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Leading order matched

  1. Sep 21, 2008 #1
    What would be the leading order match asymptotic expansion for the following differential equation?
    [tex]\epsilon y'' = f(x) - y' [/tex]where f(x) is continuous, [tex]\epsilon<<1[/tex]
    and y(0) = 0, y(1) = 1

    Thanks in advance,
  2. jcsd
  3. Sep 21, 2008 #2


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    Staff Emeritus
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    First, what is an "asymptotic expansion", both in general and for this particular problem? Then what is meant by a "leading order match"?
  4. Sep 22, 2008 #3
    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 [tex]\phi_{1}\phi_{2},...[/tex]. These functions are well-ordered, which means that [tex]\phi_{n}=o(\phi_{m}) [/tex] as some [tex]\epsilon[/tex] (epsilon)gets really really small for all m and n that satisfy m<n

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

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

    Finally, all this enables us to write f~[tex]a_{1}\phi_{1}(\epsilon)+a_{2}\phi_{2}(\epsilon)+...+a_{n}\phi_{n}(\epsilon) [/tex] as [tex]\epsilon\rightarrow 0[/tex]. 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 [tex]\epsilon[/tex] = 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) = [tex]\int f(x)dx[/tex] and now what do I do with this integral?

    Thanks again!
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