The HRU
Conceptual model
One of the underlying principles of dynamic TOPMODEL is that the landscape can
be broken up into hydrologically similar regions, or Hydrological Response
Units (HRUs), where all the area within a HRU behaves in a hydrologically
similar fashion. Further discussion and the connection of HRUs is outlined elsewhere.
This document outlines the conceptual structure and computational
implementation of the HRU. The HRU is a representation of
an area of teh catchment with, for example, similar topographic, soil and upslope area
characteristics.
The HRU is considered to be a area of catchment with outflow
occurring across a specified width of it boundary.
It is formed of four zones representing the surface water, which
passes water between HRUs and drains to the root zone. The root zone characterises
the interactions between evapotranspiration and precipitation and when full spills to
the unsaturated zone. This drains to the saturated zone which also interacts
with other HRUs. The behaviour of the saturated zone is modelled using a kinematic wave
approximation. The zones and variables used below are shown in the schematic diagram.
In the following section the governing equations of the HRU are
given in a finite volume form. Approximating equations for the solution of the
governing equations are then derived with associated implicit and semi-implicit numerical
schemes. These are valid for the wide range of surface zone and saturated zone
transmissivity profiles presented in the vignette.
Two appendices provide supporting information justifying the numerical schemes
and an alternative derivation of the governing equations based on the
infinitesimal vertical slab used in the original derivation of Dynamic TOPMODEL.
Notation
Table 1 outlines the notation used for describing an
infinitesimal slab across the hillslope HRU.
The following conventions are used:
All properties such as slope angles and time constants are considered
uniform along the hillslope.
Precipitation and Potential Evapotranspiration occur at spatially uniform
rates.
All vertical fluxes \(r_{\star \rightarrow \star}\) are considered to
spatially uniform and to be positive when
travelling in the direction of the arrow, but can, in some cases be negative.
All lateral fluxes \(q_{\star}\) and \(q_{\star}\) are considered to be positive flows travelling downslope.
The superscripts \(^+\) and \(^-\) are used to denote variables for the outflow
and inflow to the hillslope.
\(\left.y\right\rvert_t\) indicates the value of the
variable \(y\) at time \(t\)
In the derivation storage values at a given cross section of the HRU are
expressed as a depth and denoted with a \(\bar{}\). So integrating
over the length of the hillslope we find
\[\begin{equation}
\int w \bar{y} dx = y
\end{equation}\]
Table 1: Outline of notation for describing a cross
section of the hillslope HRU
| Storage |
\(s_{sf}\) |
Surface excess storage |
m\(^3\) |
|
\(s_{rz}\) |
Root zone storage |
m\(^3\) |
|
\(s_{uz}\) |
Unsaturated zone storage |
m\(^3\) |
|
\(s_{sz}\) |
Saturated zone storage deficit |
m\(^3\) |
| Vertical fluxes |
\(r_{sf \rightarrow rz}\) |
Flow from the surface excess store to the root zone |
m\(^3\)/s |
|
\(r_{rz \rightarrow uz}\) |
Flow from the root zone to the unsaturated zone |
m\(^3\)/s |
|
\(r_{uz \rightarrow sz}\) |
Flow from unsaturated to saturated zone |
m\(^3\)/s |
| Lateral fluxes |
\(q_{sf}\) |
Lateral flow in the hillslope surface zone |
m\(^3\)/s |
|
\(q_{sz}\) |
Lateral inflow to the hillslope surface zone |
m\(^3\)/s |
| Vertical fluxes to the HRU |
\(p\) |
Precipitation rate |
m\(^3\)/s |
|
\(e_p\) |
Potential Evapotranspiration rate |
m\(^3\)/s |
| Other HRU properties |
\(A\) |
Plan area |
m\(^2\) |
|
\(w\) |
Width of a hill slope cross section |
m |
|
\(\Delta x\) |
Effective length of the hillslope HRU (x = A/w$) |
m |
|
\(\Theta_{sf}\) |
Further properties and parameters for the solution of the surface zone |
:-: |
|
\(\Theta_{rz}\) |
Further properties and parameters for the solution of the root zone |
:-: |
|
\(\Theta_{uz}\) |
Further properties and parameters for the solution of the saturated zone |
:-: |
|
\(\Theta_{sz}\) |
Further properties and parameters for the solution of the saturated zone |
:-: |
Finite Volume Formulation
In this section a finite volume formulation of the Dynamic TOPMODEL equations
is derived.
Surface zone
The storage in the surface zone satisfies the mass balance equation
\[
\frac{ds_{sf}}{dt} = q^{-}_{sf} - r_{sf \rightarrow rz} - q^{+}_{sf}
\]
where surface storage is increased by lateral downslope flow from upslope HRUs.
Water flows to the root zone at a constant rate \(r_{sf \rightarrow rz}\), unless limited by
the available storage at the surface or the ability of the root zone to
receive water (for example if the saturation storage deficit is 0 and the root zone storage is full).
In cases where lateral flows in the saturated zone produce
saturation storage deficits water may be returned from the root
zone to the surface giving negative values of \(r_{sf \rightarrow rz}\).
To determine the outflow \(q^{+}_{sf}\) a second relationship between
\(q^{-}_{sf}\), \(s_{sf}\) and \(q^{+}_{sf}\) is required. This is developed using a
Muskingham type approximation to the volume. Suppose the cross-sectional area
\(a\) and flow \(q\) are related by a one-to-one invetable function \(f\) such that
\(q = f\left(a, \Theta_{sf}\right)\). Using this the storage can be approximated by
\[
s_{sf} = \Delta x \left( \eta_{sf} a^{-}_{sf} +
\left(1-\eta_{sf}\right) a^{+}_{sf} \right)
\]
We consider only that case where \(f\) is defined such that \(f\left(0\right) = 0\)
and \(\frac{dq}{da} >0\); that is positive celerity. The role of \(\eta_{sf}\) in considered futher in
Appendix A which is summarised in Table 2.and the numerical solution.
Table 2: Outline formualtions between cross sectional area
and flow for the surface zone
| linTank |
Linear Tank with stepped state dependent time constant |
\(v_sf, k_sf\) (m/s, m/s) |
|
| linTank |
Linear Tank with continuous state dependent time constant |
|
|
| kinConst |
Kinematic routing with constant celerity |
|
|
| kinShlw |
Kinematic routing with shallow depth approx for wetted perimeter |
|
|
| DifConst |
Diffuse routing with constant celerity and diffusivity |
|
|
Root zone
The root zone gains water from precipitation and the surface zone. It loses water through
evaporation and to the unsaturated zone. All vertical fluzes are
considered to be spatially uniform. The root zone storage satisfies
\[\begin{equation} 0 \leq s_{rz} \leq s_{rzmax} \end{equation}\]
with the governing ODE
\[\begin{equation}
\frac{ds_{rz}}{dt} = p - \frac{e_p}{s_{rzmax}} s_{rz} +
r_{sf\rightarrow rz} - r_{rz \rightarrow uz}
\end{equation}\]
Fluxes from the surface and to the unsaturated zone are
controlled by the level of root zone storage along with the state of the unsaturated
and saturated zones.
For \(s_{rz} \leq s_{rzmax}\) then \(r_{rz \rightarrow uz} \leq 0\). Negative
values of \(r_{rz \rightarrow uz}\) may occur only when water is returned from the
unsaturated zone due to saturation caused by lateral inflow to the saturated zone.
When \(s_{rz} = s_{rzmax}\) then
\[\begin{equation}
p - e_p + r_{sf\rightarrow rz} - r_{rz \rightarrow uz} \leq 0
\end{equation}\]
In this case \(r_{rz \rightarrow uz}\) may be positive if
\[\begin{equation}
p - e_p + r_{sf\rightarrow rz} > 0
\end{equation}\]
so long as the unsaturated zone can receive the water. If \(r_{rz \rightarrow uz}\) is ‘throttled’ by the rate at which the unsaturated zone can
receive water, then \(r_{sf\rightarrow rz}\) is adjusted (potentially becoming
negative) to ensure the equality is met.
