# Is the converse of Osgood criterion for ODEs also true?

Namely, Assuming that $$f$$ is a continuous real function and $$f(0)=0$$ , $$f(x)>0$$ when $$x\neq 0$$, Consider the differential equation $$x'= f(x)$$ with the initial value $$x(0)=0$$ , is it true that if this differential equation has a unique solution then $$\int_0^c \frac{dx}{f(x)}= \infty$$ for all $$c\in \mathbb{R}$$ ?

I can refer to my related question here https://math.stackexchange.com/questions/4029712/a-condition-for-uniqueness-of-solution

• The converse is not true, in the most dramatic way: mathoverflow.net/questions/234183/… May 28 at 19:20
• Actually the particular question in the text seems a bit different from the general question alluded in the title. May 29 at 6:10

If for $$c>0$$ one has $$b:=\int_0^c\frac{dx}{f(x)}<\infty$$, the function $$v(s):=\int_0^s\frac{dx}{f(x)}$$ is an increasing homeo $$[0,c]\to[0,b]$$ with derivative $$v'(s)= {1}/{f(s)}$$ for $$0, so the inverse homeo $$u:[0,b]\to[0,c]$$ satisfies $$u'(t)=\frac{1}{v'(u(t))}=f(u(t))$$ for $$0, and since from the equation $$u'(t)=f(u(t))\to0$$ tor $$t\to0$$, by the Mean Value Theorem there also exists the derivative of $$u'$$ at $$t=0$$, and it is $$0$$, satisfying the equation. Therefore $$u$$ (extended to be zero for $$x<0$$) is a solution of the equation with initial condition $$u(0)=0$$, different from the constant solution.

• I am not sure if I am following, f is not necessarily increasing at the right nor at the left of 0 even if the limit is 0, this would make v not necessarily a homeo no ?
– Omar
May 29 at 2:35
• But you assume f positive, so v is strictly increasing. May 29 at 5:05
• agreed thank you, and with the constant function at 0 being always a solution to this differential equation, this would mean that if this integral is infinite then there is no other solution, right ? My former answer to the other direction might be simplified?
– Omar
May 29 at 22:28
• Yes, always assuming $f$ continuous, $f(0)=0$, and $f(x)>0$ for $x>0$. If $u$ is a non-zero solution of $u'=f(u)$ with initial condition $u(0)=0$ we can assume up to translations $u(t)>0$ for $t>0$, so $u$ is actually a homeo $[0,b]\to[0,u(b)]$ for some $b>0$, and a diffeo $(0,b)\to(0,u(b))$, so it can be used to change variable and show $\int_0^{u(b)}\frac{ds}{f(s)}=b$. May 30 at 15:08

The converse is not true, in the most dramatic way:

Solutions-set first order ODE's without uniqueness

In fact we "almost always" have uniqueness but nobody knows any explicit condition which implies uniqueness and almost always holds.