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8/4/2019 Fourier Laplace Transforms

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________________________________________________________MEP 201 Advanced Engineering MathematicsFourier Transforms/Laplace Transforms

1

-p +

f x

FOURIER INTEGRAL AND FOURIER TRANSFORM

Fourier integral representation of ( )xf

The function ( )xf shown below is defined only for pxp .

The function can be represented by the Fourier series

( )

( ) ( )

( ) ( ) ( )

=

+

++

=

=

+

+++

=

=

++=

1n

p

p'dxx'x

pncos'xf

p1p

p'dx'xf

p21

1n

p

p'dx

p'xnsin

pxnsin

p'xncos

pxncos'xf

p1p

p'dx'xf

p2

1

1npxnsinnbp

xncosna2oaxf

Above is the Fourier series for ( )xf no matter how large p is.

Let p approach infinity. Provided ( )

+

'dx'xf exists, the first term vanishes as p approaches infinity.

Letpn

n= and

( )pp

np

1nn1nn

=+

=+= . Thus,

( ) ( ) ( )[ ]

( ) ( )[ ] ( )

=

=

=

+

=

=

+

=

1nnx,nF

1n

1n

p

p'dxx'xncos'xf

1

1n

p

p'dxx'xncos'xfxf

n

As p , dn so that ( ) ( )

=0

dx,nF1xf .

Fourier integral representation of ( )xf : ( ) ( ) ( )[ ]

+

=

0d'dxx'xcos'xf1xf [1]

Equation [1] is a valid representation of ( )xf provided

(a) in every finite interval, ( )xf satisfies the Dirichlet conditions, and

(b) ( )+

dxxf exists.

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________________________________________________________MEP 201 Advanced Engineering MathematicsFourier Transforms/Laplace Transforms

2

EXAMPLE 1: Obtain the Fourier integral representation of the function ( )

>

+

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________________________________________________________MEP 201 Advanced Engineering MathematicsFourier Transforms/Laplace Transforms

3

EXAMPLE 2: Given ( )

>

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( ) ( )+

= dxieg2

1xf Complex inverse transform of ( )g .

( ) ( )+

= dxxiexf

2

1g Complex Fourier transform of ( )xf ; spectrum of ( )xf .

We shall also use the symbol ( ){ }xfF to denote the complex Fourier transform of ( )xf and the symbol ( ){ } g1F

to denote the inverse transform of ( )g .

EXAMPLE 3. Obtain the Fourier integral representation of ( )

>>

and 0a > ?

EXAMPLE 4: Determine the Fourier cosine transform pair for ( )

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Partial Differential Equations

EXAMPLE 5: A slender infinite rod has its lateral surface insulated. The initial temperature is ( ) ( )xf0,xu = ,

where ( )xf is bounded. Find ( )t,xu .

SOLUTION:

Let ( ) ( ) ( )tTxXt,xu = . Substitute into the one-dimensional heat equation to get2

T'T

a

1X

''X2

==

Thus, ( )

= t22aexptT and ( ) xsin2cxcos1cxX +=

( ) [ ]xsin2cxcos1ct22aexpt,xu +

=

( ) ( ) ( )[ ]

+

= d0 xsinBxcosAt

22aexp1t,xu

Apply initial condition:

( ) ( ) ( ) ( )[ ] ( ) ( )[ ]

+

=

+

== d0 'dxx'xcos'xf

1d0

xsinBxcosA1xf0,xu

( ) ( ) ( ) +

+

= 'dx

0dx'xcost22aexp'xf1t,xu

( )( )

+

= 'dx

ta4

x'xexp

ta2'xf1

2

2

from the integration formula in EXAMPLE 4 above where x , x'xb , t2a2a .

Suppose ( )

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Refer to the formula above for the inverse sine transform of a function and observe that the right

hand side of the above equation is the inverse sine transform of ( )

2

A . Therefore,

( )( )

=

0

xdxsinxf2

2

A. Hence, ( ) ( )

=

0xdxsinxf2A .

( ) ( )( )

=

===

0 2

xdsinaB20

xdsinaB1xg0tt

y

The quantity on the right-hand side is the inverse sine transform of( )

( )

=

0

xdxsinxg2

2

aB.

Thus, ( ) ( )

=0

xdxsinxga

2B .

Therefore,

( ) ( ) ( )

+

=

0datsin

0'dx'xsin'xg

a2atcos

0'dx'xsin'xf2xsin1t,xy

EXAMPLE 7: Find the steady-state temperature in a thin semi-infinite plate bounded by the x-axis and the

positive halves of the line 0x= and 1x= . The left and the bottom edges are maintained at zero temperaturewhile the temperature at the right edge is kept at ( )yf .

