1
$\begingroup$

Question: Develop perturbation solutions to $$x^3 + (3+4\epsilon + \epsilon^2)x^2 + (3 + 9\epsilon + 7\epsilon^2 + 2\epsilon^3)x + 1 + 5\epsilon + 8\epsilon^2 + 5\epsilon^3 + \epsilon^4 = 0$$ finding the three terms in the approximation for each root, $x = x_0 + \epsilon^{\alpha}x_{\alpha} + \epsilon^{2\alpha}x_{2\alpha}$, and determining $\alpha$ along the way.

My approach: First, we make some preliminary observations. When $\epsilon = 0$, the equation in the question reduces to $(x+1)^3 = 0$. Therefore, $x = -1$ is the triple unperturbed root. Further, we can infer that the roots of the original equation would have the following form: $$x(\epsilon) = -1 + x_1\delta_1(\epsilon) + x_2\delta(\epsilon) + \cdots.$$

Comparing the above expression of $x(\epsilon)$ with the expression of each root given in the question, we infer that each root would be of the following form: $$x(\epsilon) = -1 + \epsilon^{\alpha}x_{\alpha} + \epsilon^{2\alpha}x_{2\alpha} + \cdots.$$

Since, in general, $\epsilon << 1$, neglecting the infinitesimal terms on the LHS of the equation, the equation can be approximated as follows: $$x^3 + (3+4\epsilon)x^2 + (3+9\epsilon)x + 1 + 5\epsilon \approx 0.$$

To find $\alpha$ and $x_{\alpha}$, we substitute $x=-1+\epsilon^{\alpha}x_{\alpha}$ in the above approximated equation, and obtain $$\epsilon^{3\alpha}x_{\alpha}^3 + 4\epsilon^{2\alpha+1}x_{\alpha}^2 + \epsilon x_{\alpha}\approx0.$$

We assume $x = \mathrm{ord}(1)$ and consider the following three cases: (1) $\alpha < 1$, (2) $\alpha = 1$, and (3) $\alpha > 1$.

  1. Since, $\alpha < 1$, implies that $3\alpha < 2\alpha +1$, which in turn implies that $\epsilon^{3\alpha} > \epsilon^{2\alpha+1}$. Also, $\epsilon > \epsilon^{2\alpha+1}$. Therefore, in this case, the dominant terms are $\epsilon^{3\alpha}x_{\alpha}^3$ and $\epsilon x_{\alpha}$. These would be of the same order when $\alpha = \frac{1}{3}$.

  2. When $\alpha = 1$, the approximated equation reduces to $\epsilon^3(x_1^3 + 4x_1^2) + \epsilon x_1 \approx 0$, which looks good.

  3. When $\alpha >1$, the dominant terms are $4\epsilon^{2\alpha+1}x_{\alpha}^2$ and $\epsilon x_{\alpha}$. These would be of the same order when $\alpha = 0$, which is a contradiction to our assumption that $\alpha > 1$.

Therefore, the perturbed roots are of the form: $$x = -1 + x_1\epsilon + x_2\epsilon^2+\cdots,\text{ and }$$ $$x = -1 + x_{\frac{1}{3}}\epsilon^{\frac{1}{3}} + x_{\frac{2}{3}}\epsilon^{\frac{2}{3}}+ \cdots.$$

Next, we will substitute the above-determined expressions of the roots into the original equation and find out the value of the coefficients $x_1, x_2, x_{\frac{1}{3}}$ and $x_{\frac{2}{3}}$, thus obtain all the three roots.

Can someone please check if my logic and reasoning in this entire approach is correct, and suggest if there's an alternative method to attack this problem? Thanks a lot in advance.

$\endgroup$
1
  • $\begingroup$ You could simplify the problem since there is a possible factorization. $\endgroup$ Commented Sep 19, 2023 at 4:46

0

You must log in to answer this question.

Browse other questions tagged .