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Need help to prove expressions using Fourier Expansion

  1. Jun 25, 2011 #1
    Dear all,

    Please help me to prove the Fourier expansion for three different cases as follows. I need some help to show that L.H.S = R.H.S

    Case1: when m not equal to 0 and n not equal to 0.

    dU^0/dz (1- x/a) - dU^a/dz (x/a) = -(2/m^2 Pi^2)(1- Cos[m*Pi]) [dUn^0/dz - dUn^a/dz]


    Case2: when m = 0 and n not equal to 0.

    dU^0/dz (1- x/a) - dU^a/dz (x/a) = -1/2 [dUn^0/dz + dUn^a/dz]


    Case3: when m not equal to 0 and n = 0.

    dU^0/dz (1- x/a) - dU^a/dz (x/a) = 0

    where
    dU^0/dz = dUn^0/dz Sin[ n*Pi*x/a] and dU^a/dz = dUn^a/dz Sin[ n*Pi*x/a].

    The idea is that firstly using Fourier expansion, to expand (1 - x/a), then multiply with dU^0/dz, similarly, apply Fourier expansion for (x/a), then multiply with dU^a/dz.


    Thanks for any help.
     
  2. jcsd
  3. Jun 26, 2011 #2

    hunt_mat

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    I am not too sure what you're saying here, the function 1-x/a does not have a Fourier transform, can you possibly tex up your question
     
  4. Jun 26, 2011 #3
    Apologies for the unclear statements.
    f1(x) = (1-x/a) can be transformed using Fourier expansion into
    bo +[itex]\Sigma[/itex] bm*Cos[m*Pi*x/a] and

    f2(x) = x/a can be transformed using Fourier expansion into
    ao + [itex]\Sigma[/itex]am*Cos[m*Pi*x/a]

    Iam aware that the original equation of f(x) = ao +[itex]\Sigma[/itex]am*Cos[m*Pi*x/a] + bm Sin[m*Pi*x/a].

    Hope this would helps.

    Thank you.
     
    Last edited: Jun 26, 2011
  5. Jun 26, 2011 #4

    hunt_mat

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    So:
    [tex]
    f(x)=a_{0}+\sum_{m=1}^{\infty}a_{m}\cos\left(\frac{m\pi x}{a}\right) +b_{m}\sin\left(\frac{m\pi x}{a}\right)
    [/tex]

    What is the differential equation you're trying to solve?
     
  6. Jun 26, 2011 #5
    The only available data given to me are as follows:
    -[itex]\frac{dU^{o}}{dz}[/itex] (1-[itex]\frac{x}{a}[/itex]) - [itex]\frac{dU^{a}}{dz}[/itex]([itex]\frac{x}{a}[/itex])..................................(1)

    U[itex]^{o}[/itex] = U[itex]^{o}_{n}[/itex] *Sin[[itex]\frac{n*π*x}{a}[/itex]]....................................(2)

    U[itex]^{a}[/itex] = U[itex]^{a}_{n}[/itex] *Sin[[itex]\frac{n*π*x}{a}[/itex]]...................................(3)


    Case 1: m[itex]\neq[/itex]0 and n[itex]\neq[/itex]0:

    eqn(1) = [itex]\frac{-2}{m^2*π^2}[/itex]*[1-cos(m*π)]*[[itex]\frac{dU^{0}_{n}}{dz}[/itex]-[itex]\frac{dU^{a}_{n}}{dz}[/itex]]

    Case 2: m = 0 and n[itex]\neq[/itex]0:

    eqn(1) = -[itex]\frac{1}{2}[/itex]*[[itex]\frac{dU^{0}_{n}}{dz}[/itex]+[itex]\frac{dU^{a}_{n}}{dz}[/itex]]

    Case 3: m[itex]\neq[/itex]0 and n = 0:

    eqn (1) = 0


    Using Fourier expansion, I could expand the following:

    ([itex]\frac{x}{a}[/itex]) [itex]\approx[/itex] a[itex]_{o}[/itex]+[itex]\sum[/itex] a[itex]_{m}[/itex]*Cos[[itex]\frac{m*Pi*x}{a}[/itex]]

    [itex]\int[/itex][itex]^{0}_{a}[/itex] ([itex]\frac{x}{a}[/itex])*Cos[[itex]\frac{L*m*x}{a}[/itex]] dx=[itex]\sum[/itex][itex]\int[/itex][itex]^{0}_{a} [/itex] a[itex]_{m}[/itex]*Cos[[itex]\frac{m*Pi*x}{a}[/itex]]*Cos[[itex]\frac{L*m*x}{a}[/itex]] dx

    [itex]\int[/itex][itex]^{0}_{a}[/itex] ([itex]\frac{x}{a}[/itex]) Cos[[itex]\frac{L*m*x}{a}[/itex]] dx= [itex]\sum[/itex][itex]\int[/itex][itex]^{0}_{a} [/itex] a[itex]_{m}[/itex]*[itex]\frac{1}{2}[/itex][Cos([itex]\frac{m*Pi*x}{a}[/itex]-[itex]\frac{L*Pi*x}{a}[/itex])+Cos([itex]\frac{m*Pi*x}{a}[/itex]+[itex]\frac{L*Pi*x}{a}[/itex])]dx

    Note that L = m, then, Iam stucked up to this point. I have no idea how to proceed next.

    Similarly, the above expansion can be used for (1 - [itex]\frac{x}{a}[/itex]):

    (1-[itex]\frac{x}{a}[/itex]) [itex]\approx[/itex] b[itex]_{o}[/itex]+[itex]\sum[/itex] b[itex]_{m}[/itex]*Cos[[itex]\frac{m*Pi*x}{a}[/itex]]



    Any help is greatly appreciated.
     
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