## Appendix D Proof of Proposition 3  
By combining (A4) with the control law (40), we now get the closed-loop error dynamics  
\[Jz=-k_{1}e_{\eta}-[K_{2}-V_{a}D+S(\bar{\omega}_{d})J]z+S(J(z+\bar{\omega}_{d}))z -\bar{\Delta}. \tag{46}\]  
This is similar to (A5), but with the extra estimation-error term \(-\bar{\Delta}\). Again, calculating \(V_{2}\) gives  
\[\dot{V}_{2}\leq-\kappa k_{1}\|e_{\eta}\|^{2}-\gamma\|z\|^{2}-z^{\top}\bar{ \Delta}.\]  
For some symmetric, positive definite gain matrix \(K_{3}=K_{3}^{\top}>0\), let an augmented Lyapunov function candidate be given by  
\[V_{3}=V_{2}+\frac{1}{2}\bar{\Delta}^{\top}K_{3}^{-1}\bar{\Delta}.\]  
Differentiating \(V_{3}\) gives  
\[\dot{V}_{3}=\dot{V}_{2}+\bar{\Delta}^{\top}K_{3}^{-1}\dot{\bar{\Delta}}\leq- \kappa k_{1}\|e_{\eta}\|^{2}-\gamma\|z\|^{2}+\bar{\Delta}^{\top}K_{3}^{-1}\Big{[} \dot{\bar{\Delta}}-K_{3}z\Big{]}.\]  
If the update law for \(\dot{\bar{\Delta}}\) is chosen as  
\[\dot{\bar{\Delta}}=K_{3}z,\]  
we are left with  
\[\dot{V}_{3}\leq-\kappa k_{1}\|e_{\eta}\|^{2}-\gamma\|z\|^{2}.\]  
By [PERSON]'s lemma, \(\dot{V}\) goes to zero asymptotically, and so does \(e_{\eta}\) and \(z\), and therefore also \(e_{\omega}\). From (46), we have \(J\dot{z}=a(t)+b(t)\), where  
\[a(t) \triangleq-\bar{\Delta}\] \[b(t) \triangleq-k_{1}e_{\eta}-[K_{2}-V_{a}D+S(\bar{\omega}_{d})f]z+S(J( z+\bar{\omega}_{d}))z\]  
Since \(e_{\eta}\) and \(z\) converges to zero, we know that \(b(t)\) converges to zero. The derivative of \(a(t)\) is given by  
\[a(t)=-\dot{\bar{\Delta}}=-K_{3}z\]  
which is bounded by the boundedness of \(z\). Therefore, \(a(t)\) is uniformly continuous. From Lemma A1, we get that \(\bar{\Delta}\) converges to zero as well. The rest of the proof follows closely that of Proposition 1.  
To show exponential stability, for \(\epsilon>0\), consider  
\[V_{4}=V_{3}+\epsilon\bar{\Delta}^{\top}Jz,\]  
whose time-derivative satisfies  
\[\dot{V}_{4}=\dot{V}_{3}+\epsilon\bar{\Delta}^{\top}J\dot{z}+\epsilon z^{\top} \dot{\bar{\Delta}}\leq-\kappa k_{1}\|e_{\eta}\|^{2}-\gamma\|z\|^{2}+\epsilon z ^{\top}JK_{3}z+\epsilon\bar{\Delta}^{\top}J\dot{z}\]  
From (46), we calculate the last term as  
\[\epsilon\bar{\Delta}^{\top}Jz =-\epsilon\bar{\Delta}^{\top}\bar{\Delta}+\epsilon\bar{\Delta}^{ \top}S(J(z+\bar{\omega}_{d}))z-\epsilon k_{1}\bar{\Delta}^{\top}e_{\eta}- \epsilon\bar{\Delta}^{\top}[K_{2}-V_{a}D+S(\bar{\omega}_{d})J]z\] \[\leq-\epsilon\|\bar{\Delta}\|^{2}+\epsilon\lambda_{\max}^{\max}(c _{\omega_{d}}+\kappa)\|z\|^{2}\|\bar{\Delta}\|+\epsilon k_{1}\|e_{\eta}\|\| \bar{\Delta}\|\] \[\quad+\epsilon\Big{(}\lambda_{\max}^{K_{2}}+\sigma_{\max}^{D(V_{ \alpha})}+\lambda_{\max}^{I}(c_{\omega_{d}}+\kappa)\Big{)}\|z\|\bar{\Delta}\|\]Now, for \(z\) such that \(\|z\|\leq c_{z}\), we can write \(\dot{V}_{4}\leq-\bar{x}^{\top}N\bar{x}\), where \(\bar{x}=[\|e_{\eta}\|\ \ \|z\|\ \|\ \tilde{\Delta}\|]^{\top}\) and the matrix \(N\) is given by  
\[N=\begin{bmatrix}\kappa k_{1}&0&-\frac{\kappa k_{1}}{2}\\ 0&\gamma-\epsilon\lambda^{\top}_{\max}\lambda^{K_{3}}_{\max}&-\frac{c_{\bar{x} }^{\top}}{2}\\ -\frac{\epsilon k_{1}}{2}&-\frac{\epsilon k_{2}^{\top}}{2}&\epsilon\end{bmatrix},\]  
where \(\xi=\lambda^{K_{2}}_{\max}+V_{\max}\sigma^{D}_{\max}+(1+c_{z})\lambda^{I}_{ \max}(c_{\omega_{d}}+\kappa)\). The parameter \(\epsilon\) can be chosen small enough such that \(N\) is positive definite. Requiring a positive determinant leads to a second-order inequality in \(\epsilon\) of the form \(ae^{2}-be+c>0\), where \(a,b,c\) are positive coefficients. Clearly, since \(c>0\), there exists some \(\epsilon\) (abritrarily small) that satisfies this inequality.