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90 2. Introductory applications



         which gives, with




          The solution of (2.92), which satisfies the boundary condition  is




         and   is to be determined by matching. From (2.90) we have directly that
          and from  (2.93) we see that  thus  matching requires    and the first
          term in the boundary-layer solution is




          and then a composite expansion can be written down, if that is required.



          The fundamental ideas that underpin the notion of boundary-layer-type solutions, in
          second-order ordinary differential equations, have been developed, but many variants
          of this simple idea exist; see also Q2.17–2.20. These lead to  adjustments in the for-
          mulation, or to generalisations, or to a rather different structure (with corresponding
          interpretation). We now describe a few of these possibilities, but what we present is
          far from providing a comprehensive list; rather, we present some examples which em-
          phasise the application of the basic technique of scaling to find thin layers where rapid
          changes occur. In the next section, we briefly describe a number of different scenarios,
          and present an example of each type.


          2.8 BOUNDARY LAYERS AND TRANSITION LAYERS
          Our first development from the simple notion of a boundary layer is afforded by an
          extension of our  discussion of the  position of this  layer, via equation  (2.80);  here,
          we consider the case where                Such a point is analogous to  a
          turning point (see Q2.24) and the solution valid near  takestheform of a transition
          layer. (The terminology ‘turning point’ is used to denote where the character of the
          solution changes or ‘turns’, typically from oscillatory to exponential, in equations such
          as              The  general  approach is to seek a scaling–just as for a boundary
          layer–but now at this interior point.  Let us suppose that   for given
          constants and n, then equation (2.80) becomes




          and we introduce
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