Fourier series and the heat equation

Joseph Fourier introduced the Fourier series to solve the heat equation in the 1810s. In this post, we show how the Fourier transform arises naturally in a simplified version of the heat equation.

Suppose we have the unit circle $S$ made of a metal wire. Pick an arbitrary point $A$ on the circle. Any point $P$ on the circle is identified by the distance $x$ from $A$ to $P$ along the circle in the counter-clockwise direction (i.e., $x$ is the angle of the section between $A$ and $P$ in radian). Let $u(t,x)$ represent the temperature at position $x$ and time $t$. The temperature distribution at $t=0$ is given by $u(0,x)=f(x)$. Assuming no radiation of heat out of the metal wire, $u(t,x)$ for $t>0$ and $0\le x\le 2\pi$ is determined by the following partial differential equation (PDE) called the heat equation:

$$\gamma \frac{\partial u}{\partial t}=\kappa \frac{{\partial }^{2}u}{\partial x^2}$$

and the initial condition

$$\begin{array}{}\text{(eq:finit)}& u(0,x)=f(x)\end{array}$$

where $\gamma >0$ is the heat capacity per unit length and $\kappa >0$ is the thermal conductivity via cross-section.

Naturally, the functions $u(t,x)$ and $f(x)$ have the period of $2\pi$ in $x$. This condition of periodicity (periodic boundary condition, PBC) is given either by one equation

$$u(t,x+2\pi )=u(t,x)(t\ge 0,x\in \mathbb{R}),$$

or by two equations

$$\begin{array}{rcl}u(t,2\pi )& =& u(t,0),\\ \frac{\partial u}{\partial x}(t,2\pi )& =& \frac{\partial u}{\partial x}(t,0).\end{array}$$

For simplicity, we set $\kappa =\gamma =1$. This is possible by scaling the physical units appropriately. Then the heat equation is

$$\begin{array}{}\text{(eq:heat1)}& \frac{\partial u}{\partial t}=\frac{{\partial }^{2}u}{\partial x^2}.\end{array}$$

This partial differential equation can be readily solved by using the method of separation of variables. Special solutions are given by

$$\begin{array}{}\text{(eq:fsol)}& e^{-n^{2}t}\cos (nx),e^{-n^{2}t}\sin (nx)\end{array}$$

where $n$ is a constant. Let's see that these are indeed solutions. 

$$\frac{\partial }{\partial t}\{e^{-n^{2}t}\cos (nx)\}=-n^2{e}^{-n^{2}t}\cos (nx)$$

and

$$\frac{{\partial }^2}{\partial x^2}\{e^{-n^{2}t}\cos (nx)\}=-n^2{e}^{-n^{2}t}\cos (nx).$$

Thus, we indeed have

$$\frac{\partial }{\partial t}\{e^{-n^{2}t}\cos (nx)\}=\frac{{\partial }^2}{\partial x^2}\{e^{-n^{2}t}\cos (nx)\}$$

so that $e^{-n^{2}t}\cos (nx)$ is a solution of the heat equation (eq:heat1). $e^{-n^{2}t}\sin (nx)$ is similarly shown to be a solution.

In particular, if $n$ is an integer, then these solutions satisfy the periodic boundary condition. For example,

$$e^{-n^{2}t}\cos \{n(x+2\pi )\}=e^{-n^{2}t}\cos (nx+2n\pi )=e^{-n^{2}t}\cos nx.$$

 Thus, the general solution is given by a linear combination of these functions:

$$u(t,x)=\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}{e}^{-n^{2}t}\{a_n\cos (nx)+b_n\sin (nx)\}.$$

The unknown coefficients $a_n$ and $b_n$ can be determined so that the initial condition (eq:finit) is satisfied:

$$f(x)=\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}\{a_n\cos (nx)+b_n\sin (nx)\}.$$

Thus, the problem of finding the function $u(t,x)$ is reduced to the problem of finding the Fourier expansion of $f(x)$.