! = n 2Z . The function

n=¡1

δ(x ¡2nπ) is self-dual with respect to the unbalanced form of F.

7.2.5 Trigon ometric sine and cosine functions

Consider the function f(x) = cos(!

x). As we saw above, its Fourier transform is

Ffcos(!

x)g= lim

L!1



sin[(! ¡!

)L]

! ¡!

sin[(! + !

)L]

! + !0



14 Fourier Transform Method

On the other hand, the sequences

sin[(! ¡ !

)n]

π(! ¡ !

)

and

sin[(! + !

)n]

π(! + !

)

are Dirac delta sequence

functions and approach δ(! ¡!

) and δ(! + !

) respective ly . Therefore, we obtain:

Ffc o s(!

x)g= πδ(! ¡!

) + πδ(! + !

These are two spikes at physical frequencies !

and ¡!

that we discussed in the

previous section.

Exercise 7.12. Calculate the inverses F

¡1

fδ(! ¡!

)g, and F

¡1

fδ(! + !

)g and conclude

Ffcos(!

x)g=

fδ(! ¡!

) + δ(! + !

)g:

This is depicted in the ﬁgure below:

¡!

πδ(! ¡!

)πδ(! + !

)

Find the Fourier transform Ffsin(!

x)g and draw its imaginary part in !-domain.

Exercise 7.13. Show the relation

Ffsin(!

x)g= iπδ(! ¡!

) ¡iπδ(! ¡!

7.2.6 Fourier transform of semi-bounded func tions

For functions deﬁned on [0; 1) instead of (¡1; 1), we can either use the odd or

even extension of f for Fff(x)g. This in particular is important to solve linear

partial diﬀerential equations deﬁned on semi-bounde d dom a ins.

Let f(x) be a function deﬁned on [0; 1). Deﬁning the odd extension:



f(x) x > 0

¡f(¡x) x < 0

;

we can write Fff

g as:

Fff

g= ¡i

¡1

(x) s in(!x) dx = ¡2i

f(x) sin(!x) dx:

This is called the Fourier sine transform while its imaginary part is called the Fourier

sine integral

ff(x)g:= f

(!) =

f(x) sin(!x) dx:

Note that f

(!) i s an odd function and f

(0) = 0. By the inverse Fourier transform,

we have

(x) = F

¡1

f¡2i f

(!)g=

¡1

(!) sin(!x) d!:

7.2 Fourier transform of important functions 15

Therefore, we can express the function f (x) deﬁned on x 2[0; 1) as the integral:

f(x) =

¡1

(!) sin(!x) d!;

where f

is the Fourier sine integral of f .

The Fourier cosine transform is deﬁned similarly. The even extension of a func-

tion f(x), x 2[0; 1) is deﬁned as:

(x) =



f(x) x > 0

f(¡x) x < 0

;

and thus:

Fff

(x)g:= 2

f(x) cos(!x) d x:

The Fourier cosine integral is deﬁned as:

ff(x)g= f

(!) =

f(x) cos(!x) d x:

Note that f

(!) is an eve n function in ! and f

(0) = 0. Therefore, a function f(x)

which is deﬁned on the domain [0; 1) can be expressed as:

f(x) =

¡1

(!) cos(!x) d!;

where f

(!) is the Fourier cosine of f(x).

Example 7.2. Consider the function f (x) = e

¡x

for x 2[0; 1). The following ﬁgure

shows its Fourier s ine and cosine integrals integrated over ! 2[¡10; 10]:

-10 -5 0 5 10

-0.5

0.5

0 1 2 3

0.5

-10 -5 0 5 10

0.5

0 1 2 3

0.5

16 Fourier Transform Method

The Fourier sine integral is simply derived as:

(!) =

¡x

sin(!x) dx =

1 + !

;

and the Fourier cosine integral as:

(!) =

¡x

cos(!x) dx =

1 + !

Exercise 7.14. Use the above results and determine the value of following integrals:

cos(x)

1 + x

dx;

x sin(x)

1 + x

dx:

Exercise 7.15. Draw the Fourier sine and cosine integrals of the function f (x) = e

¡x

deﬁne

din x 2[0; 1). You can use the following code in Matlab:

w=-10:0.01:10;% The range of plot of $\hat{f}_s$

X=40;%Interval for integration over x

fw=integral(@(x) exp(-x.^2).*sin(w.*x),0,X, ...

'ArrayValued',true);

plot(w,fw)

7. 3 Properties of Fourier transform

7.3.1 Linearity

The Fourier transform is a continuous li ne ar integ ral transformation that allows us

to analyze the frequency components embedded in a given function. In particular,

for any integrable functions f

and f

, and any constants c

and c

, the Fourier

transform satisﬁes the linearity property:

Ffc

+ c

g= c

(!) + c

(!):

This means that the Fourier transform of the linear combination of two functions is

equal to the corresponding linear combination of their i ndividual Fourier transforms.

In other words, the frequency components present in the sum c

+ c

are equal

to the sum of the frequency components in f

and f

, weighted by their respective

constants c

and c

. This property makes the Fourier transform a powerful tool for

analyzing the linear systems.

Problem 7.6. Prove that if f

and f

are integrable, then c

is for any constant c

; c

Exercise 7.16. Consider the function

u(x) =



1 x > 0

0 x < 0

Write u(x) as u(x) = f (x) +

, where f (x) is the function

f(x) =

x > 0

x < 0

7.3 Properties of Fourier transform 17

Show that

Fff(x)g=

; F





= πδ(!);

and thus

Ffu(x)g=

+ πδ(!):

7.3.2 Shift in frequency

Let f (x) be a function with frequency distribution f

(!). This frequency distribution

is shifted around the point !

if f is mu ltip lied by cos(!

x). This is simply veriﬁed

by the following calculation:

Fff(x) cos(!

x)g=

f(x)e

¡i!x

f(x)e

¡i!

¡i!x

The ﬁrst integral is just f

(! ¡!

) and the second one is f

(! + !

), and ﬁnally

Fff(x) cos(!

x)g=

(! ¡!

) +

(! + !

Frequency modulation i s one of the most important applications of this property.

It allow s us to eﬃciently transmit data from diﬀerent users along the s a me cable

by allocating speciﬁc frequency bands to each user's data. By multiplying each

data stre a m f

(t) by cos(k!

t), the frequency ba nd of each user's data is shifted to

! = k!

, as shown in the ﬁgure below:

This modulation technique ensures that each user's data occupies a unique fre-

quency band, preventing inter ference an d enabling simultaneous transmission of

multiple data streams through the same cable. It is wid ely used in various commu-

nication technologies, including wireless co mmunication, radio broadcasting, and

digital signal processing.

Exercise 7.17. Show the following relation



cos(!

1 + x



¡j!¡!

+ e

¡j!+!

and conclude

cos(x)

1 + x

dx =

¡1

Exercise 7.18. A band pass ﬁlter is constructed by the function

f(x) =

2 sin(ax)

cos(!

x):

for a = 1, ﬁnd !

such that the ﬁlter passes all frequencies in the rang e [99; 101].

7.3.3 Shift in spatial variable x

For a function f(x) w ith the frequency distribution f

(!), the Fourie r transform of

f(x ¡x

) is

Fff(x ¡x

)g= e

¡i!x

(!):

18 Fourier Transform Method

In particular, if Ffδ(x)g= 1, then Ffδ(x ¡x

)g= e

¡i!x

. This property is the dual

of the previou s one. Shift in x causes a phase in !, and phase in x causes a shift in !.

7.3.4 Convolution

Convolution is a fundamental operation in the analysis of linear systems, and it

arises naturally in the c o ntext of several ﬁelds of applied mathematics, engineering

and physics. The convolution between two functions f(x) and g(x) deﬁned on the

real line (¡1; 1) is deﬁned as:

(f ∗ g)(x) =

¡1

f(y) g(x ¡y) dy:

This operation combines two functions into a new function, and it can be thought of

as a “weighted average” of the two fun ctions, where the weighting is de termined by

the function g(x ¡y) as y ranges over the entire real l ine. The convolution operation

is commutative, meaning that f ∗ g = g ∗ f . It is also associative and distributive

over addition.

This special integral operation naturally arises from the ana lysis of linear sys-

tems. Consider a linear system S with the respo nse h(x) to the impu lse δ(x). This

response h(x) is called the impulse response of the system. It describes how the

system reacts to an impulse input at time x = 0.

δ(x) h(x)

The system is called translation-invariant if its response to the sh ifted input

δ(x ¡ y) is h(x ¡ y). This property means that the behavior of the system does

not change with respect to time shifts in the input signal. In other words, a delay

or shift in the input signal results in a corresponding delay or shift in the system's

response, represente d by the function h(x ¡ y):

h(x ¡y)

Invariant

Translation

δ(x ¡ y)

Now, let's consider the input to the system, denoted as f (x). The function f(x)

can be expressed in the integral form as:

f(x) =

¡1

f(y) δ(x ¡y) dy;

or equivalently as a Riemann sum as:

f(x) = lim

n!1

k=¡n

f(y

) δ

(x ¡ y

) ∆y

In thi s form, we can see that the input f(x) is represented as a summation of shifted

impulses δ(x ¡y

) multiplied by weights f(y

) ∆y

. A system is linear if its response

to the summation is a summation as:

7.3 Properties of Fourier transform 19

Translation

Invariant

k=¡n

f(y

) δ

(x ¡ y

) ∆y

k=¡n

f(y

) h

(x ¡y

) ∆y

Therefore, the response of a linear translation-invariant system with impulse

response h(x) to an input f (x) is given by the convo luti o n integral:

(f ∗h)(x) =

¡1

f(y) h(x ¡y) dy:

This integral can be challenging to compute directly, but the Fourier transform

provides an elegant method to solve this convolution integral. The property of the

Fourier transform F states that the transform of a convolution of two functions is

the product of their individual transforms:

Fff ∗gg= Fff gFfgg;

for a ny two functions f ; g with the Fourier transforms f

; g^ respectively. This

property allows u s to simplify the convolution integral and express the response of

the system to an input as the product of their individual Fourier transforms. This

property is particularly useful in simplifying calculations and solving diﬀerential

equations through Fourier transforms.

