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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


\begin{definition}
Fourier Transform

\begin{align*}
F(k) &= \dint\limits_{-\infty }^{\infty }f(x)e^{-ikx}dx \\
f(x) &= \frac{1}{2\pi }\dint\limits_{-\infty }^{\infty }F(k)e^{ikx}dk
\end{align*}
\end{definition}

\bigskip

\noindent examples

\begin{center}
\begin{equation*}
\begin{array}{ccc}
\text{Delta function} & \delta (x) & 1 \\ 
\text{Square Pulse} & H(a-|x|) & \frac{2}{k}\sin (ak) \\ 
\text{Exponential} & e^{-a|x|} & \frac{2a}{a^{2}+k^{2}} \\ 
\text{Heaviside function} & H(x) & \pi \delta (k)+\frac{1}{ik} \\ 
\text{Sign} & H(x)-H(-x) & \frac{2}{ik} \\ 
\text{Constant} & 1 & 2\pi \delta (k) \\ 
\text{Gaussian} & e^{-\frac{x^{2}}{2}} & \sqrt{2\pi }e^{-\frac{k^{2}}{2}}%
\end{array}%
\end{equation*}

\bigskip

\begin{equation*}
\begin{array}{cc}
\text{Function} & \text{Transform} \\ 
\frac{df}{dx} & ikF(k) \\ 
xf(x) & i\frac{dF}{dk} \\ 
f(x-a) & e^{-iak}F(k) \\ 
e^{iak}f(x) & F(k-a) \\ 
af(x)+bg(x) & aF(k)+bG(k) \\ 
f(ax) & \frac{1}{a}F(\frac{k}{a})%
\end{array}%
\end{equation*}
\end{center}

\newpage

\begin{center}
Heisenberg Inequality
\end{center}

\bigskip

\noindent Let $f(x)$ represent a probability density. Then%
\begin{equation*}
\dint\limits_{-\infty }^{\infty }|f(x)|^{2}dx=1
\end{equation*}

\begin{definition}
\begin{align*}
\overline{x}^{2}& =\int_{-\infty }^{\infty }|xf(x)|^{2}dx \\
\overline{k}^{2}& =\int_{-\infty }^{\infty }|kF(k)|^{2}dk
\end{align*}
\end{definition}

\bigskip

\begin{theorem}
\begin{equation*}
\overline{x} \, \overline{k}\geq \frac{1}{2}
\end{equation*}
\end{theorem}

\begin{proof}
By the Schwarz inequality 
\begin{align*}
\left\vert \int_{-\infty }^{\infty }xf(x)f^{\prime }(x)dx \right\vert &\leq 
\sqrt{\int_{-\infty}^{\infty }|xf(x)|^2 dx} \sqrt{\int_{-\infty }^{\infty
}|f^{\prime }(x)|^{2}dx} \\
&= \overline{x}\sqrt{\int_{-\infty }^{\infty }|f^{\prime }(x)|^{2}dx} \\
&= \overline{x}\overline{k} \qquad \text{by Parseval}
\end{align*}

but using integration by parts%
\begin{align*}
\left\vert \int_{-\infty }^{\infty }xf(x)f^{\prime }(x)dx \right\vert &=
\left\vert \frac{1}{2} xf^{2}(x)|_{-\infty }^{\infty } -\int_{-\infty
}^{\infty }\frac{1}{2}f^{2}(x)dx \right\vert \\
&= \left\vert 0-\frac{1}{2}\right\vert =\frac{1}{2}
\end{align*}
\end{proof}

\newpage

\begin{center}
Laplace Transform
\end{center}

\bigskip

\begin{center}
\begin{align*}
F(s) &= \mathcal{L}(f)(s)=\dint\limits_{0}^{\infty }f(t)e^{-st}dt \\
f(t) &= \frac{1}{2\pi i}\dint\limits_{\gamma -i\infty }^{\gamma +i\infty}
F(s)e^{st}ds
\end{align*}

\noindent $\gamma $ is a vertical contour chosen so that all the
singularities of $F(s) $ are to the left of it.

