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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.

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1 Answer 1

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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 $p_1< 0$.

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