From ODE I learned if $g$ is Lipchitz on $\mathbb{R}^n$ there exists a unique solution $y:\mathbb{R} \Rightarrow \mathbb{R}^n$ to the IVP \begin{eqnarray} y' &=& g(y)\\ y(t_0) &=& y_0 \end{eqnarray} where $t_0=0$
From what I understand, since $g$ is globally Lipchitz, then it is locally Lipchitz. From the fact it is locally Lipchitz, I can prove there exist a $c>0$ such that our IVP has a unique solution $y(t)$ on a small interval $[-c,c]$.
What I don't understand is, how does a Globally Lipchitz function implies there exist a unique solution on $\mathbb{R}$?
The only thing that comes to mind is this, (continuing from what we previously established) since $g$ is Globally Lipchitz and we have proven there exist a unique solution $y(t)$ on $[-c,c]$, I can pick a $t_1>0\in [-c,c]$ such that we have a new IVP \begin{eqnarray} y' &=& g(y)\\ y(t_1) &=& y_1 \end{eqnarray} Afterwards, it can be proven that this new IVP has a unique solution on some interval $[-c_1+t_1,c_1+t_1]$. Then you would continue iterating this procedure (making sure to pick $t_i>t_{i-1}$) until you construct $[c,\infty]$. You would apply a similar argument to construct the other half of our interval to get the complete $\mathbb{R}$. My only issue is, what if (let's say) after some large $n$, $[-c_n+t_n,c_n+t_n]$ becomes so small that the maximal interval converges to a fixed point and doesn't reach $\infty$? How would you prevent that issue from happening?
I'm new to the forum so I hope my question wasn't to vague. If you need me to clarify, please let me know. Thanks for the help.
Here is a different approach based on Gronwall's inequality and the fact that if the maximal interval of existence is $[0,T)$ with $t<\infty$, then $\lim_{t\to T^-}|y(t)|=\infty$.
Suppose $y$ is defined on an interval $[0,T)$ and let $L$ be $g's$ Lipschitz constant. Then $$ y(t)-y_0=\int_0^sg(y(s))\,ds=g(y_0)\,t+\int_0^t(g(y(s))-g(y_0))\,ds $$ and $$ |y(t)-y_0|\le T\,|g(y_0)|+L\int_0^t|y(s)-y_0|\,ds,\quad 0\le t<T. $$ Gronwall's inequality implies that $y$ is bounded on $[0,T)$, and this in turn implies that the solution is global.
A similar argument works on intervals $(-T,0]$.