Page 82 - Basic Structured Grid Generation

Differential geometry of surfaces in E 3 71

some formulas which are often useful in the transformation from cartesian to general
surface co-ordinates of some of the standard vector operators which appear in hosted
equations.
Here we begin by deriving the surface metric identity

∂ a 22 a 1 − a 12 a 2 ∂ −a 12 a 1 + a 11 a 11 √
√ + √ = 2κ m aN. (3.136)
∂u 1 a ∂u 2 a
The left-hand side, by eqn (3.31), is equal to
∂ √ 1 ∂ √ 2
( aa ) + ( aa ). (3.137)
∂u 1 ∂u 2
Exercise 16. Show that
1 2 1
1
a = √ (a 2 × N) and a = √ (N × a 1 ). (3.138)
a a
It follows that eqn (3.137) becomes
∂ ∂ ∂ ∂r ∂ ∂r
(a 2 × N) − (a 1 × N) = −
∂u 1 ∂u 2 ∂u 1 ∂u 2 ∂u 2 ∂u 1
∂N ∂N
×N + a 2 × − a 1 ×
∂u 1 ∂u 2
= 0 × N + a 2 × (−b 1β a βγ a γ ) − a 1 × (−b 2β a βγ a γ ),
using the Weingarten eqns (3.121). The resulting expression is
√
β1 β2 βα
−b 1β a a 2 × a 1 + b 2β a a 1 × a 2 = b αβ a a 1 × a 2 = 2κ m aN
with the aid of eqns (3.96) and (3.111), thus proving the identity.
2
1
Now suppose there exists a surface scalar function φ(u ,u ). The gradient of this
function is the surface vector
∂ϕ
α
∇ϕ = a , (3.139)
∂u α
α
where ∂ϕ/∂u are the surface covariant components. Here we shall also be concerned
with the background cartesian components of ∇ϕ.
By eqn (3.138) we can write
1 ∂ϕ ∂ϕ
∇ϕ = √ (a 2 × N) + (N × a 1 ) . (3.140)
a ∂u 1 ∂u 2
If we now assemble the three background cartesian components of (a 2 × N) and
(N × a 1 ) into the two column vectors of a 3 × 2matrix C, we can write the cartesian
components of ∇ϕ as
1 ∂ϕ
(∇ϕ) i = √ C iα α , (3.141)
a ∂u
with C represented as
√

1 2
C = (a 2 × N)(N × a 1 ) = a(a , a ), (3.142)