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My attempt:

The outer problem for leading order term: $$y_0 y_0' - y_0 = 0$$ This has solution: $y_0(x) = 0$ or $y_0(x) = x + c$.

I notice that $y(x) < 0$ on this interval, so I assume the boundary layer is at $x = 1$. So the outer solution is: $$y_0(x) = x - 1$$ Note that this alone solves the differential equation.

I approached the inner problem with the scaling: $ z = \frac{1 -x}{\epsilon}$. This leads to: $$ y_0''(z) - y_0(z) y_0'(z) = 0 $$ The solution for this is either $y_0(z) = C$ or, $$ y_0(z) = \sqrt{2} C \tan\left({\frac{(z+D)C}{\sqrt{2}}}\right)$$ With the B.C. $y_0(z = 0) = 0$, then $D = \frac{\sqrt{2} n \pi}{C}$, and the solution becomes: $$ y_0(z) = \sqrt{2} C \tan\left({\frac{zC+\sqrt{2}n\pi}{\sqrt{2}}}\right)$$ Classical matching: $$ \lim_{x \rightarrow 1} y_0(x) = \lim_{z \rightarrow \infty} y_0(z) $$ Or, $$ 0 = \sqrt{2} C \tan\left({\frac{\infty \cdot C+\sqrt{2}n\pi}{\sqrt{2}}}\right)$$ Thus, it requires $C = n = 0$. This means the outer solution is the leading approximate solution for this problem.

My question: I don't think this is correct, but I am not sure how else to approach this. ( I am not supposed to use other matching methods).

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  • $\begingroup$ Same ODE, different boundary conditions: math.stackexchange.com/q/1286926/115115 $\endgroup$ – LutzL Mar 17 '16 at 17:33
  • $\begingroup$ Check again the inner equation, the solution should be in the hyperbolical tangent, where then also the limit to infinity makes sense. $\endgroup$ – LutzL Mar 17 '16 at 17:43
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These particular boundary values are a bit special. You didn't make any mistakes in your analysis, as far as I can see. Your conclusion that the outer solution is the leading order approximate solution is also correct. In fact, the outer solution is an exact solution to the problem with these boundary conditions (try it, you'll see that $y_0$ solves the original problem exactly). So, for this special choice of boundary conditions, the boundary layer vanishes.

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