Proof. Let f and g be two functions with Fourier transforms Fff g and Ffgg. We

have

Fff ∗gg=

¡1



¡1

f(y) g(x ¡y) dy



¡i!x

dx:

Using the Fubini theorem, and changing the o rder of integrals, we obtain

Fff ∗gg=

¡1

f(y)



¡1

g(x ¡ y) e

¡i!x



dy:

The internal integral can be written by the substitution z = x ¡y as:

¡1

g(x ¡ y) e

¡i!x

dx = e

¡i!y

¡1

g(z) e

¡i!z

dz;

and thus

Fff ∗gg=



¡1

f(y) e

¡i!y



¡1

g(z) e

¡i!z



= Fff gFfgg;

and this completes the proof. 

Exercise 7.19. Use the convolution property a nd d etermine the inverse Fourier transform:

¡1

(

¡t!

1 + !

)

;

where t > 0 is a constant.

Exercise 7.20. Show the relation:

¡1

fcos(!ct)g=

fδ(x ¡ct) + δ(x + ct)g;

20 Fourier Transform Method

and conclude:

¡1

(!) cos(c!t)g=

ff (x ¡ct) + f(x + ct)g:

Problem 7.7. Show the relation:

Fff(x) g(x)g=

2π

(!) ∗ g^(!):

This equality justiﬁes the fact that multiplication in x-domain result sin the convolution in !-

domain, and convolution in x-domain results in the multiplication in !-domain.

Problem 7.8. If f and g are integrable fu nctions, prove that the integral f ∗ g converges.

7.3.5 Diﬀer e ntiation and integration

Let f (x) be an integrable function and diﬀerentiable with the derivative f

(x). The

Fourier transform Fff

(x)g is given by

Fff

g= i! f

(!): (7.6)

This is simply seen by the direct calculation:

Fff

(x)g=

¡1

(x) e

¡i!x

dx = f( x) e

¡i!x



¡1

+ i!

¡1

f(x) e

¡i!x

dx:

Since f is integrable, it vanishes at x !±1, and thus:

Fff

(x)g= i!Fff(x)g:

In a similar manner, we obtain the formula:

Fff

(x)g= i!Fff

(x)g= ¡!

(!);

as long as f is a twice diﬀerenti a ble function and Fff(x)g exists.

This property is specially useful to solve linear second-order diﬀerential equa-

tions. For example, consider the following heat problem:



= u

u(x; 0) = f(x)

By taking Fourier transform of the problem with respect to the spatial variable x,

we arrive at the following o rdinary diﬀerential equation:

(

(!; t) = ¡!

u^(!; t)

u^(!; 0) = f

(!)

;

as long as Fff(x)g exists. This ﬁrst-order ODE can be solved for u^(!; t), and thus

u(x; t) can be retrieved by the inverse Fourier F

¡1

Now, let F (x) be an anti-derivative of f(x):

F (x) =

¡1

f(t) dt:

If f(x) be an integrable func tion, th en we have

FfF

(x)g= Fff(x)g= f

(!):

7.3 Properties of Fourier transform 21

On the other hand, we have

FfF

(x)g= i!FfF (x)g;

and thus



¡1

f(t)dt



Fff(x)g: (7.7)

Problem 7.9. In one of previous exercises, we obtained the formula

Ffu( x)g=

+ πδ(!):;

where u(x) =



1 x > 0

0 x < 0

. Show the relation

u(x) =

¡1

δ(t) dt:

On the other hand, if we use the formula (7.7), we have

Ffu( x)g=

Ffδ(x)g=

What is wrong here? Show that if we use the inver se Fourier transform, the r elation Ffu(x)g=

can not be true as we should have:

¡1

fu^(! )g=

u(0

) + u(0

)

7.3.6 Multiplicati on by x

If the function g(x) = xf (x) is integrable, and if f

(!) is diﬀerentiable, we have:

Ffxf(x)g= i

d f

(!)

This demonstrates a du a lity between f

and Fff

g on one side and xf and Ffxf g

on the other side.

The following theorem provides us with a suﬃcient condition for the diﬀerentia-

bility of f

(!). The proof is given in the appendix to this chapter.

Theorem 7.3. Assume that f (x) is an integrable, and piecewise continuously dif-

ferentiable function. Moreover, assume that there is R > 0 such that the function

f(x) decay sub-exponentially outside x 2(¡R; R), i.e., there are some α; β > 0 such

that jf(x)j< αe

¡β jxj

, for jxj> R. Then f

(!) is continuously diﬀerentiable.

Exercise 7.21. Show the following relation

1 + x

= ¡ iπe

¡j!j

sign(!):

7.3.7 Expansion and shrinking

Consider an integrable function f(x), and let a > 0 be a constant. If 0< a<1, the

function f(ax) stretches in the x-domain, and if a>1, the function shrinks in the

x-domain. This behavior is reﬂected in the Fourier transform of f (ax) as follows:

Fff(a x)g=

¡1

f(ax) e

¡i!x

dx =

¡1

f(y) e

¡i!y/a

dy =





22 Fourier Transform Method

Hence, if a>1, the Fourier transform Fff(ax)g stretches in the !-domain, while if

0<a<1, Fff(ax)g shrinks in the !-domain.

The interpretation of this behavior lies in the frequency spectrum of Fff(ax)g,

which represents the angular frequency components present in the function f(ax).

The frequency spectrum indicates how fast or slow a function changes periodically.

When a function shrinks in the x-domain (a>1), it undergoes fa ster ch a nges, leading

to a wider frequency sp ectrum. In contrast, fo r 0<a<1, the function changes more

slowly, resulting in a narrower frequency spectrum that predominantly covers low

frequencies.

For instance, the ﬁgure below illustrates the graphs of functions f(x) =

sin(x)

and

f(2x). We can observe how f (2x) is contracted around x = 0:

-10 -5 0 5 10

-0.4

-0.2

0.2

0.4

0.6

0.8

-10 -5 0 5 10

-0.4

-0.2

0.2

0.4

0.6

0.8

The ﬁgure below depicts the Fourier transforms Fff(x)g an d Fff(2x)g, and

as can be seen, Fff(2x)g is wider than Fff(x)g:

-2 -1 0 1 2

-0.5

0.5

1.5

2.5

3.5

-4 -2 0 2 4

0.5

1.5

Additionally, this behavior is related to a fundamental conc ept k nown as the

uncertainty principle. As we saw, the shrinking in the x-domain results in the

expansion in the !-domain and vice versa. This expansion in !-domain introduces

uncertainty in measurement . Thus, when there is certainty in measuring a func-

tion f(x), corresponding to the shrinking in the x-domain, there will be uncertainty

in measuring f

(!), which is a consequence of the expansion of f

(!). Similarly, the

shrinking of f

(!) results in uncertainty in m easuring f(x). This uncertainty prin-

ciple is a fundamental characteristic of the Fourier transform and has far-reaching

implications in various ﬁelds, including signal processing, quantum mechanics, and

more. In the problem set, we prove an important result of this principle.

7.3 Properties of Fourier transform 23

7.3.8 Unce rtainty principle

The shrinking a nd expansion prope rty of the Fourier transform is deeply connected

to a fundamental co nc ept k nown as the uncertainty pr inciple. As we have observed,

when a function shrinks in the x-domain, it leads to an expansion in the !-domain

and vice versa. Thi s expansion in the !-domain introduces uncertainty in the mea-

surement process. Hence, when there is high certainty in measuring a function f(x),

corresponding to t he shrinking in the x-domain, there will be greater uncertainty

in measuring f

(!), a result of the expansion of f

(!) in the frequency domain.

Conversely, the shrinking of f

(!) results in uncertainty in measuring f (x).

This uncertainty principle is a fundamental characteristic of the Fourier trans-

form and has signi ﬁcant implications in various ﬁelds, including signal processing,

quantum mechanics, and more. It fundamentally describes the trade-oﬀ between the

certainty in the time or spatial domain and the certainty in the freq ue ncy domain.

To quant ify the dispersion or spread of a function f around x = 0, we deﬁne the

dispersion of f as:

D(f) =

E(f)



¡1

jxf(x)j



1/2

;

provided that both integral s exist. A similar d eﬁ nition holds for D(f

), the dispersion

of f

(!) around ! = 0. These deﬁnitions closely resemble the deﬁnition of standard

variation in probability theory.

Theorem 7.4. As long as D(f) and D(f

) are bounded, the uncertainty principle

establishes the following inequality:

D(f) D(f

) ≥

This principle clearly indicates that the more certainty we have about the local-

ization of a function in the time or spatial dom a in (i.e., higher concentration around

x = 0), the le ss certainty we will have about its localization in the frequency domain

(i.e., f

(!) will be more spread out around ! =0), and vice versa. This inherent trade-

oﬀ is why the inequality is termed the uncertainty principle. For a simple proof of

the theorem, please refer to the problem set.

7.3.9 Energy and Plancherel theorem

In physics and engineering, the energy of a function f(x) is deﬁned as the integral:

E(f) =

¡1

jf (x)j

dx;

as long as the integral is bounded. The Plancherel theorem establishes a fundamental

relationship between the energy of f(x) and the energy of its Fourier transform,

(!).

Theorem 7.5. (Plancherel) Assume that f(x) is square integrable function. Then

the following relation holds

E(f

) = 2πE(f ): (7.8)

24 Fourier Transform Method

This powerful theorem states that the energy of the Fourier transform, f

(!),

is equal to 2π times the energy of the original function, f(x). Consequently, any

ﬁltration of f

(!) will directly impact the original function f (x).

