Page 167 - Basic Structured Grid Generation

156 Basic Structured Grid Generation

and this vanishes for arbitrary variations δu in R only if the extremal function u
satisfies the second-order partial differential equation (the Euler-Lagrange equation for
this problem)
∂F ∂ ∂F ∂ ∂F
− − = 0 (6.8)
∂u ∂x ∂u x ∂y ∂u y
everywhere in R. Some care is again required in interpreting this equation, in regard
to the partial derivatives in the second and third terms. In the case of the second term,
for example, the partial derivative ∂F/∂u x differentiates F formally (as a function of
the five variables x, y, u, u x , u y ) with respect to u x . In the subsequent differentiation
∂/∂x, however, ∂F/∂u x must be regarded as a function of the two variables x and
y only.
As an example, consider
1 2 2
I = {(u x ) + (u y ) } dx dy, (6.9)
2 R
where u takes prescribed values on the boundary of R.Then
1 2 1 2 ∂F ∂F
F = (u x ) + (u y ) , = u x , = u y ,
2 2 ∂u x ∂u y

and
∂F ∂ ∂F ∂ ∂F ∂ ∂
− − = 0 − (u x ) − (u y ) =−(u xx + u yy ).
∂u ∂x ∂u x ∂y ∂u y ∂x ∂y
Thus the extremal for the variational problem δI = 0 must satisfy Laplace’s equation

2
2
∂ u ∂ u
+ = 0 (6.10)
∂x 2 ∂y 2
in R. In this problem I can take only non-negative values, and so is bounded below
by zero. This suggests that the extremal must minimize I, and in this particular case
this is indeed quite straightforward to prove, due to the quadratic nature of F.
The more general case

I = F(x, y, u, v, u x ,u y, v x ,v y ) dx dy (6.11)
R
with two functions u(x, y), v(x, y), leads to two Euler-Lagrange equations

∂F ∂ ∂F ∂ ∂F
− − = 0,
∂u ∂x ∂u x ∂y ∂u y
∂F ∂ ∂F ∂ ∂F
− − = 0, (6.12)
∂v ∂x ∂v x ∂y ∂v y

which are two simultaneous second-order partial differential equations for u and v.In
Section 6.4 of this chapter these equations will be the focus of interest.