sci.math.num-analysis SUNA09, Discretization at Quadrilaterals ======================================== In the previous article "SUNA, Discretization at Triangles", the following general formula's were found: J.dT/dx = (dy/dk.dT/dh - dy/dh.dT/dk) J.dT/dy = (dx/dh.dT/dk - dx/dk.dT/dh) where: J = dx/dh.dy/dk - dx/dk.dy/dh Here (x,y) = global coordinates, (h,k) = local coordinates, T = temperature. Let's apply the same discretization technique to a general quadrilateral. Isoparametric transformations are defined again for all nodal quantities, such as the unknown function T , and also the global coordinates (x,y) : x = (1-h).(1-k).x1 + h.(1-k).x2 + (1-h).k.x3 + h.k.x4 3 o-------o 4 y = (1-h).(1-k).y1 + h.(1-k).y2 + (1-h).k.y3 + h.k.y4 | \ / | T = (1-h).(1-k).T1 + h.(1-k).T2 + (1-h).k.T3 + h.k.T4 | X | | / \ | So: J.dT/dx = 1 o-------o 2 [ (y3 - y1).(1-h) + (y4 - y2).h ][ (T2 - T1).(1-k) + (T4 - T3).k ] - [ (y2 - y1).(1-k) + (y4 - y3).k ][ (T3 - T1).(1-h) + (T4 - T2).h ] = (1-h).(1-k). [ (y3 - y1).(T2 - T1) - (y2 - y1).(T3 - T1) ] + h .(1-k). [ (y4 - y2).(T2 - T1) - (y2 - y1).(T4 - T2) ] + (1-h). k . [ (y3 - y1).(T4 - T3) - (y4 - y3).(T3 - T1) ] + h . k . [ (y4 - y2).(T4 - T3) - (y4 - y3).(T4 - T2) ] Remember the triangle result: J.dT/dx = (y3 - y1).(T2 - T1) - (y2 - y1).(T3 - T1) And J.dT/dy = [ (x2 - x1).(1-k) + (x4 - x3).k ][ (T3 - T1).(1-h) + (T4 - T2).h ] - [ (x3 - x1).(1-h) + (x4 - x2).h ][ (T2 - T1).(1-k) + (T4 - T3).k ] = (1-h).(1-k). [ (x2 - x1).(T3 - T1) - (x3 - x1).(T2 - T1) ] + h .(1-k). [ (x2 - x1).(T4 - T2) - (x4 - x2).(T2 - T1) ] + (1-h). k . [ (x4 - x3).(T3 - T1) - (x3 - x1).(T4 - T3) ] + h . k . [ (x4 - x3).(T4 - T2) - (x4 - x2).(T4 - T3) ] Remember the triangle result: J.dT/dy = (x2 - x1).(T3 - T1) - (x3 - x1).(T2 - T1) Now replace (1,2,3) by (2,4,1) , (3,1,4) , (4,3,2) and respectively attach labels 1, 2, 3 and 4 to these _triangles_. Then for J.dT/dx and J.dT/dy , and of course for all linear combinations such as [ u.dT/dx + v.dT/dy ] too, discretizations LQ on a quadrilateral can be expressed into discretizations Lp at the triangles labeled p = 1,2,3,4 , according to the above: LQ = (1-h).(1-k).L1 + h.(1-k).L2 + (1-h).k.L3 + h.k.L4 We can go even one step further by recognizing the F.D. equivalence with: LQ = L1 + h.(L2 - L1) + k.(L3 - L1) + h.k.(L1 - L2 - L3 + L4) Replacing (L1 - L2 - L3 + L4) by (J1 - J2 - J3 + J4) learns that the latter expression must be zero (: previous SUNA). However, due to the isoparametrics, there does _not_ exist a formal difference between the J's and the L's: their algebra is exactly the same. So we must conclude that also (L1 - L2 - L3 + L4) is always zero. Right? Let's summarize: | A Finite Element discretization at a _quadrilateral_ can be expressed into | discretizations at each of all the triangles which cover the quadrilateral; | insofar as _first_ order derivatives of the unknown functions are involved, | which are multiplied by Jacobian determinants. | Let the former discretizations be denoted as LQ , and the latter be labeled | as L1, L2, L3, L4, then LQ is (not a bi)_linear_ interpolation of them: | | LQ = (1-h).(1-k).L1 + h.(1-k).L2 + (1-k).h.L3 + h.k.L4 | = L1 + h.(L2 - L1) + k.(L3 - L1) , or something equivalent. | | Here 0 < h,k < 1 are the local coordinates of the quadrilateral. Conclusion: Quadrilaterals seem to be composites of Triangles. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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