 leak}_{a}n^{a}, \\
&&\!\!\!\!\!\!\!\!
  \partial_{t}(\hat{F}_{A}) 
+ \frac{1}{x} \partial_{x}\left[ x
   \left(\frac{1}{3}\alpha \hat{E} \delta^{x}_{A} -
   \beta^{x}\hat{F}_{A} \right) \right]
+ \partial_{z}\left[
   \left(\frac{1}{3}\alpha \hat{E} \delta^{z}_{A}-\beta^{z}\hat{F}_{A} \right)\right]
  \nonumber \\
&&\!\!  \ \ \ \ \ \ \ \ 
  = -\hat{E}\partial_{A}\alpha + \hat{F}_{k}\partial_{A}\beta^{k} 
    + 2 \alpha \hat{E}\partial_{A}\phi 
    + \frac{(\hat{E}/3 - \hat{F}_{y}\beta^{y})\delta^{x}_{A}}{x} 
    + \alpha e^{6\phi} Q^{\rm leak}_{A}, \\
&&\!\!\!\!\!\!\!\!
  \partial_{t}(\hat{F}_{y}) 
- \frac{1}{x^{2}} \partial_{x}\left[ x^{2}
   \beta^{x}\hat{F}_{y}  \right]
- \partial_{z}\left[
   \beta^{z}\hat{F}_{y} \right] = \alpha e^{6\phi} Q^{\rm leak}_{y},
\end{eqnarray}
where $\hat{E} = e^{6\phi}E$ and $\hat{F}_{i} = e^{6\phi}F_{i}$, and the 
subscript $A$ again denotes $x$ or $z$ component.
The closure relation $P_{\alpha \beta} = E\gamma_{\alpha \beta}/3$ is
also substituted.


\subsection{Recover of ($\rho$, $Y_{e}$/$Y_{l}$, $T$)} \label{Reconst}

The quantities numerically evolved in the relativistic hydrodynamics
are the conserved quantities ($\rho_{*}$, $\hat{u}_{i}$, $\hat{e}$)
and the lepton fraction $Y_{e}$ or $Y_{l}$. 
The argument variables,
($\rho$, ($Y_{e}$ or $Y_{l}$), $T$), of the EOS table, 
together with the weight factor $w = \sqrt{1+\gamma^{ij}u_{i}u_{j}}$, 
should be calculated from the conserved
quantities at each time slice. 
Note that the electron ($Y_{e}$) or lepton fraction ($Y_{l}$) is readily
given by numerical evolution 
at each time slice whereas $\rho$, $u_{i}$, and $T$ are not.
This fact requires us to find an efficient method for determining $w$.

\subsubsection{Non-$\beta$-equilibrium case}
In the case that the $\beta$-equilibrium condition is not satisfied,
the argument quantities ($\rho$, $Y_{e}$, $T$) can be reconstructed
from the conserved quantities in the following straightforward manner.

\begin{enumerate}
\item Give a trial value of $w$, referred to as $\tilde{w}$. 
Then, one obtains a trial value of the rest mass density from
$\tilde{\rho} = \rho_{*}/(\tilde{w} e^{6\phi})$.
\item A trial value of the temperature, $\tilde{T}$, can be
obtained by solving the following equation:
\begin{equation}
\hat{e} = \left(1+\tilde{\varepsilon} + 
\frac{\tilde{P}}{\tilde{\rho}}\right)\tilde{w} -
\frac{\tilde{P}}{\tilde{\rho}\tilde{w}}
\equiv \tilde{e}(\tilde{\rho}, Y_{e}, \tilde{T}).
\end{equation} 
Here, one dimensional search over the EOS table is
required to obtain $\tilde{T}$.
\item The next trial value of $w$ is given by 
$\tilde{w} =
      \sqrt{1+e^{-4\phi}\tilde{\gamma}^{ij}\hat{u}_{i}\hat{u}_{j}\tilde{h}^{-2}}$, 
where the specific enthalpy was calculated
as $\tilde{h} = \tilde{h}(\tilde{\rho}, Y_{e}, \tilde{T})$ in the step 2.
\item Repeat the procedures (1)--(3) until a required degree of convergence
  is achieved. Convergent solutions of the temperature and $w$ are
  obtained typically in 10 iterations.
\end{enumerate}

