The foundation of this research is based on the principles of perturbation theory, specifically focusing on both regular and singular perturbation problems. The methods employed in this study involve the application of various algebraic and analytical techniques that are well-established in applied mathematics. These include:

1. Asymptotic Expansion: This technique involves expressing the solution as a series expansion in terms of a small parameter, typically denoted by ε. The series is then truncated to obtain an approximate solution. This method is particularly useful for regular perturbation problems.

2. Method of Matched Asymptotic Expansions: This method is employed to solve singular perturbation problems, where different scales of the problem require different asymptotic expansions. The solutions obtained in different regions (e.g., inner and outer regions) are then matched in an overlapping region to ensure a smooth transition.

3. Strained Coordinates: This technique modifies the independent variable(s) of the problem to capture the rapid changes in the solution within specific regions, often used in conjunction with other methods to handle singular perturbations.

4. Method of Multiple Scales: This method is used to tackle problems with solutions exhibiting behavior on multiple time or spatial scales. By introducing multiple independent variables, each associated with a different scale, we can derive a set of coupled differential equations that can be solved more easily.

5. Poincaré-Lindstedt Method: This perturbation technique is used to remove secular terms (terms that grow without bound) in the solution series, thereby ensuring the series remains uniformly valid over time.

The mathematical methods were implemented using symbolic computation tools such as MATLAB and Wolfram Mathematica. These tools were employed to derive, manipulate, and simplify the resulting expressions from the perturbation techniques. Additionally, numerical simulations were performed to verify the accuracy of the analytical solutions, particularly in cases where exact solutions are known.

A comparative analysis of the strengths and limitations of the aforementioned methods was conducted. This analysis involved applying each method to canonical perturbation problems, including regular and singular perturbations. The performance of each method was evaluated based on:

Several case studies were selected from physical contexts, including fluid dynamics, quantum mechanics, and elasticity theory, where singular perturbation problems naturally arise. These case studies were used to demonstrate the practical application of the methods and to highlight the importance of selecting the appropriate technique based on the nature of the problem.

The validity of the approximate solutions obtained through the various perturbation methods was confirmed by comparing them to known exact solutions (where available) or to high-precision numerical solutions. Convergence tests were conducted to ensure that the truncated series solutions provided sufficiently accurate approximations as the perturbation parameter approached zero.

A solution that can be expressed in a closed form, using a finite number of operations, including addition, subtraction, multiplication, division, roots, and elementary functions, such as trigonometric functions, exponential functions, and logarithmic functions, without any approximation or truncation. In other words, an exact solution is a precise and explicit solution that can be written down exactly, without any error or approximation, using a finite combination of basic mathematical operations and functions. For example, the solution to the equation $x^2+6x+9=0$ is $x=-3$, which is an exact solution because it can be expressed precisely and explicitly without any approximation [6] [10] .

We begin solving the quadratic equation in x containing the parameter ε, $x^2+\epsilon x-1=0$.

This equation has exact solutions using the quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ is given by

$x=\frac{-\epsilon \pm \sqrt{{\epsilon }^2+4}}{2}=-\frac{1}{2}\epsilon \pm \sqrt{\frac{{\epsilon }^2+4}{4}}=-\frac{1}{2}\epsilon \pm \sqrt{1+\frac{{\epsilon }^2}{4}}$

These solutions can be expanded using the binomial expansion for small $\epsilon$

${\left(1+x\right)}^n\approx 1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3+\cdots$

$\sqrt{1+\frac{{\epsilon }^2}{4}}={\left(1+\frac{{\epsilon }^2}{4}\right)}^{\frac{1}{2}}$, valid for $\left|n\right|<1$

$\begin{array}{c}\sqrt{1+\frac{{\epsilon }^2}{4}}={\left(1+\frac{{\epsilon }^2}{4}\right)}^{\frac{1}{2}}\\ \approx 1+\frac{{\epsilon }^2}{2\times 4}x+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}{\left(\frac{{\epsilon }^2}{4}\right)}^2+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}{\left(\frac{{\epsilon }^2}{4}\right)}^3+\cdots \\ \approx 1+\frac{{\epsilon }^2}{8}x+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}{\left(\frac{{\epsilon }^2}{4}\right)}^2\approx 1+\frac{1}{8}{\epsilon }^2-\frac{1}{128}{\epsilon }^4+\cdots \end{array}$

