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I have got the following problem, taken from Multiple Scale and singular perturbation methods, Kevorkian & Cole book, page 94, exercise 1.b.:

Find the leading order of the problem:

$\varepsilon y''+ \frac{1}{2}y^2y'-y=0 $, with $0<x<1$ and $y(0)=A$, $y(1)=B$, and A,B independent of $\varepsilon$ .

I have tried to use the boundary layer method in order to find an outer and inner solution and then make the respective matching to find some constants which appear.

The inner layer is at some point inside the interval if $B^2-A^2\ne4$.

When doing the expansion for the inner solution I have got the following equation:

$Y_0'+ \frac{1}{6}Y_0^3=K$.

This integral is not trivial but it can be solved by partial-fraction decomposition to get an implicit solution with depends in two different constants, namely $K$ and $C$.

I have got

$\frac{x}{6}+C=\frac{1}{3K^{\frac{2}{3}}}\left[-\ln\left(K^{\frac{1}{3}}-Y_{0}\right)+\frac{1}{2}\ln\left(K^{\frac{2}{3}}+K^{\frac{1}{3}}Y_{0}+Y_{0}^{2}\right)+\sqrt{3}\tan^{-1}\left(\frac{2Y_{0}}{\sqrt{3}K^{\frac{1}{3}}}+\frac{1}{\sqrt{3}}\right)\right]$

The problem here is that in order to do the matching, I should get an explicit solution $Y_0$ in order to find the constants.

Can someone give a hint?

Is there maybe another approach to solve this problem?

Thank you.

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  • $\begingroup$ Nice question. I wonder if there's a way to match solutions if they are both written as $x = f(Y_0)$. $\endgroup$ – Bennett Gardiner Aug 26 '14 at 12:50
  • $\begingroup$ Not that it matters, but I got $B^2-A^2\neq 4$. $\endgroup$ – Bennett Gardiner Aug 27 '14 at 6:17
  • $\begingroup$ Yes, you are right, I had a typo. Thank you $\endgroup$ – Ankara Aug 27 '14 at 11:30

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