Adaptive neural output feedback control for stochastic nonlinear time-delay systems with unknown control directions

Abstract

This paper first focuses on the problem of adaptive output feedback stabilization for a more general class of stochastic nonlinear time-delay systems with unknown control directions. By using a linear state transformation, the original system is transformed to a new system for which control design becomes feasible. Then a novel adaptive neural network (NN) output feedback control strategy, which only contains one adaptive parameter, is developed for such systems by combining the input-driven filter design, the backstepping technique, the NN’s parameterization, the Nussbaum gain function method and the Lyapunov–Krasovskii approach. The proposed control design guarantees that all signals in the closed-loop systems are 4-moment (or 2-moment) semi-globally uniformly bounded. Finally, two simulation examples are given to demonstrate the effectiveness and the applicability of the proposed control design.

Appendix

The Proof of Lemma 1 is given as follows

Proof

Firstly, we set

$$\begin{aligned} W(t,x)=V(t,x)e^{C_1 t}. \end{aligned}$$

(62)

It is easy to be obtained

$$\begin{aligned} \left. {EW(s,x(s))} \right| _0^t =E\int _0^t {\mathcal {L}W(s,x(s))\hbox{d}s}, \quad \forall \;\;t\in [0,t_f) \end{aligned}$$

(63)

Which is expressed, by (62), in the form

$$\begin{aligned} E\int _0^t {[C_1 V(s,x(s))e^{C_1 s}+\mathcal {L}V(s,x(s))e^{C_1 s}]\hbox{d}s} \end{aligned}$$

(64)

By the use of (4), this is bounded above by

$$\begin{aligned} E\int _0^t {(N(\zeta )\dot{\zeta }+C_2 )e^{C_1 s}\hbox{d}s} \le \frac{C_2 }{C_1 }(e^{C_1 t}-1)+\int _0^t {N(\zeta )\dot{\zeta }e^{C_1 s}\hbox{d}s}, \quad \forall \;\;t\in [0,t_f) \end{aligned}$$

(65)

From (63), we have

$$\begin{aligned} EW(t,x(t))\le EW(0,x(0))+\frac{C_2 }{C_1 }(e^{C_1 t}-1)+\int _0^t {N(\zeta )\dot{\zeta }e^{C_1 s}\hbox{d}s}, \quad \forall \;\;t\in [0,t_f ) \end{aligned}$$

(66)

By recalling that $W(t,x)=V(t,x)e^{C_1 t}$, for any $t\in [0,t_f)$, we obtain

$$\begin{aligned} EV(t,x(t))&\le EV(0,x(0))e^{-C_1 t}+\frac{C_2 }{C_1 }(1-e^{-C_1 t})+e^{-C_1 t}\int _0^t {N(\zeta )\dot{\zeta }e^{C_1 s}\hbox{d}s}\nonumber \\&\le EV(0,x(0))+\frac{C_2 }{C_1 }+\int _0^t {N(\zeta )\dot{\zeta }e^{-C_1 (t-s)}\hbox{d}s} \end{aligned}$$

(67)

Rewrite (67) as

$$\begin{aligned} 0 \le EV(t,x(t)) \le Const + \int _0^t {\left| {N(\zeta )\dot{\zeta }} \right| } \hbox{d}s,\;\forall t \in [0,t_f ) \end{aligned}$$

(68)

where $Const = EV(0,x(0)) + \frac{{C_2 }}{{C_1 }}$.

Depending on the signs of $N(\zeta )$ and $\dot{\zeta }$, (68) can be further written as

$$\begin{aligned} 0 \le EV(t,x(t)) \le Const1 \pm \int _0^{\zeta (t)} {N(\zeta )\hbox{d}\zeta },\;\;\forall t \in [0,t_f ) \end{aligned}$$

(69)

where $Const1 = Const + \int _{\zeta (0)}^0 {\left| {N(\zeta )} \right| \hbox{d}\zeta }$.

In the following, we first show that $\zeta (t)$ is bounded on $[0,t_f )$. It needs to consider the following two cases: (a) $\zeta (t)$ have upper bound, and (b) $\zeta (t)$ have lower bound.

