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Computing derivatives using Fourier transform

The Fourier transform of a function $ f(x)$ and the inverse are define respectively

$\displaystyle \tilde{f}(\omega)=F[f]=\int f(x)e^{-i\omega x}\mathrm{d}x$ (4)

and

$\displaystyle f(x)=F^{-1}[f]=\frac{1}{2\pi}\int \tilde{f}(\omega)e^{i\omega x}\mathrm{d}\omega$ (5)

where $ \tilde{f}$ is the Fourier transform of $ f$ . In spatial dimension, $ \omega$ corresponds to the wavenumber $ k$ . Therefore we have

$\displaystyle \partial_x f=\partial_x\left( \frac{1}{2\pi}\int \tilde{f}e^{ik_x...
...{d}x\right) = \frac{1}{2\pi}\int \underline{ik_x\tilde{f}}e^{ik_x x}\mathrm{d}x$ (6)

and

$\displaystyle \partial_{xx} f=\partial_{xx}\left( \frac{1}{2\pi}\int \tilde{f}e...
...\right) = \frac{1}{2\pi}\int \underline{(ik_x)^2\tilde{f}}e^{ik_x x}\mathrm{d}x$ (7)

The term $ \partial_x\left(\frac{1}{\rho}\partial_x p\right)$ will be calculated by the following precedure:

\begin{displaymath}\begin{array}{rl} p\stackrel{F}{\longrightarrow}& \tilde{p}\\...
...{-1}[ik_xF[\frac{1}{\rho}F^{-1}[ik_x\tilde{p}]]] \\ \end{array}\end{displaymath} (8)

Let us conduct a 2-D numerical simulation of acoustic wave equation with constant density. The equation is then

$\displaystyle \partial_{tt}p=c^2 (\partial_{xx}+\partial_{zz})p$ (9)

Based upon Fourier transform for spatial axis, we know the right hand side of the above equation corresponds to

$\displaystyle -(k_x^2+k_z^2)\tilde{p}$ (10)

Thus, we have the following time evolution

$\displaystyle p^{n+1}=2p^n-p^{n-1}+\Delta t^2 c^2F^{-1}[-(k_x^2+k_z^2)F p^n]$ (11)


next up previous [pdf]

Next: Fractional Laplacian Up: Fourier pseudo spectral method Previous: What is a pseudo-spectral

2021-08-31