Thread by @AlexKontorovich

As often happens in mathematics, Fourier was trying to do something completely unrelated when he stumbled on Fourier series. What was it? (thread)

He was studying the propagation of heat in a uniform medium. To keep things as simple as possible, say you have a 1-d pipe, (cont'd)

2/ with position represented by 0<x<1, and let u(x,t) be the temperature at x at time t. He reasoned that heat at the next instant changed by taking a "local average" of nearby temperatures. For reasons I won't get into here, this local average is modeled by the Laplacian, ...

3/ d^2 u / dx^2, so the equation we have to solve is:

du/dt = d^2 u/dx^2.

(You can add a multiplicative constant, but say it's 1.) The pipe has some initial heat distribution,

u(x,0)=f(x),

and the ends of the pipe are at room temperature,

u(0,t)=u(1,t)=0.

How to solve this?

4/ Notice there are no series or cosines or anything like that, just a second order PDE. Now come the fireworks!

The standard approach is to separate variables, look for a solution like: u(x,t)=X(x)T(t). The PDE becomes:

X(x)T'(t) = X''(x)T(t),

so

T'/T = X''/X = c (constant),

5/ since one side is only a function of t and the other of x. Let's play with X first. Since u(0,t)=u(1,t)=0 for all t, it must be that X(0)=X(1)=0. Trying to solve

X''-cX=0,

we guess that X=e^(a x), so a=+/- sqrt(c). The general solution is

X(x)=A e^(sqrt(c)x)+B e^(-sqrt(c)x)

6/ where A and B are constants. Now X(0)=0 forces B=-A. But X(1)=0 means:

X(1)=A(e^(sqrt(c))-e^(-sqrt(c)))=0.

If c>0, this is impossible (if A≠0). If c=0, then X''=0, so X'=const, so X=linear, and X(0)=X(1)=0, so X=0.

But! If c<0, then there are solutions if sqrt(c)=pi i n,

7/ with n some (positive, say) integer! (How did integers sneak in here?! We're just solving PDEs...) Ok, keep going. So far we've learned that c=-pi^2 n^2, and

X(x) = A sin(pi n x),

for some constant A.

Returning to T, we easily solve T'/T=-pi^2 n^2 as

T(t)=Ce^(-pi^2 n^2 t)

8/ Combining constants, we have the solution:

u(x,t)=A sin(pi n x)e^(-pi^2 n^2 t).

But there's a solution like that for every positive integer n, so the general solution is a linear combination (superposition):

u(x,t)=sum_{n>=1} A_n sin(pi n x)e^(-pi^2 n^2 t).

This is the...

9/ first step towards Fourier series, but we're not there yet. We haven't solved for the initial condition u(x,0)=f(x). This requires:

f(x)=sum_{n>=1} A_n sin(pi n x).

How do we find the A_n's that achieve this? Fourier knows that:

int_0^1 sin(pi n x) sin(pi m x)=0, unless n=m

10/ in which case =1/2. So if he fixes some m, multiplies the above equation by sin(pi m x), integrates, and reverses orders (whoops!?..), Fourier finds:

int_0^1 f(x) sin(pi m x)dx= A_m / 2.

That's it. These 10 tweets earn him a Prize in 1811 by the Institute in Paris,

11/ questioning his generality and rigor. He exchanged infinite sums and integration (for the first time, as far as I know -- because he *had to*!). He tried to recover f(x) from a sine series. (Cauchy had "proved" that limits of continuous functions were continuous... oops!)

12/ Many controversies later, we're off to the races for genuine foundations for calculus. Euler's entire lifetime and 1000s of papers did not need the foundations. Fourier presumably didn't want them either, he just wanted to solve a PDE! But once he "broke math," people had to

13/13 get their act together. So by Lebesgue's 1902 thesis, it's completely standard to talk about issues with integrating nowhere differentiable functions.

Wish I'd started learning about "why real analysis" from this neat calculation...