sci.math.num-analysis SUNA34, Escape from inconstencies ================================= It was suggested in [32] that streamline velocities should be localized in the middle of the streamline that joins upstream point with downstream vertex. But, we have the following formulas from [33]: u = - dF2/J.(x2 - x1) - dF3/J.(x3 - x1) v = - dF2/J.(y2 - y1) - dF3/J.(y3 - y1) According to [15] is: dF2 = (y1 - y3).u2 - (x1 - x3).v2 dF3 = (y2 - y1).u3 - (x2 - x1).v3 Substitute: J.u = (x2 - x1).(y3 - y1).u2 - (y2 - y1).(x3 - x1).u3 - (x2 - x1).(x3 - x1).(v2 - v3) J.v = (x2 - x1).(y3 - y1).v3 - (y2 - y1).(x3 - x1).v2 - (y2 - y1).(y3 - y1).(u3 - u2) J = (x2 - x1).(y3 - y1) - (y2 - y1).(x3 - x1) The velocities (u,v) can only be _localized_ somewhere if u does NOT depend upon v2 and v3 , and if v does NOT depend upon u2 and u3 . Because only then there can exist isoparametric relationships u = Ni.ui and v = Ni.vi . Thus: (x2 - x1).(x3 - x1).(v2 - v3) = 0 (y2 - y1).(y3 - y1).(u3 - u2) = 0 There are _two_ possibilities: 1. v2 = v3 and u2 = u3 , meaning that the velocity field is uniform (again). Mighty uninteresting indeed, if SUFED is only valid for such cases. 2. Complete degeneration of triangles has to be excluded, of course. However, the following combinations still make sense: x2 = x1 and y3 = y1 Meaning that all triangles have to be rectangular. x3 = x1 and y2 = y1 Nothing is lleft of the "generalization" that Skew Upwind Finite Element Differences pretended to be. The accompanying velocities are respectively: u = u3 and v = v2 Here we find back the good old staggered positions, u = u2 and v = v3 which were invented by the D.B. Spalding school. Summary: the classical Finite Volume approach is found back in this case. Conclusion: the "streamline velocity" in the unified SUFED/FLOTRAN scheme [33] ---------- can NOT possibly be localized somewhere in the element. Another candidate for possible inconsistency has to be removed next. The following was found in [22], as well as in [26]: / O dF = (y2 - y1).u3 - (x2 - x1).v3 / + (y3 - y2).u1 - (x3 - x2).v1 + (y1 - y3).u2 - (x1 - x3).v2 = - J.(du/dx + dv/dy) = 0 / O dP = (x2 - x1).u3 + (y2 - y1).v3 / + (x3 - x2).u1 + (y3 - y2).v1 + (x1 - x3).u2 + (y1 - y3).v2 = - J.(dv/dx - dy/dy) = 0 But interpretation differs. In [22] velocities are concentrated in the middle of triangle sides, but in [26] at triangle vertices (as is usual with finite element schemes). Moreover, vertex velocities are positioned opposite to those in the middle of triangle sides. Furthermore, there are some nasty minus signs in the expressions - J.( ... ) = 0 . But after some 3 thought, the relationship between the two pictures is o as follows. When going from the midside-node to the / \ vertex-node formulation, the following substitutions 2 * * 1 are likely to be made: / \ o ----- * ----- o f1 := (f3 + f2)/2 1 3 2 f2 := (f1 + f3)/2 where f = u,v respectively. f3 := (f2 + f1)/2 So: (y2 - y1).(u2 + u1)/2 - (x2 - x1).(v2 + v1)/2 + (y3 - y2).(u3 + u2)/2 - (x3 - x2).(v3 + v2)/2 + (y1 - y3).(u1 + u3)/2 - (x1 - x3).(v1 + v3)/2 := minus 1/2 times: (y2 - y1).u3 - (x2 - x1).v3 ^^^^^^^^^^^^^^^ + (y3 - y2).u1 - (x3 - x2).v1 + (y1 - y3).u2 - (x1 - x3).v2 / // Or: O (u.dy - v.dx) = - 1/2. - J.(du/dx + dv/dy) = || (du/dx + dv/dy).dx.dy / // .1/2.J Great relief. It is thus possible to switch from velocities at the vertices of triangles to velocities at midside nodes, without introducing further trouble. The reverse pattern, going from midside nodes to vertices, seems to be somewhat more problematic. 3 If the midpoints (*) of a triangle with vertices (o) are o joined, then a new triangle is formed with vertices (*), / \ and this triangle has the _same shape_ as the former. 2 * ----- * 1 Thus the triangle with vertices (*) is geometrically / \ / \ _equivalent_ with the triangle having (*) as midside o ----- * ----- o nodes. This explains why the equations accompanying 1 3 2 both triangles always have the same form. Differences may be in connectivity, but NOT in coefficients of the element matrix. (Of course, the scaling factor -2 doesn't matter at all.) Both kind of triangles were met before: in [22] (midside) and in [26] (vertex). It should be emphasized that fluxes dF are related to _midside_ and no never to vertex velocities. - * 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