## 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 the 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. Figure 1: Schematic of the Hill slope HRU

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
Quantity type Symbol Description unit
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 :-:

## Variable Parameter Muskingham solution

The solution of the surface and saturated zone, is based upon a variable parameter Muskingham solution. This section outlines the solution and shows how it may be used to capture a range of conceptual models.

Muskingham routing can be derived in a number of ways. The storage $$s$$ in a reach or hillslope of length $$\Delta x$$ can be expressed in terms of the representative flow $$q$$ and velocity $$v$$ as $s = \frac{q \Delta x}{v}$ Relating the representative flow to the inflow $$q^{-}$$ and outflow $$q^{+}$$ with the possibly state or time varying parameter $$\eta$$ results in $s = \frac{\Delta x}{v} \left(\eta q^{-} + \left(1-\eta\right) q^{+} \right)$ Taking vertical inflow $$r$$ gives the mass balance $\frac{ds}{dt} = q^{-} + r - q^{+}$ where by substitution $\frac{ds}{dt} = q^{-} + r - \frac{1}{1-\eta}\left(\frac{sv}{\Delta x} - \eta q^{-} \right)$

### Approximating Tank Models

Taking $$\eta=0$$ gives $\frac{ds}{dt} = q^{-} + r - \frac{sv}{\Delta x}$ which is the equation of a tank model with possibly varying time constant $$T = \frac{\Delta x}{v}$$

### Approximating Diffusion Wave Routing

Diffuse routing with lateral inflow can be expressed as the parabolic equation (see for example doi:10.1016/j.advwatres.2005.08.008)

$\frac{dq}{dt} + c\frac{dq}{dx} - D \frac{d^2 q}{dx^2} = cl - D \frac{dl}{dx}$ 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 $q = \eta q^{-} + \left(1-\eta\right) q^{+}$

Taking Taylor series expansions for $$q^{-}$$ and $$q^{+}$$ based on $$q$$ gives

$q^{+} \approx q + \eta \Delta x \frac{dq}{dx} + \frac{1}{2}\eta^2\Delta x^2 \frac{d^2 q}{dx^2}$ and $q^{-} \approx q - \left(1-\eta\right) \Delta x \frac{dq}{dx} + \frac{1}{2}\left(1-\eta\right)^2 \Delta x^2 \frac{d^2 q}{dx^2}$

Subtracting the expression for $$q^{-}$$ from that for $$q^{+}$$ gives

$q^{-} - q^{+} \approx -\Delta x \frac{dq}{dx} + \frac{1}{2}\left(1-2\eta\right) \Delta x^2 \frac{d^2 q}{dx^2}$

From the mass balance condition $\frac{ds}{dt} \approx \Delta x l -\Delta x \frac{dq}{dx} + \frac{1}{2}\left(1-2\eta\right)\Delta x^2 \frac{d^2 q}{dx^2}$

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 combining this with the approximation produces $\frac{dq}{dt} + c\frac{dq}{dx} - \frac{c}{2}\left(1-2\eta\right)\Delta x \frac{d^2 q}{dx^2} \approx c l$

Comparison to the original diffuse routing equation shows that $\eta \approx \frac{1}{2} - \frac{D}{c \Delta x}$

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 maximum dispersion of $$D = \frac{c\Delta x}{2}$$, which is given by the linear tank.

### Numerical scheme

The numerical scheme is based on the implicit solution of

$\frac{ds}{dt} = q^{-} + r - \frac{1}{1-\eta}\left(\frac{sv}{\Delta x} - \eta q^{-} \right)$

given by $s_{t+\Delta t} = s_{t} + \Delta t \left( q^{-}_{t+\Delta t} + r_{t+\Delta t} - \frac{1}{1-\eta_{t+\Delta t}}\max\left(0,q_{t+\Delta t} - \eta_{t} q^{-}_{t+\Delta t} \right) \right)$

where the maximum ensures that $$q^{+}_{t+\Delta t}$$ does not become negative. The implementation outlines is equivalent to the addition of dispersion to ensure non-negative outflows.