Thus, the exact solution

$x=\frac{-\epsilon \pm \sqrt{{\epsilon }^2+4}}{2}=-\frac{1}{2}\epsilon \pm \left(1+\frac{1}{8}{\epsilon }^2-\frac{1}{128}{\epsilon }^4+\cdots \right)$

$x=\{\begin{array}{l}+1-\frac{1}{2}\epsilon +\frac{1}{8}{\epsilon }^2-\frac{1}{128}{\epsilon }^4+O\left({\epsilon }^6\right)\\ -1-\frac{1}{2}\epsilon -\frac{1}{8}{\epsilon }^2+\frac{1}{128}{\epsilon }^4+O\left({\epsilon }^6\right)\end{array}$

These binomial expansions converge if $\epsilon <2$.

Now, suppose we did not know the exact solution, how could we generate the approximate solutions for $\epsilon \ll 1$?

$x^2+\epsilon x-1=0$

Given that $\epsilon =0$ then, we have $x^2-1=0$ and the solution is $x=\pm 1$.

Now, let’s try $x=1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots$ disregarding $O\left({\epsilon }^3\right)$.

Substituting x into $x^2+\epsilon x-1=0$ we have

${\left(1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)}^2+\epsilon \left(1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)-1=0$

${\left(1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)}^2+\epsilon \left(1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)-1=0+0\epsilon +0{\epsilon }^2+\cdots$

Equating the coefficients of each power of ε

$1+2\epsilon x_1+{\epsilon }^2\left(2x_2+x_1^2\right)+\epsilon +{\epsilon }^2{x}_1-1\cdots =0+0\epsilon +0{\epsilon }^2+\cdots$

${\epsilon }^1:2x_1+1=0,x_1=-\frac{1}{2}$

${\epsilon }^2:2x_2+x_1^2+x_1=0,x_2=-\frac{x_1^2+x_1}{2}=-\frac{\frac{1}{4}-\frac{1}{2}}{2}=\frac{1}{8}$

Thus $x=1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots$, $x=1-\frac{1}{2}\epsilon +\frac{1}{8}{\epsilon }^2+\cdots$.

Now, let’s try $x=-1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots$ disregarding $O\left({\epsilon }^3\right)$.

Substituting x into $x^2+\epsilon x-1=0$ we have

${\left(-1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)}^2+\epsilon \left(-1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)-1=0$

${\left(-1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)}^2+\epsilon \left(-1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)-1=0+0\epsilon +0{\epsilon }^2+\cdots$

Equating the coefficients of each power of ε

$1-2\epsilon x_1-2x_2{\epsilon }^2+{\epsilon }^2{x}_1^2-\epsilon +{\epsilon }^2{x}_1-1\cdots =0+0\epsilon +0{\epsilon }^2+\cdots$

${\epsilon }^1:-2x_1-1=0,x_1=-\frac{1}{2}$

${\epsilon }^2:-2x_2+x_1^2+x_1=0,x_2=\frac{x_1^2+x_1}{2}=\frac{\frac{1}{4}-\frac{1}{2}}{2}=-\frac{1}{8}$

Thus $x=1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots$

$x=-1-\frac{1}{2}\epsilon -\frac{1}{8}{\epsilon }^2+\cdots$

$x=\{\begin{array}{l}+1-\frac{1}{2}\epsilon +\frac{1}{8}{\epsilon }^2+\cdots \\ -1-\frac{1}{2}\epsilon -\frac{1}{8}{\epsilon }^2+\cdots \end{array}$

But we need to guess the form $x=x_0+\epsilon x_1+{\epsilon }^2{x}_2+\cdots$ which is the asymptotic sequence.

