# Global existence theorem and bound on the solution

Consider the following initial value problem:

$$\begin{cases} y'(t) = f(t, y(t)) \\ y(t_0) = y_0 \end{cases}$$

where the flux function $$f$$ is continuous and locally $$y$$-Lipschitz on a vertical strip $$]a,b[ \times \Bbb R$$.

I know that by Cauchy's Existence and Uniqueness Theorem and corollaries there exists a unique maximal solution $$y \in C^1(]t_{\min}, t_{\max}[, \Bbb R)$$ such that $$t_0 \in ]t_{\min}, t_{\max}[ \subseteq ]a,b[$$.

Now, my teacher told us that if $$\exists \alpha \in C^0(]a,b[): \forall t \in ]t_{\min}, t_{\max}[, |y(t)| \leq \alpha(t)$$ then $$t_{\min} = a$$ and $$t_{\max}=b$$ so that $$y$$ is also a global solution.

I would like to know if anyone knows this result and its proof or any reference concerning it for I have not been able to find anything on the internet so far.

As always, any comment or answer is welcome and let me know if I can explain myself clearer!

This is a special case of the fact that a nonextendable solution has to leave (to the left and to the right) each compact subset of $$(a,b) \times \mathbb{R}$$. I will outline the proof of $$t_{max} = b$$ in your setting: Assume by contradiction that $$t_{max} < b$$. Consider $$y$$ restricted to $$[t_0, t_{max})$$. Then $$|y|$$ is bounded on $$[t_0, t_{max})$$ by $$\gamma:=\max_{t \in [t_0, t_{max}]} \alpha(t)$$. Hence $$|y'|$$ is bounded on $$[t_0, t_{max})$$ (since $$f$$ is bounded on the compact set $$[t_0, t_{max}]\times [-\gamma,\gamma]\subseteq (a,b) \times \mathbb{R}$$). So $$y$$ is Lipschitz continuous on $$[t_0, t_{max})$$, hence $$y_1:=\lim_{t \to t_{max}-} y(t)$$ exists. From the differential equation you get that $$\lim_{t \to t_{max}-} y'(t) = f(t_{max},y_1).$$ Now you can extend $$y$$ to the right by the solution of the IVP $$z'(t)= f(t,z(t)), \quad z(t_{max})=y_1,$$ in contradiction to the fact that $$y$$ is nonextendable to the right.