I am learning perturbation theory and would like to be able to determine where boundary layers are going to occur just by looking at the differential equation.


Let $n\in\mathbb{N}$ and $p\_i(x)$, $0\leq i< n$ some sufficiently well-behaved functions. 


Am I able to determine the boundary layers of the following problem just by looking at the $p\_i(x)$ or some other easy to see property?:


$\epsilon \frac{d^n y}{dy^n} + \sum\_{i=0}^{n-1} p\_i(x) \frac{d^i y}{dy^i} =0$


$y(0)=a$ and $y(1)=b$


(I realise that I require more constraints to get a unique solution but I don't think this effects the existence of boundary layers)


Thanks in advance. 



Where are the boundary layers?
Of course, you **do** need $n$ boundary conditions, so that your solution is unique. If you have $m$ boundary conditions, with $m < n$, then you have enough choice among the solutions of the differential equation so as to avoid a boundary layer. Here is an example.
$$\epsilon y'''+y''=0,\qquad y(0)=a, y(1)=b.$$
Given $a,b\in{\mathbb R}$, there are a lot of solutions which form a line. All of them have a boundary layer (a scalar times $\exp(-\frac{x}{\epsilon})$), but $\bar y(x)=a+(b-a)x$. 


Suppose that the third boundary condition is $y'(0)=c$, then the solution $y\_\epsilon$ tends to some $y$ such that $y''=0$, $y(0)=a$ and $y(1)=b$, which is a Dirichlet problem. There is a boundary layer at $x=0$, where a boundary condition is lost.


If instead the third boundary condition is $y'(1)=d$, then the limit of $y\_\epsilon$ still solves $y''=0$, but with $y(1)=b$ and $y'(1)=d$, which is a Cauchy problem. The boundary layer is still at $x=0$.


In general, consider an equation  

$$\epsilon y^{(n}+p\_1(x)y^{(n-1)}+\cdots=0,$$
with $\epsilon>0$ but very small. Assume that $p\_1$ does not vanish, in order that the limit equation be non-singular. Then you can decide whether the boundary layer is at rigt or at left by looking at the sign of $p\_1$. It is at left if $p\_1> 0$, at right if $p\_1< 0$. 


