I have this question:
Let $f:\mathbb{R}^n \to \mathbb{R}^n$ a $C^1$ vector field. Show that, if $V:\mathbb{R}^n \to \mathbb{R}$ is $C^1$ such that $$V(x) \geq |x|^2 \quad \text{and} \quad \langle f(x), \nabla V(x) \rangle <0$$ for all $x \in \mathbb{R}^n \setminus \{0\}$, where $\langle \cdot, \cdot \rangle$ is the usual inner product of $\mathbb{R}^n$. Show that the solution $x:I \to \mathbb{R}^n$ of $$x'(t) = f(x(t)), \quad x(t_0)=x_0$$
is defined for all $t >t_0$.
I think to put the solution of IVP in a compact set, and then show that solution is not limitated. But I not know how I can find this compact set.
Thanks for any help :)
The only way for the solution to this system not to be defined is to blow up to infinity. Let us show that this is impossible, since the norm of the solution to any initial value problem is bounded.
Consider the function $V(x(t))$, where $x(t)$ is the solution to some initial value problem except for $x(t_0)=0$. We have $$ \frac{d V(x(t))}{dt}= \sum_{i=1}^{n} \frac{\partial V}{\partial x_i}\cdot \frac{d x_i}{dt}= \sum_{i=1}^{n} \frac{\partial V}{\partial x_i}\cdot f_i(x)= \langle f(x), \nabla V(x) \rangle<0, $$ thus, $V(x(t))$ is decreasing for any solution. This means that $$ \forall t>t_0 \quad V(x(t))<V(x(t_0)). $$ But $$ \forall t>t_0 \quad \|x(t)\|^2\le V(x(t))<V(x(t_0)). $$ Now notice that $V(x(t_0))$ is constant (it does not depend on $t$).