Difference between revisions of "Blatter-Pattyn model"

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<math>\begin{align}
 
<math>\begin{align}
   & x:\quad 4\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial x} \right)+\frac{\partial }{\partial y}\left( \eta \frac{\partial u}{\partial y} \right)+\frac{\partial }{\partial z}\left( \eta \frac{\partial u}{\partial z} \right)=.-2\frac{\partial }{\partial x}\left( \eta \frac{\partial v}{\partial y} \right)-\frac{\partial }{\partial y}\left( \eta \frac{\partial v}{\partial x} \right)+\rho g\frac{\partial s}{\partial x} \\  
+
   & x:\quad 4\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial x} \right)+\frac{\partial }{\partial y}\left( \eta \frac{\partial u}{\partial y} \right)+\frac{\partial }{\partial z}\left( \eta \frac{\partial u}{\partial z} \right)=-2\frac{\partial }{\partial x}\left( \eta \frac{\partial v}{\partial y} \right)-\frac{\partial }{\partial y}\left( \eta \frac{\partial v}{\partial x} \right)+\rho g\frac{\partial s}{\partial x} \\  
  & y:\quad 4\frac{\partial }{\partial y}\left( \eta \frac{\partial v}{\partial y} \right)+\frac{\partial }{\partial x}\left( \eta \frac{\partial v}{\partial x} \right)+\frac{\partial }{\partial z}\left( \eta \frac{\partial v}{\partial z} \right)=.-2\frac{\partial }{\partial y}\left( \eta \frac{\partial u}{\partial x} \right)-\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial y} \right)+\rho g\frac{\partial s}{\partial y} \\  
+
  & y:\quad 4\frac{\partial }{\partial y}\left( \eta \frac{\partial v}{\partial y} \right)+\frac{\partial }{\partial x}\left( \eta \frac{\partial v}{\partial x} \right)+\frac{\partial }{\partial z}\left( \eta \frac{\partial v}{\partial z} \right)=-2\frac{\partial }{\partial y}\left( \eta \frac{\partial u}{\partial x} \right)-\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial y} \right)+\rho g\frac{\partial s}{\partial y} \\  
 
\end{align}</math>
 
\end{align}</math>

Revision as of 17:30, 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}


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.


We continue by noting that the 3rd (vertical) balance equation above can be integrated through the depth of the ice sheet to give an expression for the pressure, P,


P=\rho g\left( s-z \right)+\tau _{zz}(z)


This expression can be substituted into the horizontal pressure gradient terms above to remove pressure from the equations. For example, for the 'x' component of velocity we have


\begin{align}
  & x:\quad \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 \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=\frac{\partial }{\partial x}\left[ \rho g\left( s-z \right)+\tau _{zz}(z) \right] \\ 
 & x:\quad \frac{\partial \tau _{xx}}{\partial x}-\frac{\partial \tau _{zz}}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=\rho g\frac{\partial s}{\partial x} \\ 
\end{align}


Now we can use the incompressibility and stress-strain colinearity constraints to rewrite the vertical normal deviatoric stress in terms of horizontal normal deviatoric stresses. This allows us to write the horizontal balance equations in terms of horizontal stress gradients only. Again, for the x direction we have


\begin{align}
  & \tau _{zz}=-\tau _{xx}-\tau _{yy} \\ 
 & -\frac{\partial \tau _{zz}}{\partial x}=-\frac{\partial }{\partial x}\left( -\tau _{xx}-\tau _{yy} \right)=\frac{\partial \tau _{xx}}{\partial x}+\frac{\partial \tau _{yy}}{\partial y} \\ 
 & \frac{\partial \tau _{xx}}{\partial x}-\frac{\partial \tau _{zz}}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=\rho g\frac{\partial s}{\partial x}\quad ...\text{becomes}... \\ 
 & 2\frac{\partial \tau _{xx}}{\partial x}+\frac{\partial \tau _{yy}}{\partial y}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=\rho g\frac{\partial s}{\partial x}\quad  \\ 
\end{align}


Note that we have removed the vertical balance equation entirely; it is incorporated into the horizontal balance equations, which are


\begin{align}
  & x:\quad 2\frac{\partial \tau _{xx}}{\partial x}+\frac{\partial \tau _{yy}}{\partial y}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=\rho g\frac{\partial s}{\partial x}\quad  \\ 
 & y:\quad 2\frac{\partial \tau _{yy}}{\partial y}+\frac{\partial \tau _{xx}}{\partial y}+\frac{\partial \tau _{xy}}{\partial x}+\frac{\partial \tau _{yz}}{\partial z}=\rho g\frac{\partial s}{\partial y} \\ 
\end{align}


