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\begin{center}
\Large{\bf XIII. PHANTOM ZONES}
\addcontentsline{toc}{section}{\protect\numberline{XIII.}{\bf ~~~Phantom Zones}}
\end{center}   
 
\noindent\rm 
Because the advection and diffusion algorithms have a five-point stencil,
we need to specify the physical variables in "phantom zones" outside of
the computational domain. We briefly examine these algorithms at the boundaries
to determine how many points are needed outside the domain.
 
\begin{center}
\it 1. Advection of zone-centered quantities
\end{center}   
 
\noindent\rm 
The advected zone-centered quantities of interest are $q = \rho$,
$E/\rho$,
and $(e+E/\rho)$. Inside the computational domain ($k = 2, $ \ldots \ , $N-1$) we
need $\overline{q}_{2}, $ \ldots \ , $\overline{q}_{N-1}$, which implies we need
$q_{0}$ and $q_{N+1}$. At the inner boundary ($k = 1$) we need only
$\overline{q}_{2}$, hence no new information about {\it q}. At the outer
boundary ($k = N$) we need $\overline{q}_{N}$ or (transmitting boundary)
$\overline{q}_{N+1}$. Thus we need to run the advection algorithm from 
$k = 2$ to $k= N+1$, which means we need to set $r_{0}$ and $r_{N+2}$,
and $q_{0}$ and $q_{N+2}$.
 
 
\begin{center}
\it 2. Advection of interface-centered quantities
\end{center}   
 
\noindent\rm 
The advected interface-centered quantities of interest are $q = u$
 and $F/\rho$. Inside the computational domain we need $\overline{q}_{1}, $
\ldots \ , $\overline{q}_{N}$, which implies we need $q_{0}$ and $q_{N+1}$.
At the inner boundary ($k = 1$) and the outer boundary ($k=N+1$) {\it q} is set
by boundary conditions, and no additional information is required. Thus we
need run the advection algorithm from $k = 1$ to $k=N$, which means we need
to set $r_{0}$ and $r_{N+2}$, and $q_{0}$ and $q_{N+2}$.
 
\begin{center}
\it 3. Zone-centered viscous energy dissipation: $\epsilon_{Q}$
\end{center}   
 
\noindent\rm 
Inside the computational domain ($k = 2, $ \ldots \ , $N-1$) we need
$r_{2}, $\ldots , $r_{N},$ and $u_{2}, $ \ldots \ , $u_{N}$. At the inner
boundary $(k=1)$ we need $r_{1}$ and $u_{1}$, and at the outer boundary
$(k=N-1)$ we need $r_{N+1}$ and $u_{N+1}$. Thus we need to run the dissipation
algorithm from $k = 1,$ \ldots \ , $N$, and compute $\epsilon_{Q}$ for
$k = 1,$ \ldots \ , $N$ from quantities known on the grid; no phantom zones are required.
 
\begin{center}
\it 4. Interface-centered viscous momentum deposition: $\phi_{Q}$
\end{center}   
 
\noindent\rm 
$\phi_{Q}$ is needed only inside the computational domain ($k = 2, $ \ldots \ , $N$). Thus we 
need only $r_{1}, $ \ldots \ , $r_{N+1}$; $u_{1}, $ \ldots \ , $u_{N+1}$; $qh_{1}, $ \ldots \ 
, $qh_{N+1}$; $qm_{1}, $ \ldots , $qm_{N+1}$; and $qu_{1}, $ \ldots \ , $qu_{N+1}$. We run the algorithm from $k = 2$ to $k = N$.
 
\newpage
\begin{center}
\it 5. Diffusion
\end{center}   
 
\noindent\rm 
We allow for artificial diffusion of $q = \rho$ and $e$.
Inside the computational domain ($k = 2, $ \ldots \ , $N-1$) we need
$r_{1},$ \ldots \ , $r_{N+1},$ and $q_{1}, $ \ldots \ , $q_{N}$.
At the inner boundary $(k=1)$ we
need $r_{0}$ and $q_{0}$; at the outer boundary $(k=N)$ we need
$r_{N+2}$ and $q_{N+1}$. Thus we need to run the diffusion algorithm
from $k = 1$ to $k = N$, and set $r_{0},\  r_{N+2}$, and $q_{0},\  q_{N+1}$.
 