\subsubsection{The $\beta$-equilibrium case}
On the other hand,
in the case that the $\beta$-equilibrium condition is satisfied,
one has to reconstruct the argument quantities ($\rho, Y_{e}, T$) from 
the conserved quantities and $Y_{l}$, under the assumption of the 
$\beta$-equilibrium. In this case, two-dimensional recover
$(Y_{l}, \hat{e}) \  \Longrightarrow \ (Y_{e}, T)$
would be required for a given value of $\tilde{w}$. A serious problem is 
that in this case, there may be more than one combination 
of ($Y_{e}$, $T$) which gives the same values of $Y_{l}$ and $\hat{e}$.
Therefore, we have to adopt a different method 
to recover ($\rho, Y_{e}, T$).
Under the assumption of the $\beta$-equilibrium, 
the electron fraction is related to the total lepton fraction: 
$Y_{e} = Y_{e}(\rho, Y_{l}, T)$. Using this relation, the EOS table
can be rewritten in terms of the argument variables of 
($\rho$, $Y_{l}$, $T$). 
Then, the same strategy as in the non-$\beta$-equilibrium
case can be adopted. Namely,
\begin{enumerate}
\item Give a trial value $\tilde{w}$. 
Then one obtains a trial value of the rest mass density.
\item A trial value of the temperature can be
obtained by solving 
$ \hat{e} = \tilde{e}(\tilde{\rho}, Y_{l}, \tilde{T})$,
with one dimensional search over the EOS table.
\item The next trial value of $w$ is given by 
$\tilde{w} = \sqrt{1+e^{-4\phi}\tilde{\gamma}^{ij}\hat{u}_{i}\hat{u}_{j}\tilde{h}^{-2}}$. 
\item Repeat the procedures (1)--(3) until a required degree of convergence
  is achieved. The electron fraction is given as 
  $Y_{e} = Y_{e}(\rho, Y_{l}, T)$ in the (new) EOS table.
\end{enumerate}

In the case of a simplified or analytic EOS, 
the Newton-Raphson method may be applied to recover the primitive
variables. In the case of a tabulated EOS, by contrast,
the Newton-Raphson method may not be a good approach because it requires
derivatives of thermodynamical quantities which in general cannot be 
calculated precisely from a tabulated EOS by the finite differentiating 
method.


\subsection{Grid Setting}\label{Grid}

In numerical simulations, 
we adopt a nonuniform grid, in which 
the grid spacing is increased as
\begin{equation}
d x_{j+1} = (1 + \delta) d x_{j}, \ \ \ \ d z_{l+1} = (1 + \delta) d z_{l}
\end{equation}
where $d x_{j} \equiv x_{j+1} - x_{j}$, $d z_{l} \equiv z_{l+1} - z_{l}$, 
and $\delta$ is a constant.
In addition, a regridding technique\cite{Shibata02,Sekiguchi05} is adopted
to assign a sufficiently large number of grid points inside the
collapsing core, saving the CPU time efficiently.
The regridding is carried out whenever the characteristic radius
of the collapsing star decreases by a factor of a 2--3. 
At each regridding, the minimum grid spacing is decreased by a factor
of $\sim 2$ while the geometrical factor $\delta$ is unchanged 
(see Table \ref{regrid}).

All the quantities on the new grid are calculated
using the fifth-order Lagrange interpolation. 
To avoid discarding the matter in the outer region, we also increase the 
grid number at each regridding.
For the regridding, we define 
a relativistic gravitational potential
$\Phi_c \equiv 1 -\alpha_c~ (\Phi_c>0)$ where $\alpha_c$ is the central value
of the lapse function.
Because $\Phi_c$ is approximately proportional to $M/R$ where $M$ and $R$
are characteristic mass and radius of the core