To illustrate this theorem, let's consider th e function f(x)=e

¡jxj

. By applying

the Plancherel theorem, we ﬁnd that E(f)=1. Furthermore, a f ter calculati ng the

Fourier transform of f (x), which is f

(!)=

1 + !

, we determine that E



1 + !



= 2π.

In the graph of f

(!), we observe the interval [¡1.838; 1.838], which contains 95

percent o f the energy of f

(!). Engineers often use this cutoﬀ value to reconstruct

an approximate function f

(x) from the original function f(x).

−

1 1 2 3 4

(

)

The reconstruction f

(x) can be deﬁned as:

(x) =

2π

¡1.838

1.838

(!) e

i!x

d!:

Similarly, a higher f requency band for f

(!) will result in a more accurate reconstruc-

tion of f(x). This technique plays a crucial role in various applications, including

digital communication and signal processing, where obtaining an accurate representa-

tion of the original signal is essential. The graph depicts the results of reconstructions

0.95

and f

0.99

using the !-interval [¡3.373; 3.373] for the latter. As evident, the

broader frequency band results in a more precise reconstruction.

−

1 1 2 3

−

Exercise 7.22. Use Plancherel theorem and prove

sin(ax) sin(bx)

aπ

a < b

bπ

b < a

Hint: use the relation



sin(ax)





π ¡a < ! < a

0 otherwise

;

for any a > 0.

7.3 Properties of Fourier transform 25

Problem 7.10. Suppose f ; g ar e real admissible functions.

a) Use the relation:

Fff(x)g(x)g=

2π

(!) ∗ g^(!):

and prove the following relation

¡1

f(x)g(x)dx =

2π

¡1

(!) g^(!) d!:

This proves the Plancherel theorem.

¡1

jf (x)j

dx =

2π

¡1

(!)jd!:

7.3.10 Nyquist-Shannon rate

The Nyquist-Shannon theorem plays a crucial role in digital signal processing, par-

ticularly when transmitting electrical signals over transmission lines. In practical

applications, we often transmit a sampled signal (digital signal) rather than the

continuous function f(t) itself. To reconstruct the original signal at the destina-

tion without any loss of information, the sampling rate should satisfy the Nyquist-

Shannon rate.

Assuming that the signal f(t) has a bounded frequency band, i.e., ¡!

≤! ≤!

Nyquist and later Shannon showed that the original function can be accurately

reconstructed from the sampled signal if the sampling rate is at least

. This result

is fundamental in digital signal processing, as sampling a signal with the Nyquist

rate ensures the preservation of all information in the signal.

Since f

(!) is bounded with spectrum band [¡!

; !

], we can express the Fourier

transform f

(!) using the complex form of the Fourier series as follows:

(!) =

n=¡1

¡i

nπ

;

where c

is obtained through the inner product:

¡!

(!) e

nπ

d! =



2π

¡!

(!) e

nπ





nπ



Let's denote

nπ

by t

which can be considered as the sampling times of the function

f(t). This means that the original function f(t) is sampled at a rate of T

equivalently, at a rate of

. If we rep lace the angular frequency !

by 2πf

, we

obtain T

, and the sampling rate f

=2f

. Thus, we can write:

(!) =

n=¡1

f(t

) e

¡it

: (7.9)

This implies that to reconstruct f

(!), we only need the data set ff(t

n=¡1

Since we can reconstruct f (t) from its Fourier tran sform f

(!) by the inverse Fourier

transform, we on ly need to sample f (t) at the rate f

=2f

. This rate is known as

the Nyquist rate.

26 Fourier Transform Method

It's important to note that the Fourier series of f

(!) given in formula (7.9) is

periodic with a period of 2!

¡!

(!) P

To ﬁlter out the periodic copies of the principal part shown with the blue line

in the ﬁgure, we need to multiply this Fourier series with a low-pass ﬁlter:

p^(!) =



1 ! 2( ¡!

; !

)

0 otherwise

Therefore, the original function f(t) can be reconstructed as:

f(t) = F

¡1

(!) p^(!)g=

n=¡1

f(t

)

¡!

¡it

i!t

d! =

n=¡1

f(t

)

sin[!

(t ¡t

)]

(t ¡t

)

The last expression represents the discrete convolution of f(t) and

sin[!

t)]

at instance

nπ

. This de mon strates how the Nyquist rate and the use of low-pass ﬁlters

ensure accurate reconstruction of the original signal from the sampled one.

Example 7.3. Consider the following function

f(t) =

2(sin(t) ¡t cos(t))

πt

The Fourier transform of f is

(!) =

(

1 ¡!

¡1 ≤! ≤1

0 otherwise

For !

= 1, sequence ff(t

n=¡1

captures all information in f(t). In fact, the

reconstruction is done by the following series

f (t) =

n=¡1

f(nπ)

sin(t ¡nπ)

t ¡nπ

;

which converges fast to the original function. The ﬁgure below illustrates the original

function and three terms of the summations:

−

10 10 20

(

)

7.3 Properties of Fourier transform 27

7. 4 Highe r dimensional transf orm

The Fourier transform, as described above, can i ndeed be extended to functions of

two or more independent variables. This extension is particularly essential when

dealing w ith partial diﬀerential equations, such as the he at or wave equations, deﬁned

on a plane or in space. For a function f: R

!R that satisﬁes the condition

jf(x; y)jdxdy < 1;

its Fourier transform with frequency parameters !

and !

associated with x and y,

respectively, is deﬁ ned as f ollows:

; !

) =

f(x; y) e

¡i!

dxdy:

Similarly, the inverse Fourier transform is given by the following formula:

f(x; y) = F

¡1

4π

; !

) e

;

provided that the integral converges.

As an example, consider the Gaussian function f (x; y) = e

. The Fourier

transform of f is calculated as:

; !

) =

¡i!

dxdy =



¡i!



¡i!



Each integral on the right-hand side represents a 1D Fourier transform, which results

in:

; !

) = 2π e

This extension of the Fourier transform to multiple variables allows us to analyze

and solve complex problems in physics, engineering, and other ﬁelds where functions

are deﬁned in two o r more dimensions. It is a powerful tool for understanding the

frequency content and characteris tics of multidimensional signals and function s.

Exercise 7.23. Find the inverse Fourier transform: F

¡1

¡t(!

)

Exercise 7.24. Find the Fo ur ier transform: Ffe

¡tjxj¡sjyj

g for positive constants t; s.

Exercise 7.25. If f(x; y) is an odd function with respect to x, show

(¡!

; !

) = ¡f

; !

The same relation holds for the y variable. If f is even function with respect to x, then

(¡!

; !

) = f

; !

The extension of the Fourier transform to functions of n independent variables

is indeed deﬁned in a similar manner. For an integrable function f : R

!R, the n-

dimensional Fourier transform is given by:

Fff g= f

; :::; !

) =

f(x

; :::; x

) e

¡i!

···e

¡i!

dV ;

28 Fourier Transform Method

where dV is the volume element in R

, and !

; :::; !

are frequency parameters

associated with coordinate s x

; :::; x

. The inverse Fourier transform, allowing us to

reconstruct the original function, is deﬁned as:

¡1

g= f(x

; :::; x

) =

(2π)

; :::; !

) e

···e

d!:

Just like the prope rties of the 1D Fourier transform, the higher-dimensional Fourier

transform also exhibits similar properties. For instance, if f is smooth enough with

respect to its arguments x and y, then we have the following relationships:

Ff@

f g= i!

; !

); Ff@

f g= ¡!

; !

The n-dimensional Fourier transform plays a crucial role in va rious ﬁelds, especially

when dealing with functions deﬁned in multiple dimensions, such as in solving partial

diﬀerential equations in physics and engineering, image processing, and many other

applications.

Proposition 7.1. Assume that u(x; t) is continuously diﬀerentiable function with

respect to t, and u

and u are integrable fu nctions with respect to x for any t, then

we have

(!; t) =

¡1

(x; t) e

¡i!x

dx:

Proof. Since u(x; t) is integrable, for any t in x, we can deﬁne the Fourier transform

of u(x; t) as

u^(!; t) =

¡1

u(x; t) e

¡i!x

dx:

For " =/ 0, we have:

u^(!; t + ") ¡u^(!; ")

¡1

u(x; t + ") ¡u(x; t)

¡i!x

dx:

By the mean value theorem, we can write

u(x; t + ") ¡u(x; t)

= u

(x; τ

);

for some τ

in the interval t and t + " (if " > 0). Therefore, we have

u^(!; t + ") ¡u^(!; t)

¡1

(x; τ

) e

¡i!x

dx:

Now, let " approaches zero:

lim

!!0

u^(!; t + ") ¡u^(!; t)

= lim

"!0

¡1

(x; τ

) e

¡i!x

dx:

Since u

is integrable, we can pass the limit inside the integral, and write

lim

"!0

u^(!; t + ") ¡u^(!; t)

¡1

lim

"!0

(x; τ

) e

¡i!x

dx:

7.4 Higher dimensional transform 29

Since u

is continuou s is continuous, we obtain

(!; t) :=

¡1

(x; t) e

i!x

dx:

The continuity of u^

(!; t) with respect to t follows from the continuity and the

integrability of u

(x; t) and the dominant convergence theorem. 

Proble ms

Problem 7.11. Find the Fourier transform of following functions

a) f (x) =

+ x

b) f( x) = e

for jxj< 1 and f (x) = 0 in jxj> 1

c) f (x) =

(

¡x

x > 0

0 x < 0

d) f( x) =

(

¡x

x > 0

0 x < 0

e) f (x) =



1 a < x < b

0 otherwise

Problem 7.12. Find the inverse Fourier transfor m of the following functions

a) f

(!) =

¡2! + 2

. Hint: a shift in !.

b) f

(!) = e

¡j!j¡i!