\vspace*{5mm}
{\flushleft\large{\bf A. Inner Boundary}}
\addcontentsline{toc}{subsection}{\protect\numberline{XIII.A.}{ ~~~Inner Boundary}}
\vspace*{3mm}
 
\begin{center}
\it(1) Eulerian with zero flux: ($\Phi_{L}^{0,1,2} \equiv  0$)
\end{center}   
 
\noindent\rm 
Here we must demand symmetry with respect to the boundary ($k=1$); then

\begin{flushright}
{$r_{0} = r_{1} - (r_{2} - r_{1})$\ \ \ \ \ (PZ1)\hspace{13mm} $\delta r_{0} \equiv \ \ \ \ \ \ \ 0$\ \ \ \ \ \ (PZ2)}\\
{$m_{0} = 2m_{1} - m_{2}$\ \ \ \ \ \ \ \ \ (PZ3)\hspace{12mm} $\delta m_{0} =\   -\delta m_{2}$\ \ \ \ \ (PZ4)}\\
{$\rho_{0} = \ \ \rho_{1}$\hspace{22mm} (PZ5)\hspace{14mm} $\delta\rho_{0} =\  \ \ \ \delta \rho_{1}$\ \ \ \ \ (PZ6)}\\
{$u_{0} = - u_{2}$\hspace{22mm} (PZ7)\hspace{13mm} $\delta u_{0} =\ -\ \delta u_{2}$\ \ \ \ \ (PZ8)}\\
{$e_{0} = \ \ e_{1}$\hspace{22mm} (PZ9)\hspace{14mm} $\delta e_{0} =\  \ \ \ \delta e_{1}$\ \ \ \ (PZ10)}\\
\end{flushright}
 
\begin{center}
\it(2) Eulerian with nonzero flux: ($\Phi_{L}^{0,1,2} > 0$)
\end{center}   
 
\noindent We choose an arbitrary $\Delta r_{L}$, so that
\begin{flushright}
{$r_{0} = r_{1} - \Delta r_{L}$\hspace{43mm} (PZ11)\hspace{5mm} $\delta r_{0} \equiv \ \ \ \ 0$\ \ \ \ \ (PZ12)}\\
{$m_{0} = m_{1} - \rho_{L} (r_{1}^{\mu + 1} - r_{0}^{\mu +1}) / (\mu + 1)$\hspace{9mm}(PZ13)\hspace{4mm} $\delta m_{0} = \delta m_{1}$\ \ \ \ \ (PZ14)}\\
{$\rho_{0} = \rho_{L}$\hspace{54mm} (PZ15)\hspace{5mm} $\delta\rho_{0} =\ \ \ \ 0$\ \ \ \ \ (PZ16)}\\
{$u_{0} = u_{L}$\hspace{54mm} (PZ17)\hspace{5mm} $\delta u_{0} =\ \ \  \ 0$\ \ \ \ \ (PZ18)}\\
{$e_{0} = e_{L}$\hspace{54mm} (PZ19)\hspace{5mm} $\delta e_{0} =\ \ \ \ 0$\ \ \ \ \ (PZ20)}\\
\end{flushright}
 
\begin{center}
\it(3) Lagrangean
\end{center}   
 
\noindent In cell zero demand $\Delta m_{L} =$ constant, and that all thermodynamic properties are the same as in cell one. Then
\begin{flushright}
{$r_{0}^{\mu+1} = r_{1}^{\mu + 1} - (\mu + 1) \Delta m_{L} / \rho_{0}$\ (PZ21)\hspace{7mm} $\delta r_{0} = (\frac{r_{1}}{r_{0}})^{\mu} \delta r_{1} +\frac{\Delta m_{L}}{r_{0}^{\mu} \rho_{0}^{2}} \delta \rho_{0}$\ (PZ22)}\\
{$m_{0} \ = m_{1} - \Delta m_{L}$\hspace{19mm} (PZ23)\hspace{6mm} $\delta m_{0} = 0$ \hspace{29mm}(PZ24)}\\
{$\rho_{0} \ = \rho_{1}$\hspace{34mm} (PZ25)\hspace{7mm} $\delta\rho_{0} = \delta \rho_{1}$\hspace{25mm} (PZ26)}\\
%{$u_{0}^{n+\theta} = (r_{0}^{n+1} - r_{0}^{n}) / dt$ \hspace{12mm} (PZ27)\hspace{1mm} $\theta \delta u_{0}^{n+1} = \delta r_{0}^{n+1} / dt$ \hspace{16mm}(PZ28)}\\
{$u_{0} = u_{1}$ \hspace{34mm} (PZ27)\hspace{7mm} $\delta u_{0} = \delta u_{1}$\hspace{25mm}(PZ28)}\\
{$e_{0} = e_{1}$\hspace{34mm} (PZ29)\hspace{8mm} $\delta e_{0} =\ \delta e_{1}$\hspace{25mm}(PZ30)}\\
\end{flushright}
 
\begin{center}
\it(4) Radiation
\end{center}   
 
\begin{flushright}
{$E_{0} =\ \ \ \ E_{1}$\ \ \ \ (PZ31) $\ \ \delta E_{0} = \ \ \delta E_{1}$ \hspace{43mm} (PZ32)}\\
\vspace{2mm}

