# How to prove that the maximal solution of $x'(t) = e^{-g(x(t))^2}$ is global?

Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be a function of class $C^{\infty}$. We consider the Cauchy's problem :

\begin{cases} x'(t) = e^{-g(x(t))^2} \\ x(0)=0 & \end{cases}

I have to show that the maximal solution $u$ of this system is defined on $\mathbb{R}$. But I don't know how to study this problem, because we have not enough informations on $g$... Someone could help me (and sorry for my bad english) ?

• Given a maximal solution defined on an interval $(a,b)$, have you learned that if $b<+\infty$ then $x(t)$ can not have (finite) accumulation points as $t\rightarrow b^-$? – H. H. Rugh May 23 '17 at 12:37
• No, I did'nt know that ! Thank you ! – Mélanie De la Cheminée May 23 '17 at 12:45

Let $$G(x) := \int_0^x e^{g(s)^2}\, ds, \qquad x\in\mathbb{R}.$$ The function $G$ is of class $C^1(\mathbb{R})$ and $G'(x) = e^{g(x)^2} \geq 1$ for every $x$, so that $G$ is strictly increasing and $G(\mathbb{R}) = \mathbb{R}$. In particular, its inverse function $G^{-1}$ is defined on all $\mathbb{R}$. The solution of your Cauchy problem is then given by $$x(t) = G^{-1}(t)$$ and it is defined for every $t\in\mathbb{R}$.
Quick answer: the r.h.s. is a $C^1(\mathbb{R})$ bounded function, hence solutions to Cauchy problems are global (i.e. defined on all $\mathbb{R}$).