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(This difficulty in the solution can be discussed separately—but not here—or the
discontinuity can be replaced by a smooth function that takes from at
Now, since

we must find a solution of (3.57), subject to

To proceed with the solution of this problem, we write and
so

this is solved very simply by introducing the similarity solution. Set
for some m, then we find directly that m = –1 /3 and

where A and a are arbitrary constants, and then all the boundary conditions are satisfied
by

Thus the solution in the boundary layer, R = O(1), is

in this special case. We have the first terms in each of the outer region (away from
the pipe wall) and the region close to the pipe wall (the boundary layer): (3.55b) and
(3.58), respectively.
We have completed all the detailed calculations that we will present, although we
make a few concluding comments that highlight some intriguing issues. First, knowing
the temperature in the fluid near the pipe wall enables us to find the heat transferred
to or from the fluid, if that is a property of particular interest.
Second, a more technical matter: what is the form of the solution in the outer region,
i.e. what terms in the asymptotic expansion must be present in order to match to
the boundary-layer solution? To answer this, we need to know the behaviour of the
boundary-layer solution as (because we have for 1 – r =
O(1) as For our similarity solution, (3.58), this can be obtained by using a