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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
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\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
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\begin{document}


\begin{center}
Distributions
\end{center}

\bigskip

\noindent Returning to the Green's function.

\noindent Consider the ODE%
\begin{equation*}
Ly=-f(x)
\end{equation*}

\noindent Define the Green's functions as the solution to%
\begin{equation*}
LG(x,\xi )=-\delta (x-\xi )
\end{equation*}%
Then the solution to the ODE is given by%
\begin{equation*}
y(x)=\int G(x,\xi )f(\xi )d\xi
\end{equation*}

\noindent Since%
\begin{align*}
Ly &= \int L\left[ G(x,\xi )\right] f(\xi )d\xi \\
&= -\int \delta (x-\xi )f(\xi )d\xi \\
&= -f(x)
\end{align*}

\noindent {\Large Problem: } How to we properly define $\delta (x)$

\noindent Also not all practical functions have derivatives or Fourier
transforms. Point forces are very useful

\bigskip

\noindent We wish to extend the definition of a function and its derivative.

\bigskip

\noindent Fear: Extending allows for existence but endangers uniqueness

\begin{notation}
support ($f(x)$) = suppf(x)= $\overline{\left\{ x|f(x)\neq 0\right\} }$
\end{notation}

\begin{definition}
$C_{0}^{\infty }=$ functions with compact support in $C^{\infty }(R)$
\end{definition}

\begin{definition}
$D(R)$= space of test functions with the following properties

\begin{itemize}
\item $\varphi \in D\implies \varphi \in C_{0}^{\infty }$

\item topology: $\varphi _{n}\rightarrow \varphi $ in $D$ if

\begin{itemize}
\item $\{\varphi _{n}\}$ have support in the same compact set $K$

\item $\frac{\partial ^{r}\varphi _{n}}{\partial x^{r}}\rightarrow \frac{%
\partial ^{r}\varphi }{\partial x^{r}}$ uniformly as $n\rightarrow \infty $
\end{itemize}
\end{itemize}
\end{definition}

\bigskip

\noindent example:

\begin{align*}
\varphi _{n}& \rightarrow 0 \\
\text{if \quad }supp\text{ }\varphi _{n}& \subset K \\
\text{and }\lim_{n\rightarrow \infty }\left\vert \frac{\partial ^{r}\varphi
_{n}}{\partial x^{r}}\right\vert & =0
\end{align*}%
example%
\begin{equation*}
\phi (x)=%
\begin{cases}
0 & |x|\geq a \\ 
e^{\frac{1}{x^{2}-a^{2}}} & |x|<a%
\end{cases}%
\end{equation*}

\noindent Then $\phi (x)\in C_{0}^{\infty }$ but is not analytic

\bigskip

\begin{definition}
A linear function $q$ on $D(R)$ is a distribution if for every compact set $%
K\quad \exists C_{k}>0$ and nonegative integer $m$ such that%
\begin{equation*}
|q(\phi )|\leq C_{k}\sup_{\substack{ k\leq m  \\ x\in K}}|\frac{\partial
^{r}\varphi _{n}}{\partial x^{r}}|
\end{equation*}

\noindent i.e. $q$ is a continuous linear functional on $D$
\end{definition}

\bigskip

\noindent application: For D'Alembart's formula to be a solution of the wave
equation we require that $u\in C^{2}.$

\noindent If the initial condition is smooth except for a finite number of
points then the solution has discontinuities along the characteristics.

\begin{definition}
Given linear spaces $U$ and $V$ the space of all continuous linear
operations $U \rightarrow V$ is a linear space $\mathcal{L}(U, V)$. \newline
Dual Space: $U^{\prime} = \mathcal{L} (I,R) $ \medskip \noindent We denote $%
q(\phi) = <q,\phi >$.
\end{definition}

\begin{definition}
$q_{n}\rightarrow q$ in the weak topology $^{\ast }$ if 
\begin{equation*}
\lim_{n\rightarrow \infty }<q_{n},\phi >=<q,\phi >\quad \text{all \ }\phi
\in D(R)
\end{equation*}
\end{definition}

\noindent example of weak convergence%
\begin{align*}
S_{n}(x) &= \frac{1}{\pi }\frac{n}{1+n^{2}x^{2}} \\
b_{n} &= \frac{n}{\sqrt{\pi }}e^{-n^{2}x^{2}}
\end{align*}

