# Difference between revisions of "Solution of the Blatter-Pattyn model"

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<math>\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial y} \right)=\underset{{}}{\mathop{\frac{\partial }{\partial \hat{x}}\left( \eta \frac{\partial u}{\partial \hat{y}} \right)}}\,+\underset{{}}{\mathop \frac{\partial \sigma }{\partial \hat{x}}\frac{\partial }{\partial \sigma }\left( \eta \frac{\partial u}{\partial \hat{y}} \right)}\,+\underset{{}}{\mathop \frac{\partial \sigma }{\partial \hat{y}}\frac{\partial }{\partial \hat{x}}\left( \eta \frac{\partial u}{\partial \sigma } \right)}\,+\underset{{}}{\mathop \frac{\partial \sigma }{\partial \hat{x}}\frac{\partial \sigma }{\partial \hat{y}}\frac{\partial }{\partial \sigma }\left( \eta \frac{\partial u}{\partial \sigma } \right)}\,+\underset{{}}{\mathop \frac{\partial _{{}}^{2}\sigma }{\partial \hat{x}\partial \hat{y}}\eta \frac{\partial u}{\partial \sigma }}\,</math> | <math>\frac{\partial }{\partial x}\left( \eta \frac{\partial u}{\partial y} \right)=\underset{{}}{\mathop{\frac{\partial }{\partial \hat{x}}\left( \eta \frac{\partial u}{\partial \hat{y}} \right)}}\,+\underset{{}}{\mathop \frac{\partial \sigma }{\partial \hat{x}}\frac{\partial }{\partial \sigma }\left( \eta \frac{\partial u}{\partial \hat{y}} \right)}\,+\underset{{}}{\mathop \frac{\partial \sigma }{\partial \hat{y}}\frac{\partial }{\partial \hat{x}}\left( \eta \frac{\partial u}{\partial \sigma } \right)}\,+\underset{{}}{\mathop \frac{\partial \sigma }{\partial \hat{x}}\frac{\partial \sigma }{\partial \hat{y}}\frac{\partial }{\partial \sigma }\left( \eta \frac{\partial u}{\partial \sigma } \right)}\,+\underset{{}}{\mathop \frac{\partial _{{}}^{2}\sigma }{\partial \hat{x}\partial \hat{y}}\eta \frac{\partial u}{\partial \sigma }}\,</math> | ||

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+ | This looks pretty scary eh? Luckily, there is a lot of symmetry here. Notice that ... blah blah blah. |

## Revision as of 23:22, 12 August 2009

The final form of the equations we'd like to solve is:

Again, note that for the *x* equation we've moved all the terms containing gradients in *v* to the right-hand side (RHS).

We've set it up this way in order to solve the equations using an **operator splitting** approach; for the *x* equation, we treat *v* as known (where we take the values of *v* from the previous iteration) and solve for *u*, and vice versa when we solve they *y* equation for *v*. The "splitting" refers to the fact that we are breaking the multi-dimensional divergence operation into multiple steps. Rather than solving one big matrix equation for *u* and *v* simultaneously we solve two smaller matrix equations in sequence with one of the unknowns treated as a known "source" term.

As with the 0-order model, we need to change from Cartesian to sigma coordinates. The first normal-stress term first term on the left-hand side becomes

where hatted values refer to the coordinate directions in sigma coordinates. Similarly, the first cross-stress term on the RHS is given by

This looks pretty scary eh? Luckily, there is a lot of symmetry here. Notice that ... blah blah blah.