Page 134 - Basic Structured Grid Generation

Differential models for grid generation 123

for some complex function f .Inthe case where f has no dependence on z and,

moreover, a derivative f (z) can be defined (so that f(z) is an analytic function), the
mapping is conformal. In this case the functions ξ(x, y), η(x, y) satisfy the Cauchy-
Riemann equations
∂ξ ∂η ∂ξ ∂η
= and =− . (5.22)
∂x ∂y ∂y ∂x
It follows from eqns (1.158), (1.160), and (1.162) that for a grid generated by a con-
formal transformation we have g 11 = g 22 ,and g 12 = 0, which means that the grid is
orthogonal. Furthermore, not only do the functions ξ(x, y), η(x, y) satisfy Laplace’s
eqns (5.1), but it follows from eqns (5.3) that the inverse functions x(ξ, η),y(ξ,η) also
satisfy Laplace’s equations in the form (5.5). However, although conformal mapping
techniques are well-established for solving Laplace’s equations in two dimensions in
numerous areas of science and engineering (see for example Nehari (1975)), they have
the disadvantages in the area of grid generation that, firstly, they are essentially appli-
cable in two dimensions only, and, secondly, that grid density in the physical domain
is not controlled.
The same disadvantages apply to quasiconformal mapping,which canbeusedto
generate a wider range of grids. A positive feature of conformal and quasiconformal
mapping, however, is that the Jacobian of the transformation is always positive, so that
the grids generated are non-overlapping (not folded).
In quasiconformal mapping the complex function f(z, z) in eqn (5.21) is taken to
satisfy the Beltrami Equation
∂f ∂f
− H(z, z) = 0, (5.23)
∂z ∂z
where the function H(z, z) has real and imaginary parts µ(x, y), ν(x, y),sothat
H(z, z) = µ(x, y) + iν(x, y). (5.24)

The case of conformal mapping occurs when we take µ(x, y) = ν(x, y) = 0.
To substitute for f from eqn (5.21) into eqn (5.23), we need

∂f ∂ ∂x ∂ ∂y ∂
= (ξ + iη) = (ξ + iη) + (ξ + iη)
∂z ∂z ∂z ∂x ∂z ∂y
1 ∂ξ ∂η 1 ∂ξ ∂η
= + i − + i
2 ∂x ∂x 2i ∂y ∂y
1 ∂ξ ∂η 1 ∂ξ ∂η
= − + i + . (5.25)
2 ∂x ∂y 2 ∂y ∂x
Similarly,
∂f 1 ∂ξ ∂η 1 ∂ξ ∂η
= + + i − + . (5.26)
∂z 2 ∂x ∂y 2 ∂y ∂x
Exercise 2. Show that equating real and imaginary parts in eqn (5.23) gives the pair
of equations
∂ξ ∂ξ ∂η ∂η
(1 − µ) + ν − + − (1 + µ) = 0, (5.27)
∂x ∂y ∂x ∂y