Fourier Transforms: Solving PDE's

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SUMMARY

The discussion focuses on the application of Fourier transforms to solve partial differential equations (PDEs) involving the function u(x,t). It clarifies that the Fourier transform \(\widehat{u}(\omega,t)\) is indeed a function of \(\omega\) when t is fixed. Additionally, it addresses the differentiation of the Fourier transform with respect to time, confirming that while t is treated as a constant during the Fourier integral, the differentiation with respect to t is valid and does not yield zero.

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  • Understanding of Fourier transforms, specifically in the context of functions of multiple variables.
  • Knowledge of partial differential equations (PDEs) and their properties.
  • Familiarity with the concept of treating variables as constants during integration.
  • Basic calculus, particularly differentiation and integration techniques.
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  • Explore the implications of differentiating under the integral sign in Fourier analysis.
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Mathematicians, physicists, and engineers working on solving partial differential equations, as well as students studying advanced calculus and Fourier analysis.

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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 [tex]\widehat{u}(x,t)[/tex]:

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


****************
Questions:

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

2: The author says that we have the following:

[tex] F(\frac{d}{dt}u(x,t))(\omega) = \frac{d}{dt}\widehat{u}(\omega,t),[/tex]

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

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