diff options
| -rw-r--r-- | notes/FourierOnS2.tex | 180 |
1 files changed, 109 insertions, 71 deletions
diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex index ced6b8d..8840f98 100644 --- a/notes/FourierOnS2.tex +++ b/notes/FourierOnS2.tex @@ -7,7 +7,7 @@ %% styling for this document \usepackage{tex/docstyle} -\usepackage[firstpageonly, color={[gray]{0.9}}]{draftwatermark} +%\usepackage[firstpageonly, color={[gray]{0.9}}]{draftwatermark} %% Theorems, TODO: move to docstyle \usepackage{amsthm} @@ -56,26 +56,26 @@ To start we first pick \(\Omega = \mathbb{R}^2 / \mathbb{Z}^2\), i.e. the set of where \(g^*\) denotes the complex conjugate of \(g\). \end{definition} -With this construction, now we just need a suitable set of basis function for the decomposition. Again, recall that in the 1D Fourier analysis the basis functions are complex exponentials. Here it is no different, we just have two dimensions instead of one. Therefore, we let +With this construction, now we just need a suitable set of basis function for the decomposition. Again, recall that in the 1D Fourier analysis the basis functions are sines and cosines. Here it is very similar, we just have two dimensions instead of one. Therefore, we let \[ - B_{m, n}(\mu, \nu) = e^{i2\pi m\mu} e^{i2\pi n\nu} = e^{i2\pi (m\mu + n\nu)}, + B_{m, n}(\mu, \nu) = \sin(\pi m \mu)\sin(\pi n \nu) \] 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 \(f(\mu, \nu) \in C(\mathbb{R}^2/\mathbb{Z}^2; \mathbb{C})\). The numbers - \[ - c_{m, n} = \langle f, B_{m, n} \rangle - = \iint_{[0, 1]^2} f(\mu, \nu) e^{-i2\pi (m\mu + n\nu)} d\mu d\nu, - \] + \begin{align*} + c_{m, n} &= \langle f, B_{m, n} \rangle \\ + &= \iint_{[0, 1]^2} f(\mu, \nu) \sin(\pi m \mu)\sin(\pi n \nu) d\mu d\nu, + \end{align*} are called the Fourier coefficients of \(f\). \end{definition} And finally by the Fourier theorem we can reconstruct the original function using a Fourier series: -\[ +\begin{equation}\label{eqn:fourier-theorem} f(\mu, \nu) = \sum_{m\in\mathbb{Z}} \sum_{n\in\mathbb{Z}} c_{m, n} B_{m, n}(\mu, \nu). -\] +\end{equation} % TODO: keep or remove? <=> discuss fourier theorem? % \begin{definition}[\(L^2\) norm] @@ -93,89 +93,127 @@ And finally by the Fourier theorem we can reconstruct the original function usin % \] % \end{theorem} -\subsection{Why complex exponentials?} +\subsection{Why sines?} -A important question now is: Why did we choose \(B_{m, n}\) to be complex exponentials? -The answer has to do with solving other problems. That is because originally Fourier developed its theory to solve some difficult problems in thermodynamics, where he wanted to solve (among many other equations) -\begin{equation} \label{eqn:laplace-cartesian} - \nabla^2 f(\mu, \nu) = \frac{\partial^2 f}{\partial \mu^2} - + \frac{\partial^2 f}{\partial \nu^2} = 0. -\end{equation} +A important question now is: Why did we choose \(B_{m, n}\) to be a multiplication of two sines? +The answer has to do with solving other problems. That is because originally Fourier developed its theory to solve some difficult problems in thermodynamics. + +The problem we want to start with is to find the eigenfunctions of the Laplace operator on the two-dimensional square domain $\Omega \in [0,1]^2$ with zero Dirichlet boundary conditions, namely +\begin{align} + \nabla^2 f &= \lambda f \label{eqn:laplace-cartesian},\\ + f(\mu, \nu) &= 0, \quad \mu, \nu \in \partial \Omega. \label{eqn:laplace-cartesian-bound} +\end{align} +What we need are orthogonal functions, so that we can expand a series as the one shown in Eq.\eqref{eqn:fourier-theorem}. If the notion of ``orthogonality'' reminds you the world of linear algebra you are on the right way. + +Eq.