Project Student

8/2/2019 Project Student

1/22

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

i


2/22

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


3/22

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.

1


4/22

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.


5/22

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


6/22


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


7/22


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.


8/22

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)

6


9/22

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.


10/22

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.

8


11/22

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 .


f(x) = ex, < x <


12/22

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.


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.


13/22

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.


14/22

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.


15/22

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.

13


16/22

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)


17/22


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)


18/22


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


19/22

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)


20/22

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)


21/22

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.


22/22

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.

20

Documents

Project Student