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Solving DE using Fourier Transform

  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data

    2me31hl.png

    Part (a): State inverse fourier transform. Show fourier transform is:
    Part (b): Show fourier transform is:
    Part (c): By transforming LHS and RHS, show the solution is:
    Part(d): Using inverse fourier transform, find an expression for T(x,t)

    2. Relevant equations



    3. The attempt at a solution

    Part(d)
    [tex]T_{(x,t)} = \int_{-\infty}^{\infty} T e^{-Dk^2t}e^{ikx} dk [/tex]
    = [tex] \int_{-\infty}^{\infty} (T_0 + \sum_{m=1}^{\infty} Tm cos(\frac {m\pi x}{L}) e^{-Dk^2t + ikx} dk [/tex]
    = [tex]\int_{-\infty}^{\infty} T_0 e^{-Dk^2t + ikx} dk + \sum_{m=1}^{\infty} T_m cos(\frac {m \pi x}{L}) \int_{-\infty}^{\infty} e^{-Dk^2t + ikx} dk [/tex]

    Attempt at evaluating the integral, letting [tex] a^2 = \frac {1}{Dt} [/tex]

    [tex] \int_{-\infty}^{\infty} e^{-(\frac{k^2}{a^2} - ikx)} dx [/tex]
    [tex] = \int_{-\infty}^{\infty} e^{-\frac{({k - \frac{ixa^2}{2}})^2}{a^2}} e^{-\frac {x^2a^2}{4}} dk [/tex]
    [tex] = e^-{\frac{x^2a^2}{4}} \int_{-\infty - i\frac{x^2a^2}{4}}^{\infty - i\frac {x^2a^2}{4}} e^{- \frac{k^2}{a^2}} dk [/tex]
    [tex] = e^{\frac {-x^2a^2}{4}} \sqrt{\pi a^2} = \sqrt {\frac{\pi}{Dt}} e^{\frac {-x^2a^2}{4}}[/tex]

    But this appears to be wrong as their final expression do not have the factor of √(1/t) in their coefficients..
     
    Last edited: Jan 9, 2014
  2. jcsd
  3. Jan 9, 2014 #2

    vanhees71

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    First you have to evaluate [itex]\tilde{T}(k,0)[/itex] from the initial condition which is given in the position domain as [itex]T(x,0)[/itex]! Then you can plug it into the general solution and Fourier transform back!
     
  4. Jan 9, 2014 #3
    Ha ha, that's a silly mistake I made!

    So I must fourier transform the initial condition, then plug it back in, then inverse fourier transform everything?
     
  5. Jan 10, 2014 #4

    vanhees71

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