For known $$\eta_{t + \Delta t}$$ it is clear that the left hand side of the expression is increasing and the right hand side decreasing so long as $$\frac{dq}{ds} = \frac{1}{\Delta x}\frac{dq}{da} \geq 0$$. If $$\eta_{t+ \Delta t}$$ is a function of $$s_{t+ \Delta t}$$ then it is required that $\frac{1}{\left(1-\eta\right)^{2}}\left(\frac{d\eta}{ds}\left(q-q^{-}\right) + \left(1-\eta\right)\frac{dq}{ds}\right) \geq 0$ A simpler, more general, semi-implicit solution is implemented in the code where $$\eta_{t + \Delta t}$$ is taken to be a function of $$s_{t}$$, hence a known value.

## Finite Volume Formulation

In this section a finite volume formulation of the Dynamic TOPMODEL equations is derived for the evolution of the states over the time step $$t$$ to $$t + \Delta t$$. The lateral flux rates are considered at the start and end of the time step. The vertical flux rates are considered as constant throughout the time step, allowing for a semi-analytical solution of some storage zones.

### 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 given by the Muskingham formulation where $\mathcal{F}\left(s_{sf},\Theta_{sf}\right) = q_{sf} = \eta_{sf}q_{sf}^{-} + \left(1-\eta_{sf}\right)q_{sf}^{+}$ and $\mathcal{F}_{\eta}\left(s_{sf},\Theta_{sf}\right) = \eta_{sf}$

The different formulation available in the model are given in Table @ref{tbl:surface} with their parameters outlined in Table @ref{tbl:surface-param}

(#tbl:surface) Outline formulations for the storage-flow relationship in the surface zone.
Type Description Parameters (s_{sf},_{sf}) {}(s{sf},_{sf})
cnst Linear Tank coupled below a constant parameter diffusive wave solution $$v_sf, k_sf$$ [q_{sf} ={ \begin{array}{cl} s_{sf}/t_{raf} & s_{sf} s_{raf} \ s_{raf}/t_{raf} + (s_{sf}-s{raf}
kin Linear tank with Kinematic Wave for remaining flow
(#tbl:surface-param) Parameters used in the surface zone storage-flow relationships.
Symbol Description unit
$$T_0$$ 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 -

### 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 fluxes 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 $\begin{equation} \frac{ds_{uz}}{dt} = r_{rz \rightarrow uz} - \frac{A s_{uz}}{T_d s_{sz}} \end{equation}$

If water is able to pass freely to the saturated zone, then it flows at the rate $$\frac{A s_{uz}}{T_d s_{sz}}$$. In the 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 the downward flux 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 implemented through the Muskingham approximation. This gives a mass balance equation considered in terms of storage deficits as:

$\frac{ds_{sz}}{dt} = q^{+}_{sz} - r_{sf \rightarrow rz} - q^{-}_{sz}$ with the relationships between the flows and storage given by and $q_{sz} = \mathcal{G}\left(s_{sz},\Theta_{sz}\right) \\ q_{sz}^{+} = 2q_{sz} - q_{sz}^{-}$ Not that the assumption of increasing flow with storage means that $- \frac{d}{ds_{sz}} \mathcal{G}\left(s_{sz},\Theta_{sz}\right) \geq 0$

The function $$\mathcal{G}\left(s_{sz},\Theta_{sz}\right)$$ is related to the HRU width through the transmissivity function. Table 2 gives $$\mathcal{G}\left(s_{sz},\Theta_{sz}\right)$$ for various the transmissivity profiles present as options within the dynatop package, the corresponding value to use for the type value in the saturated zone specification and the additional parameters required. The additional parameters are defined in Table 3.

