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SUNA07, Specializations of 3D SUFED
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This article is part of the "Series on Unified Numerical Approximations".
SKEW UPWIND FINITE ELEMENT DIFFERENCES (SUFED)
FOR PURE STEADY UNIFORM CONVECTION IN THREE DIMENSIONS
PART II : SPECIAL CASES
In order to make the meaning of the equations clear, a unit cube is subdivided
into tetrahedra. These tetrahedra act like "funnels", which conduct the flow
towards the vertex with index (0). The funnels must be such that the whole flow
in the first octant is "catched up" by them; nothing "leaks through".
At first, a proper local numbering of the unit cube vertices should be defined.
As follows. The vertices are numbered in binary, with 1 bit per dimension:
Numbering bit pattern x y z
0 000 (0,0,0)
1 001 (1,0,0)
2 010 (0,1,0)
3 011 (1,1,0) (1)
4 100 (0,0,1)
5 101 (1,0,1)
6 110 (0,1,1)
7 111 (1,1,1)
Note that the unit coordinates conform to the reversed bit pattern of the node
numbering. Also note that the numbering starts with 0 instead of 1.
Then there exist several possibilities for arranging funnels in a cube: four to
be precise. All four arrangements correspond to a special SUFED scheme.
0. Unskewed scheme
The zero'th possibility is denoted as such, because it is nothing new.
In the first octant, it involves 4 vertices and 1 tetrahedron, namely the unit
tetrahedron. Its base vectors, according to table (1), are not transformed:
(0,1,2,4) : (1,0,0) (0,1,0) (0,0,1)
The upwind scheme corresponding to this is the well known equation (Part I,3):
U = u , V = v , W = w . The complete scheme has a 7 node computational molecule.
1. First skewed scheme
The first skewed scheme involves 5 nodes and 3 tetrahedra in the first octant.
(0,1,2,7) , (0,1,4,7) , (0,2,4,7)
The base vectors are transformed according to table (1):
Unit tetrahedron: (1,0,0) (0,1,0) (0,0,1)
(0,1,2,7) : (1,0,0) (0,1,0) (1,1,1)
(0,1,4,7) : (1,0,0) (0,0,1) (1,1,1)
(0,2,4,7) : (0,1,0) (0,0,1) (1,1,1)
Which induces a transformation of the velocities, according to (Part I,4):
(0,1,2,7) : u = U + W , v = V + W , w = W
(0,1,4,7) : u = U + W , v = W , w = V + W
(0,2,4,7) : u = W , v = U + W , w = V + W
Determine the inverses of these transformations:
1 : U = u  w , V = v  w , W = w
2 : U = u  v , V = w  v , W = v
3 : U = v  u , V = w  u , W = u
Substitute this into the classical upwind scheme (Part I,3).
Also the upwind conditions U <= 0 , V <= 0 , W <= 0 must be transformed.
This finally results in:
1 : (u  w).(T0  T1) + (v  w).(T0  T2) + w.(T0  T7) = 0
2 : (u  v).(T0  T1) + (w  v).(T0  T4) + v.(T0  T7) = 0
3 : (v  u).(T0  T2) + (w  u).(T0  T4) + u.(T0  T7) = 0
1 : for u,v <= w <= 0
2 : for u,w <= v <= 0
3 : for v,w <= u <= 0
Similar equations can be derived for the other octants. The completed scheme
has a 15 node computational molecule, and covers all possible (u,v,w) cases.
2. Second skewed scheme
The second skewed scheme involves 7 nodes and 4 tetrahedra in the first octant:
(0,3,5,6) , (0,1,3,5) , (0,2,3,6) , (0,4,5,6)
The base vectors are transformed according to table (1):
Unit tetrahedron: (1,0,0) (0,1,0) (0,0,1)
(0,3,5,6) : (1,1,0) (1,0,1) (0,1,1)
(0,1,3,5) : (1,0,0) (1,1,0) (1,0,1)
(0,2,3,6) : (0,1,0) (1,1,0) (0,1,1)
(0,4,5,6) : (0,0,1) (1,0,1) (0,1,1)
Which induces a transformation of the velocities, according to (Part I,4):
(0,3,5,6) : u = U + V , v = U + V , w = V + W
(0,1,3,5) : u = U + V + W , v = V , w = W
(0,2,3,6) : u = V , v = U + V + W , w = W
(0,4,5,6) : u = V , v = W , w = U + V + W
Determine the inverses of these transformations:
1 : U = (u + v  w)/2 , V = (u + w  v)/2 , W = (v + w  u)/2
2 : U = u  v  w , V = v , W = w
3 : U = v  u  w , V = u , W = w
4 : U = w  u  v , V = u , W = v
Substitute this into the classical upwind scheme (Part I,3).
