Is local Lipschitz continuity sufficient for an ODE to have a unique solution?

Question

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?

Answer 1

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 $t \to 1^-$.

Answer 2

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 $t=x_0^{-1}$.

Answer 3

Consider an IVP problem $$\dot{x}(t)=f(x, t)$$ $$x(0)=x_0$$

Cauchy-Lipschitz theorem (locally): If

1. $U$ is an open set
2. $f: U\times[0, T]\to \mathbb{R}^n$ is continuous.
3. $f$ is Lipschitz with respect to $x$.
4. $x_0\in U$

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

1. $I=[0,T]$, if $x(t)\in U, \forall t \in [0, T]$.
2. $I=[0,T_0)$, 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

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

1. $I=\mathbb{R}$, if $x(t)$ is bounded everywhere.
2. Smaller interval $I\subset \mathbb{R}$, if $x(t)$ blow up somewhere.

Reference: Cauchy-Lipschitz_theorem