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

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

Revision as of 09:44, 12 August 2009

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


\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}


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.