# Computing steady-state subglacial water flux¶

The “routing” subglacial hydrology model is described by equations

(86)\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  (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 (86) using a method that is computationally cheaper than using the explicit in time approximation of (86) described by .

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 (86) to zero), integrating over $$B$$, and applying the divergence theorem gives

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

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

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

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 (86) 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,

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

Combining (87) and (89) 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:

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

Here the right hand side of (90) can be estimated by advancing an explicit-in-time approximation of (88) 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 (88) will automatically distribute water inputs from the ice surface (or melting) along the ice margin.

## The algorithm¶

Using an explicit time stepping approximation of (88) 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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