Also the upwind conditions U <= 0 , V <= 0 , W <= 0 must be transformed.
This finally results in:
1 : (u + v  w).(T0  T3) + (u + w  v).(T0  T5) + (v + w  u).(T0  T6) = 0
2 : (u  v  w).(T0  T1) + v.(T0  T3) + w.(T0  T5) = 0
3 : (v  u  w).(T0  T2) + u.(T0  T3) + w.(T0  T6) = 0
4 : (w  u  v).(T0  T4) + u.(T0  T5) + v.(T0  T6) = 0
1 : for u + v <= w , u + w <= v , v + w <= u
2 : for u <= v + w , v <= 0 , w <= 0
3 : for v <= u + w , u <= 0 , w <= 0
4 : for w <= u + v , u <= 0 , v <= 0
Similar equations can be derived for the other octants. The completed scheme
has a 19 node computational molecule, and covers all possible (u,v,w) cases.
3. Third skewed scheme
The third skewed scheme involves 8 nodes and 6 tetrahedra in the first octant:
(0,1,3,7) , (0,1,5,7) , (0,2,3,7) , (0,2,6,7) , (0,4,5,7) , (0,4,6,7)
The base vectors are transformed according to table (1):
Unit tetrahedron: (1,0,0) (0,1,0) (0,0,1)
(0,1,3,7) : (1,0,0) (1,1,0) (1,1,1)
(0,1,5,7) : (1,0,0) (1,0,1) (1,1,1)
(0,2,3,7) : (0,1,0) (1,1,0) (1,1,1)
(0,2,6,7) : (0,1,0) (0,1,1) (1,1,1)
(0,4,5,7) : (0,0,1) (1,0,1) (1,1,1)
(0,4,6,7) : (0,0,1) (0,1,1) (1,1,1)
Substitute the vertex coordinates into the velocitytransformations (Part I,4):
(1,3,7): u = U + V + W , v = V + W , w = W
(1,5,7): u = U + V + W , v = W , w = V + W
(2,3,7): u = V + W , v = U + V + W , w = W
(2,6,7): u = W , v = U + V + W , w = V + W
(4,5,7): u = V + W , v = W , w = U + V + W
(4,6,7): u = W , v = V + W , w = U + V + W
And determine the inverse transformations:
1 : U = u  v , V = v  w , W = w
2 : U = u  w , V = w  v , W = v
3 : U = v  u , V = u  w , W = w
4 : U = v  w , V = w  u , W = u
5 : U = w  u , V = u  v , W = v
6 : U = w  v , V = v  u , W = u
Substitute this into the classical upwind scheme (Part I,3) and simplify. Also
the upwind conditions U < 0 , V < 0 , W < 0 must be transformed. This finally
results in:
u.(T0  T1) + v.(T1  T3) + w.(T3  T7) = 0 for u <= v <= w <= 0
u.(T0  T1) + w.(T1  T5) + v.(T5  T7) = 0 for u <= w <= v <= 0
v.(T0  T2) + u.(T2  T3) + w.(T3  T7) = 0 for v <= u <= w <= 0
v.(T0  T2) + w.(T2  T6) + u.(T6  T7) = 0 for v <= w <= u <= 0
w.(T0  T4) + u.(T4  T5) + v.(T5  T7) = 0 for w <= u <= v <= 0
w.(T0  T4) + v.(T4  T6) + u.(T6  T7) = 0 for w <= v <= u <= 0
In fact, this is the "skewest" SUFED scheme that exists in 3D.
It is remarked that special cases occur for u = v , v = w , u = w , assuring
a smooth transition between the different parts of the scheme. For u = v = w
all parts of the scheme are identical, simply resulting in:
T0 = T7
Similar equations can be derived for the other octants. The completed scheme
has a 27 node computational molecule, and covers all possible (u,v,w) cases.
Acknowledgement:
The skewed schemes (1.) and (2.) were discovered by David Paterson.
To be continued later.

* 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
* Email: Han.deBruijn@RC.TUDelft.NL  Fax: +31 15 78 37 87 * Blood