\documentclass[11 pt, journal,onecolumn]{IEEEtran}
\hyphenation{op-tical net-works semi-conduc-tor}
\begin{document}
%
% paper title
\title{Homework 5}

\markboth{CME 306, Spring 2006} {Shell \MakeLowercase{\textit{et
al.}}: Bare Demo of IEEEtran.cls for Journals} \maketitle

\section{Linear Hyperbolic Systems}

\noindent Consider the Linearized Gas Dynamics equations, given by
\begin{displaymath}
\left( \begin{array}{c} \rho \\ u \end{array} \right)_t + \left( \begin{array}{cc} 0 & b \\ \frac{a^2}{b} & 0 \end{array} \right) \left( \begin{array}{c} \rho \\ u \end{array} \right)_x = 0.
\end{displaymath}

\subsection{}

\noindent Find the eigenvalues and right and left eigenvectors of the Jacobian for the system.

\subsection{}

\noindent Find the solution of the above system given the initial conditions
\begin{displaymath}
\left\{
\begin{array}{l}
\rho(x,0) = \rho_0(x) \\
u(x,0) = u_0(x)
\end{array}
\right.
\end{displaymath}


\section{ENO-LLF for Systems}

\noindent The shallow water equations are
$$
\left(\begin{array}{c}h \\ hu \end{array}\right)_t+\left(\begin{array}{c}hu \\ hu^2+\frac{1}{2}gh^2 \end{array}\right)_x=\mathbf{0}.
$$
Recall from class that the associated Jacobian eigenvalues are
$$
\lambda_1=u+\sqrt{gh},\lambda_2=u-\sqrt{gh}
$$
with left eigenvectors
$$
\mathbf{L}_1=\left(\begin{array}{r} \frac{\sqrt{gh}-u}{2\sqrt{gh}},\frac{1}{2\sqrt{gh}}\end{array}\right),\mathbf{L}_2=\left(\begin{array}{r}
\frac{\sqrt{gh}+u}{2\sqrt{gh}},-\frac{1}{2\sqrt{gh}}\end{array}\right)
$$
and right eigenvectors
$$
\mathbf{R}^1=\left(\begin{array}{c} 1 \\ u+\sqrt{gh}\end{array}\right),\mathbf{R}^2=\left(\begin{array}{c} 1 \\ u-\sqrt{gh}\end{array}\right)
$$
Now, assume we want to update point $h$ and $u$ at $x_j$. Fill in the missing steps in the following algorithm.

\noindent We first compute $\mathbf{F}_{j+\frac{1}{2}}$. We define $s1$ and $f1$ as
$$
s1=???,f1=???
$$
and similarly $s2$ and $f2$ as
$$
s2=???,f2=???.
$$
Then, to perform ENO-LLF we need to evaluate
$$
{D1}_i^1H^+=\frac{1}{2}f1_i+\frac{1}{2}{\alpha}1_{j+\frac{1}{2}}s1_i
$$
$$
{D1}_{i}^1H^-=\frac{1}{2}f1_i-\frac{1}{2}{\alpha}1_{j+\frac{1}{2}}s1_{i}
$$
where
$$
{\alpha}1_{j+\frac{1}{2}}=???.
$$
And similarly,
$$
{D2}_i^1H^+=\frac{1}{2}f2_i+\frac{1}{2}{\alpha}2_{j+\frac{1}{2}}s2_i
$$
$$
{D2}_{i}^1H^-=\frac{1}{2}f2_i-\frac{1}{2}{\alpha}2_{j+\frac{1}{2}}s2_{i}
$$
where
$$
{\alpha}2_{j+\frac{1}{2}}=???.
$$
We then set
$$
{F1}^+_{j+\frac{1}{2}}={D1}_j^1H^+,{F1}^-_{j+\frac{1}{2}}={D1}_{j+1}^1H^-
$$
and
$$
{F2}^+_{j+\frac{1}{2}}={D2}_j^1H^+,{F2}^-_{j+\frac{1}{2}}={D2}_{j+1}^1H^-
$$
and finally
$$
{F1}_{j+\frac{1}{2}}=F1^+_{j+\frac{1}{2}}+F1^-_{j+\frac{1}{2}}
$$
$$
{F2}_{j+\frac{1}{2}}=F2^+_{j+\frac{1}{2}}+F2^-_{j+\frac{1}{2}}.
$$
These fluxes are then used as
$$
\mathbf{F}_{j+\frac{1}{2}}=???.
$$
$\mathbf{F}_{j-\frac{1}{2}}$ is computed similarly (e.g. just plug in $j-1$ in place of $j$ in the previous calculation).
$\mathbf{F}_{j+\frac{1}{2}}$ and $\mathbf{F}_{j-\frac{1}{2}}$ can then be used to update $u^n_j$ and $h^n_j$ as
$$
\left(\begin{array}{c} u^{n+1}_j \\ h^{n+1}_ju^{n+1}_j \end{array}\right)=\left(\begin{array}{c} u^{n}_j \\ h^{n}_ju^{n}_j
\end{array}\right)-\frac{\Delta{t}}{\Delta{x}}(\mathbf{F}_{j+\frac{1}{2}}-\mathbf{F}_{j-\frac{1}{2}}).
$$

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