Let’s guess the asymptotic sequence $\left\{1,\epsilon ,{\epsilon }^2,{\epsilon }^3,\cdots \right\}$ by substituting $x=x_0+\epsilon x_1+{\epsilon }^2{x}_2+\cdots$ into the equation $x^2+\epsilon x-1=0$

${\left(x_0+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)}^2+\epsilon \left(x_0+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)-1=0$

${\left(x_0+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)}^2+\epsilon \left(1+\epsilon x_1+{\epsilon }^2{x}_2+\cdots \right)-1=0+0\epsilon +0{\epsilon }^2+\cdots$

$x_0^2+2\epsilon x_0{x}_1+2x_0{x}_2{\epsilon }^2+{\epsilon }^2{x}_1^2+\epsilon +{\epsilon }^2{x}_1-1\cdots =0+0\epsilon +0{\epsilon }^2+\cdots$

$x_0^2-1+2\epsilon x_0{x}_1+{\epsilon }^2\left(2x_0{x}_2+x_1^2\right)+\epsilon +{\epsilon }^2{x}_1\cdots =0+0\epsilon +0{\epsilon }^2+\cdots$

Equating coefficients

${\epsilon }^0:x_0^2-1=0,x_0=\pm 1$

${\epsilon }^1:2x_0{x}_1+1=0,x_1=-\frac{1}{2}$

${\epsilon }^2:2x_0{x}_2+x_1^2+x_1=0,x_2=-\frac{x_1^2+x_1}{2}=-\frac{\frac{1}{4}-\frac{1}{2}}{2}=\frac{1}{8}$

Thus $x=x_0+\epsilon x_1+{\epsilon }^2{x}_2+\cdots$

$x=1-\frac{1}{2}\epsilon +\frac{1}{8}{\epsilon }^2+\cdots$

We begin with the quadratic equation in x containing the parameter ε,

$\epsilon x^2+x-1=0,\epsilon \ll 1$

If $\epsilon =0$, then $x-1=0$, $x=1$.

There are two roots if $\epsilon \ne 0$. This equation has exact solutions using the quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm \sqrt{1+4\epsilon }}{2\epsilon }=-\frac{1}{2\epsilon }\pm \frac{1}{2\epsilon }\sqrt{1+4\epsilon }$

These solutions can be expanded using the binomial expansion for small ε

$\begin{array}{c}{\left(1+x\right)}^n\approx 1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3\\ +\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{x}^4+\cdots \end{array}$

$\begin{array}{c}{\left(1+4\epsilon \right)}^{\frac{1}{2}}=1+\frac{1}{2}\left(4\epsilon \right)+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}{\left(4\epsilon \right)}^2+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}{\left(4\epsilon \right)}^3\\ +\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\left(\frac{1}{2}-3\right)}{4!}{\left(4\epsilon \right)}^4+\cdots \end{array}$

${\left(1+4\epsilon \right)}^{\frac{1}{2}}=1+2\epsilon -2{\epsilon }^2+4{\epsilon }^3-10{\epsilon }^4+\cdots$

Therefore,

$\begin{array}{c}x=\frac{-1\pm \sqrt{1+4\epsilon }}{2\epsilon }=-\frac{1}{2\epsilon }\pm \frac{1}{2\epsilon }\sqrt{1+4\epsilon }\\ =-\frac{1}{2\epsilon }\pm \frac{1}{2\epsilon }\left[1+2\epsilon -2{\epsilon }^2+4{\epsilon }^3-10{\epsilon }^4+\cdots \right]\end{array}$

Thus, the exact solution

$x=\{\begin{array}{l}+1-2\epsilon +2{\epsilon }^2-5{\epsilon }^3+\cdots \\ -\frac{1}{\epsilon }-1+\epsilon -2{\epsilon }^2+5{\epsilon }^3+\cdots \end{array}$

Realize the singular second root evaporates off to $x=\infty$ in the limit $\epsilon =0$.

