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\newpage
\mychapter{Inhomogeneous ODEs; Fourier Transforms}

\newpage
\section{The Dirac Delta Function}

\subsection{Definition of $\delta(x)$}

Define a generalised function $\delta(x - \xi)$
with the following properties:
\begin{equation*}
    \boxed{
    \begin{aligned}
        \delta(x - \xi) = 0, \quad \forall x \neq
        \xi \\
        \int_{-\infty}^\infty \delta(x - \xi) \dd
        x = 1
    \end{aligned}
    }
    \tag{6.1}
\end{equation*}
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9021edf669ca11ed.png}
\end{center}
This acts as a linear operator $\int \dd x
\delta(x - \xi)$ on an arbitrary function $f(x)$
to produce a number $f(\xi)$, that is,
\[ \boxed{\int_{-\infty}^\infty \dd x \delta(x -
\xi) f(x) = f(\xi)} \tag{6.2} \]
provided $f(x)$ is `well-behaved' at $x = \xi$ and
$\pm \infty$.

\subsubsection*{Notes}

\begin{itemize}
    \item The delta function $\delta(x)$ is
        classified as a distribution (not a
        function). See lecture notes of Jozsa and
        Skinner section 6.1 (optional).
    \item $\delta(x)$ always appears in an
        integrand as a linear operator where it is
        well-defined.
    \item Represents a unit point source (for
        example mass, charge) or an impulse.
\end{itemize}

\subsubsection*{Some limiting approximations}

Discrete:
\[ \lim_{n \to \infty} \delta_n(x) = \begin{cases}
    0 & x > \frac{1}{n} \\
    \frac{n}{2} & |x| \le \frac{1}{n} \\
    0 & x < -\frac{1}{n}
\end{cases} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/5813f61069cb11ed.png}
\end{center}
\myskip
Continuous:
\[ \lim_{\eps \to 0} \delta_\eps(x) =
\frac{1}{\eps \sqrt{\pi}} e^{-x^2/\eps^2}
\tag{6.3} \]
verify (6.2):
\begin{align*}
    \int_{-\infty}^\infty f(x) \delta(x) \dd x
    &= \lim_{\eps \to 0} \int_{-\infty}^\infty
    \frac{1}{\eps\sqrt{\pi}} e^{-x^2/\eps^2} f(x)
    \dd x \\
    &= \lim_{\eps \to 0} \int_{-\infty}^\infty
    \frac{1}{\sqrt{\pi}} e^{-y^2} f(\eps y) \dd y
    \\
    &= \lim_{\eps \to 0} \int_{-\infty}^\infty \dd
    y \frac{1}{\sqrt{\pi}} (f(0) + \eps y f'(0) +
    \cdots ) \\
    &= f(0)
\end{align*}
$\forall f$. `well-behaved' at $x = 0$ so that we
can take the Taylor expansion, and also need well
behaved at $\pm \infty$ so that it doesn't grow
faster than $1/e^{-x^2/\eps^2}$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d9d3216269cb11ed.png}
\end{center}
Further examples: ($\lim n\to \infty$)
\[ \delta_n(x) = \frac{\sin nx}{\pi x} =
\frac{1}{2\pi} \int_{-n}^n e^{ikx} \dd k \tag{6.4} \]
\[ \delta_n(x) = \frac{n}{2} \sech^2 nx \tag{6.5} \]

\subsection{Properties of $\delta(x)$}

\subsubsection*{Heaviside function $H(x)$}

The unit step function,
\[ H(x) = \begin{cases}
    1 & x \ge 0 \\
    0 & x < 0
\end{cases} \tag{6.6} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/2cf15a3069cc11ed.png}
\end{center}
is the \emph{integral of $\delta(x)$}.
\[  H(x) = \int_{-\infty}^x \delta(x) \dd x
\tag{6.7} \]
and we can identify $H'(x) = \delta(x)$.

