In the course of my studies, I'm looking at at the ODE:

\begin{equation} (f^3(x))'''=\frac{1}{6}xf(x),\quad f(0)=1,\,\,f'(0)=0 \end{equation} Where $f''(0)$ is a parameter left undetermined. In looking at it numerically, for $-0.11<f''(0)<0$ or so, the solution curves down for a while until reaching a minimum before heading off to infinity. However, around the critical value $f''(0) \approx -0.11$ it touches down on the $x$-axis and the solution 'stops' as the first derivative, $f'(x)$ becomes singular there.

Firstly, I haven't been able to really understand why this singularity occurs - when I expand out the third derivative term I end up with \begin{equation} 3f^2f'''+18ff'f''+6(f')^3-\frac{1}{6}xf=0 \end{equation} which to me seems to suggest that as $f \to 0$, $f'$ should tend to zero as well, as all the terms except the $6(f')^3$ get small. Why am I wrong with this thinking?

Secondly, I really would like to know how the solutions behave beyond this point as the ODE arises as the unperturbed problem for a full problem which has solutions on the whole real line. What I expect is that if we can find such an extension, this solution will exhibit decaying oscillations for $f''(0)<-0.11$ but I have no idea how to show this or even if it's right.

Many thanks.


The problem is that the highest derivative, $f'''$, appears with the factor $3f^2$. When the coefficient of the highest derivative vanishes, singular behavior is to be expected.

How to fix this: let $g=f^3$. The ODE for $g$ is $g'''(x) =\frac16 xg(x)^{1/3}$, which is much more reasonable. The initial conditions should be adjusted, since $g''(0) = 3f''(0)$. So, the critical value is now around $-0.33$. But unlike $f$, the solution $g$ crosses the horizontal axis without too much fuss and goes on:


Produced in Maple, with the command

dsolve([diff(g(x),x$3)=x*abs(g(x))^(1/3)*signum(g(x))/6, g(0)=1, D(g)(0)=0, D(D(g))(0)=-0.33], g(x), numeric, range=0..6);

The graph never comes back into the positive territory. Indeed, you can see that around $x=5$, both $g'$ and $g''$ are negative. Since $g'''$ is also negative from the equation, the second derivative will keep on decreasing, which means it'll stay negative, which means $g'$ will keep on decreasing... the solution goes down pretty fast, $\approx -x^6$.

Unfortunately, the above suggests that this ODE is not really useful for your purposes (I understand from your other questions that you were interested in a 5th order ODE with oscillatory behavior).

  • $\begingroup$ Thanks for answering this (and commenting on some of my other questions) - however, I think you've made a mistake: $k=3k-3$ yields $k=3/2$, not $k=2/3$. Thus by this reckoning $f \sim (x-a)^{3/2}$ and $f'$ does not blow up at $a$. Is the issue the $xf(x)$ term? I'm not convinced it is truly of order $k$. $\endgroup$ – Baron Mingus May 28 '14 at 9:32
  • $\begingroup$ @BaronMingus Indeed, I made a mistake. See the revised answer. $\endgroup$ – user147263 May 28 '14 at 18:26
  • $\begingroup$ Brilliant! Thank you. I'll investigate tomorrow, but I wouldn't be surprised if there are some more negative values of $g''(0)$ for which oscillatory behaviour does manifest. $\endgroup$ – Baron Mingus May 28 '14 at 20:22
  • $\begingroup$ @BaronMingus I would be surprised. I tried several other values; the more negative $g''(0)$, the faster the curve drops into "negative and concave" territory, from which there is no way back. $\endgroup$ – user147263 May 28 '14 at 20:27
  • $\begingroup$ Hmmm, puzzling. Could this be related to somehow taking the 'wrong' branch of the cube root? $\endgroup$ – Baron Mingus May 28 '14 at 20:37

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.