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\begin{document}

\title{An illustrated guide to IUH/GIUH estimation}



\author[1]{Riccardo Rigon}%
\author[2]{Silvia Franceschi}%
\author[1]{Giuseppe Formetta}%
\author[3]{Marialaura Bancheri}%
\author[1]{Niccolò Tubini}%
\affil[1]{University of Trento}%
\affil[2]{Hydrologis}%
\affil[3]{ISAFOM}%


\vspace{-1em}



  
  \date{May 25, 2024}


\begingroup
\let\center\flushleft
\let\endcenter\endflushleft
\maketitle
\endgroup





\selectlanguage{english}
\begin{abstract}
This is an informal treatment of the instantaneous unit hydrograph
IUH/GIUH~\hyperref[csl:1]{(Rigon et al., 2015)} estimation under the simplifying assumption
that rainfall inputs are given as a discrete sequence of rain impulses
(that we will call from now, rain record) This, more than being a
simplification, is related to the discrete nature of rain at small time
scales of aggregation and has a physical basis
(e.g.~\hyperref[csl:2]{(Lovejoy \& Mandelbrot, 1985)}).~ This manuscript is distributed according to
the \href{https://creativecommons.org/licenses/by/4.0/}{CC v4 license.}~%
\end{abstract}%



\sloppy


Let's have a record of precipitation of duration~\(\Delta t\ \ll\ t_c\)
where~\(t_c\) is
the~\emph{concentration~time}~\hyperref[csl:3]{(Rigon et al., 2011)} of the catchment we are
analyzing~ (\(t_c\) can be thought as the length of the
domain over which the IUH is defined or a conventional time where the
contribution to the discharge given by a particular rainfall record is
not anymore detectable).~ Let~\(t\) be the clock time
and~\(t_{in}\) the time at which the precipitation record
happens. Finally let~\(p_{t_{in}}\) the precipitation intensity (for
instance in mm/h, {[}L/T{]}) during the time interval~\(\left(t_{in},t_{in}+\Delta t\right]\)\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-1/IUH-1}
\caption{{Representation of the rainfall record described in the text.
{\label{961889}}%
}}
\end{center}
\end{figure}

The precipitation is well represented in Figure 1 above. In this case,
according to the IUH theory, the discharge at any
time~\(t\) is given by:

\par\null

\begin{equation}
q(t,t_{in}) =A\, p_{t_{in}} \Theta(t_{in}) \int_{t_{in}}^t f(t-x) H(x)H(t-(x+\Delta t))dx
\label{eq1}
\end{equation}

where:

\begin{itemize}
\tightlist
\item
  \(q\left(t,t_{in}\right)\) {[}m\textsuperscript{3}s\textsuperscript{-1}{]} is
  the discharge generated by the single rain record
  called~\emph{age-ranked discharge}.
\item
  \(f\left(t-t_{in}\right)\ \left[s^{-1}\right]\) is the IUH (here assumed time invariant with
  respect to the injection time)
\item
  \(A\left[m^2\right]\)is the area of the catchment we are analyzing
\item
  \(H\left(x\right)\) is the Heaviside step function~ such that~
\end{itemize}

\par\null

\begin{equation}
H(x) = \left\{ 
\begin{array}{ll}
0 & x \leq 0 \\
1 &  x > 0
\end{array}
\right.
\label{eq2}
\end{equation}

\begin{itemize}
\tightlist
\item
  \(\Theta\left(t_{in}\right)\) is a~ coefficient of partition that identifies
  which portion of the rain input (at time~\(t_{in}\) goes into
  discharge).~\(\Theta\left(t_{in}\right)\le1\)~ depending if other outputs (as
  evapotranspiration) are present, but in the present case, for
  simplicity, will assume~\(\Theta\left(t_{in}\right)=1\)~
\end{itemize}

For simplicity, we will assume in the following that~\(A\ \)
is included in~\(f\) definition, which now
become~\(\left[m^2s^{-1}\right]\).~ The notation used above is congruent with
the one used in~\hyperref[csl:4]{(Rigon et al., 2016)} where, however, more general cases
are treated.~

