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Fourier Transform of an ODE

  1. Apr 5, 2012 #1
    I haven't had differential equations yet, so I am struggling in your math methods class. I understand what a Fourier Transform is, but I'm having trouble with this particular problem.

    1. The problem statement, all variables and given/known data
    Here's a screenshot. Better than I can write it.

    http://i.imgur.com/PQ6tB.png

    3. The attempt at a solution

    Here's what I did:

    http://i.imgur.com/JuUzu.jpg

    The capital letters have already been transformed, so if I take the inverse transformation, I should end up with what I had to begin with.

    Where I get stuck is with the [itex]Q, \frac{1}{D}, \frac{1}{(w^{2}+k^{2})}[/itex]. Is it possible to split the [itex]\Delta[w][/itex] up from the fraction, because that would just be back to [itex]\delta[x][/itex].

    If I'm completely wrong and beyond hope, just tell me and I will go cry in a corner.

    Thanks!
     
    Last edited: Apr 5, 2012
  2. jcsd
  3. Apr 6, 2012 #2
    Oh hey, I think the issue may be that ##\delta[x]## is actually the dirac delta function, which has the property that
    $$
    \int_{-\infty}^\infty f(x)\delta(x)\, \mathrm{d}x = f(0)
    $$
    This would mean that
    $$
    \mathbb{F}[Q\delta(x)]=Q
    $$
    See if this fixes things. I tried the problem and still had a good deal of trouble with it :confused:, but you may be able to swing the rest from there.

    PS: I'd be interested to see the rest of your solution when you get it. I suspect it may have to do with absolute values and or the step function.
     
  4. Apr 6, 2012 #3
    Using what you said (which makes sense; can't believe I didn't see that), I got this:

    http://i.imgur.com/lBkuj.jpg

    There were absolute values, however, I omitted them because I didn't think they were necessary.
     
  5. Apr 6, 2012 #4
    Hmmm....
    You know how ##|x|## has slope -1 until ##x=0##, and then it has slope 1? I think you might be able to make this claim:
    $$
    \frac{\mathrm{d}|x|}{\mathrm{d}x}=2\mathbb{H}(x)-1\text{, where H is the Heaviside step function. Note also that}\\
    \frac{\mathrm{d}\mathbb{H}}{\mathrm{d}x}=\delta(x)
    $$
    I have a suspicion this might be somehow related. I still haven't figured it out, but now it's starting to bother me.
     
  6. Apr 6, 2012 #5
    That's the first time I've heard of the Heaviside. What are it's uses (not only to this problem)?
     
  7. Apr 7, 2012 #6
    I don't know! In fact, I don't even know why or where I heard of it! But check this page out, it's pretty cool: http://mathworld.wolfram.com/HeavisideStepFunction.html
    I guess if you wanted to, you could also represent the T=0 Fermi distribution using the Heaviside.... Not sure that would actually be useful, though.
     
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