sci.math.numanalysis
SUNA09, Discretization at Quadrilaterals
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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 = (1h).(1k).x1 + h.(1k).x2 + (1h).k.x3 + h.k.x4 3 oo 4
y = (1h).(1k).y1 + h.(1k).y2 + (1h).k.y3 + h.k.y4  \ / 
T = (1h).(1k).T1 + h.(1k).T2 + (1h).k.T3 + h.k.T4  X 
 / \ 
So: J.dT/dx = 1 oo 2
[ (y3  y1).(1h) + (y4  y2).h ][ (T2  T1).(1k) + (T4  T3).k ] 
[ (y2  y1).(1k) + (y4  y3).k ][ (T3  T1).(1h) + (T4  T2).h ]
= (1h).(1k). [ (y3  y1).(T2  T1)  (y2  y1).(T3  T1) ]
+ h .(1k). [ (y4  y2).(T2  T1)  (y2  y1).(T4  T2) ]
+ (1h). 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).(1k) + (x4  x3).k ][ (T3  T1).(1h) + (T4  T2).h ] 
[ (x3  x1).(1h) + (x4  x2).h ][ (T2  T1).(1k) + (T4  T3).k ]
= (1h).(1k). [ (x2  x1).(T3  T1)  (x3  x1).(T2  T1) ]
+ h .(1k). [ (x2  x1).(T4  T2)  (x4  x2).(T2  T1) ]
+ (1h). 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 = (1h).(1k).L1 + h.(1k).L2 + (1h).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 = (1h).(1k).L1 + h.(1k).L2 + (1k).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
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