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Fourier Transform

Table of Contents

1. Introduction

The Fourier Transform is a generalization of the Fourier Series. It has applications in solving differential equations and has applications in many different fields, including quantum mechanics, radio, planetary motion, and even the study of the heat equation. In this article we will study the heat equation, the Fourier Series, and the Fourier transform.

1.1. The Heat Equation

The heat equation is the study of how heat travels in a conductor with the unknown function in question being f(r→,t)f(\vec{r}, t), giving the temperature at position r→\vec{r} at time tt. Now we want to describe the time rate of change of this function, so we use a partial derivative:

∂tf=?f\begin{aligned}\label{Heat equation 1}\partial_{t}f = ?f\end{aligned}

the left hand side should intuitively be some operator on ff dealing with how heat changes in space. A single spacial derivative operator ∂xf\partial_{x}f might give you the rate of change of the temperature over space. The continuity equation for heat might look like this:

∂tf+∇→⋅j→=0\begin{aligned}\label{}\partial_{t}f + \vec{\nabla} \cdot \vec{j} = 0\end{aligned}

where j→\vec{j} is the heat current. It makes sense that the heat equation ought to satisfy some continuity equation, because heat can't just teleport without leaving some enclosed area (in order for this theory to be physical, it must respect local conservation of energy). The heat current vector field and a scalar heat field might have a similar relation to that of the potential-vector field relation in conservative force fields:

j→=−∇→u\begin{aligned}\label{}\vec{j} = -\vec{\nabla}u\end{aligned}

substituting:

∂tf=∇2u\begin{aligned}\label{}\partial_{t}f = \nabla^{2}u\end{aligned}

and heat uu is proportional to temperature ff, so we have:

∂tf=α∇2f\begin{aligned}\label{}\partial_{t}f = \alpha\nabla^{2}f\end{aligned}

The result is a PDE that we can solve analytically in one dimension.

1.1.1. Solving the Heat Equation in One Dimension

In one dimension, the heat equation looks like this:

∂tf=α∂xxf\begin{aligned}\label{}\partial_{t}f = \alpha\partial_{xx}f\end{aligned}

we assume this is a separable differential equation i.e. the solution can be written:

f(x,t)=X(x)T(t)\begin{aligned}\label{}f(x, t) = X(x)T(t)\end{aligned}

Plugging this in:

X(x)dTdt=αT(t)d2Xdx21TdTdt=α1Xd2Xdx2\begin{aligned}\label{}X(x)\frac{dT}{dt} = \alpha T(t)\frac{d^{2}X}{dx^{2}} \\\frac{1}{T}\frac{dT}{dt} = \alpha \frac{1}{X}\frac{d^{2}X}{dx^{2}}\end{aligned}

we see that each side doesn't depend on any variables from the other side, so we can separate this into two ODEs, because each side looks like a constant to the other side:

1TdTdt=α1Xd2Xdx2=−k\begin{aligned}\label{}\frac{1}{T}\frac{dT}{dt} = \alpha \frac{1}{X}\frac{d^{2}X}{dx^{2}} = -k\end{aligned}

Now we solve for T(t)T(t), by reducing this problem to being a homogeneous differential equation:

dTdt+kT=0λ+k=0λ=kT(t)=Ae−kt\begin{aligned}\label{}\frac{dT}{dt} + kT = 0 \\\lambda + k = 0 \\\lambda = k \\T(t) = Ae^{-kt}\end{aligned}

Solving for X(x)X(x) is also a homogeneous case:

d2Xdx2+kαX=0λ2+kα=0λ=±−kαX(x)=Bex−kα+Ce−x−kα\begin{aligned}\label{}\frac{d^{2}X}{dx^{2}} + \frac{k}{\alpha}X = 0 \\\lambda^{2} + \frac{k}{\alpha} = 0 \\\lambda = \pm \sqrt{-\frac{k}{\alpha}} \\X(x) = Be^{x\sqrt{-\frac{k}{\alpha}}} + Ce^{-x\sqrt{-\frac{k}{\alpha}}}\end{aligned}

re-using λ\lambda to mean something else and casting kα=λ\sqrt{\frac{k}{\alpha}} = \lambda:

X(x)=Beiλx+Ce−iλxT(t)=Ae−λ2αtf(x,t)=ABeiλx−λ2αt+ACe−iλx−λ2αtf(x,t)=A1eiλx−λ2αt+A2e−iλx−λ2αt\begin{aligned}\label{}X(x) = Be^{i\lambda x} + Ce^{-i\lambda x} \\T(t) = Ae^{-\lambda^{2}\alpha t} \\f(x, t) = ABe^{i\lambda x - \lambda^{2}\alpha t} + ACe^{-i\lambda x - \lambda^{2}\alpha t} \\f(x, t) = A_{1}e^{i\lambda x - \lambda^{2}\alpha t} + A_{2}e^{-i\lambda x - \lambda^{2}\alpha t}\end{aligned}

we know that by the superposition principle:

f(x,t)=∑i=0Nfi(x,t)\begin{aligned}\label{}f(x, t) = \sum_{i=0}^{N}f_{i}(x, t)\end{aligned}

Now, in order to proceed, we need to formulate this as an initial value problem with boundary conditions. A classic example would be a wire of dimension 1 and length LL. Thus we set a couple of boundary conditions:

2. The Fourier Transform

Taking the Fourier series and letting T→∞T \rightarrow \infty:

f(x)=∑n=−∞∞cneinxTTcn=limT→∞∫−TTf(x)einxTdxF(ω):=Tcnω:=nTF(ω)=∫−∞∞f(x)eiωxdx\begin{aligned}\label{}f(x) = \sum_{n=-\infty}^{\infty}c_{n}e^{\frac{inx}{T}} \\Tc_{n} = \lim_{T\rightarrow\infty} \int_{-T}^{T}f(x)e^{\frac{inx}{T}}dx \\F(\omega) := Tc_{n} \\\omega := \frac{n}{T} \\F(\omega) = \int_{-\infty}^{\infty}f(x)e^{i\omega x}dx\end{aligned}

F(ω)F(\omega) is the Fourier Transform of f(x)f(x). Putting different functions as arguments gives you a Fourier Transform table, and the inverse transform is same as the forward transform, scaled by a constant. This makes Fourier Transforms useful tools for solving differential equations.