Method for finding roots of perturbation singular problems

Question

So I am confident with finding roots of regular perturbation problems, e.g. $x^{2}+2\epsilon x-4=0$. However, I am getting confused when I have a singular problem, where $\epsilon$ affects the leading behaviour. What is the best method to use when finding roots to singular perturbation problems such as $\epsilon x^{2}+2x-4\epsilon=0$ (I don't know if that example actually works, by the way)?

Many thanks in advance :)

Answer 1

You may set $x=1/y$ and multiply the equation $$\epsilon x^{2}+2x-4=0\tag{1}$$

by $y^2$ and obtain: $$\epsilon +2y-4 y^2=0\tag{2}$$

This probably answers your question.

But in your original example, $\epsilon$ appears both in the leading order term and trailing order term. So the above method do not work.

We first rewrite it as: $$\epsilon (x^{2}-4)+2x=0\tag{4}$$

Then we can solve $$x^2-4=y^2+4y$$ for x and obtain $$x=y+2,x=-(y+2)$$ We then substitute $x=y+2$ into (4) and obtain $$\epsilon ((y+2)^{2}+4(y+2))+2(y+2)=0\tag{5}$$

You may apply the method of treating (1) to treat (5). You may then do it for the second solution $x=-(y+2)$.

Answer 2

Part of the problem lies in realizing which roots are regular and which roots are singular. In this problem, one of the roots is actually regular: this root is a perturbation of the one root of $2x=0$, i.e. of $x=0$. This occurs in a regime where $\epsilon x^2$ is of smaller order than $2x$, so that the dominant term in the expression $\epsilon x^2+2x-4\epsilon$ is $2x$.

But for fixed $\epsilon$, we can take $x$ so large that the dominant term in $\epsilon x^2+2x-4\epsilon$ is $\epsilon x^2$. In this situation, the equation is badly scaled, so we rescale it, by setting $x=\epsilon^p y$. Here $p<0$ and $y$ is bounded (away from $\pm \infty$ and also away from $0$) as $\epsilon \to 0$. This means that the root is getting larger (in magnitude) as $\epsilon$ is going to zero. If you consider $p=-1$, then you get

$$\epsilon^{-1} y^2 + 2 \epsilon^{-1} y - 4 \epsilon = 0$$

which is equivalent to

$$y^2 + 2y - 4 \epsilon^2 = 0.$$

(We chose $p=-1$ so that the quadratic and linear terms would be of the same order.) The relevant solution to this is a regular perturbation perturbation of the root at $-2$ of the equation $y^2+2y=0$. So the relevant solution is $x=-2\epsilon^{-1}$ plus higher order terms.