sci.math.research

SUNA25, Repeatable Elements
===========================

Least Squares Finite Elements are produced in our theory by taking the square
of Finite Difference like equations:

                      E(i,j) = A(i).A(j)

Here  A(i)  are the coefficients    + (y2 - y1).u3 - (x2 - x1).v3 + ... = 0
of the following equation system:   + (x2 - x1).u3 + (y2 - y1).v3 + ... = 0

Suppose the problem is transformed by an orthogonal transformation, for example
a rotation:            ^^^^^^^^^^^

      x = + cos(phi).x + sin(phi).y      ;     u = + cos(phi).u + sin(phi).v
      y = - sin(phi).x + cos(phi).y      ;     v = - sin(phi).u + cos(phi).v

Then the first equation of the above two becomes something like:

[ - (x2 - x1).sin(phi) + (y2 - y1).cos(phi) ].[ + u3.cos(phi) + v3.sin(phi) ] +
[ - (x2 - x1).cos(phi) - (y2 - y1).sin(phi) ].[ - u3.sin(phi) + v3.cos(phi) ] =
+ (y2 - y1).u3.[ cos^2 + sin^2 ] - (x2 - x1).v3.[ sin^2 + cos^2 ] +
+ (x2 - x1).u3.[ - sin.cos + cos.sin ] + (y2 - y1).v3.[ cos.sin - sin.cos ] =
+ (y2 - y1).u3 - (x2 - x1).v3

And the second:

[ + (x2 - x1).cos(phi) + (y2 - y1).sin(phi) ].[ + u3.cos(phi) + v3.sin(phi) ] +
[ - (x2 - x1).sin(phi) + (y2 - y1).cos(phi) ].[ - u3.sin(phi) + v3.cos(phi) ] =
+ (x2 - x1).u3.[ cos^2 + sin^2 ] + (y2 - y1).v3.[ sin^2 + cos^2 ] +
+ (x2 - x1).v3.[ cos.sin - sin.cos ] + (y2 - y1).u3.[ sin.cos - cos.sin ] =
+ (x2 - x1).u3 + (y2 - y1).v3

Conclusion: _both_ equations of the F.D. system are _invariant_ for rotations.
Now it is trivial that they are also invariant for translations (: differences
of all coordinates). So both F.D. equations are independent of the observer's
coordinate system as a whole.

Now we know from the "Weight Tracing" poster that L.S. elements allways can be
normed. So multiplying the  A(i)  with one and the same factor has no influence
whatsoever upon the element-matrix. But coefficients  A(i)  depend only upon
coordinate-differences. Consequently, the element-matrix must remain the same
if these distances are all multiplied with a constant factor. Therefore, two
element-matrices mus be the same if their element geometries only differ by a
scaling factor. In case of the Labrujere problem, this can be accomplished as
follows. Start with the circular cylinder at radial position R0; suppose the
mesh ends at radial position Rn. Then we have the little theorem:
|
| A mesh consisting of repeatable elements is given by radial positions:
|                      ^^^^^^^^^^
|      Ri = R0.(1 + dR/R0)^i      where:      dR = R0.(Rn/R0)^(1/n) - 1
|
Proof of the first relation: R(i+1) - R(i) = R0.(1 + dR/R0)^i.(1 + dR/R0 - 1) =
= Ri.dR/R0 . The other sides of the quadrilateral are:  2.Ri.sin(phi/2)  and
2.R(i+1).sin(phi/2) = 2.Ri.sin(phi/2).dR/R0 . Thus all sides scale up with only
one variable factor  Ri . Hence all quadrilateral (super)elements have the same
geometric form. The second relation follows from: Rn = R0.(1 + dR/R0)^n . QED .

The little theorem is BASICally implemented by the following program fragment.
Here it is assumed that the circular cylinder where the flow goes around has a
radius R = R0 = 1 . The outer boundary of the mesh is limited at R = Rn = 10 .

10 REM    The Labrujere Problem
20 REM    =====================
30 REM    Series on Unified Numerical Approximations
40 REM
50 REM   MESH GENERATOR
60 REM   --------------
70 INPUT "Number of radius meshlines (7) : ",NDR
80 INPUT "Number of angle  meshlines (5) : ",NPH
90 IF NDR<=0 THEN NDR=7
100 IF NPH<=0 THEN NPH=5
110 REM
---------------------------
230 PRINT "Coordinates ..."
240 OPEN "R",#1,"coord31.bin",8
250 FIELD #1, 4 AS X$, 4 AS Y$
260 PI=4*ATN(1) : DPHI=.5*PI/(NPH-1)
275 R=10 : P=R^(1/(NDR-1))-1 : REM Last meshline at R
280 FOR I=1 TO NDR
290 R=(1+P)^(I-1) : REM Gives equivalent elements
300 FOR J=1 TO NPH
310 PHI=(J-1)*DPHI : K=(I-1)*NPH+J
320 X=R*COS(PHI) : Y=R*SIN(PHI)
330 LSET X$=MKS$(X) : LSET Y$=MKS$(Y)
340 PUT #1,K : NEXT J : NEXT I
350 CLOSE #1
360 REM

It should hardly be defended that repeatable elements have the immence advance
that their element-matrix must be evaluated only _once_. Especially at a slow
PC, where I do my homework on, this saves me a whole lot of coffee & beer ;-).

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
* E-mail: Han.deBruijn@RC.TUDelft.NL --| Fax: +31 15 78 37 87 ----* Blood