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%                    B O F / A N A L Y T I C . T E X 
%                    doc: Thu Mar 16 16:00:27 2000
%                    dlm: Thu Jun  8 00:22:41 2000
%                    (c) 2000 A.M. Thurnherr
%                    uE-Info: 119 7 NIL 0 0 72 3 2 4 ofnI
%======================================================================

The steady-state solution of \rexpr{BOF-model} yields the equilibrium
interface height
 %
\begin{equation}
 %
\lsym{Bheq} = \left(\frac{2(\BQi-\BQu)\Bw f}{\Bgamma\gp}\right)^{1/2},
														\labelEqn{BOF-heq}
 %	
\end{equation}
 %
which is real for $\BQi\geq\BQu$. To investigate the model response to
abrupt changes from an initial condition specified by \lsym{Bhzero}
\rexpr{BOF-model} is solved analytically for two cases of steady
forcing: \begin{description} 

\REM{\item[$\BQi-\BQu\leq0$] (i.e.\ $\BQu\geq\BQi$): The inflow does not
suffice to balance the loss across the interface, implying that the
basin is drained to $\Bh=0$ whence the outflow ceases. The solution of
the initial-value problem with $\Bho(0)=\lsym{Bhzero}$ is given by
 %
\begin{equation}
 %
\Bho(t) = \frac{2\Bl\Bw f \Bhzero}{\Bgamma\gp t\Bhzero+2\Bl\Bw f}
		- \frac{\BQu-\BQi}\Bl t
		= \left(\frac{\Bgamma\gp}{2\Bl\Bw f}t+\frac{1}{\Bhzero}\right)^{-1}
		- \frac{\BQu-\BQi}\Bl t.
														\labelEqn{BOF-drain}
\end{equation}ENDREM}

\item[$\BQi-\BQu\geq\Bgamma\gp\Bhzero^2/(2f\Bw)$:] The interface
adjusts upwards to accommodate an increased inflow (i.e.\
$\Bheq\geq\Bhzero$). The solution of the initial-value problem is
 %
\begin{equation}
 %
\Bh(t) = \Bheq \frac{\exp(t/\Btau+\Bs)-1}{\exp(t/\Btau+\Bs)+1},
														\labelEqn{BOF-up}
\end{equation}
 %
where
 %
\begin{equation}
 %
\lsym{Btau} = \Bl\left(\frac{\Bw f}{2\Bgamma(\BQi-\BQu)\gp}\right)^{1/2}
														\labelEqn{BOF-tau}
 %
\end{equation}
 %
is a timescale of adjustment, and
 %
\begin{equation}
 %
\lsym{Bs} = \ln\frac{\Bheq+\Bhzero}{\Bheq-\Bhzero}		\labelEqn{BOF-s}
 %	
\end{equation}
 %
is a measure of the amount of adjustment required.

\item[$0<\BQi-\BQu\leq\Bgamma\gp\Bhzero^2/(2f\Bw)$:] The interface
adjusts downwards to accommodate a decreased inflow (i.e.\
$\Bheq\leq\Bhzero$). The solution of the initial-value problem is
 %
\begin{equation}
 %
\Bh(t) = \Bheq \frac{1+\exp(\Bsp-t/\Btau)}{1-\exp(\Bsp-t/\Btau)},
														\labelEqn{BOF-down}
 %	
\end{equation}
 %
with
 %
\begin{equation}
 %
\lsym{Bsp} = \ln\frac{\Bhzero-\Bheq}{\Bhzero+\Bheq}.	\labelEqn{BOF-sp}
 %
\end{equation}

\end{description}


%--------------------
\begin{figure}[t]

\placeFig{BOF/analytic}

\caption[Analytical model solutions (steady forcing)]{Non-dimensional
analytical solutions of \protect\rexpr{BOF-model} for steady forcing
$\BQi-\BQu=1$; the three curves show the steady state
($\Bheq=\Bhzero$), \rexpr{BOF-down} with $\Bhzero=2\Bheq$, and
\rexpr{BOF-up} with $\Bhzero=0$.}

\labelFig{BOF-analytic}

\end{figure}
%--------------------

The model response is expressed in terms of \BQo. Volume fluxes are
non-dimensionalized with the steady-state flux observed at the
outflow, i.e.\ with $\BQi-\BQu=\Qo/\Bw$. Time is non-dimensionalized
with \Btau calculated from the volume-flux scale. With these scalings
the non-dimensional model solutions become independent of \Bgamma,
i.e.\ they are fully specified by \Bhzero and $\BQi-\BQu$.
\rFigNP{BOF-analytic} shows analytic solutions to steady-state forcing
$\BQi-\BQu=1$ for three initial conditions. The asymmetry of the
solutions (i.e.\ the faster downward adjustment) reflects the
non-linear dependence of \BQo on \Bh in \rexpr{BOF-w+74-2D}.

To estimate the time scales \Btau appropriate for the \A segments,
values for $\BQi-\BQu$ and \gp are taken from the hydraulic
calculations at the outflows \rsecs{hydraulics}{mix-2000}. If the
volume-flux deficits \rsec{BOF-model} are caused by ``entrainment''
into the overlying water column the time scales become \d{89} for \sA
and \d{20} for \A. If the deficits are caused by additional outflows
the corresponding time scales become \d{73} for \sA ($\Bgamma=1.5$)
and \d{17} for \A ($\Bgamma=1.3$).


