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
-4 -2 0 2 4
-5
0
5
10
15
20
-4 -2 0 2 4
-10
0
10
20
30
40
Now, consider an electrical or mechanical signal composed of various sub-signals
with different angular frequencies:
f(x) = c
1
cos(x) + c
2
co(2x) + ···+ c
n
cos(nx):
The function Fff(x)g has spikes at physical freque nc ies ! = 1; 2; :::; n weighted
by the values c
1
; c
2
; :::; c
n
. 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
0
1
2
3
-4 -2 0 2 4
-150
-100
-50
0
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) =
1
2π
Z
¡1
1
f
^
(!) 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) =
Z
¡1
1
f
^
(!) 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
1
(x) = e
¡0.2jxj
cos(x)
and f
2
(x) = e
¡0.2jxj
cos(2x). The function f
2
changes twice faster than f
1
. Draw these two
functions in x-domain. Now draw the functions f
^
1
(!) and f
^
2
(!) 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
2
Z
0
40
f
1
(x) cos(!x) dx;
instead of the integral
Z
¡40
40
f
1
(x) e
¡i!x
dx:
Your figures should be like the following ones:
-10 -5 0 5 10
-0.5
0
0.5
1
-10 -5 0 5 10
-0.5
0
0.5
1
-2 -1 0 1 2
0
1
2
3
4
5
6
-4 -2 0 2 4
0
1
2
3
4
5
6
Exercise 7.3. The function f (x) = 1 is not integrable. Its Fourier transform is
Ff1g= lim
L!1
Z
¡L
L
e
¡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(!
0
x) when !
0
approaches 0. Compare this result to the Fourier
transform of cos(!
0
x) when !
0
!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(!
0
x). This function is not integrable. Plot the imaginary
part of Ffsin(!
0
x)g. What do you observe at !
0
and ¡!
0
? Use the following code in Matlab
to draw Ffsin(!
0
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):
u
t
= ku
xx
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) =
1
2π
Z
¡1
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!
L
x
g or equivalently, trigonometric
functions
cos
¡
nπ
L
x
; sin
¡
nπ
L
x
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
u
t
= ku
xx
and o btain the ordinary differential equation:
u^
t
(!; t) = ¡k!
2
u^(!; t):
This differential equation is with respect to time t, and it can be solved as:
u^(!; t) = C(!) e
¡k!
2
t
;
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!
2
t
. Consequently, the integral
solution for u(x; t) is retrieved by the inverse Fourier transform:
u(x; t) = F
¡1
ff
^
(!) e
¡k!
2
t
g: (7.4)
This solution can be represented in integral form as:
u(x; y) =
1
2π
Z
¡1
1
f
^
(!) e
¡k!
2
t
e
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]:
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
u
t
= ku
xx
¡L < x < L
u(¡L; t) = u(L; t)
u
x
(¡L; t) = u
x
(L; t)
u(x; 0) = f (x)
:
The solution to this problem in complex form and in terms of the eigenfunctions
fe
in!x
g for ! =
π
L
is given as:
u(x; t) =
X
n=¡1
1
F
n
e
¡kn
2
!
L
2
t
e
in!
L
x
:
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
t
= u
xx
+ 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=
Z
¡a
a
e
¡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)
z
.
7.2 Fourier transform of important functions 9
-20 -10 0 10 20
-0.5
0
0.5
1
1.5
2
Now, let's investigate the inverse Fourier transform of
2 sin(a!)
!
. The expression
for the inverse transform is given as:
F
¡1
2sin(a!)
!
=
1
π
Z
¡1
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
0.2
0.4
0.6
0.8
1
-4 -2 0 2 4
-0.2
0
0.2
0.4
0.6
0.8
1
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
x
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
Z
¡1
1
sin(az)
z
dz = π:
Now, let f be a continuous function and integrable. Show th e relation
lim
a!1
Z
¡1
1
sin(az)
πz
f(z) dz = f (0):
Now, with these results, we can symbolically write:
Ff1
x
g= 2πδ(!):
10 Fourier Transform Method
Note also that
F
¡1
f2πδ(!)g=
1
2π
Z
¡1
1
2πδ(!) e
i!x
d! = 1
x
:
Using the inverse transform formula, we can find the transform of the function
f(x) =
sin(ax)
x
. We have
Z
¡1
1
sin(a!)
!
e
i!x
d! =
π ¡a < x < a
0 otherwise
Changing ! !x and x !¡!, we obtain
Z
¡1
1
sin(ax)
x
e
¡i!x
dx =
π ¡a < ! < a
0 otherwise
;
and thus
F
sin(ax)
x
=
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
f
a
(x) =
(
1
2a
¡a < x < a
0 otherwise
;
for a > 0. Let a approaches 0, find Fff
a
(x)g. What is limiting function f
a
(x) when a !0?
7.2.2 Exponential functions
Let f be given by: f
a
(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
a
(x)g=
Z
¡1
1
e
¡ajxj
e
¡i!x
dx =
Z
¡1
0
e
(a¡ i!) x
+
Z
0
1
e
¡(a+i!)x
=
2a
a
2
+ !
2
:
The figure below illustrates the reconstruction of f
a
(x) for a = 1 by its Fourier
transform:
-2 -1 0 1 2
0
0.2
0.4
0.6
0.8
1
7.2 Fourier transform of important functions 11
When a approaches zero, f
a
(x) approaches the constant function f ≡ 1. Then,
we expect that Fff
a
(x)g approaches the Dirac delta function 2πδ(!).
Problem 7.3. Show first that
Z
¡1
1
a
a
2
+ x
2
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
Z
¡1
1
ah(x)
a
2
+ x
2
dx = πh(0):
Exercise 7.6. Use the inverse Fourier transform and show the following relation
Z
0
1
cos(ax)
1 + x
2
dx =
π
2
e
¡jaj
:
Exercise 7.7. Use the inverse Fourier transform and show the relation
F
1
a
2
+ x
2
=
π
a
e
¡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)=
(
e
¡"x
x > 0
¡e
"x
x < 0
. Find Fff
"
g
and let " approaches zero.
Problem 7.4. Let g
a
(x) =
1
2
af
a
(x). Show that g
a
(x) is a Dirac delta sequence function, i.e.,
lim
a!1
Z
¡a
a
g
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
2
. Its Fourier tran sform i s:
Ffe
¡ax
2
g=
Z
¡1
1
e
¡ax
2
e
¡i!x
dx = e
¡
!
2
4a
Z
¡1
1
e
¡a
x¡
i!
2
2
dx:
The last integral in the above equality is
Z
¡1
1
e
¡a
x¡
i!
2
2
dx =
π
a
r
;
and thus
Ffe
¡ax
2
g=
π
a
r
e
¡
!
2
4a
:
In particular, if a =
1
2
, we obtain
F
n
e
¡
x
2
2
o
= 2π
p
e
¡
!
2
2
:
12 Fourier Transform Method
Exercise 7.9. Show the relation
F
¡1
fe
¡t!