\eqref{eqn:laplace-cartesian} is indeed very reminiscent of the problem of finding the eigenvectors of a matrix. In fact, if you replace the Laplace operator $\nabla^2$ with a matrix $\mathbf{A}$ and the function $f$ with a vector $\vec{v}$, you get the well-known matrix equation +\begin{equation*} +\mathbf{A} \vec{v} = \lambda \vec{v}, +\end{equation*} +which, when solved, leads to the eigenvalues and associated eigenvectors. + +This intuition makes perfect sense, since as we know from the \emph{spectral decomposition theorem}, we can reconstruct any (diagonizable) matrix $\mathbf{A}$ by only knowing its eigenvalues and their respective eigenvectors. This can be performed according to the following equation: +\begin{equation*} +\mathbf{A} \vec{v} = \mathbf{Q} \mathbf{\Lambda} \mathbf{Q}^{-1}, +\end{equation*} +where the columns of $\mathbf{Q}$ are the eigenvectors and the matrix $\mathbf{\Lambda}$ consists only of zeros except for the diagonal, in which the eigenvalues are contained. -\texttt{[FIXME: Should be an eigenvalue problem \(\nabla_2^2 f = \lambda f\)]} -This PDE is known as Laplace's equation, and can be solved by separation using the ansatz +So if we reconsider our problem, once we find a set of orthogonal functions, they can be used to reconstruct any ``nice'' function, supported on the domain $\Omega$. This is analogous to the previous case, where it was the matrix $\mathbf{A}$ that could be reconstructed using a set of orthogonal vectors. + +In order to get the mentioned set, we have to solve the eigenfunction equation \eqref{eqn:laplace-cartesian} for the laplacian operator. This corresponds to solve the following PDE: \[ - f(\mu, \nu) = M(\mu)N(\nu), + \frac{\partial^2 f}{\partial \mu^2} + + \frac{\partial^2 f}{\partial \nu^2} = \lambda f. \] -which when substituted into \eqref{eqn:laplace-cartesian} yields +The following ansatz can be performed \[ - \frac{d^2 M}{d\mu^2} N(\nu) - + \frac{d^2 N}{d\nu^2} M(\mu) = 0. + f(\mu, \nu) = M(\mu)N(\nu), \] -Notice that the partial derivatives have been simplified to normal derivatives. Continuing the separation method, we divide by \(M(\mu)N(\nu)\), obtaining: +which when substituted into \eqref{eqn:laplace-cartesian} yields to \[ - \underbrace{\frac{d^2 M}{d\mu^2} \frac{1}{M(\mu)}}_{w} - + \underbrace{\frac{d^2 N}{d\nu^2} \frac{1}{N(\nu)}}_{-w} = 0. + \frac{d^2 M}{d\mu^2} N(\nu) + + \frac{d^2 N}{d\nu^2} M(\mu) = \lambda M(\mu)N(\nu). \] -We let \(w\) be the separation constant, and we see that this results in two almost identical problems of the form +Notice that the partial derivatives have been simplified to normal derivatives. Continuing the separation method, we divide by $M(\mu)N(\nu)$ \texttt{TODO: ASSUMPTION $f \neq 0$ (?)}, obtaining: \[ - \frac{d^2 X}{d\xi^2} = \pm w X(\xi). + \frac{d^2 M}{d\mu^2} \frac{1}{M(\mu)} + + \frac{d^2 N}{d\nu^2} \frac{1}{N(\nu)} = \lambda. \] +We now let +\begin{align} +\alpha^2 &:= \frac{d^2 N}{d\nu^2} \frac{1}{N(\nu)} -\lambda \quad \text{and} \label{eqn:def1-simple-ode}\\ +\beta^2 &:= \frac{d^2 M}{d\mu^2} \frac{1}{M(\mu)} -\lambda. \label{eqn:def2-simple-ode} +\end{align} +We see that this results in two almost identical problems of the form +\begin{align*} +\frac{d^2 M}{d\mu^2} = -\alpha^2 M(\mu), \\ +\frac{d^2 N}{d\nu^2} = -\beta^2 N(\nu) . +\end{align*} % 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 +The solutions to these elementary ODEa are a linear combination of sines and cosines, which can be written as +\begin{align*} +M(\mu) &= A_1 \sin(\alpha \mu) + A_2 \cos(\alpha \mu) \quad \text{and} \\ +N(\nu) &= B_1 \sin(\beta \nu) + B_2 \cos(\beta \nu). +\end{align*} +The unknown parameters of $M(\mu)$ can be estimated using the Dirichlet boundary conditions as follows: +\begin{align*} +M(0) \stackrel{!}{=} 0 \quad &\Rightarrow A_2 = 0, \\ +N(1) \stackrel{!}{=} 0 \quad &\Rightarrow \alpha = n\pi, \quad n \in \mathbb{N}. +\end{align*} +The same method can be used to compute the coefficients for $N(\nu)$, which are +\begin{align*} +B_2 &= 0, \\ +\beta &= m\pi, \quad m \in \mathbb{N}. +\end{align*} +This means that we have just found an infinite number of eigenfunctions in the form +\begin{align*} +f_{m,n}(\mu,\nu)&= M_m(\mu)N_n(\nu) \\ + &= c_{m,n} \sin(\pi m \mu)\sin(\pi n \nu). +\end{align*} +The whole solution can be built up by a linear combination of these functions +\begin{equation*} +f(\mu,\nu) = \sum_{m=0}^\infty \sum_{n=0}^\infty c_{m,n} \sin(\pi m \mu)\sin(\pi n \nu) +\end{equation*} +of which, however, the multiplication coefficient $c_{m,n}$ remains to be found. + +This problem will not be adressed because it is beyond the scope of this chapter. An intuition to solve it is to use other conditions for the function $f(\mu,\nu)$, in order to turn it into a ``Fourier-style'' problem, i.e. to find the coefficients of some basis functions $B_{m,n}(\mu,\nu)$ (in this case $B_{m,n}(\mu,\nu)=\sin(\pi m \mu)\sin(\pi n \nu)$), in order to reconstruct $f(\mu,\nu)$. + +We can observe that the eigenfunctions which solve Eq.