I am facing difficulties trying to show that an ODE has a unique solution. The problem is given by: $$\begin{cases} \frac{du}{dt} = F(u) & t\in [0,1/2] \\ u(0) = u_0\end{cases}$$ where $F(x):= \begin{cases} x \sin( \frac{1}{x}) & x\in \mathbb{R}\backslash \{0\} \\0 & x=0 \end{cases}$, which does not fulfill the Lipschitz's criterion. We have not introduced any other criteria. I have tried using Banach fixed-point theorem, which has not worked yet. Any hints will be appreciated.

Thanks in advance.

  • $\begingroup$ AFAIK, the Picard-Lindelof theorem (which states the existence and uniqueness of ODE's in which $F$ is Lipschitz continuous) is a consequence of the Banach fixed-point theorem. $\endgroup$
    – rafa11111
    Apr 28 '18 at 17:33
  • $\begingroup$ Possibly duplicate to math.stackexchange.com/questions/3555720/… $\endgroup$ Nov 26 '21 at 17:28

It's a few years late, but note that $F$ vanishes on the countable set accumulating at $0$ given by $\{\frac{1}{k\pi} : k \in \mathbb{Z} \}$, and that away from $0$, $F$ is $C^1$ and thus locally lipschitz, so that the solutions $x_k(t) = \frac{1}{k\pi}$ are all unique in the sense that no other solution $\phi$ can satisfy $\phi(t_0) = \frac{1}{k\pi}$ for some $t_0$, unless $\phi(t) = \frac{1}{k\pi}$ for all $t$. In particular, by the intermediate value theorem, this shows that $\phi(t) = 0$ for all $t$ is the only solution satisfying $\phi(0) = 0$. This gives uniqueness of solutions to the ODE.

This same sort of trick shows $F_\alpha(x) = x^\alpha \sin(\frac{1}{x})$ has a unique solution for all $\alpha \in (0,2]$ even if for such $\alpha$ the function is not locally lipschitz.

I personally find it rather surprising, because the equation $x' = x^\alpha$ does not have a unique solution for $x \in (0,1)$, but after multiplication by $\sin(1/x)$ it does!

It is worth noting as well, that for $\alpha \in (0,2]$ that solutions exist for all time as well, by using comparison theorems and noting $\lim_{x\rightarrow \infty} \sin(\frac{1}{x})x = 1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.