- #1

Roo2

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## Homework Statement

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

http://www.physics.ucf.edu/~schellin/teaching/phz3113/lec13-3.pdf

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?

∫u(x,t)*e^(ikx)dx

## Homework Equations

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

## 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.