\eqref{eqn:laplace-cartesian} are 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 \[ \langle \nabla^2 f, B_{m, n} \rangle \] is easy to compute. This is shown in the next lemma. \begin{lemma} - Let \(f \in C(\mathbb{R}^2/\mathbb{Z}^2; \mathbb{C})\), then - \[ - \langle \nabla^2 f, B_{m, n} \rangle - = (2\pi i)^2 \left( m^2 + n^2 \right) \langle f, B_{m, n} \rangle. - \] + Let the eigenfunction $f(\mu,\nu) \in C(\mathbb{R}^2/\mathbb{Z}^2; \mathbb{C})$, then + \begin{equation}\label{eqn:lemma1-eigenfunction} + \langle \nabla^2 f, B_{m,n} \rangle = -\pi^2 \left( m^2 + n^2 \right) \langle f, B_{m,n} \rangle + \end{equation} \end{lemma} \begin{proof} - To start, we first expand the left side of the statement: - \begin{gather} - \nonumber - \langle \nabla^2 f, B_{m, n} \rangle - = \iint_{[0, 1]^2} \nabla^2 f B_{m,n} d\mu d\nu \\ - = \iint_{[0, 1]^2} \left( - \frac{\partial^2 f}{\partial \mu^2} - + \frac{\partial^2 f}{\partial \nu^2} - \right) e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu. - \label{eqn:laplacian-coeffs-expanded} - \end{gather} - Since the integrand is a sum of partial derivatives, we now have 2 integrals. Notice that inside each integral we have an expressions of the form: - \begin{equation} \label{eqn:inner-by-parts} - \int_{[0, 1]} \frac{\partial^2 f}{\partial \xi^2} e^{-i2\pi x\xi} d\xi, - \end{equation} - once with \(x = m, \xi = \mu\) and the second time with \(x = n, \xi = \nu\). - The integral \eqref{eqn:inner-by-parts} can be integrated by parts twice resulting in this ugly expression: - \[ - e^{-i2\pi x\xi} \left( - \frac{\partial f}{\partial \xi} - i2\pi f - \right)\Bigg|_0^1 - + (i2\pi x)^2 \int_{[0, 1]} f e^{-i2\pi x\xi} d\xi. - \] - However, actually this is not too bad. That is because once we substitute the bounds two things happen: the exponential in the front always equals 1 and what is inside of the parenthesis can be rewritten as - \[ - \frac{\partial f}{\partial\xi}(1) - \frac{\partial f}{\partial\xi}(0) - + i2\pi \left[ f(1) - f(0) \right], - \] - which equals zero, since \(f\) and its derivative are continuous and periodic. Hence, we are left with two integrals, that when substituted back into \eqref{eqn:laplacian-coeffs-expanded} give: + We know from Eq.\eqref{eqn:laplace-cartesian} that + \begin{equation*} + \nabla^2 f = \lambda f. + \end{equation*} + This implies + \begin{equation*} + \langle \nabla^2 f, B_{m,n} \rangle = \langle \lambda f, B_{m,n} \rangle. + \end{equation*} + Given the linearity of the inner product we can now write \begin{align*} - \langle \nabla^2 & f, B_{m, n} \rangle = \\ - &(i2\pi m)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu \\ - &\quad + (i2\pi n)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu. + \langle \lambda f, B_{m,n} \rangle &= \iint_{[0, 1]^2} \lambda f B_{m,n}^* d\mu d\nu \\ + &= \lambda \iint_{[0, 1]^2} f B_{m,n}^* d\mu d\nu \\ + &= \lambda \langle f, B_{m,n} \rangle. \end{align*} - Finally, to complete the proof we rewrite the right side using the compact notation: - \[ - (i2\pi m)^2 \langle f, B_{m, n} \rangle - + (i2\pi n)^2 \langle f, B_{m, n} \rangle. - \qedhere - \] + Proving Eq.\eqref{eqn:lemma1-eigenfunction} is therefore equivalent to proving the following relationship: + \begin{equation*} + \lambda = -\pi^2 \left( m^2 + n^2 \right). + \end{equation*} + From the two definitions in Eqs.\eqref{eqn:def1-simple-ode}\eqref{eqn:def2-simple-ode}, one can write + \begin{align*} + \lambda &= -\alpha^2 -\beta^2 = -\left(\alpha^2 + \beta^2\right) \\ + &= -\pi^2 \left(m^2 + n^2\right) + \end{align*} + \end{proof} % TODO: closing words: this is why fourier is useful |