Problem 7.13. Find the Fourier cosine integral of the following functions

a) f (x) = e

¡x

b) f( x) = xe

¡x

Problem 7.14. Use the deﬁnition of F

¡1

and show the relation:

¡1



+ πδ(!)



(x) =

1 x > 0

x = 0

0 x < 0

Hint: note that

¡1

sin(!x)

d! =

π x > 0

0 x = 0

¡π x < 0

Problem 7.15. Show the following relation

Ffcos(x

)g= π

cos



¡π



Find a relation for Ffsin(x

)g. H int: May you wish to use the formula

¡1

dx = iπ

Problem 7.16.

a) Show the following relation for t > 0

(

¡tjxj

jxj

)

= 2π

t + t

+ !

1/2

Hint: you may wish to use the formula

¡1

¡αz

dz =

;

for Refαg> 0.

30 Fourier Transform Method

b) Conclude the following relation

(

jxj

)

2π

j!j

In particular, the fun ction f( x) =

jxj

is self-dual with res pect to the balanced version

of the Fourier transform.

c) Find the following integral

cos(x)

dx:

Problem 7.17. Prove the relation:

lim

s!1

¡1

cos(sx)

1 + x

dx = 0:

Hint: use the relation for Ffcos(!

x) f(x)g.

Problem 7.18. Consider the function g(x) = 2

¡2jxj

a) Find the energy of g(x) and verify the Plancherel identity.

b) Find the frequency band of the 95 p ercent cut oﬀ energy of g(x).

c) Draw g(x) and the reconstructed function based on the band you found i n the part (b).

Problem 7.19. We give a proof for the uncertainty principle.

jf (x)j

(!)j

jf (x)j

(!)j

≥

a) Without loss of generality, assume kf k = 1 (why is this plausible?) Then, according

to the Plancherel identity, we have kf

k= 2π

. On the other hand, by the relation

Fff

g= i! f

(!), we obtain

(!) =

(x)j

We show ﬁrst the following inequality



jf (x)j



1/2



(x)j



1/2

≥

2π

b) Use the Cauchy-Schwarz inequality and show the following inequality



jf(x)j



1/2



(x)j



1/2

≥

xf (x) f

(x):

Show also that the equality holds only if xf(x) = λf

(x) for some λ 2R.

c) Show t ha t

xf (x)f

(x) dx ≥

and conclude the uncertainty principle. Show that the equality holds only if f (x) =ce

where c; a are some constants.

Problem 7.20.

a) Verify that the tent function

f(x) =

x + 1 ¡1 ≤x ≤0

1 ¡x 0 ≤x ≤1

0 otherwise

is the convolution of the following pulse function with itself:

p(x) =



1 ¡1/2 ≤x ≤1/2

0 otherwise

7.4 Higher dimensional transform 31

b) Use the property of the Fourier transform of convolutions to determine Fff(x)g.

Problem 7.21. Assume that f (x) is a smooth function with the Fourier transform f

(w), and

furthermore

n=¡1

f(n) < 1:

a) Prove the following relation which is known the Poisson summation

n=¡1

f(n) =

n=¡1

(2nπ):

Hint: note that

f(n) =

2π

¡1

(!) e

in!

d!;

and that for any m 2Z, we have

(2mπ) =

n=¡1

n;2m

;

where

n;2m

2π

(2m¡1)π

(2m+1)π

(!) e

¡in!

d!:

b) Use the result in part a) to prove the following identity:

n=¡1

1 + n

= π cothπ:

Problem 7.22. Find the Fourier transform of the following two-variable functions:

a) f (x; y) = e

¡jxj¡jy j

b) f( x; y) = e

¡tjxj¡sjyj

c) f (x; y) = e

¡x

¡y

d) f( x; y) =

+ x

)(b

+ y

)

Problem 7.23. Assume that f(x; y) is a smooth and integrable function. Furthermore, assume

that f is of order 2, i.e.,

f(λx; λy) = λ

f(x; y);

for any λ =/ 0. Find the formula of the Fourier transform Ff∆f g.

7. 5 Fou rier tra nsform and PDEs

The Fourier method is a powerful technique for so lv ing real-world problems involving

diﬀerential equations. The general idea of this method is as follows:

1. Start with the linear problem described in the x-domain.

2. Apply the Fourier transform to convert the problem into the !-domain.

This step invo lves transforming the diﬀerential equation and any initial or

boundary conditions.

3. Pe rform calculations and manipulations in the !-domain, where th e problem

becomes simpler due to the properties of the Fourier transform.

4. Use the inverse Fourier transform to obtain the solution back in the x-domain.

This step gives us the ﬁnal solution to the original problem.

32 Fourier Transform Method

linear problem

Inverse Fourier transform

-domain

Calculation in

Fourier Transform

-domain

Results in

-domain

linear problem

Working with the Fourier method in the frequency domain often allows us to

tackle complex problems more eﬃciently, as it unveils the frequency components

and their beh avior, which can be advantageous for various applications in physics,

engineering, signal processing, and many other ﬁelds.

7.5.1 1D heat equations

Let Ω be a conductive rod extending from ¡1 to 1 (in fact, it model s a very long

and thin hom o g en eous conductive rod). Consider the following he a t problem on Ω

for t > 0:



= u

x x

;

u(x; t) = f(x)

; (7.10)

where f(x) is an integrable function. Note that there is no boundary condition for

this problem; however, the boundedness condition should be satisﬁed by the solution

as x a pproaches ±1. By utilizing the Fourier transform in x, the problem reduces

to the following “ordinary” diﬀerential equation in t:

(

(!; t) = ¡!

u^(!; t)

u^(!; 0) = f

(!)

; (7.11)

where u

(!; t) represents the Fourier transform of u(x;t) denoted by Ffu(x;t)g. This

ordinary diﬀerential equation is then solved to ﬁnd the solution in the frequency

domain:

u^(!; t) = f

(!) e

¡!

: (7.12)

Now, we can return to the x-domain by performing the inverse Fourier transform.

We have two choices for th e inverse Fourier transform in this case:

1. Using the deﬁnition of the inverse Fourier transform and representing the

solution u as:

u(x; t) =

2π

¡1

(!) e

¡!

i!x

d!:

This is an integral formula over the frequency spectrum !. This integral

formula is very similar to the series s o lutions to the heat problems on bounded

domains.

2. Using the convolution as to express u as

u(x; t) = F

¡1

(!) e

¡!

g= f (x ) ∗F

¡1

¡!

7.5 Fourier transform and PDEs 33

As we saw in the previous section, we have

¡1

¡!

4πt

This function is called the heat kernel and is denoted by Φ(x; t) for t > 0.

Therefore, we can write u as the convolution

u(x; t) =

¡1

f(z) Φ(x ¡z; t) dz =

4πt

¡1

f(z) e

(x¡ z)

dz:

Both representations of u(x; t) are valid solutions to the given heat problem.

Example 7.4. Let f(x) be the rectangle function

f(x) =



1 ¡1 < x < 1

0 otherwise

;

and let us solve the following heat problem on x 2(¡1; 1)



= u

u(x; 0) = f(x)

The ﬁrst form of the inverse Fourier transform results in

u(x; t) =

¡1

sin(!)

¡!

i!x

d!:

The integral at the right-hand side at t = 0 is equal to F

¡1

2 sin(!)

which is

the rectangle function f(x). The ﬁgure below depicts the solution u by numerical

integration in the !-band [¡20; 20]. For a faster calculation, we used the following

equivalent formula:

u(x; t) =

sin(!)

¡!

cos(!x) d!:

-3 -2 -1 0 1 2 3

0.2

0.4

0.6

0.8

Notice that for any t>0, the so luti o n u(x; t) becomes smooth, even if the i nitial

condition u(x; 0) is discontinuous . This smoothing eﬀect is a characteristic behavior

of the heat equation, which helps in smoothi ng out the initial conditions. This fact

becomes m o re apparent w he n we use the second representation of u(x; t), namely

the convolution form:

u(x; t) = f (x) ∗Φ(x; t):

34 Fourier Transform Method

The function Φ(x; t) is smooth for all t >0, and when it is convoluted with any

piecewise continuous function, it eﬀectively removes discontinuities and imparts

smoothness to the function.

For the given f (x), we have

u(x; t) = f (x) ∗Φ(x; t) =

4πt

¡1

(x¡ z)

dz:

The ﬁgure below depicts the solution using the convolution:

-3 -2 -1 0 1 2 3

0.2

0.4

0.6

0.8

Remark 7.1. (Heat kernel) As previously discussed, the heat kernel Φ(x; t) plays

a fundamental role in solving the h eat equation with a point source δ(x) as the initial

condition. For t > 0, Φ(x; t) represents the temperature distribution throughout

space as time evolves from the point source:

= 0

= 1

As depicted in the graph, the behavior of Φ(x; t) for various instances of time

shows how he a t diﬀuses and spreads over space. Interestingly, as time approaches

zero, th e heat kernel approaches a Dirac delta function, which is a mathematical rep-

resentation of an instantane ous point source. Additionally, the convolution of Φ(x;t)

with any piecewise continuous function results in a smooth function, reﬂecting its

smoothing eﬀect. The heat kernel's remarkable properties make it a valuable tool in

understanding heat conduction, diﬀusion processes, and solving heat-related prob-

lems in various scientiﬁc and engineering applications.

