# Leading order matched

1. Sep 21, 2008

### 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

A.Haywood

2. Sep 21, 2008

### HallsofIvy

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

3. Sep 22, 2008

### 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