sci.math.numanalysis
SUNA10, Discretization at F.D. Stars
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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 =
[ (1/2k).(y1y4) + (1/2+k).(y5y1) ].[ (1/2h).(T1T2) + (1/2+h).(T3T1) ] 
[ (1/2h).(y1y2) + (1/2+h).(y3y1) ].[ (1/2k).(T1T4) + (1/2+k).(T5T1) ] =
= (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 =
[ (1/2h).(x1x2) + (1/2+h).(x3x1) ].[ (1/2k).(T1T4) + (1/2+k).(T5T1) ] 
[ (1/2k).(x1x4) + (1/2+k).(x5x1) ].[ (1/2h).(T1T2) + (1/2+h).(T3T1) ] =
= (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
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