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Variational methods and adaptive grid generation  157

                           6.3 One-dimensional grid generation


                        6.3.1 Variational approach


                        According to the analysis of Section 5.7, the transformation x = x(ξ) between the one-
                        dimensional computational domain 0   ξ   1 and the physical domain a   x   b,
                        where the grid-point density is governed by a weight function ϕ(ξ), must satisfy
                        eqn (5.83). This equation is actually the Euler-Lagrange equation of the variational
                        problem δI = 0, where the functional I is given by
                                                     1     1  1    dx    2
                                                 I =                 dξ.                   (6.13)
                                                     2  0 ϕ(ξ)  dξ
                          In fact, because the integrand is reasonably simple, we can reach without difficulty
                        the stronger conclusion that the solution of (5.83), subject to the end conditions x(0) =
                        a, x(1) = b, minimizes I. For suppose that x =ˆx(ξ) satisfies eqn (5.83) and the end
                        conditions. If we consider a varied function x =ˆx(ξ) + v(ξ), also satisfying the end
                        conditions, so that v(0) = v(1) = 0, we obtain, substituting into eqn (6.13),
                             1     b  1     dˆx  dv    2      1  1 dˆx dv    1     1  1    dv    2
                                                        ˆ
                         I =               +      dξ = I +              dξ +                 dξ,
                             2  a ϕ(ξ)  dξ   dξ             0 ϕ(ξ) dξ dξ     2  0 ϕ(ξ)  dξ
                               ˆ
                        where I is the value of I when x =ˆx. The middle term vanishes since, on integration
                        by parts, we get
                                                      1     1
                                              1 dˆx          d    1 dˆx
                                                   v   −                 v dξ,
                                             ϕ(ξ) dξ  0   0 dξ   ϕ(ξ) dξ
                        where the integrated part is zero because of the end conditions satisfied by v(ξ) and
                        the integral is zero because ˆx(ξ) satisfies eqn (5.83).
                          Hence we have the exact equation

                                                     1     1  1    dv    2
                                                 ˆ
                                                                          ˆ
                                             I = I +                 dξ   I,
                                                     2  0 ϕ(ξ)  dξ
                        since ϕ(ξ) takes positive values. So the minimizing property of ˆx(ξ) is established.
                          On the other hand, if we consider the functional
                                                    1     1  1    dx    2
                                                ˜
                                                I =                   dξ,                  (6.14)
                                                    2  0 [ϕ(x)] 2  dξ
                        the extremal which makes I stationary must satisfy the Euler-Lagrange equation
                               	    
                                       2
                             d   ∂I ˜   ∂I ˜  d     x ξ        1   dϕ  dx
                                      −    =              +
                                                                 3
                            dξ  ∂x ξ    ∂x   dξ   [ϕ(x)] 2  [ϕ(x)] dx  dξ
                                                     2
                                                1   d x     2      dx    2  dϕ  1  dϕ    dx    2
                                           =            −                 +
                                                   2
                                                                                  3
                                             [ϕ(x)] dξ 2  [ϕ(x)] 3  dξ  dx   [ϕ(x)] dx  dξ
                                                     2
                                                1   d x     1    dϕ    dx    2
                                           =            −                 = 0,
                                                   2
                                                               3
                                             [ϕ(x)] dξ 2  [ϕ(x)] dx  dξ
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