sci.math.num-analysis SUNA10, Discretization at F.D. Stars ==================================== In a 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 F.E. discretization technique to a Finite Difference Star. Isoparametric transformations are defined again for all nodal quantities, such as the unknown function T , and also the global coordinates (x,y) . The following derivatives were already found (Re: SUNA, Five Point Star): dx/dh = (1/2 - h).(x1 - x2) + (1/2 + h).(x3 - x1) 5 dy/dh = (1/2 - h).(y1 - y2) + (1/2 + h).(y3 - y1) / | \ dx/dk = (1/2 - k).(x1 - x4) + (1/2 + k).(x5 - x1) / | \ dy/dk = (1/2 - k).(y1 - y4) + (1/2 + k).(y5 - y1) 2 --- 1 --- 3 \ | / Paerial derivatives of the unknown function T must \ | / have analogous shape, of course: 4 dT/dh = (1/2 - h).(T1 - T2) + (1/2 + h).(T3 - T1) dT/dk = (1/2 - k).(T1 - T4) + (1/2 + k).(T5 - T1) So: J.dT/dx = [ (1/2-k).(y1-y4) + (1/2+k).(y5-y1) ].[ (1/2-h).(T1-T2) + (1/2+h).(T3-T1) ] - [ (1/2-h).(y1-y2) + (1/2+h).(y3-y1) ].[ (1/2-k).(T1-T4) + (1/2+k).(T5-T1) ] = = (1/2 - h).(1/2 - k).[ (y1 - y4).(T1 - T2) - (y1 - y2).(T1 - T4) ] + (1/2 + h).(1/2 - k).[ (y1 - y4).(T3 - T1) - (y3 - y1).(T1 - T4) ] + (1/2 - h).(1/2 + k).[ (y5 - y1).(T1 - T2) - (y1 - y2).(T5 - T1) ] + (1/2 + h).(1/2 + k).[ (y5 - y1).(T3 - T1) - (y3 - y1).(T5 - T1) ] Remember the triangle result: J.dT/dx = (y3 - y1).(T2 - T1) - (y2 - y1).(T3 - T1) And J.dT/dy = [ (1/2-h).(x1-x2) + (1/2+h).(x3-x1) ].[ (1/2-k).(T1-T4) + (1/2+k).(T5-T1) ] - [ (1/2-k).(x1-x4) + (1/2+k).(x5-x1) ].[ (1/2-h).(T1-T2) + (1/2+h).(T3-T1) ] = = (1/2 - h).(1/2 - k).[ (x1 - x2).(T1 - T4) - (x1 - x4).(T1 - T2) ] + (1/2 + h).(1/2 - k).[ (x3 - x1).(T1 - T4) - (x1 - x4).(T3 - T1) ] + (1/2 - h).(1/2 + k).[ (x1 - x2).(T5 - T1) - (x5 - x1).(T1 - T2) ] + (1/2 + h).(1/2 + k).[ (x3 - x1).(T5 - T1) - (x5 - x1).(T3 - T1) ] Remember the triangle result: J.dT/dy = (x2 - x1).(T3 - T1) - (x3 - x1).(T2 - T1) Replace (1,2,3) by (1,3,5) , (1,5,2) , (1,2,4) , (1,4,3) . 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 LS on a F.D. Star can be expressed into discretizations Lp at the triangles labeled p = 1,2,3,4 , according to the above: LS = N1(h,k).L1 + N2(h,k).L2 + N3(h,k).L3 + N4(h,k).L4 Where: Np(h,k) = (1/2 -|+ h).(1/2 -|+ k) ("|" = "xor") Recognized is the F.D. equivalence with: LS = 1/4.(L1 + L2 + L3 + L4) + h/2.(L2 - L1 + L4 - L3) + k/2.(L3 + L4 - L1 - L2) + h.k.(L1 + L4 - L2 - L3) Summarizing: | A Finite Element discretization at a _F.D. star_ can be expressed into | discretizations at each of the triangles which are spanned by its arms; | insofar as first order derivatives of the unknown functions are involved, | which are multiplied by Jacobian determinants. | Let the former discretizations be denoted as LS , and the latter be labeled | as L1, L2, L3, L4, then LS is a _bilinear_ interpolation of L1,L2,L3,L4 . Conclusion: F.D. Stars 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