{(a) Optically transmitting:\hfill}

{$F_{0} =\ \ \ \ F_{1}$\ \ \ \ \ (PZ33)$\ \ \ \delta F_{0} =\ \ \delta F_{1}$ \hspace{43mm} (PZ34)} \\
\vspace{2mm}

{(b) Optically reflecting:\hfill}

{$F_{0} =\ -\ F_{2}$\ \ \ \ \ (PZ35) $\ \ \delta F_{0} = -\delta F_{2}$ \hspace{43mm} (PZ36)}\\
\vspace{2mm}

{(c) Imposed flux:\hfill}

{$F_{0}= (\frac{r_{1}}{r_{0}})^{\mu} F_{1}$ \hspace{2mm} (PZ37) \hspace{2mm} $\delta F_{0} = (\frac{r_{1}}{r_{0}})^{\mu} \delta F_{1} + \mu (\frac{\delta r_{1}}{r_{1}} - \frac{\delta r_{0}}{r_{0}})(\frac{r_{1}}{r_{0}})^{\mu} F_{1}$\ (PZ38)}
\end{flushright}
 
 
\vspace*{5mm}
{\flushleft\large{\bf B. Outer Boundary}}
\addcontentsline{toc}{subsection}{\protect\numberline{XIII.B.}{ ~~~Outer Boundary}}
\vspace*{3mm}
 
\begin{center}
\it(1) Eulerian with zero flux: ($\Phi_{R}^{0,1,2} \equiv  0$)
\end{center}   
 
\noindent Again, demand symmetry with respect to the boundary ($k=N+1$); then
 
\begin{flushright}
{$r_{N+2} = r_{N+1} + (r_{N+1} - r_{N})$\ \ \ \ \ (PZ39)\hspace{13mm} $\delta r_{N+2} \equiv \ \ \ \ \ \ \ 0$\ \ \ \ \ \ (PZ40)}
{$m_{N+2} = 2m_{N+1} - m_{N}$ \hspace{16mm}(PZ41)\hspace{13mm} $\delta m_{N+2} = -\delta m_{N}$\ \ \ \ \ (PZ42)}
{$\rho_{N+1} = \ \ \rho_{N}$\hspace{31mm} (PZ43)\hspace{14mm} $\delta\rho_{N+1} =\  \ \ \ \delta \rho_{N}$\ \ \ \ (PZ44)}
{$u_{N+2} = - u_{N}$\hspace{31mm} (PZ45)\hspace{13mm} $\delta u_{N+2} =\ -\ \delta u_{N}$\ \ \ \ (PZ46)}
{$e_{N+1} = \ \ e_{N}$\hspace{31mm} (PZ47)\hspace{14mm} $\delta e_{N+1} =\  \ \ \ \delta e_{N}$\ \ \ \ (PZ48)}
\end{flushright}

\begin{center}
\it(2) Eulerian with nonzero flux: ($\Phi_{R}^{0,1,2} > 0$)
\end{center}   
 
\noindent We choose an arbitrary $\Delta r_{R}$, so that
\begin{flushright}
{$r_{N+2} = r_{N+1} + \Delta r_{R}$\hspace{25mm} (PZ49)\hspace{6mm} $\delta r_{N+2} \equiv \ \ \ \ 0$\ \ \ \ \ (PZ50)}
{$m_{N+2} = m_{N+1} + \frac{\rho_{R} (r_{N+2}^{\mu + 1} - r_{N+1}^{\mu +1})}{(\mu + 1)}$\hspace{5mm} (PZ51)\hspace{5mm} $\delta m_{N+2} = \delta m_{N+1}$\ (PZ52)}
{$\rho_{N+1} = \rho_{R}$\hspace{40mm} (PZ53)\hspace{5mm} $\delta\rho_{N+1} =\ \ \ \ 0$\ \ \ \ \ (PZ54)}
{$u_{N+2} = u_{R}$\hspace{40mm} (PZ55)\hspace{5mm} $\delta u_{N+2} =\ \ \  \ 0$\ \ \ \ \ (PZ56)}
{$e_{N+1} = e_{R}$\hspace{40mm} (PZ57)\hspace{5mm} $\delta e_{N+1} =\ \ \ \ 0$\ \ \ \ \ (PZ58)}
\end{flushright}
 
 
\begin{center}
\it(3) Transmitting Eulerian
\end{center}   
 
\begin{flushright}
{$r_{N+2} = r_{N+1} + \Delta r_{R}$\hspace{25mm} (PZ59)\hspace{5mm} $\delta r_{N+2} \equiv \ \ \ \ 0$\ \ \ \ \ (PZ60)}
 
