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\begin{document}

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\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}
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\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}
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\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}
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\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}
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\end{document}



