Fourier Transformation

Fourier transformation is a transformation in space. If we have a one dimensional function $c(x)$, which is a function of $x$. It can be transformed to the Fourier space, $\hat{c}(k_x)$: \begin{equation} \hat{c}(k_x)=\int_{-\infty}^{\infty}\exp(-ik_xx)c(x)dx \end{equation} With the knowledge of $\hat{c}(k_x)$, it can also be transformed back to the real space: \begin{equation} c(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(ik_xx)\hat{c}(k_x)dk_x \end{equation} For example, if $c(x)=\exp(-x^2)$: \begin{align} \hat{c}(k_x)&=\int_{-\infty}^{\infty}\exp(-ik_xx)\exp(-x^2)dx\\ &=e^{-\frac{k_x^2}{4}}\int_{-\infty}^{\infty}\exp[-(x+\frac{ik_x}{2})^2]dx\\ &=\sqrt{\pi}\exp(-\frac{k_x^2}{4}) \end{align} With $\hat{c}(k_x)=\exp(-k_x^2/4)$, we can transform that back to the real space: \begin{align} c(x)&=\frac{\sqrt{\pi}}{2\pi}\int_{-\infty}^{\infty}\exp(ik_xx)\exp(-k_x^2/4)dk_x\\ &=\frac{\exp(-x^2)}{2\sqrt{\pi}}\int_{-\infty}^{\infty}\exp[-\frac{1}{4}(k_x-2ix)^2]dk_x\\ &=\exp(-x^2) \end{align}

3D Fourier Transformation

In 3D space, the Fourier transformation of function $c(x,y,z)$ can be written as follows: \begin{align} \hat{c}(k_x,k_y,k_z)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-ik_xx-ik_yy-ik_zz)c(x,y,z)dxdydz\\ &=\int_{-\infty}^{\infty}\exp(-i\vec{k}\cdot\vec{r})c(\vec{r})d\vec{r}\\ \end{align} If the function $c(\vec{r})$ is uniform regardless of the orientation of the vector $\vec{r}$: \begin{align} &=\int_0^{2\pi}d\psi\int_0^{\infty}\int_{-1}^{1}\exp(-ikr\cos\theta)c(r) r^2d\cos\theta dr\\ &=4\pi\int_0^{\infty}\frac{\sin(kr)}{kr}c(r)r^2dr \end{align} For which, the integral has been rewritten in terms of spherical coordinate. And with the knowledge of $\hat{c}(k)$, we can transform the function back to real space: \begin{align} c(r)=\frac{1}{2\pi^2}\int_0^{\infty}\frac{\sin(kr)}{kr}\hat{c}(k)k^2dk \end{align} With these, we can solve the diffusion equation with ease!

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