I'm stuck on this problem. I don't even know how to start:
If $(f_1,...,f_n) : \mathbb{R}^n\longrightarrow{\mathbb{R}^n}$ with $f \in C^1 $ is a vector field and $V:\mathbb{R}^n\longrightarrow{\mathbb{R}}$ is a differentiable function such that $\displaystyle\sum_{i=1}^n{\frac{dV}{dx_i}f_i(x)}\leq{0}$ and $V(x)\geq{\left |{x}\right |}^2,\forall{x}\in{\mathbb{R}^n}$, prove that all solution of $\dfrac{dx}{dt}=f(x)$ is defined for all $t>0$
Thanks in advance.
For $f$ continuous, $$ \frac{dx}{dt} = f(x) $$ has a local (in $t$) solution for a given initial condition $x(0) = (x_1(0), \dots, x_n(0))$. If $f$ is Lipschitz, then this local solution is unique. For these results, and for extending a local solution to a global one, refer to any book on the theory of ODEs (for example, Chapter 1 of Theory of Ordinary Differential Equations by Coddington).
Suppose $x(t)$ is a local solution for $t\in [0,T]$ with initial condition $x(0)$. To prove its uniqueness, dot both sides by $\nabla V$ to get $$ \frac{d}{dt}V(x(t)) = \nabla V \cdot f \leq 0. $$ Hence $V(x(t))$ is a decreasing function. That is, $|x(t)|^2 \leq V(x(t)) \leq V(x(0))$. The point is that $x(t)$ remains bounded. Since $f \in C^1$, it is locally Lipschitz (due to bounded first derivative in compact sets), which gives the uniqueness of local solution.