sci.math.research
SUNA24, Staggered Quadrilaterals
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It was explained in the "Streamlines" SUNAposter why the velocity nodes are
located differently compared with the nodes where the coordinates are defined.
This was established again in "Staggered Triangles". Adopting quadrilaterals,
coordinate and velocity nodes will be arranged in an elementform as follows:
4 4 3
o  *  o o = coordinates;
 \ / 
5 * * * 3 * = velocities; node in the middle = number (1)
 / \ 
o  *  o o and * are numbered separately.
1 2 2
This arrangement is conceived as a superelement ("Substructuring" SUNAposter)
which in turn consists of the following mutually overlapping triangular parts:
4 4 4 3 4 4 3 3 3
o o  *  o o  *  o o o
 \ \   / /  / \
5 * * 1 1 * * 3 5 * * 1 1 * * 3 : 2 * * 1
 \ \   / /  / \
o  *  o o o o  *  o o  *  o
1 2 2 2 1 1 2 2 1 3 2
I II III IV
The last triangle is the generic basic triangle. It's equations are, as derived
in "Staggered Triangles":
(y2  y1).u3  (x2  x1).v3 (x2  x1).u3 + (y2  y1).v3
(y3  y2).u1  (x3  x2).v1 (x3  x2).u1 + (y3  y2).v1
(y1  y3).u2  (x1  x3).v2 = 0 (x1  x3).u2 + (y1  y3).v2 = 0
Counting the equations: 4 triangles * 2 = 8 . At first sight. But wait!
The equations are not all independent. Adding together for triangle I and II,
or III and IV, results in one and the same set, namely:
 (y2  y1).u2 + (x2  x1).v2 + (x2  x1).u2 + (y2  y1).v2
 (y3  y2).u3 + (x3  x2).v3 + (x3  x2).u3 + (y3  y2).v3
 (y4  y3).u4 + (x4  x3).v4 + (x4  x3).u4 + (y4  y3).v4
 (y1  y4).u5 + (x1  x4).v5 = 0 + (x1  x4).u5 + (y1  y4).v5 = 0
This set could also be derived directly for the quadrilateral (1,2,3,4) alone.
Therefore, the _rank_ of the equation system appears to be 6 instead of 8 .
Here the velocities at (*) node (1) were in fact _eliminated_, leaving us with
an element without a node in the middle:
4 4 3
o  *  o
 
5 * * 3
 
o  *  o
1 2 2
Using now the procedure outlined in "Substructuring", this will be done also in
the _final_ algorithm which produces the element matrix: the velocities at the
"midpoints" of the quadrilaterals can be entirely eliminated. Such is possible,
because (1) has (2,3,4,5) as a kind of boundaries. The velocities are recovered
later on by backsubstitution, as soon as the velocities in the midside nodes
have become known. Remember, however, that all of the 6 independent equations
are yet taken into account.
Counting the unknowns : 5 * 2 = 10 . Taking the superelement as a problem in
itself, we can apply the above four boundary conditions, giving:
6 equations + 4 boundaries = 10 unknowns . Which seems to be OK.
Suppose we have a mesh with N such unknowns. Suppose that we still have then
the same number of equations as we have unknowns. What happens if we extend the
mesh by adding _one_ more element?
   o  *  o o  *  o
  \ / 
1 * * * *
  / \ 
o o  *  o

In such a case, where previously was a boundary condition (1), there will be
nothing. The addon element will bring with it 10 additional unknowns, minus
2 unknowns which are common with the previous problem, due to connectivity,
giving a total of 8 additional unknowns. There are 6 additional equations
plus 3 additional boundary conditions. Summing up everything:
N  1 + 6 + 3 = N + 10  2 = N + 8
equations unknowns
For one element there are as many equations as there are unknowns. Suppose now
that for M elements there are as many equations as there are unknowns, then
we can prove that for M + 1 elements the same still holds. Thus, by induction
we may conclude that the proposed (super)element always gives the same number
of equations and unknowns, for arbitrary meshes which are built up with this
thing as the basic buiding block.
To be continued ...

* Han de Bruijn; Applications&Graphics  "A little bit of Physics * No
* TUD Computing Centre; P.O. Box 354  would be NO idleness in * Oil
* 2600 AJ Delft; The Netherlands.  Mathematics" (HdB). * for
* Email: Han.deBruijn@RC.TUDelft.NL  Fax: +31 15 78 37 87 * Blood