\noindent The Laplace transform is linear 
\begin{equation*}
\mathcal{L}(af+bg)=a\mathcal{L}(f)+b\mathcal{L}{(g)}
\end{equation*}
\end{center}

\begin{definition}
f is of exponential order if there exist positive numbers a and M such that 
\begin{equation*}
\vert f(t) \vert \le M s^{a t}
\end{equation*}
\end{definition}

\begin{theorem}
If f is piecewise continuous on $0 \le t < \infty $ and is of exponential
order then $\mathcal{L}(f)$ exists for all $s > a$.
\end{theorem}

\begin{proof}
\begin{align*}
\vert \dint\limits_{0}^{\infty }f(t)e^{-st}dt\vert &\leq
\dint\limits_{0}^{\infty }|f(t)|e^{-st}dt \\
&\leq M\dint\limits_{0}^{\infty }e^{at}e^{-st}dt \\
&= M\dint\limits_{0}^{\infty }e^{-(s-a)t}dt \\
&=\frac{M}{s-a}<\infty
\end{align*}
\end{proof}

\begin{center}
\noindent Note: $\sqrt{\frac{1}{t}}$is not of exponential order because of
its behavior near the origin. However, its Laplace transform exists. Hence
the above condition is sufficient but not necessary.

Examples

\begin{equation*}
\mathcal{L}\text{(e}^{at})=\dint\limits_{0}^{\infty }e^{-(s-a)t}dt=\frac{1}{%
s-a}<\infty
\end{equation*}

\bigskip

\begin{equation*}
\mathcal{L}\left( t^{a}\right) =\dint\limits_{0}^{\infty }t^{a}e^{-st}dt
\end{equation*}

\noindent change variables \ $p=st$. Then%
\begin{equation*}
\mathcal{L}\left( t^{a}\right) =\frac{1}{s^{a+1}}\dint\limits_{0}^{\infty
}p^{a}e^{-p}dp=\frac{1}{s^{a+1}}\Gamma (a+1)
\end{equation*}

\noindent So%
\begin{equation*}
\mathcal{L}\left( t^{n}\right) =\frac{\Gamma (n+1)}{s^{n+1}}=\frac{n!}{%
s^{n+1}}
\end{equation*}

\begin{align*}
\mathcal{L}\left( \cos (kt)+i(\sin kt)\right) &= \dint\limits_{0}^{\infty}
\left( \cos (kt)+i(\sin kt)\right) e^{-st}dt \\
&= \dint\limits_{0}^{\infty }e^{-t(s-ik)}dt= \frac{e^{-(s-ik)t}}{s-ik}%
|_{0}^{\infty }=\frac{1}{s-ik} \\
&= \frac{s}{s^{2}+k^{2}}+\frac{ik}{s^{2}+k^{2}}
\end{align*}

\noindent So%
\begin{align*}
\mathcal{L}\left( \cos (kt)\right) & =\frac{s}{s^{2}+k^{2}} \\
\mathcal{L}\left( \sin (kt)\right) & =\frac{k}{s^{2}+k^{2}}
\end{align*}
\end{center}

\begin{theorem}
Suppose $f$ is continuous on $0\leq t<\infty $ and is of exponential order
then $\mathcal{L}(f)$ exists for all $s>a$. Also if $f^{\prime }$ is
piecewise continuous on $0\leq t<\infty $ and is of exponential order then 
\begin{align*}
\mathcal{L}\left( f^{\prime }\right) &= s\mathcal{L}\left( f\right) -f(0) \\
\mathcal{L}\left( f^{(n)}\right) &= s^{n}\mathcal{L}\left( f\right)
-s^{n-1}f(0)-s^{n-2}f^{\prime }(0)\ldots -f^{(n-1)}
\end{align*}
\end{theorem}

\begin{proof}
integration by parts
\end{proof}

\begin{center}
\noindent Similarly 
\begin{equation*}
\mathcal{L}\text{(}t^{n})=\frac{\Gamma (n+1)}{s^{n+1}}=\frac{n!}{s^{n+1}}
\end{equation*}
\end{center}

\begin{theorem}
Suppose $f$ is of exponential order. Let $\alpha $ be a real number and $a$
the exponential bound. Then for $s>a+\alpha $%
\begin{align*}
F(s)& =\mathcal{L}\left( f(t)\right) \\
\mathcal{L}\left( e^{\alpha t}f(t)\right) & =F(s-\alpha )
\end{align*}
\end{theorem}

