Difference between revisions of "Blatter-Pattyn model"

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There are a number of ways to argue that because of the "shallowness" of ice sheets - that is because the ratio of <big>''H''/''L''</big>, where ''H'' is the thickness and ''L'' is a relevant horizontal length scale, is small - the equations above can be reduced to the following "first-order approximation"
 
There are a number of ways to argue that because of the "shallowness" of ice sheets - that is because the ratio of <big>''H''/''L''</big>, where ''H'' is the thickness and ''L'' is a relevant horizontal length scale, is small - the equations above can be reduced to the following "first-order approximation"
 +
  
  
 
<math>\begin{align}
 
<math>\begin{align}
   & \frac{\partial \tau _{xx}}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=\frac{\partial P}{\partial x} \\  
+
   & x:\quad \frac{\partial \tau _{xx}}{\partial x}-\frac{\partial P}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=0 \\  
  & \frac{\partial \tau _{yy}}{\partial y}+\frac{\partial \tau _{xy}}{\partial x}+\frac{\partial \tau _{yz}}{\partial z}=\frac{\partial P}{\partial y} \\  
+
  & y:\quad \frac{\partial \tau _{yy}}{\partial y}-\frac{\partial P}{\partial y}+\frac{\partial \tau _{xy}}{\partial x}+\frac{\partial \tau _{yz}}{\partial z}=0 \\  
  & \frac{\partial \tau _{zz}}{\partial z}=\rho g+\frac{\partial P}{\partial z} \\  
+
  & z:\quad \frac{\partial \tau _{zz}}{\partial z}-\frac{\partial P}{\partial z}=\rho g \\  
 
\end{align}</math>
 
\end{align}</math>
 
 
Unfortunately, the arguments that support this reduction are complex and we can't go into them in detail here. Two papers given in the references below (''Schoof and Hindmarsh'' (in press) and ''Dukowicz et al.'' (submitted)) provide more details on the mathematical background that allows us to state that these equations are "first order accurate" approximations to the Stokes equations.
 
 
 
A constitutive law, describing the relation between stress and strain rate, is then used to write stresses in terms of an effective viscosity, η, and strain rates,
 
 
 
where blah is blah and strain rate tensor is given by
 
 
 
Writing the equations out in terms of velocity gradients and the effective viscosity we have
 
 
 
For solution of the u equation, it is convenient to move all of the terms containing v gradients to the right-hand side (and vice versa for the v equation). We'll now look at Solution of the Blatter-Pattyn equations.
 
 
 
 
where u and v are the depth-independent x and y components of velocity, \bar{\eta } is the depth-averaged effective viscosity, H is the ice thickness, ρ is the ice density, g is the acceleration due to gravity, and s=s(x,y) is the ice surface elevation.
 
 
Notice the symmetry in the equations. This means that, computationally, many of the same subroutines can be used for discretization.
 

Revision as of 16:47, 30 July 2009

The starting point for the Blatter-Pattyn model is the full Stokes equations


\begin{align}
  & x:\quad \frac{\partial \tau _{xx}}{\partial x}-\frac{\partial P}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=0 \\ 
 & y:\quad \frac{\partial \tau _{yy}}{\partial y}-\frac{\partial P}{\partial y}+\frac{\partial \tau _{xy}}{\partial x}+\frac{\partial \tau _{yz}}{\partial z}=0 \\ 
 & 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},


where P is the pressure and τ is the deviatoric stress tensor. The latter is given by


\tau _{ij}=\sigma _{ij}+P\delta _{ij},


where σ is the full stress tensor.


There are a number of ways to argue that because of the "shallowness" of ice sheets - that is because the ratio of H/L, where H is the thickness and L is a relevant horizontal length scale, is small - the equations above can be reduced to the following "first-order approximation"


\begin{align}
  & x:\quad \frac{\partial \tau _{xx}}{\partial x}-\frac{\partial P}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=0 \\ 
 & y:\quad \frac{\partial \tau _{yy}}{\partial y}-\frac{\partial P}{\partial y}+\frac{\partial \tau _{xy}}{\partial x}+\frac{\partial \tau _{yz}}{\partial z}=0 \\ 
 & z:\quad \frac{\partial \tau _{zz}}{\partial z}-\frac{\partial P}{\partial z}=\rho g \\ 
\end{align}