$\underset{\epsilon \to 0}{\lim }-\frac{1}{\epsilon }-1+\epsilon -2{\epsilon }^2+5{\epsilon }^3+\cdots =\infty$

Now, suppose we did not know the exact solution, how could we generate the approximate solutions for $\epsilon \ll 1$?

when an exact solution is impossible. The next two examples demonstrate this method.

$\epsilon x^2+x-1=0$

If $\epsilon =0$ then $x-1=0$, $x=1$.

But for $\epsilon \ne 0$ even $\epsilon \ll 1$, we have two roots. There is a big qualitative change. Typical of singular perturbation. The other root $\to \infty$ as $\epsilon \to 0$

Dominant balance: Maybe the term $\epsilon x^2$ is big for this missing root.

We now find the first root by letting $x=a+b\epsilon +c{\epsilon }^2+d{\epsilon }^4+\cdots$

Substituting into the perturbation problem, we obtain

$\epsilon {\left(a+b\epsilon +c{\epsilon }^2+d{\epsilon }^4+\cdots \right)}^2+\left(a+b\epsilon +c{\epsilon }^2+d{\epsilon }^4+\cdots \right)-1=0$

$\begin{array}{l}\epsilon \left(a^2+2ab\epsilon +2ac{\epsilon }^2+2ad{\epsilon }^3+b^2{\epsilon }^2+2bc{\epsilon }^3+\cdots \right)\\ +\left(a+b\epsilon +c{\epsilon }^2+d{\epsilon }^3+e{\epsilon }^4+\cdots \right)-1=0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

${\epsilon }^0:a-1=0,a=1$

${\epsilon }^1:a^2+b=0,b=-1$

${\epsilon }^2:2ab+c=0,c=2$

${\epsilon }^3:2ac+b^2+d=0,d=-5$

Thus, the first root is $x=1-\epsilon +2{\epsilon }^2-5{\epsilon }^3+\cdots$.

Rescaling is a technique to simplify the analysis of a perturbed problem by introducing new variables or parameters that absorb the small perturbation parameter ε. Rescaling involves multiplying or dividing the original variables or parameters by appropriate powers of ε, effectively “rescaling” them to reduce the complexity of the equations or expressions. We find the second root of the equation (1.3) by

rescaling, we let substituting $x=\frac{X}{{\epsilon }^n}$ into the perturbation problem, and we obtain

$\epsilon x^2+x-1=0$

$\epsilon {\left(\frac{X}{{\epsilon }^n}\right)}^2+\frac{X}{{\epsilon }^n}-1=0$

${\epsilon }^{1-2n}{X}^2+{\epsilon }^{-n}X-1=0$

Extracting the exponents of the first two terms, we have

$1-2n=-n$

$1=n$

Therefore this ${\epsilon }^{1-2n}{X}^2+{\epsilon }^{-n}X-1=0$ becomes,

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

Multiply by ε, we have

$X^2+X-\epsilon =0$

$\epsilon =0$

$X^2+X=0,X\left(X+1\right)=0,X=0,X=-1$

Let $X=-1+a\epsilon +b{\epsilon }^2+\cdots$.

Substituting into $X^2+X-\epsilon =0$ , we obtain,

${\left(-1+a\epsilon +b{\epsilon }^2+c{\epsilon }^3+d{\epsilon }^4+\cdots \right)}^2+\left(-1+a\epsilon +b{\epsilon }^2+c{\epsilon }^3+d{\epsilon }^4+\cdots \right)-\epsilon =0$

$\begin{array}{l}\left(1-2a\epsilon -2b{\epsilon }^2+a^2{\epsilon }^2+2ab{\epsilon }^3-2c{\epsilon }^3+2ac{\epsilon }^4+b^2{\epsilon }^4-2d{\epsilon }^4\cdots \right)\\ +\left(-1+a\epsilon +b{\epsilon }^2+c{\epsilon }^3+d{\epsilon }^4+\cdots \right)-\epsilon =0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

${\epsilon }^0:1-1=0$

${\epsilon }^1:-2a+a-1=0,a=-1$

${\epsilon }^2:-2b+b+a^2=0,b=1$

${\epsilon }^3:2ab+c-2c=0,c=-2$

${\epsilon }^4:2ac+b^2-2d+d=0,d=5$

Therefore, the first root is $X=-1-\epsilon +{\epsilon }^2-2{\epsilon }^3+5{\epsilon }^4+\cdots$.

