Page 166 - Basic Structured Grid Generation

Variational methods and adaptive grid generation 155

One more integration by parts gives
b n
2
∂F d ∂F d ∂F
0 = δI = − + 2 δx i dt
a ∂x i dt ∂ ˙x i dt ∂ ¨x i
i=1
with another integrated part vanishing.
Since this must hold for arbitrary independent variations δx i , it follows straight-
forwardly (using a proof by contradiction) that the following equations must hold for
all t:
∂F d ∂F d 2 ∂F
− + = 0, i = 1, 2,...,n. (6.6)
∂x i dt ∂ ˙x i dt 2 ∂ ¨x i
These are the n fourth-order ordinary differential equations (the Euler-Lagrange
equations for this case), with 4n end conditions, whose solutions are the extremals
for the variational problem δI = 0 with I given by eqn (6.5).
Another type of generalization arises from considering functionals

I = F(x, y, u, u x ,u y ) dx dy, (6.7)
R
which involve a double integral over a domain R of the xy-plane, where the integrand
depends explicitly on a function u(x, y) and its partial derivatives u x , u y . We assume
that the value of u is prescribed on the boundary C of R.Tomake I stationary we
equate to zero the first-order expression for the increment δI
∂F ∂F ∂F

0 = δI = δF dx dy = δu + δu x + δu y dx dy
R R ∂u ∂u x ∂u y
corresponding to a variation δu (a function of x and y).
We also assume that δu x = ∂(δu)/∂x, so that integration of the second term by
parts with respect to x with y fixed gives
∂F ∂F

δu x dx dy = δu x dx dy
R ∂u x ∂u x

x 2 x 2
∂F ∂ ∂F
= δu − δu dx dy,
∂u x ∂x ∂u x
x 1 x 1
where x 1 and x 2 are values of x on C at given values of y.Since u is prescribed on
C,we musthave δu = 0on C, and hence

∂F ∂ ∂F
δu x dx dy =− δu dx dy.
R ∂u x R ∂x ∂u x
Similarly we get
∂F ∂ ∂F

δu y dx dy =− δu dx dy.
R ∂u y R ∂y ∂u y
Consequently

∂F ∂ ∂F ∂ ∂F
δI = − − δu dx dy = 0,
R ∂u ∂x ∂u x ∂y ∂u y