According to the definition of the Heaviside step function the integral
giving the age-ranked discharge can be considered for various time
steps.

\begin{itemize}
\tightlist
\item
  For~\(t\ <t_{in}\) is clearly:
\end{itemize}

\(q\left(t,t_{in}\right)=0\)

\begin{itemize}
\tightlist
\item
  For~\(t_{in\ }<\ t\ \le t_{in}+\Delta t\):~
\end{itemize}

\(q\left(t,t_{in}\right)\ =\ p_{t_{in}}\int_{_{t_{in}}}^tf\left(t-x\right)\ dx\)

because~~\(H\left(x\right)H\left(t-x\right)\ \equiv1\)

\begin{itemize}
\tightlist
\item
  For~~\(t\ >t_{in}+\ \Delta t\):
\end{itemize}

\(q\left(t,t_{in}\right)\ =\ p_{t_{in}}\int_{_{t_{in}}}^{t_{in}+\Delta t}f\left(t-x\right)\ dx\)

because~\(H\left(x\right)H\left(t-x\right)\ \equiv1\) for~\(t_{in}<\ x\le t_{in}+\Delta t\ \)~
and~\(H\left(x\right)H\left(t-x\right)\ \equiv0\ \) for \(t\ >\ t_{in}+\ \Delta t\).~

Let's~ now define the so called S-function (which, as it is known,
correspond to the probability or residence/travel
times~\hyperref[csl:4]{(Rigon et al., 2016)}, in this case multiplied by the catchment
area~\(A\):~

\begin{equation}
s_f(u):=\int_0^u f(y) dy
\end{equation}

~we can show that the previous integral can~ be expressed as function
of~\(s_f\left(\ \right)\) which is convenient when it is known in advance
(for instance by assuming or deriving it from some geomorphic
procedure~\hyperref[csl:1]{(Rigon et al., 2015)}).~

\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-3/IUH-3}
\caption{{A pictorial representation of the function~\(s_f\left(T\right)\). After the
inclusion of the catchment area in the definition, it varies from 0 to
the maximum value of~\(A\).
{\label{685077}}%
}}
\end{center}
\end{figure}

Let's then define the travel time~\(T\)as the difference
between the clock time and the injection time:

\par\null

\begin{equation}
T := t -t_{in}
\end{equation}

and use the integrating variable~\(z\ =\ t-t_{in}\ \) , so that,~
~\(t_{in}<t\ \le t_{in}+\Delta t\) becomes~\(0\ <\ T\ \le\Delta t\)~,~ ~\(dz=-dx\)~
and the extremes of integration move such~\(x=t_{in\ }\ \to z=t-t_{in}\ =T\)
and~\(x=t\ \to z=0\)~ :~

\par\null

\begin{equation}
q(t, t-T) = p_{t-T}\int_0^{t-t_{in} =:T} f(z) dz =: p_{t-T} s_f(T)
\label{eq5}
\end{equation}

therefore, the age-ranked discharge s-function, in this case depends
only on the travel time, nor on the injection time,~ neither on the
clock time alone while the dependance of the injection time is retained
only in the intensity of precipitation.~ Analogously~
~\(t\ >\ t_{in}+\Delta t\) becomes~\(T\ >\ \Delta t\)~ and~
then~\(x=t_{in}\ \to z=t_{ }-t_{in}\ =T\) and~\(x=t_{in}+\Delta t\ \to z=t-t_{in}-\Delta t\ =\ T\ -\ \Delta t\) while~\(dz\ =\ -dx\)
such that:

\par\null

\begin{equation}
q(t,t-T) = p_{t-T} \int_{T - \Delta t}^T f(z) dz = p_{t-T} \left( s_f(T) - s_f (T-\Delta t)\right)
\label{eq6}
\end{equation}

This shows that for getting the age ranked discharges for any
time,~\(t\), we need the values of~\(s_f\left(T\right)\ \ \ \forall T\)~.
Please also observe that, for any~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~\(t\ \ge t_c+\Delta t\),- it is~\(\left(s_f(T)-s_f(T-\Delta t)\right)\ =0\) and
therefore~\(q\left(t,t-T\right)\equiv0\), as expected.~ To illustrate better the
topic let's rework Figure 1 into Figure 3 below.

