Suppose we have a system $$\dot x = Ax$$
And a Lyapunov function $$V (x(t)) \geq 0, \forall t $$
My question is, does the fact $\dot V(x) = \nabla V(x)^\top \dot x < 0$ imply that $V(x(t))$ is monotonically decreasing along with time?
Ok, so intuitively, integrating bothsides we have $V(x(t)) < V(x(0))$, $\forall t$.
So it is decreasing relative to the initial time.
But what about the time afterwards, can we guarantee
$V(x(t_2)) < V(x(t_1)) $ where $t_2>t_1>0$?
Yes, $V(x(t))$ decreases with increasing $t$ along any trajectory $x(t)$, strictly monotonically in fact.
For if
$t_2 > t_1 \tag 1$
we have
$V(x(t_2)) - V(x(t_1))$ $= \displaystyle \int_{t_1}^{t_2} \dot V(x(s)) \; ds = \int_{t_1}^{t_2} \nabla V(x(s))^T \dot x(s) \; ds < 0, \tag 2$
since the integrand
$\nabla V(x(s))^T \dot x(s) < 0 \tag 3$
everywhere; thus
$V(x(t_2)) < V(x(t_1)) \tag 4$
along $x(t)$ whenever (1) binds.