\begin{center}
\newpage

Some other Laplace transforms

\begin{equation*}
\begin{array}{cc}
f(t) & F(s) \\ 
e^{at} & \frac{1}{s-a} \\ 
\cos (\omega t) & \frac{s}{s^{2}+\omega ^{2}} \\ 
\sin (\omega t) & \frac{\omega }{s^{2}+\omega ^{2}} \\ 
\cosh (\omega t) & \frac{s}{s^{2}-\omega ^{2}} \\ 
\sinh (\omega t) & \frac{\omega }{s^{2}-\omega ^{2}} \\ 
t^{n} & \frac{n!}{s^{n+1}}=\frac{\Gamma (n+1)}{s^{n+1}} \\ 
tf(t) & -\frac{d}{ds}\mathcal{L}\left( f\right) (s) \\ 
te^{at} & -\frac{d}{ds}\left( \frac{1}{s-a}\right) =\frac{1}{\left(
s-a\right) ^{2}} \\ 
t^{n}f(t) & (-1)^{n}\frac{d^{n}}{ds^{n}}\mathcal{L}\left( f\right) (s) \\ 
t^{n}e^{at} & \frac{n!}{\left( s-a\right) ^{n+1}}\qquad s>a \\ 
e^{at}\sin (kt) & \frac{k}{(s-a)^{2}+k^{2}} \\ 
e^{at}\cos (kt) & \frac{s-a}{(s-a)^{2}+k^{2}}\qquad s>a \\ 
e^{at}\sinh (kt) & \frac{k}{(s-a)^{2}-k^{2}}\qquad s>a \\ 
e^{at}\cosh (kt) & \frac{s-a}{(s-a)^{2}-k^{2}}\qquad s>a \\ 
H(t-a) & \frac{1}{s}e^{-as} \\ 
\delta (t-a) & e^{-as} \\ 
\frac{a}{\sqrt{4\pi t^{3}}}e^{-\frac{a^{2}}{4t}} & e^{-a\sqrt{s}} \\ 
\frac{1}{\sqrt{\pi t}}e^{-\frac{a^{2}}{4t}} & \frac{1}{\sqrt{s}}e^{-a\sqrt{s}%
} \\ 
1-erf\left( \frac{a}{\sqrt{4t}}\right)  & \frac{1}{s}e^{-a\sqrt{s}}%
\end{array}%
\end{equation*}

\newpage

Inverse Laplace Transform

\bigskip

\noindent Instead of using the exact formula we "guess" the inverse
transform.

\medskip

Examples:
\end{center}

\begin{enumerate}
\item 
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{2}{4+(s-1)^{2}}\right)
\end{equation*}

\noindent Comparing with formula for $e^{at}\sin (kt)$ we see that%
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{2}{4+(s-1)^{2}}\right) =e^{t}\sin (2t)
\end{equation*}

\item 
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{1}{s^{2}+2s+3}\right)
\end{equation*}

\noindent We know that%
\begin{equation*}
\frac{1}{s^{2}+2s+3}=\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\left( s+1\right)
^{2}+\left( \sqrt{2}\right) ^{2}}
\end{equation*}

\noindent So 
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{1}{s^{2}+2s+3}\right) =\frac{1}{\sqrt{2}}%
e^{-t}\sin (\sqrt{2}t)
\end{equation*}

\bigskip

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad Rational
functions

\bigskip

\begin{itemize}
\item First approach

\begin{equation*}
\mathcal{L}^{-1}\left( \frac{1}{s^{2}+2s-3}\right)
\end{equation*}

Split into its partial fractions%
\begin{align*}
\frac{1}{s^{2}+2s-3} &= \frac{1}{(s+3)(s-1)} \\
&= \frac{A}{s+3}+\frac{B}{s-1} \\
&= \frac{(A+B)s+3B-A}{(s+3)(s-1)}
\end{align*}

So%
\begin{equation*}
\frac{1}{s^{2}+2s-3}=-\frac{1}{4}\frac{1}{s+3}+\frac{1}{4}\frac{1}{s-1}
\end{equation*}

and%
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{1}{s^{2}+2s-3}\right) =-\frac{1}{4}e^{-3t}+%
\frac{1}{4}e^{t}
\end{equation*}

\item Second approach: Using above formulae%
\begin{align*}
\mathcal{L}^{-1}\left( \frac{1}{s^{2}+2s-3}\right) &= \frac{1}{2}%
e^{-t}\sinh(2t) \\
&= \frac{1}{2}e^{-t}\frac{e^{2t}-e^{-2t}}{2} \\
&= -\frac{1}{4}e^{-3t}+\frac{1}{4}e^{t}
\end{align*}
\end{itemize}

