% Homework for Physical Mathematics I
% Ewan Stewart
% KAIST
% 2011/9/29

\documentclass[a4paper,12pt]{article}

\usepackage{pm1}

\setcounter{homework}{8}

\begin{document}

\begin{homework}

\question{parity}
Show that
\begin{enumerate}
\item
\begin{equation} \label{21}
\fn{\delta}{-x} = \fn{\delta}{x}
\end{equation}
\item
\begin{equation} \label{22}
\fn{\delta'}{-x} = - \fn{\delta'}{x}
\end{equation}
\item
\begin{equation} \label{23}
x \fn{\delta}{x} = 0
\end{equation}
\item
\begin{equation} \label{24}
x \fn{\delta'}{x} = - \fn{\delta}{x}
\end{equation}
\end{enumerate}

\answer{parity}

\begin{enumerate}

\item
Using Eq.~(3.3.18),
\begin{equation}
\int_{-\infty}^\infty \d{y} \fn{\delta}{-x+y} \fn{f}{y}
= \int_{-\infty}^\infty \d{y} \fn{\delta}{-x-y} \fn{f}{-y}
= \fn{f}{x}
= \int_{-\infty}^\infty \d{y} \fn{\delta}{x-y} \fn{f}{y}
\end{equation}
hence \eq{21}.

\item
Taking the derivative of \eq{21} gives \eq{22}.

\item
\begin{equation}
\int_{-\infty}^\infty \d{x} x \fn{\delta}{x} \fn{f}{x}
= 0 \times \fn{f}{0}
= 0
\end{equation}
hence \eq{23}.

\item
Taking the derivative of \eq{23} gives \eq{24}.

\end{enumerate}

\question{derivative}
Show that
\begin{enumerate}
\item
\begin{equation} \label{11}
\fn{\delta}{x} = \fn{\theta'}{x}
\end{equation}
where
\begin{equation}
\fn{\theta}{x} = \left\{
\begin{array}{ccc}
0 & \textrm{for} & x < 0
\\
1 & \textrm{for} & x > 0
\end{array}
\right.
\end{equation}
\item
\begin{equation}
\int_{-\infty}^\infty \d{y} \fn{\delta'}{x-y} \fn{f}{y} = \fn{f'}{x}
\end{equation}
\end{enumerate}

\answer{derivative}

\begin{enumerate}

\item
Integrating by parts
\begin{eqnarray}
\int_{-\infty}^\infty \d{x} \fn{\theta'}{x} \fn{f}{x}
& = & \left[ \fn{\theta}{x} \fn{f}{x} \right]_{-\infty}^\infty - \int_{-\infty}^\infty \d{x} \fn{\theta}{x} \fn{f'}{x}
\\
& = & \fn{f}{\infty} - \int_0^\infty \d{x} \fn{f'}{x}
\\
& = & \fn{f}{0}
\end{eqnarray}
hence \eq{11}.

\item
Integrating by parts
\begin{eqnarray}
\int_{-\infty}^\infty \d{y} \fn{\delta'}{x-y} \fn{f}{y}
& = & \left[ - \fn{\delta}{x-y} \fn{f}{y} \right]_{-\infty}^\infty + \int_{-\infty}^\infty \d{y} \fn{\delta}{x-y} \fn{f'}{y}
\nonumber \\ \\
& = & \fn{f'}{x}
\end{eqnarray}

\end{enumerate}

\question{diagrams}
Use PGF to draw diagrams illustrating $\fn{\delta}{x}$ and $\fn{\delta'}{x}$.