Unsaturated Zone
The unsaturated zone acts as a non-linear tank subject to the constraint
\[\begin{equation} 0 \leq s_{uz} \leq s_{sz} \end{equation}\]
The governing ODE is written as
\[\begin{equation}
\frac{ds_{uz}}{dt} = r_{rz \rightarrow uz} - r_{uz \rightarrow sz}
\end{equation}\]
If water is able to pass freely to the saturated zone, then it flows at the rate
\(\frac{A s_{uz}}{T_d s_{sz}}\).
TODO: not sure if needed If \(\bar{s}_{sz}=\bar{s}_{uz}=0\) this is interpreted to mean
that the flow rate is \(\frac{1}{T_d}\).
In this situation where \(s_{sz}=s_{uz}\) the subsurface below
the root zone can be considered saturated, as in there is no further available
storage for water, but separated into parts: an upper part with vertical
flow and a lower part with lateral flux.
It is possible that \(r_{uz \rightarrow sz}\) is
constrained by the ability of the saturated zone to receive water. If
this is the case \(r_{uz \rightarrow sz}\) occurs at the maximum
possible rate and \(r_{rz \rightarrow uz}\) is limited to ensure that
\(s_{uz} \leq s_{sz}\).
Saturated Zone
For a HRU the alteration of the storage in the saturated zone is
given by the kinematic equation implimented through the muskingham
approximation (see Appendix A). This gives similar equations to the Surface
zone, but considered in terms of storage deficits; that is
\[
\frac{ds_{sz}}{dt} = q^{+}_{sz} - r_{sf \rightarrow rz} - q^{-}_{sz}
\]
and
\[
s_{sz} = \frac{\Delta x}{2} \left( a^{-}_{sz} + a^{+}_{sz} \right)
\]
The Kinematic approximation then requires a relationship between \(q_{sz}\) and
the cross sectionsal flow area, given in terms of the cross sectional storage
defict \(a_{sz}\). This relationship is the transmissivity
profile which defines a relationship through a one-to-one, continuously differentiable function
\(g: \left[0,D\right] \rightarrow \mathcal{R}^{+}\) which returns the lateral flow on
a unit width such that \(q_{sz} = w g\left(a_{sz}/w,\Theta_{sz}\right) \geq 0\). The function \(g\) is considered to satisify
\[\begin{equation}
-\frac{dq_{sz}}{da_{sz}} = -w\frac{d}{da_{sz}}g\left(a_{sz}/w,\Theta_{sz}\right) \geq 0
\end{equation}\]
and
\[\begin{equation}
g\left(a_{sz}/w,\Theta_{sz}\right) \rightarrow 0 \quad \mathrm{as} \quad
a_{sz}\rightarrow \infty
\end{equation}\]
The first condition ensures a non negative celerity (wave speed) which is in
keeping with the conceptualisation of the model. This combined with
the second condition ensures that the model represents a state where no
lateral inflow is generated.
Numerical Solution
No analytical solution yet exists for simultaneous integration of the system of
ODEs outlined above. In the following an implicit scheme, where fluxes
between stores are considered constant over the time step and gradients are
evaluated at the final state, is presented.
The basis of the solution is that a gravity driven system will maximise
the downward flow of water within each timestep of size \(\Delta t\). Hence on
the first downward pass through the stores teh maximum rates of the vertical
fluxes $_{* } is determined. This then moderated in the
solution of the saturated zone, before the upward pass gives the final states.
Downward Pass
Surface excess
SEARCH WITH \(a^{+}\)
Consider first the case with:
- a known constant value of \(\eta\)
- a known positive inflow \(q\left(a^{-}_{t+\Delta t}\right)\) giving a value of
\(a^{-}_{t+\Delta t}\)
- a known function \(q\) such that \(\frac{dq\left(a\right)}{da}\geq 0\) and
\(q\left(0\right)=0\)
An implicit
solution gives the joint equations
\[
s_{t+\Delta t} = \Delta x \left(\eta a^{-}_{t+\Delta t} + \left(1-\eta\right)
a^{+}_{t+\Delta t} \right)
\]
and
\[
s_{t+\Delta t} = s_{t} + \Delta t \left( q\left(a^{-}_{t+\Delta t}\right) +
r_{t+\Delta t} - q\left(a^{+}_{t+\Delta t}\right) \right)
\]
The unknown outflow crossection satisfies \(a^{+}_{t+\Delta t} \geq 0\). The first
expression for \(s_{t+\Delta t}\) is strictly increasing with respect to
\(a^{+}_{t+\Delta t}\) while the second is decreasing, hence there is a unique
solution for \(a^{+}_{t+\Delta t}\) so long as
\[
\Delta x \eta a^{-}_{t+\Delta t} \leq s_{t} + \Delta t \left( q\left(a^{-}_{t+\Delta t}\right) +
r_{t+\Delta t}\right)
\]
Failure of this condition represents a breakdown of the trapeziod
approximation. This is handled by taking \(q\left(a^{+}_{t+\Delta t}\right)=0\)
and
\[
s_{t+\Delta t} = s_{t} + \Delta t \left( q\left(a^{-}_{t+\Delta t}\right) +
r_{t+\Delta t}\right)
\]
From this we see that
\[
-r \leq \frac{1}{\Delta t} s_{t} + q\left(a^{-}_{t+Δt}\right) - \frac{\Delta
x}{\Delta t} \eta a^{-}_{t+\Delta t}
\]
A more general numeric scheme is the following
process
- select trial \(\hat{s}\)
- compute \(\hat{a} = \max\left(0, \frac{1}{\Delta x \left(1-\eta\right)} \left( \hat{s} - \Delta x \eta a^{-}_{t+\Delta t}\right)\right)\)
- evaluate \(e = \hat{s} - s_{t} - \Delta t \left(q\left(a^{-}_{t+\Delta t}\right) + r_{t+\Delta t} - q\left(\hat{a}\right)\right)\)
then search to find \(e=0\)
Note that \(\frac{de}{d\hat{s}} = 1 + \frac{dq}{da}\frac{da}{d\hat{s}} > 0\) and
\(\hat{s}=0\) implies \(\hat{a} = 0\) hence \(e \leq 0\) so there is a single
solution.
Building on the generic solution in Appendix A the implicit scheme gives
\[
\left.q^{+}_{sf}\right.\rvert_{t + \Delta t} =
\frac{ 1 }{ \kappa\left(1-\eta\right) + \Delta t }
\left. s_{sf} \right.\rvert_{t} +
\frac{ \Delta t - \kappa\eta}{ \kappa\left(1-\eta\right) + \Delta t }
\left. q^{-}_{sf} \right.\rvert_{t + \Delta t} -
\frac{ \Delta t }{ \kappa\left(1-\eta\right) + \Delta t }
\left. r_{sf \rightarrow rz} \right.\rvert_{t + \Delta t}
\]
The maximum downward flux is that which minimises \(\left. s_sf \right.\rvert_{t + \Delta t}\). Since for any positive lateral inflow there
must be surface storage the maximum downward flux is that which sets
\(\left.q^{+}_{sf}\right.\rvert_{t + \Delta t} = 0\).