2
g=
1
4πt
p
e
¡
x
2
4t
;
for t > 0. Show also the equality
Z
¡1
1
1
4πt
p
e
¡
x
2
4t
dx = 1:
Furthermore, show that the function f
t
(x) =
1
4πt
p
e
¡
x
2
4t
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
Z
¡1
1
f
t
(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:
Z
¡1
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:
Z
¡1
1
h(x) δ(x ¡x
0
) dx = h(x
0
):
Additionally, we can define the Dirac delta function sequence a s follows: A sequence
of functions δ
n
(x) is called a Dirac delta sequence if for any continuous and bounded
function h, we have:
lim
n!1
Z
¡a
a
h(x) δ
n
(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
δ
n
(x):
Exercise 7.10. Prove that the following functions are Dirac delta sequences:
a) The sequence
δ
n
(x) =
(
n
2
¡
1
n
< x <
1
n
0 otherwise
:
b) The sequence
δ
n
(x) =
8
>
>
>
>
<
>
>
>
>
:
n
2
x + n ¡
1
n
< x < 0
¡n
2
x + n 0 < x <
1
n
0 otherwise
7.2 Fourier transform of important functions 13
The Fourier transform of the function f(x) = δ(x ¡x
0
) is defined as
Ffδ(x ¡x
0
)g=
Z
¡1
1
δ(x ¡x
0
) e
¡i!x
dx = e
¡i!x
0
Z
¡1
1
δ(x ¡x
0
) dx = e
¡i!x
0
:
In particular, if x
0
= 0, then Ffδ(x)g= 1
!
. On the other hand, by the definition of
the inverse Fourier transform, we have
F
¡1
f1
!
g=
1
2π
lim
L!1
Z
¡L
L
e
i!x
d! =
1
π
lim
L!1
sin(xL)
x
:
The sequence
1
π
sin(nx)
x
is a Dirac delta sequence for L !1, and thus:
F
¡1
f1
!
g= δ(x):
Exercise 7.11. Show that the sequence of functions
1
π
sin(nx)
x
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=
1
2π
p
Z
¡1
1
f(x) e
¡i!x
dx;
and
F
¡1
ff
^
(!)g=
1
2π
p
Z
¡1
1
f
^
(!) e
i!x
d!:
It is simply seen that f (x) = 2π
p
e
¡
x
2
2
is a self-dual function with the above definitions as:
F
n
2π
p
e
¡
x
2
2
o
= 2π
p
e
¡
!
2
2
= f (!):
Consider the function defined as:
f(x) = 2π
p
X
n=¡1
1
δ(x ¡2nπ):
Show that its Fourier transform with respect to the balanced form of F is:
f
^
(!) = 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
0
and assume that
h is function continuous in the interval ( n
0
¡"; n
0
+ ") for small " > 0 and zero outside of this
interval. Show the relation
lim
n!1
Z
n
0
¡"
n
0
+"
sin[(2n + 1)π!]
sin(π!)
h(x) = h(n
0
):
This implies that f
^
(!) behaves like a Dirac delta sequence function at ! = n 2Z . The function
P
n=¡1
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(!
0
x). As we saw above, its Fourier transform is
Ffcos(!
0
x)g= lim
L!1
sin[(! ¡!
0
)L]
! ¡!
0
+
sin[(! + !
0
)L]
! + !0
:
14 Fourier Transform Method
On the other hand, the sequences
sin[(! ¡ !
0
)n]
π(! ¡ !
0
)
and
sin[(! + !
0
)n]
π(! + !
0
)
are Dirac delta sequence
functions and approach δ(! ¡!
0
) and δ(! + !
0
) respective ly . Therefore, we obtain:
Ffc o s(!
0
x)g= πδ(! ¡!
0
) + πδ(! + !
0
):
These are two spikes at physical frequencies !
0
and ¡!
0
that we discussed in the
previous section.
Exercise 7.12. Calculate the inverses F
¡1
fδ(! ¡!
0
)g, and F
¡1
fδ(! + !
0
)g and conclude
Ffcos(!
0
x)g=
1
2
fδ(! ¡!
0
) + δ(! + !
0
)g:
This is depicted in the figure below:
!
0
¡!
0
πδ(! ¡!
0
)πδ(! + !
0
)
!
Find the Fourier transform Ffsin(!
0
x)g and draw its imaginary part in !-domain.
Exercise 7.13. Show the relation
Ffsin(!
0
x)g= iπδ(! ¡!
0
) ¡iπδ(! ¡!
0
):
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
o
=
f(x) x > 0
¡f(¡x) x < 0
;
we can write Fff
o
g as:
Fff
o
g= ¡i
Z
¡1
1
f
o
(x) s in(!x) dx = ¡2i
Z
0
1
f(x) sin(!x) dx:
This is called the Fourier sine transform while its imaginary part is called the Fourier
sine integral
F
s
ff(x)g:= f
^
s
(!) =
Z
0
1
f(x) sin(!x) dx:
Note that f
^
s
(!) i s an odd function and f
^
s
(0) = 0. By the inverse Fourier transform,
we have
f
o
(x) = F
¡1
f¡2i f
^
s
(!)g=
1
π
Z
¡1
1
f
^
s
(!) 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
π
Z
¡1
1
f
^
s
(!) sin(!x) d!;
where f
^
s
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:
f
e
(x) =
f(x) x > 0
f(¡x) x < 0
;
and thus:
Fff
e
(x)g:= 2
Z
0
1
f(x) cos(!x) d x:
The Fourier cosine integral is defined as:
F
c
ff(x)g= f
^
c
(!) =
Z
0
1
f(x) cos(!x) d x:
Note that f
^
c
(!) is an eve n function in ! and f
c
0
(0) = 0. Therefore, a function f(x)
which is defined on the domain [0; 1) can be expressed as:
f(x) =
1
π
Z
¡1
1
f
^
c
(!) cos(!x) d!;
where f
^
c
(!) 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
0.5
0 1 2 3
0
0.5
1
-10 -5 0 5 10
0
0.5
1
0 1 2 3
0
0.5
1
16 Fourier Transform Method
The Fourier sine integral is simply derived as:
f
^
s
(!) =
Z
0
1
e
¡x
sin(!x) dx =
!
1 + !
2
;
and the Fourier cosine integral as:
f
^
c
(!) =
Z
0
1
e
¡x
cos(!x) dx =
1
1 + !
2
:
Exercise 7.14. Use the above results and determine the value of following integrals:
Z
0
1
cos(x)
1 + x
2
dx;
Z
0
1
x sin(x)
1 + x
2
dx:
Exercise 7.15. Draw the Fourier sine and cosine integrals of the function f (x) = e
¡x
2
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
1
and f
2
, and any constants c
1
and c
2
, the Fourier
transform satisfies the linearity property:
Ffc
1
f
1
+ c
2
f
2
g= c
1
f
^
1
(!) + c
2
f
^
2
(!):
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
1
f
1
+ c
2
f
2
are equal
to the sum of the frequency components in f
1
and f
2
, weighted by their respective
constants c
1
and c
2
. This property makes the Fourier transform a powerful tool for
analyzing the linear systems.
Problem 7.6. Prove that if f
1
and f
2
are integrable, then c
1
f
1
+c
2
f
2
is for any constant c
1
; c
2
.
Exercise 7.16. Consider the function
u(x) =
1 x > 0
0 x < 0
:
Write u(x) as u(x) = f (x) +
1
2
, where f (x) is the function
f(x) =
8
<
:
1
2
x > 0
¡
1
2
x < 0
:
7.3 Properties of Fourier transform 17
Show that
Fff(x)g=
1
i!
; F
1
2
= πδ(!);
and thus
Ffu(x)g=
1
i!
+ 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 !
0
if f is mu ltip lied by cos(!
0
x). This is simply verified
by the following calculation:
Fff(x) cos(!
0
x)g=
1
2
Z
R
f(x)e
i!
