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In the limit as $\epsilon \to \infty$, obtain a three-term asymptotic solution to the roots of the following equation.

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

What I've done so far

I've assumed $x=O(\epsilon^r)$ as $\epsilon \to \infty$ therefore,

$\epsilon x^3=O(\epsilon^{1+3r})$

$x^2=O(\epsilon^{2r})$

$x=O(\epsilon^{r})$

$1=O(\epsilon^{0})$

I'm not too sure what needs to be done from here. any help would be much appreciated.

smith
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  • For the leading terms to be of the same order in $\varepsilon$, you need $1+3r = 2r$ or $r = -1$. , Therefore you should now try $x \sim \varepsilon^{-1}x_1 + \varepsilon^{-2}x_2 + \varepsilon^{-2}x_2 ...$, substitute and compare terms of the same order. – Hans Engler Oct 07 '15 at 14:04
  • when I'm comparing terms of the same order, whats the highest order I go to? (if that makes sense) sorry this stuff confuses me :S – smith Oct 07 '15 at 14:14
  • @HansEngler aren't the two leading terms $\epsilon x^3 $ and $1$ (not $\epsilon x^3 $ and $x^2$), since we must have $x\to 0$? – Clement C. Oct 07 '15 at 14:56
  • You're right, I was too fast :) – Hans Engler Oct 07 '15 at 14:58
  • Set $\eta=1/\epsilon$ and you have a simple regular perturbation problem $x^3+\eta x^2-2\eta x+\eta=0$ for small $\eta$. – David Oct 07 '15 at 23:21

2 Answers2

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Outline (rather messy: is there a more elegant approach?):

  • First, prove that you must have $x=x(\epsilon)\xrightarrow[\epsilon\to\infty]{} 0$.

  • Then, you can get the expansion one term at a time...

Now, the second bullet (up to order two) below -- if I have not screwed up. $$ 1-2x + x^2+\epsilon x^3 = 1+\epsilon x^3 + o(1) $$ so you need $\epsilon x^3 \operatorname*{\sim}_{\epsilon\to\infty} -1$. I.e., $x = -\frac{1}{\epsilon^{1/3}} + o\left(\frac{1}{\epsilon^{1/3}}\right)$.

Write $x = -\frac{1}{\epsilon^{1/3}} + h$, so that $x^3 = -\frac{1}{\epsilon} + \frac{3h}{\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right)$. Plugging it back, $$ 1-2x + x^2+\epsilon x^3 = 1 -1 + 3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}} -2h + x^2 = 3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}} + o(h\epsilon^{1/3}) $$ To get that equal to $0$, this implies in particular $3h\epsilon^{1/3} +2\frac{1}{\epsilon^{1/3}}=o(1)$, so that $$ h = \frac{-2}{3\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right) $$ i.e. $$ x = -\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + o\left(\frac{1}{\epsilon^{2/3}}\right) $$ Now, for the final step, write $$ x = -\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g $$ with $g=o\left(\frac{1}{\epsilon^{2/3}}\right)$.

$$ 1-2x + x^2+\epsilon x^3 = 1 -2(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g)+\left(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g\right)^2+\epsilon\left(-\frac{1}{\epsilon^{1/3}} - \frac{2}{3\epsilon^{2/3}} + g\right)^3 $$ and expand (possibly with Taylor expansions to the right order if this helps). To satisfy the equation, this must be $o(1)$: with this constraint, you will get $g$.

Clement C.
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  • when you say "First, prove that you must have $x=x(\epsilon)\xrightarrow[\epsilon\to\infty]{} 0 $" how would I prove this? – smith Oct 07 '15 at 15:26
  • By contradiction, for instance. (this is a sketch, hopefully it works): suppose $x\not\to 0$, i.e. there exists a sequence $\epsilon_n \to \infty$ and the corresponding $(x_n)$ satisfying the equation, such that $x_n\to \ell\in \mathbb{R}\cup{\pm\infty}\setminus{0}$. Then $x_n^3 = \frac{-1}{\epsilon_n} + \frac{2}{\epsilon_n}x_n - \frac{1}{\epsilon_n}_n^2$. If $\ell \in \mathbb{R}$, take the limit on both sides (remember that $\epsilon_n\to \infty$). But if $|\ell| = \infty$, the LHS has magnitude much larger than the RHS when $n\to\infty$ (as $|x^3_n| \gg x^2_n/\epsilon_n$): contradiction. – Clement C. Oct 07 '15 at 15:36
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Formally, we are looking for $x(\epsilon) \sim \epsilon^{-1/3}x_1 + \epsilon^{-2/3}x_2 + \epsilon^{-1}x_3 + \epsilon^{-4/3}x_4 + \dots$. This suggests the change of the parameter $s = \epsilon^{-1/3}$, and we now are interested in the behavior of solutions of $x(s)^3 + s^{3}(x(s)^2 - 2x(s) + 1) = 0$ as $s \to 0$.

Now clearly $x(s)$ cannot go to infinity as $s \to 0$. So the $x(s)$ remain bounded as $s \to 0$. Take any subsequence $s_n \to 0$ such that $\lim_n x(s_n) = c$ exists, then $c^3 + 0\dot(c^2-2c+1) = 0$ and hence $c=0$. Therefore $x(s) \to 0$ as $s \to 0$..

Now substitute the expression for $x(s)$ into the equation, expand, and collect terms. You will obtain $$ 0 = (1+x_1^3) + s(-2x_1+3x_1^2x_2) + s^2(x_1^2-2x_2+3x_1x_2^2+3x_1^2x_3) + \dots $$ The constant term must be zero, so $x_1 = -1$. The term with $s$ must also be zero, so $x_2 = -2/3$. And so on. You will get $x_3 = -1/3, x_4 = -10/81, x_5 = -7/243$, using Mathematica.

José Carlos Santos
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Hans Engler
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