Page 164 - Essentials of applied mathematics for scientists and engineers
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book   Mobk070    March 22, 2007  11:7








                     154  ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
                       Equation (10.2) can be expanded for small values of ε in the rapidly convergent series

                                                           ε   ε 2   ε 4
                                                   x = 1 −   +    −      + ···                      (10.3)
                                                           2    8    128
                       or
                                                            ε   ε 2   ε 4
                                                  x =−1 −     −    +     − ···                      (10.4)
                                                            2    8   128
                       To apply perturbation theory we first note that if ε = 0 the two roots of the equation, which
                       we will call the zeroth-order solutions, are x 0 =±1. We assume a solution of the form

                                                               2      3     4
                                              x = x 0 + a 1 ε + a 2 ε + a 3 ε + a 4 ε + ···         (10.5)
                       Substituting (10.5) into (10.1)

                                               2            2                    3
                            1 + (2a 1 + 1)ε + a + 2a 2 + a 1 ε + (2a 1 a 2 + 2a 3 + a 2 )ε + ··· − 1 = 0  (10.6)
                                              1
                                                                               n
                       where we have substituted x 0 = 1. Each of the coefficients of ε must be zero. Solving for a n
                       we find
                                                                   1
                                                            a 1 =−
                                                                   2
                                                                 1
                                                            a 2 =                                   (10.7)
                                                                 8
                                                            a 3 = 0

                       so that the approximate solution for the root near x = 1is

                                                             ε   ε 2
                                                                          4
                                                     x = 1 −   +    + O(ε )                         (10.8)
                                                             2    8
                                      4
                       The symbol O(ε ) means that the next term in the series is of order ε 4
                            Performing the same operation with x 0 =−1

                                              2
                                                           2
                                                                                 3

                            1 − (1 + 2a 1 )ε + a − 2a 2 + a 1 ε + (2a 1 a 2 − 2a 3 + a 2 )ε + ··· − 1 = 0  (10.9)

                                              1
                                                    n
                       Again setting the coefficients of ε equal to zero
                                                                   1
                                                            a 1 =−
                                                                   2
                                                                   1                               (10.10)
                                                            a 2 =−
                                                                   8
                                                            a 3 = 0
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