# Asymptotic Methods - Boundary Layer Problems

I am currently studying a course in Asymptotic and Perturbation Methods and we have recently started discussing "Boundary Layer problems". It is not clear to me, however, exactly what form "Boundary Layer problems" typically take. In class, we have calculated approximate solutions for second order differential equations, where a small parameter multiplies the highest order derivative in the equation and the solution is subject to two boundary conditions. Why though do the equations I am presented with have to be solvable? Why does there have to be a a solution for any value of the small parameter and, further, why does the solution have to suffer a rapid change at some point? What other situations are classed as "boundary layer problems"?

Thank you in anticipation of your help.

-

First, I have to clarify your question.

1. Exactly what form "Boundary Layer Problems" typically take?
2. Why does there have to be a a solution for any value of the small parameter?
3. why does there have to suffer a rapid change at some point?

I think you have some example in class. Looks like this. $$\epsilon y''+2y'+2y=0, for\ 0<x<1,$$ $$where\ y(0)=0\ and\ y(1)=1$$

This form is "Boundary Layer Problems" typically take, exactly! The perturbation problems in differential equations with boundary, usually one is inner and another is outer. With this problem, we usually calculate first term because only first term is already messy to solve and near to the exact solution enough.

Another "Boundary Layer Problems" for you.$$\epsilon y''+3y'-y^4=0,\ for\ 0<x<1$$ $$where\ y(0)=y(1)=1$$ This picture is the solution of that problem when $\epsilon=0.1$. You can see the point at $(0,1)$ and $(1,1)$ same as the boundary condition. That is my answer of the first question.

And second question, I think your question is quite vague to give an anwser. In the viewpoint of the solution with respect to the value of $\epsilon$, you can see the solution of that problem moves when you change the value of $\epsilon$. And we can talk about the convergence of the composite expansion to the solution of the problem, as $\epsilon$ decreases.

Finally, Third question. I cite 'Introduction to Perturbation methods' page 8. It is quite long, but really helpful to give intuition.

Suppose $f=x+e^{-x/\epsilon}$, where $0<x<1$ is fixed. In this case $f~x$ for small $\epsilon$. However, it is natural to ask how well this does in approximating the functions for $0<x<1$. If we plot both togather, we obtain the curves shown in Fig. 1.2. It is apparent that the approximation is quite good away from $x=0$. It is also clear that we do not do so well near $x=0$. This is true no matter what we value of $\epsilon$ we choose since $f(0)=1$. It should be remembered that with an asymptotic approximation, given a value of $x$, the approximation is a good one if $\epsilon$ is close to $\epsilon_0=0$. what we are seeing in this example is that exactly how small $\epsilon$ must be can depend on $x$ (the closer we are to $x=0$, the smaller $\epsilon$ must be)

I think this citation give the answer. So, I apply this with certain example. We will focus on the first example I gave. Through the calculus, we obtain $$y_0'+y_0=0$$ and the general solution of this is $$y_0(x)=ae^{-x}$$ But you already know this equation cannot satisfy both boundary condition, in other words, this equation cannot describing the solution over the entire domain $0<x<1$. And this equation is weak to suffer a rapid change near the $x=0$. This is the reason why we treat this equation with outer boundary $(x=1)$ which is far from the oscillation, and do the rescale to stretch the region near $x=0$ as $\epsilon$ becomes small.

I'll talk breafly about the 'boundary-layer coordinate' given as $$s=\frac{x}{\epsilon^\alpha}$$ The reason why we can stretch the region near $x=0$ is that we need fixed $s$ when $\epsilon$ becomes small(in the same way that x is held fixed in the outer expansion) The equation above is sometimes referred to as a stretching transformation. And we use this fixed $s$ to substitute independent variable like $$\epsilon^{1-2\alpha}\frac{{d^2}y}{ds^2}+2\epsilon^{-\alpha}\frac{dy}{ds}+2y=0$$ I think you can do the next step to find composite expansion.

I apologize that I didn't answer the whole question you gave. Because I cannot understand the exact meaning of your question. Hope this answer will help you guys. Thanks!

-