Deriving expressions for Fourier Transforms of Partial Derivatives

[N00813]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I attempted to write the FT of the derivative with respect to time as \lim_{dt \to 0} \frac{u(x,t+dt) - u(x,t)}{dt} and then Fourier Transform it. Pulling the limit outside the integral, I got FT of (\frac{\partial u}{\partial t}) = \tilde{\frac {\partial u}{\partial t}}(k,t).Now, with the partial derivative of x, I'm not so sure. The partial derivative wrt x is \lim_{dx \to 0} \frac{u(x + dx,t) - u(x,t)}{dx}. 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.

 

[pasmith]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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


Now, with the partial derivative of x, I'm not so sure. The partial derivative wrt x is \lim_{dx \to 0} \frac{u(x + dx,t) - u(x,t)}{dx}. 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.

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 \delta x or another letter.

You want to calculate
<br /> \int_{-\infty}^\infty \lim_{h \to 0} \frac{u(x+h, t) - u(x,t)}h e^{-ikx}\,dx<br /> = \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)<br />

 

[N00813]

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 \delta x or another letter.

You want to calculate
<br /> \int_{-\infty}^\infty \lim_{h \to 0} \frac{u(x+h, t) - u(x,t)}h e^{-ikx}\,dx<br /> = \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)<br />

Thanks, I must have derped out last night.

I'm thinking about turning the 2nd term into \lim_{h \to \infty} \frac{1}{h} \tilde{u}(k,t) and substituting the y = x + h into the first term, to give a final result of:
<br /> \lim_{h \to 0} \frac{1}{h} ((e^{ik(0+h)}-e^{ik0}) \tilde{u}) = ike^{0} \tilde{u}<br />