This is the strange moment that I get when I solve this equation:

$$ \frac{w^4}{4} - \frac{w^3}{3} = \varepsilon, $$

where $\varepsilon$ is a small parameter. If I plot the graph $ w \mapsto \frac{w^4}{4} - \frac{w^3}{3}$, I can see that there are two real solutions to this equation. I am interested in the solution that is close to $0$ and I want to find its asymptotics in terms of $\varepsilon$.

(I have to warn that I have never used perturbation techniques before, so forgive me if I am using wrong terminology or if I do stupid mistakes)

It seems that finding solution in form $\hat{w}(\varepsilon) = a_0 + a_1 \varepsilon + \dots$ doesn't work here. I am using this text as a reference and this is an expected result according to it . My equation looks similar to Example $5.3$ on page 8, so I've decided to check substitution $w = \varepsilon^{p} \cdot y$ using considerations from this example. This lead me to the substitution $\hat{w} = \varepsilon^{\frac{1}{3}} \cdot y(\varepsilon), \; y(0) \neq 0$, and I got an equation:

$$ \frac{\varepsilon^{\frac{4}{3}}\cdot y^4}{4} - \frac{\varepsilon \cdot y^3}{3} = \varepsilon .$$

It seems to be the right thing to divide both sides by $\varepsilon$, but then I get

$$ \frac{\varepsilon^{\frac{1}{3}}\cdot y^4}{4} - \frac{y^3}{3} = 1 .$$

This looks like that now solution needs singular perturbation methods. But the original equation didn't seem to need them! This is the moment where I am kind of lost, because I expected to get an equation similar to the equation from Example $5.3$.

So I have two questions:
1. Did I use the wrong method?
2. If I used the right method, why did it lead from regular perturbation problem to singular perturbation problem?

Thanks in advance for your help!

  • 2
    $\begingroup$ You're almost there. Write out $y=y_0+\epsilon^{1/3}y_1+\epsilon^{2/3}y_2+\ldots$, and each power of $\epsilon$ will give you an equation. Your first equation will be $y_0^3=-3$, and the second will be $y_0^4/4-y_1^3/3=0$ and so on. $\endgroup$ – David Jun 20 '16 at 8:33
  • $\begingroup$ @David Thank you, I will try this method! However, I don't quite understand one moment: since this is a singular perturbation problem, shouldn't the regular perturbation method fail here? I haven't done full calculations yet, but this is an impression that I've got from examples in the cited text. $\endgroup$ – Evgeny Jun 20 '16 at 8:41
  • $\begingroup$ I don't think it's singular---let $\epsilon\rightarrow0$ in the original equation, you don't lose any solutions. $\endgroup$ – David Jun 20 '16 at 22:45

From your original equation, you can do a regular perturbation expansion $w=w_0+\epsilon w_1+\epsilon^2w_2+\ldots$ to get, at $O(1)$, $$w_0^4-\frac{4}{3}w_0^3=0,$$ which has three solutions $w_0=0$ and one solution $w_0=4/3$. At $O(\epsilon)$, you get $w_0^3w_1-w_0^2w_1=1$, which is not consistent for the roots $w_0=0$, so consider $w_0=4/3$ to give, $$ \left(\frac{4}{3}\right)^3w_1-\left(\frac{4}{3}\right)^3w_1=1$$ so $w_1=27/16$. Then, the $O(\epsilon^2)$ equation gives $w_2=1215/2048$. So the root near $4/3$ is $$w=\frac{4}{3}+\epsilon\frac{27}{16}+\epsilon^2\frac{1215}{2048}+O\left(\epsilon^3\right).$$

For the roots near zero, our naive expansion $w=w_0+\epsilon w_1+\epsilon^2w_2+\ldots$ failed, so, as you did, set $w=\epsilon^py$, from which you determine $p=1/3$, and so $$ \frac{\epsilon^{1/3}y^4}{4}-\frac{y^3}{3}=1.$$ Now, write $y$ as a series in one-third powers in $\epsilon$ (so that you can deal with the $\epsilon^{1/3}$ in the equation above), $ y=y_0+\epsilon^{1/3}y_1+\epsilon^{2/3}y_2$. The $O(1)$ equation is $$-\frac{y_0^3}{3}=1,$$ so $y_0=-3^{1/3}$ is the only real solution. The next equation is $$\frac{y_0^4}{4}-y_0^2y_1=0,$$ so $y_1=9^{1/3}/4=(3/8)^{2/3}$. The $O(\epsilon^{2/3})$ equation is $$y_0^3y_1-\frac{4y_0^2y_1}{3}+2y_0^2y_1^2=0,$$ with solution $$y_2=\frac{3^2\left(3^{1/3}-2\right)}{2^5}.$$ So, the root near zero is $$w=-\left(3\epsilon\right)^{1/3}+\left(\frac{3\epsilon}{8}\right)^{2/3}+\frac{3^2\left(3^{1/3}-2\right)}{2^5}\epsilon+O\left(\epsilon^{4/3}\right).$$

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  • $\begingroup$ Thank you very much for this answer! There is only one thing that I can't get my head around: we started from regular perturbation problem (when we let $\epsilon \rightarrow 0$, degree of polynomial doesn't change), but we arrive at singular perturbation problem (when $\epsilon = 0$ we have a poly of lower degree). So this is my question now: does the dichotomy "singular/regular" apply to the polynomial or to the number of solutions near the root of the interest? $\endgroup$ – Evgeny Jun 21 '16 at 8:43
  • $\begingroup$ I see. At $\epsilon=0$ there is a triple root at 0, but for $\epsilon\neq0$, this becomes a real root and two imaginary roots. The polynomial still has the same number of solutions, because it's degree doesn't change as $\epsilon\rightarrow0$. So I don't think this is a singular/regular thing. $\endgroup$ – David Jun 21 '16 at 23:38
  • $\begingroup$ While I still can't form the solid understanding behind the general rules of choosing the correct expansion in perturbative methods (and I hoped that it would become more clear to me), your solution helped me a lot and there are no other solutions, so I accept it. Thank you very much again! :) $\endgroup$ – Evgeny Jun 23 '16 at 14:30
  • $\begingroup$ At which step? The initial $w=w_0+\epsilon w_1+\ldots$, finding $p$ in $w=\epsilon^py$, or working out that $y$ should be $y_0+\epsilon^{1/3}+\ldots$? $\endgroup$ – David Jun 23 '16 at 22:48

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