# Asymptotic expansions for the roots of $\epsilon^2x^4-\epsilon x^3-2x^2+2=0$

I’m trying to compute the asymptotic expansion for each of the four roots to the following equation, as $\epsilon \rightarrow 0$:

$\epsilon^2x^4-\epsilon x^3-2x^2+2=0$

I’d like my expansions to go up through terms of size $O(\epsilon^2)$.

I´ve made the change of variables $x=\delta y$, performed dominant balance and found out that the only two valid options are: (rescaled eq: $\epsilon^2\delta^4y^4-\epsilon \delta^3 y^3-2\delta^2 y^2+2=0$)

• $\delta^2 \sim 2 \Rightarrow \delta=O(1)$
• $\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta=O(\epsilon^{-1})$

How do I proceed?

#### Solutions Collecting From Web of "Asymptotic expansions for the roots of $\epsilon^2x^4-\epsilon x^3-2x^2+2=0$"

Your analysis is correct.

Let’s look at the original equation
$${\epsilon ^2}{x^4} – \epsilon {x^3} – 2{x^2} + 2 = 0$$
and plot the Newton-Kruskal diagram:

There are only two possible placements of straight lines passing through two or more points with all remaining points “above the line”; those two lines correspond exactly to

• $\delta^2 \sim 2 \Rightarrow \delta \sim 1$, and
• $\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta \sim \epsilon^{-1}.$

Before we continue, let’s rescale the original equation by making the change of variables $x=\delta y$ to arrive at

$${\epsilon ^2}{\delta ^4}{y^4} – \epsilon {\delta ^3}{y^3} – 2{\delta ^2}{y^2} + 2 = 0.$$

From the first balance, $\delta^2 \sim 2 \Rightarrow \delta \sim 1$, we immediately recover the original equation:
$$\tag{1} {\epsilon ^2}{y^4} – \epsilon {y^3} – 2{y^2} + 2 = 0.$$

Letting $\epsilon \rightarrow 0$, we get
$$2{y^2} + 2 = 0 \Rightarrow y = \pm 1$$

Let´s check if there are roots close to $y=-1$ and $y=1$.
Plugging an asymptotical expansion of the form $y \sim \pm 1 + {a_1}\epsilon + {a_2}{\epsilon ^2} + {a_3}{\epsilon ^3} + {a_4}{\epsilon ^4} + \ldots$ into (1), and then matching coefficients, we arrive at

$${x_a} \sim {y_a} \sim – 1 – \frac{\epsilon }{4} – \frac{{13}}{{32}}{\epsilon ^2} + O\left( {{\epsilon ^3}} \right),$$

$${x_b} \sim {y_b} \sim 1 – \frac{\epsilon }{4} + \frac{{13}}{{32}}{\epsilon ^2} + O\left( {{\epsilon ^3}} \right).$$

Regarding the second possibility of balancing, $\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta \sim \epsilon^{-1}$, we get the following equation

$$\tag{2}{y^4} – {y^3} – 2{y^2} + 2{\epsilon ^2} = 0$$

Letting $\epsilon \rightarrow 0$, we get
$${y^4} – {y^3} – 2{y^2} = 0 \Rightarrow {y^2}\left( {{y^2} – y – 2} \right) = 0 \Rightarrow y = 0 \vee y = – 1 \vee y = 2.$$

Let´s check if there are roots close to $y=-1$ and $y=2$ ($y=0$ is of no interest, as we just recover one of the roots previously found). Plugging an asymptotical expansion of the form $y \sim {a_0} + {a_1}\epsilon + {a_2}{\epsilon ^2} + {a_3}{\epsilon ^3} + {a_4}{\epsilon ^4} + \ldots$ (with ${a_0} = – 1 \vee {a_0} = 2$) into (2), and then matching coefficients, we arrive at

$${x_c} \sim {\epsilon ^{ – 1}}{y_c} \sim – {\epsilon ^{ – 1}} + \frac{2}{3}\epsilon + O\left( {{\epsilon ^3}} \right),$$

$${x_d} \sim {\epsilon ^{ – 1}}{y_d} \sim 2{\epsilon ^{ – 1}} – \frac{1}{6}\epsilon + O\left( {{\epsilon ^3}} \right).$$

Thus, we’ve determined all the four roots of the original equation.

-EDIT-

Here’s the Mathematica code I used to do my calculations, more specifically, the last one.

   n = 4;
a[0] = 2;
x = a[0] + Sum[a[i] \[Epsilon]^i, {i, 1, n}] + O[\[Epsilon]]^(n + 1);
x^4 - x^3 - 2 x^2 + 2 \[Epsilon]^2 == 0;

LogicalExpand[%]

Solve[%]


Two of the roots will be near the roots $1$ and $-1$ of $-2x^2 + 2$, the other two will be near $\infty$. Solving numerically for $\epsilon = .01$,
those roots are approximately $-100$ and $200$.
So it looks like you want $x \sim c \epsilon^{-1}$. Indeed with $x = y/\epsilon$ the equation becomes $y^4 – y^3 – 2 y^2 + 2 \epsilon^2 = 0$.

EDIT: At $\epsilon = 0$ the second equation becomes $y^4 – y^3 – 2 y^2 = 0$, which has roots $y = -1$ and $y = 2$ as well as $y=0$ (which corresponds to $x=1$ and $x=-1$). For the solution near $y=-1$, for example, take $y = -1 + v$ and expand to get $v^4 – 5 v^3 + 7 v^2 – 3 v + 2 \epsilon^2 = 0$ where $v$ should be small. To first order, $v = (2/3) \epsilon^2$. Take $v = (2/3) \epsilon^2 + w$
and the equation becomes
$${w}^{4}+ \left( \dfrac{8 \epsilon^2}{3}-5 \right) {w}^{3}+ \left(\dfrac{8 { \epsilon}^{4}}{3}-10\,{\epsilon}^{2}+7 \right) {w}^{2}+ \left( {\frac {32 \,{\epsilon}^{6}}{27}}-{\frac {20\,{\epsilon}^{4}}{3}}+{\frac {28\,{ \epsilon}^{2}}{3}}-3 \right) w+{\frac {16\,{\epsilon}^{8}}{81}}-{ \frac {40\,{\epsilon}^{6}}{27}}+{\frac {28\,{\epsilon}^{4}}{9}} = 0$$
from which $w = (28/27) \epsilon^4 + \ldots$. Thus this solution is
$$x = -\epsilon^{-1} + \dfrac{2}{3} \epsilon + \dfrac{28}{27} \epsilon^3 + \ldots$$

I would make the substitution $y=1/x$ to get $\epsilon \,(y-\epsilon ) -2 \, y^2(y^2-1)=0$ .

Then write $y=y_0 + a \epsilon + b \epsilon^2 +O(\epsilon ^3)$ and replace for each of the limit roots ($y_0=1$, $y_0=-1$, $y_0=0$) to get the respective coefficients. For example, for the first root I get $a=1/4$