Asymptotic expansion of a cubic equation

Question

I am asked to find the first two terms of the asymptotic expansion of the cubic equation $$\varepsilon^4 x^3 - 6 \varepsilon^3 x^2 + (4-3\varepsilon^2) x - 12 \varepsilon - 2 \varepsilon^2 = 0$$

as $\varepsilon\rightarrow 0$. I am given the hint that one root is $O(\varepsilon)$ and two are $O(\varepsilon^{-2})$. I am not sure how to adjust the equation (by dividing by $\varepsilon$ or powers of $\varepsilon$, etc.) and proceed. I know to use $x \sim x_0 + \varepsilon x_1$ as I'm asked to find the first two terms of the asymptotic expansion. But I have not seen equations before with this complexity and powers of $\varepsilon$ before the highest power of $x$. I usually would discard the $O(\varepsilon^2)$ and above terms and compare coefficients of $O(1)$ and $O(\varepsilon)$, etc., but I am really struggling with this particular example.

Answer 1

The question asks about the roots of the cubic equation $$ \varepsilon^4 x^3 - 6 \varepsilon^3 x^2 + (4-3\varepsilon^2) x - 12 \varepsilon - 2 \varepsilon^2 = 0 \tag{1} $$ as $\varepsilon\to 0.\,$ One step is to make the cubic monic by dividing by its coefficient. Thus, $$ x^3 \!-\! 6 \varepsilon^{-1} x^2 \!+\! (4\varepsilon^{-4}\!-\!3\varepsilon^{-2}) x \!-\! (12 \varepsilon^{-3} \!+\! 2 \varepsilon^{-2}) \!=\! 0. \tag{2} $$ Another simplifying step is to define a variant variable $$ \delta := \varepsilon/2. \tag{3} $$ Using this variable in the cubic makes it $$ x^3 \!-\!\frac3{\delta}x^2 \!+\! \left( \frac1{4\delta^4} -\frac3{4\delta^2}\right)x \!-\! \left(\frac1{2\delta^2} + \frac3{2\delta^3}\right) \!=\! 0. \tag{4} $$

Use a small positive value for $\,\delta\,$ such as $\,.000001\,$ and solve this cubic. There is one small positive root and two big conjugate imaginary roots. They are $$ r_1 \approx .000006000002,\;\; r_2 \approx 1.5\cdot 10^6 - 0.5\cdot 10^{12}i,\;\; r_3 = {r_2}^*. $$ Thus, the leading terms appear to be $$ r_1 \approx 6\delta + 2\delta^2,\;\; r_2 \approx \frac{i}{2\delta^2} + \frac3{2\delta},\;\; r_3 \approx \frac{-i}{2\delta^2} + \frac3{2\delta}. \tag{5} $$ Check, using Veita's formulas, that the elementary symmetric functions of the three roots in $(5)$ match the coefficients of the cubic in equation $(4)$.

Answer 2

We can solve excatly the cubic equation.

As @Somos already wrote, there is only one real root since $$\Delta=-4\epsilon^4\left(64-288 \epsilon ^2-972 \epsilon ^4-216 \epsilon ^5+4428 \epsilon ^6+918 \epsilon^7+27 \epsilon ^8 \right)$$ Using the hyperbolic method, the real root is given by $$r_1=\frac 2{3\epsilon}\Bigg[3+\frac{\sqrt{12-45 \epsilon ^2}}{\epsilon } \sinh \left(\frac{1}{3} \sinh ^{-1}\left(3 \sqrt{3}\,\frac{ \epsilon \left(\epsilon ^3+17 \epsilon ^2-4\right)}{\left(4-15 \epsilon ^2\right)^{3/2}}\right)\right)\Bigg]$$ Expanded as a series $$r_1=3 \epsilon +\frac{1}{2}\epsilon ^2+\frac{9}{4}\epsilon ^3+O\left(\epsilon ^4\right)$$ Now, using deflation $$r_{2,3}=\Big[ \frac{3}{\epsilon }-\frac{3 \epsilon }{2}-\frac{\epsilon ^2}{4}-\frac{9 \epsilon ^3}{8}+O\left(\epsilon ^4\right)\Big] \pm i\,\Big[\frac{2}{\epsilon ^2}-3-\frac{9 \epsilon ^2}{2}-\frac{3 \epsilon ^3}{8}+O\left(\epsilon ^4\right) \Big]$$