Page 18 - Basic Structured Grid Generation

Mathematical preliminaries – vector and tensor analysis 7

The cofactors of the matrix L in eqn (1.22) are the various background carte-
sian components of (g j × g k ), which may be expressed, with the notation used in
eqn (1.18), as
α 1 = y η z ς − y ς z η , α 2 = x ς z η − x η z ς , α 3 = x η y ς − x ς y η

β 1 = y ς z ξ − y ξ z ς , β 2 = x ξ z ς − x ς z ξ , β 3 = x ς y ξ − x ξ y ς (1.35)
γ 1 = y ξ z η − y η z ξ , γ 2 = x η z ξ − x ξ z η , γ 3 = x ξ y η − x η y ξ
so that
g 2 × g 3 = α 1 i + α 2 j + α 3 k, g 3 × g 1 = β 1 i + β 2 j + β 3 k, g 1 × g 2 = γ 1 i + γ 2 j + γ 3 k
(1.36)
and
1 1 1
1 2 3
g = √ (α 1 i+α 2 j+α 3 k), g = √ (β 1 i+β 2 j+β 3 k), g = √ (γ 1 i+γ 2 j+γ 3 k).
g g g
(1.37)
Since M = L −1 , we also have, in the same notation, the matrix elements of M:
√ √ √
ξ x = α 1 / g, ξ y = α 2 / g, ξ z = α 3 / g
√ √ √
η x = β 1 / g, η y = β 2 / g, η z = β 3 / g (1.38)
√ √ √
ς x = γ 1 / g, ς y = γ 2 / g, ς z = γ 3 / g.
Exercise 3. Using eqn (1.29) and standard determinant expansions, derive the follow-
ing formulas for the determinant g:
g = g 11 G 1 + g 12 G 4 + g 13 G 5 = (α 1 x ξ + β 1 x η + γ 1 x ς ) 2

= g 22 G 2 + g 12 G 4 + g 23 G 6 = (α 2 y ξ + β 2 y η + γ 2 y ς ) 2 (1.39)
2
= g 33 G 3 + g 13 G 5 + g 23 G 6 = (α 3 z ξ + β 3 z η + γ 3 z ς ) .
From eqn (1.32) it follows that
1
ip
i
p
g = g · g = (g j × g k ) · (g q × g r ),
g
where {i, j, k} and {p, q, r} are in cyclic order {1, 2, 3}. Using the standard Lagrange
vector identity
(A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C), (1.40)
we have
1
ip
g = {(g j · g q )(g k · g r ) − (g j · g r )(g k · g q )}
g
1
= (g jq g kr − g jr g kq ). (1.41)
g
For example,
1
13
g = (g 21 g 32 − g 22 g 31 ).
g