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Contents
1 Introduction 1
1.1 Background of study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Statement of Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Significance of Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.4 Aims and Ob jectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.5 Scope of Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.6 Definition of Terms and Concepts . . . . . . . . . . . . . . . . . . . . . . 3
1.6.1 Integral transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.6.2 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.6.3 Inverse Fourier Transform . . . . . . . . . . . . . . . . . . . . . . 3
1.6.4 Partial Differential Equation (PDE) . . . . . . . . . . . . . . . . . 4
1.6.5 Linear Partial Differential Equation . . . . . . . . . . . . . . . . . 4
1.6.6 Classification of second order linear partial differential equation . 4
2 Literature Review 6
2.1 The Fourier Transform and the Inverse Transform . . . . . . . . . . . . . 6
2.2 Partial Differential Equations (PDEs) . . . . . . . . . . . . . . . . . . . . 7
3 Fourier Transforms and Inverse Fourier transforms 8
3.1 Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
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CONTENTS ii
3.2 Inverse Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4 Application of Fourier Transform to PDEs 134.1 The Fourier transform and the Wave equation . . . . . . . . . . . . . . . 14
4.2 The Fourier Transform and the Heat Equation . . . . . . . . . . . . . . . 17
4.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
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Chapter 1
Introduction
1.1 Background of study
Among the tools that are useful for solving linear differential equations are integral
transforms. An integral transform is a relation of the form
F(s) =
K(s, t)f(t) dt (1.1)
where K(s, t) is a given function called the kernel of the transformation and the limits
of integration and are also given. It is possible that = and = or both.The relation (1.1) transforms f into another function F, which is called the transform of
f. The general idea in using an integral transform to solve a differential equation is as
follows: use the relation in (1.1) to transform a problem for an unknown function f into
a simpler problem for F, then solve this simple problem to find F and finally recover
the desired function f from its transform f. This last step is known as inverting the
transform.
There are several integral transforms that are useful in applied mathematics, but in
this study we will consider only the Fourier transform.
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1.2 Statement of Problem 2
1.2 Statement of Problem
In this project work, we seek to define the Fourier transform and compute the Fourier
transform of several functions. We will utilize the Fourier transform in solving some
linear Partial Differential Equations (PDEs).
1.3 Significance of Study
This study is important because of the widespread applications of partial differential
equations in science and engineering. The Fourier transform is also a powerful and
useful tool in mapping functions from Euclidean space to Fourier space. A Differential
Equation transformed into Fourier space may be easier to solve.
1.4 Aims and Objectives
The aims and objectives of the study are as follows:
1. To detail the definition of the Fourier transform and inverse Fourier transform and
compute the Fourier transform and inverse Fourier transform of several functions.
2. To utilize the Fourier to solve some linear partial differential equations.
1.5 Scope of Study
This project work is written to find the solution of partial differential equations using
the Fourier transform method only.
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1.6 Definition of Terms and Concepts 3
1.6 Definition of Terms and Concepts
1.6.1 Integral transform
An integral transform is a relation of the form
F(s) =
K(s, t)f(t) dt (1.2)
where K(s, t) is a given function called the kernel of the transform and the limits of
integration and are also given. The relation (1.2) transforms the function into
another function F which is called the transform of f.
1.6.2 Fourier Transform
Given a function f(x) with derivative f(x) where f(x) and f(x) are continuous in every
finite interval and f(x) is integrable in (,), the function
F() = F[f(x)] =
f(x)eix dx (1.3)
is referred to as the Fourier transform of f(x).
1.6.3 Inverse Fourier Transform
The inverse Fourier transform recovers the original function f(x) from F(w). Given as
F1, it is defined as
f(x) = F1[F()] = 12
=
f()eix d
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1.6 Definition of Terms and Concepts 4
1.6.4 Partial Differential Equation (PDE)
The general form of PDE for a function u(x1, x2, . . . , xn) is
F(x1, x2, . . . , xn, u , ux1, ux2, . . . , uxn , . . .) = 0 (1.4)
where x1, x2, . . . , xn are the independent variables, u is the unknown function and uxi
denotes the partial derivative uxi
.