0
x
e
¡i!x
+
1
2
Z
R
f(x)e
¡i!
0
x
e
¡i!x
:
The first integral is just f
^
(! ¡!
0
) and the second one is f
^
(! + !
0
), and finally
Fff(x) cos(!
0
x)g=
1
2
f
^
(! ¡!
0
) +
1
2
f
^
(! + !
0
):
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
k
(t) by cos(k!
0
t), the frequency ba nd of each user's data is shifted to
! = k!
0
, as shown in the figure below:
!
4!
0
3!
0
2!
0
!
0
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
F
cos(!
0
x)
1 + x
2
=
π
2
fe
¡j!¡!
0
j
+ e
¡j!+!
0
j
g;
and conclude
Z
0
1
cos(x)
1 + x
2
dx =
π
2
e
¡1
:
Exercise 7.18. A band pass filter is constructed by the function
f(x) =
2 sin(ax)
x
cos(!
0
x):
for a = 1, find !
0
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
0
) is
Fff(x ¡x
0
)g= e
¡i!x
0
f
^
(!):
18 Fourier Transform Method
In particular, if Ffδ(x)g= 1, then Ffδ(x ¡x
0
)g= e
¡i!x
0
. 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) =
Z
¡1
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)
S
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)
S
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) =
Z
¡1
1
f(y) δ(x ¡y) dy;
or equivalently as a Riemann sum as:
f(x) = lim
n!1
X
k=¡n
1
f(y
k
) δ
k
(x ¡ y
k
) ∆y
k
:
In thi s form, we can see that the input f(x) is represented as a summation of shifted
impulses δ(x ¡y
k
) multiplied by weights f(y
k
) ∆y
k
. A system is linear if its response
to the summation is a summation as:
7.3 Properties of Fourier transform 19
Translation
Invariant
S
X
k=¡n
n
f(y
k
) δ
k
(x ¡ y
k
) ∆y
k
X
k=¡n
n
f(y
k
) h
k
(x ¡y
k
) ∆y
k
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) =
Z
¡1
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=
Z
¡1
1
Z
¡1
1
f(y) g(x ¡y) dy
e
¡i!x
dx:
Using the Fubini theorem, and changing the o rder of integrals, we obtain
Fff ∗gg=
Z
¡1
1
f(y)
Z
¡1
1
g(x ¡ y) e
¡i!x
dx
dy:
The internal integral can be written by the substitution z = x ¡y as:
Z
¡1
1
g(x ¡ y) e
¡i!x
dx = e
¡i!y
Z
¡1
1
g(z) e
¡i!z
dz;
and thus
Fff ∗gg=
Z
¡1
1
f(y) e
¡i!y
dy
Z
¡1
1
g(z) e
¡i!z
dz
= Fff gFfgg;
and this completes the proof.
Exercise 7.19. Use the convolution property a nd d etermine the inverse Fourier transform:
F
¡1
(
e
¡t!
2
1 + !
2
)
;
where t > 0 is a constant.
Exercise 7.20. Show the relation:
F
¡1
fcos(!ct)g=
1
2
fδ(x ¡ct) + δ(x + ct)g;
20 Fourier Transform Method
and conclude:
F
¡1
ff
^
(!) cos(c!t)g=
1
2
ff (x ¡ct) + f(x + ct)g:
Problem 7.7. Show the relation:
Fff(x) g(x)g=
1
2π
f
^
(!) ∗ 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
0
(x). The
Fourier transform Fff
0
(x)g is given by
Fff
0
g= i! f
^
(!): (7.6)
This is simply seen by the direct calculation:
Fff
0
(x)g=
Z
¡1
1
f
0
(x) e
¡i!x
dx = f( x) e
¡i!x
¡1
1
+ i!
Z
¡1
1
f(x) e
¡i!x
dx:
Since f is integrable, it vanishes at x !±1, and thus:
Fff
0
(x)g= i!Fff(x)g:
In a similar manner, we obtain the formula:
Fff
00
(x)g= i!Fff
0
(x)g= ¡!
2
f
^
(!);
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
t
= u
xx
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:
(
u^
t
(!; t) = ¡!
2
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) =
Z
¡1
x
f(t) dt:
If f(x) be an integrable func tion, th en we have
FfF
0
(x)g= Fff(x)g= f
^
(!):
7.3 Properties of Fourier transform 21
On the other hand, we have
FfF
0
(x)g= i!FfF (x)g;
and thus
F
Z
¡1
x
f(t)dt
=
1
i!
Fff(x)g: (7.7)
Problem 7.9. In one of previous exercises, we obtained the formula
Ffu( x)g=
1
i!
+ πδ(!):;
where u(x) =
1 x > 0
0 x < 0
. Show the relation
u(x) =
Z
¡1
x
δ(t) dt:
On the other hand, if we use the formula (7.7), we have
Ffu( x)g=
1
i!
Ffδ(x)g=
1
i!
:
What is wrong here? Show that if we use the inver se Fourier transform, the r elation Ffu(x)g=
1
i!
can not be true as we should have:
F
¡1
fu^(! )g=
u(0
+
) + u(0
¡
)
2
=
1
2
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
^
(!)
d!
:
This demonstrates a du a lity between f
0
and Fff
0
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
F
n
x
1 + x
2
o
= ¡ 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=
Z
¡1
1
f(ax) e
¡i!x
dx =
1
a
Z
¡1
1
f(y) e
¡i!y/a
dy =
1
a
f
^
!
a
:
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)
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
0.2
0.4
0.6
0.8
1
-10 -5 0 5 10
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
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
0.5
1
1.5
2
2.5
3
3.5
-4 -2 0 2 4
0
0.5
1
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) =
1
E(f)
p
Z
¡1
1
jxf(x)j
2
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
^
) ≥
1
2
:
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) =
Z
¡1
1
jf (x)j
2
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,
f
^
(!).
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
^
(!)=
2
1 + !
2
, we determine that E
2
1 + !
2
= 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).
−
4
−
3
−
2
−
1 1 2 3 4
ω
1
2
ˆ
f
(
ω
)
The reconstruction f
~
(x) can be defined as:
f
~
(x) =
1
2π
Z
¡1.838
1.838
f
^
(!) 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
f
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.
−
3
−
2
−
1 1 2 3
0
.
2
0
.
4
0
.
6
0
.
8
1
.
0
e
−
x
f
0
.
99
f
0
.
95
Exercise 7.22. Use Plancherel theorem and prove
Z
0
1
sin(ax) sin(bx)
x
2
=
8
<
:
aπ
2
a < b
bπ
2
b < a
:
Hint: use the relation
F
sin(ax)
x
=
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=
1
2π
f
^
(!) ∗ g^(!):
and prove the following relation
Z
¡1
1
f(x)g(x)dx =
1
2π
Z
¡1
1
f
^
(!) g^(!) d!:
This proves the Plancherel theorem.
Z
¡1
1
jf (x)j
2
dx =
1
2π
Z
¡1
1
jf
^
(!)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., ¡!
0
≤! ≤!
0
,
Nyquist and later Shannon showed that the original function can be accurately
reconstructed from the sampled signal if the sampling rate is at least
!
0
π
. 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 [¡!
0
; !
0
], we can express the Fourier
transform f
^
(!) using the complex form of the Fourier series as follows:
f
^
(!) =
X
n=¡1
1
c
n
e
¡i
nπ
!
0
!
;
where c
n
is obtained through the inner product:
c
n
=
1
2!
0
Z
¡!
0
!
