Trying to get my head around the generic method of handling boundary layer problems. For example,

$$\epsilon y''(x) - x^2y'(x) -y = 0, \qquad y(0)=y(1)=1.$$

After following lecture notes and some of the examples on Stackexchange (e.g. Boundary value problem with rescaling or Thickness of the Boundary Layer), I've a rough idea of what's going on and can do the scaling at the boundary layers using $x = x_0 + e^\alpha X$ type subtitutions but still not sure about some aspects.

In the above example, the notes I have show that there is a boundary layer at x=0 because the outer layer is singular at x=0 (which I understand). It also says that there is a boundary layer at x=1 because $-x^2$ is monotonically decreasing on the interval $[0,1]$

i. Is there a general rule for identifying boundary layers such as at $x=1$ a priori (or for that matter from the outer layer)?

ii. Why does the coefficient of $y'$ being negative on the interval cause the boundary layer at $x=1$?

iii. Will the boundary layer(s) always be at the end points or, in general, could there be a boundary layer in the middle of the interval (assuming 2nd order linear ODE).

For doing the asymptotic matching, I'd got that $y_{out}(x) = c_1 e^{1/x}$, $Y_{inner,right}(X) = c_3 + c_4 e^X$, and $Y_{inner,left}(X) = c_5 e^X + c_6 e^{-X}$.

iv. What's the proper way of doing the asymptotic matching? I'd managed to handle the right side with the outer layer (assuming $X \to -\infty$)but the left side with the outer layer seems to require ignoring the limit going to infinity on both sides of the equation.

  • $\begingroup$ This discussion in this question might help. $\endgroup$ – David Nov 28 '16 at 15:41
  • $\begingroup$ at $x=1$, $y\sim y'x^2$ are of the same order which signals that something is going on at this point $\endgroup$ – tired Nov 28 '16 at 16:24

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