# First order, inner, outer and uniform approximations of boundary layer problem

$\epsilon \frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=0, y(0) = \alpha, y(1) = \beta$

Since we have the general form:

$\epsilon \frac{d^2y}{dx^2}+a(x)\frac{dy}{dx}+b(x)y=0$

we can see that $a(x) = 2 >0$ so our boundary layer is near $x=0$.

Outer Solution:

Setting $\epsilon = 0$ we get

$2y' + 2y = 0, y(1) = \beta$

$y(x) = \beta e^{1-x}$

Inner Solution:

Here I am unsure how to make the following transformation

let $y(x) = Y(X)$, $x = \delta X$, $\delta = \delta(\epsilon)$

Following my lecture notes they went from

$\epsilon y'' + x^2y' - y = 0$ -> $\frac{\epsilon}{\delta^2}Y'' + \delta X^2Y' - Y = 0$

What you are doing is going from $x$, which when changed a little bit in the boundary layer makes $y$ change a lot, to a variable $X$, which is like $x$, but stretched out so that $y$ doesn't change as quickly.
So, you have $x=\delta(\epsilon)X$, so you have to work out how the derivatives change. Use the chain rule $$\frac{\mathrm d}{\mathrm dx}=\frac{\mathrm dX}{\mathrm dx}\frac{\mathrm d}{\mathrm dX}=\frac{1}{\delta}\frac{\mathrm d}{\mathrm dX}.$$ So the new equation is $$\frac{\epsilon}{\delta^2}Y''(X)+\frac{2}{\delta}Y'(X)+2Y=0$$ and the boundary condition is $Y(0)=0$. You next have to use the method of dominant balance to work out what $\delta(\epsilon)$ should be.