Uniqueness of solution of an ODE in an interval implies uniqueness of solution in a subinterval

Question

I'm studying the book Lições de Equações Diferenciais Ordinárias by Jorge Sotomayor. There is a proof of a theorem about maximum solutions of first order ODE in page 17 where I can't understand a detail. If I assume that the function is locally Lipschitz the problem goes away but the book treats a slightly more general case. So I was wondering if the following theorem is true.

Let $\Omega\subseteq\mathbb{R}\times\mathbb{R}^n$ be an open set, $(t_0,x_0)$ an element of $\Omega$ and $f:\Omega\to \mathbb{R}^n$ a continuous function. Suppose there exists an open interval $I$ such that $x'=f(t,x), x(t_0)=x_0$ has a unique solution in $I$. Suppose $J$ is an open subinterval of $I$ containing $t_0$. Does the equation $x'=f(t,x), x(t_0)=x_0$ have a unique solution in $J$?

If the previous theorem is not true, assume that the hypothesis holds for every $(t_0,x_0)\in \Omega$, i.e. that for each $(t_0',x_0')\in \Omega$ there exists an open interval $I(t_0',x_0')$ such that the $x'=f(t,x), x(t_0')=x_0'$ has a unique solution in $I(t_0,x_0)$, does the conclusion hold now?

This is the actual theorem in Sotomayor's book

Let $f$ be continuous in an open set $\Omega\subseteq \mathbb{R}\times\mathbb{R}^n$. Suppose that for each $(t_0,x_0)\in \Omega$ there exists a unique solution of $x'=f(t,x),x(t_0)=x_0$ defined in an open interval $I=I(t_0,x_0)$ (for example, if $f$ is locally Lipschitz this condition is satisfied). Then, for each $(t_0,x_0)\in \Omega$ there exists a unique solution $\varphi=\varphi(t,t_0,x_0)$ of $x'=f(t,x),x(t_0)=x_0$, defined in an open interval $M(t_0,x_0)=(w_-(t_0,x_0),w_+(t_0,x_0))$ with the property that every solution $\psi$ of $x'=f(t,x),x(t_0)=x_0$ in an interval $I$ satisfies $I\subseteq M(t_0,x_0)$ and $\psi=\varphi|I$

Answer 1

The theorem as formulated appears not to be true. Consider the IVP $x' = f(x)$, $x(0) = 0$, where $$ f(x) = \begin{cases} 0 & \text{ for } x \le 0 \\ \sqrt{x} & \text{ for } 0 < x < 1 \\ x^2 & \text{ for } x \ge 1. \end{cases} $$ Clearly, $\varphi(t) \equiv 0$ is a solution of the IVP. Another solution of the IVP is $$ \psi(t) = \begin{cases} 0 & \text{ for } t < 0 \\[1ex] \displaystyle \frac{1}{4}t^2 & \text{ for } 0 \le t \le 2 \\[1ex] \displaystyle \frac{1}{3 - t} & \text{ for } 2 < t < 3. \end{cases} $$ Both solutions are nonextendible. Further, all nonextendible solutions, besides $\varphi$, are of the form $\psi_{\tau}(\cdot) := \psi(\cdot - \tau)$, $\tau \in [0,\infty)$.

Take $I = (-\infty, \infty)$. $\varphi$ is the only solution of the IVP defined on $I$. For any open interval $J$ of the form $(-\infty, a)$, where $0 < a < \infty$, $\psi_{\tau}$, where $\tau \in [\max\{0, a- 3\}, a)$, is a solution of the IVP on $J$, not equal to $\varphi$ on $J$.

On the other hand, if we have local uniqueness then the uniqueness of the solution on a subinterval is a trivial consequence of the uniqueness on the whole interval of existence.