This provides the condition
\[
\left. r_{sf \rightarrow rz} \right.\rvert_{t + \Delta t} \leq
\frac{ 1 }{ \Delta t }
\left. s_{sf} \right.\rvert_{t} +
\frac{ \Delta t - \kappa\eta}{ \Delta t }
\left. q^{-}_{sf} \right.\rvert_{t + \Delta t}
\]
\[\begin{equation}
\left. \bar{s}_{sf} \right\rvert_{\Delta t}
=
\left. \bar{s}_{sf} \right\rvert_{0}
+ \frac{\Delta t}{\Delta x} \left(
\hat{l}_{sf}^{-} - c_{sf} \max\left(0, \left. \bar{s}_{sf}
\right\rvert_{\Delta t} - s_{raf}\right)
\right)
- \frac{\Delta t}{t_{raf}} \min\left(s_{raf},\left. \bar{s}_{sf}
\right\rvert_{\Delta t}\right)
- \Delta t \hat{r}_{sf \rightarrow rz}
\end{equation}\]
To ensure that the surface storage does not become negative
\[\begin{equation}
\hat{r}_{sf \rightarrow rz} \leq \min\left( k_{sf},
\left. \frac{1}{\Delta t}\left(\bar{s}_{sf} \right\rvert_{0} + \frac{\Delta
t}{\Delta x} \hat{l}_{sf}^{-}
\right)\right)
\end{equation}\]
The surface outflow over the time step is evaluated as
\[\begin{equation}
\hat{q}_{sf}^{+} = w^{+}c_{sf} \max\left(0, \left. \bar{s}_{sf}
\right\rvert_{\Delta t} -s_{raf}\right) +
\frac{A}{t_{raf}} \min\left(s_{raf},\left. \bar{s}_{sf}
\right\rvert_{\Delta t}\right)
\end{equation}\]
The solution for \(\left. \bar{s}_{sf} \right\rvert_{\Delta t}\) is conditional upon which side of \(s_{raf}\)
it falls.
If
\[
s_{raf} \lt
\left. \bar{s}_{sf} \right\rvert_{0}
+ \frac{\Delta t}{\Delta x} \hat{l}_{sf}^{-}
- \frac{\Delta t}{t_{raf}} s_{raf}
- \Delta t \hat{r}_{sf \rightarrow rz}
\]
then \(\left. \bar{s}_{sf} \right\rvert_{\Delta t} \gt s_{raf}\) and is given by
\[\begin{equation}
\left. \bar{s}_{sf} \right\rvert_{\Delta t}
= \left(1 + \frac{c_{sf}\Delta t}{\Delta x}\right)^{-1}
\left( \left. \bar{s}_{sf} \right\rvert_{0}
+ \frac{\Delta t}{\Delta x} \hat{l}_{sf}^{-}
+ \frac{c_{sf}\Delta t}{\Delta x} s_{raf}
- \frac{\Delta t}{t_{raf}} s_{raf}
- \Delta t \hat{r}_{sf \rightarrow rz}
\right)
\end{equation}\]
Otherwise
\[\begin{equation}
\left. \bar{s}_{sf} \right\rvert_{\Delta t}
= \left(1 + \frac{\Delta t}{t_{raf}}\right)^{-1}
\left( \left. \bar{s}_{sf} \right\rvert_{0}
+ \frac{\Delta t}{\Delta x} \hat{l}_{sf}^{-}
- \Delta t \hat{r}_{sf \rightarrow rz}
\right)
\end{equation}\]
To ensure that the surface storage does not become negative
\[\begin{equation}
\hat{r}_{sf \rightarrow rz} \leq \min\left( k_{sf},
\left. \frac{1}{\Delta t}\left(\bar{s}_{sf} \right\rvert_{0} + \frac{\Delta
t}{\Delta x} \hat{l}_{sf}^{-}
\right)\right)
\end{equation}\]
The surface outflow over the time step is evaluated as
\[\begin{equation}
\hat{q}_{sf}^{+} = w^{+}c_{sf} \left. \bar{s}_{sf} \right\rvert_{\Delta t}
\end{equation}\]
Root Zone
The implicit formulation for the root zone gives
\[\begin{equation}
\left. \bar{s}_{rz} \right.\rvert_{\Delta t} =
\left( 1 + \frac{e_{p}\Delta t}{s_{rzmax}} \right)^{-1}
\left(
\left. \bar{s}_{rz} \right.\rvert_{0}
+ \Delta t \left(
p + \hat{r}_{sf \rightarrow rz} - \hat{r}_{rz \rightarrow uz} \right)
\right)
\end{equation}\]
Since flow to the unsaturated zone occur only to keep
\(\bar{s}_{rz} \leq s_{rzmax}\) then
\[\begin{equation}
\hat{r}_{rz \rightarrow uz} \leq \max\left(0,
\frac{1}{\Delta t}\left(
\left. \bar{s}_{rz} \right.\rvert_{0}
+ \Delta t \left(
p + \hat{r}_{sf \rightarrow rz} - e_{p} \right)
- s_{rzmax}
\right)\right)
\end{equation}\]
Unsaturated Zone
The expression for the unsaturated zone is given in terms of
\(\left. \bar{s}_{uz} \right.\rvert_{\Delta t}\) and
\(\left. \bar{s}_{sz} \right.\rvert_{\Delta t}\) as
\[\begin{equation}
\left. \bar{s}_{uz} \right.\rvert_{\Delta t} =
\frac{T_{d}\left. \bar{s}_{sz} \right.\rvert_{\Delta t}}{
T_{d}\left. \bar{s}_{sz} \right.\rvert_{\Delta t} + \Delta t}
\left( \left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \hat{r}_{rz
\rightarrow uz}\right)
\end{equation}\]
Since \(\left. \bar{s}_{uz} \right.\rvert_{\Delta t} \leq \left. \bar{s}_{sz} \right.\rvert_{\Delta t}\) this places an additional condition of
\[\begin{equation}
\hat{r}_{rz \rightarrow uz} \leq \frac{1}{\Delta t} \left(
\left. \bar{s}_{sz} \right.\rvert_{\Delta t} + \frac{\Delta t}{T_d} - \left. \bar{s}_{uz} \right.\rvert_{0}\right)
\end{equation}\]
This condition also implies that \(\hat{r}_{uz \rightarrow sz} \leq \frac{1}{T_d}\)
Saturated Zone
Building on the generic solution in Appendix A the implicit scheme gives
\[
\left.q^{+}_{sz}\right.\rvert_{t + \Delta t} =
\frac{ 1 }{ \kappa\left(1-\eta\right) + \Delta t }
\left( D - \left. s_{sz} \right.\rvert_{t} \right) +
\frac{ \Delta t - \kappa\eta}{ \kappa\left(1-\eta\right) + \Delta t }
\left. q^{-}_{sz} \right.\rvert_{t + \Delta t} +
\frac{ \Delta t }{ \kappa\left(1-\eta\right) + \Delta t }
\left. r_{uz \rightarrow sz} \right.\rvert_{t + \Delta t}
\]
The maximum downward flux is that which minimises \(\left. s_sf \right.\rvert_{t + \Delta t}\). Since for any positive lateral inflow there
must be surface storage the maximum downward flux is that which sets
\(\left.q^{+}_{sf}\right.\rvert_{t + \Delta t} = 0\).
This provides the condition
\[
\left. r_{sf \rightarrow rz} \right.\rvert_{t + \Delta t} \leq
\frac{ 1 }{ \Delta t }
\left. s_{sf} \right.\rvert_{t} +
\frac{ \Delta t - \kappa\eta}{ \Delta t }
\left. q^{-}_{sf} \right.\rvert_{t + \Delta t}
\]
Initial evaluation and substitution of the flux from the unsaturated zone gives
\[\begin{equation}
\left. \bar{s}_{sz}\right.\rvert_{\Delta t} =
\left. \bar{s}_{sz}\right.\rvert_{\Delta 0} +
\frac{\Delta t}{\Delta x} \left(g\left(\left.\bar{s}_{sz}\right\rvert_{\Delta t},\Theta_{sz}\right) -
l_{sz}^{-}\right) - \Delta t \frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \hat{r}_{rz \rightarrow uz} }
{ T_{d}\left. \bar{s}_{sz} \right.\rvert_{\Delta t} + \Delta t}
\end{equation}\]
Limiting the inflow by reaching the saturation level gives
\[\begin{equation}
\hat{r}_{rz \rightarrow uz} \leq \frac{1}{\Delta t}\left(
\left. \bar{s}_{sz}\right.\rvert_{\Delta 0} +
\frac{\Delta t}{\Delta x} \left(g\left(\left.\bar{s}_{sz}\right\rvert_{\Delta t},\Theta_{sz}\right) -
l_{sz}^{-}\right) - \left. \bar{s}_{uz} \right.\rvert_{0} \right)
\end{equation}\]
Selecting \(\bar{s}_{sz}\) and \(\hat{r}_{rz \rightarrow uz}\) to satisfy the above equation describes a non-linear
problem with two unknowns a multiple solutions. A further condition, that the
downward flux is maximised, is imposed to ensure a unique solution.