Table 2: Transmissivity profiles available with the dynatop package.
Name $$G\left(\bar{s}_{sz},\Theta_{sz}\right)$$ $$\Theta_{sz}$$ type value Notes
Exponential $$T_{0}w\sin\left(\beta\right)\exp\left(-\frac{\cos\beta}{Am}s_{sz}\right)$$ $$T_0,\beta,m$$ exp Originally given in Beven & Freer 2001
Bounded Exponential $$T_{0}w\sin\left(\beta\right)\left(\exp\left(-\frac{\cos\beta}{Am}s_{sz}\right) - \exp\left(-\frac{\cos\beta}{m}D\right)\right)$$ $$T_0,\beta,m,D$$ bexp Originally given in Beven & Freer 2001
Constant Celerity $$\frac{c_{sz}w}{A}\left(AD-s_{sz}\right)$$ $$c_{sz},D$$ cnst
Double Exponential $$T_{0}w\sin\left(\beta\right)\left( \omega\exp\left(-\frac{\cos\beta}{Am}s_{sz}\right) + \left(1-\omega\right)\exp\left(\frac{\cos\beta}{Am_2}s_{sz}\right)\right)$$ $$T_0,\beta,m,m_2,\omega$$ dexp
Table 3: Additional parameters used in the transmissivity profiles.
Symbol Description unit
$$T_0$$ 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
$$D$$ Storage deficit at which zero lateral flow occurs m
$$\omega$$ weighting parameter -

## 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 the maximum downward volumes of the vertical fluxes $$\hat{v}_{* \rightarrow *} = \Delta t \hat{r}_{* \rightarrow *}$$ are determined. These then moderated by the solution of the saturated zone, with an upward pass gives the final states.

### Downward Pass

#### Surface excess

The implicit solution of the Muskingham methodology gives

$\left. s_{sf} \right\rvert_{t+\Delta t} = \left. s_{sf} \right\rvert_{t} + \Delta t \left( \left. q_{sf}^{-} \right\rvert_{t+\Delta t} - \frac{1}{1 - \left. \eta_{sf} \right\rvert_{t+\Delta t} } \max\left(0, \left. q_{sf} \right\rvert_{t+\Delta t} - \left. \eta_{sf} \right\rvert_{t+\Delta t} \left. q_{sf}^{-} \right\rvert_{t+\Delta t} \right) \right) - \left. v_{sf \rightarrow rz} \right\rvert_{t+\Delta t}$

So the maximum downward flux volume is $\left. \hat{v}_{sf \rightarrow rz} \right\rvert_{t+\Delta t} = \left. s_{sf} \right\rvert_{t} + \Delta t \left. q_{sf}^{-} \right\rvert_{t+\Delta t}$

#### Root Zone

The implicit formulation for the root zone gives $\begin{equation} \left. s_{rz} \right.\rvert_{t + \Delta t} = \left( 1 + \frac{e_{p}\Delta t}{s_{rzmax}} \right)^{-1} \left( \left. s_{rz} \right.\rvert_{0} + \Delta t p + v_{sf \rightarrow rz} - v_{rz \rightarrow uz} \right) \end{equation}$

Since flow to the unsaturated zone occur only to keep $$s_{rz} \leq s_{rzmax}$$ then $\begin{equation} \hat{v}_{rz \rightarrow uz} = \max\left(0, \left. s_{rz} \right.\rvert_{0} + \Delta t \left( p -e_{p}\right) + \hat{v}_{sf \rightarrow rz} - s_{rzmax} \right) \end{equation}$

#### Unsaturated Zone

The implicit expression for the unsaturated zone is given in terms of $$\left. s_{uz} \right.\rvert_{t+\Delta t}$$ and $$\left. s_{sz} \right.\rvert_{t+\Delta t}$$ as

$\begin{equation} \left. s_{uz} \right.\rvert_{t+\Delta t} = \frac{T_{d}\left. s_{sz} \right.\rvert_{t+\Delta t}}{ T_{d}\left. s_{sz} \right.\rvert_{t+\Delta t} + A \Delta t} \left( \left. s_{uz} \right.\rvert_{t} + v_{rz \rightarrow uz}\right) \end{equation}$

Since $$\left. s_{uz} \right.\rvert_{t+\Delta t} \leq \left. s_{sz} \right.\rvert_{t+\Delta t}$$ this places an additional condition of $\begin{equation} \left. s_{uz} \right.\rvert_{t} + v_{rz \rightarrow uz} \leq \left. s_{sz} \right.\rvert_{t+\Delta t} + \frac{A \Delta t}{T_d} \end{equation}$

By mass balance this gives the maximum downward flux volume as $\hat{v}_{uz \rightarrow sz} = A\Delta t \min\left( \frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)} {T_{d}\left. s_{sz} \right.\rvert_{t+\Delta t} + A \Delta t}, \frac{1}{T_{d}} \right)$

This is a function of $$\left. s_{sz} \right.\rvert_{t+\Delta t}$$.