\item 
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{1}{(s^{2}+1)(s+1)^{2}}\right)
\end{equation*}

\noindent Split into its partial fractions

\begin{equation*}
\frac{1}{(s^{2}+1)(s+1)^{2}}=\frac{A_{1}}{s+1}+\frac{A_{2}}{\left(
s+1\right) ^{2}}+\frac{Bs+C}{s^{2}+1}
\end{equation*}

\noindent To find constants we have two ways

\begin{itemize}
\item find common denominator%
\begin{align*}
\frac{A_{1}}{s+1}& +\frac{A_{2}}{\left( s+1\right) ^{2}}+\frac{Bx+C}{s^{2}+1}%
=\frac{\left[ A_{1}\left( s+1\right) +A_{2}\right] \left( s^{2}+1\right)
+\left( Bx+C\right) \left( s+1\right) ^{2}}{(s^{2}+1)(s+1)^{2}} \\
& =\frac{\left( A_{1}+B\right) s^{3}+\left( A_{1}+A_{2}+C+2B\right)
s^{2}+\left( A_{1}+B+2C\right) s+A_{1}+A_{2}+C}{(s^{2}+1)(s+1)^{2}}
\end{align*}

\noindent So%
\begin{align*}
A_{1}+A_{2}+C& =1 \\
A_{1}+B+2C& =0 \\
A_{1}+A_{2}+C+2B& =0 \\
A_{1}+B& =0
\end{align*}

Therefore%
\begin{align*}
A_1 &= A_2=\frac{1}{2} \\
B &= -\frac{1}{2} \\
C &= 0
\end{align*}

So%
\begin{equation*}
\frac{1}{(s^{2}+1)(s+1)^{2}}=\frac{1}{2\left( s+1\right) }+\frac{1}{2\left(
s+1\right) ^{2}}-\frac{s}{2\left( s^{2}+1\right) }
\end{equation*}

and%
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{1}{(s^{2}+1)(s+1)^{2}}\right) =\frac{1}{2}%
e^{-t}+\frac{1}{2}te^{-t}-\frac{1}{2}\cos (t)
\end{equation*}

\begin{equation*}
\frac{1}{(s^{2}+1)(s+1)^{2}}=\frac{A_{1}}{s+1}+\frac{A_{2}}{\left(
s+1\right) ^{2}}+\frac{Bs+C}{s^{2}+1}
\end{equation*}

multiple both sides by $(s+1)^{2}$ and evaluate at $s=-1\implies A_{2}=\frac{%
1}{2}.$

multiple both sides by $(s+1)^{2}$and differentiate with respect to $s$ and
evaluate at $s=-1\implies A_{1}=\frac{1}{2}$

Substitute the values for $A_{1}$ and $A_{2}$ and evaluate at $s=0$ $%
\implies C=0$

\noindent multiply both sides by $s^{2}$ and evaulate at $s=\infty $ $%
\implies B=-\frac{1}{2}$
\end{itemize}

\begin{equation*}
\mathcal{L}^{-1}\left( \frac{4s}{(s^{2}+1)^{2}(s-1)}\right)
\end{equation*}

\noindent As before%
\begin{equation*}
\frac{1}{(s^{2}+1)(s+1)^{2}}=\frac{A}{s-1}+\frac{B_{1}s+C_{1}}{s^{2}+1}+%
\frac{B_{2}s+C_{2}}{\left( s^{2}+1\right) ^{2}}
\end{equation*}

\noindent Again comparing coefficients we find%
\begin{align*}
\frac{1}{(s^{2}+1)(s+1)^{2}} &= \frac{1}{s-1}-\frac{s+1}{s^{2}+1}+\frac{-2s+2%
}{\left( s^{2}+1\right) ^{2}} \\
&= \frac{1}{s-1}-\frac{s}{s^{2}+1}-\frac{1}{s^{2}+1}-2\frac{s}{\left(
s^{2}+1\right) ^{2}}+2\frac{1}{\left( s^{2}+1\right) ^{2}} \\
&= \frac{1}{s-1}-\frac{s}{s^{2}+1}-\frac{s^{2}+4s-1}{\left( s^{2}+1\right)
^{2}} \\
&= \frac{1}{s-1}-\frac{s}{s^{2}+1}-\frac{s^{2}-1}{\left( s^{2}+1\right) ^{2}}%
-\frac{4s}{\left( s^{2}+1\right) ^{2}}
\end{align*}