Problem 7.24. Show that if f(x) is a piecewise continuous and integrable function, the

convolution f (x) ∗Φ(x; t) for any t > 0 is a smooth function. Show that

¡1

Φ(x; t) dx = 1;

7.5 Fourier transform and PDEs 35

for any t > 0, and that:

lim

t!0

¡1

f(x) Φ(x; t) dx = f(0);

for any integrable and continuous function f (x). Show that the following r elation holds for

any integrable function f (x) and if x is a continuity point of f .:

lim

t!0

f(x) ∗Φ(x; t) = f(x):

Problem 7.25. Let u(x; t) be the solution of the equation



= u

u(x; 0) = f (x)

Use the convolution form of the solution and show the relation:

¡1

u(x; t) dx =

¡1

f(x) dx:

Exercise 7.26. Show that Φ(x; t) is the solution of the following heat problem



= u

u(x; 0) = δ(x)

;

where δ(x) is t he Dir ac delta f unct ion. Thus, Φ(x; t) is the impulse response of the heat system

to the initial condition δ(t). Due to the relation f (x) = f (x) ∗δ(x), the response of the heat

system to the input f (x) is u = f(x) ∗Φ(x; t) as shown in the ﬁgure below:

Heat system

δ(x) u = Φ(x; t)

Heat system

u =

¡1

f(z) Φ(x ¡z) dz

f(x) =

¡1

f(z) δ(x ¡z) dz

Exercise 7.27. The convolution form of the representing solution is specially useful if the

Fourier transform of the initial condition is not eas y to calculate. Consider the following heat

problem

= u

u(x; 0) =



1 x < 0

2 x > 0

Write down the convolution solution and show that u(0; t) =

for all t > 0. Sketch the graph

of u(x; t) at time t = 0.1; 1.

You may use the following code in Matlab:

x=-3:0.01:3;

f=@(x) 1*(x<=0)+2*(x>0);

t=0.1;

ux1=integral(@(z) f(z).*exp(-(x-z).^2/(4*t)),-20,20, ...

'ArrayValued',true)/sqrt(4*pi*t);

t=1;

ux2=integral(@(z) f(z).*exp(-(x-z).^2/(4*t)),-20,20, ...

'ArrayValued',true)/sqrt(4*pi*t);

plot(x,f(x),x, ux1,x,ux2)

36 Fourier Transform Method

You will obtain a grap h similar to below:

-3 -2 -1 0 1 2 3

1.2

1.4

1.6

1.8

Exercise 7.28. Find and integral solution for the following problem deﬁned on x 2(¡1; 1):

(

= ku

u(x; t) = e

¡jxj

;

and draw the solution u(0; t) for t ≥0.

The solutions to n o n-hom o g eneous equations follow a similar approach. Let's

consider the heat problem on the domain (¡1; 1):



= ku

x x

+ h(x; t)

u(x; 0) = 0

Taking th e Fourier transform of the equation leads to the f o llowing ordinary diﬀer-

ential equation:

+ k!

u^ = h

(!; t);

where h

is equal to Ffh(x; t)g. This ordinary diﬀerential equation can be solved

using the methods dis cu ssed in the textbook on ordi nary d ifferential equations

(ODEs).

Example 7.5. Let's solve the following problem:

(

= u

δ(t ¡ 1)

1 + x

u(x; 0) = 0

The ordinary diﬀerential equation for u^(!; t) is:

(

(!; t) = ¡!

u^(!; t) + πe

¡j!j

δ(t ¡1)

u^(!; 0) = 0

To solve t his ordinary diff erential equation , we will use the Laplace transform

method. The unilateral Laplace transform for a function f (t), for t>0, is deﬁned as:

Lff g=

f(t) e

¡st

dt:

In particular, we have:

Lfδ(t ¡1)g= e

¡s

7.5 Fourier transform and PDEs 37

For a detailed discussion on this subject, the reader can refer to the textbook on ordi-

nary diﬀerent ial equations (ODEs). Utilizing the Laplace transform fo r the derived

ﬁrst-order ODE, we obtain:

(s + !

(s; !) = πe

¡j!j

¡s

;

where U

(!; s) is equal to Lfu^(!; t)g. By inverse Laplace transform L

¡1

, we obtain

u^(!; t) as

(!; t) = L

¡1

(!; s)g= πe

¡j!j

¡1



¡s

s + !



= πe

¡j!j

¡!

(t¡ 1)

u(t ¡1);

where u(t¡1) is the Heaviside function u(t) shifted to t¡1. The solution u in t he

x-domain can be obtained by taking the inverse Fourier transform as:

u(x; t) = F

¡1

fu^(!; t)g= F

¡1

fπe

¡j!j

¡!

(t¡1)

gu(t ¡1):

The convolution form of u is given by:

u(x; t) =

1 + x

∗Φ(x; t ¡1) u(t ¡1):

Problem 7.26. Let f (x) be an integrable function. Show that the solution of the problem:



= u

+ δ(t) f (x)

u(x; 0) = 0

;

is equal to the solution to the following homogeneous equation:



= u

u(x; 0) = f(x)

Exercise 7.29. Solve the following heat problem

(

= u

+ δ(t) e

¡x

u(x; 0) = e

¡jxj

Problem 7.27. Find an integral solution for the following partial diﬀerential equation

= u

+ u

u(x; 0) =

cos(x)

1 + x

Problem 7.28. Solve the following heat pr ob lem

= u

¡jxj

1 + x

u(x; 0) = 0

;

38 Fourier Transform Method

and sketch the graph of the solution at t = 1.

7.5.2 1D Wave pro blems

The method to solve a wave equation is completely similar to solve a heat problem.

Consider the following problem deﬁned on x 2(¡1; 1):

= c

u(0; x) = f(x)

u(0; x) = g(x)

Taking Fourier transform of the problem results in the following second-order ordi-

nary diﬀerential equation:

= ¡c

u^(!; 0) = f

(!)

(!; 0) = g^(!)

This diﬀerential equation is solved for:

u^(t; !) = f

(!) cos(c!t) +

g^(!) sin(c!t);

and thus:

u(t; x) = F

¡1

(!) cos(c!t)g+

¡1



g^(!) sin(c!t)



By the deﬁnition o f invers e Fourier transform, we have

u(x; t) =

2π

¡1

(!) cos(c!t) e

i!x

d! +

2πc

¡1

g^(!)

sin(c!t) e

i!x

d!:

Here again we have two choices to represent the inverse Fourier transform F

¡1

1. To use the deﬁnition of the inverse Fourier transform and represent the solu-

tion u as:

u(x; t)=

2π

¡1

(!) cos(c!t) e

i!x

d! +

2πc

¡1

g^(!)

sin(c!t)e

i!x

d!: (7.13)

2. Use the convolution and write u as:

u(x; t) = f (x) ∗F

¡1

fcos(c!t)g+

¡1



g^(!)



∗F

¡1

fsin(c!t)g: (7.14)

Recall the relation



g(τ ) dτ



g^(!);

and hence:

¡1



g^(!)



= i

g(τ ) dτ

Problem 7.29. Show the relations

¡1

fcos(c!t)g=

[δ(x ¡ct) + δ(x + ct)];

7.5 Fourier transform and PDEs 39

and

¡1

fsin(c!t)g=

[δ(x ¡ct) ¡δ(x + ct)]:

The result of the above problem leads to the following formula for u:

u(x; t) =

[f(x ¡ct) + f(x + ct)] +

x¡ct

x+ct

g(τ )dτ:

This formula is called D'Alembert formula for the 1D wave problem.

Exercise 7.30. With the formula (7.13), use the Euler formula for cos(c!t) and sin(c!t) and

conclude the convolution formula (7.14). Hence, two formula for u(x; t) are equivalent.

Example 7.6. Let's solve the following wave equation

= c

x x

u(x; 0) = f (x)

(x; 0) = 0

;

where f(x) i s the rectangle function f (x) =



1 ¡1 < x < 1

0 otherwise

. The inverse Fourier trans-

form can be written as

u(x; t) =

sin(!)

cos(c!t) cos(!x) dx:

The ﬁgure below uses the numerical integration on the frequency band [¡20; 20] for

c = 1 to represent the solution at times t = 0and t = 4:

-10 0 10

-0.2

0.2

0.4

0.6

0.8

-10 0 10

-0.1

0.1

0.2

0.3

0.4

0.5

The convolution solution does not need the numerical integration and yields the

exact solution:

-10 0 10

0.2

0.4

0.6

0.8

-10 0 10

-0.1

0.1

0.2

0.3

0.4

0.5

40 Fourier Transform Method

Example 7.7. Let us solve the following damped wave equation

+ 2ξu

= c

u(x; 0) = f (x)

(x; 0) = 0

;

where f(x) is the function f (x) =

(

(1 ¡ x

)/4 ¡1 ≤ x ≤ 1

0 othersise

. Taking the Fourier transform,

the given problem reduces to the following ordinary diﬀerential equation:

(

+ 2ξu^

= ¡c

u^(!; 0) = f

(!); u^

(!; 0) = 0

;

which is solved for the function:

u^(t; !) = e

¡ξt

(!)



cos(µt) +

sin(µt)



;

where µ = !

¡ξ

. The ﬁgure below shows the solution u(x; t ) for c = 1 and ξ = 0.5.

Observe that two branches are not separated completely due to the damping factor ξ.

−

2 2 4

= 1

= 2

= 3

Exercise 7.31. Find the solution of the following wave equation

= c

u(x; 0) = δ(x)

(x; 0) = 0

This solutio n is the res ponse of the wave system to the impulse displacement at time t = 0.

What will be the solution if the initial conditions changes to the following one:



u(x; 0) = 0

(x; 0) = δ(x)

Find the resp on se of the damped wave system

+ 2ξu

= u

;

to the impulse displacement exercised at t = 0.

Exercise 7.32. Consider the following wave equation on the domain x 2(¡1; 1):

= u

+ δ(t ¡1) f (x)

u(x; 0) = 0

(x; 0) = 0

;

7.5 Fourier transform and PDEs 41

where f (x) is the rectangle function

f(x) =



1 ¡1 < x < 1

0 otherwise

a) Use the Fourier transform and ﬁnd the function u^(!; t).

b) Show that u(x; t) can be determined as the convolution

u(x; t) =

f(x) ∗ g

(x):

What is g

(x)?

