Some questions on outer region and outer solution of a boundary problem

There is this problem below that I have some doubts and confusions, I will appreciate if anyone could provide some clarifications and explanations. I am new to the Boundary Layer Theory, this question involves the Boundary Layer Theory, it is a collection of perturbation methods for solving differential equations whose solutions exhibit a boundary layer structure.

Consider the boundary value problem $\epsilon y''+(1+\epsilon)y'+y=0$, $y(0)=0,y(1)=1$. Keeping $x$ fixed and taking the limit $\epsilon\to0$ (these are called outer regions) we approximate the solution of the differential equation by the solution to the outer equation $eq_{out}=y_{out}'(x)+y_{out}(x)=0$. Since this is a first order equation it cannot satisfy both boundary conditions. The outer solution satisfies only $y_{out}(1)=1$.

My understanding is that an outer region is a region outside the boundary layer. Now comes my question, by taking the limit $\epsilon\to0$, we get $y'+y=0$, then how can we tell where the outside region lies? How can we know the outer solution satisfies only $y_{out}(1)=1$?

I think understanding this step is crucial to proceed, and that is the reason I want to get the ideas correct.

Thanks in advance for any helps!

1 Answer

Let's pretend we know nothing at all about boundary layers and just try to solve this equation by expanding $y$ in powers of $\epsilon$, i.e.

$$y(x) = y_{0}(x) + \epsilon y_{1}(x) + \epsilon^{2}y_{2}(x) + \ldots$$

Substitute this expression for $y(x)$ into the given differential equation and then collect the like powers of $\epsilon$ to recover the following:

$$\begin{array}{ll} \epsilon^{0}: & y_{0}' + y_{0} = 0 \\ \epsilon^{1}: & y_{0}'' + y_{0}' + y_{1}' + y_{1} = 0 \\ \epsilon^{2}: & y_{1}'' + y_{1}' + y_{2}' + y_{2} = 0 \end{array}$$ and so on. We solve the first equation and find that $y_{0}(x) = Ae^{-x}$. Substitute into the second equation and solve to find that $y_{1}(x) = Be^{-x}$. Substitute into the third equation to find that $y_{2}(x) = Ce^{-x}$ and so on.

We realise that all these solutions are effectively the same, so we confidently claim that $y(x) = Ae^{-x}$ and now we just have to fix $A$ by using the boundary conditions. Unfortunately, we experience a sinking feeling when we realise that we have two boundary conditions, $y(0) = 0$ and $y(1) = 1$, but only one free variable.

Our solution cannot satisfy the boundary condition at $x = 0$ (unless $A = 0$, but then we will recover the trivial solution), so we pick $A = e$ so that $y(1) = 1$. To satisfy the boundary condition at zero we introduce a boundary layer of thickness $\delta$ at zero, i.e. we rescale $x = \delta X$ and $y(x) = Y(X)$. Substitute into our equation

$$\frac{\epsilon}{\delta^{2}} Y'' + \frac{1+\epsilon}{\delta}Y' + Y = 0.$$

To ensure that the terms in this equation are balanced we choose $\delta = \epsilon$, then

$$Y'' +(1+\epsilon) Y' + \epsilon Y = 0.$$

Note that the first two terms are both order one, and the third term is negligibly small, i.e. the first two terms can balance each other out so that the entire expression is equal to zero.

If we had chosen some other width for the boundary layer, e.g. $\delta = \sqrt{\epsilon}$ then our equation would become

$$Y'' + \frac{1+\epsilon}{\sqrt{\epsilon}} Y' + Y = 0.$$

This is unbalanced - although the $Y''$ and $Y$ terms are both order one, the middle term is no longer negligible, instead it blows up as $\epsilon \rightarrow 0$.

What remains is to solve the rescaled equation by expanding $Y(X) = Y_{0}(X) + \epsilon Y_{1}(X) + \ldots$, and then to match the rescaled (or inner solution) to the outer solution using Van Dyke's rule. I may add those details later.