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Consider the equation

$(x+1-\epsilon)\frac{dy}{dx}+(1-\frac{1}{4}\epsilon^2y)y=2(1-\epsilon x)$

with $y(1)=1$.

I am interested in finding an asymptotic expansion for the inner solution so I put $x= \epsilon^{\alpha} X$. My question here is how I do manage to determine the value of $\alpha$? Or equivalently, what is the thickness of the inner layer?

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Basically, I know that I need to rewrite the ODE in terms of X and then do some kind of dominant balance but I get the equation $(\epsilon^{\alpha}X+1-\epsilon) \epsilon^{-\alpha}\frac{dY}{dX} + (1-\frac{1}{4}\epsilon^2 Y)Y=2(1-\epsilon \epsilon^{\alpha} X)$ and it doesn't look right. –  chango Dec 24 '11 at 15:50
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1 Answer

I believe that you may be confused over the concept of inner/outer expansions. This terminology is not used consistently in the literature, so the mistake is not uncommon. I suggest that we talk about "regular" vs. "boundary layer" expansions instead. The regular expansion is the Taylor series-type expansion in small parameter $\epsilon$ that you would expect to hold in many situations: $$ y(x)=y_0(x)+\epsilon y_1(x)+O(\epsilon^2). \qquad (1)$$ If ansatz (1) is inserted into the original differential equation and boundary condition, you split the original boundary value problem (BVP) into a sequence of boundary value problems at various orders of $\epsilon$. Specifically, at the leading order (at $O(\epsilon^0)$) you'll get $$ (x+1)y_0'+y_0-2=0 \;\;\mbox{solved subject to}\;\; y_0(1)=1. $$ The solution of this BVP has the following form $$ y_0(x)=\frac{2x}{x+1}. \qquad (2) $$ At the next order (at $O(\epsilon)$) you need to solve another BVP: $$ (x+1)y_1'+y_1+\frac{2(x^3+2x^2+x-1)}{(x+1)^2}=0 \;\;\mbox{solved subject to}\;\; y_1(1)=0, $$ which is satisfied by $$ y_1(x)=\frac{(2-x^2-2/(x+1))}{x+1}. \qquad (3) $$ Solutions (2) and (3) can be substituted back into (1) to obtain the sought for expansion.

What you were trying to do seems similar to the derivation of the "boundary layer" expansion. Put simply, asymptotic ansatz (1) tacitly assumed that functions $\{y_n(x)\}$ are of the same asymptotic order and that their derivatives are of the same order as the original functions (this is why we could insert (1) into the differential equation). This does not have to be always true and, in principle at least, our process could have ended with an unsolvable BVP. This would indicate the presence of a "boundary layer", a rapidly varying solution that cannot be differentiated without disturbing the asymptotic orders of various terms. These solutions tend to manifest themselves in the vicinity of problem boundaries, hence the terminology. However, keep in mind that it is not unusual to find boundary layers in other parts of the problem domain.

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