- #1
Dustinsfl
- 2,281
- 5
Does every FFT have \(i\) in it?
Given \(u_t = -(u_{xxx} + 6uu_x)\).
\(f'''(x) = \mathcal{F}^{-1}\left[(ik)^3\mathcal{F}(f(x))\right]\)
\(f'(x) = \mathcal{F}^{-1}\left[(ik)\mathcal{F}(f(x))\right]\)
The only equation I have used the pseudo-spectral method on was the NLS which is
\(u_t = i(u_{xx} + |u|^2u)\). In this case, I know I will have \(i\) in the FFT.
Are my transforms for the KdV correct or do I need to remove \(i\)?
Given \(u_t = -(u_{xxx} + 6uu_x)\).
\(f'''(x) = \mathcal{F}^{-1}\left[(ik)^3\mathcal{F}(f(x))\right]\)
\(f'(x) = \mathcal{F}^{-1}\left[(ik)\mathcal{F}(f(x))\right]\)
The only equation I have used the pseudo-spectral method on was the NLS which is
\(u_t = i(u_{xx} + |u|^2u)\). In this case, I know I will have \(i\) in the FFT.
Are my transforms for the KdV correct or do I need to remove \(i\)?