Consider the boundary value problem $$ε \frac{d^2y}{ dx^2} + (1 + x) \frac{dy }{dx} + y = 0$$ subject to $y(0) = 0$, $y(1) = 1$, for $0 \le x \le 1$, $ε ≪ 1$.
By considering the rescaling $x = x_0 + ε^αX$, $y = Y$, for $X, Y = O(1)$, $α > 0$, show that a boundary layer is only possible at $x = 0$.
I am stuck on this question but I am mostly stuck on why we have to use dominant balance and how to use it in this.
We can turn the main equation into $$ε^{1-2\alpha }Y'' + ε^{-\alpha}Y'+x_0 ε^{-\alpha} Y'+ XY' +Y=0$$
Then the balance stuff comes in but I have tried reading my notes on it for ages but really don't understand how it work. I do know that we use it to find out what the form of our solution looks like.
Because $x=x_0+\epsilon^\alpha X$, $$ \frac{\mathrm d}{\mathrm dx}=\frac{\mathrm d}{\mathrm dX}\frac{\mathrm dX}{\mathrm dx}=\epsilon^{-\alpha}\frac{\mathrm d}{\mathrm dX}$$ because $X=\epsilon^{-\alpha}(x-x_0)$. Therefore, the second derivative has a power of $\epsilon^{-2\alpha}$.
For your equation $$\epsilon^{1-2\alpha }Y'' + \epsilon^{-\alpha}Y'+x_0 \epsilon^{-\alpha} Y'+ XY' +Y=0,$$ doing dominant balance means checking all the possible values of $\alpha $ that can balance two or more terms in the equation, and seeing which balances are consistent and dominant.
You have 3 different powers of $\epsilon$, $1-2\alpha$, $-\alpha$, and $0$. Thus, the balance is either $1-2\alpha=-\alpha$, $1-2\alpha=0$, or $-\alpha=0$. The first balance is $\alpha=1$, the second is $\alpha=1/2$ and the third is $\alpha=0$.
Setting $\alpha=0$ is no scaling. This gives your outer solution.
Setting $\alpha=1/2$ is not dominant, because you have an $O(\epsilon^{-1/2})$ term ($Y'$) that dominates the rest of the equation.
$\alpha=1$ gives a dominant, consistent balance, with resulting equation $$ Y''+(1+x_0)Y'=0. $$ Solving this equation will tell you what $x_0$ can or cannot be.