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I am trying to solve this boundary value problem and obtain the leading-order approximations using asymptotic matching. But I got my solution wrong and I am stuck along the way. I would really appreciate any help!

Find leading-order asymptotic approximations to the solution of
$\epsilon y'' + cosh(x) y' - y = 0, y(0)=y(1)=1$.
Compare with the numerical solution for epsilon = 0.05.

This is my work:
First, I solved the outer solution $y_{out}=e^{1-x}$.

Then I am not sure how can I find the inner solution, is it correct if I use the substitution $x=\epsilon w$ and $y[\epsilon w]=Y[w]$. But if I did that I will get $y_{in}=0$. Can anyone lend some helps? And furthermore I don't know how to find the overlapping region for the matching.

Thanks for any helps!

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To get the inner solution, rescale using $x = \epsilon X$. Then, letting $Y(X) = y(\epsilon X)$

$$ Y'' + \cosh{(\epsilon x)} Y' - \epsilon Y = 0$$

Since we are only considering solutions for small $\epsilon$, we can throw away the $Y$ term as well as the cosh for the lowest-order approximation. Thus we have

$$Y'' + Y' = 0 \implies Y(X) = A + B e^X$$

By using the condition at $x=0$, we get $A=1-B$. Thus

$$y_{\text{inner}}(x) = Y\left(\frac{x}{\epsilon}\right) = 1-B \left ( 1-e^{-x/\epsilon}\right)$$

We find $B$ by asymptotic matching; that is

$$\lim_{X \to \infty} Y(X) = \lim_{x \to 0} y_{\text{outer}}(x) \implies 1-B=e$$

To get the uniform approximation, we add the inner and outer solutions and subtract off this common value of $e$ in the matched region (kind of like the inclusion-exclusion principle). The result is

$$y_{\text{unif}}(x) = e^{1-x} - (e-1) e^{-x/\epsilon}$$

Here is a plot for $\epsilon=0.05$. I guess you need to compare with a numerical solution of the diff eq'n.

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  • $\begingroup$ Thanks for your answer. Just wondering whether you meant $Y(X)=y(\epsilon X)$ ? $\endgroup$ – user71346 Jun 13 '13 at 16:06
  • $\begingroup$ @user71346: yes I did, thanks - too easy to make typos with this notation. $\endgroup$ – Ron Gordon Jun 13 '13 at 16:10
  • $\begingroup$ Thanks. I still don't quite understand how did you get $y_{inner}(x)=Y(\frac{x}{\epsilon})=1-B(1-e^{-x/\epsilon})$, is there a typo somewhere, I am bit confused :( Also, is it suppose to be $cosh(\epsilon X)$? $\endgroup$ – user71346 Jun 13 '13 at 16:17
  • $\begingroup$ @user71346: I neglected the cosh piece because it is essentially one for small $\epsilon$ - it makes a difference for higher-order approximations. No typo there, you need to work out the steps in between. $\endgroup$ – Ron Gordon Jun 13 '13 at 16:34
  • $\begingroup$ I see. I mean $cosh(\epsilon X)$ instead of $cosh(\epsilon x)$, but maybe that's just a minor thing. Regarding the uniform approximation, if you add in and out then subtract e, then you will get $e^{1-x}+e(1-e^{-x/\epsilon})-e=e^{1-x}+e-e^{1-x/\epsilon}-e\neq e^{1-x}-(e-1)e^{-x/\epsilon}$ ? Had I made any mistake? $\endgroup$ – user71346 Jun 13 '13 at 22:34

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