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I learned two "different" Fourier transforms in the real analysis course.

  • For any function $f\in C({\bf R/Z;C})$, and any integer $n\in{\bf Z}$, we define the $n$th Fourier coefficient of $f$, denoted $\hat{f}(n)$, by the formula $$\hat{f}(n):=\langle f,e_n\rangle=\int_{[0,1]}f(x)e^{-2\pi inx}dx.$$ Here $C({\bf R/Z;C})$ donotes the space of complex-valued continuous ${\bf Z}$-periodic functions. The function $\hat{f}:{\bf Z}\to {\bf C}$ is called the Fourier transform of $f$.
  • If $f\in L^1({\mathbb R}^n)$, its Fourier transform $\hat{f}$ is a bounded function on ${\mathbb R}^n$ defined by $$\hat{f}(\xi)=\int_{{\mathbb R}^n} e^{-2\pi ix\cdot\xi}f(x)dx.$$

I learned that this technique is usually needed in PDE.

Here are my questions (which may be vague):

How do I know which transform I should use when solving a PDE? Does it totally depends on the properties of $f$? Is there a rule of thumb?

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Well, depends on the domain of your PDE... – Jonas Teuwen Jul 19 '11 at 18:25
Informally: Periodic functions <=> Fourier series, L2 functions <=> Fourier transform… – leonbloy Jul 19 '11 at 18:52
up vote 3 down vote accepted

The reason that the Fourier transform is useful in PDEs (of constant coefficients) is that it transforms differential equations to algebraic equations via the correspondance $\partial_i \leftrightarrow i \xi_i$.

But for this correspondance to make rigorous sense, your function must admit sufficiently many derivatives. This is particular important when you try to use the former of your two transforms. More precisely, instead of considering functions on the interval $[0,1]$, the proper thing to think about is that the functions are functions on the unit circle $\mathbb{T}$, with sufficient periodicity built in.

Why is this important? Suppose your function $f$ is not periodic, so that $f(0) \neq f(1)$. Now try to compute the Fourier series of $\hat{f'}(n)$ and compare against the Fourier series $i n \hat{f}(n)$. (A nice example would be the saw-tooth wave: $f(x) = x$ on $(0,1)$. Its derivative is just the constant function.) They will be different. The difference comes in the fact that, formally, because of the discontinuity when you try to extend $f$ by periodicity, you are picking up an extra factor of a $\delta$ function supported at the point of discontinuity. When you multiply by $in$, this is factored into the expression. But when you compute the Fourier series of $f'$, that Dirac delta is not included!

Note that this is not just some "technical difficulty". Consider the heat equation $\partial_t u =\partial^2_{xx}u$ on $(0,1]$. If you consider it as an equation on the unit circle $\mathbb{T}$, the cirlce has no boundary, and no boundary conditions will be prescribed. An initial data of the saw-tooth wave will eventually converge to the limit $u_\infty(x) = \frac12$. But if you consider the equation as on the interval, then it would be necessary to prescribe boundary conditions. And if the boundary conditions are given to be $u(t,0) = 0$ and $u(t,1) = 1$, then the constant in time function $u(t,x) = x$ solves the heat equation with these boundary conditions. (And this solution emphatically does not converge to $u_\infty$.) (If you just apply the Fourier series without thinking about the actual underlying domain and its boundary conditions, the Fourier method will yield the same solution as you would get in the case of $\mathbb{T}$. )

To repeat what Jonas Teuwen said in his comment: the choice of whether you use the Fourier series or the Fourier transform is usually based on the domain of your function. If your function is defined on a periodic background $\mathbb{T}^n$ (the $n$-dimensional torus), then you should use the Fourier series. If your function is defined on $\mathbb{R}^n$, you should use the Fourier transform. (If your function is partially periodic, say $f:\mathbb{R}^{d_1}\times \mathbb{T}^{d_2} \to \mathbb{C}$, then you can take the Fourier transform in the first $d_1$ coordinate variables, and the Fourier series in the last $d_2$ coordinate variables. )

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BTW, a good book for getting the feel of how and when to apply Fourier methods is Tom Korner's Fourier Analysis. – Willie Wong Jul 19 '11 at 19:01
Thanks for your answer. Is the ${\mathbb T}$ in your second paragraph ${\mathbb T}:={\mathbb R}/(2\pi{\mathbb Z})$? – Jack Jul 19 '11 at 19:04
I guess the second sentence in the third paragraph can be more readable.("Suppose your function $f$ is not periodic, so that $f(1)\neq f(0)$.") – Jack Jul 19 '11 at 19:14
Yes. (You could equally use $\mathbb{R}/\mathbb{Z}$, you just have to put in an additional factor of $2\pi$ when you transform between derivatives and frequency multiplication.) – Willie Wong Jul 19 '11 at 19:17
For non-periodic boundary conditions on a finite interval, you often use other types of Fourier series (e.g. sine or cosine series), the more general concept unifying these being the eigenfunction expansion for a Sturm-Liouville boundary value problem. – Robert Israel Jul 20 '11 at 1:48

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