Page 199 - Basic Structured Grid Generation

188 Basic Structured Grid Generation

Hence
√
g = l(y b − y a ). (7.42)
Moreover,

∂y a dy b
x ξ = l, x η = 0, y ξ = l (1 − η) + η , y η = (y b − y a ). (7.43)
∂x dx
It follows from eqn (1.158) that

2
∂y a dy b
g 11 = l 2 1 + (1 − η) + η ,
∂x dx
2
2
g 22 = (y b − y a ) = g/l , (7.44)

∂y a dy b
g 12 = l (1 − η) + η (y b − y a ).
∂x dx
where η is given in terms of x, y, and t by eqn (7.40).
Also, from (1.163),
11 2
g = 1/l ,
2

∂y a dy b
g 22 = 1 + (1 − η) + η (y b − y a ) −2 , (7.45)
∂x dx
1 ∂y a dy b −1
12
g =− (1 − η) + η (y b − y a ) .
l ∂x dx
In evaluating Christoffel symbols through eqn (1.102) we have to obtain the second
partial derivatives

x ξξ = x ξη = x ηη = 0,
2
2
∂ y a d y b
y ξξ = l 2 (1 − η) + η , (7.46)
∂x 2 dx 2
dy b ∂y a
y ξη = l 2 − , y ηη = 0.
dx ∂x
Hence we get
1 = 1 = 1 = 2 = 0,
11 12 21 22
2
2
∂ y a d y b
2 11 = l 2 (1 − η) + η (y b − y a ) −1 , (7.47)
∂x 2 dx 2

2 12 = 2 21 = l 2 dy b − ∂y a (y b − y a ) −1 .
dx ∂x
Turning our attention now to the time variable, we can deduce from eqn (7.16) that

2
k ∂g i k ∂ y j ∂x k
= · g = , (7.48)
i0 i
∂t ∂t∂x ∂y j