% vim: tw=50
% 21/11/2022 10AM

\subsubsection*{Discrete Fourier Transform}

Suppose we have a finite number $N$ of samples
\[ h_m = h(t_m), \quad t_m = m\Delta, \quad m = 0, 1,
\dots, N - 1 \tag{8.45} \]
We want to approximate the Fourier Transform for
$N$ frequencies using equally spaced frequencies
$\Delta_f = \frac{1}{N\Delta}$ in the range $-f_c
\le f \le f_c$. We \emph{could} take $f_n =
n\Delta_f = \frac{n}{n\Delta}$ with $n =
-\frac{N}{2}, -\frac{N}{2} + 1, \dots, -1, 0, 1,
\dots, \frac{N}{2}$. \emph{But} this has $N + 1$
frequencies, with $f_c$ and $-f_c$ aliased (8.39).
Instead, note that $\left(\frac{N}{2} + m \right)\Delta_f =
f_c + \delta f$ is aliased back to $\left(
\frac{N}{2} - m \right) \Delta_f = -(f_c - \delta
f)$ from (8.40) so we choose
\[ f_n = \frac{n}{N\Delta} \text{ with } n = 0, 1,
2, \dots, \frac{N}{2} - 1, \frac{N}{2},
\frac{N}{2} + 1, \dots, N - 1 \]
The \emph{discrete Fourier Transform} at frequency
$f_n$ becomes (8.46)
\begin{align*}
    \tilde{h}(f_n)
    &= \int_{-\infty}^\infty
    e^{-2\pi i f_n t} \dd t \\
    &\simeq \Delta \sum_{m = 0}^{N - 1} h_m
    e^{-2\pi i f_n t_m} \\
    &= \Delta \sum_{m = 0}^{N - 1} h_m e^{-2\pi i
    mn/N} \\
    &= \Delta \tilde{h}_d(f_n) \tag{8.47}
\end{align*}
\begin{hiddenflashcard}[discrete-FT]
\prompt{Discrete FT?}
\cloze{
\begin{align*}
    \tilde{h}_n = \tilde{h}_d(f_n)
    &= \sum_{m = 0}^{N - 1} h_m e^{-2\pi i
    mn / N} \\
    h_m
    &= h(t_m) \\
    t_m
    &= m\Delta \\
    f_n
    &= \frac{n}{N\Delta}
\end{align*}
}
\end{hiddenflashcard}
Recalling section 8.2 Fourier series $\to$ Fourier
transform Riemann integral. Here $\tilde{h}_d(f_n)
\equiv \tilde{h}_n$ is the discrete Fourier
Transform. So the matrix $[DFT]_{mn} = e^{-2\pi i
mn/n}$ defines the discrete Fourier Transform for
$\bf{h} = \{h_m\}$ (data vector) as
$\tilde{\bf{h}}_d = [DFT] \bf{h}$. \\
The \emph{inverse} is its adjoint $[DFT]^{-1} =
\frac{1}{N} [DFT]^\dag$ and it's built from roots
of unity $\omega = e^{-2\pi i/N}$. For example $N
= 4$, $\omega = -i$,
\[ DFT =
\begin{pmatrix}
    1 & 1 & 1 & 1 \\
    1 & -i & -1 & i \\
    1 & -1 & 1 & -1 \\
    1 & i & -1 & -i
\end{pmatrix}
\]
The \emph{inverse DFT} is
\begin{align*}
    h_m
    &= h(t_m) \\
    &= \frac{1}{2\pi} \int_{-\infty}^\infty
    \tilde{h}(\omega) e^{i\omega t_m} \dd \omega
    \\
    &= \int_{-\infty}^\infty \tilde{h}(f) e^{2\pi
    i f t_m} \dd f \\
    &\simeq \frac{1}{N\Delta} \sum_{m = 0}^{N - 1}
    \Delta \tilde{h}_d(f_n) e^{2\pi imn/N} \\
    &= \frac{1}{N} \sum_{n = 0}^{N - 1}
    \tilde{h}_n e^{2\pi imn/N} \tag{8.48}
\end{align*}
\begin{hiddenflashcard}[inverse-dft]
\prompt{Inverse DFT?}
\[ h_m = \cloze{\frac{1}{N}
\sum_{n = 0}^{N - 1} \tilde{h}_n e^{2\pi imn/N}} \]
\end{hiddenflashcard}
or interpolating Fourier series is $h(t) =
\frac{1}{N} \sum_{n = 0}^{N - 1} \tilde{h}_n
e^{2\pi i nt/N}$. \myskip
Exercise: Establish Parseval's theorem
\[ \sum_{m = 1}^{N - 1} |h_m|^2 = \frac{1}{N}
\sum_{m = 0}^{N - 1} |\tilde{h}_m|^2 \tag{8.49} \]
The convolution theorem for $g_m, h_m$ is
\[ c_k = \sum_{m = 0}^{N - 1} g_m h_{k - m} \iff
\tilde{c}_k = \tilde{g}_k \tilde{h}_k \tag{8.50} \]

