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Consider the boundary value problem $$\varepsilon \frac{d^2y}{dx^2}+(1+x)\frac{dy}{dx}+y=0$$ subject to $y(0)=0$, $y(1)=1$, for $0 \leq x \leq 1$.

Use the method of matched asymptotic expansions to construct two-term inner and outer expansions to the problem, which should then be matched using Van Dyke's matching principle.


What i did so far is:

$y \sim y_0 +y_1 \varepsilon + O(\varepsilon ^2) $

$$\varepsilon \frac{d^2y_0}{dx^2}+\varepsilon ^2y_1+(1+x)\bigg(\frac{dy_0}{dx}+ \varepsilon \frac{dy_1}{dx} \bigg)+y_0 +\varepsilon y_1=0$$

At $O(1)$: $$ (1+x)\frac{dy_0}{dx}+y_0=0 $$ giving $y_0=2(1+x)^{-1}$

At $O(\varepsilon )$: $$\frac{d^2y_0}{dx^2}+(1+x) \frac{dy_1}{dx} +y_1=0 $$ giving $y_1=2(1+x)^{-3}-\frac12(1+x)^{-1}$

What do I do next?

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  • $\begingroup$ @David I think I fixed it now. $\endgroup$
    – snowman
    Commented Nov 19, 2015 at 23:08
  • $\begingroup$ Did you solve this? $\endgroup$
    – David
    Commented Nov 27, 2015 at 0:13
  • $\begingroup$ @David yeah took ages! $\endgroup$
    – snowman
    Commented Nov 29, 2015 at 15:05
  • $\begingroup$ I'd like to see an answer, even if you can just give a brief outline. I've never used Van Dyke's principle. $\endgroup$
    – David
    Commented Nov 29, 2015 at 21:29
  • $\begingroup$ @David is there a way I can PM you. I only have pics of the solution $\endgroup$
    – snowman
    Commented Nov 29, 2015 at 21:38

1 Answer 1

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Inner solution: $$Y \sim -B_0 +B_0 e^{-X} + \varepsilon \bigg( -\frac12 B_0 X^2 e^{-X} - ke^{-X} +B_0 X +k \bigg) + O(\varepsilon ^2)$$

Outer solution: $$y \sim 2(1+x)^{-1} +2 \varepsilon (1+x)^{-3} - \frac{\varepsilon}2(1+x)^{-1} + O(\varepsilon ^2)$$

Van Dyke:

Write inner in terms of outer: $$y \sim B_0 (x-1) + \varepsilon k + O(\varepsilon ^2)$$

Write outer in terms of inner (after expanding terms by binomial): $$ Y \sim 2+ \varepsilon (3 /2 -2X) +O(\varepsilon ^2)$$

Matching them gives $B_0 =-2$ and $k=3/2$

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  • $\begingroup$ Do you match $B_0(x-1)$ to $2$ by matching the constant parts, or do you have to take a limit? Thanks for the answer! $\endgroup$
    – David
    Commented Nov 29, 2015 at 22:23
  • $\begingroup$ @David just by matching the constant parts. No limits. $\endgroup$
    – snowman
    Commented Nov 29, 2015 at 22:25
  • $\begingroup$ How did you get the inner solution? $\endgroup$
    – Leo 254
    Commented Mar 28, 2017 at 18:44

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