Numerical Algorithm
An implementation of the approximating equations which is consistent with
maximising the downward flux produces a fully implicit numeric scheme.
The maximum downward flux from the surface is given by
\[
\check{r}_{sf \rightarrow rz} = \min\left( k_{sf},
\left. \frac{1}{\Delta t}\left(\bar{s}_{sf} \right\rvert_{0} + \frac{\Delta
t}{\Delta x} \hat{l}_{sf}^{-}
\right)\right)
\]
which results in a maximum downward flux from the root zone of
\[
\check{r}_{rz \rightarrow uz} = \max\left(0, \frac{1}{\Delta t}\left(
\left. \bar{s}_{rz} \right.\rvert_{0} + \Delta t \left( p + \check{r}_{sf
\rightarrow rz} - e_{p} \right) - s_{rzmax} \right)\right)
\]
Consider the pseudo-function
\[
f\left(z\right) =
z -
\left. \bar{s}_{sz}\right.\rvert_{0} -
\frac{\Delta t}{\Delta x} \left(g\left(z,\Theta_{sz}\right) -
l_{sz}^{-}\right) + \Delta t \min\left(\frac{1}{T_d},\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t}\right)
\]
adapted from the approximating equation for the saturated zone storage with the minima
ensuring the limit on the flux from the unsaturated zone. If \(f(0) \geq 0\) the
downward flux is enough to produce saturation in the subsurface and
\(\left. \bar{s}_{sz}\right.\rvert_{\Delta t} = 0\). If \(f(0) < 0\) then
\(\left. \bar{s}_{sz}\right.\rvert_{\Delta t}\) is given by a solution of
\(f(z) = 0\). In Appendix A it is
shown that \(f\left(z\right)\) is a monotonic increasing function of
\(z > 0\); hence if \(f\left(0\right) \leq 0\) and \(f\left(D\right) \geq 0\) a
unique solution for \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t}\) can be found.
A direct implementation of the above results takes the form of Algorithm 1.
Algorithm 1: Direct Implementation
- Compute \(\check{r}_{sf \rightarrow rz}\) and \(\check{r}_{rz \rightarrow uz}\)
- Test for saturation. If \(f\left(0\right) \geq 0\) then go to Step 3 else go to Step 4
- Saturated. Set \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t} = 0\) and
\[\hat{r}_{rz \rightarrow uz} = \frac{1}{\Delta t}\left(
\left. \bar{s}_{sz}\right.\rvert_{\Delta 0} +
\frac{\Delta t}{\Delta x} \left(g\left(0,\Theta_{sz}\right) -
l_{sz}^{-}\right) - \left. \bar{s}_{uz} \right.\rvert_{0} \right)
\]
Go to Step 5.
- Not saturated. Find \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t}\) such that
\(f\left(\left. \bar{s}_{sz}\right.\rvert_{\Delta t}\right) = 0\) and take
\[
\hat{r}_{rz \rightarrow uz} = \min\left(\check{r}_{rz \rightarrow uz},\frac{1}{\Delta t} \left(
\left. \bar{s}_{sz} \right.\rvert_{\Delta t} + \frac{\Delta t}{T_d} -
\left. \bar{s}_{uz} \right.\rvert_{0}\right)\right)\]
Go to Step 5.
- Compute
\[
\hat{r}_{sf \rightarrow rz} = \min\left(\check{r}_{sf \rightarrow
rz}, \frac{1}{\Delta t}\left( s_{rzmax} -
\left. \bar{s}_{rz} \right.\rvert_{0} - \Delta t \left( p - e_p - \hat{r}_{rz
\rightarrow uz} \right) \right)\right)\]
- Solve for \(\left. \bar{s}_{uz}\right.\rvert_{\Delta t}\),
\(\left. \bar{s}_{rz}\right.\rvert_{\Delta t}\) and
\(\left. \bar{s}_{sf}\right.\rvert_{\Delta t}\) using the computed vertical
fluxes.
The challenge with Algorithm 1 is that maintaining a water (mass) balance and
correct limits on the state values
depends upon the accuracy of the solution of \(f\left(z\right)=0\). Since
finding \(z\) requires multiple evaluations of the transmissivity profile it can
rapidly become the most expensive part of the code.
To limit the computational cost suppose that \(f\left(z\right)=0\) is solved
approximately to give \(\tilde{z}\)
such that for some numerical tolerance \(\epsilon\) we have \(\left\lvert f\left(\tilde{z}\right) \right\rvert \leq \epsilon\). Using \(\tilde{z}\) to
evaluate the outward lateral flux from the saturated zone gives the
pseudo-function
\[
h\left(z\right) =
z -
\left. \bar{s}_{sz}\right.\rvert_{0} -
\frac{\Delta t}{\Delta x} \left(g\left(\tilde{z},\Theta_{sz}\right) -
l_{sz}^{-}\right) + \Delta t \min\left(\frac{1}{T_d},\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t}\right)
\]
Since \(h\left(z\right)\) is a monotonic increasing function of
\(z > 0\) the solution for \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t}\) is
similar to that using \(f\left(z\right)\). Taking \(z \in \left[0,D\right]\) the
solution for \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t}\) is given by
- If \(h(0) \geq 0\) the inward flux is enough to produce saturation in the subsurface and
\(\left. \bar{s}_{sz}\right.\rvert_{\Delta t} = 0\).
- If \(h(z_{max}) < 0\) (possible only due to the outflow flux approximation) then
\(\left. \bar{s}_{sz}\right.\rvert_{\Delta t} = D\) and the lateral outflow is
adjusted such that
\[
\frac{\Delta t}{\Delta x} l_{sz}^{+} =
D -
\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} l_{sz}^{-} +
\Delta t \min\left(\frac{1}{T_d},\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} D + \Delta t}\right)
\]
- If \(h(0) < 0\) and If \(h(D) \geq 0\) then with \(z_{C} = \left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t\left( \check{r}_{rz \rightarrow uz} - \frac{1}{T_{d}}\right)\)
- If \(h\left( z_{C} \right) \geq 0\) then the inflow from the unsaturated
zone is \(1/T_{d}\) and
\[
\left. \bar{s}_{sz}\right.\rvert_{\Delta t} =
\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} \left(g\left(\tilde{z},\Theta_{sz}\right) -
l_{sz}^{-}\right) - \frac{\Delta t}{T_d}
\]
- else (see Appendix C)
\[
z = \frac{1}{2}\left(-b + \sqrt{ b^2 - 4c }\right) \\
= \frac{1}{2} \left( z_{C} - h\left(z_{C}\right) + \sqrt{
\left(h\left(z_{C}\right)-z_{C}\right)^2 - 4 \frac{\Delta t}{T_{d}}
h\left(z_{C}\right) } \right)
\]
The implementation of this approach is presented in Algorithm 2. it can be seem
that the exchange for decreasing the accuracy of the solution of the zero
finding problem is a potential to evaluate a power and a square root.