0
f
^
(!) e
i
nπ
!
0
!
d! =
π
!
0
1
2π
Z
¡!
0
!
0
f
^
(!) e
i
nπ
!
0
!
d!
=
π
!
0
f
nπ
!
0
:
Let's denote
nπ
!
0
by t
n
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
0
=
π
!
0
or
equivalently, at a rate of
!
0
π
. If we rep lace the angular frequency !
0
by 2πf
0
, we
obtain T
0
=
1
2f
0
, and the sampling rate f
N
=2f
0
. Thus, we can write:
f
^
(!) =
X
n=¡1
1
π
!
0
f(t
n
) e
¡it
n
!
: (7.9)
This implies that to reconstruct f
^
(!), we only need the data set ff(t
n
)g
n=¡1
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
N
=2f
0
. 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!
0
.
!
¡!
0
!
0
f
^
(!) P
!
0
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
; !
0
)
0 otherwise
:
Therefore, the original function f(t) can be reconstructed as:
f(t) = F
¡1
ff
^
(!) p^(!)g=
X
n=¡1
1
1
2!
0
f(t
n
)
Z
¡!
0
!
0
e
¡it
n
!
e
i!t
d! =
X
n=¡1
1
f(t
n
)
sin[!
0
(t ¡t
n
)]
!
0
(t ¡t
n
)
:
The last expression represents the discrete convolution of f(t) and
sin[!
0
t)]
!
0
t
at instance
t
n
=
nπ
!
0
. 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
3
:
The Fourier transform of f is
f
^
(!) =
(
1 ¡!
2
¡1 ≤! ≤1
0 otherwise
:
For !
0
= 1, sequence ff(t
n
)g
n=¡1
1
captures all information in f(t). In fact, the
reconstruction is done by the following series
f (t) =
X
n=¡1
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:
−
20
−
10 10 20
t
0
.
1
0
.
2
f
(
t
)
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
2
!R that satisfies the condition
ZZ
R
2
jf(x; y)jdxdy < 1;
its Fourier transform with frequency parameters !
1
and !
2
associated with x and y,
respectively, is defi ned as f ollows:
f
^
(!
1
; !
2
) =
Z
R
2
f(x; y) e
¡i!
1
x
e
¡i!
2
y
dxdy:
Similarly, the inverse Fourier transform is given by the following formula:
f(x; y) = F
¡1
ff
^
g=
1
4π
2
Z
R
2
f
^
(!
1
; !
2
) e
i!
1
x
e
i!
2
y
d!
1
d!
2
;
provided that the integral converges.
As an example, consider the Gaussian function f (x; y) = e
¡
x
2
+y
2
2
. The Fourier
transform of f is calculated as:
f
^
(!
1
; !
2
) =
Z
R
2
e
¡
x
2
+y
2
2
e
¡i!
1
x
e
¡i!
2
y
dxdy =
Z
R
e
¡
x
2
2
e
¡i!
1
x
dx
Z
R
e
¡
y
2
2
e
¡i!
2
y
dy
:
Each integral on the right-hand side represents a 1D Fourier transform, which results
in:
f
^
(!
1
; !
2
) = 2π e
¡
!
1
2
+!
2
2
2
:
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
fe
¡t(!
1
2
+!
2
2
)
g.
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
^
(¡!
1
; !
2
) = ¡f
^
(!
1
; !
2
):
The same relation holds for the y variable. If f is even function with respect to x, then
f
^
(¡!
1
; !
2
) = f
^
(!
1
; !
2
):
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
n
!R, the n-
dimensional Fourier transform is given by:
Fff g= f
^
(!
1
; :::; !
n
) =
Z
R
n
f(x
1
; :::; x
n
) e
¡i!
1
x
1
···e
¡i!
n
x
n
dV ;
28 Fourier Transform Method
where dV is the volume element in R
n
, and !
1
; :::; !
n
are frequency parameters
associated with coordinate s x
1
; :::; x
n
. The inverse Fourier transform, allowing us to
reconstruct the original function, is defined as:
F
¡1
ff
^
g= f(x
1
; :::; x
n
) =
1
(2π)
n
Z
R
n
f
^
(!
1
; :::; !
n
) e
i!
1
x
1
···e
i!
n
x
n
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@
x
f g= i!
1
f
^
(!
1
; !
2
); Ff@
xy
f g= ¡!
1
!
2
f
^
(!
1
; !
2
):
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
t
and u are integrable fu nctions with respect to x for any t, then
we have
u^
t
(!; t) =
Z
¡1
1
u
t
(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) =
Z
¡1
1
u(x; t) e
¡i!x
dx:
For " =/ 0, we have:
u^(!; t + ") ¡u^(!; ")
"
=
Z
¡1
1
u(x; t + ") ¡u(x; t)
"
e
¡i!x
dx:
By the mean value theorem, we can write
u(x; t + ") ¡u(x; t)
"
= u
t
(x; τ
"
);
for some τ
"
in the interval t and t + " (if " > 0). Therefore, we have
u^(!; t + ") ¡u^(!; t)
"
=
Z
¡1
1
u
t
(x; τ
"
) e
¡i!x
dx:
Now, let " approaches zero:
lim
!!0
u^(!; t + ") ¡u^(!; t)
"
= lim
"!0
Z
¡1
1
u
t
(x; τ
"
) e
¡i!x
dx:
Since u
t
is integrable, we can pass the limit inside the integral, and write
lim
"!0
u^(!; t + ") ¡u^(!; t)
"
=
Z
¡1
1
lim
"!0
u
t
(x; τ
"
) e
¡i!x
dx:
7.4 Higher dimensional transform 29
Since u
t
is continuou s is continuous, we obtain
u^
t
(!; t) :=
Z
¡1
1
u
t
(x; t) e
i!x
dx:
The continuity of u^
t
(!; t) with respect to t follows from the continuity and the
integrability of u
t
(x; t) and the dominant convergence theorem.
Proble ms
Problem 7.11. Find the Fourier transform of following functions
a) f (x) =
1
a
2
+ x
2
b) f( x) = e
i!
0
x
for jxj< 1 and f (x) = 0 in jxj> 1
c) f (x) =
(
e
¡x
x > 0
0 x < 0
d) f( x) =
(
xe
¡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
^
(!) =
1
!
2
¡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:
F
¡1
1
i!
+ πδ(!)
(x) =
8
>
>
<
>
>
:
1 x > 0
1
2
x = 0
0 x < 0
:
Hint: note that
Z
¡1
1
sin(!x)
!
d! =
8
<
:
π x > 0
0 x = 0
¡π x < 0
:
Problem 7.15. Show the following relation
Ffcos(x
2
)g= π
p
cos
!
2
¡π
4
:
Find a relation for Ffsin(x
2
)g. H int: May you wish to use the formula
Z
¡1
1
e
ix
2
dx = iπ
p
Problem 7.16.
a) Show the following relation for t > 0
F
(
e
¡tjxj
jxj
p
)
= 2π
p
t + t
2
+ !
2
p
t
2
+ !
2
!
1/2
:
Hint: you may wish to use the formula
Z
¡1
1
e
¡αz
2
dz =
π
α
r
;
for Refαg> 0.
30 Fourier Transform Method
b) Conclude the following relation
F
(
1
jxj
p
)
=
2π
j!j
r
:
In particular, the fun ction f( x) =
1
jxj
p
is self-dual with res pect to the balanced version
of the Fourier transform.
c) Find the following integral
Z
0
1
cos(x)
x
p
dx:
Problem 7.17. Prove the relation:
lim
s!1
Z
¡1
1
cos(sx)
1 + x
2
dx = 0:
Hint: use the relation for Ffcos(!
