sci.math.num-analysis

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 = (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 ...
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* Han de Bruijn; Applications&Graphics | "A little bit of Physics * No
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