Page 33 - Basic Structured Grid Generation
P. 33

22  Basic Structured Grid Generation

                        and, using eqn (1.32),
                                                         1  ∂  √   i
                                                ∇× u = √       ( gg × u).                 (1.145)
                                                          g ∂x i
                          Note that, again by eqn (1.32), eqn (1.143) can be written as

                                                                 ∂u
                                                              i
                                                     ∇× u = g ×                           (1.146)
                                                                 ∂x i
                        with summation over i, which may be directly compared with eqn (1.134).
                                                                 2                          2
                          To obtain an expression for the Laplacian ∇ ϕ of a scalar field ϕ,where ∇ ϕ =
                        ∇·(∇ϕ), using eqns (1.133) or (1.135), the contravariant component of ∇ϕ is needed.
                        This is just
                                                                ∂ϕ
                                                              ij
                                                         i
                                                     (∇ϕ) = g      ,
                                                                ∂x j
                                                                                               j
                        where the effect of the g ij  term is to ‘raise the index’ of the covariant vector ∂ϕ/∂x .
                        Then eqn (1.135) gives
                                                       1  ∂    √    ∂ϕ
                                                 2                ij
                                               ∇ ϕ = √          gg       .                (1.147)
                                                        g ∂x i      ∂x j
                          Alternatively, we have, using the expressions for div and grad in eqns (1.13)
                        and (1.134),

                                                            ∂      ∂ϕ
                                                   2     i       j
                                                 ∇ ϕ = g ·      g                         (1.148)
                                                           ∂x i   ∂x j
                        with summation over both i and j.Hence
                                                                       j
                                                           2
                                                          ∂ ϕ        ∂g ∂ϕ
                                              2     i  j          i
                                             ∇ ϕ = g · g       + g ·        .             (1.149)
                                                           i
                                                                       i
                                                        ∂x ∂x j      ∂x ∂x j
                          But since, from eqn (1.148),
                                                ∂     ∂x k        ∂             ∂g k
                                      2 k    i       j         i      j k     i
                                     ∇ x = g ·      g       = g ·    (g δ ) = g ·  ,
                                                                        j
                                               ∂x i   ∂x j       ∂x i           ∂x i
                        the identity (1.149) may be written in the form
                                                          2
                                                         ∂ ϕ           ∂ϕ
                                                2     ij           2 j
                                              ∇ ϕ = g         + (∇ x )    .               (1.150)
                                                          i
                                                       ∂x ∂x j        ∂x j
                                          k
                          Substituting ϕ = x in eqn (1.147) gives another formula
                                                 1  ∂  √           1  ∂  √
                                          2 k              ij k              ik
                                        ∇ x = √       ( gg δ ) = √       ( gg ).          (1.151)
                                                             j
                                                  g ∂x i            g ∂x i
                          Thus, eqn (1.147) also gives
                                             1  ∂     ∂  √           ∂  √
                                       2                     ij             ij
                                     ∇ ϕ = √      i    j  ( gg ϕ) − ϕ  j  ( gg )
                                              g ∂x  ∂x              ∂x
                                             1    ∂ 2  √   ij     1  ∂   √    2 i
                                          = √     i  j  ( gg ϕ) − √   i  (ϕ g(∇ x )).     (1.152)
                                              g ∂x ∂x             g ∂x
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