\documentclass{beamer} %beamer setup \usetheme[height=8mm]{Rochester} \usefonttheme[onlylarge]{structurebold} \setbeamerfont*{frametitle}{size=\normalsize,series=\bfseries} \setbeamertemplate{navigation symbols}{} %-----------packages-------------------------------------------- \usepackage{amstext} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsthm} \usepackage{amsmath} \usepackage{xmpmulti} \usepackage{movie15} \usepackage{hyperref} \usepackage[english]{babel} \usepackage[latin1]{inputenc} \usepackage{times} \usepackage[T1]{fontenc} %----------------------------------------------- \newcommand{\half}{{\frac{1}{2}}} \newcommand{\Mu}{{\cal M}} \renewcommand{\O}{{\cal O}} \renewcommand{\o}{{o}} \newcommand{\grad}{{\nabla}} \renewcommand{\div}{{\nabla \cdot}} \newcommand{\eps}{{\epsilon}} \def\L{{\cal{L}}} \def\n{\hat{n}} \def\l{\ell} \def\bra{\langle} \def\ket{\rangle} \def\t{{\bf t}} \def\that{ \hat{\bf t}} \def\x{{\bf x}} \def\v{{\bf v}} \def\y{{\bf y}} \def\g{{\bf g}} \def\A{{\bf A}} \def\vareps{\varepsilon} \def\T{{\bf T}} \def\tensor{{\otimes}} \def\R{{\mathbb{R}}} \def\D{ {\cal D}} \begin{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{} \begin{Large} \begin{centering} Distributions and distributional derivatives \end{centering} \end{Large} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Effect of a short pulse} Consider the ordinary differential equation $$ x'(t) + x(t) = f(t), \quad x(0) =0. $$ \pause Solve using integrating factor $\exp(t)$: $$ x(t) = \int_0^t \exp(s -t) f(s) ds. $$ \pause Let $f(t)$ be a pulse of duration $\eps \ll 1$ at time $t_0$. For $t > t_0 + \eps$, $$ x(t) = \int_{t_0}^{t_0+\eps} \exp(s -t) f(s) ds \approx \exp(t_0 -t) \int_{t_0}^{t_0+\eps} f(s) ds = I \exp(t_0 -t), $$ where $I$ is the the integral of $f(t)$. \pause \vspace{.1 in} Can we define a function with unit integral, supported at just a point $t_0$? \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Distributions} Motivations: \begin{enumerate} \item Ordinary functions are not enough to describe every physical situation %figure here \item We want to include non-smooth solutions of differential equations \end{enumerate} \vspace{.1 in} \pause Idea: create new class of function-like objects called {\it distributions} by defining how they ``act" on smooth functions. \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{The delta function} \begin{small} The most basic distribution (or ``generalized function") is the so-called $\delta$-function. \pause \vspace{-0.1 in} \begin{columns} \column{.6\textwidth} Approximate for $\eps>0$ \begin{equation*} \label{aprpoxdelta} \delta_{\eps}(x) = \frac{1}{\eps \sqrt{\pi}} e^{-x^2/\eps^2}. \end{equation*} \column{.4\textwidth} \includegraphics[scale=0.3]{../ARTWORK/delta.png} \end{columns} \vspace{-0.05 in} \pause For any continuous and bounded $f(x)$, $$ \int_{-\infty}^{\infty} f(x) \delta_{\eps} (x-x_0) dx \approx f(x_0) \int_{-\infty}^{\infty} \delta_{\eps} (x-x_0) dx = f(x_0), \quad \eps \to 0. $$ \pause Define the object $\delta$ to act as \begin{equation*} \label{deltadef} \int_{-\infty}^{\infty} f(x) \delta(x-x_0) dx = f(x_0). \end{equation*} The $\delta$-function ``picks out" the value of the function $f(x)$ at $x_0$. \pause \vspace{0.05 in} Also define in higher dimensions: $$ \int_{\mathbb{R}^n} f(\x) \delta(\x-\x_0) d\x = f(\x_0), \quad \x,\x_0 \in \mathbb{R}^n. $$ \end{small} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Duality and linear functionals} The operation \vspace{-.1 in} \begin{equation*} \label{deltadef} \int_{-\infty}^{\infty} f(x) \delta(x-x_0) dx = f(x_0). \end{equation*} can be regarded as a mapping from a continuous function $f(x)$ to a real number. \vspace{.1 in} \pause Definition: a {\it linear functional} is a linear mapping from a vector space to the real numbers. \vspace{.1 in} \pause Examples: \\ (1) If $g(x) \in C_1(\mathbb{R})$, $g(x) \to g'(0)$ is a linear functional. \\ \pause(2) If $q(x)$ is bounded, integrable,\\ $f(x) \to \int_{-\infty}^{\infty} q(x) f(x) \,dx$ is a linear functional. \vspace{.1 in} \pause Should $q(x)$ be regarded as a function, or a {\it functional}? \\ \pause \alert{BOTH!