Next, we want to write these equations in terms of velocities, since that is what we are ultimately solving for. The link between stress gradients and velocities is through the constitutive law for ice (here we'll assume Glen's law) - relating strain rates to stresses - and the definition of the strain rate tensor - relating strain rates to velocity gradients.


\begin{align}
  & 1.\quad \tau _{ij}=B\dot{\varepsilon }_{e}^{\frac{1-n}{n}}\dot{\varepsilon }_{ij},\quad B=B(T) \\ 
 & 2.\quad \dot{\varepsilon }_{ij}=\frac{1}{2}\left( \frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}} \right) \\ 
 & 3.\quad 2\dot{\varepsilon }_{e}=\dot{\varepsilon }_{ij}\dot{\varepsilon }_{ij} \\ 
 & 4.\quad \eta \equiv \frac{1}{2}B\dot{\varepsilon }_{e}^{\frac{1-n}{n}} \\ 
 & 5.\quad \tau _{ij}=2\eta \dot{\varepsilon }_{ij} \\ 
\end{align}


In order, the four expressions above give

  1. Glen's flow law (actually, the inverse flow-law here, giving stress as a function of strain rate)
  2. The definition of the strain rate tensor in terms of velocity gradients
  3. The definition of the effective strain rate, \dot{\varepsilon }_{e}, a norm of the strain rate tensor
  4. A definition for the "effective viscosity" (rearranging some terms in (1))
  5. ... which allows us to write the relationship between stress and strain in a more standard way


Now taking these definitions into our stress balance equations above and expanding in terms of strain rates, velocity gradients, and effective viscosities, we have (for the 'x' direction only)


\begin{align}
  & x:\quad 2\frac{\partial \tau _{xx}}{\partial x}+\frac{\partial \tau _{yy}}{\partial x}+\frac{\partial \tau _{xy}}{\partial y}+\frac{\partial \tau _{xz}}{\partial z}=\rho g\frac{\partial s}{\partial x} \\ 
 & x:\quad 2\frac{\partial }{\partial x}\left( 2\eta \dot{\varepsilon }_{xx} \right)+\frac{\partial }{\partial x}\left( 2\eta \dot{\varepsilon }_{yy} \right)+\frac{\partial }{\partial y}\left( 2\eta \dot{\varepsilon }_{xy} \right)+\frac{\partial }{\partial z}\left( 2\eta \dot{\varepsilon }_{xz} \right)=\rho g\frac{\partial s}{\partial x} \\ 
 & x:\quad 4\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial x} \right)+2\frac{\partial }{\partial x}\left( \eta \frac{\partial v}{\partial y} \right)+\frac{\partial }{\partial y}\left[ \eta \left( \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} \right) \right]+\frac{\partial }{\partial z}\left( \eta \frac{\partial u}{\partial z} \right)=\rho g\frac{\partial s}{\partial x} \\ 
\end{align}


... with an analogous expression for the y balance equation. For actually solving the system of equations, it is advantageous to combine all v gradient terms for the x balance equation and move them to the right-hand side (and vice versa for the y balance equation). This gives the final form of the equations to be discretized and solved:


\begin{align}
  & x:\quad 4\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial x} \right)+\frac{\partial }{\partial y}\left( \eta \frac{\partial u}{\partial y} \right)+\frac{\partial }{\partial z}\left( \eta \frac{\partial u}{\partial z} \right)=-2\frac{\partial }{\partial x}\left( \eta \frac{\partial v}{\partial y} \right)-\frac{\partial }{\partial y}\left( \eta \frac{\partial v}{\partial x} \right)+\rho g\frac{\partial s}{\partial x} \\ 
 & y:\quad 4\frac{\partial }{\partial y}\left( \eta \frac{\partial v}{\partial y} \right)+\frac{\partial }{\partial x}\left( \eta \frac{\partial v}{\partial x} \right)+\frac{\partial }{\partial z}\left( \eta \frac{\partial v}{\partial z} \right)=-2\frac{\partial }{\partial y}\left( \eta \frac{\partial u}{\partial x} \right)-\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial y} \right)+\rho g\frac{\partial s}{\partial y} \\ 
\end{align}