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                                           Non-dimensional temperature  0.8 1  Ritz method  THE FINITE ELEMENT METHOD
                                                      Exact
                                             0.6
                                             0.4

                                             0.2
                                              0
                                               0     0.2   0.4   0.6   0.8    1
                                                        Distance from fin tip
                              Figure 3.25 Comparison between the Ritz method and the exact solution


                        3.3.2 Rayleigh–Ritz method (Variational method)

                        In the case of the variational method, we make use of an important theorem from the
                        theory of the calculus of variations, which states, ‘The function T(x) that extremises the
                        variational integral corresponding to the governing differential equation (called Euler or
                        Euler–Lagrange equation) is the solution of the original governing differential equation
                        and boundary conditions’. This implies that the solution obtained is unique, which is the
                        case for well-posed problems. Thus, the first step is to determine the variational integral
                        ‘I’, which corresponds to the governing differential equation and its boundary conditions.
                        The differential equation is, Equation 3.172,

                                                       2
                                                      d θ    2
                                                          − µ θ = 0                        (3.185)
                                                      dζ 2
                        with the following boundary conditions:

                                                 dθ(0)
                                                       = 0and θ(1) = θ b                   (3.186)
                                                  dζ
                           Using the differential equation as the Euler–Lagrange equation, we can write

                                                         2
                                                      1     d θ
                                                               2
                                               δI =       2  − µ θ δθdζ = 0                (3.187)
                                                    0   dζ
                           Integrating by parts gives

                                              1     1                    1
                                        dθ           dθ   d          2
                                          δθ  −             (δθ)dζ − µ    θδθdζ = 0        (3.188)
                                        dζ           dζ  dζ
                                             0    0                     0
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