{\hspace{6mm}$m_{N+2} = m_{N+1} + \rho_{N+1} \Delta V_{N+1}$\hfill (PZ61)}

{\hfill $\delta m_{N+2} = \delta \rho_{N+1} \Delta V_{N+1} + \delta m_{N+1}$\ (PZ62)}
\end{flushright}
 
\begin{flushright}
{$\rho_{N+1} = \rho_{N}$\hspace{40mm} (PZ63)\hspace{5mm} $\delta\rho_{N+1} = \delta\rho_{N}$\ \ \ \ \ (PZ64)}
 
{$u_{N+2} = u_{N+1}$\hspace{36mm} (PZ65)\hspace{5mm} $\delta u_{N+2} = \delta u_{N+1}$\ \ (PZ66)}
 
{$e_{N+1} = e_{N}  $\hspace{40mm} (PZ67)\hspace{5mm} $\delta e_{N+1} = \delta e_{N}  $\ \ \ \ \ (PZ68)}
\end{flushright}
 
\begin{center}
\it(4) Lagrangean or Transmitting Lagrangean
\end{center}   
 
\noindent In cell $N+1$ demand $\Delta m_{R} =$ constant, and that all thermodynamic properties are the same as in cell $N$. Then
 
\begin{flushright}
{\hspace{14mm}$r_{N+2}^{\mu+1} = r_{N+1}^{\mu + 1} - \frac{(\mu + 1) \Delta m_{R}}{\rho_{N+1}}$\ \ \ \ \ (PZ69)\hfill}

{$\delta r_{N+2} = (\frac{r_{N+1}}{r_{N+2}})^{\mu} \delta r_{N+1} -\frac{\Delta m_{R}}{r_{N+2}^{\mu} \rho_{N+1}^{2}} \delta \rho_{N+1}$\ \ (PZ70)}
 
{$m_{N+2} \ = m_{N+1} - \Delta m_{R}$\hspace{10mm} (PZ71)\hspace{4mm} $\delta m_{N+2} = 0$ \hspace{14mm}(PZ72)}
{$\rho_{N+1} \ = \rho_{N}$\hspace{30mm} (PZ73)\hspace{5mm} $\delta\rho_{N+1} = \delta \rho_{N}$\hspace{9mm} (PZ74)}
 
%{$u_{N+2}^{n+\theta} = (r_{N+2}^{n+1} - r_{N+2}^{n}) / dt$ \hspace{4mm} (PZ75)\hspace{2mm} $\theta \delta u_{N+2}^{n+1} = \delta r_{N+2}^{n+1} / dt$\hspace{2mm}(PZ76)}
{$u_{N+2} = u_{N+1}$\hspace{29mm} (PZ75)\hspace{5mm} $\delta u_{N+2} = \delta u_{N+1}$\hspace{9mm}(PZ76)}
 
{$e_{N+1} = e_{N}$\hspace{30mm} (PZ77)\hspace{4mm} $\delta e_{N+1} =\ \delta e_{N}$\hspace{10mm}(PZ78)}
\end{flushright}
 
\begin{center}
\it(5) Radiation
\end{center}   
 
\noindent{(a) Optically transmitting:}

\begin{flushright}
{$E_{N+1} = E_{N}$\hspace{30mm}(PZ79)\hspace{4mm}$\delta E_{N+1} = \delta E_{N}$\hspace{10mm}(PZ80)} \\
{$F_{N+2} = F_{N+1}$\hspace{26mm}(PZ81)\hspace{4mm}$\delta F_{N+2} = \delta F_{N+1}$\hspace{6mm} (PZ82)}\\
\end{flushright}

\noindent{(b) Optically reflecting:}

\begin{flushright}
{$E_{N+1} = \hspace{3mm}E_{N}$\hspace{27mm}(PZ83)\hspace{4mm}$\delta E_{N+1} = \hspace{3mm}\delta E_{N}$\hspace{7mm}(PZ84)} \\
{$F_{N+2} =\ -F_{N}$\hspace{27mm}(PZ85) $\hspace{4mm}\delta F_{N+2} = -\delta F_{N}$\hspace{7mm}(PZ86)} \\
\end{flushright}

\noindent{(c) Imposed flux:}

\begin{flushright}
{$E_{N+1} = E_{N}$\hspace{30mm}(PZ87)\hspace{4mm}$\delta E_{N+1} = \delta E_{N}$\hspace{10mm}(PZ88)} \\
{\hspace{14mm}$F_{N+2}= (\frac{r_{N+1}}{r_{N+2}})^{\mu}\  F_{N+1}$\hspace{12mm} (PZ89)\hfill}

{\hspace{13mm}$\delta F_{N+2} = (\frac{r_{N+1}}{r_{N+2}})^{\mu} \delta F_{N+1} + \mu (\frac{\delta r_{N+1}}{r_{N+1}} - \frac{\delta r_{N+2}}{r_{N+2}})(\frac{r_{N+1}}{r_{N+2}})^{\mu} F_{N+1}$\hfill \rm(PZ90)}
\end{flushright}
 
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