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54 Basic Structured Grid Generation
3.4 Geodesic curves
Another aspect of the intrinsic geometry of surfaces arises from the problem of deter-
mining the curve of minimum length which joins two given points on the surface.
3
For example, when the surface is a plane in E , the shortest distance between two
points will be a straight line. If the general problem is approached through the usual
calculus of variations, differential equations are obtained which any solution must sat-
isfy. Curves which satisfy these equations are called geodesics, although in fact not all
solutions are necessarily curves of minimum length.
1
According to eqn (3.22) the length of a curve C on a surface with co-ordinates u ,u 2
and covariant metric tensor a αβ is given by
t 2 t 2
α β 1 2 1 2
L = a αβ ˙u ˙u dt = f(u ,u , ˙u , ˙u ) dt, (3.66)
t 1 t 1
1
2
where t is used to parametrize the curve, each a αβ in general is a function of u and u ,
and we assume that t takes the fixed values t 1 ,t 2 at the given end-points. Moreover, f
is formally regarded as a function of four independent variables. Neighbouring curves,
α
α
α
α
having first-order variations δu and δ ˙u in the values of u and ˙u at corresponding
values of t, but still having the same end-points, have a first-order variation in length
t 2 t 2 ∂f α ∂f α
δL = δf dt = δu + δ ˙u dt
∂u α ∂ ˙u α
t 1 t 1
with summation over α. Integration by parts on the last term gives
∂f α t 2 ∂f d ∂f α
t 2
δL = δu + − δu dt,
∂ ˙u α ∂u α dt ∂ ˙u α
t 1 t 1
and the integrated part vanishes because of the fixed end-point requirement.
If L is to be a minimum for C, the first-order variation δL must be zero for arbitrary
α
variations δu . Thus
t 2
∂f d ∂f α
− δu dt = 0
∂u α dt ∂ ˙u α
t 1
2
1
for arbitrary (first-order) variations δu ,δu . A standard argument of the calculus of
variations then leads to the conclusion that the two differential equations
∂f d ∂f
− = 0, α = 1, 2, (3.67)
∂u α dt ∂ ˙u α
must hold everywhere on C. Curves for which these equations hold are geodesics.
Equations (3.67) are the Euler-Lagrange equations (or just the Euler equations)for
the variational problem δL = 0 with L given by (3.66).
2
1
α β
Now since f = a αβ (u ,u )˙u ˙u ,wehave
∂f 1 ∂a αβ α β ∂f 1 β
= ˙ u ˙u and = a γβ ˙u .
∂u γ 2f ∂u γ ∂ ˙u γ f