sci.math.numanalysis
SUNA51, Superelement Details
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A Least Squares Finite Element for Incompressible and Irrotational Flow has
been devised according to the method in Re: SUNA24, Staggered Quadrilaterals.
The Finite Difference equations for an elementary triangle are repeated:
[  y32 + x32  y13 + x13  y12 + x12 ]  u1 
[ + x32 + y32 + x13 + y13 + x12 + y12 ]  v1 
 u2  = 0
Here: xji = xj  xi ; yji = yj  yi ; i,j = 1,2,3 .  v2 
 u3 
The Least Squares Finite Element matrix is constructed  v3 
by multiplying the transpose of the above 6*2 matrix
with itself.
Let: X1 = x3  x2 ; X2 = x1  x3 ; X3 = x2  x1
Y1 = y3  y2 ; Y2 = y1  y3 ; Y3 = y2  y1 : circular.
And: Iij = Xi.Xj + Yi.Yj ; Uij = Xi.Yj  Xj.Yi ; i,j = 1,2,3 .
It is easily seen that U12 = U23 = U31 , since all these quantities equalize
2 times the area of the triangle element. Therefore Uij is replaced by U .
A special case occurs for i = j : Uii = 0 , giving zero matrix entries.
The result of the above proposed matrix multiplication can be written now as:
 I11 0 I12 + U I13  U 
 I11  U I12 + U I13 
 I22 0 I23 + U  = L.S. element matrix
 I22  U I23 
 symmetric I33 0 
 I33 
The velocity xcomponents as well as the velocity ycomponents are coupled by
admittances (ui,uj) or (vi,vj). These are _inner_ products Iij of coordinate
differences of the triangle sides. The coupling is strongest for the identical
velocity components (ui,ui) or (vi,vi). The inner products can be written as:
Iij =  Ri  Rj .cos(ij) where ij = 0 for i = j : Iii =  Ri ^2
By drawing a picture, it can be seen that inner products with i <> j will be
negative most of the time. Provided that all angles of the triangle element are
< 90 degrees. An inner product equals zero if the triangle element has a right
angle = 90 degrees. This might well occur with rectangular grids, the classical
Finite Difference case. Obtuse angle give rise to positive offdiagonal matrix
coefficients. According to one of Patankar's "Five Basic Rules", this should be
avoided, assuming these rules are applicable here too. There is some evidence
for this in the fact that inner products of the above kind also play a role in
finite element matrices which are derived for: "SUNA42, Discretization for 2D
diffusion". So there exist kind of _DIFFUSIONlike couplings_ between velocity
components of the same kind. At the same time, it is well known that u and v
indeed separately obey Laplace's equations, analytically.
The velocity x and ycomponents are also coupled by admittances (ui,vj). These
are all equal to plus or minus 2 times the area of the triangle element: +/ U.
Consequently, positive offdiagonal coefficients between velocity components of
_different_ kind can NOT be avoided. Because the quantity U denotes the area
the triangle element, and it is also in the denominator of the matrix elements
for diffusion, it is advantageous to divide everything by this quantity U .
We can make the above more explicit by writing the element matrix as follows:
 I11 I12 I13  u1   I11 I12 I13  v1 
1/U. I22 I23  u1  + 1/U. I22 I23  v2  +
 symm. I33  u1   symm. I33  v3 
 . 0 . + .   u1 
 .  . + .  v1 
+ 1. . 0 . +  u2 
 .  .  v2  = 0 due to taking out diffusion
 symmetric . 0  u3 
 .  v3 
The last matrix is a bit difficult to interpret at this moment. Suppose for a
moment that there are no "diffusion" contributions, then setting the remaining
equations to zero implies that:
+ v2  v3 = 0 ;  u2 + u3 = 0 ; Basically:
 v1 + v3 = 0 ; + u1  u3 = 0 ;
+ v1  v2 = 0 ;  u1 + u2 = 0 . u1 = u2 = u3 ; v1 = v2 = v3 .
Thus the xcomponents as well as the ycomponents of the velocities across the
triangle tend to be equalized. Which can be understood, anyway.
Consider the following patch of two mutually overlapping triangles, which is
actually used in defining the quadrilateral superelement for ideal flow:
____*____ Concentrate upon the velocity node indicated as (1).
\ 4 / When assembling the matrices belonging to the different
 \ /  triangles for that node, it receives contributions
5* 1 *3 (u1,v2) = + 1 , (v1,u2) =  1 from triangle (1,2,3) ;
 / \  (u1,v2) =  1 , (v1,u2) = + 1 from triangle (1,5,2) .
/___*___\ The net result is that node (1) receives NO contribution
2 from the node (2). The same argument holds if we replace
(2) by (3) or (4) or (5). All nondiffusive contributions
simply cancel out. (Note that the quadrilateral in general is not rectangular.)
Consequently, node (1) is attached to the others nodes (2,3,4,5) by diffusion
_only_. In the special case of a _square_ superelement, this means that:
4.u1 = u2 + u3 + u4 + u5 ; 4.v1 = v2 + v3 + v4 + v5 : mean values (1).

* 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