But $x=\frac{X}{{\epsilon }^n}$, $n=1$.

Therefore $x=\frac{X}{{\epsilon }^n}=\frac{-1-\epsilon +{\epsilon }^2-2{\epsilon }^3+5{\epsilon }^4+\cdots }{\epsilon }=-\frac{1}{\epsilon }-1+\epsilon -2{\epsilon }^2+5{\epsilon }^3+\cdots$.

Thus, the solution from the expansion method equals the exact solutions.

$x=\{\begin{array}{l}+1-2\epsilon +2{\epsilon }^2-5{\epsilon }^3+\cdots \\ -\frac{1}{\epsilon }-1+\epsilon -2{\epsilon }^2+5{\epsilon }^3+\cdots \end{array}$

The same as the exact solution.

Solving the perturbation problem, we find the first root by letting $x=a+b\epsilon +c{\epsilon }^2+\cdots$.

Substituting into the perturbation problem, we obtain

$\epsilon {\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)}^3+{\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)}^2+\left(2+\epsilon \right)\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)+1=0$

$\begin{array}{l}\epsilon \left(a^3+3a^{2}b\epsilon +3a^{2}c{\epsilon }^2+3a{b}^2{\epsilon }^2+6abc{\epsilon }^3+b^3{\epsilon }^3+\cdots \right)\\ +\left(a^2+2ab\epsilon +2ac{\epsilon }^2+b^2{\epsilon }^2+2bc{\epsilon }^3+\cdots \right)+\left(2+\epsilon \right)\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)+1=0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

${\epsilon }^0:a^2+2a+1=0,a=-1,-1$

${\epsilon }^1:a^3+a+2b+2ab=0,b=\frac{-a\left(a^2+1\right)}{2\left(1+a\right)}=\infty$

Thus, we now let $x=-1+\delta +{\delta }^2+\cdots$, where $\delta =\delta \left(\epsilon \right)$.

$\epsilon {\left(-1+\delta +{\delta }^2+\cdots \right)}^3+{\left(-1+\delta +{\delta }^2+\cdots \right)}^2+\left(2+\epsilon \right)\left(-1+\delta +{\delta }^2+\cdots \right)+1=0$

$\left(-1+3\delta -5{\delta }^2+\cdots \right)+{\left(-1-2\delta -{\delta }^2+\cdots \right)}^2+\left(2+\epsilon \right)\left(-1+\delta +{\delta }^2+\cdots \right)+1=0$

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

Extracting the dominant terms, we have

$-2\epsilon +{\delta }^2=0,\delta =\pm \sqrt{2}{\epsilon }^{\frac{1}{2}}$

Therefore, we let $x=-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}+2c\epsilon +\cdots$ and $x=-1-\sqrt{2}{\epsilon }^{\frac{1}{2}}-2c\epsilon +\cdots$.

Substituting $x=-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}+2c\epsilon +\cdots$ into the equation

$\begin{array}{l}\epsilon {\left(-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}+2c\epsilon +\cdots \right)}^3+{\left(-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}+2c\epsilon +\cdots \right)}^2\\ +\left(2+\epsilon \right)\left(-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}+2c\epsilon +\cdots \right)+1=0\end{array}$

$\begin{array}{l}\epsilon \left(-1+3\sqrt{2}{\epsilon }^{\frac{1}{2}}-6\epsilon +2\sqrt{2}{\epsilon }^{\frac{3}{2}}+6c\epsilon +\cdots \right)+\left(1-2\sqrt{2}{\epsilon }^{\frac{1}{2}}+2\epsilon -4c\epsilon +4c\sqrt{2}{\epsilon }^{\frac{3}{2}}+\cdots \right)\\ +\left(2+\epsilon \right)\left(-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}+2c\epsilon +\cdots \right)+1=0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

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

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

${\epsilon }^{\frac{3}{2}}:3\sqrt{2}+4c\sqrt{2}+\sqrt{2}=0,c=-1$

Thus, one root is $x=-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}-2\epsilon +\cdots$.

