sci.math.research
SUNA26, Labrujere Revised
=========================
It was suggested in previous postings that "ordinary" linear triangle elements,
with their velocities concentrated at vertices, do _not_ seem to work so well.
I've revised this theory while enjoying 1991 Christmas vacation. The objective
was to give a proof of this hitherto unsupported conviction. I couldn't find
any, however. Instead, evidence accumulated that (kind of a modified) assembly
procedure for Labrujere's triangles would turn it into a usefull least squares
finite element scheme, as well.
Recapitulating. The Cauchy equations for ideal flow are:
du/dx + dv/dy = 0
dv/dx  du/dy = 0
(u,v) = velocities; (x,y) = coordinates; d = partial differentiation.
Isoparametrics as usual. Local coordinates (h,k) are introduced. The chain rule
for partial derivatives which relates local to global is:
3
df/dh = df/dx.dx/dh + df/dy.dy/dh o
df/dk = df/dx.dx/dk + df/dy.dy/dk Or: / \
/ \
f2  f1 = df/dx.(x2  x1) + df/dy.(y2  y1) / \
f3  f1 = df/dx.(x3  x1) + df/dy.(y3  y1) 1 o  o 2
Solving for df/dx , df/dy has become a regular exercise in this Series:
J.df/dx = + (y3  y1).(f2  f1)  (y2  y1).(f3  f1)
=  (y3  y2).f1  (y1  y3).f2  (y2  y1).f3
J.df/dy =  (x3  x1).(f2  f1) + (x2  x1).(f3  f1)
= + (x3  x2).f1 + (x1  x3).f2 + (x2  x1).f3
where: J = (x2  x1).(y3  y1)  (x3  x1).(y2  y1)
= + x1.y2  x2.y1 + x2.y3  x3.y2 + x3.y1  x1.y3
(Gee, seems that I can't get enough of these patterns :) Consequently:
 J.(du/dx + dv/dy) = + (y3  y2).u1  (x3  x2).v1 +
+ (y1  y3).u2  (x1  x3).v2 +
+ (y2  y1).u3  (x2  x1).v3 = 0
 J.(dv/dx  du/dy) = + (x3  x2).u1 + (y3  y2).v1 +
+ (x1  x3).u2 + (y1  y3).v2 +
+ (x2  x1).u3 + (y2  y1).v3 = 0
These equations were found before (: Staggered Triangles), but with a different
geometrical interpretation. This time, velocities are NOT concentrated in the
middle of the triangle sides, but at triangle vertices, as is usual with finite
element schemes. Moreover, these velocities are positioned _opposite_ to those
in the middle of triangle sides.
Consider a patch of such linear triangles, one which is as simple as possible:
4 The accompanying discretizations are:
o
/  \ I : u2  u5 + v4  v5 = 0 ; v2  v5  u4 + u5 = 0
/  \
/ II  I \ II : u5  u1 + v4  v5 = 0 ; v5  v1  u4 + u5 = 0
1 o  5  o 2
\ III  IV / III: u5  u1 + v5  v3 = 0 ; v5  v1  u5 + u3 = 0
\  /
\  / IV : u2  u5 + v5  v3 = 0 ; v2  v5  u5 + u3 = 0
o
3 (a) (b)
The system of 8 equations is simplified as follows.
Substract IV(a) from I(a) giving: v4  2.v5 + v3 = 0 > 2.v5 = v3 + v4
Substract II(b) from I(b) giving: v2  2.v5 + v1 = 0 > 2.v5 = v1 + v2
Substract I(b) from IV(b) giving: u3  2.u5 + u4 = 0 > 2.u5 = u3 + u4
Substract II(a) from I(a) giving: u2  2.u5 + u1 = 0 > 2.u5 = u1 + u2
Hence the two velocity components at node (5) can be eliminated. If this is
actually accomplished, by substituting the above expressions for u5 and v5
back into the system I  IV (a,b) then the following two equations are the
one and only result:
u2  u1 + v4  v3 = 0 (1)
v2  v1  u4 + u3 = 0 (2)
Together with: 2.u5 = u1 + u2 = u3 + u4 (3 & 4)
2.v5 = v1 + v2 = v3 + v4 (5 & 6)
It is concluded at first that the RANK of the 8 equations system is in fact 6 .
If we eliminate the node in the middle, then 4 equations are left. Of these the
first two (1,2) are recognized again as the Cauchy equations, discretized upon
the quadrilateral (1,2,3,4). The other two (3 & 4,5 & 6) can be interpreted
with a little help from previous theory. A function f is discretized upon a
quadrilateral, upon its F.D. base, and with an adapted numbering, as follows:
f = f1 + h.(f3  f1) + k.(f4  f1) + h.k.(f1 + f2  f3  f4)

According to the above, the underlined term in this expression is zero for both
the velocity components. This in fact means that function behaviour upon the
quadrilateral, as assembled from four triangles, must be restricted to _linear_
instead of bilinear, for u and v . Not necessarily a bad thing ...
Let's complete now the patch test, by carrying out a complete flow calculation
for the above quadrilateral. Assume boundary conditions as follows:
u1 = U ; v1 = 0
v3 = 0 ; v4 = 0 Then:
v1 + v2 = v3 + v4 > v2 = 0
u2  u1 + v4  v3 = 0 > u2 = U
v2  v1  u4 + u3 = 0 > u4 = u3 
u1 + u2 = u3 + u4 > 2.U = u3 + u4  > u3 = u4 = U . OK.
So Th.E. Labrujere's setup still runs through _everything_ we tried so far !
Yet we know that it does _not_ work for the problem at hand: ideal flow around
a circular cylinder.
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