Fourier transform

Introduction

Fourier transform of a complex-valued function $f (t)$ is a another complex-valued function $\hat{f} (w)$ that codifies the same information that $f (z)$ in the following way: for every $w \in R$ , the value $\hat{f} (w)$ represent the weight of the function $e^{2 π i w t}$ in the decomposition of $f (t)$ as a sum of exponential

f (t) = \dots + \hat{f} (1) e^{2 π i t} + \hat{f} (1.1) e^{2 π i 1.1 t} + \hat{f} (2) e^{2 π i 2 t} + \dots

\dots + \hat{f} (w) e^{2 π i w t} + \dots =

= \int_{- \infty}^{+ \infty} \hat{f} (w) e^{2 π i w t} d w

The transform can be obtained with $\hat{f} (w) = \int_{- \infty}^{+ \infty} f (t) e^{- 2 π i w t} d t$ .

A great explanation of the intuition behind Fourier transform: this video. It also explains why a real function possesses a conjugate symmetric Fourier transform ( $\hat{f} (- w) = \overset{―}{\hat{f} (w)}$ ).

It is a particular case of the Laplace transform. Related Mellin transform. See also translation and momentum operators.

Using Fourier Series to Solve Linear PDEs with Initial Data

Given a linear PDE $u_{t} = u_{x x}$ with initial data $u (x, 0) = f (x)$ , the first step in the solution process is to decompose the initial data function $f (x)$ into its Fourier series representation. This is achieved by expressing $f (x)$ as an infinite sum of complex exponential functions.

f (x) = \sum_{n = - \infty}^{+ \infty} c_{n} e^{2 π i n x}

where the coefficients $c_{n}$ are given by:

c_{n} = \int_{- \infty}^{+ \infty} f (x) e^{- 2 π i n x} d x

Also, we decompose

u (x, t) = \sum_{n = - \infty}^{+ \infty} u_{n} (t) e^{2 π i n x}

where $u_{n} (t)$ represents the time-dependent coefficients of the Fourier series expansion of $u (x, t)$ .

The next step is to substitute the Fourier series representation of $u (x, t)$ into the given PDE. By doing so, we obtain a set of ordinary differential equations (ODEs) for the coefficients $u_{n} (t)$ .

For the given PDE $u_{t} = u_{x x}$ , the substitution yields:

\frac{d u_{n} (t)}{d t} = - 4 π^{2} n^{2} u_{n} (t)

This is a simple first-order ODE which can be solved for $u_{n} (t)$ . The general solution is:

u_{n} (t) = u_{n} (0) e^{- 4 π^{2} n^{2} t}

Where $u_{n} (0)$ is the initial value of the coefficient, which can be obtained from the Fourier series representation of the initial data $u (x, 0) = f (x)$ . Specifically, $u_{n} (0) = c_{n}$ , the Fourier coefficient of $f (x)$ .

Finally, substituting the solution for $u_{n} (t)$ into the Fourier series representation of $u (x, t)$ , we obtain the solution of the PDE in terms of the initial data $f (x)$ :

u (x, t) = \sum_{n = - \infty}^{+ \infty} c_{n} e^{- 4 π^{2} n^{2} t} e^{2 π i n x}

Using Fourier Transform to Solve Linear PDEs with Given Initial Data

For problems defined on an infinite domain, the Fourier series might not be the most suitable tool. Instead, the Fourier Transform is employed to handle such cases. Given a linear PDE $u_{t} = u_{x x}$ with initial data $u (x, 0) = f (x)$ , the solution process begins by taking the Fourier Transform of the initial data function $f (x)$ . This transforms the function from the spatial domain to the frequency domain. The Fourier Transform of $f (x)$ is given by:

f (x) = \int_{- \infty}^{+ \infty} \hat{f} (ω) e^{2 π i ω x} d ω

where $\hat{f} (ω)$ represents the function in the frequency domain.

Similarly,

u (x, t) = \int_{- \infty}^{+ \infty} \hat{u} (ω, t) e^{2 π i ω x} d ω

Substituting the Fourier Transformed functions into the PDE, we obtain a differential equation in the frequency domain. For the PDE $u_{t} = u_{x x}$ , this yields:

\int_{- \infty}^{+ \infty} \frac{d \hat{u} (ω, t)}{d t} e^{2 π i ω x} d ω = \int_{- \infty}^{+ \infty} - 4 π^{2} ω^{2} \hat{u} (ω, t) e^{2 π i ω x} d ω

\frac{d \hat{u} (ω, t)}{d t} = - 4 π^{2} ω^{2} \hat{u} (ω, t)

This equation is a first-order ODE in the frequency domain, which can be solved for $\hat{u} (ω, t)$ . The general solution is:

\hat{u} (ω, t) = \hat{u} (ω, 0) e^{- 4 π^{2} ω^{2} t}

Where $\hat{u} (k, 0)$ is the initial value in the frequency domain, which is simply $\hat{f} (ω)$ , the Fourier Transform of the initial data $f (x)$ .

To obtain the solution in the spatial domain, we take the inverse Fourier Transform of $\hat{u} (ω, t)$ . This gives us the solution of the PDE in terms of the initial data $f (x)$ .