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This question is taken from Bender & Orszag "perturbation methods"

$y' = (1 + X^{-2}/100)y^2 - 2y + 1$ ,$y(1)=1$

first we can see that if we set $\epsilon=100x^{2}$ we can translate the above to the boundary layer problem

$\epsilon y' = (\epsilon+1)y^{2}-\epsilon y+\epsilon$

my question is what is the outer and the inner solution on [0,1]?

Thanks

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  • $\begingroup$ Could you work out the setup of the inner and outer equations? $\endgroup$ – Ian May 6 '15 at 16:04
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This is not a boundary layer perturbation problem. The proposed $ϵ=100x^2$ is not a small term around $x=1$.

One reduced equation is $$ y'=y^2-2y+1=(y-1)^2 $$ with solution $$ y\equiv 1\text{ or }y=1-\frac{1}{x-c}. $$ Now one can add perturbation terms either to $y$ or to $(y-1)^{-1}$

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  • $\begingroup$ How did you arrive to your first line? Can you just drop the $100/x^2$ term? $\endgroup$ – Jade Mar 4 '18 at 5:34
  • $\begingroup$ If there were such a term, it would be large, not small. But I can drop $0.01/x^2$ against $1$, as that is a relatively small perturbation. $\endgroup$ – LutzL Mar 4 '18 at 6:20
  • $\begingroup$ So you've dropped the small term, solved it exactly, and then how would you determine the perturbed terms to either the $y$ or $(y-1)^{-1}$ term? Would you sub in the solution you found back into the DE with an extra term resulting from the perturbation? $\endgroup$ – Jade Mar 4 '18 at 6:45
  • $\begingroup$ Yes. You can write the equation as $$y'-(y-1)^2=ϵ\frac{y^2}{x^2}$$ where you can then apply standard perturbation methods. $\endgroup$ – LutzL Mar 4 '18 at 7:16

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