\noindent So%
\begin{equation*}
\mathcal{L}^{-1}\left( \frac{4s}{(s^{2}+1)^{2}(s-1)}\right) =e^{t}-\cos
(t)-t\cos (t)-2t\sin (t)
\end{equation*}
\end{enumerate}

\begin{center}
\newpage

Other properties of the Laplace Transform

\bigskip

\bigskip
\end{center}

\begin{definition}
Convolution 
\begin{equation*}
f\ast g=\dint\limits_{0}^{t}f(t-\tau )g(\tau )d\tau
\end{equation*}
\end{definition}

\begin{center}
\bigskip
\end{center}

\begin{theorem}
If $f$ and $g$ are piecewise continuous and of exponential order, then%
\begin{equation*}
\mathcal{L}(f\ast g)=\mathcal{L}(f)\mathcal{L}(g)
\end{equation*}
\end{theorem}

\begin{definition}
Heaviside step function

\begin{equation*}
H(t-a) = 
\begin{cases}
1 & \text{if } t \ge a \\ 
0 & \text{if } t < a%
\end{cases}%
\end{equation*}
\end{definition}

\begin{center}
\noindent So 
\begin{equation*}
H(t-a)f(t-a) = 
\begin{cases}
f(t-a) & \text{if } t \ge a \\ 
0 & \text{if } t < a%
\end{cases}%
\end{equation*}
\end{center}

\begin{theorem}
Shift 
\begin{align*}
\mathcal{L}\left( H(t-a)f(t-a)\right) (s)& =\dint\limits_{0}^{\infty
}H(t-a)f(t-a)e^{-st}dt=\dint\limits_{a}^{\infty }f(t-a)e^{-st}dt \\
\text{let }p=t-a\text{ then }\mathcal{L}\left( F(t-a)\right) &
=\dint\limits_{0}^{\infty }f(t)e^{-s(p+a)}dp \\
& =e^{-as}F(s)
\end{align*}
\end{theorem}

\begin{definition}
delta function

Suppose f is continuous in $[a,b]$ then 
\begin{equation*}
\dint\limits_{a}^{b}f(t)\delta (t-t_{0})dt=%
\begin{cases}
f(t_{0}) & \text{if }t_{0}\text{ is in }[a,b] \\ 
0 & \text{otherwise}%
\end{cases}%
\end{equation*}
\end{definition}

\begin{center}
\noindent Hence 
\begin{equation*}
\mathcal{L}(\delta (t-t_{0}))=\dint\limits_{0}^{\infty }\delta
(t-a)e^{-st}dt=e^{-as}\qquad \text{if }0\leq a<\infty
\end{equation*}

\newpage

Applications to ODE/PDE

\bigskip

\noindent Consider the heat equation%
\begin{align*}
\frac{\partial u}{\partial t} &= k\frac{\partial ^{2}u}{\partial x^{2}}%
\qquad 0<x<\infty \quad 0<t<\infty \\
u(x,0) &= 0 \\
u(0,t) &= f(t)
\end{align*}

\noindent We Laplace transform in time to get%
\begin{align*}
sU(x,s)-u(x,0) &= k\frac{\partial ^{2}U}{\partial x^{2}} \\
\text{or \ \ }k\frac{\partial ^{2}U}{\partial x^{2}}-sU(x,s) &= 0
\end{align*}

\noindent This has the solution%
\begin{equation*}
U(x,s)=A(s)e^{\sqrt{\frac{s}{k}}x}+B(s)e^{-\sqrt{\frac{s}{k}}x}
\end{equation*}

\noindent Since the Laplace transform is bounded at infinity we have%
\begin{equation*}
U(x,s)=B(s)e^{-\sqrt{\frac{s}{k}}x}
\end{equation*}

\noindent Using the boundary condition at $x=0$ and denoting the Laplace
transform of $f(t)$ by $F(s)$%
\begin{equation*}
U(0,s)=F(s)=B(s)
\end{equation*}

\noindent So%
\begin{equation*}
U(x,s)=F(s)e^{-\sqrt{\frac{s}{k}}x}
\end{equation*}

