4
$\begingroup$

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!

$\endgroup$
  • 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
2
$\begingroup$

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 $$w=-\left(3\epsilon\right)^{1/3}+\left(\frac{3\epsilon}{8}\right)^{2/3}+\frac{3^2\left(3^{1/3}-2\right)}{2^5}\epsilon+O\left(\epsilon^{4/3}\right).$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.