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!
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.