\noindent Using the convolution theorem we have%
\begin{align*}
u(x,t)& =f(t)\ast \mathcal{L}^{-1}\left( e^{-\sqrt{\frac{s}{k}}x}\right)  \\
& =f(t)\ast \frac{x}{\sqrt{4k\pi }t^{\frac{3}{2}}}e^{-\frac{x^{2}}{4kt}} \\
\text{or}\quad u(x,t)& =\frac{x}{\sqrt{4k\pi }}\dint\limits_{0}^{t}\frac{%
f(\tau )}{\left( t-\tau \right) ^{\frac{3}{2}}}e^{-\frac{x^{2}}{4k\left(
t-\tau \right) }}d\tau 
\end{align*}

\noindent If $f(t)=T_{0}$ then%
\begin{equation*}
u(x,t)=\frac{x}{\sqrt{4k\pi }}T_{0}\dint\limits_{0}^{t}\frac{1}{\left(
t-\tau \right)^{\frac{3}{2}}}e^{-\frac{x^2}{4k\left( t-\tau \right)}} d\tau
\end{equation*}

\noindent Define $z=\frac{x}{2\sqrt{k}\left( t-\tau \right) }$ and $dz=\frac{%
x}{4\sqrt{k}\left( t-\tau \right) }d\tau $ and%
\begin{equation*}
u(x,t)=\frac{2}{\sqrt{\pi }}T_{0}\dint\limits_{\frac{x}{2\sqrt{kt}}%
}^{t}e^{-z^{2}}dz=T_{0}\func{erf}\!c\left( \frac{x}{2\sqrt{kt}}\right) 
\end{equation*}

\newpage

\bigskip

\noindent Now consider the wave equation

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

\noindent Again Laplace transform in time yielding%
\begin{align*}
s^{2}U(x,s)-su(x,0)-\frac{\partial u}{\partial t}(x,0)& =c^{2}\frac{\partial
^{2}U}{\partial x^{2}}+F(s) \\
\text{using the IC we get }\quad -c^{2}\frac{\partial ^{2}U}{\partial x^{2}}%
\text{\ }+s^{2}U(x,s)& =F(s)
\end{align*}

\noindent We now have an inhomogenous second order ODE with constant
coefficients. The general solution is the general solution to the homogenous
part plus a particular solution to the inhomogenous equation.

\noindent So the homogenous part gives:%
\begin{equation*}
U_{\text{homo}}=A(s)e^{-\frac{s}{c}x}+B(s)e^{\frac{s}{c}x}
\end{equation*}

\noindent However, $F(s)$ is independent of $x$. so a particular solution is 
$U=\frac{F(s)}{s^{2}}.$So the general solution is%
\begin{equation*}
U(x,s)=A(s)e^{-\frac{s}{c}x}+B(s)e^{\frac{s}{c}x}+\frac{F(s)}{s^{2}}
\end{equation*}

\noindent As before the transform is bounded at infinity and so $B(s)=0$

\begin{equation*}
U(x,s)=A(s)e^{-\frac{s}{c}x}+\frac{F(s)}{s^{2}}
\end{equation*}

\noindent Using the BC at $x=0$%
\begin{equation*}
0=U(0,s)=A(s)+\frac{F(s)}{s^{2}}\implies A(s)=-\frac{F(s)}{s^{2}}
\end{equation*}

\noindent So%
\begin{equation*}
U(x,s)=\frac{F(s)}{s^{2}}\left( 1-e^{-\frac{s}{c}x}\right)
\end{equation*}

\noindent Again using the convolution theorem%
\begin{align*}
u(x,t) &= f(t)\ast \mathcal{L}^{-1}\left( \frac{1-e^{-\frac{s}{c}x}}{s^{2}}%
\right) \\
&= f(t)\ast \left[ t-(t-\frac{x}{c})H(t-\frac{x}{c})\right] \\
&= \dint\limits_{0}^{t}f(t-\tau )\left[ \tau -(\tau -\frac{x}{c})H(\tau -%
\frac{x}{c})\right] d\tau
\end{align*}

\noindent where H is the Heaviside step function

\newpage

Vibrations of a string subject to gravity

\bigskip

\noindent If the only force on a string is gravity we have

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

\noindent So

\begin{align*}
u(x,t) &= -g\dint\limits_{0}^{t}\left[ \tau -(\tau -\frac{x}{c})H(\tau -%
\frac{x}{c})\right] d\tau \\
&= -g\frac{t^{2}}{2}+g\dint\limits_{0}^{t}(\tau -\frac{x}{c})H(\tau -\frac{x%
}{c})d\tau
\end{align*}

\noindent If $x>ct$ then $\tau -\frac{x}{c}$ is always negative and the
Heaviside function is zero.