} We call this situation {\it duality}. \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Distributions in general} \begin{small} \vspace{.03 in} \alert{Idea:} create new set of objects not as functions, but as continuous linear functionals. \vspace{.03 in} \pause Precise definition: A {\it distribution} is a continuous linear functional on the set of infinitely differentiable functions with bounded support (Notated $C^{\infty}_0$ or simply $\D$). Write $$ d [\phi]: \D \to \mathbb{R} $$ \vspace{-.15 in} \pause Some facts: \begin{itemize} \item A continuous function $g(x)$ can be regarded as a distribution by setting $ g[\phi] = \int _{-\infty}^{\infty} g(x) \phi(x) dx. $ \pause \item Distributions can be approximated by usual functions: there exists a sequence $d_n(x) \in \D$ so that $$ d [\phi] = \lim_{n \to \infty} \int d_n(x) \phi(x) dx, \quad \text{for all $\phi \in \D$}. $$ \pause \item \vspace{-.15 in} Distributions have integrals: $$ \int_{-\infty}^{\infty} d(x) \phi(x) dx \equiv d [\phi], \quad \text{for any $\phi \in \D$}. $$ \end{itemize} \end{small} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Some examples} \begin{small} Ex. \#1: Linear combinations of delta functions are distributions: If $d = 3 \delta(x -1) + 2 \delta(x)$ then corresponding linear functional is $$ d [\phi] = \int _{-\infty}^{\infty} d(x) \phi(x) dx = 3 \phi(1) + 2 \phi(0). $$ \vspace{.05 in} \pause Ex. \#2: $ d [\phi] = \int_0^{\infty} x \phi'(x) dx. $ (check that it's linear!) \pause Can it be written in the form $\int_{-\infty}^{\infty} g(x) \phi(x) dx$? \pause Integrate by parts: $$ d [\phi] = x \phi(x)\Big|_0^{\infty} - \int_0^{\infty} \phi(x) dx = \int_{-\infty}^{\infty} \Big(-H(x) \Big) \phi(x) dx, $$ where the ``step function" is \begin{equation*} H(x) = \begin{cases} 0 & x \le 0 \\ 1 & x > 0. \end{cases} \end{equation*} \pause Abuse of notation: $d = -H$. \end{small} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Distributions as derivatives} \begin{small} Can a non-differentiable function or distribution be differentiated? \pause We define the {\it distributional derivative} of $d[\phi]$ to be the distribution $d'[\phi]$ so that $$ d'[\phi] \equiv - d[\phi']. $$ \pause Idea behind this: if $d[\phi] = \int_{-\infty}^{\infty} g(x) \phi(x) dx$ for smooth $g(x)$, integration by parts gives $$ g'[\phi] = \int_{-\infty}^{\infty} g'(x) \phi(x) dx \equiv - \int_{-\infty}^{\infty} g(x) \phi'(x) dx = -g[\phi'], \quad \text{for all $\phi \in \D$}. $$ \vspace{.05 in} \pause Ex. \#1: Compute derivative of step function $H(x)$. \pause The distributional derivative $H'[\phi]$ of $H$ is the rule \vspace{-.05 in} $$ H'[\phi] = - \int_{-\infty}^{\infty} H(x) \phi'(x) dx = - \int_0^{\infty} \phi'(x) dx= \phi(0) = \int_{-\infty}^{\infty} \delta(x) \phi(x) dx. $$ Thus $H' = \delta$. \vspace{.05 in} \pause Ex. \#2: What is the rule implied by the derivative of the $\delta$-function? \pause By definition, $\delta'[\phi] = - \delta[\phi'] = - \phi'(0)$. \end{small} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Distributions as derivatives: more examples} \begin{small} The $n$-th derivative of a distribution $d$ is defined to be $$ (-1)^n d[\phi^{(n)}]. $$ (this follows from integration by parts n-times). \vspace{.05 in} \pause Example: Find rule representing second derivative of $f(x) =|x|$. \pause By definition, $$ f''[\phi] = (-1)^2 f[\phi''] = \int_{-\infty}^{\infty} |x| \phi''(x) dx. $$ \pause Now integrate by parts and integrate again: \begin{equation*} \begin{split} f''[\phi] =& -\int_{-\infty}^0 x \phi''(x) dx + \int_0^{\infty} x \phi''(x) dx \\ =& -x \phi'(x)|_{-\infty}^0 + x \phi'(x)| _0^{\infty} + \phi(x)|_{-\infty}^0 - \phi(x)| _0^{\infty} = 2 \phi(0). \end{split} \end{equation*} So the second derivative of $|x|$ in the distributional sense is $2 \delta(x)$. \end{small} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{frame}{Distributions as solutions of equations} \begin{small} Returning to the first differential equation $$ x'(t) + x(t) = f(t), \quad x(0) =0, $$ with $f(t) = \delta(t-t_0)$ the solution is $$ x(t) = \int_0^t \exp(s -t) \delta(t-t_0) ds =\begin{cases} 0 & t< t_0, \\ \exp(t_0-t) & t \ge t_0. \end{cases} $$ This solution is discontinuous, so how does is solve the equation? \pause Computing a distributional derivative, $$ x'[\phi] = - \int_{-\infty}^\infty x(t) \phi'(t) dt = - \int_{t_0}^\infty x(t) \phi'(t) dt = x(t_0) \phi(t_0) + \int_{t_0}^\infty x'(t) \phi(t) dt. $$ The last integral can be written as $$ - \int_{-\infty}^\infty H(t-t_0) \exp(t_0-t) \phi(t) dt, $$ so $$ x'(t) = \delta(t-t_0) - H(t-t_0) \exp(t_0-t), $$ and $x'(t) + x(t) = \delta(t-t_0)$ as desired. \end{small} \end{frame} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document}