I have $y'(t) = e^{-[g(y)]^2}$, with initial value $y(0) = y_0$, $g: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ and g $\in C^1$. I want to find the limits of $t \to \pm \infty$ taking into account the possible values of $y_0$.
Now, I know that there exists a unique local solution because f(y) = $e^{-[g(y)]^2}$ is locally Lipschzit. the integral of an exponential is always exponential times something defined on all $\mathbb{R}$ so there exists unique a global solution which is maximal because defined on all $\mathbb{R}$.
That means that I can actually take those limits to plus and minus infinity. However I am stuck on how to actually calculate those. What theorem should I use?
I will consider the case $t\to+\infty$, the other being similar.
Since $0<e^{-g(y)^2}\le1$, solutions are strictly increasing, global, and satisfy the inequality $$ y(t)\le y_0+t,\quad t\ge0. $$ Claim: $\lim_{t\to+\infty}y(t)=+\infty$.
Proof: Suppose that $y$ is bounded, that is, $y_0\le y(t)\le M$ for some $M$. Since $g$ is regular, $e^{-g(y)^2}$ is bounded below on $[y_0,M]$, that is, $e^{-g(y(t))^2}\ge c$ for some constant $c>0$ and all $t\ge0$. This implies that $y(y)\ge y_0+c\,t$, in contradiction with the assumption that $y$ is bounded.