\noindent So we only need consider the case $x<ct$ . Then%
\begin{equation*}
\dint\limits_{0}^{t}(\tau -\frac{x}{c})H(\tau -\frac{x}{c})d\tau
=\dint\limits_{\frac{x}{c}}^{t}(\tau -\frac{x}{c})d\tau =\frac{1}{2}(t-\frac{%
x}{c})^{2}
\end{equation*}

\noindent So%
\begin{equation*}
u(x,t)=%
\begin{cases}
-\frac{g}{2}\left( t^{2}-(t-\frac{x}{c})^{2}\right) & \text{if }0<x<ct \\ 
-\frac{g}{2}t^{2} & \text{if }x>ct%
\end{cases}%
\end{equation*}

\newpage

Hankel Transform

\begin{equation*}
H_{\upsilon }(f)(s)=\int_{0}^{\infty }f(r)J_{\upsilon }(sr)rdr\qquad s\geq 0
\end{equation*}

\noindent If%
\begin{equation*}
\int_{0}^{\infty }|f(r)|\sqrt{r}dr<\infty
\end{equation*}

\noindent Then%
\begin{equation*}
H_{\upsilon }(H_{\upsilon }(f))(s)=f(x)
\end{equation*}

\noindent Thus the Hankel transform is its own inverse !

\bigskip

\noindent Properties%
\begin{align*}
-sH_{0}(f)(s) &= H_{1}(f^{\prime })(s) \\
H_{0}\left( f^{\prime }+\frac{f}{r}\right) &= sH_{1}(f)(s) \\
H_{0}\left( f^{\prime \prime }+\frac{f^{\prime }}{r}\right) &=
-s^{2}H_{0}(f)(s)
\end{align*}

\bigskip

\noindent example:

\noindent Consider an infinitely long hanging chain fastened at $\infty$. 
\\[0pt]
We label the vertical axis as $x$ pointing up.

\noindent The free motion of the chain is described by

\begin{align*}
\frac{\partial ^{2}u}{\partial t^{2}}& =g\left[ x\frac{\partial ^{2}u}{%
\partial x^{2}}+\frac{\partial u}{\partial x}\right] \qquad 0<x<\infty \quad
0<t<\infty \\
u(x,0)& =f(x) \\
\frac{\partial u}{\partial t}(x,0)& =v(x)
\end{align*}
\noindent Change variables%
\begin{equation*}
z^2=x\qquad \text{so } \quad 2zdz=dx
\end{equation*}

\noindent Then%
\begin{align*}
\frac{\partial ^{2}u}{\partial t^{2}} &= \frac{g}{4}\left[ \frac{\partial
^{2}u}{\partial z^{2}}+\frac{1}{z}\frac{\partial u}{\partial z}\right]
\qquad 0<x<\infty \quad 0<t<\infty &  \\
u(z^{2},0) &= f(z^{2}) &  \\
\frac{\partial u}{\partial t}(z^{2},0) &= v(z^{2}) & 
\end{align*}

\bigskip

\noindent Denote the zeroeth order Hankel transform of $u(z^{2},t)$ by $%
U(s,t)$. \\[0pt]
Then we transform the equations to get%
\begin{align*}
\frac{\partial ^{2}U}{\partial t^{2}} &= -\frac{s^{2}g}{4}\frac{\partial^2 u%
}{\partial x^2} \\
U(s,0) &= H_{0}(f(z^2)) \\
\frac{\partial U}{\partial t}(s,0) &= H_{0}(v(z^2))
\end{align*}

\noindent As in the previous cases we solve the second order ODE to get%
\begin{align*}
U(s,t) &= A(s)\cos (\frac{\sqrt{g}}{2}st)+B(s)\sin (\frac{\sqrt{g}}{2}st) \\
A(s) &= H_{0}(f(z^{2}))\qquad B(s)=\frac{2}{\sqrt{g}s}H_{0}(v(z^{2}))
\end{align*}

\noindent Using the inverse transform we get%
\begin{equation*}
u(x,t)=\int_{0}^{\infty }\left[ A(s)\cos (\frac{\sqrt{g}}{2}st)+B(s)\sin (%
\frac{\sqrt{g}}{2}st)\right] J_{0}(s\sqrt{x})sds
\end{equation*}
\end{center}

\end{document}
