Page 17 - Basic Structured Grid Generation

6 Basic Structured Grid Generation

∂y j ∂x i
and L ij = i , M ij = ; it may be seen directly from eqn (1.6) that
∂x ∂y j
T
LM = I, (1.23)
where I is the 3 × 3 identity matrix. Thus L and M T are mutual inverses. Moreover

det L ={g 1 · (g 2 × g 3 )}= V (1.24)
T
as previously defined in eqn (1.8). Since M = L −1 , it follows that
2
3
1
det M ={g · (g × g )}= V −1 . (1.25)
ij
It is easy to see that the symmetric matrix arrays (g ij ) and (g ) for the associated
metric tensors are now given by
ij
T
T
(g ij ) = LL , (g ) = MM . (1.26)
T −1
T
Since M = L −1 and M = (L ) , it follows that
ij −1
(g ) = (g ij ) . (1.27)
In component form this is equivalent to
jk j
g ik g = δ . (1.28)
i
From the properties of determinants it also follows that

2 2
g = det(g ij ) = (det L) = V , (1.29)
ij −1
det(g ) = g , (1.30)
and
√
V ={g 1 · (g 2 × g 3 )}= g, (1.31)
where g must be a positive quantity.
Thus in place of eqn (1.8) we can write
1 1 1
1 2 3
g = √ g 2 × g 3 , g = √ g 3 × g 1 , g = √ g 1 × g 2 . (1.32)
g g g
From eqn (1.27) and standard 3 × 3 matrix inversion, we can also deduce the fol-
lowing formula:
 
1 G 1 G 4 G 5
ij
g =  G 4 G 2 G 6  , (1.33)
g
G 5 G 6 G 3
where the co-factors of (g ij ) are given by

2 2 2
G 1 = g 22 g 33 − (g 23 ) , G 2 = g 11 g 33 − (g 13 ) , G 3 = g 11 g 22 − (g 12 )
G 4 = g 13 g 23 − g 12 g 33 , G 5 = g 12 g 23 − g 13 g 22 , G 6 = g 12 g 13 − g 23 g 11 . (1.34)