\answer{diagrams}
\begin{tikzpicture}[x=0.1\textwidth,y=0.1\textwidth,baseline=(delta.base)]
% axes
\draw[->] (-2.2,0) -- (2.2,0) node[right]{$x$};
\draw[->] (0,-2.2) -- (0,2.2) node[above](delta){$\fn{\delta}{x}$};
\path (0,0) node[below left]{$0$};
% delta
% Note that one has to be careful about numerical errors, hence the exp(4) factor and limited domain.
\draw[thick,blue!25,domain=-2:2,samples=200] plot (\x,{0.5*exp(-4)*exp(4-0.5*6.25*(\x)^2)});
\draw[thick,blue!50,domain=-1:1,samples=200] plot (\x,{exp(-4)*exp(4-0.5*25*(\x)^2)});
\draw[thick,blue,domain=-0.5:0.5,samples=200] plot (\x,{2*exp(-4)*exp(4-0.5*100*(\x)^2)});
\end{tikzpicture}
\begin{tikzpicture}[x=0.1\textwidth,y=0.1\textwidth,baseline=(deltaprime.base)]
% axes
\draw[->] (-2.2,0) -- (2.2,0) node[right]{$x$};
\draw[->] (0,-2.2) -- (0,2.2) node[above](deltaprime){$\color{blue} \fn{\delta}{x}$ , $\color{red} \fn{\delta'}{x}$};
\path (0,0) node[below left]{$0$};
% delta
\draw[thick,blue!50,domain=-0.9:0.9,samples=200] plot (\x,{0.1*exp(-3.5)*exp(4-0.5*25*(\x)^2)});
\draw[thick,blue,domain=-0.45:0.45,samples=200] plot (\x,{2*0.1*exp(-3.5)*exp(4-0.5*100*(\x)^2)});
% \delta'
\draw[thick,red!50,domain=-0.9:0.9,samples=200] plot (\x,{-1*2.5*\x*exp(-3.5)*exp(4-0.5*25*(\x)^2)});
\draw[thick,red,domain=-0.45:0.45,samples=200] plot (\x,{-2*10*\x*exp(-3.5)*exp(4-0.5*100*(\x)^2)});
\end{tikzpicture}

\question{fourier}
Show that
\begin{enumerate}
\item
\begin{equation}
\fn{\delta}{x} = \lim_{\epsilon \to 0^+} \frac{1}{2\pi} \int_{-\infty}^\infty \d{k} e^{ikx - \epsilon k^2}
\end{equation}
\item
\begin{equation}
\fn{\delta}{x} = \lim_{\Lambda \to \infty} \frac{1}{2\pi} \int_{-\Lambda}^\Lambda \d{k} e^{ikx}
\end{equation}
\end{enumerate}

\answer{fourier}

\begin{enumerate}

\item
Define
\begin{eqnarray}
\fn{\delta_\epsilon}{x} & \equiv & \frac{1}{2\pi} \int_{-\infty}^\infty \d{k} e^{ikx - \epsilon k^2}
\\
& = & \frac{1}{2\pi} e^{-\frac{x^2}{4\epsilon}} \int_{-\infty}^\infty \d{k} e^{- \epsilon \left( k - \frac{ix}{2\epsilon} \right)^2}
\\
& = & \frac{1}{2\pi} e^{-\frac{x^2}{4\epsilon}} \int_{-\infty}^\infty \d{k} e^{- \epsilon k^2}
\\
& = & \frac{1}{\sqrt{4\pi\epsilon}\,} e^{-\frac{x^2}{4\epsilon}}
\end{eqnarray}
Therefore
\begin{equation}
\lim_{\epsilon \to 0^+} \fn{\delta_\epsilon}{x} = \left\{
\begin{array}{ccc}
\infty & \textrm{if} & x = 0 \\
0 & \textrm{if} & x \neq 0
\end{array}
\right.
\end{equation}
and $\fn{\delta_\epsilon}{x}$ has a simple spike around $x=0$ with
\begin{equation}
\int_{-\infty}^\infty \d{x} \fn{\delta_\epsilon}{x} = 1
\end{equation}
Therefore
\begin{equation}
\lim_{\epsilon \to 0^+} \fn{\delta_\epsilon}{x} = \fn{\delta}{x}
\end{equation}

\item
Define
\begin{eqnarray}
\fn{\delta_\Lambda}{x} & \equiv & \frac{1}{2\pi} \int_{-\Lambda}^\Lambda \d{k} e^{ikx}
\\
& = & \frac{\sin(\Lambda x)}{\pi x}
\end{eqnarray}
Therefore
\begin{eqnarray}
\int_{-\infty}^\infty \d{x} \fn{\delta_\Lambda}{x} \fn{f}{x}
& = & \int_{-\infty}^\infty \d{x} \frac{\sin(\Lambda x)}{\pi x} \fn{f}{x}
\\
& = & \int_{-\infty}^\infty \d{x} \frac{\sin x}{\pi x} \fn{f}{\frac{x}{\Lambda}}
\end{eqnarray}
Therefore, using the results of Homework~4 Question~1,
\begin{eqnarray}
\lim_{\Lambda \to \infty} \int_{-\infty}^\infty \d{x} \fn{\delta_\Lambda}{x} \fn{f}{x}
& = & \int_{-\infty}^\infty \d{x} \frac{\sin x}{\pi x} \fn{f}{0}
\\
& = & \fn{f}{0}
\end{eqnarray}
Therefore
\begin{equation}
\lim_{\Lambda \to \infty} \fn{\delta_\Lambda}{x} = \fn{\delta}{x}
\end{equation}

\end{enumerate}

\end{homework}

\end{document}

