# If $y(t)$ cannot be extended to a solution on $[t_0,\infty)$, then $|y(t)|\rightarrow\infty$

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be a $C^1$ function and let $(t_0,y_0)\in\mathbb{R}^2$. Suppose $y(t)$ is a solution to the initial-value problem

$\frac{dy}{dt}=f(t,y)$, $y(t_0)=y_0$.

in a neighborhood of the point $t_0$.

Prove that if $y(t)$ cannot be extended to a solution $dy/dt=f(t,y)$ on the interval $[t_0,\infty)$, then there exists a $t_1>t_0$ such that $|y(t)|\rightarrow\infty$ as $t\rightarrow t_1$ from below.

My attempt: Intuitively, this seems to make sense--if the solution doesn't exist, then $y$ must blow up near some point $t_1$. I'm imagining you have to use uniqueness of solutions to ODE's, but I'm not sure how to go about this.

Any help appreciated!

Note that the smoothness of $f$ implies that it satisfies a Lipschitz condition an any closed interval (in the second variable). We can thus use the Picard-Lindeloeff theorem to bound the lifetime of any solution from below.
More concretely, if we assume that the maximal solution $y$ lives on $[0,t_1)$ but stays bounded as $t\to t_1-$ we can find a rectangle $$[0,t_1)\times y([0,t_1))\subset R:=[0,t_2]\times I.$$ Picard-Lindeloeff assures that for any fixed $t'\in [0,t_1)$ the problem of propagating $y(t')$ into the future admits a solution at least on $[t',t'+b]$, where $b$ is determined by the size of the rectangle and the supremum of the values of $f$ on it (the concrete formulation somewhat differs by source). Now we can simply move $t'$ closer to $t_1$ so that $t'+b>t_1$. This extension then glues with $y$, which is bogus.