## 3 Adaptive Robust Control Design
### Deign of Adaptive Robust Control for Delta Robot
\begin{array}{l}If\quad\|\dot{r}+Ar\|\left\| \frac{\partial\Gamma}{\partial\rho}(\hat{\rho},\delta,\dot{\delta},t)\right\| >\epsilon,\\ v_{1}\left\|\dot{r}+Ar\right\|\frac{\partial\Gamma}{\partial\rho}(\hat{\rho}, \delta,\dot{\delta},t)-\left(v_{2}e^{-\|t+Ar\|}+v_{3}\right)\hat{\rho};\\ If\quad\|\dot{r}+Ar\|\left\|\frac{\partial\Gamma}{\partial\rho}(\hat{\rho}, \delta,\dot{\delta},t)\right\|\leq\epsilon,\\ -\left(v_{2}e^{-\|t+Ar\|}+v_{3}\right)\hat{\rho},\end{array} \right. \tag{24}\]  
where \(v_{i}\in\mathbf{R}^{k\times k}\), each element in \(v_{i}\) is non-negative, \(i=1,2,3\), \(\hat{\rho}_{i}(t_{0})>0\) (\(\hat{\rho}_{i}\) is the \(i\)-th element in \(\hat{\rho}\)), \(i=1,2,\ldots,k\), \(\epsilon\in\mathbf{R}\), \(\epsilon>0\). **Remark 6**. : _The adaptive law (24) and (25) is constructed to mimic the bound of \(\rho\) (\(\hat{\rho}\) represents the estimated value of \(\rho\)), which possesses two types: the leakage term and the dead-zone. The first term in the (24) is designed to be non-negative to compensate the uncertainty in Delta robot. The second term in the (24), which is referred to as the leakage term, is designed to be a negative exponential form to make \(\hat{\rho}\) decay to zero. Notice that \(\hat{\rho}_{i}(t)>0\) when \(\hat{\rho}_{i}(t_{0})\) is chosen to be strictly positive for all \(t>t_{0}\), \(i=1,2,\ldots,k\). The dead-zone portion (25) is actually an option, when it combines with (24), the adaptive law will simplify the calculation and the algorithm._  
**Theorem 1**. : _Let \(\varrho=\left[r^{T},\dot{r}^{T},(\hat{\rho}-\rho)^{T}\right]^{T}\in\mathbf{R} ^{6+k}\). Based on Assumptions 1-4, the proposed control (20) for Delta robot (3) renders \(\varrho\) the deterministic performance [35]:_  
_(i) Uniform boundedness: For any \(y>0\) with \(\|\varrho(t_{0})\|\leq y\), there exists a \(d(y)>0\) such that \(\|\varrho(t)\|\leq d(y)\) for all \(t\geq t_{0}\);_  
_(ii) Uniform ultimate boundedness: For any \(y>0\) with \(\|\varrho(t_{0})\|\leq y\), there exists a \(\underline{d}>0\) such that \(\|\varrho(t)\|\leq\bar{d}\) for any \(\bar{d}>\underline{d}\) as for all \(t\geq t_{0}+T(\bar{d},y)\), where \(T(\bar{d},y)<\infty\)._Proof of Theorem 1.: The Lyapunov function is given by  
\[V=\frac{1}{2}(\dot{r}+Ar)^{T}M(\dot{r}+Ar)+\frac{1}{2}r^{T}(B+AD)r+\frac{1}{2}( \dot{\rho}-\rho)^{T}(v_{1})^{-1}(\dot{\rho}-\rho). \tag{26}\]  
The function \(V\) is needed to be (globally) positive definite and decrescent. By Assumption 1, we have  
\[V \geq \frac{1}{2}\sigma\|\dot{r}+Ar\|^{2}+\frac{1}{2}r^{T}(B+AD)r+\frac{ 1}{2}(\dot{\rho}-\rho)^{T}(v_{1})^{-1}(\dot{\rho}-\rho) \tag{27}\] \[= \frac{1}{2}\sigma\sum_{i=1}^{n}\left(\dot{r}_{i}^{2}+2a_{i}\dot{r }_{i}r_{i}+a_{i}^{2}r_{i}^{2}\right)+\frac{1}{2}\sum_{i=1}^{n}(b_{i}+a_{i}d_{i })r_{i}^{2}+\frac{1}{2v_{1}}\sum_{i=1}^{n}(\dot{\rho}_{i}-\rho_{i})^{2}\] \[= \frac{1}{2}\sum_{i=1}^{n}[r_{i}\quad\dot{r}_{i}\quad\dot{\rho}_{ i}-\rho_{i}]\Theta_{i}[r_{i}\quad\dot{r}_{i}\quad\dot{\rho}_{i}-\rho_{i}]^{T}\] \[\geq \frac{1}{2}\sum_{i=1}^{n}\lambda_{\min}(\Theta_{i})\left[r_{i}^{ 2}+\dot{r}_{i}^{2}+(\dot{\rho}_{i}-\rho_{i})^{2}\right]\] \[\geq \underline{\Psi}\|\varrho\|^{2}.\]  
Here, \(\underline{\Psi}=\frac{1}{2}\sum_{i=1}^{n}\lambda_{\min}(\Theta_{i})\), and  
\[\Theta_{i}= \begin{bmatrix}\sigma a_{i}^{2}+b_{i}+a_{i}d_{i}&\sigma a_{i}&0\\ \sigma a_{i}&\sigma&0\\ 0&0&(v_{1})^{-1}\end{bmatrix} . \tag{28}\]  
Since \(\Theta_{i}>0,\forall i\), we prove that \(V\) is a positive definite function. By Assumption 2, we have  
\[V\leq \frac{1}{2}\kappa_{0}\|\dot{r}+Ar\|^{2}+\frac{1}{2}r^{T}(B+AD)r+ \frac{1}{2}(\dot{\rho}-\rho)^{T}(v_{1})^{-1}(\dot{\rho}-\rho). \tag{29}\]  
For the first term on the right-hand side,  
\[\frac{1}{2}\kappa_{0}\|\dot{r}+Ar\|^{2}\] \[= \frac{1}{2}\kappa_{0}(\dot{r}+Ar)^{T}(\dot{r}+Ar)\] \[= \frac{1}{2}\kappa_{0} \begin{bmatrix}r&\dot{r}\end{bmatrix} \begin{bmatrix}A^{2}&A\\ A&1\end{bmatrix}\begin{bmatrix}r\\ \dot{r}\end{bmatrix} \] \[\leq \frac{1}{2}\kappa_{0}\bar{A}\|\underline{r}\|^{2}, \tag{30}\]  
where \(\bar{A}=\lambda_{\max}\left(