Consider the following statement:

If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function, for the autonomous equation $$x' = f (x)$$ the "Peano phenomenon" can arise only at those values of $\bar x$ for which $f(\bar x) = 0$.

(The "Peano phenomenon" = Cauchy problems associated to the above equation can admit more than one solution.)

Can someone provide me with a (proof or a) reference for a proof ? According to this paper, this statement is "well known", so a proof of it should be found in textbooks, but I couldn't find any. A paper that contains the proof is mentioned in the linked paper above, but I don't have access to that paper.


1 Answer 1


I'm not sure this is "well known" (it wasn't to me). We can do this with a Lyapunov function $V(x,y)=g(y)-g(x)$, with $g(x)=\int_a^x (1/f(u))\, du$. This is well defined near an $a$ with $f(a)\not= 0$.

Now if $x(t),y(t)$ both solve $x'=f(x)$, $x(0)=a$, then $W(t)= V(x(t),y(t))$ satisfies $W(0)=0$, $$ W'(t)= g'(y)f(y)-g'(x)f(x)= 0 , $$ by our choice of $g$. It follows that $W\equiv 0$, and this gives uniqueness because $g$ is strictly monotone.

  • $\begingroup$ Why not just define $V(x,y) = \left( \int_x^y 1/f(u) ~\mathrm{d}u \right)$ and assume $f|_{[x,y]} \neq 0$? (In regards to your edits.) Also, last sentence do you really mean $g$ is monotone? $\endgroup$ Jun 15, 2016 at 19:25
  • $\begingroup$ @WillieWong: Yes, that works too, thanks ($V=(\ldots )^2$ is a trace of my earlier attempts with functions such as $(y-x)^2$). And yes, I do mean $g$ is strictly increasing or decreasing (depending on the sign of $f$). $\endgroup$ Jun 15, 2016 at 19:27
  • $\begingroup$ Ah, I see what you mean now. Okay. $\endgroup$ Jun 15, 2016 at 19:29
  • 1
    $\begingroup$ In fact, one can also just observe that the inverse function $t(x)$ exists locally and recover it (and thus also $x(t)$) from $dt/dx = 1/f(x)$ by integrating. $\endgroup$ Jun 16, 2016 at 16:23
  • 1
    $\begingroup$ Some information and references can be found here: mathoverflow.net/questions/234183/… $\endgroup$ Jun 17, 2016 at 18:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .