1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Deriving expressions for Fourier Transforms of Partial Derivatives

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Using the formal limit definition of the derivative, derive expressions for the Fourier Transforms with respect to x of the partial derivatives [itex] \frac{\partial u}{\partial t} [/itex] and [itex] \frac {\partial u}{\partial x} [/itex].

    2. Relevant equations

    The Fourier Transform of a function [itex] u(x,t) [/itex] is:
    \tilde{u}(k,t) = \int_{-\infty}^{\infty} u(x,t) e^{-ikx} dx

    3. The attempt at a solution

    I attempted to write the FT of the derivative with respect to time as [itex] \lim_{dt \to 0} \frac{u(x,t+dt) - u(x,t)}{dt} [/itex] and then Fourier Transform it. Pulling the limit outside the integral, I got [itex] FT of (\frac{\partial u}{\partial t}) = \tilde{\frac {\partial u}{\partial t}}(k,t). [/itex]

    Now, with the partial derivative of x, I'm not so sure. The partial derivative wrt x is [itex] \lim_{dx \to 0} \frac{u(x + dx,t) - u(x,t)}{dx} [/itex]. When placed in the Fourier transform, I don't know what I can do. I don't think the dx at the end of the integral cancels out the dx in the definition of the derivative.
  2. jcsd
  3. Mar 24, 2014 #2


    User Avatar
    Homework Helper

    You should avoid using "dx" and so forth as anything other than part of a derivative or integral (or as a differential form). Otherwise you run into this sort of confusion. When taking limits, use either [itex]\delta x[/itex] or another letter.

    You want to calculate
    \int_{-\infty}^\infty \lim_{h \to 0} \frac{u(x+h, t) - u(x,t)}h e^{-ikx}\,dx
    = \lim_{h \to 0} \frac1h \left( \int_{-\infty}^\infty u(x + h, t)e^{-ikx}\,dx - \int_{-\infty}^\infty u(x, t)e^{-ikx}\,dx \right)
  4. Mar 24, 2014 #3
    Thanks, I must have derped out last night.

    I'm thinking about turning the 2nd term into [itex] \lim_{h \to \infty} \frac{1}{h} \tilde{u}(k,t) [/itex] and substituting the y = x + h into the first term, to give a final result of:
    \lim_{h \to 0} \frac{1}{h} ((e^{ik(0+h)}-e^{ik0}) \tilde{u}) = ike^{0} \tilde{u}
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted