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Applying the fourier transform to a PDE

  1. Nov 19, 2013 #1
    I have a tutorial question for maths involving the heat equation and fourier transform.

    [itex] {\frac{∂u}{∂t}} = {\frac{∂^2u}{∂x^2}} [/itex]

    you are given the initial condition:

    [itex]u(x,0) = 70e^{-{\frac{1}{2}}{x^2}} [/itex]

    the answer is:

    [itex]u(x,t) = {\frac{70}{\sqrt{1+2t}}}{e^{-{\frac{x^2}{2+4t}}}} [/itex]

    In this course the definition of a fourier transform is:

    [itex] U(k) = F[/itex]{[itex]u(x)[/itex]}[itex] = {\frac{1}{\sqrt{2π}}}{\int_{-∞}^{∞}{{u(x)}{e^{-ikx}}}\,dx}[/itex]

    [itex] u(x) = F^{-1}[/itex]{[itex]U(k)[/itex]}[itex] = {\frac{1}{\sqrt{2π}}}{\int_{-∞}^{∞}{{U(k)}{e^{+ikx}}}\,dk}[/itex]

    MY SOLUTION:

    So far I can use the transform with the pde to obtain
    [itex] U(k,t) = A(k).e^{-k^{2}t}[/itex]
    [itex] U(k,0) = A(k)[/itex]

    From the initial condition and a transform table:
    [itex] U(k,0) = 70.e^{-{\frac{1}{2}}k^2}[/itex]
    [itex] U(k,t) = 70.e^{-{\frac{1}{2}}k^2}.e^{-k^{2}t}[/itex]

    from here it looks like I would have to apply the inverse fourier transform to get u(x,t). This is a difficult integration an I don't remember it being this complicated the first time. So basically:

    Is there an easier way to do this question?
    Any tips for the inverse fourier integration below?

    [itex]u(x,t) = {\frac{1}{\sqrt{2π}}}{\int_{-∞}^{∞}{{70.e^{-{\frac{1}{2}}k^2}.e^{-k^{2}t}}{e^{+ikx}}}\,dk}[/itex]
     
  2. jcsd
  3. Nov 21, 2013 #2

    vela

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    You have a Fourier transform of a Gaussian
    $$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\alpha k^2} e^{ikx}\,dk$$ where ##\alpha = t+\frac{1}{2}##, so you could just look up the answer in a table of Fourier transform pairs. If you want to derive the result, combine the exponentials and then complete the square.
     
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