Barbalat's lemma for Stability Analysis Good day,
We have: 
Lyapunov-Like Lemma: If a scaler function V(t, x) satisfies the
following conditions:


*

*$V(t,x)$ is lower bounded

*$\dot{V}(t,x)$ is negative semi-definite

*$\dot{V}(t,x)$ is uniformly continuous in time
then $\dot{V}(t,x) \to 0$ as $t \to \infty $.


Now if we have the following system:
$\dot{e} = -e + \theta w(t) \\
 \dot{\theta} = -e w(t)$
and assume that $w(t)$ is a bounded function, then we can select the following Lyapunov function:
$V(x,t) = e^2 + \theta^2$
Taking the time derivative:
$\dot{V}(x,t) = -2e^2 \leq 0$
Taking the time derivative again:
$\ddot{V}(x,t) = -4e(-e+\theta w)$
Now $\ddot{V}(x,t)$ satisfies condition (3) when $e$ and $\theta$ are bounded, but how can I be sure that these two variables are indeed bounded? Should I perform some other test, or...?
 A: Notice that you can decouple your system by writing the equations governing the "polar coordinates" $(e,\theta)=(R\cos\phi,R\sin\phi)$. In particular, (if my calculations are correct, and please comment if you find an error): \begin{align}\dot{\phi}= &\cos\phi \sin\phi -w ,\\\dot{R} =& -R \cos^2\phi .  \end{align}
Since $V=R^2$, you are trying to show that $\dot{V}=2R\dot{R}\to 0$ as $t\to \infty$. For an initial value problem for $(e,\theta)$ with $(e(0),\theta(0))\neq (0,0)$, we can choose a corresponding initial value problem for $(R,\phi)$ with $R(0)>0$. In such a case, on the interval of existence of a solution $R$ will be a monotone decreasing positive function (although potentially not strictly decreasing!). If either $\dot{R}\to 0$ or $R\to 0$ as $t\to \infty$ then $R\dot{R} \to 0$, since in the first case we know $|R|$ is bounded (as any positive decreasing function must be), and in the second case we have the formula $|R\dot{R}|=|-R^2\cos^2\phi|\leq R^2$. 
Finally, notice that for an equation like $\dot{y} = a(t) y$ with $a(t)\leq 0$ and $a$ continuous, either $a(t)\to 0$ as $t\to T$ for some $T\in [0,\infty]$ (note I include $\infty$ here) or else,  $a(t)<-m<0$ for all $t$ and by Gromwall's inequality $y(t)<y(0) e^{-mt}$. In particular, in this situation (assuming positive initial value) either $\dot{y}\to 0$ as $t\to T$ for some $T\in[0,\infty]$ or $y\to 0$ as $t\to \infty$. Since $\phi$ is meant to be a solution some (generally speaking) integral equation, it and $-\cos^2\phi$ is continuous, and we have that $\dot{V}\to 0$ as $t\to \infty$ or as $t\to T$ if there exists a values $T$ such that $\cos^2\phi(T)=0$.
There are some cautionary statements however. It's generally possible (although I'm not sure about this particular system), that $\dot{R}=0$ in finite time (like in our case, if $w=0$ and $\phi=\pi/2$). Solving explicitly for $R$ you have $R(t)=R_0 \exp(-\int_{0}^{t} \cos^2\phi(\tau)d\tau)$, so it's obvious that $R$ won't reach zero in finite time (with $\cos^2 \phi(\tau)$ being a bounded function, afterall). Also, I haven't shown anything about existence, but I have assumed existence throughout. 
EDIT: Let me sheepishly add that it's clear that $(e,\theta)$ are bounded since $V=e^2+\theta^2$ is decreasing in time. (So in particular, $e(t),\theta(t)\leq \sqrt{V(t)}\leq \sqrt{V(0)}$ for all $t\geq 0$).
A: Short answer. $V$ is not increasing, therefore $e$ and $\theta$ are bounded since their derivatives are well defined functions are $w$ is bounded.
A: Since $\dot{V} = -2e^2 \leq 0$, from the Lyapunov stability theory, one concludes that the system states $(e,\theta)$ are bounded. 
Observe above that $\dot{V}$ is a function of only one state ($e$); if it were a function of the two states $(e,\theta)$ and $\dot{V} < 0$ (except in $e=\theta = 0$, in that case $V=0$), one would conclude asymptotic stability. 
