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\subsubsection*{Eigenfunction expansion of $G(x,
\xi)$}

Suppose $\mathcal{L}$ is in Sturm Liouville form
(2.7) with eigenfunctions $y_n(x)$ and eigenvalues
$\lambda_n$, then seek
\[ G(x, \xi) = \sum_{n = 1}^\infty A_n y_n(x) \]
satisfying $\mathcal{L} G = \delta(x - \xi)$.
\begin{align*}
    \mathcal{L} G
    &= \sum_n A_n \mathcal{L} y_n(x) \\
    &= \sum_n A_n \lambda_n \omega(x) y_n(x) \\
    &= \delta(x - \xi) \\
    &= \omega(x) \sum_n \frac{y_n(\xi)
    y_n(x)}{\mathcal{N}_n}
\end{align*}
with $\mathcal{N}_n = \int \omega y_n^2 \dd x$. So
$A_n(\xi) = \frac{y_n(\xi)}{\lambda_n
\mathcal{N}_n}$ by orthogonality (2.13). Thus
\[ \boxed{G(x, \xi) = \sum_{n = 1}^\infty
\frac{y_n(\xi) y_n(x)}{\lambda_n \mathcal{N}_n} =
\sum_{n = 1}^\infty \frac{Y_n(\xi)
Y_n(x)}{\lambda_n}} \tag{7.23} \]
which obtained without $\delta(x - \xi)$ in
(2.31): refer to section 2.6 Sturm Liouville
theory.

\subsection{Constructing $G(t, \tau)$: Initial
Value Problem}

Solve $\mathcal{L} y = f(t)$ for $t \ge a$ with
$y(a) = y'(a) = 0$ using $G(t, c)$ satisfying
$\mathcal{L} G = \delta(t - \tau)$ with same
boundary conditions.
\begin{itemize}
    \item For $t < \tau$, $G_1 = A y_1(t) + B
        y_2(t)$ with $W(a) \neq a$, $y_1, y_2$
        independent.
        \[ A y_1(a) + B y_2(a) = 0\]
        and
        \[ A y_1'(a) + B y_2'(a) = 0 \]
        implies
        \[ y_1 y_2' - y_2 y_1' = 0 \]
        unless $A = B = 0$. So $G_1(t, \tau)
        \equiv 0$, $a \le t < \tau$, i.e. no
        change until impulse at $t = \tau$.
    \item For $t > \tau$, by $G$ continuity
        (7.12), $G_2(\tau, \tau) = 0$ so choose
        $G_2 = D y_+(t)$, $y_+(t) = A y_1(t) + B
        y_2(t)$ such that $y_+(\tau) = 0$.
\end{itemize}
But by discontinuity in $G'$ (7.13):
\[ [G']_{\tau_-}^{\tau_+} = G_2'(\tau, \tau) -
G_1'(\tau, \tau) = D y_+'(\tau) =
\frac{1}{\alpha(\tau)} \]
i.e. $A y_1'(\tau) + B y_2'(\tau) =
\frac{1}{\alpha(\tau)}$ so
\[ D(\tau) = \frac{1}{\alpha(\tau) y_+'(\tau)} \]
or solve for $A, B$. Hence, we have
\[ \boxed{G(t, \tau) = \begin{cases}
    0 & t < c \\
    \frac{y_+(t)}{\alpha(\tau) y_+'(\tau)} & t >
    \tau
\end{cases}} \tag{7.25} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/137c992a61ac11ed.png}
\end{center}
The initial value problem is
\begin{align*}
    y(t)
    &= \int_a^t G(t, \tau) f(\tau) \dd \tau \\
    &= \int_a^t \frac{y_+(t) f(\tau)}{\alpha(\tau)
    y_+'(\tau)} \dd \tau
\end{align*}
Causality is ``built in'', as only forces acting
prior to $t$ affect the solution at $t$.

\begin{example*}
    Solve
    \[ y'' - y = f(t) \]
    with $y(0) = y'(0) = 0$.
    \begin{enumerate}
        \item[1\&2] Homogeneous solutions and
            initial conditions
            \begin{itemize}
                \item $t < \tau$, $G_1 = 0$,
                \item $t > \tau$, $G_2 = A e^t + B
                    e^{-t}$
            \end{itemize}
        \item[3] Continuity implies $G_2(\tau,
            \tau) = 0 \implies G_2 = D \sinh(t -
            \tau)$
        \item $[G'] = \frac{1}{\alpha} = 1
            \implies G_2'(\tau, \tau) = D \cosh(0)
            = D = 1$. 
    \end{enumerate}
    Hence, solution (7.26) is
    \[ y(t) = \int_0^t f(\tau) \sinh(t - \tau) \dd
    t \]
\end{example*}