\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-2/IUH-2}
\caption{{We select two time instant,~\(t_1\) and~\(t_2\)
and, in the text below, we describe how to get the corresponding
discharge.
{\label{900583}}%
}}
\end{center}
\end{figure}

Figure 2 shows the new axes represented by~\(T\). It goes
in the opposite direction of~\(t\). For~\(t=t_1\)
we are in the case~\(T\ <\ \Delta t\). Therefore the solution is given
by ({\ref{eq5}}). However, because~\(t_1\)
is fixed,~ the extremes of the integral vary just between 0
and~\(T_1\). We can repeat the procedure for
any~\(T_1\) in this interval.~ Upon the knowledge
of~\(f\), the s-function is known in advance and we have
just to pick its values at~\(T_1\) (and the values of the
precipitation at \(t_2-T_2\)) to know the discharge, as shown in
Figure 4 below.

\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-4/IUH-4}
\caption{{A pictorial representation od the~\(s_f\) function with two
time instants highlighted.
{\label{376185}}%
}}
\end{center}
\end{figure}

For~\(T_2\), since, it is~~\(T_2>\Delta t\), we have
instead to take the difference of the~ s-function between the two
instant~\(T_2=t_2-t_{in}\) and~\(T_2-\Delta t\ =\ t_2-t_{in}-\Delta t\).

Please observe that in the previous derivation Figure 3 can be
misleading. In fact the integration in
equation~{\ref{eq1}} is over~\(t_{in}\) while
Figure~{\ref{900583}} represent the IUH vs clock
time~\(t\). Therefore, in principle, our integral is not
the marron area below the IUH (it is just in a special case, which is
actually the one we are dealing with) and the change of variable we did
(exchanging~\(t_{in\ }\)with the travel time~\(T\))
hides this fact. A right representation of the integration is, instead
shown in Figure X below.

\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-5-0/IUH-5-0}
\caption{{When integrating the IUH for obtaining the discharges we are integrating
over~\(t_{in}\) not overt~\(t\). This mean we have
to consider a set of (infinitely many)~\(f\), each one for
any available ~\(t_{in}\). We actually integrate for
any~\(t\) the intersection of these curves with the
plane~\(\left(q,t_{in}\right)\)
{\label{997988}}%
}}
\end{center}
\end{figure}

In Figure~{\ref{997988}}~ we show three of the
(infinite) curves that correspond to an instant of rainfall
between~\(t\ \) and~\(t\ +\ \Delta t\) . For
any~\(t\ =t^{\bullet}\) we actually integrate over these curves, not just
one of them. The procedure we followed can be used to show that any of
these sections in the third dimension is actually equal to the forefront
maroon curve. But this is the case because we
assumed~\(f=f\left(t-t_{in}\right)\), depending just on the difference between the
clock and injections time. In the most general case, as seen
in~\hyperref[csl:4]{(Rigon et al., 2016)} , any of the~\(f\) is time dependent,
i.e.~\(f=f\left(t,t_{in}\right)\) and neglecting the right integration direction
can drive to conceptual mistakes easily.~

\section*{Examples}

{\label{528954}}\par\null

\subsection*{The exponential travel time
distribution}

{\label{801369}}

For making clearer the example let's consider first the simple case of~

\begin{itemize}
\tightlist
\item
  \(t_{in}=0\),~~\(i.e.\ T=t\)
\item
  \(p_{t_{in}}=p\)
\end{itemize}

and of:

\begin{equation}
f(t - 0) = \frac{1}{\lambda} e^{-(t-0)/\lambda}  
\end{equation}

where~\(\lambda\) is a parameter that can be proven to be the
mean travel time.~ Therefore, by integration is easy to prove that:

\begin{equation}
s_f(T) =A \left(1 - e^{-T/\lambda}\right) 
\end{equation}

Therefore for any~\(T\ \leq\Delta t\)~ ~(i.e.~\(t\ \leq\Delta t\)), it is

\begin{equation}
q(t,0) = q(T=t)= A\, p \left(1-e^{-t/\lambda}\right)
\end{equation}