\newpage
\mychapter{PDEs on Unbounded Domains}

\newpage
\section{Characteristics}

\subsection{Well-posed Cauchy Problems}

Solving PDEs depends on the nature of the
equations in combination with the boundary and /
or initial data. A \emph{Cauchy problem} is the
PDE for $\phi$ together with this auxillary data
(i.e. $\phi$ and its derivatives) specified on a
surface (or curve in 2D), which is called
\emph{Cauchy data}. \\
A Cauchy problem is \emph{well-posed} if:
\begin{enumerate}[(i)]
    \item a solution exists
    \item the solution is unique
        item the solution depends continuously on
        auxillary data.
\end{enumerate}

\subsection{Method of Characteristics}

Consider a parametrised curve $C$ given by $(x(S),
y(s))$ with tangent vector $\bf{v} = \left(
\dfrac{x}{s}(s), \dfrac{y}{s}(s) \right)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9acfbde6698911ed.png}
\end{center}
For a function $\phi(x, y)$ we can define a
directional derivative along $C$
\[ \left. \dfrac{\phi}{s} \right|_C =
\dfrac{x(s)}{s} \pfrac{\phi}{x} +
\dfrac{y(s)}{s} \pfrac{\phi}{y} = \bf{v} \cdot
\nabla \phi |_C \tag{9.1} \]
If $\bf{v} \cdot \nabla \phi = 0$, then
$\dfrac{\phi}{s} = 0$ and $\phi = \text{constant}$
along $C$. \\
Now suppose we have a vector field
\[ \bf{u} = (\alpha(x, y), \beta(x, y)) \tag{9.2} \]
with its family of integral curves $C$
non-intersecting and filling $\RR^2$ (i.e. at a
point $(x, y)$ the integral curve has tangent
$\bf{u}$).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/2330daee698a11ed.png}
\end{center}
Define a curve $B$ by $((x(t), y(t))$ transverse
to $\bf{u}$, such that its tangent $\bf{\omega} =
\left( \dfrac{x(t)}{t}, \dfrac{y(t)}{t} \right)$
is nowhere parallel to $\bf{u}$. Label each
integral curve $C$ of $\bf{u}$ using $t$ at the
intersection point with $\bf{B}$, then use $s$
to parametrise along the curve (i.e. take $s = 0$
at $B$). \\
Our integral curves $(x(s, t), y(s, t))$ satisfy:
\[ \dfrac{x}{s} = \alpha(x, y), \qquad
\dfrac{y}{s} = \beta(x, y) \tag{9.3} \]
Solve these to find a family of characteristic
curves along which $t$ remains constant (i.e. new
coordinates $(s, t)$).

\subsection{Characteristics of a 1st order PDE}

Consider 1st order linear PDE
\[ \alpha(x, y) \pfrac{\phi}{x} + \beta(x, y)
\pfrac{\phi}{y} = 0 \tag{9.4} \]
with specified Cauchy data on an initial curve $B$
$(x(t), y(t))$:
\[ \phi(x(t), y(t)) = f(t) \tag{9.5} \]
