Assume $F\in C^2(\mathbb{R}^n, \mathbb{R})$, for each initial value $x$, consider the IVP: $$ \dot{\phi}(t)=-\omega(\phi(t)) \nabla F(\phi(t)), \ \phi(0)=x, $$where $\omega(y)=\dfrac{|\nabla F(y)|}{1+|\nabla F(y)|^2}$. Prove that the solution of the IVP can be defined on $[0,+\infty)$.
I am currectly learning mountain pass theorem and the textbook use the proposition above without proof. I have some diffculty in solving the problem by the extensibility theorem. The $\omega$ is complicated so I only want to use its boundness. Is my thoughts ok and how to go on? If not, please suggest your method.
Any hint or complete proof can be very helpful!
Suppose that there is $t_0$ such that $$\lim_{t\to t_0^-}|\phi(t)|=\infty. \tag1$$ Integrating the equation from $0$ to $t$ ($t<t_0$) gives $$ \phi(t)-x=-\int_0^t\omega(\phi(s)) \nabla F(\phi(s))\mathrm{d}s. $$ So $$ |\phi(t)|\le |x|+\int_0^t\omega(\phi(s)) |\nabla F(\phi(s))|\mathrm{d}s=|x|+\int_0^t\dfrac{|\nabla F(\phi(s))|^2}{1+|\nabla F(\phi(s)|^2}ds\le|x|+t. \tag{2} $$ Letting $t\to t_0^-$ in (2) will be against (2). So the solution $\phi(t)$ is well defined in $[0,\infty)$.