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\subsection{Eigenfunction expansions of
$\delta(x)$}

\subsubsection*{Fourier series (complex)}

For $-1 \le x < L$, represent
\[ \delta(x) = \sum_{n = -\infty}^\infty c_n
e^{-n\pi x/L} \]
Fourier series coefficient (1.15):
\[ c_n = \frac{1}{2L} \int_{-L}^L \delta(x)
e^{-n\pi x/L} \dd x = \frac{1}{2L} \]
so
\[ \delta(x) = \frac{1}{2L} \sum_{n =
-\infty}^\infty e^{in\pi x/L} \tag{6.14} \]
Take $f(x) = \sum_{n = -\infty}^\infty d_n
e^{in\pi x/L}$, then (using section 2.2)
\begin{align*}
    \int_{-L}^L f^*(x) \delta(x) \dd x
    &= \frac{1}{2L} \sum_n d_n \int_{-L}^L
    e^{-in\pi x/L} e^{in\pi x/L} \dd x \\
    &= \sum_n d_n \\
    &= f(0)
\end{align*}
The \emph{Diract count} comes from extending
periodically to all $\RR$:
\[ \sum_{m = 0\infty}^\infty \delta(x - 2mL) =
\frac{1}{2L} \sum_{n = -\infty}^\infty e^{in\pi
x/L} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/cc083d845e8511ed.png}
\end{center}

\subsubsection*{General eigenfunctions}

Suppose $\delta(x - \xi) = \sum_{n = 1}^\infty a_n
y_n(x)$, $a \le x \le b$ with coefficients (2.17):
\begin{align*}
    a_n
    &= \frac{\int_a^b \omega(x) y_n(x) \delta(x -
    \xi) \dd x}{\int_a^b \omega y_n^2 \dd x} \\
    &= \frac{\omega(\xi)y_n(\xi)}{\int_a^b \omega
    y_n^2 \dd x} \\
    &= \omega(\xi) Y_n(\xi)
\end{align*}
for unit norm $Y_n$ (2.18). Then
\begin{align*}
    \delta(x - \xi)
    &= \omega(\xi) \sum_{n = 1}^\infty Y_n(\xi)
    Y_n(x) \\
    &= \omega(x) \sum_{n = 1}^\infty Y_n(\xi)
    Y_n(x)
\end{align*}
since
\[ \frac{\omega(x)}{\omega(\xi)} \delta(x - \xi) =
\delta(x - \xi) \]
by (6.13). Hence
\begin{flashcard}[eigenfunction-expansion-of-delta]
\prompt{What is the eigenfunction expansion of
$\delta$ (SL form)?}
\[ \boxed{\delta(x - \xi) = \cloze{\omega(x) \sum_{n =
1}^\infty \frac{y_n(\xi) y_n(x)}{\mathcal{N}_n}}}
\tag{6.15} \]
\end{flashcard}
where $\mathcal{N}_n = \int_a^b \omega y_n^a \dd
x$.

\begin{example*}
    Consider Fourier series $y(0) = y(1) = 0$ with
    $y_n(x) = \sin n\pi x$. Here, from (1.11) we
    have
    \[ \delta(x - \xi) = 2\sum_{n = 1}^\infty \sin
    n\pi \xi \sin n\pi x \tag{$*$} \]
    Exercise:
    \begin{enumerate}[(i)]
        \item Integrate both sides to show
            \[ \sum_{m = 1}^\infty \frac{(-1)^{m +
            }}{2m - 1} = \frac{1}{4} \]
            when $\xi = \half$.
        \item Integrate twice and compare with
            $G(x, \xi)$ (1.25) or (2.31).
    \end{enumerate}
\end{example*}

