Fourier Transforms: Solving PDE's

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In summary, When considering the function u(x,t) with x ranging from -infinity to infinity and t being greater than 0, we can take its Fourier transform with respect to x, denoted by \widehat{u}(x,t). For a fixed t, this transform becomes a function of the spatial variable, \omega. The author also explains that when taking the Fourier transform of the derivative of u(x,t) with respect to t, denoted by F(\frac{d}{dt}u(x,t))(\omega), the result is equivalent to differentiating \widehat{u}(\omega,t) with respect to t, even though t is considered a constant over space.
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Niles
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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.
 

1. What is a Fourier Transform?

A Fourier Transform is a mathematical tool used to decompose a function into its constituent frequencies. It converts a function from its original domain (usually time or space) to a representation in the frequency domain.

2. How are Fourier Transforms used to solve PDE's?

Fourier Transforms are used to solve PDE's by transforming the PDE into an algebraic equation in the frequency domain. This makes it easier to solve the equation and then use the inverse Fourier Transform to obtain the solution in the time or space domain.

3. Can Fourier Transforms be used to solve any type of PDE?

Yes, Fourier Transforms can be used to solve a wide range of PDE's, including linear and non-linear equations. However, it is not always the most efficient method and other techniques may be more suitable depending on the specific problem.

4. Are there any limitations to using Fourier Transforms in solving PDE's?

One limitation of using Fourier Transforms in solving PDE's is that it assumes the function is continuous and has a continuous derivative. This may not always be the case in real-world applications, and in these situations, other methods may be more appropriate.

5. Are there any practical applications of using Fourier Transforms to solve PDE's?

Yes, there are many practical applications of using Fourier Transforms to solve PDE's, such as in engineering, physics, and finance. It can be used to model and analyze a wide range of phenomena, including heat transfer, fluid dynamics, and signal processing.

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