#### Saturated Zone

Building on the generic solution the implicit scheme for the Muskingham model gives

$\left. s_{sz} \right\rvert_{t+\Delta t} = \left. s_{sz} \right\rvert_{t} - \Delta t \left. q_{sz}^{-} \right\rvert_{t+\Delta t} - \left. v_{uz \rightarrow sz} \right\rvert_{t+\Delta t} + \Delta t \min\left(q_{sz}^{max},\max\left( 0, 2 \left. q_{sz} \right\rvert_{t+\Delta t} - \left. q_{sz}^{-} \right\rvert_{t+\Delta t} \right)\right)$

where the maxima stops the outflow becoming negative. The minima derives from a further condition on the saturated zone; based on the maximum outflow give by $$f_{sz}\left(0\right) = q_{sz}^{max}$$. Applying this flow limit to every cross section of the saturated zone mean $$q_{sz}^{-}\leq q_{sz}^{max}$$ and $$q_{sz}^{+}\leq q_{sz}^{max}$$. The inflow inequality to handled by reassigning an inflow greater then the limit to the surface inflow. The outflow inequality results in the minima.

Alternatively the upper flow limit could be evaluated by finding $$s_{sz}^{max}$$ for which $\mathcal{G}\left(s_{sz}^{max},\Theta_{sz}\right) = \frac{ q_{sz}^{max} + \left. q_{sz}^{-}\right\rvert_{t+\Delta t} }{2}$ and treating this as the saturated deficit at which point water is returned from the saturated to unsaturated zone. In contrast to this the minima solution holds the water in the saturated zone. This is preferred both conceptually, since the water remains immediately available for future lateral flow, and also computationally since it avoids the need to invert the function relating storage to flow.

Consider the pseudo-function $\mathcal{H}\left(z\right) = z - \left. s_{sz}\right.\rvert_{t} + \Delta t \left. q_{sz}^{-} \right\rvert_{t+\Delta t} + A\Delta t \min\left( \frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)} {T_{d}z + A \Delta t}, \frac{1}{T_{d}} \right) - \\ \Delta t \min\left(q_{sz}^{max},\max\left( 0, 2 \mathcal{G}\left(z,\Theta_{sz}\right) - \left. q_{sz}^{-} \right\rvert_{t+\Delta t} \right)\right)$

adapted from the approximating equation for the saturated zone storage. In Appendix A it is shown that $$\mathcal{H}\left(z\right)$$ is a monotonic increasing function of $$z > 0$$ and $$\mathcal{H}\left(D\right) \geq 0$$. Hence if $$\mathcal{H}\left(0\right) \lt 0$$ unique solution for $$\left. s_{sz}\right.\rvert_{\Delta t}$$ can be found. If $$\mathcal{H}\left(0\right) \geq 0$$ then the incoming fluxes are enough to produce saturation in the subsurface and $$\left. s_{sz}\right.\rvert_{\Delta t} = 0$$.

Maintaining a water (mass) balance and correct limits on the state values depends upon the accuracy of the solution of $$\mathcal{H}\left(z\right)=0$$. Since each evaluation of $$\mathcal{H}\left(z\right)$$ requires an evaluation of the transmissivity profile the search for $$z$$ such that $$\mathcal{H}\left(z\right)=0$$ can rapidly become the most expensive part of the code.

Use of a numerical scheme which allows mass balance to maintained regardless of the accuracy of the solution of $$\mathcal{H}\left(z\right)=0$$ allows scope for compromises based on computational time. If $$\mathcal{H}\left(z\right)=0$$ is solved by a bracketing algorithm then at each iteration the lower and upper limits of the bracket satisfy $$z^{l} < \left. s_{sz}\right.\rvert_{\Delta t} < z^{u}$$. Iterating until; for some tolerance $$\epsilon>0$$; $$z^{u} - z^{l} < \epsilon$$ take $$\left. s_{sz}\right.\rvert_{\Delta t} = z^{u}$$. In both is and the saturated case $$\mathcal{H}\left(\left. s_{sz}\right.\rvert_{\Delta t}\right)>0$$ and the upward pass can proceed as outlined in the following section without compromising the mass conservation.