Algorithm 2: Stable Implementation
- Compute \(\check{r}_{sf \rightarrow rz}\) and \(\check{r}_{rz \rightarrow uz}\)
- Compute the lateral outflow
- If \(f\left(0\right) \geq 0\) then \(l_{sz}^{+} = l_{szmax}\)
- else find \(\tilde{z}\) such that \(\left\lvert f\left(\tilde{z}\right) \right\rvert \leq \epsilon\). Set \(l_{sz}^{+} = g\left(\tilde{z},\Theta\right)\)
- Solve for \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t}\)
- If \(h(0) \geq 0\) then
\(\left. \bar{s}_{sz}\right.\rvert_{\Delta t} = 0\).
- If \(h(z_{max}) < 0\) then \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t} = D\) and
\[
\frac{\Delta t}{\Delta x} l_{sz}^{+} =
D -
\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} l_{sz}^{-} +
\Delta t \min\left(\frac{1}{T_d},\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} D + \Delta t}\right)
\]
- else with \(z_{C} = \left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t\left( \check{r}_{rz \rightarrow uz} - \frac{1}{T_{d}}\right)\)
- If \(h\left( z_{C} \right) \geq 0\) then
\[
\left. \bar{s}_{sz}\right.\rvert_{\Delta t} =
\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} \left(g\left(\tilde{z},\Theta_{sz}\right) -
l_{sz}^{-}\right) - \frac{\Delta t}{T_d}
\]
- else
\[
\left. \bar{s}_{sz}\right.\rvert_{\Delta t} =
= \frac{1}{2} \left( z_{C} - h\left(z_{C}\right) + \sqrt{
\left(h\left(z_{C}\right)-z_{C}\right)^2 - 4 \frac{\Delta t}{T_{d}}
h\left(z_{C}\right) } \right)
\]
- Solve for \(\hat{r}_{rz \rightarrow uz}\)
- If \(\left. \bar{s}_{sz}\right.\rvert_{\Delta t} = 0\) then
\[\hat{r}_{rz \rightarrow uz} = \frac{1}{\Delta t}\left(
\left. \bar{s}_{sz}\right.\rvert_{\Delta 0} +
\frac{\Delta t}{\Delta x} \left(g\left(0,\Theta_{sz}\right) -
l_{sz}^{-}\right) - \left. \bar{s}_{uz} \right.\rvert_{0} \right)
\]
- else
\[
\hat{r}_{rz \rightarrow uz} = \min\left(\check{r}_{rz \rightarrow uz},\frac{1}{\Delta t} \left(
\left. \bar{s}_{sz} \right.\rvert_{\Delta t} + \frac{\Delta t}{T_d} -
\left. \bar{s}_{uz} \right.\rvert_{0}\right)\right)\]
- Compute
\[
\hat{r}_{sf \rightarrow rz} = \min\left(\check{r}_{sf \rightarrow
rz}, \frac{1}{\Delta t}\left( s_{rzmax} -
\left. \bar{s}_{rz} \right.\rvert_{0} - \Delta t \left( p - e_p - \hat{r}_{rz
\rightarrow uz} \right) \right)\right)\]
- Solve for \(\left. \bar{s}_{uz}\right.\rvert_{\Delta t}\),
\(\left. \bar{s}_{rz}\right.\rvert_{\Delta t}\) and
\(\left. \bar{s}_{sf}\right.\rvert_{\Delta t}\) using the computed vertical
fluxes.
Transmissivity Profiles
The solution and numerical scheme have been presented with a general
transmissivity function \(g\left(\bar{s}_{sz},\Theta_{sz}\right)\). Table
3 present the transmissivity profiles present as options within the dynatop
package, the corresponding value to use for the transmissivity_profile value
in the model options vector and the additional parameters or properties
required. The additional parameter and properties are defined in Table 4.
Table 3: Transmissivity profiles available with the
dynatop package.
| Exponential |
\(T_{0}\sin\left(\beta\right)\exp\left(-\frac{\cos\beta}{m}s_{sz}\right)\) |
\(\frac{\cos\beta}{m}g\left(\bar{s}_{sz},\Theta_{sz}\right)\) |
\(T_0,\beta,m\) |
"exp" |
Originally given in Beven & Freer 2001 |
| Bounded Exponential |
\(T_{0}\sin\left(\beta\right)\left(\exp\left(-\frac{\cos\beta}{m}s_{sz}\right) - \exp\left(-\frac{\cos\beta}{m}D\right)\right)\) |
\(\frac{cos \beta}{m}T_{0}\sin\left(\beta\right)\exp\left(-\frac{\cos\beta}{m}s_{sz}\right)\) |
\(T_0,\beta,m,D\) |
"bexp" |
Originally given in Beven & Freer 2001 |
| Constant Celerity |
\(c_{sz}\left(D-s_{sz}\right)\) |
\(c_{sz}\) |
\(c_{sz},D\) |
"cnst" |
|
| Double Exponential |
\(T_{0}\sin\left(\beta\right)\left( \omega\exp\left(-\frac{\cos\beta}{m}s_{sz}\right) + \left(1-\omega\right)\exp\left(\frac{\cos\beta}{m_2}s_{sz}\right)\right)\) |
\(\omega\frac{cos \beta}{m}T_{0}\sin\left(\beta\right)\exp\left(-\frac{\cos\beta}{m}s_{sz}\right) + \left(1-\omega \right)\frac{cos \beta}{m_2}T_{0}\sin\left(\beta\right)\exp\left(-\frac{\cos\beta}{m_2}s_{sz}\right)\) |
\(T_0,\beta,m,m_2,\omega\) |
“dexp” |
|
Table 4: Additional parameters used in the transmissivity profiles.
| \(T_0\) |
Transmissivity at saturation |
m\(^2\)/s |
| \(m\),\(m_2\) |
Exponential transmissivity constants |
m\(^{-1}\) |
| \(\beta\) |
Angle of hill slope |
rad |
| \(c_{sz}\) |
Saturated zone celerity |
m/s |
| \(D\) |
Depth at which zero lateral flow occurs |
m |
| \(\omega\) |
weighting parameter |
- |
test appendix VPMC v2
Muskingham routing can be derived in a numer of ways. Consistent with the
wedge derivation the evolution of the storage \(s\) in a reach of length \(\Delta x\) can be expressed in terms of the inflow \(q^{-}\)
and outflow \(q^{+}\) and state or time varying parameter \(\nu\) and \(\eta\) as
\[
s = \frac{\Delta x}{\nu} \left(\eta q^{-} + \left(1-\eta\right) q^{+} \right)
\]
Let the representave area \(a\) be given by \(a=\frac{s}{\Delta x}\). If \(q = \eta q^{-} + \left(1-\eta\right) q^{+}\) is a representative
flow then \(\nu\) is the representative velocity since \(\nu=\frac{q}{a}\).
Below we show that in many hydrological settings that \(\eta\) and \(q\) are functions of
\(a\) and hence \(s\). In the following we therfore consider that
\[
q^{+} = \mathcal{F}\left(s,q^{-}\right)
\]
where
\[
\frac{\partial q^{+}}{\partial s} = \frac{\partial}{\partial
s}\mathcal{F}\left(s,q^{-}\right) \geq 0
\]
and to maintain positive flows the value of \(s\) is bounded below by \(s^{min}\)
such that
\[
0 = \mathcal{F}\left(s^{min},q^{-}\right)
\]
Taking flux to be a function of cross sectional area and lateral inflow \(r\)
gives the mass balance
\[
\frac{ds}{dt} = q^{-} + r - q^{+}
\]
If the inputs an values at time \(t\) are known the mass error for a four point solution with
time weighting parameter \(\omega\) is given by
\[
\mathcal{E}\left(s_{t+\Delta t},\omega\right) =
s_{t+\Delta t} +
\Delta t \left( \omega q^{+}_{t+\Delta t} + \left(1-\omega\right)q^{+}_{t}
\right) -
s_{t} - v_{t+\Delta t}^{-} - v_{t+\Delta t}^{r}
\]
where the ouflow volume \(v^{+}_{t + \Delta t}\) is given by
\[
v^{+}_{t + \Delta t} =
\Delta t \left( \omega
q^{+}_{t+\Delta t} + \left(1-\omega\right)q^{+}_{t} \right)
\]
to solve for the storage at time \(t+\Delta t\) we seek the value of
\(s_{t+\Delta t}\) for which
\(\mathcal{E}\left(s_{t+\Delta t},\omega\right) = 0\).
