## Appendix B Proof of Proposition 1
### Equilibrium Solutions  
When \(\eta=\pm\eta_{d}\), then \(e_{\eta}=0\). By substituting \((e_{\eta},e_{\omega})=(0,0)\) into (19) and (17), we see that \((\eta(t),e_{\omega}(t))=(\pm\eta_{d}(t),0)\) indeed represent equilibrium solutions of the closed-loop dynamics. To see that all solution converge to either of these equilibria, consider the Lyapunov function candidate  
\[V(\eta,\eta_{d},e_{\omega})=k_{p}\Psi(\eta,\eta_{d})+\frac{1}{2}e_{\omega}^{ \top}Je_{\omega}\geq 0, \tag{20}\]  
whose time-derivative along the solutions of (19), (4) and (11) satisfies  
\[\dot{V}=k_{p}e_{\omega}^{\top}e_{\eta}+e_{\omega}^{\top}J\dot{e}_{\omega}=-e_ {\omega}^{\top}[K_{d}-V_{a}D+S(\omega_{d})J]e_{\omega}.\]  
By Assumption 4 and the gain condition (29) (inspired by [93]), we get  
\[\dot{V}\leq-\Big{(}\lambda_{\min}^{K_{d}}-\lambda_{\max}^{I}c_{\omega_{d}} \Big{)}\|e_{\omega}\|^{2}\leq-\gamma\|e_{\omega}\|^{2}\leq 0. \tag{21}\]  
From \(V\geq 0\) and \(\dot{V}\leq 0\) we get that \(e_{\omega}\) is bounded. In addition, the limit \(V_{\infty}=\lim\limits_{t\to\infty}V(t)\) exists and is finite ([94], Lemma 3.2.3). This means that \(\int_{t_{0}}^{\infty}\dot{V}d\tau=V_{\infty}-V(t_{0})\). Therefore, the function \(W=\gamma\|e_{\omega}\|^{2}\) satisfies \(\int_{t_{0}}^{\infty}Wd\tau\leq-\int_{t_{0}}^{\infty}\dot{V}d\tau=V(t_{0})-V_{ \infty}<\infty\), so the limit \(\lim\limits_{t\to\infty}\int_{t_{0}}^{t}Wd\tau\) exists and is finite. By definition, \(e_{\eta}\) is bounded. Since \(e_{\omega}\) is bounded, \(e_{\omega}\) is bounded. This follows from (19) and the boundedness of \(\omega_{d}\) and \(V_{a}\). Now, since \(\dot{W}=2\gamma e_{\omega}^{\top}e_{\omega}\) is bounded, it follows that \(W(t)\) is a uniformly continuous function. From [PERSON]'s Lemma ([94], Lemma 3.2.6), \(W(t)\) (and thus \(e_{\omega}(t)\)) converges to zero as \(t\to\infty\).  
To show convergence of \(e_{\eta}\) to zero we make use of the following lemma:  
**Lemma A1**.: _Let \(x(t)\) denote a solution to the differential equation \(\dot{x}=a(t)+b(t)\) with \(a(t)\) a uniformly continuous function. Assume that \(\lim_{t\to\infty}x(t)=c\) and \(\lim_{t\to\infty}b(t)=0\), with \(c\) a constant value. Then, \(\lim_{t\to\infty}\dot{x}(t)=0\)[48; 92]._  
From (19) we can write \(J\dot{e}_{\omega}=a(t)+b(t)\), where  
\[a(t) \triangleq-k_{p}e_{\eta}\] \[b(t) \triangleq-[K_{d}-V_{a}D+S(\omega_{d})J]e_{\omega}+S(J(e_{\omega} +\omega_{d}))e_{\omega}\]  
Since \(e_{\omega}\) converges to zero, we know that \(b(t)\) converges to zero. From (17), the derivative of \(a(t)\) is given by  
\[\dot{a}(t)=-k_{p}\dot{e}_{\eta}=k_{p}\Big{[}S(\omega_{d}^{\perp})e_{\eta}+S( \eta_{d})S(\eta)e_{\omega}\Big{]},\]  
which is bounded because \(e_{\eta},e_{\omega}\) and \(\omega_{d}^{\perp}\) are bounded. Therefore, \(a(t)\) is uniformly continuous, and convergence of \(e_{\eta}\) to zero follows from Lemma A1.  
To summarize, all solutions converge to one of the two equilibria given by \((\eta,e_{\omega})=(\pm\eta_{d},0)\).  
The Lyapunov function (A2) is quadratic, and as we will show, leads to exponential stability, which in general leads to good performance and robustness to perturbations [85]. In [95], it is shown that non-quadratic Lyapunov functions could lead to better performance. Therefore, future work based on the general approach presented in this paper might explore whether different Lyapunov functions than (A2), possibly non-quadratic ones, could lead to better performance.