0
x) f(x)g.
Problem 7.18. Consider the function g(x) = 2
p
e
¡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.
1
Z
R
jf (x)j
2
Z
R
jf
^
(!)j
2
Z
R
x
2
jf (x)j
2
Z
R
!
2
jf
^
(!)j
2
≥
1
4
:
a) Without loss of generality, assume kf k = 1 (why is this plausible?) Then, according
to the Plancherel identity, we have kf
^
k= 2π
p
. On the other hand, by the relation
Fff
0
g= i! f
^
(!), we obtain
Z
R
!
2
f
^
(!) =
Z
R
jf
0
(x)j
2
:
We show first the following inequality
Z
R
x
2
jf (x)j
2
1/2
Z
R
jf
0
(x)j
2
1/2
≥
2π
p
2
:
b) Use the Cauchy-Schwarz inequality and show the following inequality
Z
R
x
2
jf(x)j
2
1/2
Z
R
jf
0
(x)j
2
1/2
≥
Z
R
xf (x) f
0
(x):
Show also that the equality holds only if xf(x) = λf
0
(x) for some λ 2R.
c) Show t ha t
Z
R
xf (x)f
0
(x) dx ≥
1
2
and conclude the uncertainty principle. Show that the equality holds only if f (x) =ce
ax
2
where c; a are some constants.
Problem 7.20.
a) Verify that the tent function
f(x) =
8
<
:
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
X
n=¡1
1
f(n) < 1:
a) Prove the following relation which is known the Poisson summation
X
n=¡1
1
f(n) =
X
n=¡1
1
f
^
(2nπ):
Hint: note that
f(n) =
1
2π
Z
¡1
1
f
^
(!) e
in!
d!;
and that for any m 2Z, we have
f
^
(2mπ) =
X
n=¡1
1
f
n;2m
;
where
f
n;2m
=
1
2π
Z
(2m¡1)π
(2m+1)π
f
^
(!) e
¡in!
d!:
b) Use the result in part a) to prove the following identity:
X
n=¡1
1
1
1 + n
2
= 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
2
¡y
2
d) f( x; y) =
1
(a
2
+ x
2
)(b
2
+ y
2
)
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
2
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
!
x
linear problem
x
Inverse Fourier transform
-domain
-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
t
= 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:
(
u^
t
(!; t) = ¡!
2
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
¡!
2
t
: (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) =
1
2π
Z
¡1
1
f
^
(!) e
¡!
2
t
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
ff
^
(!) e
¡!
2
t
g= f (x ) ∗F
¡1
fe
¡!
2
t
g:
7.5 Fourier transform and PDEs 33
As we saw in the previous section, we have
F
¡1
fe
¡!
2
t
g=
1
4πt
p
e
¡
x
2
4t
:
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) =
Z
¡1
1
f(z) Φ(x ¡z; t) dz =
1
4πt
p
Z
¡1
1
f(z) e
¡
(x¡ z)
2
4t
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
t
= u
xx
u(x; 0) = f(x)
:
The first form of the inverse Fourier transform results in
u(x; t) =
1
π
Z
¡1
1
sin(!)
!
e
¡!
2
t
e
i!x
d!:
The integral at the right-hand side at t = 0 is equal to F
¡1
n
2 sin(!)
!
o
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) =
2
π
Z
0
1
sin(!)
!
e
¡!
2
t
cos(!x) d!:
-3 -2 -1 0 1 2 3
0
0.2
0.4
0.6
0.8
1
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) =
1
4πt
p
Z
¡1
1
e
¡
(x¡ z)
2
4t
dz:
The figure below depicts the solution using the convolution:
-3 -2 -1 0 1 2 3
0
0.2
0.4
0.6
0.8
1
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
.
2
0
.
4
0
.
6
0
.
8
x
0
t
= 0
.
1
t
= 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
Z
¡1
1
Φ(x; t) dx = 1;
7.5 Fourier transform and PDEs 35
for any t > 0, and that:
lim
t!0
Z
¡1
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
t
= u
xx
u(x; 0) = f (x)
:
Use the convolution form of the solution and show the relation:
Z
¡1
1
u(x; t) dx =
Z
¡1
1
f(x) dx:
Exercise 7.26. Show that Φ(x; t) is the solution of the following heat problem
u
t
= u
xx
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 =
Z
¡1
1
f(z) Φ(x ¡z) dz
f(x) =
Z
¡1
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
8
<
:
u
t
= u
xx
u(x; 0) =
1 x < 0
2 x > 0
:
Write down the convolution solution and show that u(0; t) =
3
2
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
1.2
1.4
1.6
1.8
2
Exercise 7.28. Find and integral solution for the following problem defined on x 2(¡1; 1):
(
u
t
= ku
xx
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):
u
t
= 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:
u^
t
+ k!
2
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
t
= u
xx
+
δ(t ¡ 1)
1 + x
2
u(x; 0) = 0
:
The ordinary differential equation for u^(!; t) is:
(
u^
t
(!; t) = ¡!
2
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=
Z
0
1
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 + !
2
)U
^
(s; !) = πe
¡j!j
e
¡s
;
where U
^
(!; s) is equal to Lfu^(!; t)g. By inverse Laplace transform L
¡1
, we obtain
u^(!; t) as
u
^
(!; t) = L
¡1
fU
^
(!; s)g= πe
¡j!j
L
¡1
e
¡s
s + !
2
= πe
¡j!j
e
¡!
2
(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
e
¡!
2
(t¡1)
gu(t ¡1):
The convolution form of u is given by:
u(x; t) =
1
1 + x
2
∗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
t
= u
xx
+ δ(t) f (x)
u(x; 0) = 0
;
is equal to the solution to the following homogeneous equation:
u
t
= u
xx
u(x; 0) = f(x)
Exercise 7.29. Solve the following heat problem
(
u
t
= u
xx
+ δ(t) e
¡x
2
u(x; 0) = e
¡jxj
:
Problem 7.27. Find an integral solution for the following partial differential equation
8
<
:
u
t
= u
xx
+ u
x
u(x; 0) =
cos(x)
1 + x
2
:
Problem 7.28. Solve the following heat pr ob lem
8
<
:
u
t
= u
xx
+
e
¡jxj
1 + x
2
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):
8
>
>
<
>
>
:
u
tt
= c
2
u
xx
u(0; x) = f(x)
@
t
u(0; x) = g(x)
:
Taking Fourier transform of the problem results in the following second-order ordi-
nary differential equation:
8
>
>
<
>
>
:
u^
tt
= ¡c
2
!
2
u^
u^(!; 0) = f
^
(!)
u^
t
(!; 0) = g^(!)
:
This differential equation is solved for:
u^(t; !) = f
^
(!) cos(c!t) +
1
c!
g^(!) sin(c!t);
and thus:
u(t; x) = F
¡1
ff
^
(!) cos(c!t)g+
1
c
F
¡1
1
!
g^(!) sin(c!t)
:
By the definition o f invers e Fourier transform, we have
u(x; t) =
1
2π
Z
¡1
1
f
^
(!) cos(c!t) e
i!x
d! +
1
2πc
Z
¡1
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)=
1
2π
Z
¡1
1
f
^
(!) cos(c!t) e
i!x
d! +
1
2πc
Z
¡1
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
c
F
¡1
g^(!)
