Page 89 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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81
                        THE FINITE ELEMENT METHOD
                           Let the governing equations be represented by
                                                     L(T ) = 0in 
                         (3.199)
                           Let
                                                             n

                                                    T ≈ T =    a i N i (x)                 (3.200)
                                                            i=1
                           Substitution of the above equation into Equation 3.199 results in
                                                    L(T)  = 0
                                                         = R(residual)                     (3.201)
                           The method of weighted residual requires that the parameters a 1 ,a 2 ,... ,a n be deter-
                        mined by satisfying

                                              w i (x)R dx = 0 with i = 1, 2,... ,n         (3.202)

                        where the functions w i (x) are the n arbitrary weighting functions. There are an infinite
                        number of choices for w i (x) but four particular functions are most often used. Depending
                        on the choice of the weighting functions, different names are given

                        Collocation: w i = δ(x − x i )

                                                                      = 0                  (3.203)
                                                   Rδ(x − x i )dx = R x=x i

                        Sub-domain: w i = 1 (Note the sub-domain 
 i in the integration)

                                                 R dx = 0 with i = 1, 2,... ,n             (3.204)
                                              
 i
                        Galerkin: w i (x) = N i (x), that is, the same trial functions as used in T(x)

                                              N i (x)R dx = 0 with i = 1, 2,... ,n         (3.205)

                        Least Squares: w i = ∂R/∂a i

                                                ∂R
                                                   dx = 0 with i = 1, 2,... ,n             (3.206)
                                              
 ∂a i
                           For illustration purposes the fin problem is re-solved with each of the above methods.


                        Collocation method
                        The weight is w i = δ(x − x i )
                           Let ζ i = 1/2 as there is only one unknown in the fin problem. Rewriting the equation
                        in collocation form in the non-dimensional coordinates gives the following:

                                                    2
                                                 1  d θ
                                                          2
                                                      − µ θ δ(ζ − ζ i )dζ = 0              (3.207)
                                               0   dζ 2
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