Also substituting $x=-1-\sqrt{2}{\epsilon }^{\frac{1}{2}}-2c\epsilon +\cdots$ into the equation

$\begin{array}{l}\epsilon {\left(-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}-2c\epsilon +\cdots \right)}^3+{\left(-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}-2c\epsilon +\cdots \right)}^2\\ +\left(2+\epsilon \right)\left(-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}-2c\epsilon +\cdots \right)+1=0\end{array}$

$\begin{array}{l}\epsilon \left(-1-3\sqrt{2}{\epsilon }^{\frac{1}{2}}-6\epsilon -2\sqrt{2}{\epsilon }^{\frac{3}{2}}-6c\epsilon +12c{\epsilon }^{\frac{3}{2}}+\cdots \right)\\ +\left(1+2\sqrt{2}{\epsilon }^{\frac{1}{2}}+2\epsilon +4c\epsilon +4c\sqrt{2}{\epsilon }^{\frac{3}{2}}+\cdots \right)\\ +\left(2+\epsilon \right)\left(-1-\sqrt{2}{\epsilon }^{\frac{1}{2}}-2c\epsilon +\cdots \right)+1=0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

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

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

${\epsilon }^{\frac{3}{2}}:-3\sqrt{2}+4c\sqrt{2}-\sqrt{2}=0,c=1$

Thus, the second root is $x=-1-\sqrt{2}{\epsilon }^{\frac{1}{2}}-2\epsilon +\cdots$.

We find the third root by using the method of rescaling.

We let $x=\frac{X}{{\epsilon }^n}$.

Substituting into the perturbation problem, we obtain

$\epsilon {\left(\frac{X}{{\epsilon }^n}\right)}^3+{\left(\frac{X}{{\epsilon }^n}\right)}^2+\left(2+\epsilon \right)\left(\frac{X}{{\epsilon }^n}\right)+1=0=0$

${\epsilon }^{1-3n}{X}^3+{\epsilon }^{-2n}{X}^2+\left(2+\epsilon \right){\epsilon }^{-n}X+1=0$

Extracting the exponents of the first two terms, we have

$1-3n=-2n,n=1$

Therefore this ${\epsilon }^{1-3n}{X}^2+{\epsilon }^{-2n}{X}^2+\left(2+\epsilon \right){\epsilon }^{-n}X+1=0$ becomes.

${\epsilon }^{-2}{X}^3+{\epsilon }^{-2}{X}^2+\left(2+\epsilon \right){\epsilon }^{-1}X+1=0$

Multiply by ${\epsilon }^2$, we have

$X^3+X^2+\left(2+\epsilon \right)\epsilon X+{\epsilon }^2=0$

Setting $\epsilon =0$, we have

$X^3+X^2=0,X^2\left(X+1\right)=0,X=0,X=-1$

Let $X=-1+a\epsilon +b{\epsilon }^2+\cdots$.

Substituting into $X^3+X^2+\left(2+\epsilon \right)\epsilon X+{\epsilon }^2=0$ we obtain

${\left(-1+a\epsilon +b{\epsilon }^2+\cdots \right)}^3+\left(-1+a\epsilon +b{\epsilon }^2+\cdots \right)+{\epsilon }^2=0$

$\begin{array}{l}\left(-1+3a\epsilon +3b-3a^2{\epsilon }^2+3b^3{\epsilon }^3-6ab{\epsilon }^3+\cdots \right)\\ +\left(1-2a\epsilon -2b{\epsilon }^2+a^2{\epsilon }^2+2ab{\epsilon }^3+\cdots \right)+{\epsilon }^2=0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

${\epsilon }^0:-1+1=0$

${\epsilon }^1:3a-2a-2=0,a=2$

${\epsilon }^2:-2b+b+a^2=0,b=1$

${\epsilon }^3:3b-3a^2-2b+a^2-1+1=0,b=4$

Therefore, the first root is $X=-1+2\epsilon +4{\epsilon }^2+\cdots$.

