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Just curious if anyone had any thoughts on does non-uniqueness of solutions to ODEs always imply that the solutions blow up in a finite time?

Thanks!

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Not necessarily: consider the classical example: $$f'(x) = \sqrt{f(x)}$$ with$ f(0) = 0$ The solutions are of type $0$ for $x \leq C$ and $\frac{(x-C)^2}{4}$ for $x > C$ where $C$ is an arbitrary non-negative number. As you see, there is no blow up in finite time, yet the solution is not unique. Unfortunately, on the top of the head I can't refer to a book, where it's proven all solutions are of this type, but I'm sure I've seen it.

Nevertheless, for this example it's easy to see that no solution $f$ has finite time blow-up: if $f(x_0) > 0$ for any $x_0 > 0$ the Picard-Lindeloff theorem gives us uniqueness of solution and it's straightforward to find the closed form $\frac{1}{4}(x-x_0)^2$, which does not have finite time blow up.

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  • $\begingroup$ Hi Milen. This is very helpful, thanks! $\endgroup$ – Dr. Ikjyot Singh Kohli Dec 24 '15 at 17:04
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Consider the ivp: $$ y'=\frac{2}{x}(y-1) \qquad y(0)=1. $$ Clearly, this problem has infinite solutions $$ y=1+kx^{2},\qquad k\in R $$ which non of them dose not blow up in a finite time.

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  • $\begingroup$ Hi. Okay, thanks. So, I think then solutions that blow up are non-unique, but the converse, by your example, is clearly not true! $\endgroup$ – Dr. Ikjyot Singh Kohli Dec 24 '15 at 16:48
  • $\begingroup$ A solutions that blow up may be unique. For example $y'=-y^{2}$ has a unique solution $y=1/x$ which blows up in $x=0$. $\endgroup$ – Albert Dec 24 '15 at 16:54
  • $\begingroup$ I don't mean blow-up in that sense. I mean blow up in a finite time. $\endgroup$ – Dr. Ikjyot Singh Kohli Dec 24 '15 at 16:57

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