Exercise 7.33. Determine the integral solution of the wave equation

+ 0.6u

= u

u(x; 0) = f(x)

(x; 0) = 0

;

where f (x) is the function

f(x) =

(

sin(x)

¡π < x < π

0 otherwise

and draw the solution at t = 4. You can use the following code in Matlab to draw the solution:

x=-5:0.01:5;

xi=0.3;t=4;

f=@(x) sin(x).*(x>-pi & x<pi)./x;

fw=@(w) 2*integral(@(x) f(x).*cos(w.*x),0,pi);

u=exp(-xi*t)*integral(@(w) cos(w.*x).*fw(w).*(cos(sqrt(w.^2-

xi^2)*t)+xi*sin(sqrt(w.^2-xi^2)*t)./sqrt(w.^2-xi^2)),0,40, ...

'ArrayValued',true)/pi;

plot(x,u)

7.5.3 1D problems in semi-unbounded domains

The diﬀerential equations deﬁned on semi-unbounded domains such a s (0; 1) can

be eﬀectively solved using the Fourier sine or cosine integrals. As introduced before,

the Fourier sine transform of a function f(x) deﬁned on [0; 1) is deﬁned as:

ff(x)g:= f

(!) =

f(x) sin(!x) dx;

with the inverse

¡1

(!)g=

¡1

(!) sin(!x) d!:

It is important to note that F

¡1

(!)gj

x=0

=0, me a ning that the inverse Fourier

transform converges to zero at x = 0. This transform is particularly useful in solving

Dirichlet problems, where boundary conditions are s peciﬁed at one end of the semi-

unbou nded domain.

Example 7.8. Let's consider the heat problem deﬁned on the domain x 2(0; 1):

= u

; 0 < x < 1; t > 0

u(0; t) = 0; t > 0

u(x; 0) = f(x)

: (7.15)

42 Fourier Transform Method

Since the boundary condition at x=0 is of the Dirichlet type, we will use the Fourier

sine series to solve this problem:

(!; t) = ¡!

(!; t)

(!; 0) = f

(!)

;

where u^

is equal to the Fourier sine transform of u. This ordinary diﬀerential

equation is then solved for the function:

(!; t) = f

(!) e

¡!

By performing the invers e Fourier transform, we obtain the expression for u as:

u(x; t) =

¡1

(!) e

¡!

sin(!x) d!:

It is important to note that u(0; t)=0 for all t > 0, which satisﬁes the Dirichlet

condition at x=0. The ﬁgure below illustrates the solution u(x; t) for the initial

condition f(x) = e

¡x

0 0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

If the boundary condition at x = 0 is Neumann, the solution can be expressed in

terms of the Fo urier cosine transform:

ff(x)g:= f

(!) =

f(x) cos(!x) dx;

with its inverse given by:

¡1

(!)g=

¡1

(!) cos(!x) d!:

If we modify the boundary condition for the above example to u

(0; t)=0, the

integral solution changes to:

u(x; t) =

¡1

(!) e

¡!

cos(!x) d!:

For the g iven piecewise function f(x) deﬁned as: f(x) =



1 0 < x < 1

, we can calculate

its Fo urier cosine transform as:

(!) =

cos(!x)dx =

sin(!)

7.5 Fourier transform and PDEs 43

Consequently, the solution to the heat problem with the Neumann boundary con-

dition at time t > 0 is:

u(x; t) =

¡1

sin(!)

¡!

cos(!x) d!:

The ﬁgure below illustrates the solution at time t = 0.1. By comparing this solution

to the solution of the p revious example (Dirichlet boundary condition), we can

observe that the Neumann boundary condition enforces the derivative of the solution

at x = 0 to be zero for all t > 0:

0 0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

Remark 7.2. An alternative a pproach to solve Dirichlet problems is to extend the

functions involved as odd functions over the entire domain (¡ 1; 1). For example,

the odd extension of the problem (7.15) is formu lated as:



= u

; ¡1< x < 1; t > 0

u(x; 0) = f

(x)

;

where f

is the odd extension of f, deﬁned as:

(x) =



f(x) x > 0

¡f(¡x) x < 0

Taking the Fourier transform of this e xtended problem results in the solution:

(!; t) = f

(!) e

¡!

;

and ﬁnally u(x; t) is given by the convolution as:

u(x; t) = f

(x) ∗

4πt

¡1

(z)

(x¡ z)

dz:

It can be observed that u(0; t) is equal to zero as:

u(0; t) =

¡1

(z)

dz = 0:

44 Fourier Transform Method

This approach is particularly useful to represent the s o lutions in convolution form.

Both methods, either extending the functions as odd over the entire domain or using

the Fourier sine integral, yield the same results bu t in diﬀerent ex pressions. T he

choice of method depends on the convenience of the problem and the desired form

of the solution.

Exercise 7.34. Find and integral solution for the following heat problem deﬁned on [0; 1):

= u

+ tf (x); 0 < x < 1; t > 0

u(0; t) = 0 t > 0

u(x; 0) = 0

;

where f (x) =



1 0 < x < 1

Exercise 7.35. Consider the following wave equation deﬁned on [0; 1):

= u

+ δ(t ¡1)e

¡x

(0; t) = 0

u(x; 0) = 0

(x; 0) = 0

Use even extension of the function involved and then employ the Fourier transform to solve

the problem. Verify that your solution satisﬁes the b oundary condition u

(0; t) = 0.

Exercise 7.36. Let Ω be the interval [1; 1 ). Solve the following problem on Ω

= u

+ δ(t ¡1)e

¡x

(1; t) = 0

u(x; 0) = 0

Exercise 7.37. Consider the following heat problem on a semi-in ﬁnite domain

= u

u(0; x) = 1

u(x; 0) = f(x)

The boundary condition at x=0 is non-homogeneous. To solve this problem, we assume the

solution u(x; t) can be writ ten as the sum of a steady-state solution V (x) and a transient

solution w(x; t), i.e., u(x; t)=V (x)+w(x; t). The s teady-s tate solution V (x) satisﬁes the non-

homogeneous b oundary condition u(0;x)=1. Determine this function, and then ﬁnd an integral

solution for w.

7. 6 Highe r dimensional problems

n higher dimensional problems, we extend the Fourier transform method to solve

various partial diﬀerential equations on unbounded domains in R

. Speciﬁcally, we

focus on solving the heat equation, wave equation, Poisson equation, and Laplace

equation.

These problems a re considered over the entire space R

, as well as on speciﬁc

regions such as strips in the plane and half-planes. The Fourier transform provides

an eﬃcient and powerful tool for transforming these partial d iﬀerential equations

into simpler ordinary diﬀerential equations in the frequency domain.

7.6 Higher dimensional problems 45

7.6.1 Equations deﬁned on R

Consider the h eat equation u

=∆u with the initial condition u(x; y; 0)= f(x; y). To

solve this partial diﬀerential equation in two dimensions, we employ the two-variable

form of the Fourier tran sform. This results in the following ordinary diﬀerential

equation:

; !

; t) = ¡(!

+ !

) u^(!

; !

; t);

where !

and !

are the f requency variables associated with the x and y directions,

respectively. The Fourier transform of the function u(x; y; t) is deﬁned as:

; !

; t) =

u(x; y; t) e

¡i!

dxdy:

Using the deﬁnition of the Laplacian operator ∆, we can express the Fourier trans-

form ∆ as:

Ff∆ug= Ffu

+ u

g= ¡(!

+ !

) u^(!

; !

; t):

The resulting ordinary diﬀerential equation, along with the given initial c o ndition,

is solved to obtain the solution:

u^(!

; !

; t) = f

; !

) e

¡(!

Consequently, the soluti o n u in the xy-domain can be obtained by taking the inverse

Fourier transform:

u(x; y; t) = F

¡1

; !

) e

¡(!

This inverse transform can be expressed either as a double integral:

¡1

; !

) e

¡(!

4π

; !

) e

¡(!

;

or using the convolution:

¡1

; !

) e

¡(!

g= f(x; y) ∗F

¡1

¡(!

The Fourier transform method provides an eﬀective approach to solve the heat

equation in two dimensions, enabling us to study the evolution of temperature dis-

tributions over time in complex spatial domains.

It turns out that F

¡1

¡(!

g is given by

4πt

, which is known as

the 2D heat kernel and is denoted by Φ(x; y; t). This heat kernel shares similar

properties with the one-dimensional heat kernel we encountered in problems deﬁned

on unbounded one-dimensional conductive rods. Just like its one-dimensional coun-

terpart, the 2D heat kernel has the property of smoothing out initial conditions. As

time progresses (t>0), the heat kernel acts as a convolution operator with the initial

condition function f(x; y). It diﬀuses the initial information and spreads it over the

entire domain, leading to a smooth solution at each time instance. Additionally,

for t!0, the heat kernel a pproaches the Dirac delta function centered at the initial

point (x; y)=(0; 0). The ﬁgure below illustrates Φ(x; y; t) for t = 0.1 and t = 1:

46 Fourier Transform Method

Problem 7.30. Show the relation

¡1

¡(!

4πt

+ y

This is called the 2D hea t kernel:

Φ(x; y; t) =

4πt

+ y

Show the relation

Φ(x; y; t) d xdy = 1:

Furthermore, show that the following relation holds:

lim

t!0

f(x; y) Φ(x; y; t) dxdy = f (0; 0);

for any integrable and continuous fun ction f (x; y):

Exercise 7.38. Consider the heat pr ob lem



= ∆u

u(x; y; 0) = f (x)

Show that the solution of the problem is

u(x; y; t) = f(x) ∗Φ(x; t):

Exercise 7.39. Write an integral solution for the following heat problem

(

= ∆u

u(x; y; 0) = e

¡jxj¡jyj

Exercise 7.40. Solve the following heat equation on R

(

= 4∆u

u(x; y; 0) = e

¡x

¡y

7.6 Higher dimensional problems 47

Exercise 7.41. Solve the following heat problem on R

(

= ∆u + r

¡x

¡y

u(x; y; 0) = 0

Exercise 7.42. Solve the following heat problem on R

= ∆u +

δ(t ¡1)

(1 + x

)(1 + y

)

u(x; y; 0) = 0

Exercise 7.43. Solve the following heat problem on R



= ∆u + δ(t ¡1) δ(x) δ(y)

u(x; y; 0) = 0

Now, let's consider wave equation u

= c

∆u in R

. For simplicity, we consider

the following problem:

= c

∆u

u(x; y; 0) = f (x; y)

u(x; y; 0) = 0

Indeed, solving the wave equation in two dimensions using Fourier transform leads

us to the ordinary diﬀerential equation:

+ c

u^ = 0

where ! = !

