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So long as solver lives we have 3 field size equivalent memory with us all the time. It makes sense because after all we need to do a timestep after transeq such that $u_i = u_i + dt \cdot du_i$
and then after pressure is sorted we need to do a pressure correction such that $u_i = u_i - p_x \cdot \epsilon ??$
So obviously we can't get rid of them. But inside transeq we can play a trick to limit the total amount of temporary storage transeq requires.
We first use the velocity field components here, and obtain all the derivatives we need in x direction.
And this is the key part. Because at this stage we have a copy of velocity field in y orientation as well as the original arrays in x orientation, we can actually get rid of the memory of the original arrays.
If we do this then the reordering below will have to go from y to z.
And then at the very end we'll have to reorder z->x so that the original u, v, and w fields are restrored before returning back to run method in solver.
@semi-h Can you summarise here?
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