### Upward Pass

Given an estimate of $$\left. s_{sz}\right.\rvert_{\Delta t}$$ such that $$\mathcal{H}\left(\left. s_{sz}\right.\rvert_{\Delta t}\right)>0$$ compute the vertical flux volumes in the following fashion

1. $$\left. q_{sz}^{+}\right.\rvert_{t+\Delta t} = \min\left(q_{sz}^{max},\max\left( 0, 2 \mathcal{G}\left(\left. s_{sz}\right.\rvert_{t+\Delta t},\Theta_{sz}\right) - \left. q_{sz}^{-} \right\rvert_{t+\Delta t} \right)\right)$$
2. $$v_{uz \rightarrow sz} = \left. s_{sz}\right.\rvert_{t} + \Delta t \left(\left. q_{sz}^{+}\right.\rvert_{t+\Delta t} - \left. q_{sz}^{-} \right\rvert_{t+\Delta t}\right) - \left. s_{sz}\right.\rvert_{t+\Delta t}$$
3. $$\left. s_{uz}\right.\rvert_{t+\Delta t} = \min\left(\left. s_{sz}\right.\rvert_{t+\Delta t},\left. s_{uz}\right.\rvert_{t} + \hat{v}_{rz \rightarrow uz} - v_{uz \rightarrow sz}\right)$$
4. $$v_{rz \rightarrow uz} = \left. s_{uz}\right.\rvert_{t+\Delta t} -\left. s_{uz}\right.\rvert_{t} + v_{uz \rightarrow sz}$$
5. $$v_{sf \rightarrow rz} = \min\left(\hat{v}_{sf \rightarrow rz}, s_{rzmax} - \left. \bar{s}_{rz} \right.\rvert_{0} - \Delta t \left( p - e_p \right) + {v}_{rz \rightarrow uz} \right)$$
6. Solve for $$\left. s_{rz}\right.\rvert_{t+\Delta t}$$ using the equation already given.
7. Compute the volume loss to evapotranspiration as $$\left. v_{e_{p}}\right.\rvert_{t+\Delta t} = \left. s_{rz}\right.\rvert_{t} + v_{sf \rightarrow rz} - v_{rz \rightarrow uz} + \Delta t p - \left. s_{rz}\right.\rvert_{t+\Delta t}$$

This leaves the solution of the surface zone

#### Surface Zone

For the surface zone we take a semi implicit scheme where $$\eta_{sf}$$ is evaluated with the initial state giving $\left. \eta_{sf} \right\rvert_{t+\Delta t} = \mathcal{F}_{\eta}\left(\left. s_{sf} \right\rvert_{t},\Theta_{sf}\right)$

Proceeding in a similar fashion to the saturated zone consider the pseudo-function $\mathcal{S}\left(w\right) = \left. s_{sf} \right\rvert_{t} + \Delta t \left( \left. q_{sf}^{-} \right\rvert_{t+\Delta t} - \frac{1}{1 - \left. \eta_{sf} \right\rvert_{t+\Delta t} } \max\left(0, \mathcal{F}\left(w,\Theta_{sf}\right) - \left. \eta_{sf} \right\rvert_{t+\Delta t} \left. q_{sf}^{-} \right\rvert_{t+\Delta t} \right) \right) - \left. v_{sf \rightarrow rz} \right\rvert_{t+\Delta t} - w$ which is monotonically decreasing for $$w \geq 0$$ with $$\mathcal{S}\left(0\right)\geq 0$$.

