Consider a non-dimensional pendulum, $$ \frac{d^{2}}{d t^{2}} \theta{\left(t \right)} = - u \cos{\left(\theta{\left(t \right)} \right)} + \sin{\left(\theta{\left(t \right)} \right)}, $$ where $\theta$ is the angle from the vertical and $u$ is the control input. The pendulums energy is $$ E = \cos{\left(\theta{\left(t \right)} \right)} + \frac{\left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2}}{2} - 1 $$
We want to swing the pendulum to $\theta(t) = \frac{d}{d t} \theta{\left(t \right)} = 0$ so that $E = 0$.
Åström and Furuta provided the Lyapunov function $$ V = \frac{E^2}{2} = V = \frac{\left(\cos{\left(\theta{\left(t \right)} \right)} + \frac{\left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2}}{2} - 1\right)^{2}}{2}. $$ Taking the time-derivative of this, we get $$ \frac{d}{d t} V{\left(t \right)} = \frac{\left(- 2 \sin{\left(\theta{\left(t \right)} \right)} \frac{d}{d t} \theta{\left(t \right)} + 2 \frac{d}{d t} \theta{\left(t \right)} \frac{d^{2}}{d t^{2}} \theta{\left(t \right)}\right) \left(\cos{\left(\theta{\left(t \right)} \right)} + \frac{\left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2}}{2} - 1\right)}{2} $$ Substituting in $\frac{d^{2}}{d t^{2}} \theta{\left(t \right)}$ from the first equation, we get $$ \frac{d}{d t} V{\left(t \right)} = \frac{u \left(- 2 \cos{\left(\theta{\left(t \right)} \right)} - \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2} + 2\right) \cos{\left(\theta{\left(t \right)} \right)} \frac{d}{d t} \theta{\left(t \right)}}{2}. $$
Now let the control signal be $$ u = k \left(\cos{\left(\theta{\left(t \right)} \right)} + \frac{\left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2}}{2} - 1\right) \cos{\left(\theta{\left(t \right)} \right)} \frac{d}{d t} \theta{\left(t \right)}, $$ where $k > 0$ is a signal gain.
Substituting in the control signal, we get $$ \frac{d}{d t} V{\left(t \right)} = - \frac{k \left(2 \cos{\left(\theta{\left(t \right)} \right)} + \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2} - 2\right)^{2} \cos^{2}{\left(\theta{\left(t \right)} \right)} \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2}}{4}. $$ All the terms in $\frac{d}{d t} V{\left(t \right)}$ are squared, thus $$ V < 0 \quad \forall \theta{\left(t \right)}, \frac{d}{d t} \theta{\left(t \right)} ~~ \in \mathbb{R} \setminus \left\{0\right\}, $$ i.e. the control signal is globally asymptotically stabilizing.
Now put the controlled system into state-space form with $x = [\theta(t), \frac{d}{d t} \theta{\left(t \right)}]$: $$ \dot{x} = f(x) = \left[\begin{matrix}\frac{d}{d t} \theta{\left(t \right)} \\ \frac{d^{2}}{d t^{2}} \theta{\left(t \right)} \end{matrix}\right] $$
Now taking the Jacobian of $f$ with respect to $x$, we get: $$ \left[\begin{matrix}0 & 1\\\left(k \left(2 \cos{\left(\theta{\left(t \right)} \right)} + \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2} - 2\right) \sin{\left(\theta{\left(t \right)} \right)} \frac{d}{d t} \theta{\left(t \right)} + \frac{k \sin{\left(2 \theta{\left(t \right)} \right)} \frac{d}{d t} \theta{\left(t \right)}}{2} + 1\right) \cos{\left(\theta{\left(t \right)} \right)} & \frac{k \left(- 2 \cos{\left(\theta{\left(t \right)} \right)} - 3 \left(\frac{d}{d t} \theta{\left(t \right)}\right)^{2} + 2\right) \cos^{2}{\left(\theta{\left(t \right)} \right)}}{2}\end{matrix}\right]. $$
Plugging in the desired final state $x^{\star} = [0,0]$, we get: $$ \left[\begin{matrix}0 & 1\\1 & 0\end{matrix}\right], $$ whose Eigen values are $-1$ and $1$ (multiplicity of one). Thus, this is an unstable saddle point.
How do we reconcile that the controlled system is asymptotically stable to the point, but at the same time the point is a saddle point?