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