Proof of asymptotic stability of quadrotor attitude error with nonlinear controller

Question

I'm looking at the paper "Multicopter attitude control for recovery from large disturbances" (pdf here).

There's a short proof of the asymptotic stability of the following skew-symmetric controller using Barbalat's lemma, on page 3: $$\alpha_e = -K_\omega \omega_e - \frac{1}{2}K_Rv(R_e - R_e^T) \\ = -K_\omega \omega_e - K_R\sin\rho_e\mathbf{n_e}$$

$R_d$ is the desired attitude and $R$ is the current attitude. $R_e$ is the error between these relative to the body frame, $R_e := R_d^TR_e$

$\mathbf{n_e}$ and $\rho_e$ are the axis and angle of rotation of $R_e$, so that $\omega_e = \dot{\rho_e}\mathbf{n_e}$, $\alpha = \ddot{\rho_e}\mathbf{n_e}$ and $\dot{R_e} = R_e[\omega_e]_\times$

$v$ is the inverse cross product matrix operator, so that $\mathbf{x} = v([\mathbf{x}]_\times)$.

I'm on board with the proof up to and including the conclusion that $\omega_e \rightarrow 0$ as $t \rightarrow \infty$. My trouble is that I can't find a proof for the assertion that this also implies that $R_e \rightarrow I$.

One possible answer is to show that $\omega_e \rightarrow 0$ implies $\alpha_e \rightarrow 0$, but I don't think we can just swap the derivative and limit of $\omega_e$ in this case. Another thing I tried is substituting $\omega_e = -K_\omega^{-1} \alpha_e - \frac{1}{2}K_\omega^{-1}K_Rv(R_e - R_e^T)$ into the Lyapunov function derivative $\frac{dJ}{dt}$ in the hopes that I would get something of the form $\alpha_e^TM\alpha_e + n_e^TNn_e$ from which the result should follow assuming positive definiteness of the matrix products, but there are multiple terms of the form $\alpha_e^TKn_e$ which get in the way of this approach. Any ideas on proceeding?

Answer 1

In order to conclude that for certain initial conditions it holds that $R_e\to I$ one can use the fact that $d/dt\,J^{SS}\leq0$ and by substituting the asymptotic result of $\omega_e\to0$ into the dynamics of $\omega_e$ itself, using the proposed control law (assuming that $\alpha_{e,des}^{SS}$ is the same as $\alpha_e$)

$$ \frac{d}{dt} \omega_e = -K_\omega\,\omega_e - K_R\,\sin\rho_e\,\mathbf{n}_e. \tag{1} $$

Namely in the limit of $t\to\infty$ the Lyapunov function, together with its two derivatives and Barbalat’s lemma, shows that $\omega_e\to0$ and $d/dt\,\omega_e\to0$. Substituting this into $(1)$ yields that as $t\to\infty$

$$ 0 = 0 - K_R\,\sin\rho_e\,\mathbf{n}_e, \tag{2} $$

which can only be true if $\sin\rho_e\to0$. When only considering $-\pi<\rho_e\leq\pi$ implies that either $\rho_e\to0$ or $\rho_e\to\pi$.

The author does skips a bit by saying that if $\rho_e\neq\pi$ then $\rho_e\to0$ and thus $R_e\to I$. Namely it can be shown that $\rho_e=\pi$ is a saddle point so, similar to an inverted pendulum, there are initial conditions (also with $\rho_e\neq\pi$) that still do converge to $\rho_e=\pi$. But from the authors argument it can de deduced that $\rho_e=0$ and $\omega_e=0$ is a locally asymptotically stable equilibrium point.

Answer 2

We can show that $\alpha_e$ is uniformly continuous:

$$\dot{\alpha_e} = -K_{\omega}\alpha_e - K_R \cos(\rho_e) \dot{\rho_e} \mathbf{n_e} = -K_{\omega}(-K_{\omega}\omega_e - K_R \sin(\rho_e)\mathbf{n_e}) - K_R \cos(\rho_e) \omega_e$$

$$\dot{\alpha_e} = K_{\omega}^2 \omega_e + K_{\omega} K_R \sin(\rho_e)\mathbf{n_e} - K_R \cos(\rho_e) \omega_e $$

$\omega_e$, $\sin(\rho_e)$ and $\cos(\rho_e)$ are all bounded, so $\dot{\alpha_e}$ is bounded and therefore $\alpha_e$ is uniformly continuous.

$\omega(t)$ has the finite limit $\omega(t) \rightarrow 0$ as $t \rightarrow \infty$ and so by Barbalat's lemma, $\alpha(t) \rightarrow 0$ .