Welcome back. In this week’s blog, we will explore the Fourier transform, which is a valuable tool across engineering disciplines. Like most tools, its primary power is to make complicated problems simpler. To demonstrate this, we will first define the Fourier transform and establish some of its most important properties. We will then apply it to solve the one-dimensional heat equation (∂u/∂t = a ∂^{2}u/∂x^{2}) on the entire real number line. More details on the specifics of this problem will appear in the following paragraphs.

Let us suppose that we have a real-valued function f(x) that is defined for all real number inputs. We will call F(ω) the *Fourier transform* of f(x), where

F(ω) = 1/sqrt(2π) ∫ f(x) exp(-iωx) dx.

The integral here is understood to be from -∞ to +∞ and exp(-iωx) is shorthand for the number e (2.71828…) to the power of “-iωx.” We will write “exp(something)” in place of e to the power of “something” because it is easier to type “exp(something)” on one line. Because we are integrating with respect to x, F(ω) will have no “x’s” in its expression.

If we start with a function f(t) in the time domain and take its Fourier transform, then we will end up with a function F(ω), which is f(t) moved into the frequency domain. To illustrate this point, we can consider

f(t) = sqrt(2/π) cos(2πt)

and its Fourier transform, which is

F(ω) = δ_{1}(ω) + δ_{-1}(ω).

We can see that f(t) has a frequency of 1, meaning that it completes one cycle every unit of time. We can also think of f(t) as having a frequency of -1, because it also completes one cycle if we move backward one unit in time. Accordingly, the graph of F(ω) consists of one infinitely tall spike at ω = 1 (represented by the *Dirac delta function *δ_{1}(ω)) and a second infinitely tall spike at ω = -1 (represented by δ_{-1}(ω)). We can thus see that F(ω) achieves its highest values when its input ω corresponds with the frequencies that most closely match up with the frequencies of f(t).

There are four key properties of the Fourier transform that we will state here (without proof) that will come into play later on. Let us suppose that the Fourier transform of some function f(x) is F(ω).

First, the Fourier transform of f(x – a) is equal to exp(-iaω)F(ω). In other words, shifting the graph of f(x) to the left or right corresponds with multiplying the Fourier transform of f(x) by a complex exponential.

Second, if b is a constant, then the Fourier transform of f(bx) is equal to 1/b F(ω/b). This property is saying that if we scale the input of a function f(x) by a certain number b, then the input of its Fourier transform will be scaled by 1/b and the magnitude of its Fourier transform will be scaled by 1/b.

Third, if the fourier transform of a second function g(x) is G(ω), then the Fourier transform of 1/sqrt(2π) f(x)*g(x) is equal to F(ω)G(ω). Note here that f(x)*g(x) is not multiplication. Rather, f*g is the *convolution* of f(x) with g(x), where convolution is defined as follows, where the integral is again understood to be from -∞ to +∞:

f(x)*g(x) = ∫ f(x-y) g(y) dy.

Since we are integrating with respect to a dummy variable y, f(x)*g(x) is indeed only a function of the original variable x.

Fourth, the Fourier transform of f’’(x), the second derivative f(x), is -ω^{2} F(ω). This property is extremely powerful because it turns differential equations in x into algebraic equations in ω.

Now we have developed the framework to solve the 1D heat equation on the entire real line. In particular, we wish to solve the following problem.

Suppose we have an infinitely long rod that sits on the x-axis. At time t = 0, the temperature distribution across the rod follows the function f(x). We want to solve for the function u(x,t), which gives the temperature of any point along the infinite rod at any future point in time.

Mathematically speaking, the parameters of our problem are

(1) ∂u/∂t = a ∂^{2}u/∂x^{2} (a is some constant we already know) and

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

What we will do is take the Fourier transform on both sides of the heat equation with respect to the spatial variable x to yield

∂/∂t[U(ω,t)] = -aω^{2} U(ω,t),

where U(ω,t) is the Fourier transform of u(x,t) with respect to x.

The solution to this new differential equation (which we typically learn in a first course in ordinary differential equations) is

U(ω,t) = U(ω,0) exp(-aω^{2}t) = F(ω) exp(-aω^{2}t).

If we can find a function g(x,t) such that the fourier transform of g(x,t) with respect to x is

G(ω,t) = exp(-aω^{2}t),

then we can write the solution to our problem u(x,t) as the convolution of the following two functions with respect to x:

u(x,t) = 1/sqrt(2π) g(x,t)*f(x).

Using the second property of the Fourier transform that we went over, we can show that if

g(x,t) = 1/sqrt(2at) exp[-x^{2}/(4at)],

then G(ω,t) = exp(-aω^{2}t).

Thus, u(x,t) = 1/sqrt(4πat) ∫ f(y) exp[-(x-y)^{2}/(4at)] dy is the solution to our problem (this integral too is going from -∞ to +∞).

There are several lessons we can learn from using the Fourier transform to solve the heat equation. One is that solutions to partial differential equations are not pretty. Unfortunately, our final answer for u(x,t) is actually one of the nicest answers to a heat equation problem that I have seen. However, on the bright side, this equation would have been even more cumbersome to solve if we did not use Fourier transforms. Therefore, Fourier analysis is our friend even if it does not initially appear to be the nicest tool to work with. Lastly, solving this problem allows us to observe how Fourier transforms turn derivatives into algebra, because once we took the Fourier transform of both sides of the heat equation, we no longer had to worry about ∂^{2}u/∂x^{2}.

I will do my best to cook up something interesting to discuss next week. Until then, take care.

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