But $x=\frac{X}{{\epsilon }^n}$, $n=1$.

Therefore $x=\frac{X}{{\epsilon }^n}=\frac{-1+2\epsilon +4{\epsilon }^2+\cdots }{\epsilon }=-\frac{1}{\epsilon }+2+4\epsilon +\cdots$.

Thus, the three roots are $x=\{\begin{array}{l}-1+\sqrt{2}{\epsilon }^{\frac{1}{2}}-2\epsilon +\cdots \\ -1-\sqrt{2}{\epsilon }^{\frac{1}{2}}-2\epsilon +\cdots \\ -\frac{1}{\epsilon }+2-2\epsilon +\cdots \end{array}$

$\epsilon x^3+x^2+\left(2-\epsilon \right)x+1=0$

$\epsilon x^3-\epsilon x+x^2+2x+1=0$

$\epsilon x\left(x^2-1\right)+x^2+2x+1=0$

$\epsilon x\left(x^2-1\right)+{\left(x+1\right)}^2=0$

$\epsilon x\left(x-1\right)\left(x+1\right)+{\left(x+1\right)}^2=0$

$\left(x+1\right)\left\{x\epsilon \left(x-1\right)+\left(x+1\right)\right\}=0$

$x+1=0,x=-1$

$x\epsilon \left(x-1\right)+\left(x+1\right)=0$

$\epsilon x^2-x\epsilon +x+1=0$

$\epsilon x^2+\left(1-\epsilon \right)x+1=0$

Thus, the first root is $x=-1$.

Solving for the second and third roots by using the quadratic formula, we obtain

$\begin{array}{c}x=\frac{\left(\epsilon -1\right)\pm \sqrt{{\left(1-\epsilon \right)}^2-4\epsilon }}{2\epsilon }=\frac{\left(\epsilon -1\right)\pm \left(1-\epsilon \right){\left[1-\frac{4\epsilon }{{\left(1-\epsilon \right)}^2}\right]}^{\frac{1}{2}}}{2\epsilon }\\ =\frac{\left(\epsilon -1\right)\pm \left(1-\epsilon \right){\left[1-4\epsilon {\left(1-\epsilon \right)}^{-2}\right]}^{\frac{1}{2}}}{2\epsilon }\end{array}$

By using the Binomial Expansion

$\begin{array}{c}{\left(1+x\right)}^n\approx 1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3\\ +\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{x}^4+\cdots \end{array}$

${\left(1-\epsilon \right)}^{-2}=1+2\epsilon +3{\epsilon }^2+4{\epsilon }^3+\cdots$

Therefore,

$x=\frac{\left(\epsilon -1\right)\pm \left(1-\epsilon \right){\left[1-4\epsilon \left(1+2\epsilon +3{\epsilon }^2+4{\epsilon }^3+\cdots \right)\right]}^{\frac{1}{2}}}{2\epsilon }$

$x=\frac{\left(\epsilon -1\right)\pm \left(1-\epsilon \right){\left[1-4\epsilon -8{\epsilon }^2\right]}^{\frac{1}{2}}}{2\epsilon }$

Using the binomial expansion once again, we obtain

${\left(1-4\epsilon -8{\epsilon }^2\right)}^{\frac{1}{2}}=1-2\epsilon -6{\epsilon }^2+\cdots$

$x=\frac{\left(\epsilon -1\right)\pm \left(1-\epsilon \right)\left(1-2\epsilon -6{\epsilon }^2\right)}{2\epsilon }$

$x=\frac{\epsilon -1+1-3\epsilon -4{\epsilon }^2+\cdots }{2\epsilon }=-1-2\epsilon +\cdots$

$x=\frac{\epsilon -1-1+3\epsilon +4{\epsilon }^2+\cdots }{2\epsilon }=-\frac{1}{\epsilon }+2+2\epsilon +\cdots$

To solve the perturbed problem, we find the first root by letting $x=a+b\epsilon +c{\epsilon }^2+\cdots$.

