Page 135 - Basic Structured Grid Generation

124 Basic Structured Grid Generation

∂ξ ∂η ∂ξ ∂η

ν + − (1 + µ) − (1 − µ) = 0, (5.28)
∂x ∂y ∂y ∂x
which are equivalent to
∂ξ ∂ξ ∂η
β + γ =− , (5.29)
∂x ∂y ∂x
∂ξ ∂ξ ∂η
α + β = , (5.30)
∂x ∂y ∂y
2
2
(1−µ) +ν 2 −2ν (1+µ) +ν 2
where α = , β = , γ = . Show that
2
2
2
1−µ −ν 2 1−µ −ν 2 1−µ −ν 2
2
αγ − β = 1. (5.31)
∂f ∂f
We also have, since H(z, z) = ,
∂z ∂z
2
∂f

∂z
2 2 2
|H(z, z)| = µ + ν =
2
∂f

∂z

2 2
∂ξ ∂η ∂ξ ∂η

− + + √
∂x ∂y ∂y ∂x g 11 + g 22 − 2 g
= √ ,
2 2
=
∂ξ ∂η ∂ξ ∂η g 11 + g 22 + 2 g

+ + − +
∂x ∂y ∂y ∂x
using eqns (5.25), (5.26), (1.158), (1.160), and (1.162). The identity
2
2
2(1 + µ + ν ) g 11 + g 22
α + γ = = √ (5.32)
2
1 − µ − ν 2 g
then follows.
In the case of conformal mapping, we have α = γ = 1, and β = 0, and eqns (5.29),
(5.30) reduce to the Cauchy-Riemann equations. More generally, these equations form
a pair of first-order partial differential equations which could be used to generate a grid
for some choice of µ, ν. From eqns (1.162) the inverse equations can immediately be
written as
∂x ∂y ∂x
= α − β , (5.33)
∂ξ ∂η ∂η
∂y ∂y ∂x
= β − γ . (5.34)
∂ξ ∂η ∂η
Exercise 3. Show that the Jacobian of the transformation from (ξ, η) to (x, y) can be
expressed as

2 2
2 2
√ ∂x ∂y ∂x ∂y 1 ∂x ∂x 1 ∂y ∂y
J = g = − = + = + .
∂ξ ∂η ∂η ∂ξ α ∂ξ ∂η γ ∂ξ ∂η
(5.35)