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Fourier Transform of Undefined Function

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm trying to derive the result on slide 1 of this link:


    Unfortunately, I'm not sure how to integrate the Fourier transform when my u(x,t) function is undefined. Could someone help me get the following?


    2. Relevant equations

    The Fourier transform of a function u(x,t) = ∫u(x,t)*e^(ikx)dx

    3. The attempt at a solution

    I tried integration by parts but it doesn't seem to work. First, I tried:

    u = u(x,t)
    du = u'(x,t)
    dv = e^(ikx)
    v = 1/(ik) e^(ikx)

    uv - ∫v du =[itex]\frac{u(x,t)}{ik}[/itex] * [itex]e^{ikx}[/itex] -∫[itex]\frac{u'(x,t)}{ik}[/itex]* [itex]e^{ikx}[/itex]

    = [itex]\frac{u(x,t)}{ik}[/itex] * [itex]e^{ikx}[/itex] -[itex]\frac{1}{ik}[/itex] ∫u'(x,t)* [itex]e^{ikx}[/itex]

    And now I seem to be in a repeating loop of integration by parts because the[itex]e^{ikx}[/itex] will always be in the integral as dv. I then tried to do it the other way:

    u = [itex]e^{ikx}[/itex]
    du = [itex]\frac{1}{ik}[/itex][itex]e^{ikx}[/itex]
    dv = u(x,t)
    v = ∫u(x,t)

    uv-∫vdu doesn't work because it requires ∫u(x,t), which I can't define. I don't understand how to take the integral, and I therefore don't understand the slide linked above. If I take the derivative property of Fourier transforms on faith, I should get the following:

    [itex]\frac{d^{2}u}{dx^2} = \frac{1}{α^2}\frac{du}{dt}[/itex]

    If the FT of y'(x) = ik*Y(k), then the above should become:

    [itex]ik\frac{dU(k,t)}{dx} = \frac{ik}{a^2}U(k,t)[/itex]

    In which case the ik's on each side should cancel out. I therefore don't understand how they got a factor of -k^2 in their solution.

    Please remedy my confusion.
  2. jcsd
  3. Jan 29, 2013 #2


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    Homework Helper

    You can't calculate the Fourier transform of an arbitrary function, except in a trivial sense: you can define a function ##\hat{u}(k,t)## as the fourier transform with respect to x of u(x,t), or vice versa (u(x,t) is the inverse fourier transform of ##\hat{u}(k,t)## with respect to k):

    $$u(x,t) = \int_{-\infty}^\infty \frac{dk}{2\pi}~\hat{u}(k,t)e^{ikx}.$$

    Note that the first slide has a typo: the integration variable should be k, not x. Also, I used a hat (^) to denote that ##\hat{u}(k,t)## is a different function than ##u(x,t)##. I also included the conventional factor of ##2\pi## in the inverse fourier transform definition.

    Anyways, so what is the point of this if this is only a definition? Well, although you can't say what ##\hat{u}(k,t)## is for arbitrary ##u(x,t)##, or vice versa, you can compute the fourier transforms of derivatives of u(x,t) in terms of ##\hat{u}(k,t)##.

    So, what the professor does in those lecture notes is plug the definition of ##u(x,t)## as a fourier integral into the differential equation and, applying the derivatives through the integrals, derives an equation that looks like

    $$\int_{-\infty}^\infty dk~(\mbox{something}) = \int_{-\infty}^\infty dk~(\mbox{something else}),$$

    and concludes that ##\mbox{something} = \mbox{something else}##, where I leave it to you to figure out what the 'something' and 'something else' are. I'll also ask you to explain why we can just 'drop the integrals' in this case. Do you know why we can do that?

    The point of this exercise is that you will find that by using the fourier transform expression of u(x,t) we are able to convert a partial differential equation for u(x,t) into an ordinary differential equation for ##\hat{u}(k,t)##.
  4. Jan 29, 2013 #3
    Thanks for your reply. I'm starting to get it, but I still need some help (it's been quite a while since I've done math; I'm a bit slow on the uptake). I think you can drop the integrals because if ∫x dk = ∫y dk, then x = y. However, I'm not terribly sure about that; is it not possible to have two different functions with the same area under the curve?

    Also, could you please show me the process of differentiating a more simple function via Fourier transform? I'm not seeing how you can apply a derivative through an integral (I think that I just never learned the operation or the rules for it).

    Finally, why is U(k,t) an ODE while u(x,t) is a PDE, when they're both functions of two variables?
  5. Feb 4, 2013 #4


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    Homework Helper

    Hey, sorry I hadn't replied yet. I've been out of town for a few days.

    In general, ∫x dk = ∫y dk does not imply x = y. For example, ##\int_0^{2\pi} dk \cos(k) = \int_0^{2\pi} dk \sin(k) = 0##, but we know that ##\cos(k) \neq \sin(k) \neq 0##. The cases in which you can deduce that ∫x dk = ∫y dk implies x = y typically have some sort of arbitrary parameter in both integrals, such that the equality has to hold for any value of the parameter. For example, in the fourier case you are considering in the notes, the variable x is arbitrary. The equation has to hold true for every possible value of x. This generally implies that the two integrands have to be equal. You can easily show this to be the case for fourier integrals. If

    $$\int_{-\infty}^\infty dk~f(k)e^{ikx} = \int_{-\infty}^\infty dk~g(k)e^{ikx},$$
    then you can conclude that f(k) = g(k) by multiplying both equations by ##\exp(iqx)## and integrating over x. Try it out. (You will need the identity ##\int_{-\infty}^\infty dx \exp(i(k-q)x) = \delta(k-q)##).

    The general rule for differentiating an integral is

    $$\frac{\partial}{\partial x} \left[ \int_{\alpha(x)}^{\beta(x)} dt~f(x,t)\right] = f(x,\beta(x))\frac{\partial \beta}{\partial x} - f(x,\alpha(x))\frac{\partial \alpha}{\partial x} + \int_{\alpha(x)}^{\beta(x)} dt~\frac{\partial f(x,t)}{\partial x}.$$

    In your case, ##\alpha## and ##\beta## are constants (infinite, but constants), so only the last term remains. Basically, for constant limits of integration, a derivative of an integral (with respect to a variable that is not the integration variable) will pass through the integral.

    Just to be clear on terminology: you have an ODE for ##\hat{u}(k,t)## and you have a PDE for u(x,t). The functions themselves are not ODEs or PDEs.

    When insert the fourier transform of u(x,t) into ##\partial u(x,t)/\partial x##, you will find it gives you ##\hat{u}(k,t)## times some k-dependent factor - no derivatives leftoever. Since your PDE only had x or t derivatives, after plugging in the fourier representation of u(x,t) and dropping the integrals, only t derivatives will remain. Since you only have one kind of derivative, you have an ODE.

    Similarly, say you had an ODE for u(x) - only x, no t. If you insert the fourier transform of u(x), all of the x derivatives would go away, leaving you with a purely algebraic equation for the fourier transform ##\hat{u}(k)##.

    Also, for completeness, I should point out that you can also derive the ODE for ##\hat{u}(k,t)## by fourier transforming the PDE directly.
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