Is local Lipschitz continuity sufficient for an ODE to have a unique solution? I have learned that for an ordinary differential equation of the form:
\begin{align}
\dot{x}(t)&=f(x,t) \\
x(t_{0})&=x_{0}
\end{align}
If $\;\;f:\mathbb{R}^{n}\rightarrow{}\mathbb{R}^{n}$ is globally Lipschitz continuous on $\mathbb{R}^{n}$, then there exists a unique solution to the ODE.
My question is: since only global Lipschitz continuity is sufficient for this ODE to have a unique solution, does this mean that local Lipschitz continuity is not sufficient? If this is true, can someone please provide an example where $f$ is locally, but not globally, Lipschitz continuous, and there does NOT exist a unique solution to the ODE?
 A: Local Lipschitz continuity suffices to have uniqueness. What it does not suffice to is global existence. That is, if $f$ is local but not global Lipschitz, then the (unique) solution to the Cauchy problem might cease to exist (blow up) in finite time. The prototypical example is the Cauchy problem 
\begin{equation*}
\begin{cases}
\dot x = x^2 \\
x(0)=x_0
\end{cases}
\end{equation*}
which has the unique solution 
\begin{equation}
x(t)=\frac1{x_0^{-1}-t}, 
\end{equation}
that ceases to exist at $t=x_0^{-1}$.
A: Uniqueness isn't the issue, it is that there may not be a solution for all $t$.  For instance, take $f(x,t) = x^2$.  The unique solution to $\dot x(t) = x(t)^2$, $x(0) = 1$ is $x(t) = \dfrac{1}{1-t}$ which blows up at $t \to 1^-$.
A: Consider an IVP problem
$$\dot{x}(t)=f(x, t)$$
$$x(0)=x_0$$
Cauchy-Lipschitz theorem (locally): If

*

*$U$ is an open set

*$f: U\times[0, T]\to \mathbb{R}^n$ is continuous.

*$f$ is Lipschitz with respect to $x$.

*$x_0\in U$
then for some interval $I \subset [0, T]$ there is a unique solution $x(t): I\to U$. In particular, there are only two cases for interval $I$. Either

*

*$I=[0,T]$, if $x(t)\in U, \forall t \in [0, T]$.

*$I=[0,T_0)$, if $x(t)\in U, \forall t \in [0, T_0)$, where $T_0 \leq T$ and $x(t)$ approaches the boundary of $U$ as $t\to T_0$.

Corollary: If

*

*$f: \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}$ is continuous.

*$f$ is locally Lipschitz with respect to $x$.

then for some interval $I \subset \mathbb{R}$ there is a unique solution $x(t): I\to U$. In particular, there are only two cases for interval $I$. Either

*

*$I=\mathbb{R}$, if $x(t)$ is bounded everywhere.

*Smaller interval $I\subset \mathbb{R}$, if $x(t)$ blow up somewhere.

Reference: Cauchy-Lipschitz_theorem
