2 added 864 characters in body

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 boudary 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 $p_1< 0$.

1

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 boudary layer, but $\bar y(x)=a+(b-a)x$.