I think Raymond pretty much covered all the bases. The way I would write
this is using
time = Variable(0.)
eqn = someTerms + mesh.x * mesh.y * t
for step in range steps:
eqn.solve(...)
t.setValue(t.value + dt)
The key here (due to FiPy pathologies) is to realize that mesh.x and mesh.y
are CellVariables, while using the alternative "x, y = mesh.cellCenters"
syntax does not result in CellVariables, but only numpy arrays and hence
can't be used in the source term unless these are explicitly cast to a
CellVariable. This is because the source term must be either an object of
no length or a CellVariable. Since a "Variable" of no length multiplied by
a CellVarialbe results in a new CellVarialbe "mesh.x * mesh.y * t" can be
used as the souce term function and will know when t changes. Note that the
time "t" has to be explicitly updated in the loop.
Hope the above helps.
On Tue, Mar 19, 2013 at 1:26 PM, Raymond Smith <[email protected]> wrote:
> Hi Serbulent,
>
> I was curious about the same thing, so I gave it a shot. I'm not sure if
> what I've done is a good way to do things, but I think I managed to get a
> source term that's a function of x, y. I modified your script in a number
> of ways, and wrote it so that myFunc takes positions as arguments:
>
> 1) dx should be a float (dx = 1., not dx =1)
> 2) As far as I know, it's better to use fipy's numerix package than numpy.
> 3) I think for a source term that's dependent on position, you can use a
> CellVariable. So I declared a CellVariable, which I then modify values of
> at each time step.
> 4) Rather than loop through x and y and then trying to find the indexes
> that correspond to the a value close to that, you can just loop through all
> the indexes of the grid points in the problem. You can get the X and Y
> arrays which are indexed by this total number of points from
> mesh.CellCenters, then loop through all your points (nx*ny in the
> rectangular 2D case).
> 5) For your time dependence, when you're calling myFunc, you need to have
> calculated what the current time is before you call it (t =
> step*timeStepDuration)
> 6) I made an arbitrary function of x, y, t for myFunc. The way I cast it,
> it takes _values_ of position and time as arguments. The way you wrote it,
> x and y need to be indexes of your "tmp" array.
>
> File at http://pastebin.com/sZLHBTRV
>
> Cheers,
> Ray
>
>
> On Tue, Mar 19, 2013 at 9:43 AM, Serbulent UNSAL <[email protected]>wrote:
>
>> Hi,
>>
>> First thanks to FiPy community for such a practical tool. I'm a newbie
>> in FiPy and try to solve a problem based on source term.
>>
>> My equation is a 2D diffusion equation which has a source term as a
>> function depends on x,y cordinates and time t
>>
>> \frac{ \partial c( \vec{x},t)}{\partial t} =
>> D_{c}\bigtriangledown ^{2} c( \vec{x},t)} - f_{c}( \vec{x},t)
>>
>> I am googleing about 3 days found some clues [1] but they could'nt
>> solve my problem. My code is working with a constant source term but I
>> still couldn't find how can I represent it as a function depends on
>> x,y and t.
>>
>> I'm trying to get my source term values from a mock up matrix for sake
>> of simplicity. I'll be appriciate for any ideas/solutions. My code can
>> be seen at http://pastie.org/6605575
>>
>> Thanks,
>>
>> Serbulent
>>
>>
>> [1]
>>
>> http://bb10.com/python-fipy/2012-12/msg00016.html
>> http://osdir.com/ml/python.fipy/2007-10/msg00009.html
>> http://bb10.com/python-fipy/2011-05/msg00030.html
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>
>
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