Page 187 - Basic Structured Grid Generation

176 Basic Structured Grid Generation

1
A harmonic map will take M onto a unit square N in the (ξ, η) plane, with ξ = ξ,
2
ξ = η. Straight lines ξ = const., η = const.in N are what results from mapping ξ,
η curvilinear co-ordinate curves in M. The ‘induced’ covariant metric components are
given by
2
2
g 11 = (x ξ ) + (y ξ ) + (z ξ ) 2
g 12 = x ξ x η + y ξ y η + z ξ z η
2
2
2
g 22 = (x η ) + (y η ) + (z η ) .
However, since z = f (x, y), we have, using Chain Rules,
z ξ = f x x ξ + f y y ξ and z η = f x x η + f y y η , (6.91)

giving
2
2
2
g 11 =[1 + (f x ) ](x ξ ) + 2f x f y x ξ y ξ +[1 + (f y ) ](y ξ ) 2
2 2
g 12 =[1 + (f x ) ]x ξ x η + f x f y (x ξ y η + x η y ξ ) +[1 + (f y ) ]y ξ y η
2 2 2 2
g 22 =[1 + (f x ) ](x η ) + 2f x f y x η y η +[1 + (f y ) ](y η ) . (6.92)
Exercise 7. Verify that

2
2
2
2
g = det(g ij ) =[1 + (f x ) + (f y ) ](x ξ y η − x η y ξ ) = γJ , (6.93)
i
i
where J is the Jacobian (x ξ y η − x η y ξ ) of the mapping from the ξ sto the x s.
We now have
g 11 = g 22 /g, g 12 =−g 12 /g, g 22 = g 11 /g.
Substituting into eqns (6.87) gives

L(x) = g 22 x ξξ − 2g 12 x ξη + g 22 x ηη

g ∂ 1 + (f y ) 2 ∂ f x f y
−√ √ − √ = 0,
γ ∂x γ ∂y γ

L(y) = g 22 y ξξ − 2g 12 y ξη + g 22 y ηη
g ∂ f x f y ∂ 1 + (f x ) 2
−√ − √ + √ = 0 (6.94)
γ ∂x γ ∂y γ
with γ , g 11 , g 12 , g 22 ,and g given by eqns (6.89), (6.92), and (6.93).
To solve these equations numerically, the second-order accurate finite difference
approximations (4.3) for first derivatives and (4.4), (4.5), and (4.6) for second deriva-
tives may be used, taking uniform differences h, k in ξ and η, respectively, in the
computational domain N. To calculate f x and f y we need to invert eqns (6.91), so that
1 1
f x = (z ξ y η − z η y ξ ) and f y = (z η x ξ − z ξ x η ). (6.95)
J J