2

we have

$$ y'' = \sqrt{x} y $$

I want to find the leading asymptotic behavior as $x \to \infty$

I tried substituting $y = e^{S(x)}$ and then after obtaining $S(x) \sim \pm \frac{4}{3} x^{3/4} $ and doing the correction $S = \pm \frac{4}{3} x^{3/4} + C(x) $ where $C << x^{3/4}$, I end up obtaining that

$$ y \sim \pm \frac{1}{x^{1/4}} e^{1/3 x^{3/2} } $$

How do I check that this is indeed the leading asymptotic behavior as $x \to \infty$?

  • Why is the dominant term in your exponential not the same as what it was before the correction? The form you stated for your correction suggests that it should still be the same. – Ian Nov 27 '16 at 05:55
  • because once I do the correction I get, $$ C(x) \sim \frac{1}{3} x^{3/2} \mp \frac{4}{3} x^{3/4} \pm \frac{1}{4} \ln x $$

    So, the $x^{3/4}$ terms cancel out

     –  Nov 27 '16 at 05:58
  • But you said $C(x) \ll x^{3/4}$, and then you computed $C(x)$ and it turned out to be $\gg x^{3/4}$. What gives? (The answer is that your dominant term was wrong in the first place, presumably because your dominant balance was inconsistent.) – Ian Nov 27 '16 at 12:54

1 Answers1

2

That the correction you obtain cancelled the dominant term is a sign that you did not obtain the correct asymptotic expression.

If you substitute $y=\exp(S)$ then you obtain the equation $$S'' + S'^2 =\sqrt{x}.$$

Now you have to test the different cases, in order to see if they are consistent:

a) $ S'^2 \gg S''$ and thus $S'^2 \sim\sqrt{x}$ with the solution $$S= \frac{4}{5} x^{5/4}.$$ In this case, we have $S'^2 =\sqrt{x} \gg S''= O(x^{-3/4})$, so this case is consistent.

b) $S'' \gg S'^2$ and thus $S''\sim \sqrt{x}$ with the solution $$S= \frac{4}{15} x^{5/2}.$$ In this case, we have $S''=\sqrt{x} \not\gg S'^2= O(x^3)$.

So a) is correct. Let us find the correction: we set $S= (4/5) x^{5/4}+C$ and with $x^{1/4} C' \gg C'^2 , C''$, we obtain the result $$ C(x) = \ln c-\frac{\ln x}{8}.$$

As a result, the asymptotic expansion is given by $$y\sim \frac{c}{x^{1/8}} \exp\left[(4/5) x^{5/4}\right].$$

Fabian
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  • I agree with you, but you only obtain one asymptotic solution. we should get two solutions since it is a second order equation. –  Nov 27 '16 at 06:28
  • I think it should be $\pm c $ right? –  Nov 27 '16 at 06:29
  • $\pm c$ does not help, as this is the same solution. The asymptotic expansion gives you the dominant solution. The second solution is $y \sim (c/x^{1/8}) \exp[-(4/5) x^{5/4}]$ that only appears when finetuning the initial conditions. – Fabian Nov 27 '16 at 06:36
  • I get it now, but I dont understand how you obtain $\ln c $? is this a arbitrary constant? –  Nov 27 '16 at 06:37
  • @Jibarito: yes that constant is arbitrary and is set by the initial conditions. – Fabian Nov 27 '16 at 06:43
  • thank you very much! by the way, here is a problem which I have put a 500 bounty on whihc is similar to this problem: http://math.stackexchange.com/questions/1168976/leading-behaviour-of-de-at-infinity –  Nov 27 '16 at 06:46