## 2 Materials and Methods
### Stability Analysis
Let \(y(t)\) be a Filippovsolution to the differential inclusion \(\dot{y}\in K[h](y)\) for all \(t\in[t_{0},\infty)\), where \(K[\cdot]\) is defined in [42] and \(h:\mathbb{R}^{2n+q}\to\mathbb{R}^{2n+q}\) is defined as \(h(y)\triangleq\left[ \begin{array}{cc}t^{T},&\epsilon_{f}^{T},&\hat{\theta}^{ T}\end{array} \right]^{T}\). Consider times when \(\phi=w\) for any arbitrary \(w\in\mathcal{S}\) (i.e., \(t\in\left[t_{i}^{w,\text{om}},t_{i}^{w,\text{off}}\right)\) for some \(i\in\mathbb{N}\)). During these times, the dynamic functions \(Y_{\phi}\), \(\hat{\phi}_{\phi}\), and \(\hat{g}_{\phi}\) are continuous; thus, \(K[\hat{g}_{\phi}]=\hat{g}_{\phi}\) and \(K[\hat{Y}_{\phi}]=Y_{\phi}\). Taking the generalized time derivative of (19), using the calculus of \(K[\cdot]\) from [43], and using (6), (13), (16) and (17) yields  
\[ \begin{array}{rcl}\dot{V}_{L}&\subseteq&r^{T}\Big{(}K[\chi_{\sigma}]+Y_{ \phi}\bar{\theta}-(k_{3}+\alpha k_{2})r-\alpha k_{1}K[\text{sgn}(r)]+k_{2}e_{f }\Big{)}\\ &&+e_{f}^{T}\Big{(}-k_{1}K[\text{sgn}(r)]-k_{2}r-\beta e_{f}\Big{)}-\bar{ \theta}^{T}\Gamma^{-1}\Big{(}\Gamma Y_{\phi}^{T}r\Big{)}.\end{array}  \tag{26}\]  
Implementing (9) and (25) into (26), applying Young's Inequality, completing the squares, canceling common terms, and using the facts that \(\dot{V}_{L}(y)\stackrel{{\text{a.e. }}}{{\in}}\dot{V}_{L}(y)\), \(\|K[\text{sgn}(r)]\|\leq\sqrt{n}\), and \(\|y\|\geq\|z\|\) yields  
\[\dot{V}_{L}\ \ \ \stackrel{{\text{a.e.}}}{{\leq}}\ \ \ -k_{3}\|r\|^{2}-\Big{(}\beta-\tfrac{\alpha k_{1}}{2}\Big{)}\Big{\|}e_{f} \Big{\|}^{2}+\tfrac{1}{4\alpha k_{2}}\rho^{2}(\|y\|)\|z\|^{2}+\tfrac{nk_{1}}{2 \varepsilon}. \tag{27}\]  
Additionally, using (10) and (21), \(v\triangleq\delta v_{1}+\tfrac{nk_{1}}{2\varepsilon}\), and \(\delta\triangleq\min\Big{(}k_{3},\ \beta-\tfrac{\alpha k_{1}}{2}\Big{)}\) with (27) yields the following bound:  
\[\dot{V}_{L}\ \ \stackrel{{\text{a.e.}}}{{\leq}}\ \ \ -\delta V_{L}+v, \tag{28}\]  
\(\forall t\in\left[t_{i}^{w,\text{om}},t_{i}^{w,\text{off}}\right)\) provided that \(y(t)\in\mathcal{D}\), \(\forall t\in\left[t_{i}^{w,\text{om}},t_{i}^{w,\text{off}}\right)\). Since \(\dot{V}_{L}\) is bounded for any arbitrary \(\phi\in\mathcal{S}\), it can likewise be bounded for all \(\phi\in\mathcal{S}\). Hence, (19) is a common Lyapunov-like function across all \(t\in[t_{0},\infty)\), which means that (28) holds true for all \(t\in[t_{0},\infty)\), provided that \(y(t)\in\mathcal{D}\), \(\forall t\in[t_{0},\infty)\). The differential equation in (28) can be further solved to yield  
\[V_{L}(t)\leq\Big{(}V_{L}(t_{0})-\frac{v}{\bar{\delta}}\Big{)}\exp(-\delta(t-t_ {0}))+\frac{v}{\bar{\delta}}, \tag{29}\]  
\(\forall t\in[t_{0},\infty)\), provided that \(y(t)\in\mathcal{D}\), \(\forall t\in[t_{0},\infty)\). The solution in (24) in the theorem statement can be determined using (20) and (29). Provided that \(y(t_{0})\) lies within \(\mathcal{S}_{D}\) and the gain condition \(\sqrt{\frac{v}{\lambda_{1}\delta}}\leq\gamma\) is satisfied, it can be proven that \(y(t)\in\mathcal{D}\), \(\forall t\in[t_{0},\infty)\). It can be observed from (24) in the theorem statement that as \(t\to\infty\), the first term in (24) converges exponentially to \(0\) and the second term in (24) converges to \(\frac{v}{\lambda_{1}\delta}\). Therefore, the state vector \(y(t)\) converges to the ultimate bound \(\sqrt{\frac{v}{\lambda_{1}\delta}}\), which establishes a neighborhood about the origin defined by \(\{\,y\mid\|y\|\leq\sqrt{\frac{v}{\lambda_{1}\delta}}\}\) that will ultimately reside within. Moving on, the proper selection of control gains \(\alpha\), \(k_{1}\), \(k_{2}\), \(k_{3}\), \(\beta\), and \(\Gamma\) and the selectable constant \(\varepsilon\) can guarantee that \(y(t_{0})\) lies within \(\mathcal{S}_{D}\). The common Lyapunov-like function in (19) can be used with (29) to prove that \(r,e_{f},\bar{\theta}\in\mathcal{L}_{\infty}\). Noting that \(x_{d}\) is user-defined and is bounded and using (4) and (5), it can be shown that \(e,x\in\mathcal{L}_{\infty}\).