!
∗F
¡1
fsin(c!t)g: (7.14)
Recall the relation
F
Z
0
x
g(τ ) dτ
=
1
i!
g^(!);
and hence:
F
¡1
g^(!)
!
= i
Z
0
x
g(τ ) dτ
Problem 7.29. Show the relations
F
¡1
fcos(c!t)g=
1
2
[δ(x ¡ct) + δ(x + ct)];
7.5 Fourier transform and PDEs 39
and
F
¡1
fsin(c!t)g=
1
2i
[δ(x ¡ct) ¡δ(x + ct)]:
The result of the above problem leads to the following formula for u:
u(x; t) =
1
2
[f(x ¡ct) + f(x + ct)] +
1
2c
Z
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
8
>
>
<
>
>
:
u
tt
= c
2
u
x x
u(x; 0) = f (x)
u
t
(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) =
2
π
Z
0
1
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
0.2
0.4
0.6
0.8
1
-10 0 10
-0.1
0
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
0.2
0.4
0.6
0.8
1
-10 0 10
-0.1
0
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
8
>
>
<
>
>
:
u
tt
+ 2ξu
t
= c
2
u
xx
u(x; 0) = f (x)
u
t
(x; 0) = 0
;
where f(x) is the function f (x) =
(
(1 ¡ x
2
)/4 ¡1 ≤ x ≤ 1
0 othersise
. Taking the Fourier transform,
the given problem reduces to the following ordinary differential equation:
(
u^
tt
+ 2ξu^
t
= ¡c
2
!
2
u^
u^(!; 0) = f
^
(!); u^
t
(!; 0) = 0
;
which is solved for the function:
u^(t; !) = e
¡ξt
f
^
(!)
cos(µt) +
ξ
µ
sin(µt)
;
where µ = !
2
¡ξ
2
p
. 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.
−
4
−
2 2 4
t
= 1
t
= 2
t
= 3
Exercise 7.31. Find the solution of the following wave equation
8
>
>
<
>
>
:
u
tt
= c
2
u
xx
u(x; 0) = δ(x)
u
t
(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
u
t
(x; 0) = δ(x)
:
Find the resp on se of the damped wave system
u
tt
+ 2ξu
t
= u
xx
;
to the impulse displacement exercised at t = 0.
Exercise 7.32. Consider the following wave equation on the domain x 2(¡1; 1):
8
<
:
u
tt
= u
xx
+ δ(t ¡1) f (x)
u(x; 0) = 0
u
t
(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) =
1
2
f(x) ∗ g
t
(x):
What is g
t
(x)?
Exercise 7.33. Determine the integral solution of the wave equation
8
<
:
u
tt
+ 0.6u
t
= u
xx
u(x; 0) = f(x)
u
t
(x; 0) = 0
;
where f (x) is the function
f(x) =
(
sin(x)
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:
F
s
ff(x)g:= f
^
s
(!) =
Z
0
1
f(x) sin(!x) dx;
with the inverse
F
¡1
ff
^
s
(!)g=
1
π
Z
¡1
1
f
^
s
(!) sin(!x) d!:
It is important to note that F
¡1
ff
^
s
(!)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):
8
<
:
u
t
= u
xx
; 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:
8
<
:
@
@t
u^
s
(!; t) = ¡!
2
u^
s
(!; t)
u^
s
(!; 0) = f
^
s
(!)
;
where u^
s
is equal to the Fourier sine transform of u. This ordinary differential
equation is then solved for the function:
u^
s
(!; t) = f
^
s
(!) e
¡!
2
t
:
By performing the invers e Fourier transform, we obtain the expression for u as:
u(x; t) =
1
π
Z
¡1
1
f
^
s
(!) e
¡!
2
t
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
0.2
0.4
0.6
0.8
1
If the boundary condition at x = 0 is Neumann, the solution can be expressed in
terms of the Fo urier cosine transform:
F
c
ff(x)g:= f
^
c
(!) =
Z
0
1
f(x) cos(!x) dx;
with its inverse given by:
F
¡1
ff
^
c
(!)g=
1
π
Z
¡1
1
f
^
c
(!) cos(!x) d!:
If we modify the boundary condition for the above example to u
x
(0; t)=0, the
integral solution changes to:
u(x; t) =
1
π
Z
¡1
1
f
^
c
(!) e
¡!
2
t
cos(!x) d!:
For the g iven piecewise function f(x) defined as: f(x) =
1 0 < x < 1
0
, we can calculate
its Fo urier cosine transform as:
f
^
c
(!) =
Z
0
1
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
π
Z
¡1
1
sin(!)
!
e
¡!
2
t
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
0.2
0.4
0.6
0.8
1
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
t
= u
xx
; ¡1< x < 1; t > 0
u(x; 0) = f
o
(x)
;
where f
o
is the odd extension of f, defined as:
f
o
(x) =
f(x) x > 0
¡f(¡x) x < 0
:
Taking the Fourier transform of this e xtended problem results in the solution:
u^
t
(!; t) = f
^
o
(!) e
¡!
2
t
;
and finally u(x; t) is given by the convolution as:
u(x; t) = f
o
(x) ∗
1
4πt
e
¡
x
2
4t
=
Z
¡1
1
f
o
(z)
e
¡
(x¡ z)
2
4t
dz:
It can be observed that u(0; t) is equal to zero as:
u(0; t) =
Z
¡1
1
f
o
(z)
e
¡
z
2
4t
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):
8
<
:
u
t
= u
xx
+ tf (x); 0 < x < 1; t > 0
u(0; t) = 0 t > 0
u(x; 0) = 0
;
where f (x) =
1 0 < x < 1
0
.
Exercise 7.35. Consider the following wave equation defined on [0; 1):
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
u
tt
= u
xx
+ δ(t ¡1)e
¡x
u
x
(0; t) = 0
u(x; 0) = 0
u
t
(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
x
(0; t) = 0.
Exercise 7.36. Let Ω be the interval [1; 1 ). Solve the following problem on Ω
8
>
>
<
>
>
:
u
t
= u
xx
+ δ(t ¡1)e
¡x
u
x
(1; t) = 0
u(x; 0) = 0
Exercise 7.37. Consider the following heat problem on a semi-in finite domain
8
<
:
u
t
= u
xx
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
2
. 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
2
, 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
2
Consider the h eat equation u
t
=∆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:
u^
t
(!
1
; !
2
; t) = ¡(!
1
2
+ !
2
2
) u^(!
1
; !
2
; t);
where !
1
and !
2
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:
u
^
(!
1
; !
2
; t) =
ZZ
R
2
u(x; y; t) e
¡i!
1
x
e
¡i!
2
y
dxdy:
Using the definition of the Laplacian operator ∆, we can express the Fourier trans-
form ∆ as:
Ff∆ug= Ffu
xx
+ u
yy
g= ¡(!
1
2
+ !
2
2
) u^(!
1
; !
2
; t):
The resulting ordinary differential equation, along with the given initial c o ndition,
is solved to obtain the solution:
u^(!
1
; !
2
; t) = f
^
(!
1
; !
2
) e
¡(!
1
2
+!
2
2
)t
:
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
ff
^
(!
1
; !
2
) e
¡(!
1
2
+!
2
2
)t
g:
This inverse transform can be expressed either as a double integral:
F
¡1
ff
^
(!
1
; !
2
) e
¡(!
1
2
+!
2
2
)t
g=
1
4π
2
ZZ
R
2
f
^
(!
