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            The solution of (3.98),  with  (3.94),  produces exactly the solution obtained earlier,
          (3.88), but now expressed in terms of  rather than x:





          We now turn to the vexing issue of solving (3.93)—and it is vexing because this is one
          equation in two unknown functions:  and   How do we proceed?
            The aim of this new technique is to obtain a uniformly valid asymptotic expansion—
          if that is at all possible—for   1],  although we have yet to  determine
          (which corresponds to x = 0).  If there is to be any chance of success,  then we must
          remove any  terms that generate  non-uniformities in  the  asymptotic expansion for
                     from  Q3.18, it is  clear that the  only  such term  here is   i.e.
          Thus  we  define   so as to  remove  this term  from the  equation for   it is
          sufficient to  remove  such singular terms in any suitable manner,  but if it is possible
          to choose    so as to leave an homogeneous equation for   then this  is the
          usual move.  (Other choices produce different asymptotic representations of the same
          solution, but all equivalent to  a given order in   Here, therefore, we  elect to write
          the equation for   as





          leaving

          An immediate response to this procedure is to observe that the term that causes all
          the difficulty,   has now appeared as a forcing term in the equation for   —so
          all we  have  succeeded in doing is moving the  non-uniformity  from one  asymptotic
          expansion to another! As we shall see, this is indeed the case, but the non-uniformity
          in the expansion for the strained coordinate is not as severe as that in the expansion for
          y. Before we address this critically important issue, we may note, from (3.97), that





          where B is an arbitrary constant; but from (3.94) we see that we require B = 0 and so
                Further, if this same strategy for selecting the equations for each   is adopted,
          then        for every n = 1,  2, ..., and the exact solution, in terms of  becomes





            It is typical of these problems that it is not necessary to solve completely the equations
          for each   it is sufficient to examine the nature of the solutions as  Thus  we
          will employ the same approach as described in Q3.18. First, from (3.92), we substitute
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