\newpage
\section{Green's Function}

\subsection{Physical motivation: Static forces
on a string}

Consider a \emph{massive} static string (tension
$T$, density $\mu$) with fixed ends
\[ y(0) = y(1) = 0 \tag{7.1} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/20322c845e8711ed.png}
\end{center}
By resolving forces, we have (3.3)
\[ T \pfrac[2]{y}{x} - \mu g = 0 \]
(time independent). So solve inhomogeneous ODE
subject to (7.1) with $f(x) = -\frac{\mu g}{T}$.
\[ -\dfrac[2]{y}{x} = f(x) \tag{7.2} \]
Solution 1: Direct integration for uniform mass
density ODE (7.2) implies:
\[ -y = -\frac{\mu g}{2T} x^2 + k_1 x + k_2 \]
Boundary conditions (7.1) implies
\[ y(x) = \left( -\frac{\mu g}{T} \right) \half x
(1 - x) \tag{7.3} \]
Solution 2: Superposition of point masses on light
string $\tilde{\mu} \to 0$. Consider point mass
$\delta m$ ($= \mu \delta x$) suspended at $x =
\xi$:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/1bfe06f05e8811ed.png}
\end{center}
Resolve in $y$ dimension to find $y_i(\xi_i)$:
\begin{align*}
    0
    &= T(\sin \theta_1 + \sin\theta_2) - \delta mg
    \\
    &= T \left( \left( -\frac{y_i}{\xi_i} \right)
    + \left( \frac{-y_i}{1 - \xi_i} \right)
    \right) - \delta mg \\
    \implies -T(y_i(1 - \xi_i) + y_i \xi_i)
    &= \delta mg \xi_i (1 - \xi_i)
\end{align*}
so
\[ y_i(\xi_i) = \left( -\frac{\delta mg}{T}
\right) \xi_i(1 - \xi_i) \]
Hence solution
\[ y_i(x) = \left( -\frac{\delta mg}{T} \right)
\begin{cases}
    x(1 - \xi_i) & x < \xi_i \\
    \xi_i (1 - x) & x < \xi_i
\end{cases} = f_i G(x, \xi) \]
where $f_i$ is the source (here, $(-\frac{\delta
mg}{T}$) and $G(x, \xi)$ is the solution for unit
point mass (Green's function). Now sum $N$ point
masses $\delta m$ at $x = \{\xi_i\}$ by linearity
\[ y(x) = \sum_{i = 1}^N f_i G(x, \xi_i) \]
or in \emph{continuum limit} with
\[ f_i = -\frac{\delta mg}{T} = -\frac{\mu \delta
xg}{T} \equiv f(x) \dd x \]
$f(x) = -\frac{\mu g}{T}$. We have ($x \to \xi$):
\begin{align*}
    y(x)
    &= \int_0^1 f(\xi) G(x, \xi) \dd \xi \tag{7.5} \\
    &= \left( -\frac{\mu g}{T} \right) \left[
    \int_0^x \xi(1 - x) \dd \xi  + \int_x^1 x(1 -
    \xi) \dd \xi \right] \\
    &= \left( -\frac{\mu g}{T} \right) \left(
    \left[ \frac{\xi^2}{2} (1 - x) \right]_0^x +
    \left[ x \left( \xi - \frac{\xi^2}{2} \right)
    \right]_x^1 \right) \\
    &= \left( -\frac{\mu g}{T} \right) \left(
    \frac{x^2}{2}(1 - x) - 0 + \frac{x}{2} -
    x \left(x -
    \frac{x^2}{2} \right) \right) \\
    &= \left( \frac{-\mu g}{T} \right) \half x
    (1 - x)
\end{align*}
so it matches (7.3).

\subsection{Definition of Green's function}

We wish to solve inhomogeneous ODE (section 2.1)
on $a \le x \le b$.
\[ \mathcal{L} y \equiv \alpha(x) y'' + \beta(x)
y' + \gamma(x) y = f(x) \tag{7.6} \]
$f(x)$ is a source. With $\alpha \neq 0$, $\beta,
\gamma$ continuous and bounded. Homogeneous
boundary conditions $y(a) = y(b) = 0$. The Green's
function for the operator $\mathcal{L}$ is the
solution for a unit point source (or impulse) at
$x = \xi$.
\[ \mathcal{L} G(x, \xi) = \delta(x - \xi)
\tag{7.7} \]
which satisfies $G(a, \xi) = G(b, \xi) = 0$ (or
similar). By linearity, we construct solutions by
integrating over source $f(x)$ with $G$:
\[ y(x) = \int_a^b G(x, \xi) f(\xi) \dd \xi
\tag{7.8} \]
Formally verify this:
\[ \mathcal{L} y = \int \mathcal{L}_(x) G(x, \xi)
f(\xi) \dd \xi = \int \delta(x - \xi) f(\xi) \dd
\xi = f(x) \]
so the solution (7.8) is given by the inverse
operator $\mathcal{L}^{-1} = \int \dd x G(x,
\xi)$.

\subsubsection*{Defining properties (summary)}

The Green's function splits into two parts:
\[ G(x, \xi) = \begin{cases}
    G_1(x, \xi) & a \le x < \xi \\
    G_2(x, \xi) & \xi < x \le b
\end{cases} \tag{7.9} \]
such that:
\begin{enumerate}[(1)]
    \item Homogeneous solutions: $G$ solves
        homogeneous equation for all $x \neq \xi$.
        So
        \[ \mathcal{L} G_1 = 0, \quad \mathcal{L}
        G_2 = 0 \tag{7.10} \]
    \item Homogeneous boundary conditions: $G$
        satisfies homogeneous boundary conditions
        so
        \[ G_1(a, \xi) = 0, \quad G_2(b, \xi) = 0
        \tag{2.11} \]
    \item Continuity condition: $G$ is continuous
        at $x = \xi$ so
        \[ G_1(\xi, \xi) = G_2(\xi, \xi) \]
    \item Jump condition: Derivative discontinuous
        at $x = \xi$ with
        \[ [G']_{\xi_-}^{\xi_+} = \left.
        \dfrac{G_2}{x} \right|_{x = \xi_+} -
        \left. \dfrac{G_1}{x} \right|_{x = \xi_-}
        = \frac{1}{\alpha(\xi)} \tag{7.13} \]
        where $\alpha(x)$ is defined in (7.6).
\end{enumerate}