For~\(t\ >\ \Delta t\) is instead:

\begin{equation}
q(t,0) = A\, p \left(1 - e^{-t/\lambda}\right)- \left(1-e^{-(t-\Delta t)/\lambda}\right)= A\, p \left(e^{-(t-\Delta t)/\lambda} -e^{-t/\lambda}\right) 
\end{equation}

~Studying the function it is easy to show that the discharge has a
maximum at~\(t\ =\ \Delta t\) equal to~\(A\ p\ \left(1-\ e^{-\frac{\Delta t}{\lambda}}\right)\) and then
decrease asymptotically to 0. In this case the concentration time is not
defined but for any practical purpose this can be set
when~\(t\ \ge\lambda\ \ln\left(\frac{1}{\alpha}\left(e^{\frac{\Delta t}{\lambda}}-1\right)^{^{^{-1}}}\right)^{ }\) with~\(\alpha>>1\). The discharge
qualitative behavior is shown in Figure 5 below.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
~ ~ ~\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-5/IUH-5}
\caption{{The behavior of the exponential IUH for a rainfall record
at~\(t_{in}=0\)
{\label{452579}}%
}}
\end{center}
\end{figure}

The generalization to cases for which~\(t_{in}\ne0\) is left for
exercise.

\subsection*{The width function case}

{\label{837202}}

Another interesting and classical case is the case in which the IUH is
given by a width function~\hyperref[csl:1]{(Rigon et al., 2015)}. This is defined as the
area of the catchment at a certain distance from the outlet measured
along the drainage directions. A width function is usually derived from
the analysis of a digital elevation model (DEM) and given at discrete
time distance steps, according to the resolution of the DEM. To obtain
from it a IUH, the geomorphic width function has to be properly
normalised by the catchment area and space (distance to outlet) has to
be mapped to time. Procedures to do it are illustrated, for instance
in~\hyperref[csl:5]{(``{The uDig Spatial Toolbox for hydro-geomorphic analysis}'', 2014)} and in~\hyperref[csl:1]{(Rigon et al., 2015)}. The resulting IUH looks
like the one in Figure {\ref{973251}} below:\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-7/IUH-7}
\caption{{A pictorial representation of a width function and the corresponding
s-function.
{\label{973251}}%
}}
\end{center}
\end{figure}

Once we get the~\(s_f\) (remind that here t stands
for~\(T\)) that can be considered continuous when
interpolated appropriately, the previous theory applies verbatim.

\par\null

\section*{Numerics}

{\label{203336}}\par\null

We have now all the tools for implementing the numerics of the IUH, at
least in the case of a single rain record.~ The main choice to do is to
select a discretization of the time axis. A reasonable choice is to
select a~\(\delta t\ <\ \Delta t\) and such not to loose the information
derived from the~\(f\) function, especially if it is given
as a discrete set of points. In this case the continuous variables are
set to:

\begin{itemize}
\tightlist
\item
  \(t\ =\ j\ \delta t\ \ \ \ j=0,\ ...,\ n\) with~\(n\ \delta t\ =\ \Delta t\)
\item
  \(t_{in}=i\ \delta t\)
\item
  \(T\ =t-t_{in}=\left(j-i\right)\delta t\)
\item
  \(t_c=m\ \delta t\)~
\end{itemize}

With these choices, equations~{\ref{eq5}}
and~{\ref{eq6}} becomes:

\par\null

\begin{equation}
q_{j,i} = p_i \left\{
\begin{array}{ll}
s_f|_{j-i} & 0 < j-i \leq n \\
\left( s_f|_{j-i}-s_f|_{j-i-n}\right) & j-i  > n
\end{array}
\right.
\end{equation}

where~\(s_f\left|_k\right|\) is the s-function estimated at
the~\(k-est\) point.

\subsection*{Examples}

{\label{313968}}

To illustrate the example let's consider the simple case in Figure 8
below.

