Difference between revisions of "Kees' assignment"

Revision as of 14:08, 5 August 2009

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

where

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

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

$D(x_{j+1/2}) = C \left(\frac{H_j + H_{j+1}^5}{2}\right) \left(\frac{h_{j+1} - h_j}{\Delta x}\right)^2.$

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 12: / Line 12: @@
 ==Model parameters==
 *<math>\frac{\partial b}{\partial x} = -0.1</math>
-* <math>M(x) = M_0 - x M_1 = 4.0 - 2\times \textrm{10}^{-4} x</math>
+* <math>M(x) = M_0 - x M_1 = 4.0\times 10^{-3} - 0.2 x</math> km/yr
 * <math>\rho</math> = 920 <math>kg/m^3</math>
 *g=9.8 <math>m/s^2</math>