A bracketing solution algorithm gives at any iteration $$w^{l} < \left. s_{sf} \right\rvert_{t+\Delta t} < w^{u}$$. An initial value of $$w^{u}$$ can be computed as $$\left. s_{sf} \right\rvert_{t} + \Delta t\left. q_{sf}^{-} \right\rvert_{t+\Delta t} - \left. v_{sf \rightarrow rz} \right\rvert_{t+\Delta t}$$. Iterating until; for some tolerance $$\epsilon>0$$; $$w^{u} - w^{l} < \epsilon$$ take $$\left. s_{sf}\right.\rvert_{\Delta t} = w^{l}$$. In this case $$\mathcal{S}\left(\left. s_{sf}\right.\rvert_{\Delta t}\right) > 0$$ hence mass conservation can be maintained by taking $\left. q_{sf}^{+} \right\rvert_{t+\Delta t} = \left. q_{sf}^{-} \right\rvert_{t+\Delta t} + \frac{1}{\Delta t}\left( \left. s_{sf} \right\rvert_{t} - \left. v_{sf \rightarrow rz} \right\rvert_{t+\Delta t} - \left. s_{sf}\right.\rvert_{t + \Delta t}\right)$

## Appendix A - Monotonicity of $$\mathcal{H}\left(z\right)$$

By definition $$\left. s_{uz} \right.\rvert_{t}$$, $$\Delta t$$, A, $$T_d$$, $$\hat{v}_{rz \rightarrow uz}$$ and $$z$$ are all greater or equal to 0. To show that the gradient of

$\mathcal{H}\left(z\right) = z - \left. s_{sz}\right.\rvert_{t} + \Delta t \left. q_{sz}^{-} \right\rvert_{t+\Delta t} + A\Delta t \min\left( \frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)} {T_{d}z + A \Delta t}, \frac{1}{T_{d}} \right) - \\ \Delta t \min\left(q_{sz}^{max},\max\left( 0, 2 \mathcal{G}\left(z,\Theta_{sz}\right) - \left. q_{sz}^{-} \right\rvert_{t+\Delta t} \right)\right)$

is strictly positive consider first the gradient of the last term which takes the value

$\frac{d}{dz} \min\left(q_{sz}^{max},\max\left( 0, 2 \mathcal{G}\left(z,\Theta_{sz}\right) - \left. q_{sz}^{-} \right\rvert_{t+\Delta t} \right)\right) = \left\{ \begin{array}{cl} 0 & 2 \mathcal{G}\left(z,\Theta_{sz}\right) < \left. q_{sz}^{-} \right\rvert_{t+\Delta t}\\ 0 & 2 \mathcal{G}\left(z,\Theta_{sz}\right) - \left. q_{sz}^{-} \right\rvert_{t+\Delta t}> q_{sz}^{max} \\ 2 \frac{d}{dz} \mathcal{G}\left(z,\Theta_{sz}\right) & \mathrm{otherwise} \end{array} \right.$

By definition $$-\frac{d}{dz} \mathcal{G}\left(z,\Theta_{sz}\right) \geq 0$$ hence

$\frac{d}{dz} \mathcal{H}\left(z\right) \geq 1 + A\Delta t \frac{d}{dz} \min\left( \frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)} {T_{d}z + A \Delta t}, \frac{1}{T_{d}} \right)$

Separating the range of $$z$$ into two sections gives

$\frac{d}{dz} \mathcal{H}\left(z\right) \geq \left\{ \begin{array}{cl} 1 & \frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)} {T_{d}z + A \Delta t} \geq \frac{1}{T_{d}} \\ 1 - T_{d} A\Delta t \frac{\left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}} {\left(T_{d}z + A \Delta t\right)^2} & frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)}{T_{d}z + A \Delta t} \leq \frac{1}{T_{d}} \end{array} \right.$

Further using the range of $$z$$ valid for the second case

$\frac{d}{dz} \mathcal{H}\left(z\right) \geq \left\{ \begin{array}{cl} 1 & \frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)} {T_{d}z + A \Delta t} \geq \frac{1}{T_{d}} \\ 1 - \frac{A\Delta t}{T_{d}z + A \Delta t} & \frac{\left( \left. s_{uz} \right.\rvert_{t} + \hat{v}_{rz \rightarrow uz}\right)}{T_{d}z + A \Delta t} \leq \frac{1}{T_{d}} \end{array} \right.$

From this it can be seen that $$\frac{d}{dz} \mathcal{H}\left(z\right) \geq 0$$ with equality possible when $$z=0$$.