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A motivational problem for the study of Fourier series

March 23, 2012

In this post, I will start describing the Dirichlet problem for the Laplacian on the disk and show how this motivates the study of Fourier series. Then, in the next 3 posts, I will develop the necessary tools to completely solve the problem and post the solution 4 posts from now.

First things first, the heat equation is a partial differential equation used to model the heat distribution in some region over time. Explicitly, imagine \textbf{R}^2 as an infinite metal plate. If \Omega \subset \textbf{R}^2 is some region of that plate and u(x,y,t) denotes the temperature of (x,y) \in \Omega at time t, then the heat equation tells us that

{\alpha\frac {\partial u}{\partial t} =\frac {\partial ^2u}{\partial x^2} + \frac {\partial ^2u}{\partial y^2}}

where \alpha is a constant dependent on the material of the plate. (See Stein’s book or google for how to derive this equation). If we let t \rightarrow \infty, the distribution of heat on \Omega will reach an equilibrium so that no more heat exchange occurs, i.e. \frac {\partial u}{\partial t} \rightarrow 0 as t \rightarrow \infty. The heat equation then reduces to the so-called steady-state heat equation

{\bigtriangleup u = 0}

where \bigtriangleup :=\frac {\partial ^2}{\partial x^2} + \frac {\partial ^2}{\partial y^2} is the Laplace operator or the Laplacian. The solutions to this equation are referred to as the harmonic functions.

Now, let D be the unit disk and C the unit circle. The Dirichlet problem for the Laplacian on the unit disk is the problem of finding a function u defined on D, satisfying the steady-state heat equation, such that the restriction to C of u be a given function f. In other words, what we do is we fix the temperature distribution on the unit circle (so that u(x,y,t) = f(x,y) for all (x,y) \in C at all time t), then we wait long enough for the heat distribution to reach equilibrium, then we look inside the disk to see how the heat is distributed.

Since we’re working on a disk, let’s switch to polar coordinates so that the boundary condition is given by u(1,\theta) = f(\theta). Rewriting the Laplacian in polar coordinates, we get

{\bigtriangleup u = \frac {\partial ^2u}{\partial r^2}+\frac {1}{r}\frac {\partial u}{\partial r}+\frac {1}{r^2}\frac {\partial ^2 u}{\partial \theta ^2}}

(I’ll post that derivation as exercise 1 later). Since \bigtriangleup u = 0, we can write

{r^2 \frac {\partial ^2u}{\partial r^2} + r \frac {\partial u}{\partial r} = - \frac {\partial ^2u}{\partial \theta ^2}}.

A common technique used for solving differential equations is the so-called separation of variables: we look for a solution of the form u(r, \theta ) = F(r)G(\theta ). Substituting in our last equation, we get

{\frac {r^2 F''(r) + rF'(r)}{F(r)} = - \frac {G''(\theta)}{G(\theta)}}.

But say we have g(x) = h(y) \quad \forall x,y for two functions g, h of independent variables. Then fixing y tells us that g is constant while fixing x tells us h is constant. This is what happens in our last equation, so both sides must be equal to some constant \lambda. So what we get is two equations :

{G''(\theta) + \lambda G(\theta) = 0}, and

{r^2F''(r) + rF'(r) - \lambda F(r) = 0}.

Since G depends on \theta, we certainly want it to be 2\pi-periodic. But for G to be periodic, we must have \lambda \geq 0. Indeed, with an argument analogous to what is used in the solution of exercice 2, we can see that if \lambda is negative, then the equation for G implies G(\theta) = A\cosh(\theta) + B\sinh(\theta) for some A,B \in \textbf{R} which is not a periodic function.

Let \lambda = m^2 with m \in \mathbf{N}. In exercice 2, we’ll see that we can now deduce that we have

{G(\theta) = Ae^{im \theta} + Be^{-im \theta}}

for some A, B \in \mathbf{C}. Furthermore, if m = 0, we’ll see in exercice 3 that the only possibility for F is to be a linear combination of 1 and \log r but if F(r) = c \log r, then F(r) \rightarrow \infty as r \rightarrow 0 which is counter-intuitive since then the heat distribution would be unbounded at the origin. We’ll also see that if m \neq 0, then F must be given by a linear combination of r^n and r^{-n} and again, if F(r) = cr^{-n}, u would be unbounded at the origin.

We are thus led to consider the solutions u(r, \theta) = a_mr^{|m|}e^{im \theta} for some m \in \mathbf{Z} and since the steady-state heat equation is linear, we obtain, by superposing these solutions,

{\sum_{m = - \infty}^{\infty}a_mr^{|m|}e^{im \theta} }.

The boundary condition u(1, \theta) = f(\theta) now gives us f(\theta) = \sum_{m = - \infty}^{\infty} a_me^{im \theta}. ie f is given by a Fourier series.

Let’s review what we’ve done here. We fixed a heat distribution on the circle with the aid of a periodic function f(\theta). We then waited long enough for the heat distribution inside the circle to stabilize, then looked at how that heat naturally distributed itself. This process somehow gave us a representation of our function f as an infinite sum of exponentials. This is serious indication that any reasonable periodic function can be expanded in Fourier series.

It is by studying problems in heat diffusion that Fourier gave rise to the theory of Fourier Analysis. Others have since then made very important discoveries in numerous areas of analysis in order to prove to what extent exactly can an arbitrary function be represented as a Fourier series. Basically all our technology which uses signal processing is based on Fourier Analysis so it is definitely an interesting thing to know about.

In the next posts, I will develop the necessary tools to satisfyingly solve the Dirichlet problem mentioned here. The solution is in this post. But before that, I will post the solution to the exercises that I mentioned in this post.


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