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

    >

    +

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    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: