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\begin{document}

\begin{center}
{\bf EML 4930/5061 \hfill Analysis in M.E. II \hfill 04/27/10}\\
{\it Closed book \hfill Van Dommelen \hfill 3-5 pm}
\end{center}

Show all reasoning and intermediate results leading to your answer, or
credit will be lost. One book of mathematical tables, such as Schaum's
Mathematical Handbook, may be used, as well as a calculator and one
handwritten letter-size single formula sheet.

\begin{enumerate}

\item (20\%) Solve the following PDE and boundary condition
  \begin{displaymath}
    2 t u_t + x u_x = x \quad \mbox{ for } -1<x<1,\; t \ge 0
    \qquad u(1,t) = \frac{t^2}{1+t^2}
  \end{displaymath}
  Draw at least four characteristics very neatly in the $xt$-plane.
  Explain why there is no boundary condition at $x=-1$.  Explain why
  there is no initial condition given.

\begin{rawhtml}
<A HREF="figures/1.jpg">Solution.</A>
\end{rawhtml}

\item (20\%) Use D{\rq}Alembert to find the pressure $u$ to the
  problem of acoustics in a pipe of length $\frac12$ with one end
  closed,
  \begin{displaymath}
    u_{tt} = a^2 u_{xx} \qquad u_x(0,t)= u(\tfrac12,t)=0
  \end{displaymath}
  if the initial conditions are
  \begin{displaymath}
    u(x,0)=\sin(\pi x)\quad u_t(x,0) = \cos(\pi x).
  \end{displaymath}
  Use your results to evaluate $u$ at $x=2.1$ and $t=1.1$ if $a=2$.
  Exact value only.

\begin{rawhtml}
<A HREF="figures/2.jpg">Solution.</A>
\end{rawhtml}

\item (40\%) Solve the following wave propagation problem of
  acoustics in a pipe that is forced at its ends:
  \begin{displaymath}
    u_{tt} = a^2 u_{xx}
    \qquad u(x,0)=u_t(x,0)=0
    \quad u_x(0,t)=t^3 \quad u(\tfrac12\pi,t)=\tfrac12\pi t^3
  \end{displaymath}
  Show all reasoning.  List the eigenfunctions and eigenvalues
  completely.  Write out the fully worked-out solution precisely and
  completely.  Hint: for brevity, write $\bar q_n$ as $t\bar p_n$.

\begin{rawhtml}
<A HREF="figures/3a.jpg">Solution.</A>
<A HREF="figures/3b.jpg">Solution.</A>
\end{rawhtml}

\item (20\%) Solve the following problem of supersonic flow
  along a wavy plate using the Laplace transform method:
  \begin{displaymath}
    u_{xx} - a^2 u_{yy} = 0 \qquad
    u(0,y)=u_x(0,y)=0\quad u_y(x,0)=f(x) \quad u(x,\infty)=0
  \end{displaymath}
  where $f(x)$ is to be considered to be some given function and
  $x\ge0,y\ge0$.  Show all steps and reasoning.  Your answer should
  not have weird mathematics, but be in simple terms that anyone with
  a basic understanding of calculus can understand.  Use convolution
  only if it is unavoidable.  If $u_x(x,0)$ gives the pressure on the
  surface, then what is it in terms of $f(x)$?

\begin{rawhtml}
<A HREF="figures/4.jpg">Solution.</A>
\end{rawhtml}

\end{enumerate}

\newpage
\begin{center}
  \renewcommand{\arraystretch}{2}
  {\bf Table 6.3: Properties of the Laplace Transform}\\
  \begin{tabular}{|l|l|}
    \hline\hline
    \multicolumn{1}{|c|}{$\displaystyle f(t)_{\strut}^{\strut}$} &
    \multicolumn{1}{c|}{$\displaystyle \hat f(s)
      =\int_{\strut0}^{\strut\infty}f(t)e^{-st}\D t$} \\
    \hline\hline
    {\bf 1.} $\displaystyle C_1 f_1(t) + C_2 f_2(t) $ &
    $\displaystyle C_1 \hat f_1(s) + C_2 \hat f_2(s)$ \\
    \hline
    {\bf 2.} $\displaystyle f(at)$ &
    $\displaystyle a^{-1} \hat f(s/a)\qquad (a>0)$ \\
    \hline
    {\bf 3.} $\displaystyle f^{(n)}(t)$ &
    $\displaystyle s^n \hat f(s) - s^{n-1} f(0) - \ldots -
     f^{(n-1)}(0) \qquad (n=1,2,\ldots)$ \\
    \hline
    {\bf 4.} $\displaystyle t^n f(t)$ &
    $\displaystyle (-1)^n
    \hat f^{(n)}(s)\qquad (n=1,2,\ldots)$ \\
    \hline
    {\bf 5.} $\displaystyle e^{ct} f(t)$ &
    $\displaystyle \hat f(s-c)\qquad (s = \mbox{const.})$ \\
    \hline
    {\bf 6.}
    \begin{tabular}[t]{l}
      $\displaystyle H(t-b) f(t-b)$, where \\
      {\renewcommand{\arraystretch}{1.5}
        $\displaystyle \quad H(t) =
        \bigg\{_{\strut}^{\strut}
        \begin{array}{lr} 0\ & t<0 \\ 1\ & t>0 \end{array}$}
    \end{tabular} &
    $\displaystyle e^{-bs} \hat f(s)\qquad (b>0)$ \\
    \hline
    {\bf 7.} $\displaystyle f * g(t)\equiv
    \int_{{\strut}0}^{{\strut}t} f(t-\tau)g(\tau)\D\tau$ &
    $\displaystyle \hat f(s)\hat g(s)$ \\
    \hline\hline
  \end{tabular}
  \\
  \vspace{5pt}
  {\bf Table 6.4: Laplace Transform Pairs}\\
  \begin{tabular}{|l|l|}
    \hline\hline
    \multicolumn{1}{|c|}{$\displaystyle f(t)$} &
    \multicolumn{1}{c|}{$\displaystyle \hat f(s)
      =\int_{\strut0}^{\strut\infty}f(t)e^{-st}\D t$} \\
    \hline\hline
    {\bf 1.} $\displaystyle 1 $ &
    $\displaystyle \frac{1}{s}_{\strut}$ \\
    \hline
    {\bf 2.} $\displaystyle t^n$ &
    $\displaystyle \frac{n!}{s^{n+1}}_{\strut}\qquad(n=1,2,\ldots)$ \\
    \hline
    {\bf 3.} $\displaystyle e^{kt}$ &
    $\displaystyle \frac1{s-k}_{\strut}$ \\
    \hline
    {\bf 4.} $\displaystyle \sin(at)$ &
    $\displaystyle \frac{a}{s^2+a^2}_{\strut}$ \\
    \hline
    {\bf 5.} $\displaystyle \cos(at)$ &
    $\displaystyle \frac{s}{s^2+a^2}_{\strut}$ \\
    \hline
    {\bf 6.} $\displaystyle \frac{1}{\sqrt{\pi t}}_{\strut}$ &
    $\displaystyle \frac{1}{\sqrt{s}}$ \\
    \hline
    {\bf 7.} $\displaystyle \frac{1}{\sqrt{\pi t}}_{\strut} e^{-k^2/4t}$ &
    $\displaystyle \frac{1}{\sqrt{s}} e^{-k\sqrt{s}}\qquad(k>0)$ \\
    \hline
    {\bf 8.} $\displaystyle \frac{k}{\sqrt{4\pi t^3}}_{\strut} e^{-k^2/4t}$ &
    $\displaystyle e^{-k\sqrt{s}}\qquad(k>0)$ \\
    \hline
    {\bf 9.} 
    \begin{tabular}[t]{l}
      $\displaystyle \hbox{erfc}\left(k/2\sqrt{t}\right)_{\strut}$,
      where\\
      $\displaystyle \hbox{erfc}(z) \equiv
      \frac2{\sqrt{\pi}}\int_{\strut z}^\infty e^{-u^2} \D u$
    \end{tabular} &
    $\displaystyle \frac1s e^{-k\sqrt{s}}\qquad(k>0)$ \\
    \hline\hline
  \end{tabular}
\end{center}

\end{document}
