3
$\begingroup$

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

Improve this question
$\endgroup$
2
  • $\begingroup$ The converse is not true, in the most dramatic way: mathoverflow.net/questions/234183/… $\endgroup$
    – Alexandre Eremenko
    Commented May 28, 2021 at 19:20
  • $\begingroup$ Actually the particular question in the text seems a bit different from the general question alluded in the title. $\endgroup$
    – Pietro Majer
    Commented May 29, 2021 at 6:10

2 Answers 2

Reset to default
3
$\begingroup$

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 $0<s<b$, so the inverse homeo $u:[0,b]\to[0,c]$ satisfies $u'(t)=\frac{1}{v'(u(t))}=f(u(t))$ for $0<t<c$, 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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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 ? $\endgroup$
    – NotaChoice
    Commented May 29, 2021 at 2:35
  • $\begingroup$ But you assume f positive, so v is strictly increasing. $\endgroup$
    – Pietro Majer
    Commented May 29, 2021 at 5:05
  • $\begingroup$ 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? $\endgroup$
    – NotaChoice
    Commented May 29, 2021 at 22:28
  • 1
    $\begingroup$ 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 $\int_0^{u(b)}\frac{ds}{f(s)}=b$. $\endgroup$
    – Pietro Majer
    Commented May 30, 2021 at 15:08
0
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

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