Can someone please help with the Fourier Transform of :

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Thank you in advance!

::Edit:: This is what I am trying to solve:

$\frac{\partial(p(x, t))}{\partial t} = -A\frac{\partial(xp(x, t))}{\partial x}+\frac{B}{2}\frac{\partial^{2}(xp(x, t))}{\partial x^{2}}$

Where:[{A, B} = Constants]

Define: $FT\{p(x, t)\}(\omega) = \int_{-\infty }^{\infty }p(x, t)e^{-2\pi ix\omega }\,dx$ and $FT^{-1}\{\bar{p}(\omega , t)\}(x) = \int_{-\infty }^{\infty }p(\omega , t)e^{2\pi ix\omega }\,d\omega$

The next step is to convert each term so I can reduce the order but I started reading about Fourier Transforms two days ago, so I do not know all the tricks.

I did use the properties below to get rid of the derivative - but I have no idea how to convolve x with p(x,t) - since p(x,t) is unknown.

p(x,t) is a density function - so it goes to zero in the infinities (if this is important)

Thank you again!

  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Dec 31 '14 at 10:22
  • $\begingroup$ (Edited.Thanks!) $\endgroup$ – Edv Beq Dec 31 '14 at 19:25
  • $\begingroup$ There is a nice post here : link but I need min[50 points] to post there. The question is very similar to mine. How can I get some more points please? $\endgroup$ – Edv Beq Jan 3 '15 at 3:06

So let's try and tackle the first order derivative, $\partial_x(xp)=p+x\partial_x p$, let's look at the Fourier of the second term:

$F(x\partial_x p)=\int_{\mathbb R}x\partial_xpe^{-2\pi i xw}\,dx$

Notice that $\partial_we^{-2\pi i xw}=-2\pi ixe^{-2\pi i xw}$ so

$F(x\partial_x p)=\frac{i}{2\pi}\partial_wF(\partial_xp)=-w\partial_wF(p)$. By the way I may be out by some constants, but use the magic $2\pi=1$ formula and its all good.

  • $\begingroup$ The $2\pi=1$ formula is a joke by the way, there just tends to be a lot of constants floating around when it comes to Fourier transforms $\endgroup$ – Ellya Jan 3 '15 at 17:34
  • $\begingroup$ Can you look at my reply please - My negative signs are conflicting. Thank you $\endgroup$ – Edv Beq Jan 5 '15 at 3:42
  • $\begingroup$ @edvbeq, yours is right, I added a minus sign In when taking the Fourier of the derivative $\endgroup$ – Ellya Jan 5 '15 at 7:54
  • $\begingroup$ Ok thank you so much. $\endgroup$ – Edv Beq Jan 5 '15 at 10:33
  • $\begingroup$ Can you please help with this math.stackexchange.com/questions/2288576/… $\endgroup$ – Edv Beq May 21 '17 at 22:49

Well it appears that I am getting some neg signs wrong but mostly my findings agree with yours. Here is what I have:

I have gathered these online - I have not done the integration myself - so I may be wrong.

Identity 1: $FT\left\{\frac{\partial^{n}p(x, t)}{\partial x^{n}}\right\} = (i2\pi w)^{n}\bar{p}(w, t)$ Note: No negative sign.

Identity 2: $FT\{xp(x, t)\} = \frac{i}{-2\pi }\frac{\partial\bar{p}(w, t)}{\partial w}$ Note: Negative sign in the denominator.

Let: $g(x, t) = x\cdot p(x, t)$

Then: By ID 1: $FT\left\{\frac{\partial g(x, t)}{\partial x}\right\} = (i2\pi w)\bar{g}(w, t)$

So: By ID 2: $\bar{g}(w, t) = \int_{-\infty }^{\infty }g(x, t)e^{-i2\pi wx}\,dx$ $ = \int_{-\infty }^{\infty }x\cdot p(x, t)e^{-i2\pi wx}\,dx = \frac{i}{-2\pi }\frac{\partial\bar{p}(w, t)}{\partial w}$

Finally, by adding the two parts: $FT\left\{\frac{\partial xp(x, t)}{\partial x}\right\} = (i2\pi w)\frac{i}{-2\pi }\frac{\partial\bar{p}(w, t)}{\partial w}=w\frac{\partial\bar{p}(w, t)}{\partial w}$

I think its correct! Right?


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