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Diffstat (limited to '')
| -rw-r--r-- | notes/FourierOnS2.tex | 12 |
1 files changed, 7 insertions, 5 deletions
diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex index aeae7ef..2dc30c1 100644 --- a/notes/FourierOnS2.tex +++ b/notes/FourierOnS2.tex @@ -61,7 +61,7 @@ With this construction, now we just need a suitable set of basis function for th where \(m, n \in \mathbb{Z}\), be our basis functions in the space of ``nice'' functions from \(\mathbb{R}^2\) to \(\mathbb{C}\). Like in the one dimensional Fourier analysis, we can now define the Fourier coefficients. \begin{definition}[Fourier coefficients] - Let \(f : \mathbb{R}^2 \to \mathbb{C}\) be a ``nice'' function. The numbers + Let \(f(\mu, \nu) \in C(\mathbb{R}^2/\mathbb{Z}^2; \mathbb{C})\). The numbers \[ c_{m, n} = \langle f, B_{m, n} \rangle = \iint_{[0, 1]^2} f(\mu, \nu) e^{-i2\pi (m\mu + n\nu)} d\mu d\nu, @@ -69,12 +69,13 @@ where \(m, n \in \mathbb{Z}\), be our basis functions in the space of ``nice'' f are called the Fourier coefficients of \(f\). \end{definition} -And finally by the Fourier theorem we can reconstruct the original function using an unoriginally named Fourier series: +And finally by the Fourier theorem we can reconstruct the original function using a Fourier series: \[ f(\mu, \nu) = \sum_{m\in\mathbb{Z}} \sum_{n\in\mathbb{Z}} c_{m, n} B_{m, n}(\mu, \nu). \] +% TODO: keep or remove? <=> discuss fourier theorem? % \begin{definition}[\(L^2\) norm] % \[ % \|f\|_2 = \sqrt{\langle f, f \rangle} @@ -115,8 +116,9 @@ Notice that the partial derivatives have been simplified to normal derivatives. \] We let \(w\) be the separation constant, and we see that this results in two almost identical problems of the form \[ - \frac{d^2 X}{d\chi^2} = \pm w X(\chi). + \frac{d^2 X}{d\xi^2} = \pm w X(\xi). \] +% TODO: prove / discuss that w (=m or n) must be integers because of the periodicity of f The solutions to this elementary ODE are of course complex exponentials, the same we used to build the Fourier theory. This is not a coincidence, in fact quite the opposite: the basis functions of the Fourier decomposition were chosen such that the Laplacian operator is easy in the frequency domain. In other words, such that the expression \[ \langle \nabla^2 f, B_{m, n} \rangle @@ -124,14 +126,14 @@ The solutions to this elementary ODE are of course complex exponentials, the sam is easy to compute. This is shown in the next lemma. \begin{lemma} - Let \(f \in C(\mathbb{R}^2/\mathbb{Z}^2)\), then + Let \(f \in C(\mathbb{R}^2/\mathbb{Z}^2; \mathbb{C})\), then \[ \langle \nabla^2 f, B_{m, n} \rangle = (2\pi i)^2 \left( m^2 + n^2 \right) \langle f, B_{m, n} \rangle. \] \end{lemma} \begin{proof} -To begin this proof, we first expand the left side of the statement: +To start, we first expand the left side of the statement: \begin{gather} \nonumber \langle \nabla^2 f, B_{m, n} \rangle |