Page 185 - Basic Structured Grid Generation

174 Basic Structured Grid Generation

3
When M is a two-dimensional surface in E , we can use surface co-ordinates u α
ij
i
in place of x , and the surface metric a αβ in place of γ . With N again a unit square
with G ij = δ ij ,weget
k
k
1 αβ γ ∂ξ ∂ξ √ 1 2
E = a (u ) a du du , (6.80)
α
2 M ∂u ∂u β
where all indices are summed from 1 to 2. Thus we have a variational problem of the
form (6.11). The two Euler-Lagrange eqns (6.12) may be expressed here as
√ √
∂(e a) ∂ ∂(e a)
− = 0, i = 1, 2, (6.81)
∂ξ i ∂u γ ∂ ∂ξ /∂u γ
i
i
with summation over γ . Since no terms in e depend explicitly on ξ ,and
k
∂ ∂ξ ∂ξ k γ ∂ξ k ∂ξ k γ ∂ξ i γ ∂ξ i γ
= δ ik δ + δ ik δ = δ + δ ,
i γ α β α β α β β α α β
∂ ∂ξ /∂u ∂u ∂u ∂u ∂u ∂u ∂u
we obtain the equations
1 ∂ √ αβ ∂ξ i γ ∂ξ i γ 1 ∂ √ γβ ∂ξ i αγ ∂ξ i
aa δ + δ = a a + a = 0.
2 ∂u γ ∂u β α ∂u α β 2 ∂u γ ∂u β ∂u α
αβ
Using the symmetry of a , we have, finally,
∂ √ γβ ∂ξ i
aa = 0, i = 1, 2. (6.82)
∂u γ ∂u β
√
Pre-multiplying by 1/ a shows by eqn (3.160) that the harmonic mapping in this
case must satisfy the pair of Beltramian equations
1
2
B ξ = B ξ = 0 (6.83)
which are the same as eqns (5.104) in the absence of the control functions P , Q.
Returning to the more general form (6.76) with N a unit square or cube and G kl =
δ kl , a similar procedure shows that the Euler-Lagrange equations are

1 ∂ √ jk ∂ξ i
√ j γγ k = 0, i = 1,...,n, (6.84)
γ ∂x ∂x
which can be expressed as
2 i
∂ ξ 1 ∂ξ i ∂ √ jk
jk
γ + √ ( γγ ) = 0. (6.85)
k
j
∂x ∂x k γ ∂x ∂x j
This equation can be inverted to produce a different partial differential equation
i
k
satisfied by x (ξ ), by making use of the identity
2 l
2 i
i
∂ ξ ∂x l ∂ x ∂ξ ∂ξ m
=− ,
j
k
j
i
m
∂x ∂x ∂ξ i ∂ξ ∂ξ ∂x ∂x k