Page 26 - Basic Structured Grid Generation

Mathematical preliminaries – vector and tensor analysis 15

Evaluating the scalar products in eqns (1.98) and (1.99) on background cartesians
gives the formulas

2 2 k
∂ y l ∂y l k ∂ y l ∂x
[ij, k]= , = . (1.102)
ij
j
i
j
i
∂x ∂x ∂x k ∂x ∂x ∂y l
Exercise 7. Using eqns (1.100), (1.102), with (1.33) and (1.34), verify the formula
2
2
2
∂ x k ∂ y k ∂ z
k
k
= K + L + M , (1.103)
ij i j i j i j
∂x ∂x ∂x ∂x ∂x ∂x
where (in the obvious notation)
(G 1 x ξ + G 4 x η + G 5 x ς ) (G 4 x ξ + G 2 x η + G 6 x ς )
1 2
K = ; K = ;
g g
(G 5 x ξ + G 6 x η + G 3 x ς )
3
K =
g
(G 1 y ξ + G 4 y η + G 5 y ς ) (G 4 y ξ + G 2 y η + G 6 y ς )
1 2
L = ; L = ;
g g
(G 5 y ξ + G 6 y η + G 3 y ς )
3
L =
g
(G 1 z ξ + G 4 z η + G 5 z ς ) (G 4 z ξ + G 2 z η + G 6 z ς )
1 2
M = ; M = ;
g g
(G 5 z ξ + G 6 z η + G 3 z ς )
3
M = .
g
i
Expressions for the derivatives of contravariant base vectors g may be obtained by
k
differentiating eqn (1.6) with respect to x , which gives
i ∂g j ∂g i
g · + · g j = 0.
∂x k ∂x k
Hence
∂g i i ∂g j i l i l i
· g j =−g · =−g · g l =−δ jk =− . (1.104)
jk
l
jk
∂x k ∂x k
Comparison with eqn (1.99) now shows that
∂g i i k
=− g . (1.105)
jk
∂x j
The metric tensor g ij can also be differentiated:
∂g ij ∂ ∂g j ∂g i l l
= (g i · g j ) = g i · + · g j = g i ·[jk, l]g +[ik, l]g · g j
∂x k ∂x k ∂x k ∂x k
l
l
=[jk, l]δ +[ik, l]δ =[jk, i]+[ik, j]. (1.106)
j
i
Exercise 8. By differentiating both sides of eqn (1.28), and using eqns (1.101) and
(1.106), show that
∂g lm jl m jm l
=−g jk − g . (1.107)
jk
∂x k