# Computing steady-state subglacial water flux¶

The “routing” subglacial hydrology model is described by equations

(85)\begin{align}\begin{aligned}\diff{W}{t} + \diff{\Wtill}{t} + \Div \bq = \frac{m}{\rho_w}\\\diff{\Wtill}{t} = \frac{m}{\rho_w} - C_d\\\bq = -k W^{\alpha} |\nabla \psi|^{\beta - 2} \nabla \psi\\\psi = P_o + \rho_w g (b + W)\end{aligned}\end{align}

on a an ice covered area $$\Omega$$. We assume zero flux boundary conditions on the inflow part of the boundary and no boundary condition on the outflow boundary. See [110] (equations 1, 2, 6, 16, 26) for details and the notation. Here we also assume that $$m \ge 0$$.

Our goal is to estimate $$Q = \bq \cdot \n$$, the flux through the outflow part of the boundary of $$\Omega$$ corresponding to the steady state of (85) using a method that is computationally cheaper than using the explicit in time approximation of (85) described by [110].

Pick a contiguous section $$\omega$$ of $$\partial \Omega$$ (the terminus of an outlet glacier, for example). Let $$B$$ be the union of all the trajectories of the vector field $$\bq$$ in $$\Omega$$ that pass through $$\omega$$. The area $$B$$ is the “drainage basin” corresponding to $$\omega$$.

Let $$\gamma = \partial B \setminus \omega$$. Note that if a point $$P$$ is in $$\gamma$$ then one of the following conditions is satisfied.

1. $$|\bq| = 0$$ (it is the origin of a trajectory that starts within $$\Omega$$) or

2. $$P \in \partial \Omega$$ (specifically, $$P$$ is a part of the inflow part of the boundary of $$\Omega$$)

3. $$\bq \cdot \n = 0$$ ($$P$$ is not at the end of a trajectory, and so the normal to the boundary is orthogonal to $$\bq$$).

Therefore $$\bq \cdot \n = 0$$ on $$\gamma$$ and

\begin{align}\begin{aligned}\oint_{\partial B} \bq \cdot \n\; ds &= \int_{\omega} \bq \cdot \n\; ds + \int_{\gamma} \bq \cdot \n\; ds\\&= \int_{\omega} \bq \cdot \n ds.\end{aligned}\end{align}

Assuming the steady state (and setting time derivatives in (85) to zero), integrating over $$B$$, and applying the divergence theorem gives

(86)$\int_{\omega} \bq \cdot \n\; ds = \int_{B} \frac{m}{\rho_w},$

i.e. in a steady state the flux through a terminus is equal to the total rate at which water is added to the corresponding drainage basin due to the source term.

Next, consider a related initial boundary value problem

(87)$\diff{u}{t} = -\Div (\V u)$

on $$B$$ with $$u(x, y, 0) = u_0(x, y)$$ ($$u_0 \ge 0$$), $$\V = -k(x, y) \nabla \psi$$, zero flux on the inflow boundary, and no boundary condition on the outflow boundary.

Here $$\psi$$ is the hydraulic potential corresponding to the steady state of (85) and $$k(x, y)$$ is a strictly positive but otherwise arbitrary conductivity function.

Note that since $$\psi$$ is a steady state hydraulic potential all trajectories of the vector field $$\V$$ leave $$B$$ and for $$\epsilon > 0$$ there is a time $$T > 0$$ such that

$\int_B u(T) = \epsilon \int_B u_0.$

Integrating over time from $$0$$ to $$T$$, we get

$\int_0^T \diff{u}{t}\, dt = - \int_0^T \Div (\V u),\, \text{or}$
$u_0 = u(T) + \int_0^T \Div (\V u).$

Integrating over $$B$$ and using the divergence theorem gives

\begin{align}\begin{aligned}\int_B u_0 &= \int_B u(T) + \int_B \int_0^T \Div (\V u)\\&= \epsilon \int_B u_0 + \int_0^T \int_B \Div (\V u)\\&= \epsilon \int_B u_0 + \int_0^T \oint_{\partial B} (\V u) \cdot \n\\&= \epsilon \int_B u_0 + \int_0^T \int_{\omega} (\V u) \cdot \n.\end{aligned}\end{align}