1.6.5 Linear Partial Differential Equation
An equation is called a linear partial differential equation if in
F(x1, x2, . . . , xn, u , ux1, ux2, . . . , uxn , . . .) = 0
F is a linear function of the unknown function u and its derivatives. Thus, the equation,
x7
ux + exy
uy + sin(x2
+ y2
)u = x3
, is linear equation.
1.6.6 Classification of second order linear partial differential
equation
A second order linear Partial Differential Equation for functions in two independent
variables x and y has the form
L[u] = auxx + 2buxy + cuyy + dux + euy + f u = g. (1.5)
The factor 2 in front of the coefficient b is for convenience. We assume the coefficients
a, b, c, do not vanish simultaneously. The operator
L0[u] = auxx + 2buxy + cuyy
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1.6 Definition of Terms and Concepts 5
that consists of the second order terms of the operator L is called the principal part of
L. Many fundamental properties of the solution of (1.5) are determined by its principal
part, and more precisely, by the sign of the discriminant (L) := b2ac of the equation.We classify the equation according to the sign of (L).
Definition 1.1 Equation (1.5) is said to be hyperbolic at a point (x, y) if (L) := b2 ac > 0, it is said to be parabolic at (x, y) if (L) = 0 and it is said to be elliptic at(x, y)
if (L) < 0.
For example, the heat equation given by ut = k 2
ux2 is a parabolic equation. The wave
equation given by 2ut2
= c2 2u
x2is a hyperbolic equation, while the Laplace equation given
by 2u
x2+
2uy2
= 0 is an example of elliptic equation.
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Chapter 2
Literature Review
To successfully understand the application of Fourier transform to Partial Differential
Equations, it is important to understand the basics of the Fourier transform and Partial
Differential Equations as described by certain authors.
2.1 The Fourier Transform and the Inverse Trans-
form
According to Wikipedia (2010), the Fourier transform and its generalization form the
subject of Fourier analysis. It lists out the basic properties of the Fourier transform
such as linearity, translation, and conjugation. Tung (2004) provides a clear definition
for the Fourier transform and the Inverse Fourier transform. He also demonstrates howFourier transform can be used in solving differential equations. He defines the Fourier
transform as
F() = F[f(x)] =
f(x)eix dx (2.1)
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2.2 Partial Differential Equations (PDEs) 7
and the inverse Fourier transform as;
f(x) = F1[F()] = 12
F()eix d (2.2)
Yosida (1980) shows clearly how Fourier transforms are used in solving differential equa-
tion in an abstract sense.
2.2 Partial Differential Equations (PDEs)
According to Boyce and Diprima (2000), the study of differential equations has attracted
the attention of many of the worlds greatest mathematicians during the past three
centuries. Nevertheless, it remains a dynamic field of inquiry today with many interesting
open questions, so before embarking on a serious study of differential equation one
needs to know that a differential equation describes some physical process often called
mathematical model. Furthermore, the simplest differential equations provide useful
models of important physical processes.
Pinchover and Rubinstein (2005) give a clear definition and classification of PDEs.
They also demonstrate how the method of separation of variables and several other
methods can be used in solving PDEs. Tung (2004) shows how integral transforms can
be used in solving PDEs. Burden & Faires (1997) shows how the finite difference method
can be used as an effective numerical method for solving PDEs.
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Chapter 3
Fourier Transforms and Inverse
Fourier transforms
This chapter deals with some simple examples of Fourier and Inverse Fourier transforms
of certain functions. It also shows the procedures which will make Fourier transform and
its inverse very suitable for the solution of Partial Differential Equations. An integraltransform is a relation of the form
F(s) =
K(s, t)f(t) dt
where K(s, t) is a given function called the kernel of the transform and the limits of
integration and are given. The relation above transforms the function f into another
function F which is called the transform of f and it is a tool used in solving linear
differential equations.
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3.1 Fourier transform 9
3.1 Fourier transform
Given a function f(x) with derivative f
(x) where f(x) and f
(x) are continuous in every
finite interval and f(x) is integrable in (,), it is defined as
F() = F[f(x)] =
f(x)eix dx.
Example 3.1 Find the Fourier transform of
f(x) = e|x|, < x <
Solution
F() = F[e|x|] =
e|x|eix dx
=0
ex+ix dx +0
ex+ix dx
=1
(1 + i)e(1i)x|0 +
1
(1 + i)e(1+i)x|0
=1
1 i +1
1 + i=
2
1 + 2
Note its decay as .