Since
\[
\frac{\partial}{\partial
s}\mathcal{E}\left(s,\omega\right) > 0
\]
a solution can be found so long as \(\mathcal{E}\left(s^{min},\omega\right) \leq 0\)
The solution is numerically stable so long as \(0.5 \leq \omega \leq 1\) with second order accuray optained when $= 0.5. Since increasing
\(\omega\) adds numerical dispersion the solution uses the following cases:
- If \(\mathcal{E}\left(s^{min},0.5\right) \leq 0\) find \(s_{t+\Delta t}\) by
numerical search
- If \(\mathcal{E}\left(s^{min},0.5\right) > 0\) and
\(\mathcal{E}\left(s^{min},1\right) \leq 0\) set \(s_{t+\Delta t}=s^{min}\) and
compute
\[
\omega =
\frac{ s^{min} + \Delta t q^{+}_{t} - s_{t} - v_{t+\Delta t}^{-} -
v_{t+\Delta t}^{r}}{\Delta t q^{+}_{t}}
\]
- If \(\mathcal{E}\left(s^{min},1\right) > 0\) take \(\omega = 1\), \(q_{t+\Delta t}=0\) and \(s_{t+\Delta t} = s_{t} + v_{t+\Delta t}^{-} + v_{t+\Delta t}^{r}\)
Case 3 represents the breakdown of the trapesoid assumption, resulting from a
numeric shock to the system. The
solution outlines offers a “smooth” mass conservative way of handling this,
although it does not avoid suprious osscilations.
Flux limiters for TVD
To avoid osscilations flux limiters can be proposed based upon the TVD
principle (ref), specifically we constrain the solution such that \(q^{l}\leq q_{t+\Delta t} \leq q^{u}\) where \(q^{l}\) and \(q^{u}\) are based on the known
values at the start of the time step.
\[
q^{l} = \max\left(0, \min\left(q^{+}_{t},q^{-}_{t},q^{-}_{t+1}\right)
+\min\left(v^{r}_{t+\Delta t}/\Delta t, 0\right)\right)
\]
\[
q^{u} = \max\left(q^{+}_{t},q^{-}_{t},q^{-}_{t+1}\right)
+\max\left(v^{r}_{t+\Delta t}/\Delta t, 0\right)
\]
Corresponding storages
\[
q^{l} = \mathcal{F}\left(s^{l},q^{-}\right)
\]
\[
q^{u} = \mathcal{F}\left(s^{u},q^{-}\right)
\]
Solve with multple cases
- If \(\mathcal{E}\left(s^{u},1\right) \leq 0\) let \(q_{t+\Delta t}=q^{u}\), \(\omega = 1\) and take
\[
s_{t+\Delta t} = s_{t} + v_{t+\Delta t}^{-} + v_{t+\Delta t}^{r} - \Delta t q^{u}
\]
- If \(\mathcal{E}\left(s^{u},0.5\right) \leq 0\) and
\(\mathcal{E}\left(s^{u},1\right) > 0\) let
\(q_{t+\Delta t}=q^{u}\),
\(s_{t+\Delta t}=s^{u}\) and compute
\[
\omega =
\frac{ s^{u} + \Delta t q^{+}_{t} - s_{t} - v_{t+\Delta t}^{-} -
v_{t+\Delta t}^{r}}
{\Delta t \left( q^{+}_{t} - q^{u}\right)}
\]
- If \(\mathcal{E}\left(s^{l},0.5\right) < 0\) and
\(\mathcal{E}\left(s^{u},0.5\right) > 0\) find \(s_{t+\Delta t}\) by
numerical search
- If \(\mathcal{E}\left(s^{l},0.5\right) \geq 0\) and
\(\mathcal{E}\left(s^{l},1\right) \leq 0\) set \(s_{t+\Delta t}=s^{l}\) and
compute
\[
\omega =
\frac{ s^{l} + \Delta t q^{+}_{t} - s_{t} - v_{t+\Delta t}^{-} - v_{t+\Delta t}^{r}}{q^{+}_{t}}
\]
- If \(\mathcal{E}\left(s^{l},1\right) > 0\) take \(\omega = 1\), \(q_{t+\Delta t}=q^{l}\) and \(s_{t+\Delta t} = s_{t} + v_{t+\Delta t}^{-} + v_{t+\Delta t}^{r} - \Delta t q^{l}\)
Appendix A - Variable Parameter Muskingham solution
Muskingham routing can be derived in a numer of ways. Consistent with the
wedge derivation the evolution of the storage \(s\) in a reach of length \(\Delta x\) can be expressed in terms of the cross sectional area of the inflow \(a^{-}\)
and outflow \(a^{+}\) and possibly state or time varying parameter \(\eta\) as
\[
s = \Delta x \left(\eta a^{-} + \left(1-\eta\right) a^{+} \right)
\]
Taking flux to be a function of cross sectional area and lateral inflow \(r\)
gives the mass balance
\[
\frac{ds}{dt} = q\left(a^{-}\right) + r - q\left(a^{+}\right)
\]
Approximating Tank Models
A tank model with possibly varying time constant \(T\) has
\[
q\left(a^{+}\right) = \frac{1}{T}s
\].
This is a special case of the solution with \(\eta=0\) so that
\(s=a^{+} \Delta x\)
Approximating Diffusion Wave Routing
Diffuse routing with lateral inflow can be expressed as the parabolic
equation
\[
\frac{dq}{dt} + c\frac{dq}{dx} - D \frac{d^2 q}{dx^2} = cl - D \frac{dl}{dx}
\]
where \(l\) is lateral inflow per unit length. Simplify this by
considering that the lateral inflow \(r\) uniformly
distributed so that \(l=\frac{r}{\Delta x}\) and \(\frac{dl}{dx}=0\) to give
\[
\frac{dq}{dt} + c\frac{dq}{dx} - D \frac{d^2 q}{dx^2} = cl
\]
Let us relate this to the Muskingham Solution using the relationship
\[
a = \eta a^{-} + \left(1-\eta\right) a^{+}
\]
Taking Taylor series expansions for \(q\left(a^{-}\right)\) and
\(q\left(a^{+}\right)\) based on \(a\) gives
\[
q\left(a^{+}\right) \approx q\left(a\right) + \eta \Delta x \frac{dq\left(a\right)}{dx} + \frac{1}{2}\eta^2\Delta x^2 \frac{d^2 q\left(a\right)}{dx^2}
\]
and
\[
q\left(a^{-}\right) \approx q\left(a\right) - \left(1-\eta\right) \Delta x \frac{dq\left(a\right)}{dx} + \frac{1}{2}\left(1-\eta\right)^2 \Delta x^2 \frac{d^2 q\left(a\right)}{dx^2}
\]
Subtracting the expression for \(q\left(a^{-}\right)\) from that for \(q\left(a^{+}\right)\) gives
\[
q\left(a^{-}\right) - q\left(a^{+}\right) \approx
-\Delta x \frac{dq\left(a\right)}{dx} +
\frac{1}{2}\left(1-2\eta\right)^2\Delta x^2 \frac{d^2 q\left(a\right)}{dx^2}
\]
From the mass balance condition
\[
\frac{ds}{dt} \approx \Delta x l -\Delta x \frac{dq\left(a\right)}{dx} +
\frac{1}{2}\left(1-2\eta\right)\Delta x^2 \frac{d^2 q\left(a\right)}{dx^2}
\]
Using the expression for storage and relationship between flow and area
\[
\frac{dq}{dt} = \frac{dq}{da} \frac{da}{dt} = \frac{c}{\Delta x} \frac{ds}{dt}
\]
Then combinign this with the approximation produces
\[
\frac{dq\left(a\right)}{dt} + c\frac{dq\left(a\right)}{dx} -
\frac{c}{2}\left(1-2\eta\right)\Delta x \frac{d^2 q\left(a\right)}{dx^2}
\approx c l
\]
Comparision to the orginal diffuse routing equation shows that
\[
\eta = \frac{1}{2} - \frac{D}{c \Delta x}
\]
The value of \(\eta\) is limited to between 0 and \(\frac{1}{2}\). Firstly \(\eta = 1/2\) results from \(D=0\); that is the Kinematic wave equation. Taking \(\eta=0\)
produces the maxium dispersion of \(D = \frac{c\Delta x}{2}\), which is given by
the linear tank.
