- #1
trelek2
- 88
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Hi!
I'm trying to implement an implicit scheme for the continuity equation.
The scheme is the following:
http://img28.imageshack.us/img28/3196/screenshot20111130at003.png
With [TEX]\rho[/TEX] being the density, [TEX]\alpha[/TEX] is a weighing constant. d is a parameter that relates the grid spacing to the velocity of flow. j is the spatial grid and n is the time grid.
The problem with this scheme is the fact that it is implicit and effectively I have no idea how to successfully implement it. I tried by doing the following:
Assume have initial time n=0 spatial (j) grid full. Also assume I know spatial boundaries (j=-1/2 and j=max) at time n+1.
Then I set j=1/2 equal to a variable x.
Next rearrange so that have j=3/2 in terms of variable x and boundary.
Do likewise for all the (half)grid points so that they can be written in terms of x. then when I reach the opposite boundary of grid (j=max) solve for x. So then I have j=1/2 and hence can substitute x into all the other equations to fill the grid points with data.
This can't work since:
When writing expression for any grid point in terms of x, i divide by [tex]0.5d(1-\alpha)[/tex]. Which is roughly 0.5. So going through all the grid points I end up with 0.5^400 (400 grid points) in the denominator which is bound to kill the calculation.
Can anyone tell me how to deal with this implicit scheme?
I'm trying to implement an implicit scheme for the continuity equation.
The scheme is the following:
http://img28.imageshack.us/img28/3196/screenshot20111130at003.png
With [TEX]\rho[/TEX] being the density, [TEX]\alpha[/TEX] is a weighing constant. d is a parameter that relates the grid spacing to the velocity of flow. j is the spatial grid and n is the time grid.
The problem with this scheme is the fact that it is implicit and effectively I have no idea how to successfully implement it. I tried by doing the following:
Assume have initial time n=0 spatial (j) grid full. Also assume I know spatial boundaries (j=-1/2 and j=max) at time n+1.
Then I set j=1/2 equal to a variable x.
Next rearrange so that have j=3/2 in terms of variable x and boundary.
Do likewise for all the (half)grid points so that they can be written in terms of x. then when I reach the opposite boundary of grid (j=max) solve for x. So then I have j=1/2 and hence can substitute x into all the other equations to fill the grid points with data.
This can't work since:
When writing expression for any grid point in terms of x, i divide by [tex]0.5d(1-\alpha)[/tex]. Which is roughly 0.5. So going through all the grid points I end up with 0.5^400 (400 grid points) in the denominator which is bound to kill the calculation.
Can anyone tell me how to deal with this implicit scheme?
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