## 3 Main Results
### Stability Analysis  
The properties of the main result are presented in Theorem 1. Theorem 1: _For the ship heading control system (3) with asymmetric dead-zone, via \(ADT\)\(\tau_{a}>(\log\mu/p_{2})\) with \(\mu=\max\{\frac{\lambda_{\max}(P_{0})}{\lambda_{\min}(P_{a})},k,q\in\Sigma\}\) and \(\sigma(t)\), the convergence of \(z_{1}\) is obtained under the proposed control scheme (29) and (39)-(41); meanwhile, the resulting signals are bounded in the probability._  
Before showing the analysis of Theorem 1, the following Lemma 3 is given to develop the proof, which can easily verify that the constructed compensating signals (28) and (38) are bounded. Lemma 3: _The system defined in (28) and (38) has states bounded by \(\|\vartheta(t)\|\leq\frac{s_{M}}{2\kappa_{0}}(1-e^{-2\kappa_{0}(t-T_{1})})\), then:_  
\[\lim_{t\to\infty}\|\vartheta(t)\|\leq\frac{s_{M}}{2\kappa_{0}}, \tag{43}\]  
_where \(\kappa_{0}=\frac{1}{2}\min_{i}(\kappa_{i}-1)\)._  
Remark 5: _The relevant proof is easily confirmed and thus is omitted here._  
Next, we begin the proof of Theorem 1 via the ADT approach. Proof. : Choosing appropriate parameters \(r,\
u_{0},\
u_{i2}(i=1,2)\) such that:  
\[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]  
defining:  
\[p_{2}= \min_{k\in\mathfrak{L}}\{4\kappa_{i},\frac{2c_{0}}{r\lambda_{\max }^{2}(P_{k})},\beta,i=1,2\},\] \[p_{1}= \max_{k\in\mathfrak{L}}\{\frac{r}{2\
u_{0}^{4}}(\theta^{2}+\xi_{ 0}^{4})+\frac{3}{2\
u_{0}}rn\sqrt{n}\omega^{2}+\frac{a^{2}}{2}+\frac{1}{4}t^{4} +\frac{\beta}{2\gamma}\theta^{2}\}>0,\]  
and defining \(V_{k}(X)\) as the whole stochastic Lyapunov with \(X=[\tilde{\varsigma}^{T},\tilde{\varsigma}^{T},\tilde{\theta}]\), then (42) can be rewritten as:  
\[\mathcal{L}V_{k}(X)\leq p_{1}-p_{2}V_{k}. \tag{45}\]  
Furthermore, it is easily obtained that \(\tilde{\upsilon}(\|X\|)\leq V_{k}(X)\leq\underline{\varrho}(\|X\|)\). Next, the following piecewise differentiable function is considered to develop the analysis:  
\[\mathcal{W}(t)=e^{p_{2}t}V_{\sigma(t)}(X(t)), \tag{46}\]  
and then, for \(k,q\in\mathfrak{L}\), it follows from the fact \(V_{k}(X(t))\leq\mu V_{q}(X(t))\), during \([t_{k},t_{k+1})\), that one furthermore obtains:  
\[E\{\mathcal{W}(t_{k+1})\}\leq\mu[E\{\int_{t_{k}}^{t_{k+1}}p_{1}e^{p_{2}t}dt\}+ E\{\mathcal{W}(t_{k})\}]. \tag{47}\]  
Then, integrating (47), we have:  
\[E\{\mathcal{W}(T^{-})\}\leq [E\{\mathcal{W}(0)\}+E\{\sum_{k=0}^{N_{\sigma}(T,0)-1}\mu^{-k} \int_{t_{k}}^{t_{k+1}}p_{1}e^{p_{2}t}dt\}\] \[+E\{\mu^{-N_{\sigma}(T,0)}\int_{t_{N_{\sigma}}(T,0)}^{T}p_{1}e^{ p_{2}t}dt\}]\mu^{N_{\sigma}(T,0)}. \tag{48}\]  
Then, \(\tau_{a}>(\log\mu/p_{2}-\delta)\) can be obtained by considering \(0<\delta<p_{2}-(\log\mu/\tau_{a})\) and ADT \(\tau_{a}>(\log\mu/p_{2})\). Combining it with (10), we have \(N_{\sigma(t)}(T_{2},T_{1})\leq\frac{(p_{2}-\delta)(T_{2}-T_{1})}{\log\mu}+N_{0}\). Taking \(k\) to be \(0,1,\cdots,N_{\sigma}(T,0)\), the following hold:  
\[E\{\int_{t_{k}}^{t_{k+1}}p_{1}e^{p_{2}t}dt\}\leq E\{e^{(p_{2}- \delta)t_{k+1}}\int_{t_{k}}^{t_{k+1}}p_{1}e^{\beta t}dt\}, \tag{49}\]  
and:  
\[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]0  
Considering \(\mathcal{W}(t)\) yields:  
\[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]1 \[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]2The following (52) holds:  
\[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]3  
where \(\tau_{a}>(\log\mu/p_{2})\) and \(0<e^{-\delta(T-t)}\leq 1\) are used. Therefore, according to Lemma 2, the boundedness can be concluded. Furthermore, for \(\forall T>0\), we obtain:  
\[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]4  
considering that \(\tau_{a}>(\log\mu/p_{2})\),  
\[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]5  
where \(e\) is a positive constant, and then, in accordance with (15) and Lemma 3, we can obtain:  
\[\Xi-\sum_{i=1}^{3}\frac{1}{4\
u_{12}^{A}}\geq c_{0}>0, \tag{44}\]6  
This completes the proof.