\noindent then%
\begin{equation*}
\lim_{n\rightarrow \infty }\int_{-\infty }^{\infty }S_{n}(x)\varphi
(x)dx=\lim_{n\rightarrow \infty }\int_{-\infty }^{\infty }b_{n}(x)\varphi
(x)dx=\varphi (0)
\end{equation*}

\bigskip\ 

\begin{definition}
Dirac delta function%
\begin{equation*}
\int_{-\infty }^{\infty }\delta (x)\varphi (x)dx=\varphi (0)
\end{equation*}
\end{definition}

\noindent and we see that the distribution $\delta(x)$ is the weak limit of
classical functions 
\begin{equation*}
\lim_{n\rightarrow \infty }(q_{n},\varphi )=\int_{-\infty }^{\infty }\delta
(x)\varphi (x)dx
\end{equation*}

\noindent note: pointwise 
\begin{equation*}
\delta (x)=%
\begin{cases}
0 & x\neq 0 \\ 
\infty & x=0%
\end{cases}%
\end{equation*}%
so $\delta (x)$ is not a classical function

\bigskip

\noindent example:%
\begin{align*}
c_{n} &= \frac{1}{2}+\tan ^{-1}\frac{nx}{\pi } \\
\text{then\quad }\lim_{n\rightarrow \infty }c_{n} &=H(x)=%
\begin{cases}
0 & x<0 \\ 
\frac{1}{2} & x=0 \\ 
1 & x>0%
\end{cases}
\\
&= \lim_{n\rightarrow \infty }\frac{1}{\pi }\int_{-\infty }^{x} \frac{%
\sin(nt)}{t}dt
\end{align*}

\begin{definition}
f is locally integrable if $\int_{a}^{b}|f(x)|dx<\infty $ for every $\left[
a,b\right] $
\end{definition}

\noindent if $F(\phi )$ is a distribution and $f(x)$ is a classical function
then%
\begin{equation*}
F(\phi )=\int_{-\infty }^{\infty }f(x)\phi (x)dx\leq \sup_{K}|\phi
(x)|\int_{K}f(x)dx
\end{equation*}

\begin{definition}
F is a regular distribution if it corresponds to a locally integrable
function \newline
Otherwise F is a singular distribution
\end{definition}

\noindent Note: If $f$ and $g$ are distributions $fg$ is not necessarily a
distribution.

\noindent If $g\in C^{\infty }$ then $fg$ is a distribution.

\noindent Assume $f^{\prime }$ is locally integrable. Then 
\begin{equation*}
<f^{\prime },\phi >=\int_{-\infty }^{\infty }f^{\prime }(x)\phi
(x)dx=-\int_{-\infty }^{\infty }f(x)\phi ^{\prime }(x)dx
\end{equation*}

\bigskip

\begin{definition}
p is the distributional derivative of q if%
\begin{equation*}
<p,\phi >=-<q,\phi ^{\prime }> \quad \text{ for all }\phi \text{ in } D(R)
\end{equation*}
\end{definition}

\bigskip

\noindent Note: If $\{q_{n}\}\rightarrow \{q\}$ then $\{q_{n}^{\prime
}\}\rightarrow \{q\prime \}$

\begin{proof}
\noindent $<q_{n}^{\prime },\varphi >=-<q_{n},\varphi ^{\prime }>$

\noindent So \ $<q_{n},\varphi ^{\prime }>\rightarrow <q,\varphi
^{\prime}>=-<q^{\prime },\varphi >$
\end{proof}

\bigskip

\noindent Hence, we can derivatives of functions that don't have derivatives
in the classical sense.

\bigskip

\begin{definition}
Dipole distribution%
\begin{equation*}
<\delta ^{\prime },\varphi >=-<\delta ,\varphi ^{\prime }>= -\varphi
^{\prime}(0)
\end{equation*}
\end{definition}

\bigskip

\begin{lemma}
If $f$ is an integrable function and we know the functional $\int f\varphi
dx $ for all test functions $\varphi (x)$ then we know $f(x)$
\end{lemma}

\begin{proof}
Assume $\int \varphi dx = 1$ (normalization)

\noindent Define%
\begin{equation*}
\varphi _{t}(x)=\frac{1}{t^{n}}\varphi (\frac{x}{t})\qquad \text{in }R^{n}
\end{equation*}

\noindent Then%
\begin{equation*}
\lim_{t\rightarrow 0}\int_{R^{n}}f(s)\varphi _{t}(x-s)ds=f(x)
\end{equation*}
\end{proof}