1
; !
2
) e
¡(!
1
2
+!
2
2
)t
e
i!
1
x
e
i!
2
y
d!
1
d!
2
;
or using the convolution:
F
¡1
ff
^
(!
1
; !
2
) e
¡(!
1
2
+!
2
2
)t
g= f(x; y) ∗F
¡1
fe
¡(!
1
2
+!
2
2
)t
g:
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
fe
¡(!
1
2
+!
2
2
)t
g is given by
1
4πt
e
¡
x
2
+y
2
4t
, 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
F
¡1
fe
¡(!
1
2
+!
2
2
)t
g=
1
4πt
e
¡
x
2
+ y
2
4t
:
This is called the 2D hea t kernel:
Φ(x; y; t) =
1
4πt
e
¡
x
2
+ y
2
4t
:
Show the relation
ZZ
R
2
Φ(x; y; t) d xdy = 1:
Furthermore, show that the following relation holds:
lim
t!0
ZZ
R
2
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
t
= ∆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
t
= ∆u
u(x; y; 0) = e
¡jxj¡jyj
:
Exercise 7.40. Solve the following heat equation on R
2
(
u
t
= 4∆u
u(x; y; 0) = e
¡x
2
¡y
2
:
7.6 Higher dimensional problems 47
Exercise 7.41. Solve the following heat problem on R
2
(
u
t
= ∆u + r
¡x
2
¡y
2
u(x; y; 0) = 0
:
Exercise 7.42. Solve the following heat problem on R
2
8
<
:
u
t
= ∆u +
δ(t ¡1)
(1 + x
2
)(1 + y
2
)
u(x; y; 0) = 0
:
Exercise 7.43. Solve the following heat problem on R
2
u
t
= ∆u + δ(t ¡1) δ(x) δ(y)
u(x; y; 0) = 0
:
Now, let's consider wave equation u
tt
= c
2
∆u in R
2
. For simplicity, we consider
the following problem:
8
>
>
<
>
>
:
u
tt
= c
2
∆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:
u^
tt
+ c
2
!
2
u^ = 0
where ! = !
1
2
+ !
2
2
p
. With the initial conditions:
(
u^(!
1
; !
2
; 0) = f
^
(!
1
; !
2
)
u^(!
1
; !
2
; 0) = 0
;
we obtain the solution:
u^(!
1
; !
2
; t) = f
^
(!
1
; !
2
) 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) =
1
4π
2
ZZ
R
2
f
^
(!
1
; !
2
) cos(ct!) e
i!
1
x
e
i!
2
y
d!
1
d!
2
:
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 !
1
2
+ !
2
2
p
t
is not straightfor-
ward and requires intricate calculations. Let us try to find this inverse transform.
To do that, we rewrite !
1
; !
2
as !
1
= ! cosγ, !
2
= ! sinγ for γ 2[¡π; π]. Therefore,
we can write
F
¡1
fcos(ct!)g=
1
4π
2
Z
¡π
π
Z
0
1
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
F
¡1
fcos(ct!)g=
1
4π
2
Z
¡π
π
Z
0
1
cos(ct!)cos (r! cos(γ ¡θ))!d!dγ:
Using the formula:
1
2π
Z
¡π
π
cos(s cosγ) dγ = J
0
(s);
where J
0
is the Bessel function of the first kind, we obtain:
F
¡1
fcos(ct!)g=
1
2π
Z
0
1
J
0
(r!) cos(ct!)!d!:
The integral at the right-hand side can be rewritten as:
1
2π
Z
0
1
J
0
(ρ!) cos(ct!)!d ! =
1
2πc
d
dt
1
ρ
Z
0
1
J
0
(!) sin
ct
ρ
!
d!:
Using the formula:
Z
0
1
J
0
(!) sin
ct
ρ
!
d! =
1
c
2
t
2
ρ
2
¡1
q
; ρ < ct;
we finally obtain:
F
¡1
fco s(ct!)g=
1
2πc
d
dt
1
c
2
t
2
¡ρ
2
p
; ρ < ct:
Finally, the solution u(t; x; y) can be written as follows:
u(t; x; y) =
1
2pc
d
dt
Z
B
ct
(x;y)
f(z
1
; z
2
)
c
2
t
2
¡(x ¡z
1
)
2
¡(y ¡z
2
)
2
p
dz
1
dz
2
;
where B
ct
(x; y) is the disk centered at (x; y) of radius ct. By a similar calculation,
it is shown that if u
t
(0; x; y) = g(x), the solution is:
u(t; x; y) =
1
2p c
d
dt
Z
B
ct
(x;y)
f(z
1
; z
2
)
c
2
t
2
¡(x ¡z
1
)
2
¡(y ¡z
2
)
2
p
dz
1
dz
2
+
1
2p c
Z
B
ct
(x;y)
g(z
1
; z
2
)
c
2
t
2
¡(x ¡z
1
)
2
¡(y ¡z
2
)
2
p
dz
1
dz
2
:
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
t
= ∆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:
f
o
(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
R
2
as:
u
t
= ∆u
u(x; y; 0) = f
o
(x; y)
:
Next, we take the Fourier transform to derive the following ordinary differential
equation:
(
u^
t
(!
1
; !
2
; t) = ¡(!
1
2
+ !
2
2
) u^(!
1
; !
2
; t)
u^(!
1
; !
2
; 0) = f
^
o
(!
1
; !
2
)
:
Solving this ordinary differential equation yields the function:
u^(!
1
; !
2
; t) = f
^
o
(!
1
; !
2
) e
¡(!
1
2
+!
2
2
)t
:
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
o
(x; y) ∗
1
4πt
e
¡
x
2
+y
2
4t
=
1
4πt
ZZ
¡1
1
f
o
(z
1
; z
2
) e
¡
(x¡ z
1
)
2
+(y ¡ z
2
)
2
4t
dz
1
dz
2
:
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) =
1
4πt
Z
¡1
1
e
¡
(x¡ z
1
)
2
4t
Z
¡1
1
f
o
(z
1
; z
2
) e
¡
z
2
2
4t
dz
2
dz
1
= 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
8
>
>
<
>
>
:
u
t
= ∆u + e
¡y
e
¡jxj
u
y
(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:
8
>
>
>
>
<
>
>
>
>
:
u
t
= ∆u
u(0; y; t) = 0
u(x; y; 0) =
e
¡x
1 + y
2
:
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:
8
<
:
u
t
= ∆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:
(
V
^
yy
(!; y) ¡!
2
V
^
(!; y) = 0
V
^
(!; 0) = f
^
(!)
:
The general solution of this equation is:
V
^
(!; y) = A(!) e
¡!y
+ B(!) e
!y
:
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:
V
^
(!; 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:
V
^
(!; 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) ∗
1
π
y
y
2
+ x
2
=
y
π
Z
¡1
1
f(z)
y
2
+ (x ¡z)
2
dz:
An alternative representation using the definition of F
¡1
is:
V (x; y) =
1
2π
Z
¡1
1
f
^
(!) e
¡yj!j
e
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
8
<
:
∆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
8
>
>
>
>
<
>
>
>
>
:
u
t
= ∆u on Ω
u(x; 0; t) =
1
1 + x
2
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
1 + x
2
is:
u(x; y) =
y + 1
(y + 1)
2
+ x
2
:
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) =
2
π
tan
¡1
1
y
:
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) =
8
>
>
<
>
>
:
1 x > 0
1
2
x = 0
0 x < 0
;
is
u(x; y) =
1
π
tan
¡1
x
y
+
1
2
:
7.6.3 Equations on strips
Consider the heat equation u
t
= ∆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:
8
>
>
<
>
>
:
u^
t
(!; y; t) = u^
yy
(!; y; t) ¡!
