Motivations of Fourier Analysis

We start with descriptions of a few physical phenomena

Simple harmonic motion describes the behavior of the most basic oscillatory system, a mass $\{m\}$ attached to a spring on a frictionless surface

An ideal spring satisfies Hooke’s law, $F=-ky(t)$
$-ky(t)=m{y}^{\prime \prime }(t)$
$y^{\prime \prime }(t)+c^{2}y(t)=0$ (with $c=\sqrt{k/m}$)

We can derive the general solution for this differential equation
$y(t)=a\cos ct+b\sin ct$

The particular solution for initial conditions is
$y(t)=y(0)\cos ct+\frac{y^{\prime }(0)}{c}\sin ct$

We can verify that $a\cos ct+b\sin ct=A\cos (ct-\varphi )$, i.e. our solution is a transformation of cosine

We conclude that an elementary oscillatory system involves $\cos t$ and $\sin t$ (and also keep in mind Euler’s identity relating these to the exponential)

We also consider the importance of the initial conditions

Standing waves are wavelike motions described by $y=u(x,t)=\varphi (x)\psi (t)$ where $\varphi (x)$ is the initial position and $\psi (t)$ scales with time $t$

Traveling waves are are wavelike motions described by $u(x,t)=F(x-ct)$ where $c$ represents velocity

Our last interesting physical phenomena is that the timbre of an instrument is made up of overtones on top of pure tones

We will later relate this to linearity

The Wave Equation

Say we stretch a string between $x=0$ and $x=L$, how does it vibrate when we release it?

We want to understand $y=u(x,t)$

View the string as a system of $N$ particles oscillating in the vertical direction, linked by their neighbors

$h=L/N$
$x_{n+1}-x_n=h$
$y_n(t)=u(x_n,t)$

We assume constant density $\rho$ and assign mass $\rho h$ to each particle
Newton’s law says the force on the $n$th particle is $\rho h{y}_n^{\prime \prime }(t)$

If we assume this force is due to the neighboring particles and proportional to $(y_{n+1}-y_n)/h$, then we get a tension from the right particle $\left(\frac{\tau }{h}\right)(y_{n+1}-y_n)$ ($\tau$ is a coefficient of tension)

The tension from the left particle is then $\left(\frac{\tau }{h}\right)(y_{n-1}-y_n)$

Together, we have $\rho h{y}_n^{\prime \prime }(t)=\frac{\tau }{h}(y_{n+1}(t)+y_{n-1}(t)-2y_n(t))$

$y_{n+1}(t)+y_{n-1}(t)-2y_n(t)=u(x_n+h,t)+u(x_n-h,t)-2u(x_n,t)$
For any reasonable function, we have $\frac{F(x+h)+F(x-h)-2F(x)}{h^2}\to F^{\prime \prime }(x)$ as $h\to 0$
So we conclude with $\rho \frac{{\partial }^{2}u}{\partial t^2}=\tau \frac{{\partial }^{2}u}{\partial x^2}$

Written similarly, $\frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}=\frac{{\partial }^{2}u}{\partial x^2}$ with $c=\sqrt{\tau /\rho }$ representing the velocity of the motion

This is the one-dimensional wave equation

We note that we can transform these functions to an interval on $[0,\pi ]$ with velocity $c=1$ without loss of generality

Solving the Wave Equation

Traveling Waves

We start with a method involving traveling waves

If $F$ is any twice differentiable function then $u(x,t)=F(x\pm t)$ solves the wave equation

This is simply a traveling wave

Using the intuition of linearity, we look at $u(x,t)=F(x+t)+G(x-t)$

We change variables $\xi =x+t,\eta =x-t$ and define $v(\xi ,\eta )=u(x,t)$
The wave equation is then $\frac{{\partial }^{2}v}{\partial \xi \partial \eta }=0$
Integrating this directly gives us our general formula, $v(\xi ,\eta )=F(\xi )+G(\eta )$

With the physical problem, we imposed $0\le x\le \pi$, $u(x,0)=f(x)$, and $u(0,t)=u(\pi ,t)=0$ for all $t$
To connect this to our equation, we extend $f$ to all of RR by making it odd on $[-\pi ,\pi ]$ and periodic in $x$ of period $2\pi$ (similarly with $u(x,t)$)