Finally,

(88)$\int_B u_0 = \frac{1}{1 - \epsilon} \int_0^T \int_{\omega} (\V u) \cdot \n$

Combining (86) and (88) and choosing $$u_0 = \tau m\, /\, \rho_w$$ for some $$\tau > 0$$1 gives us a way to estimate the flux through the outflow boundary if we know the direction of the steady state flux:

(89)$\int_{\omega} \bq \cdot \n\; ds = \frac{1}{\tau(1 - \epsilon)} \int_0^T \int_{\omega} (\V u) \cdot \n\; ds.$

Here the right hand side of (89) can be estimated by advancing an explicit-in-time approximation of (87) until $$\int_B u$$ drops below a chosen threshold.

However, the direction of the steady state flux $$\bq$$ depends on steady state distributions of $$W$$ and $$\Wtill$$ and these quantities are expensive to compute.

To avoid this issue we note that $$W \ll H$$ and so $$\psi$$ is well approximated by $$\psi_0 = P_o + \rho_w g b$$ everywhere except the vicinity of subglacial lakes. Moreover, if $$|\nabla W|$$ is small then $$\nabla \psi_0$$ is a reasonable approximation of $$\nabla \psi$$.

We approximate $$\psi$$ by $$\tilde \psi = P_o + \rho_w g b + \delta$$ where $$\delta > 0$$ is an adjustment needed to ensure that $$\tilde \psi$$ has no local minima in the interior of $$\Omega$$ and $$|\nabla \tilde \psi| > 0$$ everywhere on $$\Omega$$ except possibly on a set of measure zero (no “plateaus”).

The approximation of $$\tilde \psi$$ on a computational grid is computed as follows.

1. Set $$k = 0$$, $$\tilde \psi_{0} = \psi$$.

2. Iterate over all grid points. If a grid point $$(i, j)$$ is at a local minimum, set $$\tilde \psi_{k + 1}(i, j)$$ to the average of neighboring values of $$\tilde \psi_{k}$$ plus a small increment $$\Delta \psi$$, otherwise set $$\tilde \psi_{k + 1}(i, j)$$ to $$\tilde \psi_{k}(i, j)$$.

3. If step 2 found no local minima, stop. Otherwise increment $$k$$ and proceed to step 2.

Next, note that it is not necessary to identify the drainage basin $$B$$ for a terminus $$\omega$$: it is defined by $$\psi$$ and therefore an approximation of (87) will automatically distribute water inputs from the ice surface (or melting) along the ice margin.

## The algorithm¶

Using an explicit time stepping approximation of (87) we can estimate $$\int_{\omega} \bq \cdot \n \; ds$$ as follows.

1. Given ice thickness $$H$$ and bed elevation $$b$$ compute $$\tilde \psi$$ by filling “dips” as described above.

2. Choose the stopping criterion $$\epsilon > 0$$ and the scaling for the source term $$\tau > 0$$.

3. Set

\begin{align}\begin{aligned}u &\leftarrow \frac{\tau m}{\rho_w},\\t &\leftarrow 0,\\Q &\leftarrow (0, 0).\end{aligned}\end{align}
4. Compute the CFL time step $$\dt$$ using $$u$$ and $$\V$$.

5. Perform an explicit step from $$t$$ to $$t + \dt$$, updating $$u$$.

6. Accumulate this step’s contribution to $$Q$$:

$Q \leftarrow Q + \dt \cdot \V u.$
7. Set $$t \leftarrow t + \dt$$

8. If $$\int_{\Omega} u\; dx\, dy > \epsilon$$, go to 4.

9. Set

$Q \leftarrow \frac{1}{t (1 - \epsilon^{*})}\; Q,$

where

$\epsilon^{*} = \frac{\int_{\Omega} u}{\int_{\Omega} u_{0}}.$

Footnotes

1

The constant $$\tau$$ is needed to get appropriate units, but its value is irrelevant.

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