Fourier Series

Fourier series are a way to represent a periodic function as a superposition of elementary waves.

FOURIER SERIES

Let

f

be real-valued on

[-L,L]

. Its Fourier series is

\begin{equation*}f(x) \sim \frac{a_0}{2}+ \sum_{n=1}^{\infty}\left(a_n\cos\left(\frac{n\pi x}{L}\right)+ b_n\sin\left(\frac{n\pi x}{L}\right)\right),\end{equation*}

where

\begin{equation*}a_n=\frac{1}{L}\int_{-L}^L f(x)\cos\left(\frac{n\pi x}{L}\right)\,dx, \quad b_n=\frac{1}{L}\int_{-L}^L f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx.\end{equation*}

Motivation

Consider the step function on the interval

[-1,1]

with

L=1

\begin{equation*}f(x)=\begin{cases}1, & |x| < \frac12,\\-1, & \frac12 \leq |x| < 1.\end{cases}\end{equation*}

The introductory animation displays only the part of the resulting Fourier series on

[0,1]

For this full interval, the constant term is

\begin{align*}a_0 &=\int_{-1}^{1} f(x)\,dx =2\int_0^1 f(x)\,dx\\&=2\left(\int_0^{1/2}1\,dx-\int_{1/2}^{1}1\,dx\right) =2\left(\frac12-\frac12\right)=0.\end{align*}

For

n\geq 1

, the cosine coefficients are

\begin{align*}a_n&=\int_{-1}^{1} f(x)\cos(n\pi x)\,dx =2\int_0^1 f(x)\cos(n\pi x)\,dx =2\left(\int_0^{1/2}\cos(n\pi x)\,dx-\int_{1/2}^{1}\cos(n\pi x)\,dx\right)\\&=\frac{2}{n\pi}\left(\sin\left(\frac{n\pi}{2}\right)-\sin(n\pi)+\sin\left(\frac{n\pi}{2}\right)\right) =\frac{4}{n\pi}\sin\left(\frac{n\pi}{2}\right).\end{align*}

The sine coefficients are

\begin{equation*}b_n=\int_{-1}^{1} f(x)\sin(n\pi x)\,dx=-\int_{0}^{1} f(u)\sin(n\pi u)\,du +\int_{0}^{1} f(x)\sin(n\pi x)\,dx=0.\end{equation*}

Then every even coefficient is zero, while for

n=2k+1

we get

\begin{equation*}a_{2k+1}=\frac{4}{\pi}\frac{(-1)^k}{2k+1}. \quad \Longrightarrow \quad f(x) \sim \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{(-1)^k\cos((2k+1)\pi x)}{2k+1}.\end{equation*}

The heat equation for the step function

The step function

f(x)

can be interpreted as the initial temperature on a rod of length

1

, where the first half has temperature

1

and the second half has temperature

-1

. The temperature is evolved by the heat equation:

\begin{equation*}\begin{cases}\dfrac{\partial T}{\partial t}=\alpha\dfrac{\partial^2 T}{\partial x^2},& 0<x<1,\quad t>0,\\T(x,0)=f(x),& 0<x<1.\end{cases}\end{equation*}

where

\alpha>0

is the diffusivity. By representing the step function as the Fourier series

\begin{equation*}f(x) = \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{(-1)^k\cos((2k+1)\pi x)}{2k+1}.\end{equation*}

we can guess the form of the solution

T(x,t)

\begin{equation*}T(x,t)=\sum_{k=0}^{\infty}\underbrace{\frac{4}{\pi}\frac{(-1)^k}{2k+1}\cos((2k+1)\pi x)e^{-\alpha(2k+1)^2\pi^2t}}_{U_{2k+1}(x,t)}.\end{equation*}

Indeed, each summand

U_{2k+1}(x,t)

satisfies the heat equation:

\begin{align*}\frac{\partial U_{2k+1}}{\partial t}&=-\alpha(2k+1)^2\pi^2 U_{2k+1}(x,t),\\\frac{\partial^2 U_{2k+1}}{\partial x^2}&=-(2k+1)^2\pi^2 U_{2k+1}(x,t).\end{align*}

Since the heat equation is linear, the sum of

U_{2k+1}(x,t)

also satisfies the equation. Thus the Fourier-mode evolution shown in the animation is

\begin{equation*}T(x,t) = \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} \cos((2k+1)\pi x) e^{-\alpha(2k+1)^2\pi^2t}.\end{equation*}

The formula shown at the top of the animation writes out the first few terms:

\begin{equation*}T(x,t)=\frac{4}{\pi}\left(\frac{\cos(\pi x)}{1}e^{-\alpha\pi^2t}-\frac{\cos(3\pi x)}{3}e^{-9\alpha\pi^2t}+\frac{\cos(5\pi x)}{5}e^{-25\alpha\pi^2t}-\cdots\right).\end{equation*}

[Osgood2019]Osgood, Brad G. Lectures on the Fourier Transform and Its Applications. First edition. American Mathematical Society, 2019.