Page 91 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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THE FINITE ELEMENT METHOD
                           The result from the sub-domain method coincides with the heat balance integral solution
                        as in the present case integration is carried out over the entire domain in view of only one
                        constant being involved.                                               83
                        Galerkin method
                        This is one of the most important methods used in finite element analysis. The weight
                                              2
                        function is N i (x) = (1 − ζ ). The Galerkin formulation of the fin equation is

                                                          2
                                                  1      d θ
                                                                2
                                                   N i (x)  2  − µ θ dζ = 0                (3.217)
                                                0        dζ
                           Substituting Equation 3.177 and integrating, we obtain
                                                   2B    2     8     2µ 2
                                              2B −    + µ     B −       = 0                (3.218)
                                                    3       15       3
                        and
                                                             µ 2
                                                       B =                                 (3.219)
                                                               2  2
                                                           1 + µ
                                                               5
                           Thus, the solution is
                                                                     2
                                                                    µ
                                                θ(ζ)            2    2
                                                     = 1 − (1 − ζ )                        (3.220)
                                                 θ b                 2  2
                                                                 1 + µ
                                                                     5
                           It can be observed that the solution using Galerkin’s method is exactly the same as that
                        obtained by the variational method. It can also be shown that the variational and Galerkin
                        methods give the same results, provided the problem has a classical variational statement.
                        In fact, later we will see that when the finite element formulation is carried out on a quasi-
                        harmonic equation, using both the variational and Galerkin methods, the same results are
                        obtained since a classical variational principle does exist for a quasi-harmonic equation.
                        Least-squares method
                        In this case, the minimization of the error is carried out in a least squares sense, that is,
                                                      ∂      2
                                                           R dx = 0                        (3.221)
                                                      ∂a i
                        which can also be written as
                                                          ∂R

                                                            dx = 0                         (3.222)
                                                        
 ∂a i
                        where the weighting function is
                                                               ∂R
                                                       w i (x) =                           (3.223)
                                                               ∂a i
                        and the error E is given by
                                                        1
                                                          2
                                                 E =     R dζ
                                                       0
                                                                     2
                                                           2
                                                        1  d θ
                                                                 2
                                                   =        2  − µ θ  dζ                   (3.224)
                                                       0  dζ
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