Difference between revisions of "Kees' assignment"

Revision as of 13:13, 5 August 2009

$\frac{\partial H}{\partial t} = - \frac{\partial}{\partial x}\left(-D(x) \frac{\partial h}{\partial x}\right) + M$

where

$D(x) = C H^{n+2}\left|\frac{\partial h} {\partial x}\right| ^{n-1},$

and

$C = \frac{2 A}{n+2} \left(\rho g\right)^n$

Use a staggered grid such that the $D(x_{j+1/2})$ are computed at the centers of the grid (as opposed to the vertices, as we have been doing), so

$D(x_{j+1/2}) = C \left(\frac{H_j + H_{j+1}}{2}\right)^{n+2} \left(\frac{h_{j+1} - h_j}{\Delta x}\right)^{n-1}.$

From the diffusivity, the flux is computed

$\phi_{i+1/2} = D(x_{i+1/2}) \frac{\partial h}{\partial x}$ ,

where

$\frac{\partial h}{\partial x} = \frac{h_{i+1}-h_{i}}{\Delta x}$

and then the flux ( $\phi_i$ ) can be used to compute the rate of change of the surface from

$\frac{\partial H}{\partial t} = -\frac{\partial }{\partial x} H\bar u + M = - \frac{\phi_{i+1/2} - \phi_{i-1/2} }{\Delta x} + M$

@@ Line 28: / Line 28: @@
 Use a staggered grid such that the <math>D(x_{j+1/2})</math> are computed at the '''centers''' of the grid (as opposed to the vertices, as we have been doing), so
-:<math>D(x_{j+1/2}) = C \left(\frac{H_j + H_{j+1}^5}{2}\right)  \left(\frac{h_{j+1} - h_j}{\Delta x}\right)^2.</math>
+:<math>D(x_{j+1/2}) = C \left(\frac{H_j + H_{j+1}}{2}\right)^{n+2}  \left(\frac{h_{j+1} - h_j}{\Delta x}\right)^{n-1}.</math>
 From the diffusivity, the flux is computed