\noindent Let%
\begin{equation*}
S(x,t)=\frac{1}{\sqrt{4\pi kt}}e^{-^{\frac{x^{2}}{4kt}}}\qquad t>0
\end{equation*}

\noindent Then%
\begin{align*}
\lim_{t\rightarrow 0}\int_{-\infty }^{\infty }S(x,t)\varphi (x)dx& =\varphi
(0) \\
\text{or\qquad }S(x,t)& \rightarrow \delta (x)
\end{align*}

\noindent Similarly for the Dirichlet kernel%
\begin{equation*}
K_{n}(\theta )=1+2\sum_{n=1}^{N}\cos (n\theta )=\frac{\sin (N+\frac{1}{2}%
)\theta }{\sin (\theta )}\overset{N\rightarrow \infty }{\longrightarrow }%
2\pi \delta (x)
\end{equation*}

\newpage

\begin{center}
Derivatives
\end{center}

\bigskip

\begin{definition}
The derivative $f^{\prime}$ is defined by 
\begin{equation*}
(f^{\prime}, \varphi) = - (f, \varphi^{\prime})
\end{equation*}
\end{definition}

\noindent Note: If $f_n \rightarrow f$ then $f_n^{\prime} \rightarrow
f^{\prime}$ weakly

\bigskip

\noindent example%
\begin{align*}
(\delta ^{\prime },\varphi ) &= -(\delta ,\varphi ^{\prime })=
-\varphi^{\prime }(0) \\
(\delta ^{\prime \prime },\varphi ) &= -(\delta ^{\prime },\varphi
^{\prime})= (\delta ,\varphi ^{\prime \prime })=\varphi ^{\prime \prime }(0)
\\
(H^{^{\prime }},\varphi ) &= -(H,\varphi ^{\prime })=-\int_{-\infty}^{\infty
}H(x)\varphi ^{\prime }(x)dx \\
&= -\int_{0}^{\infty }\varphi ^{\prime }(x)dx=\varphi (0) \\
\text{So} \qquad H^{\prime }(x) &= \delta (x)
\end{align*}

\begin{theorem}
Every distribution has derivatives of all orders
\end{theorem}

\bigskip

\noindent example%
\begin{equation*}
|x|=\frac{\pi }{2}-\frac{1}{\pi }\sum_{n\text{ odd}}\frac{4}{n^{2}}\cos
(nx)\qquad -\pi <x<\pi
\end{equation*}

\noindent then%
\begin{equation*}
\int_{-\infty }^{\infty }|x|\varphi (x)dx=\frac{\pi }{2}\int_{-\infty
}^{\infty }\varphi (x)dx-\frac{4}{\pi }\sum_{n\text{ odd}}\frac{1}{n^{2}}%
\int_{-\infty }^{\infty }\cos (nx)\varphi (x)dx
\end{equation*}

\bigskip

\noindent So this is a distribution. So if we formally differentiate we get
in the distributional sense%
\begin{equation*}
|x|^{\prime }=\frac{1}{\pi }\sum_{n\text{ odd}}\frac{4}{n}\sin (nx)=%
\begin{cases}
1 & 0<x<\pi \\ 
-1 & -\pi <x<0%
\end{cases}%
\end{equation*}

\noindent formally differentiate a second time%
\begin{align*}
|x|^{\prime \prime }& =\frac{4}{\pi }\sum_{n\text{ odd}}\cos (nx)=\delta (x)
\\
\text{i.e.}\quad & \sum_{n\text{ odd}}\int_{-\pi }^{\pi }\varphi (x)\cos
(nx)dx=\frac{\pi }{2}\varphi (0)
\end{align*}

\newpage

\begin{center}
differential equations
\end{center}

\bigskip

\noindent Consider%
\begin{equation*}
x\frac{du}{dx}=0
\end{equation*}

\noindent The classical solution is $u$ is constant \newline
The distributional solution is 
\begin{equation*}
u(x)=c_{1}H(x)+c_{2}
\end{equation*}

\begin{theorem}
Ehrenpreis: \newline
Let $P$ be a linear differential operator with constant coefficients. 
\newline
Then $Pu = \delta$ has a solution in terms of distributions
\end{theorem}

\medskip \noindent This solution is called the fundamental solution and $u$
is the Green's function.

