Find the leading order uniform approximation when the conditions are not $0<x<1$

Question

$$\epsilon y''+y'\sin x+y\sin 2x = 0$$ with boundary conditions $y(0)=\pi$ and $y(\pi)=0$ as $\epsilon \rightarrow 0$. I don't know how to find out where the boundary layer is? I thought initially it was at $x=0$, but this just leads to the outer solution of $$y=Ae^{-2\sin x}$$ if I've done it correctly and applying the boundary conditions means $A=0$ or $A=\pi$. Am I right so far? What do I do now? I'm so lost. Any hints? Thanks!

Answer 1

First note that for $\epsilon = 0$ you get the outer solution

$$ y_o = A e^{- \sin x} $$

As you pointed out, $A=0$ or $A=\pi$. Let's try to put the boundary layer at $x=0$. If the Bl is at $x=0$, we get $A=0$. We introduce $\hat{x}=x/\epsilon$

$$ \epsilon \frac{1}{\epsilon^2} \frac{d^2 y}{d \hat{x}^2} + \frac{1}{\epsilon} \frac{d y}{d \hat{x}} \sin (\epsilon \hat{x})+y\sin [2(\epsilon \hat{x})] = 0 $$

$$ \frac{d^2 y}{d \hat{x}^2} + \frac{d y}{d \hat{x}} \sin (\epsilon \hat{x})+\epsilon y \sin [2(\epsilon \hat{x})] = 0 $$

The leading order solution can be obtained from $\epsilon=0$

$$ \frac{d^2 y}{d \hat{x}^2} = 0 $$

From which the inner solution is

$$ y_i = B \hat{x} + C $$

Since $y(0)=\pi$ we have $C=\pi$. If we required that the solutions match at $x=0$, i.e. $y_o(0)=0=y_i(\hat{x}=\infty)=\infty$ if $B \neq 0$, therefore we get a contradiction, we can't match them this way. Try the scaling $\hat{x}=x/\sqrt{\epsilon}$

$$ \epsilon \frac{1}{\epsilon} \frac{d^2 y}{d \hat{x}^2} + \frac{1}{\sqrt{\epsilon}} \frac{d y}{d \hat{x}} \sin (\sqrt{\epsilon} \hat{x})+y\sin [2 \sqrt{\epsilon} \hat{x}] = 0 $$

To the leading order we get

$$ \frac{d^2 y}{d \hat{x}^2} + \frac{d y}{d \hat{x}} \hat{x}= 0 $$

which has the solution

$$ y_i = \pi + \sqrt{\frac{\pi}{2}} B \cdot erf \left[ \frac{\hat{x}}{\sqrt{2}} \right]$$

We can match this inner solution to the outer solution at $x=0$ in the following way:

$$ y_o(x=0) = 0 =y_i(\hat{x}=\infty) = \pi + \sqrt{\frac{\pi}{2}} B $$

from which $B=-\sqrt{2 \pi}$. The combined solution can be written as

$$\pi \left[ 1 - erf \left[ \frac{x}{\sqrt{2 \epsilon}} \right] \right] $$