# Blatter-Pattyn Boundary Conditions

We will go through an approximate derivation of the boundary conditions that are implemented with Glimmer/CISM's higher-order scheme. By "approximate" we mean that some of the derivation is guided by physical intuition and what appear to be "reasonable" arguments, rather than through the application of rigorous mathematics. We take comfort in the fact that, in the end, we wind up with the same sets of equations that one ends up with from the more rigorous approach. We will look at the derivation in three parts, (1) the free surface boundary condition, (2) the specified basal traction boundary condition, and (3) lateral boundary conditions.

## Stress Free Surface

At the ice surface, a stress-free boundary condition is applied. The traction vector, T, must be continuous at the ice sheet surface and, assuming that atmospheric pressure and surface tension are small, we have

\begin{align} & T_{i}=-T_{i(boundary)}\approx 0 \\ & T_{i}=\sigma _{ij}n_{j}=\sigma _{i1}n_{1}+\sigma _{i2}n_{2}+\sigma _{i3}n_{3}=0\\\end{align}

where the ni are the components of the outward facing, unit normal vector in Cartesisan coordinates.

For a function F(x,y,z) = f(x,y) - z = 0, where z = f(x,y) defines the surface, the gradient of F(x,y,z) gives the components of the surface normal vector:

$a=\sqrt{\left( \frac{\partial f}{\partial x} \right)^{2}+\left( \frac{\partial f}{\partial y} \right)^{2}+1^{2}}$

For the case of the ice sheet surface, s = f(x,y) and the surface normal is given by

$n_{i}=\left( \frac{\partial s}{\partial x},\frac{\partial s}{\partial y},-1 \right)\frac{1}{a}$

and

$a=\sqrt{\left( \frac{\partial s}{\partial x} \right)^{2}+\left( \frac{\partial s}{\partial x} \right)^{2}+1^{2}}\approx \sqrt{1^{2}}=1$

This simplification comes about because, in general, the slopes on glaciers and ice sheets are small, in which case the slope squared is very small. Thus, to first order, the surface normal vector components are simply given by

$n_{i}=\left( \frac{\partial s}{\partial x},\frac{\partial s}{\partial y},-1 \right)$

The expression above for Ti=0 gives three equations that must be satisfied for a free surface boundary condition:

\begin{align} & i=x:\quad T_{x}=\sigma _{xx}n_{x}+\sigma _{xy}n_{y}+\sigma _{xz}n_{z}=0, \\ & i=y:\quad T_{y}=\sigma _{yx}n_{x}+\sigma _{yy}n_{y}+\sigma _{yz}n_{z}=0, \\ & i=z:\quad T_{z}=\sigma _{zx}n_{x}+\sigma _{zy}n_{y}+\sigma _{zz}n_{z}=0. \\ \end{align}

Expanding the final equation, including the components of the surface-normal vector, and solving for the vertical normal stress at the surface gives

$\sigma _{zz}=\tau _{zx}\frac{\partial s}{\partial x}+\tau _{zy}\frac{\partial s}{\partial y}=\tau _{zz}-P$.

To go further, we need to return to the vertical stress balance from the Stokes equations,

\begin{align} & z:\quad \frac{\partial \tau _{zz}}{\partial z}-\frac{\partial P}{\partial z}+\frac{\partial \tau _{zy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial x}=\rho g \\ \end{align},

which, when integrated with respect to the vertical coordinate z and evaluated at z = s(x,y), gives

$\left. \left. \sigma _{zz} \right|_{z=s}=\left. \rho gz \right|_{z=s}+c\left( x,y \right) \right|_{z=s}=\rho gs+c\left( x,y \right)$

Comparing the above two equations we see that

$c\left( x,y \right)=\tau _{zx}\frac{\partial s}{\partial x}+\tau _{zy}\frac{\partial s}{\partial y}-\rho gs$

and thus, at the surface,

$\sigma _{zz}=\rho g\left( z-s \right)+\tau _{zx}\frac{\partial s}{\partial x}+\tau _{zy}\frac{\partial s}{\partial y}.$

Using this expression and the definition of P,

$P=\frac{1}{3}\left( \sigma _{xx}+\sigma _{yy}+\sigma _{zz} \right)$