Hello-
I'm once again in need of guidance. As a bit of background, I'm
working on a project similar in general idea to what PECOS is doing
with Bayesian Calibration/Validation etc, but as applied to modeling
tumor growth. We are currently just trying to show proof of concept
using virtual experiments as data, i.e. use a model with a certain set
of parameters and use the resulting solution as calibration and/or
validation data. Keeping in mind that a real set of data would
probably be something like an MRI pixelated image, for the data misfit
we would like to consider as data the values per "pixel" or in this
case per element. So, we would like to calculate, if u is the solution
being used as the data and u' is the solution generated through a
sample point,
sqrt(int_{elem}u^2) - sqrt(\int_{elem}(u')^2).
This would not be a problem at all if they were on the same mesh.
However, the computational expense of such a calculation is a really
big issue, and so I'm using the PatchRecoveryErrorEstimator for
refinement. Thus, there is absolutely no reason to suspect they would
be the same mesh. I had tried using ExactErrorEstimator (as I
mentioned in my last email to the list serve), but I see that that is
calculating
sqrt(int_{elem}(u-u')^2).
So, a couple of questions:
I've just realized that in the estimate_errors routine it is assumed
that the two meshes have the same parent mesh... Is there a problem if
one is not necessarily a refined version of the other? i.e. is it ok
if they are each refined in locations where the other isn't?
Assuming it isn't a problem, I'm really needing the first norm I've
listed. After looking through the code, I'm kind of talking myself
into the idea that maybe I just need to create a child class of
ExactErrorEstimator that has its own estimate_error routine which is
exactly the same except instead of calling find_squared_element_error
called something else which was exactly the same as
find_squared_element_error but use the difference I need instead.
Would there be a better way to do it? Or would this not work at all?
Thank you!
Andrea
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