sci.math.num-analysis SUNA51, Superelement Details ============================ A Least Squares Finite Element for Incompressible and Irrotational Flow has been devised according to the method in Re: SUNA24, Staggered Quadrilaterals. The Finite Difference equations for an elementary triangle are repeated: [ - y32 + x32 - y13 + x13 - y12 + x12 ] | u1 | [ + x32 + y32 + x13 + y13 + x12 + y12 ] | v1 | | u2 | = 0 Here: xji = xj - xi ; yji = yj - yi ; i,j = 1,2,3 . | v2 | | u3 | The Least Squares Finite Element matrix is constructed | v3 | by multiplying the transpose of the above 6*2 matrix with itself. Let: X1 = x3 - x2 ; X2 = x1 - x3 ; X3 = x2 - x1 Y1 = y3 - y2 ; Y2 = y1 - y3 ; Y3 = y2 - y1 : circular. And: Iij = Xi.Xj + Yi.Yj ; Uij = Xi.Yj - Xj.Yi ; i,j = 1,2,3 . It is easily seen that U12 = U23 = U31 , since all these quantities equalize 2 times the area of the triangle element. Therefore Uij is replaced by U . A special case occurs for i = j : Uii = 0 , giving zero matrix entries. The result of the above proposed matrix multiplication can be written now as: | I11 0 I12 + U I13 - U | | I11 - U I12 + U I13 | | I22 0 I23 + U | = L.S. element matrix | I22 - U I23 | | symmetric I33 0 | | I33 | The velocity x-components as well as the velocity y-components are coupled by admittances (ui,uj) or (vi,vj). These are _inner_ products Iij of coordinate differences of the triangle sides. The coupling is strongest for the identical velocity components (ui,ui) or (vi,vi). The inner products can be written as: Iij = | Ri || Rj |.cos(ij) where ij = 0 for i = j : Iii = | Ri |^2 By drawing a picture, it can be seen that inner products with i <> j will be negative most of the time. Provided that all angles of the triangle element are < 90 degrees. An inner product equals zero if the triangle element has a right angle = 90 degrees. This might well occur with rectangular grids, the classical Finite Difference case. Obtuse angle give rise to positive off-diagonal matrix coefficients. According to one of Patankar's "Five Basic Rules", this should be avoided, assuming these rules are applicable here too. There is some evidence for this in the fact that inner products of the above kind also play a role in finite element matrices which are derived for: "SUNA42, Discretization for 2-D diffusion". So there exist kind of _DIFFUSION-like couplings_ between velocity components of the same kind. At the same time, it is well known that u and v indeed separately obey Laplace's equations, analytically. The velocity x- and y-components are also coupled by admittances (ui,vj). These are all equal to plus or minus 2 times the area of the triangle element: +/- U. Consequently, positive off-diagonal coefficients between velocity components of _different_ kind can NOT be avoided. Because the quantity U denotes the area the triangle element, and it is also in the denominator of the matrix elements for diffusion, it is advantageous to divide everything by this quantity U . We can make the above more explicit by writing the element matrix as follows: | I11 I12 I13 || u1 | | I11 I12 I13 || v1 | 1/U.| I22 I23 || u1 | + 1/U.| I22 I23 || v2 | + | symm. I33 || u1 | | symm. I33 || v3 | | . 0 . + . - || u1 | | . - . + . || v1 | + 1.| . 0 . + || u2 | | . - . || v2 | = 0 due to taking out diffusion | symmetric . 0 || u3 | | . || v3 | The last matrix is a bit difficult to interpret at this moment. Suppose for a moment that there are no "diffusion" contributions, then setting the remaining equations to zero implies that: + v2 - v3 = 0 ; - u2 + u3 = 0 ; Basically: - v1 + v3 = 0 ; + u1 - u3 = 0 ; + v1 - v2 = 0 ; - u1 + u2 = 0 . u1 = u2 = u3 ; v1 = v2 = v3 . Thus the x-components as well as the y-components of the velocities across the triangle tend to be equalized. Which can be understood, anyway. Consider the following patch of two mutually overlapping triangles, which is actually used in defining the quadrilateral superelement for ideal flow: ____*____ Concentrate upon the velocity node indicated as (1). |\ 4 /| When assembling the matrices belonging to the different | \ / | triangles for that node, it receives contributions 5* 1 *3 (u1,v2) = + 1 , (v1,u2) = - 1 from triangle (1,2,3) ; | / \ | (u1,v2) = - 1 , (v1,u2) = + 1 from triangle (1,5,2) . |/___*___\| The net result is that node (1) receives NO contribution 2 from the node (2). The same argument holds if we replace (2) by (3) or (4) or (5). All non-diffusive contributions simply cancel out. (Note that the quadrilateral in general is not rectangular.) Consequently, node (1) is attached to the others nodes (2,3,4,5) by diffusion _only_. In the special case of a _square_ superelement, this means that: 4.u1 = u2 + u3 + u4 + u5 ; 4.v1 = v2 + v3 + v4 + v5 : mean values (1). - * 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