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I have been searching in some literature and Wikipedia about the definition or explanation of outer region, outer solution, inner region and inner solution of boundary layer theory, perturbation theory and asymptotic matching. But I could not find any useful and easily understood one.

In Wikipedia, it is written An approximation in the form of an asymptotic series is obtained in the transition layer(s) by treating that part of the domain as a separate perturbation problem. This approximation is called the "inner solution," and the other is the "outer solution"

I don't find this explanation clear enough. I am a bit confused what exactly is outer/inner region and outer/inner solution? Is it correct if I interpret outer and inner as outside the boundary and inside the boundary? But again, a question arises, where is considered as outside/inside? Does it have any relationship with the order of $\epsilon$, that is $O(\epsilon)$? Such as the limit as $\epsilon\to0$ or something like that?

I would just want some easily understood and not so deep explanation since I am quite immature in this field. I apologise for such a naive question, this idea has always been so vague to me, I just want to get a better understanding. I will really appreciate if anyone would like to provide some explanations and share some insights.

Thanks!

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  • $\begingroup$ A proper answer for this will not be short, so for now I will just address a minor point of terminology: The inner solution/region refers to inside the boundary layer, not the boundary itself. And the outer solution/region is then outside the boundary layer. The inner solution has a short length scale; the outer solution, a longer one. Note that there is a fuzzy transition between the inner and outer regions, where matching takes place. $\endgroup$ – Harald Hanche-Olsen Jun 11 '13 at 14:07
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This is perhaps best illustrated by an example. Consider the ODE with boundary values $$\varepsilon y''+(1-\varepsilon)y'-y=0,\qquad y(0)=0,\quad y(1)=1.$$ The exact solution is easily found: $$ y=\frac{e^x-e^{-x/\varepsilon}}{e-e^{-1/\varepsilon}}.$$ Basically, the $e^x$ part will be the outer solution. It has the natural length scale $1$, and is approximately valid in most of the interval $[0,1]$, i.e., where $x\gg\varepsilon$. Not coincidentally, $e^{x-1}$ solves $$y'-y=0,\qquad y(1)=1$$ which is the original problem with $\varepsilon=0$ and one boundary condition thrown away. And the reason for the lost boundary condition is that it is inside the inner region, defined by $x\lesssim\varepsilon$. Inside the inner region the problem has a much shorter length scale, and in fact the behaviour solution is dominated by the $e^{-x/\varepsilon}$ part (but one must not forget the $e^x$ part – just notice that $e^x\approx1$ in the inner region, so it is quite “tame”).

Typically, one finds the inner solution by rescaling the problem to fit the natural length scale of the inner region; in this case, $\varepsilon$. So use $X=x/\varepsilon$ as the new length scale. Put $Y(X)=y(x)$ and rewrite the original problem to $$Y''+(1-\varepsilon)Y'-\varepsilon Y=0,\qquad Y(0)=0,$$ where I have thrown away the other boundary condition, which would be $Y(1/\varepsilon)=1$. That is too far outside the inner region to be useful.

To a zeroth approximation this becomes $$Y''+Y'=0,\qquad Y(0)=0,$$ with a general solution $Y(X)=C(1-e^{-X})$.

The remaining problem is to match up inner and outer solutions in order to determine $C$. This may done by appealing to $x$ in the intermediate region, where $x\gg\varepsilon$ but still $x\ll1$, where $X\gg1$, so the inner solution will be ${}\approx C$ while the outer solution is ${}\approx e^{-1}$, thus leading to $C=e^{-1}$.

So for this example, in summary: The inner region is $x\lesssim\varepsilon$, the inner solution is $e^{-1}(1-e^{-x/\varepsilon})$, the outer region is $x\gg\varepsilon$, and the outer solution is $e^{x-1}$. All this just to lowest order, of course.

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  • $\begingroup$ Thank you for your elaborate answer. There is a point that I don't quite understand, why is $e^x$ valid for $x\gg\epsilon$ ? Thanks :) $\endgroup$ – user71346 Jun 13 '13 at 10:28
  • $\begingroup$ Because the other term, $e^{-x/\varepsilon}$, is negligible in that case. $\endgroup$ – Harald Hanche-Olsen Jun 13 '13 at 10:36
  • $\begingroup$ I see thanks! Also after you substitute $X=x/\epsilon$ and $Y(X)=y(x)$ should there be a coefficient of $1/\epsilon$ in front of $Y''$ ? $\endgroup$ – user71346 Jun 13 '13 at 10:53
  • $\begingroup$ I don't think so, as I had silently multiplied the whole equation by $\varepsilon$ after doing the substitution. (Sorry about the silence; I just wanted to keep the length down.) $\endgroup$ – Harald Hanche-Olsen Jun 13 '13 at 11:41
  • $\begingroup$ Oh yes, you are right! how careless, thanks! $\endgroup$ – user71346 Jun 13 '13 at 12:08

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