sci.math.numanalysis
SUNA53, Midside quads for internal flow
=======================================
Three "good" Least Squares Finite Elements have been developed for Ideal Flow
around a circular cylinder (: the Labrujere problem). They are depicted below.
o  *  o o  *  o o  *  o
 \ /   /  \   
 \ /   /  \   
* * * *  *  * * *
 / \   \  /   
 / \   \  /   
o  *  o o  *  o o  *  o
SUNA24 SUNA28 SUNA29
Staggered Patched Linearized
It can be shown, in addition, that the three cases are more or less equivalent.
This is done at first for the Staggered and Patched elements.
3 3 3
o o o
/ \ / \ / \
2 * * 1 / \ 2 *  * 1
/ \ / \ / \ / \
o  *  o o  o o  *  o
1 3 2 1 2 1 3 2
SUNA22 SUNA26 Unified
Both SUNA22 and SUNA26 give rise to the following sets of equations:
(y3  y2).u1  (x3  x2).v1 + (x3  x2).u1 + (y3  y2).v1 +
(y1  y3).u2  (x1  x3).v2 + (x1  x3).u2 + (y1  y3).v2 +
(y2  y1).u3  (x2  x1).v3 = 0 (x2  x1).u3 + (y2  y1).v3 = 0
In SUNA22 geometry is defined at the vertices (o) while velocities are defined
at the midside nodes (*): a staggered mesh. In SUNA26 _both_ geometry and flow
velocities are defined at the vertices (o), as usually is the case with F.E.A.
It was explained in our SUNA34 poster how exactly the same set of equations can
emerge for these two different cases. Essentially, to be honest, it is a mere
consequence of the "handsom" numbering which was assigned to the midside nodes.
3 The midsidenode equations for the "big" triangle with vertices
o (o) are exactly the same as the vertexnode equations for the
 \ geometrically equivalent "small" triangle with vertices (*).
 \ The "staggered" quadrilateral is built up from the former,
2 *  * 1 while the "patched" quadrilateral is built up from the
 \  \ latter. This already completes the proof. So we are
 \  \ stucked with only TWO working elements for Ideal
1 o  *  o 2 Flow, in the end. And even less than that, for
3 it was shown in SUNA28 how the Linearized element
emerges from the Patched form. OK. So far so good.
The following is quoted (without permission ;) from SUNA26:
> 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)
A coupling to cylindrical symmetric Internal Flow can be forced by recognizing
that the mass conservation equation of the latter case can also be written as:
d(r.u)/dr + d(r.v)/dz = 0
While the P.D.E. for irrotational has exactly the same form as in the cartesian
coordinate system.
Thus substitute r := x , z := y and r.u := u , r.v := v _only_ in the mass
conservation equation. Then we get back the Cauchy equations for flow around a
circular cylinder. Reproducing the argument quoted > above with just the proper
substitutions made in it, such as u := r.u and v := r.v only in (a) , then
we get:
r2.u2  r1.u1 + r4.u4  r3.v3 = 0 (1)
v2  v1  u4 + u3 = 0 (2)
And:
2.r5.u5 = r1.u1 + r2.u2 = r3.u3 + r4.u4 (3 & 4)
2.v5 = v1 + v2 = v3 + v4 (5 & 6)
This gives us an alternative for the reduced quadrilateral. Reading again the
part of this poster, which establishes the equivalence between three possible
alternatives, due to the rectangular geometry, there isn't really much choice
left. Taking into account the rest of the geometry and velocities numbering:
REM Incompressible: 5,6
A(3)=+R2*DZ : A(7)=R1*DZ : A(6)=+RM*DR : A(2)=RM*DR o  *  o
REM Irrotational:  
A(4)=+DZ : A(8)=DZ : A(5)=DR : A(1)=+DR 7,8 * * 3,4
REM Linearized u :  
A(7)=+R1 : A(3)=+R2 : A(1)=RM : A(5)=RM o  *  o
REM Linearized v : 1,2
A(8)=+1 : A(4)=+1 : A(2)=1 : A(6)=1
It is noticed that: RM*DR = (r2^2  r1^2)/2 , which is consistent with earlier
findings (: SUNA39).
As far as the possible element forms are concerned, it seems we are finished.
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