2
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
00
¡!
2
φ = ¡λφ
φ(0) = φ(L) = 0
:
Solving f o r the eigenfunctions a nd eigenvalues yields:
φ
n
(y) = sin
nπ
L
y
; λ
n
= !
2
+
n
2
π
2
L
2
:
Hence, the solution to the resulting heat equation in (y; t) is given by:
u^(!; y; t) =
X
n=1
1
C(!) e
¡!
2
t
e
¡
n
2
π
2
L
2
t
sin
nπ
L
y
:
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
f
^
(!; y) =
X
n=1
1
C(!) sin
nπ
L
y
;
and by performing the inner product, we obtain:
C(!) =
2
L
Z
0
L
f
^
(!; y) sin
nπ
L
y
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=
X
n=1
1
F
¡1
fC(!) e
¡!
2
t
ge
¡
n
2
π
2
L
2
t
sin
nπ
L
y
:
Exercise 7.48. Let Ω be the strip (¡1; 1) ×(0; 1) in the xy-plane. Consider the following
heat problem in Ω
8
>
>
>
>
<
>
>
>
>
:
u
t
= ∆u on Ω
u(x; 0; t) = u(x; 1; t) = 0 B:Cs:
u(x; y; 0) =
sin(πy)
1 + x
2
I:C:
:
a) Find an integral solution for the problem.
b) Show that u
¡
0;
1
2
;
1
π
is equal to
u
0;
1
2
;
1
π
= e
¡π
Z
0
1
e
¡
π
4
z
2
1 + z
2
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:
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
u
tt
= ∆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:
u
t
(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:
8
<
:
∆u = 0 on Ω
u(x; 0) = f (x)
u(x; L) = g(x)
:
Taking the Fourier transform in x yie lds:
8
>
>
<
>
>
:
u^
yy
(!; y) ¡!
2
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
t
= ∆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
1 + x
2
;
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:
8
>
>
<
>
>
:
u
t
= ∆u on Ω
u(x; 0; t) = e
¡jxj
u(x; y; 0) = e
¡x
2
¡y
2
:
Problem 7.35. Solve the following heat equa tion on the upper half plane y > 0
8
>
>
<
>
>
:
u
t
= ∆u
u
y
(x; 0; t) = 0
u(x; y; 0) = e
¡jxj
e
¡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) =
y
π
Z
¡1
1
f(z)
y
2
+ (x ¡z)
2
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
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
u
t
= ∆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)
is
u(x; y) =
2
π
Z
0
1
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
8
>
>
>
>
<
>
>
>
>
:
u
t
= ∆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
f
^
(!). Consider the following heat equation
(
u
t
= u
xx
+ e
t
f(x)
u(0; x) =
1
2
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)
(
u
tt
= c
2
u
xx
+ δ(t) f (x)
u(0; x) = 0; u
t
(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) =
Z
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
f
^
(! + ") = lim
"!0
Z
R
f(x)e
¡i!x
e
¡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
^
(! + ") =
Z
R
f(x) e
¡i!x
lim
"!0
e
¡i!"
dx = f
^
(!):
Now we prove that f
^
(!) is differentiable with respect to !. Fix ! and write:
f
^
(! + ") ¡ f
^
(!)
"
=
Z
R
f(x) e
¡i!x
e
¡i"x
¡1
"
dx:
Using the mean value theorem, we can write
e
¡i"x
¡1
"
= xe
¡ix"
0
;
for some "
0
in the interval (0; ") if " > 0 (alternatively, ("; 0) is " < 0). Thus,
f
^
(! + ") ¡ f
^
(!)
"
= ¡i
Z
R
xf(x) e
¡i!x
e
¡ix"
0
dx
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
Z
R
xf(x) e
¡i!x
e
¡ix"
0
dx =
Z
R
xf(x) e
¡i!x
lim
"!0
e
¡ix"
0
dx =
Z
R
xf(x) e
¡i!x
dx:
Therefore,
lim
"!0
f
^
(! + ") ¡ f
^
(!)
"
= ¡i
Z
R
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
n
(x) = x
n
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 =
ZZ
R
2
jf (x; y)jdA < 1:
If integrals
I
1
=
Z
R
Z
R
f(x; y) dx
dy; I
2
=
Z
R
Z
R
f(x; y) dy
dx;
exist, then I
1
= I
2
.
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
Z
¡1
1
f(x) sin(nx) dx = 0:
In f a ct, we have
f
^
(n) =
Z
¡1
1
f(x) e
¡inx
dx =
Z
¡1
1
1
n
f
x
n
e
¡ix
dx:
The dominant convergence theorem allows us to write
lim
n!1
f
^
(n) = lim
n!1
Z
¡1
1
1
n
f
x
n
e
¡ix
dx =
Z
¡1
1
lim
n!1
1
n
f
x
n
e
¡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
1
2π
lim
n!1
Z
¡n
n
f
^
(!) e
i!x
d! =
1
2
[f(x
+
) + f (x
¡
)]:
For simplicity, we assume that f is C
1
.
Proof. Sinc e f
^
(!) is continuous, then for any n > 0, the integral
I
n
(x) =
Z
¡n
n
f
^
(!) e
i!x
d!;
exists. On the other hand, we have
I
n
(x) =
Z
¡n
n
Z
R
f(ξ) e
¡iξ!
dξ
e
i!x
d! =
Z
¡n
n
Z
R
f(ξ) e
i!(x¡ξ)
dξd!:
By Fubini theorem, we can switch the integrals as
I
n
(x) =
Z
R
f(ξ)
Z
¡n
n
e
i!(x¡ξ)
d!
dξ:
7.7 Appendix 57
Note that:
Z
¡n
n
e
i!(x¡ξ)
d! = 2
sin [n(x ¡ ξ)]
x ¡ ξ
;
and then
I
n
(x) = 2
Z
R
f(ξ)
sin [n(x ¡ ξ)]
x ¡ ξ
dξ:
By taking z = x ¡ ξ, we ob tain
I
n
(x) = 2
Z
R
f(x ¡z)
sin(nz)
z
dz:
We have
Z
R
f(x ¡z)
sin(nz)
z
dz =
Z
¡1
¡1
f(x ¡z)
z
sin(nz) dz +
Z
¡1
1
f(x ¡z)
sin(nz)
z
dz +
+
Z
1
1
f(x ¡z)
z
sin(nz) dz:
By the Lebesgue-Riemann lemma, we have
lim
n!1
Z
¡1
¡1
f(x ¡z)
z
sin(nz) d z = lim
n!1
Z
1
1
f(x ¡z)
z
sin(nz) dz = 0:
We have al so
Z
¡1
1
f(x ¡z)
sin(nz)
z
dz =
Z
¡1
1
[f(x ¡z) ¡ f(x)]
sin(nz)
z
dz + f (x)
Z
¡1
1
sin(nz)
z
dz
lim
n!1
I
n
(x) = 2 lim
n!1
Z
R
f
x ¡
z
n
sin z
z
dz = 2
Z
R
lim
n!1
f
x ¡
z
n
sin z
z
dz =
=2 f(x)
Z
R
sin z
z
dz = 2πf(x);
and this completes the proof.
58 Fourier Transform Method
