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There is an approach to solve the heat equation $$ \begin{align*} \frac{\partial u}{\partial t}(x,t) &= \frac{\partial^2 u}{\partial x^2}(x,t) \quad \text{for $(x,t) \in \mathbb{R} \times (0, \infty)$} \\ u(x,0) &= f(x) \quad \text{for $x \in \mathbb{R}$} \end{align*} $$ by applying Fourier Series.

What is the motivation for that? How did Fourier see that his Fourier Series will help to resolve this equation? What makes Fourier Series to special to be applicable to this problem?

BCLC
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Qi Zhu
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  • You're working on $0 \le x < \infty$, which is not an interval where the Fourier series is useful. Fourier used a limit of the Fourier series with increasingly long periods to end up with Fourier cosine and sine transforms. His arguments were painfully contorted (and not correct,) but his answers were correct. – Disintegrating By Parts Aug 29 '18 at 10:29
  • Oh yes, I see, I was not very exact, we can let $u$ be periodic or simply look at a compact interval - I only wanted to showcase the equation but you're right, of course. Yes, his arguments were not perfectly correct but still, he thought of using Fourier Series to resolve this equation - there had to be some kind of intuition behind that. – Qi Zhu Aug 29 '18 at 10:32
  • I guess it naturally arises from the desire to find a solution $u(x,t)$ where we can separate the variables? – BCLC Aug 29 '18 at 10:44
  • You need the Fourier transform to solve this problem on a semi-infinite or infinite interval. Fourier did not use complex exponentials. He used sin's and cos's. He derived these 'transform' expressions using limits of sine and cosine series on increasingly large finite intervals. – Disintegrating By Parts Aug 29 '18 at 10:49
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    It is because the differential operator is a linear operator which you can find canonical basis for. The "simplest" canonical basis (most sparse & diagonal-looking) happens to be the basis of sin and cos which occur in the Fourier Transform. – mathreadler Aug 29 '18 at 13:28

1 Answers1

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Fourier noticed that:

(a) the heat equation is linear, so if $A(x,t)$ and $B(x,t)$ are solutions than so is $rA + sB$ for any constants $r,s$.

(b) the family of functions $\sin(kx)e^{-k^2t}$ and $\cos(kx)e^{-k^2t}$ are solutions of the heat equation.

So if we can express $u(x,0)$ as the sum of sine and cosine functions $\sin(k_ix), \cos(k_ix)$ then each of these components will decay over time in a known way and we can write down an expression for $u(x,t)$ as the sum of these decaying sine and cosine components.

The "frequencies" $k_i$ are typically multiples of a "natural" frequency which is determined by the boundary conditions - for example, if we know that $u(0,t)=u(L,t)=0 \space \forall t$ then we want to express $u(x,0)$ as the sum of functions of the form $\sin(\frac{n\pi x}{L})$.

So Fourier's challenge was to find a way to express a general function as an infinite series of sines and cosines.

gandalf61
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