So $u(x,t)=F(x+t)+G(x-t)$
And $F(x)+G(x)=f(x)$
This is still very general and suggests we also need to specify the initial velocity $\frac{\partial u}{\partial t}(x,0)=g(x)$ with $g(0)=g(\pi )=0$

This gives us,
$\{\begin{array}{l}F(x)+G(x)=f(x)\\ F^{\prime }(x)-G^{\prime }(x)=g(x)\end{array}$

We can solve for,
$F(x)=\frac{1}{2}\left[f(x)+{\int }_0^{x}g(y)dy\right]+C_1$
$G(x)=\frac{1}{2}\left[f(x)-{\int }_0^{x}g(y)dy\right]+C_2$
where $C_1+C_2=0$

d’Alembert’s formula: $u(x,t)=\frac{1}{2}[f(x+t)+f(x-t)]+\frac{1}{2}{\int }_{x-t}^{x+t}g(y)dy$

One nice result is this exhibits the time reversal property of the wave equation, $u^{-}(x,t)=u(x,-t)$

Standing Waves

Based off our understanding of standing waves, we are interested in solutions of the form $\varphi (x)\psi (t)$

These solutions are called pure tones, and by the linearity of the wave equation, we hope to combine these pure tones into more complex combinations

Our final goal is to express the general solution in terms of sums of “pure tones”

$u(x,t)=\varphi (x)\psi (t)$
$\varphi (x){\psi }^{\prime \prime }(t)={\varphi }^{\prime \prime }(x)\psi (t)$
$\frac{{\psi }^{\prime \prime }(t)}{\psi (t)}=\frac{{\varphi }^{\prime \prime }(x)}{\varphi (x)}$

Because this inequality holds as either variable changes, they must be equal to a constant, so we can reduce this to,
${\psi }^{\prime \prime }(t)-\lambda \psi (t)=0$
${\varphi }^{\prime \prime }(x)-\lambda \varphi (x)=0$

This system only oscillates with $\lambda <0$, so we write $\lambda =-m^2$ and get,
$\psi (t)=A\cos mt+B\sin mt$
$\varphi (x)=\tilde{A}\cos mx+\tilde{B}\sin mx$

$\varphi (0)=\varphi (\pi )=0$
Without loss of generality, we can simplify to $\varphi (x)=\tilde{B}\sin mx$ where $m\ge 1$ is an integer

So,
$u_m(x,t)=(A_m\cos mt+B_m\sin mt)\sin mx$

We call $u(x,t)=\cos mt\sin mx$ the $m$th harmonic for positive integer values of $m$
The first harmonic is called the fundamental tone, and higher harmonics are called overtones
We call zeroes of the equation nodes, and points of maximum amplitude antinodes

Given the linearity of the wave equation, we guess that the final solution is,
$u(x,t)={\sum }_{m=1}^{\infty }(A_m\cos mt+B_m\sin mt)\sin mx$

If that’s the case, given an initial position of the string $f$ on $[0,\pi ]$ with $f(0)=f(\pi )=0$, we have $u(x,0)=f(x)$, so
${\sum }_{m=1}^{\infty }{A}_m\sin mx=f(x)$

Can we find solutions to this equation, expressing a general function $f$ as a sum of sin functions?

This is the problem that initiated the study of Fourier analysis

If that expansion is to hold, we can come up with a nice equation to solve for $A_m$,
${\int }_0^{\pi }f(x)\sin nxdx={\int }_0^{\pi }\left({\sum }_{m=1}^{\infty }{A}_m\sin mx\right)\sin nxdx$
$={\sum }_{m=1}^{\infty }{A}_m{\int }_0^{\pi }\sin mx\sin nxdx=A_n\cdot \frac{\pi }{2}$

This uses the fact that ${\int }_0^{\pi }\sin mx\sin nxdx=\{\begin{array}{ll}0& \text{if }m\ne n\\ \pi /2& \text{if }m=n\end{array}$

Therefore,
$A_n=\frac{2}{\pi }{\int }_0^{\pi }f(x)\sin nxdx$

We can expand this reasoning on to the interval $[-\pi ,\pi ]$ where $f$ is odd, or ask it about an even function $g(x)$ on $[-\pi ,\pi ]$ where $g(x)={\sum }_{m=0}^{\infty }{A}_m^{\prime }\cos mx$

