Asymptotic Expansion for a Differential Equation with a Small Parameter

[haywood]

What would be the leading order match asymptotic expansion for the following differential equation?
\epsilon y'' = f(x) - y'where f(x) is continuous, \epsilon<<1
and y(0) = 0, y(1) = 1

Thanks in advance,
A.Haywood

 

[HallsofIvy]

First, what is an "asymptotic expansion", both in general and for this particular problem? Then what is meant by a "leading order match"?

 

[haywood]

Firstly, thank you for your response.
To the best of my knowledge, an asymptotic expansion is in general defined this way:

1. First, an asymptotic sequence is formed from functions called scale, or gauge, or basis functions, denoted \phi_{1}\phi_{2},.... These functions are well-ordered, which means that \phi_{n}=o(\phi_{m}) as some \epsilon (epsilon)gets really really small for all m and n that satisfy m<n

2. Now, if \phi_{1}\phi_{2},... is an asymptotic sequence, then f(epsilon) has an asymptotic expansion to n terms, with respect to this sequence, if and only if

f = {^m}\sum_{k=1} a_{k}\phi_{k}(\epsilon)+o(\phi_{m}) for m=1,...,n as epsilon gets really really small towards like 0. The a_{k} are independent of \epsilon.

Finally, all this enables us to write f~a_{1}\phi_{1}(\epsilon)+a_{2}\phi_{2}(\epsilon)+...+a_{n}\phi_{n}(\epsilon) as \epsilon\rightarrow 0. Here, the ~ denotes asymptotic.


Whew! Now, for this particular problem... I think the procedure is to find the outer solution away from the boundary layer (reduce given original equation by setting \epsilon = 0), and then the inner solution near the boundary layer (in this case near x=0) and then push both towards each other into an overlap region in which they are supposed to match!

I have started on the outer problem thus far:
y' = f(x) so y(x) = \int f(x)dx and now what do I do with this integral?

Thanks again!
A.Haywood