Bishesh Khanal <[email protected]> writes:

> Within A, for now, I can consider mu to be constant, although later if
> possible it can be a variable even a tensor to describe anisotropy. But to
> start with I want put this a constant.
> The original equations start with mu (grad(u) + grad(u)^T) but then
> simplifications occur due to div(u) = f2

Rework that step in case of variable mu.

> I'm mostly interested in the phenomenon in A with my model, here B is the
> extension of the very irregular domain of A to get a cuboid. Here, in B I
> release the div(u) = f2 constraint and just put a regularisation to
> penalize large deformation. What is of importance here is to compensate the
> net volume expansion in domain A by corresponding contraction in domain B
> so that the boundaries of the cuboid do not move. It does not particularly
> represent any physics except probably that it gives me a velocity field
> having a certain divergence field that penalizes big deformations.

Okay, sounds like it's already an artificial equation, so you should be
able to leave in a normal equation for p, with a big mass matrix on the
diagonal,

div(mu(grad(u))) - grad(p) = f1
div(u)           - c(x) p  = f2

c(x) = 0 in domain A and c(x) is large (the inverse of the second Lamé
parameter) in domain B.

> I do not know much about FEM. But some of the reasons why I have avoided it
> in this particular problem are:  (Please correct me on any of the following
> points if they are wrong)
> 1. The inputs f1 and f2 are 3D images (in average of size 200^3) that come
> from other image processing pipeline; it's important that I constrain u at
> each voxel for div(u) = f2 in domain A. I am trying to avoid having to get
> the meshing from the 3D image(with very detailed structures), then go back
> to the image from the obtained u again because I have to use the obtained u
> to warp the image, transport other parameters again with u in the image
> space and again obtain new f1 and f2 images. Then iterate this few times.

Okay, there's nothing wrong with that.

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