Since an arbitrary function $F$ on $[-\pi ,\pi ]$ can be expressed as $f+g$ where $f$ is odd and $g$ is even, our general question is whether we can write $F(x)={\sum }_{m=1}^{\infty }{A}_m\sin mx+{\sum }_{m=0}^{\infty }{A}_m^{\prime }\cos mx$ or even $F(x)={\sum }_{m=-\infty }^{\infty }{a}_m{e}^{imx}$

By that last formulation, we would expect to be able to write,
$a_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }F(x)e^{-inx}dx$ which is the final form referred to as the nth Fourier coefficient of F

So given any reasonable function $F$ on $[-\pi ,\pi ]$ with Fourier coefficients defined above, is it true that $F(x)={\sum }_{m=-\infty }^{\infty }{a}_m{e}^{imx}$?

Joseph Fourier believed this was the case with conviction that his predecessors lacked

To solve the wave equation, we also would like to enforce the initial condition,
$\frac{\partial u}{\partial t}(x,0)=g(x)$

This is easily consistent with our previous equations, requiring $g(x)={\sum }_{m=1}^{\infty }m{B}_m\sin mx$

The Heat Equation

Let the temperature of a metal plate at time $t$ be denoted by $u(x,y,t)$ and consider a small square centered at $(x_0,y_0)$ with side length $h$

$H(t)=\sigma {\iint }_{S}u(x,y,t)dxdy$ is the total heat in $S$ at time $t$, where $\sigma >0$ is the specific heat of the material

$\frac{\partial H}{\partial t}=\sigma {\iint }_S\frac{\partial u}{\partial t}dxdy$ is the heat flow into $S$, which is approximately equal to $\sigma h^2\frac{\partial u}{\partial t}(x_0,y_0,t)$

Newton’s law of cooling states that heat flows from the higher to lower temperature at a rate proportional to the difference

The heat flow through the right side of our square is therefore $-\kappa h\frac{\partial u}{\partial x}(x_0+h/2,y_0,t)$, where $\kappa >0$ is the conductance

We can apply this logic to all 4 sides and apply the mean value theorem (with some manipulations) to come up with,
$\frac{\sigma }{\kappa }\frac{\partial u}{\partial t}=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}$

This is the time-dependent heat equation

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$ is called the steady-state equation
Also written as $\Delta u=0$
Solutions of this equation are called harmonic functions

Example:
Consider the unit disc $D=\{(r,\theta ):0\le r<1\}$ with boundary $C=\{(r,\theta ):r=1\}$

The Dirichlet problem is to solve the steady-state heat equation in the unit disc subject to the boundary condition $u=f$ on $C$

$\Delta u=\frac{{\partial }^{2}u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2}$
$-\frac{{\partial }^{2}u}{\partial {\theta }^2}=r^2\frac{{\partial }^{2}u}{\partial r^2}+r\frac{\partial u}{\partial r}$

We write $u(r,\theta )=F(r)G(\theta )$ to obtain,
$\frac{r^2{F}^{\prime \prime }(r)+r{F}^{\prime }(r)}{F(r)}=-\frac{G^{\prime \prime }(\theta )}{G(\theta )}$

As in the wave equation section, we find a system of equations,
$\{\begin{array}{l}{G}^{\prime \prime }(\theta )+\lambda G(\theta )=0\\ r^2{F}^{\prime \prime }(r)+r{F}^{\prime }(r)-\lambda F(r)=0\end{array}$

$\lambda \ge 0$
$\lambda =m^2$
$G(\theta )=\tilde{A}\cos m\theta +\tilde{B}\sin m\theta =A{e}^{im\theta }+B{e}^{-im\theta }$

We see $F(r)=r^m$ is the only solution which keeps $u$ bounded, so we are left with,
$u_m(r,\theta )=r^{|m|}{e}^{im\theta }$, $m\in \mathbb{Z}$

Again, $u$ is linear, so we can obtain a general solution
$u(r,\theta )={\sum }_{m=-\infty }^{\infty }{a}_m{r}^{|m|}{e}^{im\theta }$
$u(1,\theta )={\sum }_{m=-\infty }^{\infty }{a}_m{e}^{im\theta }=f(\theta )$

We arrive at the same question in this new context, given any reasonable function $f$ on $[0,2\pi ]$ with $f(0)=f(2\pi )$, can we find coefficients to satisfy $f(\theta )={\sum }_{m=-\infty }^{\infty }{a}_m{e}^{im\theta }$?