## 3 Main Results
### Stability of AUV System with Time-Varying Delays
\begin{bmatrix}\mathcal{O}_{1,1}^{c}&\mathcal{O}_{1,2}& \mathcal{O}_{1,3}&0&\mathcal{O}_{1,5}&\mathcal{O}_{1,6}&\mathcal{O}_{1,7}\\ *&\mathcal{O}_{2,2}&\mathcal{O}_{2,3}&\mathcal{O}_{2,4}&0&\mathcal{O}_{2,6}& \mathcal{O}_{2,7}\\ *&*&\mathcal{O}_{3,3}&\mathcal{S}&0&0&0\\ *&*&*&\mathcal{O}_{4,4}&0&0&0\\ *&*&*&*&-\gamma^{2}&\tau_{m}B_{w}^{T}&\delta B_{w}^{T}\\ *&*&*&*&*&-\mathcal{T}_{4}^{1}&0\\ *&*&*&*&*&*&-\mathcal{T}_{5}^{1}\end{bmatrix} \]  
\[\mathcal{O}_{1,1}^{c}=P\bar{A}+\bar{A}^{T}P+PBK+K^{T}B^{T}P+2\bar{b}\bar{x}_{a }PB+T_{1}+T_{2}+T_{3}-T_{4}+C^{T}C+\bar{\gamma}\bar{\chi}_{s},\;\;\mathcal{O} _{1,2}=P\bar{A}_{d},\mathcal{O}_{1,5}=PB_{w},\mathcal{O}_{1,6}=\tau_{m}(\bar{A }^{T}P+K^{T}B^{T}P),\mathcal{O}_{1,7}=\delta(\bar{A}^{T}P+K^{T}B^{T}P),\mathcal{ O}_{2,2}=-(1-\mu)T_{3}-2T_{5}+S+S^{T},\mathcal{O}_{2,3}=T_{5}-S^{T},\mathcal{O}_{2,4}=T_{5}-S,\mathcal{O}_{2,6}=\tau_{m}\bar{A}_{d}^{T}P,\mathcal{O}_{2,7}=\delta\bar{A}_ {d}^{T}P,\mathcal{O}_{3,3}=-T_{1}-T_{4}-T_{5},\mathcal{O}_{4,4}=-T_{2}-T_{5}.\]  
Pre- and post-multiplying both sides of \(\Omega\) by doing \(\{P^{1},P^{1},P^{1},P^{1},I,P^{1},I,P^{1},P^{1}\}>0\) and matrix variables denoted by \(P^{1}=\hat{P},\;\hat{P}\hat{T}_{1}\hat{P}=\hat{T}_{1},\;\hat{P}\hat{T}_{2}\hat {P}=\hat{T}_{2},\;\hat{P}\hat{T}_{3}\hat{P}=\hat{T}_{3},\;\hat{P}\hat{T}_{4} \hat{P}=\hat{T}_{4},\;\hat{P}\hat{T}_{5}\hat{P}=\hat{T}_{5},\;\hat{P}\hat{S} \hat{P}=\hat{S},K\hat{  
It is noted that (23) is not an LMI condition because of the term \(-\hat{P}\hat{T}_{k}^{-1}\hat{P},k=4,5\). In view of the inequality \(-\hat{P}\hat{T}_{k}^{-1}\hat{P}\leq-2\rho\hat{P}+\rho^{2}\hat{T}_{k},(\hat{T}_{k }>0,k=4,5)\), we have:  
\[[\hat{\Omega}]_{7\times 7}= \begin{bmatrix}\hat{O}_{1,1}^{c}&\hat{O}_{1,2}&\hat{O}_{1,3 }&0&\hat{O}_{1,5}&\hat{O}_{1,6}&\hat{O}_{1,7}\\ *&\hat{O}_{2,2}&\hat{O}_{2,3}&\hat{O}_{2,4}&0&\hat{O}_{2,6}&\hat{O}_{2,7}\\ *&*&\hat{O}_{3,3}&\hat{S}&0&0&0\\ *&*&*&\hat{O}_{4,4}&0&0&0\\ *&*&*&*&-\gamma^{2}&\tau_{m}B_{m}^{T}&\delta B_{m}^{T}\\ *&*&*&*&-2\rho\hat{P}+\rho^{2}\hat{T}_{4}&0\\ *&*&*&*&*&-2\rho\hat{P}+\rho^{2}\hat{T}_{5}\end{bmatrix} .\]  
Now, separating the measured output and uncertainties from \(\bar{A}\) and \(\bar{A}_{d}\) in \(\hat{\Omega}\), we have:  
\[[\hat{\Omega}]_{20\times 20} = [\hat{\Omega}_{1}]_{8\times 8}+\sum_{i=1}^{6}(\mathcal{M}_{i} \Delta(t)\mathcal{N}_{i}+\mathcal{N}_{i}^{T}\Delta(t)^{T}\mathcal{M}_{i}^{T}),\]  
where  
\[\mathcal{M}_{1}= \begin{bmatrix}\sqrt{\bar{\alpha}}\hat{P}M_{1}\\ [0]_{7\times 1}\end{bmatrix} ,\mathcal{N}_{1}= \begin{bmatrix}\sqrt{\bar{ \alpha}}N_{1}^{T}\\ [0]_{7\times 1}\end{bmatrix} ,\mathcal{M}_{2}= \begin{bmatrix}\sqrt{\bar{ \alpha}}\hat{P}M_{2}\\ [0]_{7\times 1}\end{bmatrix} ,\mathcal{N}_{2}= \begin{bmatrix}0\\ \sqrt{\bar{\alpha}}N_{2}^{T}\\ [0]_{6\times 1}\end{bmatrix} ,\]  
\[\mathcal{M}_{3}= \begin{bmatrix}\tau_{m}\sqrt{\bar{\alpha}}\hat{T}_{4}M_{1}\\ [0]_{7\times 1}\end{bmatrix} ,\mathcal{N}_{3}= \begin{bmatrix}[0]_{5\times 1} \\ \tau_{m}\sqrt{\bar{\alpha}}N_{1}^{T}\\ [0]_{2\times 1}\end{bmatrix} ,\mathcal{M}_{4}= \begin{bmatrix}\delta\sqrt{\bar{ \alpha}}\hat{T}_{5}M_{1}\\ [0]_{7\times 1}\end{bmatrix} ,\mathcal{N}_{4}= \begin{bmatrix}[0]_{6\times 1} \\ \delta\sqrt{\bar{\alpha}}N_{1}^{T}\\ 0\end{bmatrix} ,\]  
\[\mathcal{M}_{5}= \begin{bmatrix}0\\ \tau_{m}\sqrt{\bar{\alpha}}\hat{T}_{4}M_{2}\\ [0]_{6\times 1}\end{bmatrix} ,\mathcal{N}_{5}= \begin{bmatrix}[0]_{5\times 1} \\ \tau_{m}\sqrt{\bar{\alpha}}N_{2}^{T}\\ [0]_{2\times 1}\end{bmatrix} ,\mathcal{M}_{6}= \begin{bmatrix}0\\ \delta\sqrt{\bar{\alpha}}\hat{T}_{5}M_{2}\\ [0]_{6\times 1}\end{bmatrix} ,\mathcal{N}_{6}= \begin{bmatrix}[0]_{6\times 1} \\ \delta\sqrt{\bar{\alpha}}N_{2}^{T}\\ 0\end{bmatrix}