# Extending solutions of an ODE past a singular point

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

$$(f^3(x))'''=\frac{1}{6}xf(x),\quad f(0)=1,\,\,f'(0)=0$$ 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 $$3f^2f'''+18ff'f''+6(f')^3-\frac{1}{6}xf=0$$ 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:
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$. Unfortunately, the above suggests that this ODE is not really useful for your purposes (I understand from your other questions that you were interested in a 5th order ODE with oscillatory behavior). • Thanks for answering this (and commenting on some of my other questions) - however, I think you've made a mistake:$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$. – Baron Mingus May 28 '14 at 9:32 • @BaronMingus Indeed, I made a mistake. See the revised answer. – user147263 May 28 '14 at 18:26 • Brilliant! Thank you. I'll investigate tomorrow, but I wouldn't be surprised if there are some more negative values of$g''(0)$for which oscillatory behaviour does manifest. – Baron Mingus May 28 '14 at 20:22 • @BaronMingus I would be surprised. I tried several other values; the more negative$g''(0)\$, the faster the curve drops into "negative and concave" territory, from which there is no way back. – user147263 May 28 '14 at 20:27