MHB Fourier Transform of a function squared.

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The discussion focuses on the Fourier Transform of a function squared, specifically analyzing the equation \(u_t = -u_{nxxx} - 3(u^2)_{nx}\). It confirms that the reduction using the Inverse Fourier Transform is correct, leading to the expression involving \(ik^3\mathcal{F}^{-1}[\mathcal{F}(u)]\) and \(-ik\mathcal{F}^{-1}[\mathcal{F}(u^2)]\). Additionally, it asserts that \(\mathcal{F}(u^2) = \mathcal{F}(u\cdot u)\) cannot be simplified further. The key takeaway is the validation of the Fourier Transform operations and their implications on the function squared. Understanding these transformations is crucial for analyzing differential equations in mathematical physics.
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Consider \(u_t = -u_{nxxx} - 3(u^2)_{nx}\).

The Fourier Transform is linear so taking the Inverse Fourier transform of the Fourier Transform on the RHS we have
\begin{align}
-\mathcal{F}^{-1}\left[\mathcal{F}\left[u_{nxxx} - 3(u^2)_{nx}\right]\right] &= -\mathcal{F}^{-1} \left[\mathcal{F}\left[(ik)^3u\right]\right] - 3\mathcal{F}^{-1}\left[\mathcal{F} \left[(ik)u^2\right]\right]\\
&= ik^3\mathcal{F}^{-1}\left[\mathcal{F}(u)\right] - ik\mathcal{F}^{-1}\left[\mathcal{F}(u^2)\right]
\end{align}
  1. Is the above reduction correct?
  2. Can \(\mathcal{F}(u^2) = \mathcal{F}(u\cdot u)\) be further reduced?
 
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Yes, the reduction is correct. You cannot further reduce \(\mathcal{F}(u^2) = \mathcal{F}(u\cdot u)\).
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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