% ch07slidesa.tex

\begin{slide}
\begin{center}\large\bf
Fourier Transform for Partial Differential Equations
\end{center}

\begin{itemize}
\item {Introduction: Fourier Transform}
\item {Definition: Fourier Transform}
\item {Fourier Transform Properties}
\item {Parseval's Energy Identity}
\item{Fourier Sine and Cosine Integral Representations}
\item{Fourier Sine and Cosine Transforms}
\item{Fourier Sine and Cosine Transform Properties}
\end{itemize}

\end{slide}

\begin{slide}
\header{Introduction: Fourier Transform}

The Fourier transform creates another representation of a signal,
specifically a representation as a weighted sum of complex exponentials.
It is designed for non-periodic signals that decay at infinity, the
condition that $\int_{-\infty}^{\infty} |f(x)|dx$ is finite.

\bigskip

Because of Euler's formula
$$e^{iq} = \cos(q) + i \sin(q)$$
where $i^2 = -1$, the Fourier transform produces a representation of
a signal (or an image) as a weighted sum of sines and cosines.

\bigskip

Given a signal (or image) $a$ and its Fourier transform $A$, then the
{\bf forward Fourier transform} goes from the spatial domain, either
continuous or discrete, to the frequency domain, which is always
continuous. The {\bf inverse Fourier transform} goes from the frequency domain back
to the spatial domain.
$$\mbox{Forward}:\quad A =F(a), \quad \mbox{Inverse}:\quad a =F^{-1}(A)$$
\end{slide}

\begin{slide}
\header{Definition: Fourier Transform}

\bigskip

$$F(w)=FT[f](w)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
f(x)e^{-iwx}dx$$
$$f(x)=FT[f]^{-1}(x) =
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(w)e^{iwx}dw$$
The {\bf Reciprocity Relation} connects the two similar formulas:
$$
\begin{array}{lcl}
f(-u) &=&\dd\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(w)e^{-iwu}dw\\[2ex]
 &=&\dd\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(x)e^{-iux}dx \\[2ex]
 &=&FT[F](u)
 \end{array}
 $$
In words, setting $x=-u$ in the inverse Fourier transform equation
produces the forward Fourier transform equation for the function $F$.
\end{slide}

\begin{slide}
\header{Fourier Transform Properties}

\bigskip

\begin{tabular}{lll}
1.&Linearity& $FT[f+g]=FT[f]+FT[g]$ and $FT[cf]=c FT[f]$
\\[2ex]
2.&$x$-differentiation & $FT[f']=(iw)FT[f]$
\\[2ex]
3.&$w$-differentiation& $FT[xf(x)] = \dd i\frac{d}{dw} FT[f]$
\\[2ex]
4.&Convolution&  $FT[f*g]=FT[f]FT[g]$, where\\[1ex]
 && $f*g(x)=g*f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x-t)g(t)dt$
\\[2ex]
5.&$x$-shifting&  $FT[f(x-a)] = e^{-iaw}FT[f]$
\\[2ex]
6.&$w$-shifting&  $FT[e^{iax}f(x)](w)=FT[f](w-a)$
\end{tabular}
\end{slide}

\begin{slide}
\header{Parseval's Energy Identity}

\bigskip
The square $|f(t)|^2$ of the time signal represents how the energy
contained in the signal distributes over time t, while the spectrum
squared
$|F(w)|^2$ represents how the energy distributes over frequency
(the power density spectrum). The same
amount of energy is contained in either time or frequency domain,
because of  Parseval's formula:

$$\int_{-\infty}^{\infty} |f(t)|^2 dt = \int_{-\infty}^{\infty} |F(w)|^2 dw$$

To evaluate what this means graphically, compute both integrands and graph them on a large interval.
Use the Gaussian example
$$f(t)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{t^2}{2\sigma^2}}$$

\end{slide} 

\begin{slide}
\header{Fourier Sine and Cosine Integral Representations}

\bigskip

The theory applies to functions $f(x)$ defined only on $0<x<\infty$.

{\bf Definition}. The Fourier Cosine Integral Representation
$$f(x)=\int_0^\infty A(w)\cos(wx)dw,\quad A(w)=\frac{2}{\pi}\int_0^\infty f(t)\cos(wt)dt$$

{\bf Definition}. The Fourier Sine Integral Representation
$$f(x)=\int_0^\infty B(w)\sin(wx)dw,\quad B(w)=\frac{2}{\pi}\int_0^\infty f(t)\sin(wt)dt$$


\end{slide}

\begin{slide}
\header{Fourier Sine and Cosine Transforms}

\bigskip

{\bf Definition}. The Fourier Cosine and Sine Forward Transforms
$$FCT[f](w)=\sqrt{\frac{2}{\pi}}\int_0^\infty f(t)\cos(wt)dt,$$
$$FST[f](w)=\sqrt{\frac{2}{\pi}}\int_0^\infty f(t)\sin(wt)dt$$

{\bf Definition}. The Fourier Cosine and Sine Inverse Transforms
$$f(x)=\sqrt{\frac{2}{\pi}}\int_0^\infty FCT[f](w)\cos(wx)dw,$$
$$f(x)=\sqrt{\frac{2}{\pi}}\int_0^\infty FST[f](w)\sin(wx)dw$$
\end{slide}

\begin{slide}
\header{Fourier Sine and Cosine Transform Properties}

\bigskip
\begin{Theorem}[Properties]
\begin{itemize}
\item $FCT[f](w)=FT[f](w)$ for $w\ge 0$, provided $f$ is even on $(-\infty,\infty)$.
\item  $FST[f](w)=FT[f](w)$ for $w\ge 0$, provided $f$ is odd on $(-\infty,\infty)$. 
\item Both $FCT$ and $FST$ satisfy the Fourier transform's linearity property.
\item $FCT[f']=w\,FST[f]-\sqrt{2/\pi}f(0)$, provided $\lim_{x\to\infty} f(x)=0$.
\item $FST[f']=-w\,FCT[f]$, provided $\lim_{x\to\infty} f(x)=0$.
\item $FCT[xf(x)]=\frac{d}{dw}FST[f]$
\item $FST[xf(x)]=-\frac{d}{dw}FCT[f]$
\end{itemize}
\end{Theorem}
\end{slide}