Substituting into the perturbation problem, we obtain

$\epsilon {\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)}^3+{\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)}^2+\left(2-\epsilon \right)\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)+1=0$

$\begin{array}{l}\epsilon \left(a^3+3a^{2}b\epsilon +3a^{2}c{\epsilon }^2+3a{b}^2{\epsilon }^2+6abc{\epsilon }^3+b^3{\epsilon }^3+\cdots \right)\\ +\left(a^2+2ab\epsilon +2ac{\epsilon }^2+b^2{\epsilon }^2+2bc{\epsilon }^3+\cdots \right)+\left(2-\epsilon \right)\left(a+b\epsilon +c{\epsilon }^2+\cdots \right)+1=0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

${\epsilon }^0:a^2+2a+1=0,a=-1,-1$

${\epsilon }^1:a^3-a+2b+2ab=0,b=\frac{a\left(1-a^2\right)}{2\left(1+a\right)}=\infty$

Thus, we now let $x=-1+\delta +{\delta }^2+\cdots$, where $\delta =\delta \left(\epsilon \right)$

$\epsilon {\left(-1+\delta +{\delta }^2+\cdots \right)}^3+{\left(-1+\delta +{\delta }^2+\cdots \right)}^2+\left(2-\epsilon \right)\left(-1+\delta +{\delta }^2+\cdots \right)+1=0$

$\left(-1+3\delta -5{\delta }^2+\cdots \right)+\left(1-2\delta -{\delta }^2+\cdots \right)+\left(2-\epsilon \right)\left(-1+\delta +{\delta }^2+\cdots \right)+1=0$

$2\epsilon \delta +{\delta }^2-6\epsilon {\delta }^2=0$

Extracting the dominant terms, we have

$2\epsilon \delta +{\delta }^2=0,\delta =-2\epsilon$

Therefore, we let $x=-1-2\epsilon +4b{\epsilon }^2+\cdots$.

Substituting $x=-1-2\epsilon +4b{\epsilon }^2+\cdots$ into the equation

$\begin{array}{l}\epsilon {\left(-1-2\epsilon +4b{\epsilon }^2+\cdots \right)}^3+{\left(-1-2\epsilon +4b{\epsilon }^2+\cdots \right)}^2\\ +\left(2-\epsilon \right)\left(-1-2\epsilon +4b{\epsilon }^2+\cdots \right)+1=0\end{array}$

$\begin{array}{l}\epsilon \left(-1-6\epsilon -12{\epsilon }^2+12b{\epsilon }^2-8{\epsilon }^3+\cdots \right)+\left(1+4\epsilon +4{\epsilon }^2-8b{\epsilon }^2-16{\epsilon }^3+\cdots \right)\\ +\left(2+\epsilon \right)\left(-1-2\epsilon +4b{\epsilon }^2+\cdots \right)+1=0\end{array}$

Equating Coefficients of equal powers of ε, we obtain

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

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

${\epsilon }^2:-6+4-8b+8b+2=0$

${\epsilon }^3:-12+12b-16b-4b=0,b=-\frac{3}{2}$

Thus, one root is $x=-1-2\epsilon -6{\epsilon }^2+\cdots$.

Finding the second root, we let $x=-1+a{\epsilon }^{\frac{1}{2}}+b\epsilon +c{\epsilon }^{\frac{3}{2}}+\cdots$ and substituting into the perturbed problem.

$\begin{array}{l}\epsilon {\left(-1+a{\epsilon }^{\frac{1}{2}}+b\epsilon +c{\epsilon }^{\frac{3}{2}}+\cdots \right)}^3+{\left(-1+a{\epsilon }^{\frac{1}{2}}+b\epsilon +c{\epsilon }^{\frac{3}{2}}+\cdots \right)}^2\\ +\left(2-\epsilon \right)\left(-1+a{\epsilon }^{\frac{1}{2}}+b\epsilon +c{\epsilon }^{\frac{3}{2}}+\cdots \right)+1=0\end{array}$