+ !

. With the initial conditions:

(

u^(!

; !

; 0) = f

; !

)

u^(!

; !

; 0) = 0

;

we obtain the solution:

u^(!

; !

; t) = f

; !

) cos(ct!): (7.16)

To express the solution u(x; y; t ) in the xy-domain, we use the inverse Fourier

transform. The solution can be written as an inte g ral over the frequency domain:

u(x; y; t) =

4π

; !

) cos(ct!) e

Alternatively, the solution can also be expressed in the convolution form:

u(x; y; t) = f (x; y) ∗F

¡1

fco s(ct!)g: (7.17)

The inverse Fourier transform of the function cos



c !

+ !



is not straightfor-

ward and requires intricate calculations. Let us try to ﬁnd this inverse transform.

To do that, we rewrite !

; !

as !

= ! cosγ, !

= ! sinγ for γ 2[¡π; π]. Therefore,

we can write

¡1

fcos(ct!)g=

4π

¡π

cos(ct!)e

¡i!(xcosγ +ysinγ)

!d!dγ:

48 Fourier Transform Method

If we use the polar coordinate formula for x and y: x = r cosθ, y = r sinθ, we reach

¡1

fcos(ct!)g=

4π

¡π

cos(ct!)cos (r! cos(γ ¡θ))!d!dγ:

Using the formula:

2π

¡π

cos(s cosγ) dγ = J

(s);

where J

is the Bessel function of the ﬁrst kind, we obtain:

¡1

fcos(ct!)g=

2π

(r!) cos(ct!)!d!:

The integral at the right-hand side can be rewritten as:

2π

(ρ!) cos(ct!)!d ! =

2πc

(!) sin





d!:

Using the formula:

(!) sin





d! =

¡1

; ρ < ct;

we ﬁnally obtain:

¡1

fco s(ct!)g=

2πc

¡ρ

; ρ < ct:

Finally, the solution u(t; x; y) can be written as follows:

u(t; x; y) =

2pc

(x;y)

f(z

; z

)

¡(x ¡z

)

¡(y ¡z

)

;

where B

(x; y) is the disk centered at (x; y) of radius ct. By a similar calculation,

it is shown that if u

(0; x; y) = g(x), the solution is:

u(t; x; y) =

2p c

(x;y)

f(z

; z

)

¡(x ¡z

)

¡(y ¡z

)

2p c

(x;y)

g(z

; z

)

¡(x ¡z

)

¡(y ¡z

)

7.6.2 Equations on half-plane

Consider the domain Ω = f(x; y ); y > 0g. We are interested in solving the heat

equation u

= ∆u on Ω, subject to the bo undary condition u(x; 0; t) = 0 and the

initial condition u(x; y; 0) = f (x; y). To tackle this problem, we can make use of the

odd extension of f(x; y) with respect to y, which can be deﬁned as follows:

(x; y) =



f(x; y) y > 0

¡f (x; ¡y) y < 0

7.6 Higher dimensional problems 49

By performing the odd extension, we obtain the heat equation on the entire domain

as:



= ∆u

u(x; y; 0) = f

(x; y)

Next, we take the Fourier transform to derive the following ordinary diﬀerential

equation:

(

; !

; t) = ¡(!

+ !

) u^(!

; !

; t)

u^(!

; !

; 0) = f

; !

)

Solving this ordinary diﬀerential equation yields the function:

u^(!

; !

; t) = f

; !

) e

¡(!

To obtain the solution u(x; y; t), we apply the inverse Fourier transform and express

it in the convolution form:

u(x; y; t) = f

(x; y) ∗

4πt

¡1

; z

) e

(x¡ z

)

+(y ¡ z

)

It is important to note that the solution satisﬁes the boundary condition u(x; 0;

t) = 0 due to the symmetry argument:

u(x; 0; t) =

4πt

¡1

(x¡ z

)



¡1

; z

) e



= 0:

For problems with Neumann boundary conditions, the even extension of the func-

tions involved should be utilized.

Remark 7.3. An alternative approach to solve these problems is by using Fourier

sine o r cosine integrals, depending on the type of boundary condition involved.

Exercise 7.44. Consider the following problem

= ∆u + e

¡y

¡jxj

(x; 0; t) = 0

u(x; y; 0) = 0

;

on the half-plane y ≥0. Express the solution in the convolution form and in term of the integral

over !-domain. Verify that both expressions satisfy the prescribed boun dary condition.

Exercise 7.45. Solve the following heat problem on the half-plane x ≥0:

= ∆u

u(0; y; t) = 0

u(x; y; 0) =

¡x

1 + y

We are now a ddressing the Laplace equations on half-planes, which arise when

dealing with non-homogeneous boundary conditions f o r heat or wave equations.

Consider the following heat problem on the domain Ω = f(x; y); y ≥0g:

= ∆u

u(x; 0; t) = f (x)

u(x; y; 0) = g(x)

50 Fourier Transform Method

To solve this problem, we take the sol ution as u(x; y; t) = V (x; y) + w(x; y; t), where

V satisﬁes the Laplace equation with the boundary condition:



∆V = 0

V (x; 0) = f(x)

Since x is unbounde d in this domain, we take the Fourier transfo rm of the equation

with respect to x:

(

(!; y) ¡!

(!; y) = 0

(!; 0) = f

(!)

The general solution of this equation is:

(!; y) = A(!) e

¡!y

+ B(!) e

However, to ensure the solution remains bounded as y approaches inﬁnity, we assume

that B(!) = 0 for ! > 0 and A(!) = 0 for ! < 0. This implies that the solution V

(!;

y) can be represented as:

(!; y) = C(!) e

¡y j!j

;

where C(!) is an undetermined function. The initial condition V

(!; 0) enforc es C(!)

to be equal to f

(!), and therefore:

(!; y) = f

(!) e

¡y j! j

The function V (x; y) can be retrieved using the inverse Fourier transform F

¡1

, which

can be expressed in convolution form as:

V (x; y) = f(x) ∗

+ x

¡1

f(z)

+ (x ¡z)

dz:

An alternative representation using the deﬁnition of F

¡1

is:

V (x; y) =

2π

¡1

(!) e

¡yj!j

i!x

d!:

Exercise 7. 46. Let Ω be the domain Ω = f(x; y); x > 0; y > 0g. S olve the following Laplace

equation

∆u = 0 on Ω

u(x; 0) = f(x)

u(0; y) = 0

Express the solution in convolution form.

Exercise 7.47. Solve the following heat problem

= ∆u on Ω

u(x; 0; t) =

1 + x

u(x; y; 0) = 0

;

where Ω is the half-plane y > 0.

7.6 Higher dimensional problems 51

Problem 7. 3 1 . Let Ω be the domain Ω = f(x; y); y > 0g. Use the Fourier transform method and

show that the bounded solution to the L aplace equation ∆u = 0 on Ω subject to the boundary

condition u(x; 0) =

1 + x

is:

u(x; y) =

y + 1

(y + 1)

+ x

Problem 7. 3 2 . Let Ω be the domain Ω = f(x; y); y > 0g. Use the Fourier transform method and

show that the bounded solution to the L aplace equation ∆u = 0 on Ω subject to the boundary

condition u(x; 0) =



1 ¡1 < x < 1

0 otherwise

satisﬁes the relation:

u(0; y) =

tan

¡1





What is u(x; y)?

Problem 7.33. Show that the solution to the Laplace equation ∆u = 0 in the half-plane y > 0

subject to the bou ndary condition

u(x; 0) =

1 x > 0

x = 0

0 x < 0

;

u(x; y) =

tan

¡1





7.6.3 Equations on strips

Consider the heat equation u

= ∆u on the strip Ω = (¡1; 1) × (0; L) i n the xy-

plane, subject to the boundary conditions:



u(x; 0; t) = 0

u(x; L; t) = 0

Since the domain is unbounded in x, we take the Fourier transform of the equation

with respect to x and reach:

(!; y; t) = u^

(!; y; t) ¡!

u^(!; y; t)

u^(!; 0; t) = 0

u^(!; L; t) = 0

This leads to a heat equation in the (y; t) spac e, for 0 < y < L, which can be solved

using the method we introduced for problems in bou nded domains. The associated

eigenvalue problem is:

(

¡!

φ = ¡λφ

φ(0) = φ(L) = 0

Solving f o r the eigenfunctions a nd eigenvalues yields:

(y) = sin



nπ



; λ

= !