SOLUTION: Let ( ) ( ) ( )yYxXy,xu = and substitute into the two-dimensional steady-state heat equation to

get 2Y

''YX

''X == .

( ) ( )( )xsinh4cxcosh3cysin2cycos1cy,xu ++=

BC 1: ( ) 00,xu = Therefore, 01c = .

BC 2: ( ) 0y,0u = Therefore, 03c = .

( ) ysinxsinhy,xu = ( ) ( )

=

0ydsinxsinhCy,xu

BC 3: ( ) ( ) ( )

==0

ydsinsinhCyfy,1u

( ) ( )

= 0

ydsinsinhC2yf2

The right hand side is the inverse sine transform of ( ) sinhC . Therefore,

( ) ( )

=0

ydysinyf22sinhC

( ) ( )

=

0ydysinyf

sinh

2C

Suppose ( )

>

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EXAMPLE 8: SolveEXAMPLE 7by getting the Fourier sine transform of both sides of the PDE.

SOLUTION:

00

ydysiny

u0

ydysinx

u2

2

2

2=

+

( ) 00

dyyu

yysin

0ydysiny,xu

x22 =

+

Integrate by parts. In the second integral, let ysins = , dyyu

ydt

= ;

yut

= .

Hence,

=

0ydycos

yu

0yuysin

0dy

yu

yysin

+=

= 0 ydysinu0ycosu0 dyyuycos0

=

0

ydysinu2

Hence,

( ) ( ) 00

ydysiny,xu20

ydysiny,xudx

d2

2=

Denoting the Fourier sine transform of ( )y,xu by ( )x,sU , we get the following ordinary differential from the

above equation:

( )( ) 0x,sU

2

dx

x,Ud

2

s2

=

Thus, the effect of the Fourier sine transform is to convert the PDE to the ordinary differential equation shown

above. The general solution of this differential equation is( ) xsinh2cxcosh1cx,sU +=

BC 1: ( ) ( ) ( ) ( )0sinh2c0cosh1c00ydysiny,0u20,sU +==

= Therefore, 01c = .

BC 3: ( ) ( ) =

= sinh2c0 ydysinyf21,sU Therefore, ( )

=

0ydysinyf

sinh12

2c

( ) ( ) xsinh0

ydysinyfsinh

12x,sU

=

( )( )

=

0d

sinh

ysinxsinhydysinyf2y,xu 0

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LAPLACE TRANSFORM

Definition

Let ( )tf be defined for all 0t> . The Laplace transform of ( )tf is defined by the equation

L ( ){ } ( ) ( )

== 0 dttfstesFtf

over the range of values of s for which the integral exists.

EXAMPLE 1:

a) ( ) 1tf =

L{ } ( )s1

0s

e0

dt1ste1st

=

= =

b) ( ) atetf =

L

{ }( )

as

1

0as

e

0 dt

at

e

st

e

at

e

tas

+=

+=

=

+

c) ( ) ttf =

L{ }22

ststst

s

1

0s

e0

dts

e

0s

te0

tdtstet =

=

+

= =

Theorem: L ( ) ( ){ } 1ct2f2ct1f1c =+ L ( ){ } 2ct1f + L ( ){ }t2f

Definition ( )tf is said to be of exponential order if there exists a number 0M> and an such that( ) Mtfte

t

lim =

L{ } = 0 dttstet Let szt = sdzdt = For 0s >

L{ } ( )11 s

1

0dzzze

s

10 s

dzzeszt

++

+=

=

=

For n= (an integer), L1ns

!nt+

=

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1

a

u(t-a)

t

Theorem: Let ( )tf be sectionally continuous and of exponential order. Then, L ( ){ }tf exists when >s .

Theorem: Let ( )tf be continuous and ( )t'f be sectionally continuous in every finite interval Tt0 . Let( ){ }tf be of exponential order. Then, when >s ,

L ( ){ } st'f = L ( ){ } ( )0ftf

Similarly, L ( ){ } st''f = L ( ){ } ( )0'ft'f

[ss= L ( ){ } ( )] ( ) 2s0'ftf =0f- L ( ){ } ( ) ( )0'f0sftf

L ( ) nstnf = L ( ){ } ( ) ( ) ( )( )01nf...0'ftf 2-ns0f1-ns provided ( ) ( ),...t''f,t'f satisfy the same conditions as

( )tf .

EXAMPLE 4: