Fourier Transforms: Solving PDE's

[Niles]

Homework Statement

Hi all.

We have a function u(x,t), where x can go from (-infinity;infinity) and t>0. In my book it says:

"For fixed t, the function u(x,t) becomes a function of the spatial variable x, and as such, we can take its Fourier transform with respect to the x-variable. We denote this transform by \widehat{u}(x,t):

<br /> \widehat{u}(\omega,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{u(x,t)e^{-i\omega x}} dx<br />


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Questions:

1: First of all, \widehat{u}(x,t) is only a function of \omega, since we have fixed t, correct?

2: The author says that we have the following:

<br /> F(\frac{d}{dt}u(x,t))(\omega) = \frac{d}{dt}\widehat{u}(\omega,t),<br />

where the large F denotes the Fourier transform of u(x,t) with respect to x. Since we have fixed t, then why are we differentiating with respect to t? Doesn't this give zero?

Thanks in advance.

 

[gabbagabbahey]

The "fixed t" concept just means that when performing the Fourier integral over x, you treat t as a constant over space.