\bigskip

\begin{center}
Laplace Equation
\end{center}

\bigskip

\begin{align*}
\phi (0) &= -\diiint \frac{1}{r}\frac{\Delta \phi }{4\pi }dV \\
\text{i.e.\qquad }\Delta \left( -\frac{1}{4\pi r}\right) &= \delta (x)\qquad 
\text{in 3D}
\end{align*}

\bigskip

\begin{center}
Poisson Equation
\end{center}

\begin{align*}
\Delta u &= f\qquad \text{in }D \\
u &= 0\qquad \text{on }\partial D
\end{align*}

\noindent then%
\begin{eqnarray*}
u(x_{0}) &=&\diiint\limits_{D}\delta (x-x_{0})u(x)dx \\
&=&\diiint\limits_{D}\Delta G(x,x_{0})u(x)dx
\end{eqnarray*}

\noindent but%
\begin{equation*}
G(x,x_{0})+\frac{1}{4\pi |x-x_{0}|}\text{ \ \ \ is harmonic in \ }D\text{
including }x_{0}
\end{equation*}

\noindent So%
\begin{equation*}
\Delta \left( -\frac{1}{4\pi |x-x_{0}|}\right) =\delta (x-x_{0})
\end{equation*}

\bigskip

\begin{center}
Wave Equation
\end{center}

\bigskip

\noindent Consider the Riemann function%
\begin{align*}
\frac{\partial ^{2}S}{\partial t^{2}}& =c^{2}\Delta S\qquad -\infty
<x,y,z<\infty \\
S(x,y,z,0)& =0 \\
\frac{\partial S}{\partial t}(x,y,z,0)& =\delta (x,y,z)=\delta (x)\delta
(y)\delta (z)
\end{align*}%
Let $\varphi (x)$ be a test function and (consider for simplicity 1D)

\bigskip

\begin{equation*}
u(x,t)=\int S(x-y,t)\varphi (y)dy
\end{equation*}

\bigskip

\noindent then

\begin{align*}
\frac{\partial ^{2}u}{\partial t^{2}} &= c^{2}\Delta u\qquad -\infty
<x<\infty \\
u(x,0) &= 0 \\
\frac{\partial u}{\partial t}(x,0) &= \varphi (x)
\end{align*}

\noindent In 1D we have D'Alambert's formula. So%
\begin{align*}
\int_{-\infty }^{\infty }S(x-y,t)\varphi (y)dy &= \frac{1}{2c}%
\int_{x-ct}^{x+ct}\varphi (y)dy \\
&= \frac{1}{2c}\int_{-\infty }^{\infty } \frac{H(x+ct)-H(x-ct)}{2}%
\varphi(y)dy
\end{align*}

\noindent Thus, we see that%
\begin{align*}
S(x,y)& =%
\begin{cases}
\frac{1}{2c} & |x|<ct \\ 
0 & |x|>ct%
\end{cases}
\\
& =\frac{1}{2c}H(c^{2}t^{2}-x)sgn(t)\qquad c^{2}t^{2}\neq x
\end{align*}

\noindent and so the solution is discontinuous along the characteristics

\noindent Define%
\begin{equation*}
u(x,t)= 
\begin{cases}
\frac{H(x+ct)-H(x-ct)}{2} & t>0 \\ 
0 & t <0%
\end{cases}%
\end{equation*}

\noindent then%
\begin{equation*}
\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial
x^{2}}=\delta (x)\qquad -\infty <x<\infty
\end{equation*}

In three dimensions%
\begin{eqnarray*}
S(x,t) &=&\frac{1}{4\pi c^{2}t}\delta (ct-|x|)\qquad t>0 \\
&=&\frac{1}{2\pi c}\delta (c^{2}t^{2}-|x|^{2})sgn(t)
\end{eqnarray*}

and%
\begin{equation*}
u(x,t)=\frac{1}{4\pi c^{2}t}\diiint\limits_{|x-y|=ct}\varphi (y)dy
\end{equation*}

\bigskip

\begin{center}
Diffusion equation
\end{center}

\bigskip

\begin{align*}
\frac{\partial ^{2}S}{\partial t^{2}}& =k\Delta S\qquad -\infty <x<\infty \\
S(x,0)& =\delta (x)
\end{align*}

\noindent Let%
\begin{equation*}
R(x,t)= 
\begin{cases}
S(x-x_{0},t-t_{0}) & t >t_0 \\ 
0 & t< t_0%
\end{cases}%
\end{equation*}

\noindent then%
\begin{equation*}
\frac{\partial ^{2}R}{\partial t^{2}}-k\Delta R=\delta
(x-x_{0},t-t_{0})\qquad -\infty <x<\infty
\end{equation*}

\end{document}