\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-8/IUH-8}
\caption{{We consider here an example where there is a single record of
precipitation at time~\(i=5\) of duration~\(n=4\)
time steps (on top). The IUH is represented instead in the central
figure. It is given over four time intervals, each one of 3 time steps.
The total rainfall of the catchment is~\(A\ \) times the
volume of total rainfall~\(p_i\ \Delta t\ =\ p_i\ n\ \delta t\) where~\(n=4\) is
the number of intervals over which the rainfall duration
spans.~\(A=7\ h\ =21\) being~\(h=3\) the number of
intervals in which each IUH column is subdivided.~ On bottom there is
the s-function resulting. The whole mass of the rainfall is
then~\(84\ p_i\). The concentration time of the example
is~\(t_c=12\)
{\label{556644}}%
}}
\end{center}
\end{figure}

The caption of the Figure explains all what is expected.~ To be specific
the rainfall record has a total volume of~\(84\ p_i\). It is easy
to estimate for all discrete times the value of the s-function as

$s_f|_{j,i} = \left\{ (0,0),(1,1),(2,2),(3,3),(4,5),(5,7),(6,9),(7,12),(8,15),(9,18),(10,19),(11,20),(12,21) \right\}$

for any~\(t\ >12\ \delta t\)~ the s-function is not increasing anymore,
i.e. \(s_f\left(t\right)\ =\ 12\). For any~\(t\ \le5\ \delta t\)
,~\(s_f\left(t\right)\ =0\). Therefore\selectlanguage{english}
\begin{table}
\begin{tabular}{lll}
$q_{1,5} = p_5 s_f(-4)=0 $ & $q_{2,5} = p_5 s_f(-3)=0 $ & $q_{3,5} = p_5 s_f(-2)=0 $\\
$q_{4,5} = p_5 s_f(-1)=0 $ & $q_{5,5} = p_5 s_f(0)=0 $ & $q_{6,5} = p_5 s_f(1)= 1 p_5 $\\
$q_{7,5} = p_5 s_f(2)= 2 p_5 $ & $q_{8,5} = p_5 s_f(3)= 3 p_5 $ & $q_{9,5} = p_5 s_f(4)= 5 p_5 $\\
$q_{10,5} = p_5 (s_f(5) - s_f(1))= 6 p_5 $ & $q_{11,5} = p_5 (s_f(6) - s_f(2))= 7 p_5 $ & $q_{12,5} = p_5 (s_f(7) - s_f(3))= 9 p_5 $\\
$q_{13,5} = p_5 (s_f(8) - s_f(4))= 10 p_5 $ & $q_{14,5} = p_5 (s_f(9) - s_f(5))= 11 p_5 $ & $q_{15,5} = p_5 (s_f(10) - s_f(6))= 10 p_5 $\\
$q_{16,5} = p_5 (s_f(11) - s_f(7))= 8 p_5 $ & $q_{17,5} = p_5 (s_f(12) - s_f(8))= 6 p_5 $ & $q_{18,5} = p_5 (s_f(13) - s_f(9))= 3 p_5 $\\
$q_{19,5} = p_5 (s_f(14) - s_f(10))= 2 p_5 $ & $q_{20,5} = p_5 (s_f(15) - s_f(11))=  p_5 $ & $q_{21,5} = p_5 (s_f(16) - s_f(12))= 0  $\\
\end{tabular}
\end{table}

The table above represents all the estimable points. Taking the sum of
all the values, it does~\(84\ p_i\) as requested, meaning the the
whole water in input exit the catchment.~ The table values are
represented in Figure {\ref{493524}} below

\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-9/IUH-9}
\caption{{This is the hydrograph derived from the previous rainfall record. Still
it attends to be coloured yellow, orange, red and violet to associate
each discharge to its position in the IUH.~~
{\label{493524}}%
}}
\end{center}
\end{figure}



\section*{The case of multiple rainfall
records}

{\label{812029}}\par\null

Now, let's consider the case of multiple rainfall records, as those
shown in Figure Z. For simplification, we assume that each rainfall
records is of the same duration,~\(n\delta t\ =\ 4\ \delta t\)
where~\(\delta t\ \) is the unit of time as in the examples above.~