Example 3.2 Find the Fourier transform of
f(x) = ex, < x <
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3.1 Fourier transform 10
Solution
F() = F[ex] =
exeix dx
=1
(1 + i)e(1+i)x|
The limit at x = blows up. We say the Fourier transform of ex does not exist becausethe function f(x) = ex is not integrable such that
|f(x)| dx =
ex dx
does not have a finite value.
Example 3.3 Find the Fourier transform of
f(x) = ex2
, < x <
Solution
The value of the function decreases rapidly when x is away from x = 0 in both the
positive and negative x directions. There is a finite area under the curve ex2
, so this
function is integrable. To find its Fourier transform, we need to perform the integral
F() =
ex2
eix dx
By completing the square in exponent we get
F(w) =
e2/4.
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3.2 Inverse Fourier Transform 11
The detailed working is as follows:
x2 + ix = (x i/2)2 2/4.
So F() = e2/4
e(xi/2)2
dx = e2/4
ey2
dy where we have made a change
of variable y = x i/2 and also shifted the path of integration. The remaining integralis a standard one (Eulers integral) and is equal to
.
3.2 Inverse Fourier Transform
This recovers the original function f(x) from F(). Given as F1, it is defined as
f(x) = F1[F()] = 12
=
F()eix d
We will now compute the inverse Fourier transform of some simple functions.
Example 3.4 Find the Inverse transform of
F() =
e2/4, < < .
Solution
F1[F()] =2
e2/4ix d
=
2ex
2
e/2ix d
= ex2
We have thus recorded the original function f(x) in Example 3.3.
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3.2 Inverse Fourier Transform 12
Example 3.5 Find the inverse Fourier transform of
F() = 21 + 2
Solution
F1[F()] = 22
1
1 + 2eix d
This integral can be evaluated using residue calculus. Alternatively, using tables of
integrals, we see that
f(x) = e|x|
We have thus recovered the f(x) in Example 3.1.
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Chapter 4
Application of Fourier Transform to
PDEs
The usual difficulty with PDEs is that the solution involves more than one independent
variable. The transform method allows us to reduce one independent variable. We
commonly try to transform the x variable through a Fourier transform provided thatthe domain in is infinite such that < x < . Now consider a function u(x, t) with < x < , t > 0 and let
U(, t) = F[u(x, t)] =
u(x, t)eix dx (4.1)
be the Fourier transform of u(x, t) with respect to x. The original function u(x, t) can
then be recovered from the inverse Fourier transform
u(x, t) =1
2
U(, t)eix d (4.2)
Note that in both equations (4.1) and (4.2), t plays no role, it is arbitrary.
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4.1 The Fourier transform and the Wave equation 14
For the solution of second order linear PDEs, one has to know the following terms.
ut(x, t) = 12
Ut(, t)eix d (4.3)
utt(x, t) =1
2
Utt(, t)eix d (4.4)
ux(x, t) =1
2
(i)U(, t)eix d (4.5)
uxx(x, t) =1
2
(i)2U(, t)eix d (4.6)
In this chapter, we would use the Fourier transform to solve the heat equation and
the wave equation.
4.1 The Fourier transform and the Wave equation
The wave equation is represented as follows.
2
u(x, t)t2
= c22
u(x, t)x2
(4.7)
where c is a certain physical constant, u is the dependent variable and x and t are
two independent variables. This is considered in an infinite domain such that the wave
equation is represented below as an Initial-Boundary Value Problem (IBVP).
PDE : utt
= c2uxx
,
< x 0, (4.8)
BCs : u(x, t) 0, as x (4.9)
ICs : u(x, 0) = f(x) (4.10)
ut(x, 0) = 0, < x < (4.11)
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4.1 The Fourier transform and the Wave equation 15
We assume that the solution to be of the form of an integral
u(x, t) = 12
U(, t)eix d
which we substitute into the PDE. This yields
1
2
(Utt(, t) + c
22U(, t))
eix d = 0
which is the same as
F1[Utt + c22U] = 0 (4.12)
so
Utt + c22U = 0. (4.13)
This is an ODE; the partial derivatives 2u
x2have been converted to (i)2 an algebraic
multiplication. The ODE in t is to be solved subject to the following initial conditions:
ut(x, 0) =1
2
Ut(, 0)eix d = 0
and
u(x, 0) =1
2
U(, 0)eix d = f(x) =1
2
F()eix d.
