# 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?

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 $t \to 1^-$.
• Of course I agree with your answer, I would only change the phrase "there will not be a solution for all $t$". As you know very well, blow up might or might not occur. If you say that, one understands that blow up always occurs. – Giuseppe Negro Sep 18 '15 at 21:23
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 $$x(t)=\frac1{x_0^{-1}-t},$$ that ceases to exist at $t=x_0^{-1}$.