\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/IUH-10/IUH-10}
\caption{{The case of multiple rainfall recordss of duration~\(n\delta t\ =\ 4\ \delta t\).
As one can note there can be empty rainfall intervals
(\(p_k=0\)) without loss of generality.~
{\label{310463}}%
}}
\end{center}
\end{figure}

It is a consequence of the IUH mathematical structure that each one of
the records can be treated separately and the total discharge obtained
by~ summing over all the rainfall records (we skip the proof though).
The general expression for the discharge is then:

\par\null

\begin{equation}
Q(t) = \sum_{k=0}^l p_{i(k)} \left\{
\begin{array}{ll}
0 & t < t_{in}(k) \\
s_f(t-t_{in}(k)) &  t - t_{in}(k) \leq \Delta t\\
s_f(t-t_{in}(k)) - s_f(t-t_{in}(k)-\Delta t) & t - t_{in}(k) > \Delta t 
\end{array}    
\right.
\end{equation}

with~\(\)~~\(l:=\ \frac{t}{\Delta t}\) and where~\(i\left(k\right)\ =\ \left(k-1\right)n\)
if~\(\Delta t\ =\ n\ \delta t\) and~\(t_{in}\left(k\right)=i\left(k\right)\ \delta t\).~ It can be verified
that~\(Q\left(t\right)\ =\ 0\ \) for~\(t\ \ge t_c\). The discrete
representation of the above equation is derived then as:

\begin{equation}
Q_j = \sum_{k=0}^l p_{i(k)} \left\{
\begin{array}{ll}
0 & j < i(k) < (k-1)*n \\
s_f|_{j-i(k)} & j-i(k) \leq n \\
s_f|_{j-i(k)} - s_f|_{j-i(k)-n} & j-i(k) > n
\end{array}
    \right.
\end{equation}

From the numerics point of view, there are just minimal changes from
what happens in the case of a single rain record, with the addition of
an iteration over the rain impulse and a different evaluation
of~\(i\left(k\right)\)s, at least when the IUH is not changing shape over
time.~

In this case it is clear that the domain where the hydrograph due to a
specific rain record~\(k\) is different from zero in a
domain of length~\(t_c+\Delta t\ =\ \left(m+n\right)\ \delta t\). The total hydrograph, at any clock
time~\(t\), is easily seen to be of
length~\(\left(l\cdot n+m\right)\ \delta t\). Please notice that at any new record incoming
the knowledge of future discharges due only to past rain record is
available for all the m future time steps, being no modified by future
at all.~

In a real time perspective, when we receive a new rain record (of
duration~\(n\ \cdot\delta t\)) we have to expand the whole hydrograph
domain by~\(m\) steps forward in time.~

\selectlanguage{english}
\FloatBarrier
\section*{References}\sloppy
\phantomsection
\label{csl:1}{The geomorphological unit hydrograph from a historical-critical perspective}. (2015). \textit{Earth Surface Processes and Landforms}, \textit{41}(1), 27–37. \url{https://doi.org/10.1002/esp.3855}

\phantomsection
\label{csl:2}{Fractal properties of rain and a fractal model}. (1985). \textit{Tellus A}, \textit{37A}(3), 209–232. \url{https://doi.org/10.1111/j.1600-0870.1985.tb00423.x}

\phantomsection
\label{csl:3}{The geomorphic structure of the runoff peak}. (2011). \textit{Hydrology and Earth System Sciences}, \textit{15}(6), 1853–1863. \url{https://doi.org/10.5194/hess-15-1853-2011}

\phantomsection
\label{csl:4}{Age-ranked hydrological budgets and a travel time description of catchment hydrology}. (2016). \textit{Hydrology and Earth System Sciences}, \textit{20}(12), 4929–4947. \url{https://doi.org/10.5194/hess-20-4929-2016}

\phantomsection
\label{csl:5}(2014). In \textit{Geomorphological Techniques}. British Society for Geomorphology. \url{https://www.geomorphology.org.uk/sites/default/files/chapters/2.4.1_GISToolbox.pdf}

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