Hence, the solution to the resulting heat equation in (y; t) is given by:

u^(!; y; t) =

n=1

C(!) e

¡!

sin



nπ



52 Fourier Transform Method

The coeﬃcient fun ctions C(!) can be determined by the initial condition of the

problem i f provid ed . Let the initial condition be given by: u(x; y; 0) = f (x; y). By

taking the Fourier transform, we obtain u^(!; y; 0) = f

(!; y), an thus

(!; y) =

n=1

C(!) sin



nπ



;

and by performing the inner product, we obtain:

C(!) =

(!; y) sin



nπ



dy:

The solution u(x; y; t) can be retrieved by performing the inverse Fourier transform

as:

u(x; y; t) = F

¡1

fu^(!; y; t)g=

n=1

¡1

fC(!) e

¡!

sin



nπ



Exercise 7.48. Let Ω be the strip (¡1; 1) ×(0; 1) in the xy-plane. Consider the following

heat problem in Ω

= ∆u on Ω

u(x; 0; t) = u(x; 1; t) = 0 B:Cs:

u(x; y; 0) =

sin(πy)

1 + x

I:C:

a) Find an integral solution for the problem.

b) Show that u

;



is equal to



;



= e

¡π

1 + z

dz:

Exercise 7.49. Let Ω be the strip (0; 1) ×(¡1; 1) in the xy-plane. Determine and integral

solution for the following wave equation:

= ∆u + δ(t ¡1) sin(πx) on Ω

u(0; y; t) = u(1; y; t) = 0 B:Cs:

u(x; y; 0) = 0 I:C:

(x; y; 0) = 0 I:C:

The Laplace equation deﬁned on a strip in the xy-plan e can also be solved using

the Fourier tran sform method. Let Ω be the same strip as above, and consider the

following Laplace equation:

∆u = 0 on Ω

u(x; 0) = f (x)

u(x; L) = g(x)

Taking the Fourier transform in x yie lds:

(!; y) ¡!

u^(!; y) = 0

u^(!; 0) = f

(!)

u^(!; L) = g^(!)

7.6 Higher dimensional problems 53

This equation can be solved using the ordinary differential equations method,

resulting in:

u^(!; y) = f

(!) cosh(!y) +



g^(!)

sinh(!L)

¡f

(!) coth(!L)



sinh(!y):

The solution u(x; y) is retrieve d by performing the inverse Fourier transform F

¡1

Exercise 7.50. Solve the heat problem u

= ∆u on the strip 0 < y < 1, ¡1< x < 1 subject

to the boundary condition

(

u(x; 0; t) = 0

u(x; 1; t) =

1 + x

;

and the initial condition u(x; y; 0) = 0.

Proble ms

Problem 7.34. Let Ω be the upper half-plane y > 0. Solve the following heat problem:

= ∆u on Ω

u(x; 0; t) = e

¡jxj

u(x; y; 0) = e

¡x

¡y

Problem 7.35. Solve the following heat equa tion on the upper half plane y > 0

= ∆u

(x; 0; t) = 0

u(x; y; 0) = e

¡jxj

¡y

Problem 7.36. Let Ω be the half-plane y > 0. Consider the Laplace equation



∆u = 0 on Ω

u(x; 0) = f (x)

The solution in the convolution form is:

u(x; y) =

¡1

f(z)

+ (x ¡z)

dz:

At ﬁ rst gla nce, it may seem that lim

y!0

u(x; y) = 0. However, this assumption is not true. Sketch

the graph of the solution u(x; y) for y = 0.1; 0.05; 0.01 if f(x) = e

¡jxj

and explain the co nverges

of the solution to f(x).

Problem 7.37. Write the solution of the following heat problem deﬁned in the quadrant x > 0,

y > 0 in the integral form

= ∆u

u(0; y; t) = f(x)

u(x; 0; 0) = 0

u(x; y; 0) = g(x; y)

Problem 7.38. Let Ω be the strip f(x; y); 0 < y < 1g

a) Show that the solution to the Laplace equation



∆u = 0

u(x; 0) = 0; u(x; 1) = δ(x + 1) + δ(x ¡1)

u(x; y) =

sinh(!y)

sinh(!)

cos(!) cos(!x) d!:

54 Fourier Transform Method

b) Find a solution if the boundary condition changes to u(x; 1) = δ(x + 1) ¡δ(x ¡1).

Problem 7.39. Solve the following heat pr ob lem in the strip Ω = f(x; y); 0 < y < 1g

= ∆u

u(x; 0; 0) = u(x; 1; t) = 1

u(x; y; 0) =



1 ¡1 < x < 1

0 otherwise

Problem 7.40. Find the convolution solution to the following Poisson equation on Ω: f(x; y);

y ≥0g



∆u = f(x)

u(x; 0) = 0

;

where f (x) is the function

f(x) =



¡1 ¡1 < x < 0

1 0 < x < 1

Problem 7.41. Let f (x); x 2(¡1; 1) be a continuous function with the Fourier t ransform

(!). Consider the following heat equation

(

= u

+ e

f(x)

u(0; x) =

f(x) ∗e

¡jxj

a) Show that the solution of the above equation can be in the form

u(t; x) = f (x) ∗ g(t; x):

What is g(t; x)?

b) Assume that f (x) =



1 ¡1 < x < 1

0 otherwise

. What is u(t; x) for x > 1?

Problem 7.42. Consider the following equation on x 2(¡1; 1)

(

= c

+ δ(t) f (x)

u(0; x) = 0; u

(0; x) = 0

;

where δ is the Dirac delta function and f (x) is a continuous function.

a) Show that the solution can be written in the following convolution form (as long as the

convolution exists)

u(t; x) = α (f ∗ g)(x);

for some constant α and function g. Determine α; g(x).

b) Show that the convolution can be calculated as follow s

(f ∗ g)(x) =

a(t;x)

b(t;x)

f(y) dy:

What are functions a(t; x); b(t; x)?

7.6 Higher dimensional problems 55

7. 7 Appendi x

7.7.1 A proof of the theorem (7.3)

If f is integrable, then f

is continuous: Fix !, then fo r any " =/ 0, we h ave

lim

"!0

(! + ") = lim

"!0

f(x)e

¡i!x

¡i"x

dx:

Since f is integ rable, the dominant converge nce theorem allows us to pass the limit

inside the integral and write

lim

"!0

(! + ") =

f(x) e

¡i!x

lim

"!0

¡i!"

dx = f

(!):

Now we prove that f

(!) is diﬀerentiable with respect to !. Fix ! and write:

(! + ") ¡ f

(!)

f(x) e

¡i!x

¡i"x

¡1

dx:

Using the mean value theorem, we can write

¡i"x

¡1

= xe

¡ix"

;

for some "

in the interval (0; ") if " > 0 (alternatively, ("; 0) is " < 0). Thus,

(! + ") ¡ f

(!)

= ¡i

xf(x) e

¡i!x

¡ix"

The condition of the exponentially decay of f at inﬁnity implies that there is R > 0

such that

jf(x)j≤αe

¡β jxj

; jxj> R;

for some constants α; β > 0. N o te that this implies the function g(x) = xf (x) to be

integrable and piecewise continuously diﬀerentiable. We have

lim

"!0

xf(x) e

¡i!x

¡ix"

dx =

xf(x) e

¡i!x

lim

"!0

¡ix"

dx =

xf(x) e

¡i!x

dx:

Therefore,

lim

"!0

(! + ") ¡ f

(!)

= ¡i

xf(x) e

¡i!x

dx = ¡iFfxf (x)g:

Since xf(x) is integrable, Ffxf(x)g exists.

Exercise 7.51. If f(x) decays exponentially at inﬁnity, the function g

(x) = x

f(x) is

integrable.

7.7.2 A proof of the Fourier theorem

We need the following Fubini theo rem

56 Fourier Transform Method

Theorem 7.6. Assume that f(x; y) is an int egrable function, i.e.,

I =

jf (x; y)jdA < 1:

If integrals



f(x; y) dx



dy; I



f(x; y) dy



dx;

exist, then I

= I

We also need the following lemma.

Lemma 7.1 . (Lebesgue-Riemann) Assume that f (x) is a piecewise continuous

and integrable function. We have the following relat ion:

lim

n!1

¡1

f(x) sin(nx) dx = 0:

In f a ct, we have

(n) =

¡1

f(x) e

¡inx

dx =

¡1





¡ix

dx:

The dominant convergence theorem allows us to write

lim

n!1

(n) = lim

n!1

¡1





¡ix

dx =

¡1

lim

n!1





¡ix

dx = 0:

Now we can prove t he Fourier theorem.

Theorem 7.7. Assume that f is an integrable ans piecewise continuously diﬀeren-

tiable function. Then we have

2π

lim

n!1

¡n

(!) e

i!x

d! =

[f(x

) + f (x

)]:

For simplicity, we assume that f is C

Proof. Sinc e f

(!) is continuous, then for any n > 0, the integral

(x) =

¡n

(!) e

i!x

d!;

exists. On the other hand, we have

(x) =

¡n



f(ξ) e

¡iξ!

dξ



i!x

d! =

¡n

f(ξ) e

i!(x¡ξ)

dξd!:

By Fubini theorem, we can switch the integrals as

(x) =

f(ξ)



¡n

i!(x¡ξ)



dξ:

7.7 Appendix 57

Note that:

¡n

i!(x¡ξ)

d! = 2

sin [n(x ¡ ξ)]

x ¡ ξ

;

and then

(x) = 2

f(ξ)

sin [n(x ¡ ξ)]

x ¡ ξ

dξ:

By taking z = x ¡ ξ, we ob tain

(x) = 2

f(x ¡z)

sin(nz)

dz:

We have

f(x ¡z)

sin(nz)

dz =

¡1

f(x ¡z)

sin(nz) dz +

¡1

f(x ¡z)

sin(nz)

dz +

f(x ¡z)

sin(nz) dz:

By the Lebesgue-Riemann lemma, we have

lim

n!1

¡1

f(x ¡z)

sin(nz) d z = lim

n!1

f(x ¡z)

sin(nz) dz = 0:

We have al so

¡1

f(x ¡z)

sin(nz)

dz =

¡1

[f(x ¡z) ¡ f(x)]

sin(nz)

dz + f (x)

¡1

sin(nz)

lim

n!1

(x) = 2 lim

n!1



x ¡



sin z

dz = 2

lim

n!1



x ¡



sin z

dz =

=2 f(x)

sin z

dz = 2πf(x);

and this completes the proof. 

58 Fourier Transform Method