Numerical Solution
SEARCH WITH \(a^{+}\)
Consider first the case with:
- a known constant value of \(\eta\)
- a known positive inflow \(q\left(a^{-}_{t+\Delta t}\right)\) giving a value of
\(a^{-}_{t+\Delta t}\)
- a known function \(q\) such that \(\frac{dq\left(a\right)}{da}\geq 0\) and
\(q\left(0\right)=0\)
An implicit
solution gives the joint equations
\[
s_{t+\Delta t} = \Delta x \left(\eta a^{-}_{t+\Delta t} + \left(1-\eta\right)
a^{+}_{t+\Delta t} \right)
\]
and
\[
s_{t+\Delta t} = s_{t} + \Delta t \left( q\left(a^{-}_{t+\Delta t}\right) +
r_{t+\Delta t} - q\left(a^{+}_{t+\Delta t}\right) \right)
\]
The unknown outflow crossection satisfies \(a^{+}_{t+\Delta t} \geq 0\). The first
expression for \(s_{t+\Delta t}\) is strictly increasing with respect to
\(a^{+}_{t+\Delta t}\) while the second is decreasing, hence there is a unique
solution for \(a^{+}_{t+\Delta t}\) so long as
\[
\Delta x \eta a^{-}_{t+\Delta t} \leq s_{t} + \Delta t \left( q\left(a^{-}_{t+\Delta t}\right) +
r_{t+\Delta t}\right)
\]
Failure of this condition represents a breakdown of the trapeziod
approximation. This is handled by taking \(q\left(a^{+}_{t+\Delta t}\right)=0\)
and
\[
s_{t+\Delta t} = s_{t} + \Delta t \left( q\left(a^{-}_{t+\Delta t}\right) +
r_{t+\Delta t}\right)
\]
From this we see that
\[
-r \leq \frac{1}{\Delta t} s_{t} + q\left(a^{-}_{t+Δt}\right) - \frac{\Delta
x}{\Delta t} \eta a^{-}_{t+\Delta t}
\]
A more general numeric scheme is the following
process
- select trial \(\hat{s}\)
- compute \(\hat{a} = \max\left(0, \frac{1}{\Delta x \left(1-\eta\right)} \left( \hat{s} - \Delta x \eta a^{-}_{t+\Delta t}\right)\right)\)
- evaluate \(e = \hat{s} - s_{t} - \Delta t \left(q\left(a^{-}_{t+\Delta t}\right) + r_{t+\Delta t} - q\left(\hat{a}\right)\right)\)
then search to find \(e=0\)
Note that \(\frac{de}{d\hat{s}} = 1 + \frac{dq}{da}\frac{da}{d\hat{s}} > 0\) and
\(\hat{s}=0\) implies \(\hat{a} = 0\) hence \(e \leq 0\) so there is a single
solution.
Solution for saturated zone
Buildin on the above let the storage deficit be \(z\) and area deficit be \(v\) is
is simple to see that
solution gives the joint equations
\[
z_{t+\Delta t} = \Delta x \left(\eta v^{-}_{t+\Delta t} + \left(1-\eta\right)
v^{+}_{t+\Delta t} \right)
\]
and
\[
z_{t+\Delta t} = z_{t} - \Delta t \left( q\left(v^{-}_{t+\Delta t}\right) +
r_{t+\Delta t} - q\left(v^{+}_{t+\Delta t}\right) \right)
\]
where now \(q\left(v\right) \rightarrow 0\) as \(v \rightarrow \infty\) and
\(\frac{dq}{dv} \leq 0\).
Subsituting in the expression
\[
r_{t+\Delta{ t}} = \min\left( \frac{1}{T_{d}}, \frac{U}{T_{d}z_{t+\Delta t} +
\Delta t} \right)
\]
Gives a similar numeric scheme to that previously
- select trial \(\hat{q}\)
- compute \(\hat{v} = q^{-1}\left(\hat{q}\right)\)
- compute \(hat{z} = \Delta x \left(\eta v^{-}_{t+\Delta t} + \left(1-\eta\right) \hat{v} \right)\)
- compute \(\hat{r} = \min\left( \frac{1}{T_{d}}, \frac{U}{T_{d}\hat{z} + \Delta t} \right)\)
- evaluate \(e = \hat{z} - z_{t} +\Delta t \left(q\left(a^{-}_{t+\Delta t}\right) + \hat{r} - q\left(\hat{v}\right)\right)\)
and search for \(e=0\).
The derivative \(\frac{de}{d\hat{z}} = 1 + \frac{d\hat{r}}{dz - }\frac{dq}{dv}\frac{dv}{d\hat{z}} > 0\). TODO based on proof below
For a unique solution \(e \leq 0\) for \(hat{s}=0\). if this is not the case need
to limit \(\hat{r}\)
[
e = 0 - z_{t} +t (q(a^{-}_{t+t}) +
)$
and
\(\hat{s}=0\) implies \(\hat{a} = 0\) hence \(e \leq 0\) so there is a single
solution.
meaning that
\[
s_{t+\Delta t} \geq \Delta x \eta a^{-}_{t+\Delta t}
\]
Letting \(\frac{dq\left(a\right)}{da}\geq 0\) and \(q\left(0\right)=0\) gives
\[
s_{t} + \Delta t \left( q\left(a^{-}_{t+\Delta t}\right) +
r_{t+\Delta t} \right)
\geq \Delta x \eta a^{-}_{t+\Delta t}
\]
Assuming \(\eta\) and \(\kappa\) are constant over the time window \(\Delta t\) this
gives the solution
\[
\left. s \right.\rvert_{t + \Delta t} =
\frac{ \kappa\left(1-\eta\right) }{ \kappa\left(1-\eta\right) + \Delta t }
\left( \left. s \right.\rvert_{t} + \frac{\Delta t}{1-\eta} \left. q^{-}
\right.\rvert_{t + \Delta t} + \Delta t \left. r
\right.\rvert_{t + \Delta t} \right)
\]
By derivation this scheme is mass conservative. The final state is positive when the bracket term is positive. The outflow is given by
\[
\Delta t \left.q^{+}\right.\rvert_{t + \Delta t} =
\left. s \right.\rvert_{t} +
\Delta t \left. q^{-} \right.\rvert_{t + \Delta t} +
\Delta t \left. r \right.\rvert_{t + \Delta t} -
\left. s \right.\rvert_{t + \Delta t}
\]
Subsituting in the expression for \(\left. s \right.\rvert_{t + \Delta t}\)
gives
\[
\left.q^{+}\right.\rvert_{t + \Delta t} =
\frac{ 1 }{ \kappa\left(1-\eta\right) + \Delta t }
\left. s \right.\rvert_{t} +
\frac{ \Delta t - \kappa\eta}{ \kappa\left(1-\eta\right) + \Delta t }
\left. q^{-} \right.\rvert_{t + \Delta t} +
\frac{ \Delta t }{ \kappa\left(1-\eta\right) + \Delta t }
\left. r \right.\rvert_{t + \Delta t}
\]
Further manipulations indicates that \(\left.q^{+}\right.\rvert_{t + \Delta t} \geq 0\) requires \(\Delta t \geq \kappa\eta\). TODO - show these
Appendix A - Monotonicity of \(f\left(z\right)\)
By definition \(\left. \bar{s}_{uz} \right.\rvert_{0}\), \(\Delta t\), \(\Delta X\),
\(T_d\), \(\check{r}_{rz \rightarrow uz}\) and \(z\) are all greater or equal to 0. Rewrite the expression for \(f\left(z\right)\) as
\[
f\left(z\right) = \left\{ \begin{array}{cl}
z -
\left. \bar{s}_{sz}\right.\rvert_{\Delta 0} -
\frac{\Delta t}{\Delta x} \left(g\left(z,\Theta_{sz}\right) -
l_{sz}^{-}\right) + \frac{\Delta t}{T_d} &
\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} \geq \frac{1}{T_d} \\
z -
\left. \bar{s}_{sz}\right.\rvert_{\Delta 0} -
\frac{\Delta t}{\Delta x} \left(g\left(z,\Theta_{sz}\right) -
l_{sz}^{-}\right) + \Delta t \frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} &
\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} < \frac{1}{T_d}
\end{array}\right.
