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\begin{document}
\title{About antecedent storage}
\author[1]{Marialaura Bancheri}%
\author[2]{Riccardo Rigon}%
\affil[1]{Affiliation not available}%
\affil[2]{University of Trento}%
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\date{\today}
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In paper Rigon et al. 2016, all of the theory started at a time $t_0=0$ that we specified to be sufficiently back in the past. We made this choice for a better comparison with practical cases where information (in the perspective of the paper regarding precipitation and/or meteorological forcings) available for a limited range of times.
Therefore our storage function was defined as:
\begin{equation}
S_{t_p}(t) = \int_0^{\min(t,t_p)} s(t,t_{in}) dt_{in}
\end{equation}
Assume, however, that we realised, in simulating, that our system is not stationary and we have to add information to it starting from previous times (before $t=0$). This is easily accomplished. From the point of view of of the age-ranked equations, we have, in fact the system:
\begin{equation}
\frac{d s(t,t_{in})}{dt} = j(t,t_{in})-et(t,t_{in})-q(t,t_{in}) + r(t,t_{in})
\label{k1}
\end{equation}
where the symbols used are explained elsewhere. The r.h.s of the equation either contains forcings (given as function of time) or as a function of age-ranked storages themselves. These forcing represents a limitation to our knowledge, since usually we know them starting from a certain clock time, that we conventionally called $t=0$, and are unknown before.
To solve (\ref{k1}) initial conditions are given by $s(t_{in},t_{in})=0$ and $s(t,t_{in}}) \equiv 0$ for any $t_{in} < t < 0$. We cannot access equations for forcings happened before $t=0$ but their actions can result in a storage different from zero (no less, no more). We can observe that, for any time $t$, then the natural extension of the previously developed theory gives:
\begin{equation}
p(t-t_{in}|t):= \frac{s(t,t_{in})}{\int_{-\infty}^{t} s(t,t_{in}) dt_{in}}
\label{k3}
\end{equation}
where
\begin{equation}
S(t) = {\int_{-\infty}^{t} s(t,t_{in}) dt_{in}} = \underbrace{\int_{-\infty}^{0} s(t,\hat{t}_{in}) d\hat{t}_{in}}_{S_0(t)} + \underbrace{\int_{0}^{t} s(t,t_{in}) dt_{in}}_{S_t(t)}
\end{equation}
is the water present inside the control volume at time $t$ as results of available forcings ($S_t(t)$) and past forcings ($S_0(t)$). We indicated with $\hat{t}_{in}$ precipitations timing previous than 0.
is the initial storage (necessary also to solve the bulk water budget equation)
Please notice that on the base of the above probability in (\ref{k3}), the mean residence time of the water at $t=0$ is:
\begin{equation}
_0 = \int_{0}^\infty T_r\, p(T_r|0) dT_r
\end{equation}
and in the second integral in the r.h.s. of the equation, $T_r \equiv 0-\hat{t}_{in}$ and $\hat{t}_{in} < 0$.
Next question is how we can assign the initial conditions on storages if we do not have explicit information about the forcings. An educated guess is that, for them we can make is that we can use a probability distribution function (pdf), for instance a $\Gamma$ function or others. To stay generic let's say that $p(T_r|0) = g(T_r|0)$ where $g$ is a known pdf, and $T_r=0-\hat{t}_{in}$.
This ~g has been recently interpreted in literature (Benettin et al.,
2017) as as the ensemble average of a set of time-variant age
distribution.
The knowledge of $g$ is not enough to estend the system of age-ranked equations to the past. However some inferences can be done. In fact the set of $s(0-\hat{t}_{in}|0)$, can be known by observing that, with the appropriate extensions of the definitions is, for any precipitation input at time $\hat{t}_{in} < 0$:
\begin{equation}
s(0,\hat{t}_{in}) = p(0-\hat{t}_{in}|0) S(0) = g(T_r|0) S(0)
\end{equation}
which seems, at this point, quite obvious.
Extending the backward probability on the basis of definitions is easy. It can then be obtained for any $t \geq 0$:
\begin{equation}
p(t-t_{in}|t):= \left\{
\begin{array}{ll}
\frac{s(t,t_{in})}{\int_{-\infty}^t s(t,t_{in}) dt_{in}} & t > t_{in} >0 \\
0 & t_{in} < t < 0
\end{array}
\right.
\end{equation}
The mean residence time is estimated, according to its definition, and using $T_r=t-t_{in}$ as:
\begin{equation}
_t = \int_t^\infty T_r g(t_r-t) dT_r + \int_0^t \underbrace{(t-t_{in})}_{$T_r$} p(\underbrace{(t-t_{in})}_{$T_r$}|t) d(\underbrace{(t-t_{in})}_{$T_r$})
\label{k9}
\end{equation}
The first integral accounts for the residence time of water in input before $t=0$ and the second term is the part simply calculated through the solutions of the age-ranked system but it embeds in the definition of the probability the initial storage. XXX IT SEEMSOK BUT CHECK AGAIN CALCULATIONS XXX. It is apparent that the first integral is decreasing with time and null after an amount of time which is strictly connected with the standard deviation of the pdf $g$. While in the second integral the effect of the initial storage in the definition of the probability is also vanishing as soon as t increases. Neglecting the old waters contribution means, on the other side, to decrease the age of water at least untill $t$ is large enough. This, based on actual knowlegde, means that for a correct residence time simulation, models should simulate decades (10s of years), which could be unfeasible. At the same time, it seems not completely realistic to use a time-invariant travel time function for $g$ because, it could be guessed that a time-variant distribution would be more appropriate, since its dependence of time varying forcings. But this is probably the best we can do.
An implicit assumption used in (\ref{k9}) is that $g$ is time invariant. It allowed us to use the pdf initially assumed for $t=0$ to be valid for any time. The reader should also observe that we dealt with the residence time. Travel times can be obtained after assuming a SAS or doing appropriate equivalent hypotheses.
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