\newpage
\section{Fourier Transforms}

\subsection{Introduction}

\begin{flashcard}[FourierTransform]
\begin{definition*}
    The \emph{Fourier transform} (FT) of a
    function $f(x)$ is
    \begin{align*}
        \tilde{f}(k)
        &= \mathcal{F}(f)(k) \\
        &= \cloze{\int_{-\infty}^\infty f(x) e^{-ikx} \dd
        x} \tag{8.1}
    \end{align*}
    \fcscrap{and the \emph{inverse Fourier transform} is
    \begin{align*}
        f(x)
        &= \mathcal{F}^{-1}(\tilde{f})(x) \\
        &= \frac{1}{2\pi} \int_{-\infty}^\infty
        \tilde{f}(k) e^{ikx} \dd k \tag{8.2}
    \end{align*}
    Beware there are several conventions.}
\end{definition*}
\end{flashcard}

\begin{hiddenflashcard}[InvFourierTransform]
    \begin{definition*}
        The \emph{inverse Fourier transform} is
        \begin{align*}
            f(x)
            &= \mathcal{F}^{-1}(\tilde{f})(x) \\
            &= \cloze{\frac{1}{2\pi}
            \int_{-\infty}^\infty \tilde{f}(k)
            e^{ikx} \dd k}
        \end{align*}
    \end{definition*}
\end{hiddenflashcard}

\noindent
The \emph{Fourier inversion theorem} states that
\[ \mathcal{F}^{-1}(\mathcal{F}(f))(x) = f(x)
\tag{8.3} \]
with a sufficient condition that $f$ and
$\tilde{f}$ are absolutely integrable. That is,
\[ \int_{-\infty}^\infty |f(x)| \dd x = M < \infty \]
so $f \to 0$ as $x \to \pm \infty$.

\subsubsection*{Gaussian example}

Find the Fourier transform of
\[ f(x) = \frac{1}{\sigma \sqrt{\pi}}
e^{-x^2/\sigma^2} \tag{8.4} \]
\begin{align*}
    \tilde{f}(k)
    &= \frac{1}{\sigma \sqrt{\pi}}
    \int_{-\infty}^\infty e^{-x^2/\sigma^2}
    e^{-ikx} \dd x \\
    &= \frac{1}{\sigma \sqrt{\pi}}
    \int_{-\infty}^\infty e^{-x^2/\sigma^2} \cos
    kx \dd x
\end{align*}
Consider $\dfrac{\tilde{f}}{k}$:
\begin{align*}
    \tilde{f}'(k)
    &= -\frac{1}{\sigma\sqrt{\pi}}\int_{-\infty}^\infty x
    e^{-x^2/\sigma^2} \sin kx \dd x \\
    &= \frac{1}{\sigma \sqrt{\pi}} \left[
    \frac{\sigma^2}{2} e^{-x^2/\sigma^2} \sin kx
    \right]_{-\infty}^\infty - \frac{1}{\sigma
    \sqrt{\pi}} \int_{-\infty}^\infty \left(
    \frac{k\sigma^2}{2} \right) e^{-x^2/\sigma^2}
    \cos kx \dd x \\
    &= -\frac{k\sigma^2}{2} \tilde{f}(k)
\end{align*}
Integrate $\frac{\tilde{f}'}{\tilde{f}} =
-\frac{k\sigma^2}{2}$ to find
\[ \tilde{f}(k) = C e^{-k^2 \sigma^2/4} \]
But put $k = 0$ into (8.4), $\tilde{f}(0) = 1
\implies C = 1$
\[ \boxed{\tilde{f}(k) = e^{-k^2 \sigma^2/4}}
\tag{8.5} \]
Exercise: Show that $\mathcal{F}^{-1}(e^{-k^2
\sigma^2/4}) = f(x)$.

\subsubsection*{Exponential exercise:}

Show that $f(x) = e^{-a|x|}$, $a > 0$ has Fourier
Transform
\[ \tilde{f}(k) = \frac{2a}{a^2 + k^2} \tag{8.6} \]
in two ways:
\begin{enumerate}[(i)]
    \item Integrate $2\int_0^\infty e^{-ax} \cos
        kx \dd x$ by parts twice.
    \item Integrate $\int_0^\infty e^{-(a - ik)x}
        \dd x + \int_{-\infty}^0 e^{(a + ik)x} \dd
        x$ directly.
\end{enumerate}
Note if
\[ f(x) = \begin{cases}
    e^{-ax} & x > 0 \\
    0 & x \le 0
\end{cases} \]
($a > 0$) then
\[ \boxed{\tilde{f}(k) = \frac{1}{ik + a}}
\tag{8.6a} \]
\begin{hiddenflashcard}[inverse-FT-of-1-over-ik-plus-a]
If $\tilde{f}(x) = \frac{1}{ik + a}$ then
\[ f(x) = \cloze{
\begin{cases}
    e^{-ax} & x > 0 \\
    0 & x \le 0
\end{cases}} \]
\end{hiddenflashcard}

\subsection{Fourier Transform relation to Fourier
series}

We can write Fourier series (1.13) as
\[ f(x) = \sum_{n = -\infty}^\infty c_n e^{ik_n x}
\tag{$*$} \]
where $k_n = \frac{n\pi}{L}$, so write $k_n =
n\Delta k$ with $\Delta k = \frac{\pi}{L}$, then
\begin{align*}
    c_n
    &= \frac{1}{2L} \int_{-L}^L f(x) e^{-ik_n x}
    \dd x \\
    &= \frac{\Delta k}{2\pi} \int_{-L}^L f(x)
    e^{-ik_n x} \dd x
\end{align*}
