sci.math.research
SUNA21, Reduced Quadrilaterals
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The chain is as weak as its weakest part, that's the summary of this article.
I've used reduced quadrilaterals for quite a while. They don't have the obvious
shortcomings of convential finite element linear triangles in the first place.
Hmm, who said "obvious"? We'll see about that later.
Reduced quadrilaterals are quadrilateral elements with just _one_ integration
point in the middle. If a mesh consists of M x N reduced quadrilaterals, then
the number of F.D. equations is 2 x M x N . The boundary conditions give rise
to 2 x (M + 1) + 2 x (N + 1) additional equations. The number of unknowns in
the problem is 2 x (M + 1) x (N + 1) = 2 x M x N + 2 x (M + 1) + 2 x (N + 1)
minus 2 (these two can be neglected ;). So: number of equations = number of
unknowns. Hurrah!
Not surprisingly, a mesh of reduced quadrilaterals can give _exellent_ answers
for the Labrujere problem. Yet ... they are plain wrong!
The following argument is essentially due to prof.dr. G.W. Veltkamp (Eindhoven
University of Technology).
3 4 du/dx + dv/dy =
o  o (u2 + u4)/2  (u1 + u3)/2 + (v3 + v4)/2  (v1 + v2)/2
 
 *  du/dy  dv/dx =
  (u3 + u4)/2  (u1 + u2)/2  [(v2 + v4)/2  (v1 + v3)/2]
o  o
1 2 o = coordinates & velocities ; * = integration point .
The matrix form of the above, apart from a constant factor 1/2 , is used for
calculating the Least Squares Finite Element matrix:
u1 u2 u3 u4 v1 v2 v3 v4
 1 1   1 +1 1 +1 1 1 +1 +1   + o o  o +  o 
 +1 1   1 1 +1 +1 +1 1 +1 1  =  o +  o  o o +  u
 1 +1   o  + o + o o  
 +1 +1  Here the elements of the Least   o o + o  + o _____
 1 +1  Squares matrix are also apart  o  + o + o o  
 1 1  from a constant factor (= 2).  + o o  o +  o 
 +1 +1  Furthermore: + = +1 ,  = 1   o o + o  + o  v
 +1 1  and o = 0 . Symmetric matrix:  o +  o  o o + 
Next, the following patch of four reduced quadrilaterals is devised:
7 8 9 Contributions from the elements for u5 sum of:
o  o  o
   I :  u1 + u5  v2 + v4 The values
 III  IV  II :  u3 + u5 + v2  v6 for v
 2  1  III : + u5  u7  v4 + v8 cancel out.
4 o  5  o 6 IV : + u5  u9 + v6  v8
 4  3 
 I  II  Contributions from the elements for v5 sum of:
  
o  o  o I : + u2  u4  v1 + v5 The values
1 2 3 II :  u2 + u6  v3 + v5 for u
III : + u4  u8 + v5  v7 cancel out.
IV :  u6 + u8 + v5  v9
For u5 : 4.u5  u1  u3  u7  u9 = 0
: Laplacelike discretizations.
For v5 : 4.v5  v1  v3  v7  v9 = 0
The xcomponent and the ycomponent of the velocities are mutually uncoupled.
Moreover, u5 does not depend upon u2 , u4 , u6 , u8 . Thus, the velocities
are only coupled in diagonal directions. Uneven and even numbered velocities
don't "feel" each other. There exist in fact _two_ independent solutions upon
the mesh. Which, of course, cannot be accepted.

* 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