These imply that
Ut(, 0) = 0 (4.14)
and
U(, 0) = F() (4.15)
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4.1 The Fourier transform and the Wave equation 16
where the Fourier transform F() off(x) is known iff(x) is known. The general solution
to the PDE (4.13) is
U(, t) = A() sin(ct) + B()cos(ct).
The initial conditions (4.14) and (4.15) can be used to determine to constants A and B
to be B() = F() and A() = 0. Thus,
U(, t) = F()cos(ct) (4.16)
We recover u(x, t) by substituting (4.16) back to (4.2)
u(x, t) = F1[U(, t)] = 12
U(, t)eix d =1
2
F() cos(ct)eix d
(4.17)
Typically one cannot perform the integral explicitly unless F() is known. In the par-
ticular case of the wave equation however, progress can be made by noting that
cos(ct) =1
2(eict + eict)
and so (4.17) can be rewritten as
u(x, t) =1
2
1
2F()ei(xct) d +
1
2
1
2F()ei(x+ct) d (4.18)
=1
2f(x ct) + 1
2f(x + ct)
f(x) =1
2
F()eix d
and so
f(x + ct) =1
2
F()ei(x+ct) d
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4.2 The Fourier Transform and the Heat Equation 17
and
f(x
ct) =
1
2
F()ei(xct) d
4.2 The Fourier Transform and the Heat Equation
The heat conduction is represented as follows. The wave equation is represented as
follows.
u(x, t)
t= 2
2u(x, t)
x2(4.19)
where is a certain physical constant, u is the dependent variable and x and t are
two independent variables. This is considered in an infinite domain such that the heat
equation is represented below as an Initial-Boundary Value Problem (IBVP).
PDE : ut = 2uxx, < x < , t > 0, (4.20)
BCs : u(x, t) 0, as x (4.21)
ICs : u(x, 0) = f(x) < x < (4.22)
We assume a solution of the form of an integral (4.2) and substitute into PDE (4.21).
Using
uxx(x, t) =1
2
(i)2U(, t)eix d
this yields
12
(Ut +
22U)
eix d = 0
which implies
Ut + 22U = 0 (4.23)
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4.2 The Fourier Transform and the Heat Equation 18
The ODE (4.23) is solved subject to the initial condition
U(, 0) = F() (4.24)
which is obtained by taking the Fourier transform of (4.22).
The solution is
U(, t) = A()e22t = F()e
22t (4.25)
The final solution is obtained by substituting (4.25) into (4.2)
u(x, t) =1
2F()e
22tix d. (4.26)
For special case of
f(x) = ae(x/L)2
, < x < .
In this case,
F() = F[f(x)] = aLe(L)2/4.
Then
u(x, t) =aL
2
e(L)
2/422tix d
can be evaluated by completing the squares
u(x, t) =aL
2
e(2t+L
2
4)2ix d
=aL
42t + L2ex
2/(42t+L2)
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4.3 Conclusion 19
4.3 Conclusion
We find the Fourier transform as an effective method for solving the Initial Boundary
Value Problems (IBVPs) for the wave equation and heat equation, and it will clearly be
useful in solving other time dependent Partial Differential Equations.
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Bibliography
[1] Boyce, E. W. and Diprima, R. C. (2001), Elementary Differential Equations and
Boundary Value Problems, John Wiley and Sons, New York.
[2] Burden, R.L. and Faires, J.D. (1997), Numerical Analysis, Brookes/Cole, CA.
[3] Kammler, D. (2000), A First Course in Fourier Analysis, Prentice-Hall.
[4] Pinchover, Y. and Rubinstein, J. (2005), An Introduction to Partial Differential
Equations, Cambridge University Press.
[5] Stroud, K. A. (2003), Advanced Engineering Mathematics, Palgrave Macmillan.
[6] Tung K.K (2004), Partial Differential Equations and Fourier Analysis - A Short
Introduction.
[7] Wikipedia article (2010), Fourier Transform.
[8] Yosida, K. (1980), Functional Analysis, Springer-Verlag.
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