Cases (a): Suppose that $\zeta (t)$ has no upper bounds on $[0,t_f )$. According to the properties of Nussbaum-type function, there must exist two monotone increasing sequences $\{\zeta _n^{(j)} \}$ with $\zeta _1^{(j)} > \left| {\zeta (0)} \right|$ and $\mathop {\lim }\nolimits _{n \rightarrow \infty } \zeta _n^{(j)} = \infty$, $j=1,2$, such that

$$\begin{aligned}& \mathop {\lim }\limits _{n \rightarrow \infty } \sup \frac{1}{{\zeta _n^{(1)} }}\int _0^{\zeta _n^{(1)} } {N(\zeta )} \hbox{d}\zeta = + \infty \end{aligned}$$

(70)

$$\begin{aligned}&\mathop {\lim }\limits _{n \rightarrow \infty } \inf \frac{1}{{\zeta _n^{(2)} }}\int _0^{\zeta _n^{(2)} } {N(\zeta )} \hbox{d}\zeta = - \infty \end{aligned}$$

(71)

Since $\zeta (t)$ has no upper bounds on $[0,t_f )$, thus, there exist monotone increasing sequences ${t_n^{(j)}}, j=1,2$ such that

$$\begin{aligned} \left\{ {\begin{array}{l} {\zeta (t_n^{(j)} ) = \zeta _n^j ,\;\;j = 1,2,\;\;n = 1,2,...} \\ {\mathop {\lim }\limits _{n \rightarrow \infty } t_n^{(j)} = t_f ,\;j = 1,2} \\ \end{array}} \right. \end{aligned}$$

(72)

Dividing (69) by $\zeta (t_n^{(j)} ) = \zeta _n^{(j)},\;j = 1,2$, we obtain

$$\begin{aligned}&0\le \frac{{EV(t_n^{(1)},x(t_n^{(1)}))}}{{\zeta (t_n^{(1)} )}} \le \frac{{Const1}}{{\zeta (t_n^{(1)} )}} \pm \frac{{\int _0^{\zeta _n^{(1)}} {N(\zeta )\hbox{d}\zeta } }}{{\zeta (t_n^{(1)} )}}\end{aligned}$$

(73)

$$\begin{aligned}&0 \le \frac{{EV(t_n^{(2)},x(t_n^{(2)}))}}{{\zeta (t_n^{(2)} )}} \le \frac{{Const1}}{{\zeta (t_n^{(2)} )}} \pm \frac{{\int _0^{\zeta _n^{(2)}} {N(\zeta )\hbox{d}\zeta } }}{{\zeta (t_n^{(2)} )}} \end{aligned}$$

(74)

Thus, if $N(\zeta ) > 0$, (74) contradicts (71) and when $N(\zeta ) < 0$, (73) contradicts (70). Therefore, $\zeta (t)$ has upper bound on $[0,t_f )$.

Case (b): the following procedure will prove that $\zeta (t)$ have lower bounds on $[0,t_f )$. Firstly, we assume that $\zeta (t)$ have no lower bounds on $[0,t_f )$. According to the properties of Nussbaum-type function, there exist two monotone increasing sequences $\{ t_n^{(j)} \}$, $j=1,2$, such that $\zeta (t_n^{(j)} ) = - \zeta _n^{(j)},\;j = 1,2$. Obviously, $\mathop {\lim }\limits _{n \rightarrow \infty } t_n^{(j)} = t_f,\;j = 1,2$. Noting that the function $N(\cdot )$ is even, (69) can be further written as

$$\begin{aligned} 0 \le EV(t,x(t)) \le Const1 \mp \int _0^{ - \zeta (t)} {N(\zeta )\hbox{d}\zeta },\;\;\;4\forall t \in [0,t_f ) \end{aligned}$$

(75)

Dividing (75) by $\zeta (t_n^{(j)} ) = -\zeta _n^{(j)},\;j = 1,2$, we obtain

$$\begin{aligned}&0 \le \frac{{EV(t_n^{(1)},x(t_n^{(1)} ))}}{{ - \zeta (t_n^{(1)} )}} \le \frac{{Const1}}{{\zeta _n^{(1)} }} \mp \frac{{\int _0^{\zeta _n^{(1)}} {N(\zeta )\hbox{d}\zeta } }}{{\zeta _n^{(1)} }}\end{aligned}$$

(76)

$$\begin{aligned}&0 \le \frac{{EV(t_n^{(2)},x(t_n^{(2)} ))}}{{ - \zeta (t_n^{(2)} )}} \le \frac{{Const1}}{{\zeta _n^{(2)} }} \mp \frac{{\int _0^{\zeta _n^{(2)}} {N(\zeta )\hbox{d}\zeta } }}{{\zeta _n^{(2)} }} \end{aligned}$$

(77)

where $n=1,2,\ldots$. Noting that $t_n^{(j)} \rightarrow t_f$, $\zeta _n^{(j)} \rightarrow \infty, j=1,2$ when $n \rightarrow \infty$. Similarly, a contradiction can be found no matter what the sign of $N(\zeta )$ is. Therefore, $\zeta (t)$ has lower bound on $[0,t_f )$.

From the above analysis, we thus conclude the boundedness of $\zeta (t)$ on $[0,t_f )$, which in turns shows that $\int _0^t {N(\zeta )\dot{\zeta }\hbox{d}s}$ and $EV(t,x(t))$ must be bounded on $[0,t_f )$.