Asymptotic expansion, solving roots to an equation with dominant balance, what went wrong in my approach?

Question

So I wanted to compute the asymptotic expansion of the roots to, as $\epsilon \to 0$, $$\epsilon x^3-x^2+2x-1=0$$

Now when I tried to find $x\sim x_0+\epsilon x_1+\epsilon^2x_2+...$ I ran into trouble as at $O(\epsilon)$, I had $0=2x_1-2x_1+1.$ Now I just wanted to know why did I reach such contradiction? Is it because I implicitly balanced $x$ with $1$? In other words, is it because if I assume $x=O(1)$ then $x^2$ would also have order $1$ and thus the two terms I balanced are not dominant enough?

Moreover, I applied dominant balance to $\epsilon x^3$ and $x^2$ to reach one of the root, how could I reach the other two roots via dominant balance? This is because if I balance any other two terms in the above expression, there is always another term that is larger or have the same order as $\epsilon \to 0.$

Many thank in advance!

Answer 1

The dominant balance argument goes like this:

Let $x=\epsilon^\alpha x_0+\epsilon^\beta x_1+\ldots$ with $\alpha<\beta<\ldots$. We start by working out $\alpha$, so substitute $x=\epsilon^\alpha x_0$ (we can ignore the $\epsilon^\beta$ and smaller terms because they must be small than the $\epsilon$^\alpha$ terms).

$$\epsilon^{1+3\alpha}x_0^3-\epsilon^{2\alpha}x_0^2+2\epsilon^\alpha x_0-1=0$$

Dominant balance gives either

• $\alpha=-1$ (balancing first two terms) this is a dominant balance
• $\alpha=-1/2$ (balancing first and third terms) this is not a dominant balance, $\epsilon^{2\alpha}$ is larger than $\epsilon^\alpha$
• $\alpha=-1/3$ (first and fourth terms) this is not a dominant balance, $\epsilon^{2\alpha}$ and $\epsilon^\alpha$ are larger
• $\alpha=0$ (second, third, fourth terms) this is a dominant balance

That's all the combinations, so there are two possible balances, $\alpha=-1$ and $\alpha=0$.

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With $\alpha=0$ you get the regular expansion, $x=x_0+\epsilon^\beta x_1+\epsilon^\gamma x_2+\ldots$. The $O(1)$ equation is $$x_0^2+2x_0-1=0$$ with solutions $x_0=1,1$. Now you do the same thing again with $x=1+\epsilon^\beta x_1$, (with $\beta>0$), $$ \epsilon\left(1+3\epsilon^\beta x_1+3\epsilon^{2\beta}x_1^2+\epsilon^{3\beta}x_1^3\right)-\left(1+2\epsilon^\beta x_1+\epsilon^{2\beta}x_1^2\right)+2\left(1+\epsilon^\beta x_1\right)-1=0 $$ which simplifies to $$ \epsilon-\epsilon^{2\beta}x_1^2=0 $$ and so $\beta=1/2$ and $x_1^2=1$ so $x_1=\pm1$.

You can (usually) assume the pattern continues now, and let $x=1\pm\sqrt\epsilon+\epsilon x_2+\epsilon^{3/2}x_3+\ldots$.

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For the singular root, you have $x=\epsilon^{-1}x_0+\epsilon^\beta x_1+\ldots$ with $\beta>-1$. With this, you get $x_0=0,0,1$ (the two zeros correspond to the regular roots we found before). For the actual singular root, you find $\beta=0$ from dominant balance and the $O(\epsilon^{-1})$ equation is $$3x_1-2x_1+2=0$$ so $x_1=-2$. Continuing on, you get $x=\epsilon^{-1}-2-3\epsilon+\ldots$

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Putting it all together, we get that the roots are $$x=1+\sqrt\epsilon+O(\epsilon),\quad1-\sqrt\epsilon+O(\epsilon),\quad\frac{1}{\epsilon}-2-3\epsilon+O(\epsilon^2).$$

Answer 2

The assumed expansion is wrong. Note that the equation may be rewritten as

$$(x-1)^2=\epsilon x^3$$

$$x=1\pm\sqrt{\epsilon x^3}$$

A few iterations of fixed-point iteration then reveal that

\begin{align}x&\approx1\\x&\approx1\pm\sqrt\epsilon\\x&\approx1\pm\sqrt{\epsilon\pm3\epsilon^{3/2}+\mathcal O(\epsilon^2)}\\&=1\pm\sqrt\epsilon+\frac32\epsilon+\mathcal O(\epsilon^{3/2})\end{align}

where all $\pm$ signs are chosen to be the same (each corresponding to a root on one side of $1$).

Answer 3

Another way to obtain the expansion is by using series inversion. We have \begin{align*} \varepsilon = \frac{{(x - 1)^2 }}{{x^3 }} & = \frac{{(x - 1)^2 }}{{1 + 3(x - 1) + 3(x - 1)^2 + (x - 1)^3 }} \\ & = (x - 1)^2 (1 - 3(x - 1) + 6(x - 1)^2 - \cdots ), \end{align*} and thus \begin{align*} \pm \sqrt \varepsilon & = (x - 1)\left( {1 - \frac{3}{2}(x - 1) + \frac{{15}}{8}(x - 1)^2 - \cdots } \right) \\ & = (x - 1) - \frac{3}{2}(x - 1)^2 + \frac{{15}}{8}(x - 1)^3 - \cdots \,. \end{align*} By series inversion $$ x - 1 = \pm \sqrt \varepsilon + \frac{3}{2}\varepsilon \pm \frac{{57}}{8}\varepsilon ^{3/2} + \cdots , $$ i.e., $$ x = 1 \pm \sqrt \varepsilon + \frac{3}{2}\varepsilon \pm \frac{{57}}{8}\varepsilon ^{3/2} + \cdots \,. $$ Addendum. Following David's comment, I add the expansion for the third root. By Viète's formula and simple series manipulation, we find \begin{align*} x & = \frac{1}{\varepsilon }\frac{1}{{1 + \sqrt \varepsilon + \frac{3}{2}\varepsilon + \frac{{57}}{8}\varepsilon ^{3/2} + \cdots }}\frac{1}{{1 - \sqrt \varepsilon + \frac{3}{2}\varepsilon - \frac{{57}}{8}\varepsilon ^{3/2} + \cdots }} \\ & = \frac{1}{\varepsilon }\frac{1}{{1 + 2\varepsilon + \cdots }} = \frac{1}{\varepsilon }(1 - 2\varepsilon + \cdots ) = \frac{1}{\varepsilon } - 2 + \cdots \,. \end{align*}