\begin{example*}
    Verify using (6.5) $\delta(x) = \lim_{n \to
    \infty} \frac{n}{2} \sech^2 nx$. (You will
    find $\half (\tanh nx + 1)$ is the approximate
    step function. - also $H(0) = \half$
    (alternate definition).)
\end{example*}

\subsubsection*{Derivative of $\delta(x)$}

Define $\delta'(x)$ using integration by parts:
\begin{align*}
    \int_{-\infty}^\infty \delta'(x - \xi) f(x)
    \dd x
    &= [\delta(x - \xi) f(x)]_{-\infty}^\infty -
    \int_{-\infty}^\infty \delta(x - \xi) f'(x)
    \dd x \\
    &= -f'(\xi) \tag{6.8}
\end{align*}
for all $f(x)$ smooth at $x = \xi$.

\begin{example*}
    Consider Gaussian approximation (6.3)
    \[ \delta_\eps'(x) = -\frac{2x}{\eps^3
    \sqrt{\pi}}e^{-x^2/\eps^2} \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/d5b3d26069cc11ed.png}
    \end{center}
\end{example*}

\subsubsection*{Sampling property}

\[ \int_a^b f(x) \delta(x - \xi) \dd x = \begin{cases}
    f(\xi) & a < \xi < b \\
    0 & \text{otherwise}
\end{cases} \]

\subsubsection*{Even property}

\[ \int_{-\infty}^\infty f(x) \delta(-(x - \xi))
\dd x = \int_{-\infty}^\infty f(x) \delta(x - \xi)
\dd x \tag{6.10} \]
\begin{align*}
    LHS
    &= \int_\infty^{-\infty} f(\xi - u) \delta(u)
    (-\dd u) \\
    &= \int_{-\infty}^\infty f(\xi - u) \delta(u)
    \dd u \\
    &= f(\xi) \\
    &= RHS
\end{align*}

\subsubsection*{Scaling property}

\[ \int_{-\infty}^\infty f(x) \delta(a(x - \xi))
\dd x = \frac{1}{|a|} f(\xi) \tag{6.11} \]
Exercise: Show this using $u = ax$ (noting
integral limit order with $a < 0$).
\begin{hiddenflashcard}[delta-fn-scaling-property]
For $a \neq 0$,
\[ \delta(ax) = \cloze{\frac{1}{|a|} \delta(x)} \]
\end{hiddenflashcard}

\subsubsection*{Advanced scaling}

Suppose $g(x)$ has $n$ isolated zeros at $x_1,
x_2, \dots, x_n$ then (with $g'(x_i) \neq 0$):
\[ \delta(g(x)) = \sum_{i = 1}^n \frac{\delta(x -
x_i)}{|g'(x_i)|} \tag{6.12} \]
(Exercise: Show for $g$ has 1 root at $x = x_i$).

\begin{example*}
    \[ I = \int_{-\infty}^\infty f(x) \delta(x^2 -
    1) \dd x \]
    $x^2 - 1$ has roots $x = \pm 1$ with $g'(x) =
    2x$. So
    \begin{align*}
        I
        &= \int_{1 - \eps}^{1 + \eps} f(x)
        \frac{\delta(x - 1)}{|2x|} \dd x +
        \int_{-1 - \eps}^{-1 + \eps} f(x)
        \frac{\delta(x + 1)}{|2x|} \dd x \\
        &= \half (f(1) + f(-1))
    \end{align*}
\end{example*}

\subsubsection*{Isolation property}

If $g(x)$ is continuous at $x = 0$ then
\[ g(x) \delta(x) = g(0) \delta(x) \tag{6.13} \]
Exercise: evaluate and show
\[ \int_0^\infty \delta'(x^2 - 1) x^2 \dd x =
-\frac{1}{4} \]
using $u = x^2 - 1$ and note (6.8) and (6.12).
\begin{hiddenflashcard}[delta-fn-isolation-property]
\prompt{Isolation property of $\delta$ function?}
\[ \cloze{g(x)\delta(x) = g(0)\delta(x)} \]
\cloze{if $g(x)$ is \fcemph{continuous} at $x =
0$.}
\end{hiddenflashcard}