\]
Differentiating with respect to \(z\) gives
\[
\frac{d}{dz} f\left(z\right) = \left\{ \begin{array}{cl}
1 -
\frac{\Delta t}{\Delta x} \frac{d}{dz}g\left(z,\Theta_{sz}\right) &
\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} \geq \frac{1}{T_d} \\
1 -
\frac{\Delta t}{\Delta x} \frac{d}{dz}g\left(z,\Theta_{sz}\right)
- T_d \Delta t \frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ \left(T_{d} z + \Delta t\right)^2} &
\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} < \frac{1}{T_d}
\end{array}\right.
\]
The definition of the function \(g\) states that
\[
-\frac{d}{dz}g\left(z,\Theta_{sz}\right) \geq 0
\]
Using this relationship and the expression for the range of \(z\) gives
\[
\frac{d}{dz} f\left(z\right) \geq \left\{ \begin{array}{cl}
1 &
\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} \geq \frac{1}{T_d} \\
1 - \frac{\Delta t}{T_{d} z + \Delta t} &
\frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} < \frac{1}{T_d}
\end{array}\right.
\]
From this it can be seen that \(\frac{d}{dz} f\left(z\right) \geq 0\) with
equality possible when \(z=0\).
Appendix B - Comparison to the Infinitesimal slab derivation
Surface Zone
A similar expression for the evolution can be derived by considering the
infinitesimal slab at the downslope end of the HRU. Taking a discrete
approximation for the change of flow with length the governing ODE can be written as
\[\begin{equation}
\frac{dw^{+}s_{sf}^{+}}{dt} = \frac{q_{sf}^{-} - q_{sf}^{+}}{\Delta x} - w^{+}r_{sf \rightarrow rz}^{+}
\end{equation}\]
Dividing through by \(w^{+}\) gives
\[\begin{equation}
\frac{ds_{sf}^{+}}{dt} = \frac{q_{sf}^{-}/w^{+} - l_{sf}^{+}}{\Delta x} - r_{sf \rightarrow rz}^{+}
\end{equation}\]
Note the similarities between this and the finite volume solution. In fact if
\(r_{sf \rightarrow rz}\) is taken to be spatially uniform and
\(\Delta x = A/w^{+}\) then this solution for \(s_{sf}^{+}\) will evolve in a similar fashion to
\(\bar{s}_{sf}\)
Root Zone
The same formulation follows for the infinitesimal slab at the bottom of the
hillslope with the averaged terms (using \(\bar{}\)) replaced with those valid
at the bottom of the hillslope (using \(^{+}\)).
Unsaturated Zone
The same formulation follows for the infinitesimal slab at the bottom of the
hillslope with the averaged terms (using \(\bar{}\)) replaced with those valid
at the bottom of the hillslope (using \(^{+}\)).
Saturated Zone
Evaluating for the foot of the hillslope and taking a discrete approximation
of the spatial gradient gives
\[\begin{equation}
\frac{ds_{sz}^{+}}{dt} = \frac{l_{sz}^{+} - q_{sz}^{-}/w^{+}}{\Delta x} - r_{uz \rightarrow sz}^{+}
\end{equation}\]
As with the surface zone we see the similarity in form, indicating that,
subject to the evaluation of the temporal integral, the two formulations
should return the same results when \(\Delta x = A/w^{+}\)
Appendix C Solution to \(h\left(z\right)\) when \(h(0) < 0\) and \(h(D) \geq 0\)
In this case we see that \(0 < z < D\). Taking \(z_{C} = \left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t\left( \check{r}_{rz \rightarrow uz} - \frac{1}{T_{d}}\right)\) the function \(h\) can
be rewritten as
\[
h\left(z\right) = \left\{ \begin{array}{cl}
z -
\left. \bar{s}_{sz}\right.\rvert_{\Delta 0} -
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) + \frac{\Delta t}{T_d}
& z_{C} \geq z \\
z -
\left. \bar{s}_{sz}\right.\rvert_{0} -
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) + \Delta t \frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t}
& z_{C} < z
\end{array} \right.
\]
Using the monotonicity of \(h\left(z\right)\) we solve each part
separately. Firstly if \(h\left( z_{C} \right) \geq 0\) then the inflow from the unsaturated
zone is \(1/T_{d}\) and
\[
\left. \bar{s}_{sz}\right.\rvert_{\Delta t} =
\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} \left(g\left(\tilde{z},\Theta_{sz}\right) -
l_{sz}^{-}\right) - \frac{\Delta t}{T_d}
\]
if \(h\left( z_{C} \right) < 0\) the value of \(z\) satisfies
\[
z -
\left. \bar{s}_{sz}\right.\rvert_{0} -
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) + \Delta t \frac{
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz} }
{ T_{d} z + \Delta t} =
0
\]
Rearrangement to a quadratic gives
\[
0 = T_{d} z^2 - T_{d} z \left(\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) \right) +
\Delta t z -
\Delta t \left(\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) \right)
+
\Delta t \left(
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz \rightarrow uz}
\right)
\]
\[
0 = z^2 - z \left(\left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) - \frac{\Delta t}{T_{d}}\right) +
\frac{\Delta t}{T_{d}} \left(
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz
\rightarrow uz} -
\left. \bar{s}_{sz}\right.\rvert_{0} -
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) \right)
\]
The commonly quoted solution for the quadratic
\[
0 = z^2 + b z +c
\]
is
\[
z = \frac{-b \pm\sqrt{ b^2 - 4c }}{2}
\]
Comparing terms we see that
\[
-b = \left. \bar{s}_{sz}\right.\rvert_{0} +
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) - \frac{\Delta t}{T_{d}}
\]
and
\[
c =
\frac{\Delta t}{T_{d}} \left(
\left. \bar{s}_{uz} \right.\rvert_{0} + \Delta t \check{r}_{rz
\rightarrow uz} -
\left. \bar{s}_{sz}\right.\rvert_{0} -
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right) \right)
\]
Since \(-b = z_{C} - h\left(z_{C}\right)\) gives \(-b > 0\).
Since \(z > z_{C}\) the subsurface (\(s_{uz}\) and
\(s_{sz}\)) does not fill so
\[
\left. \bar{s}_{sz}\right.\rvert_{0} - \left. \bar{s}_{uz}
\right.\rvert_{0} \gt
\Delta t \check{r}_{rz \rightarrow uz} -
\frac{\Delta t}{\Delta x} \left(l_{sz}^{+} -
l_{sz}^{-}\right)
\]
and \(c < 0\)
Combining these conditions results in
\(\sqrt{b^2 - 4c} > \left\lvert b \right\rvert\). Since \(z>0\) only one root of
the quadratic is valid (as always positive) so
\[
z = \frac{1}{2}\left(-b + \sqrt{ b^2 - 4c }\right)
\]
