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book   Mobk070    March 22, 2007  11:7








                                                     INTRODUCTION TO PERTURBATION METHODS           155
                   so that the root near x 0 =−1is

                                                          ε   ε 2
                                                                       4
                                                x =−1 −     −    + O(ε )                       (10.11)
                                                          2    8
                   The first three terms in (10.8) give x = 0.951249219, accurate to within 1.16% of the exact
                   value while (10.11) gives the second root as x =−1.051249219, which is accurate to within
                   1.05%.
                        Next suppose the small parameter occurs multiplied by the squared term,

                                                       2
                                                     εx + x − 1 = 0                            (10.12)
                   Using the quadratic formula gives the exact solution.


                                                        1       1    1
                                                  x =−     ±       +                           (10.13)
                                                        2ε     4ε 2  ε
                   If ε = 0.1 (10.13) gives two solutions:

                                                    x = 0.916079783

                   and

                                                   x =−10.91607983

                   We attempt to follow the same procedure to obtain an approximate solution. If ε = 0 identically,
                   x 0 = 1. Using (10.5) with x 0 = 1 and substituting into (10.12) we find

                                                                    2
                                                        2
                                                                           3


                                   (1 + a 1 )ε + (2a 1 + a 2 )ε + 2a 2 + a + a 3 ε + ··· = 0   (10.14)
                                                                    1
                                           n
                   Setting the coefficients of ε = 0,solvingfor a n , and substituting into (10.5)
                                                             2     3
                                               x = 1 − ε + 2ε − 5ε + ···                       (10.15)
                   gives x = 0.915, close to the exact value. However Eq. (10.12) clearly has two roots, and the
                   method cannot give an approximation for the second root.
                        The essential problem is that the second root is not small. In fact (10.13) shows that as
                                 1                 2
                   ε → 0, |x| →   so that the term εx is never negligible.
                                2ε

                   10.1.2 Singular Perturbation
                   Arranging (10.12) in a normal form
                                                     x − 1
                                                 2
                                                x +        = 0   (10.12a)
                                                       ε
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