Adaptive fractional-order switching-type control method design for 3D fractional-order nonlinear systems

Abstract

In this paper, an adaptive sliding mode technique based on a fractional-order (FO) switching-type control law is designed to guarantee robust stability for uncertain 3D FO nonlinear systems. A novel FO switching-type control law is proposed to ensure the existence of the sliding motion in finite time. Appropriate adaptive laws are shown to tackle the uncertainty and external disturbance. The calculation formula of the reaching time is analyzed and computed. The reachability analysis is visualized to show how to obtain a shorter reaching time. A stability criterion of the FO sliding mode dynamics is derived based on indirect approach to Lyapunov stability. Advantages of the proposed control scheme are illustrated through numerical simulations.

Appendix: Proof of Lemma 2.1

For $\forall t > 0$, there exists a time interval $(t_k,t_{k + 1} ]$ such that $t \in (t_k,t_{k + 1} ]$ and $\sigma (t')\ge 0, \forall t' \in (t_k,t_{k + 1} ]$ if $\sigma (t)>0$, or $\sigma (t') \le 0, \forall t'\in (t_k,t_{k + 1} ]$ if $\sigma (t)<0$. Furthermore, there exists a finite partition given by $0=t_0<t_1<t_2<\cdots <t_{k-1 } < t_{k}$, such that: 1) for every interval $(t_i,t_{i + 1} ],(i=0,1,\cdots ,k-1)$, $\sigma (t') \le 0, \forall t'\in (t_i,t_{i + 1} ]$ or $\sigma (t') \ge 0, \forall t'\in (t_i,t_{i + 1} ]$; and 2) $\sigma (t') \le 0,\forall t' \in (t_{i+1} ,t_{i + 2} ]$ if $\sigma (t') \ge 0,\forall t' \in (t_{i },t_{i+1} ]$ or $\sigma (t') \ge 0,\forall t' \in (t_{i+1},t_{i + 2} ]$ if $\sigma (t') \le 0,\forall t' \in (t_{i },t_{i+1} ]$, for every two adjacent intervals $(t_{i },t_{i+1} ]$ and $(t_{i+1},t_{i + 2} ]$. Moreover, we require that the initial time in every $(t_i,t_{i+1}]$ is not equal to zero. From the integral properties, denoting $t_0=0$, one has

$$\begin{aligned} D_t^{\bar{\beta }} {\mathop {\mathrm{sgn}}} (\sigma (t))= & {} \frac{\left[ {f_0 (t) + f_1 (t) + \cdots + f_k (t)} \right] }{{\varGamma (1 - {\bar{\beta }} )}}, \end{aligned}$$

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where $f_i (t) = \frac{\mathrm{d}}{{\mathrm{d}t}}\int _{t_i }^{t_{i + 1} } {\frac{{{\mathop {\mathrm{sgn}}} (\sigma (\tau ))}}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau },(i = 0,1,2, \cdots ,k - 1)$, $f_k (t) = \frac{\mathrm{d}}{{\mathrm{d}t}}\int _{t_k }^t {\frac{{{\mathop {\mathrm{sgn}}} (\sigma (\tau ))}}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau }$.

First, we consider $\sigma (t)>0$. From the above analysis, one has $\sigma (t') \ge 0,\forall t' \in (t_k,t_{k + 1} ]$. There exists $t_k=t_{k0} <t_{k1} <t_{k2}< \cdots <t_{kl_k - 1} <t_{kl_k } =t$ in $(t_k,t]$ such that $(t_k,t] = (t_{k0},t_{k1} ] \cup (t_{k1} ,t_{k2} ] \cup \cdots \cup (t_{kl_k - 1},t_{kl_k } ]$. Moreover, $\sigma (t')\ge 0, \forall t' \in (t_{k0},t_{k1} ]$ in which certain $\sigma (t') = 0$ just happen at some isolate points $t'$; $\sigma (t')\equiv 0, \forall t' \in (t_{k1},t_{k2} ]$; $\sigma (t')\ge 0, \forall t' \in (t_{k2},t_{k3} ]$ in which certain $\sigma (t') = 0$ just happen at some isolate points $t'$; $\cdots$; $\sigma (t')\equiv 0, \forall t' \in (t_{kl_k-2},t_{kl_k-1} ]$; $\sigma (t')\ge 0, \forall t' \in (t_{kl_k-1},t_{kl_k} ]$ in which $\sigma (t') = 0$ just happen at isolate points $t'$. In addition, we also require that the initial time in $(t_{kj},t_{kj+1}],(j=0,1,\cdots ,l_{k}-1)$ is not zero. Thus, one has

$$\begin{aligned}&\!\!\!\frac{\mathrm{d}}{{\mathrm{d}t}}\int _{t_{kl_k - 1} }^t {\frac{{{\mathop {\mathrm{sgn}}} (\sigma (\tau ))}}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau } = (t - t_{kl_k - 1} )^{-{\bar{\beta }}},\nonumber \\&\!\!\! \frac{\mathrm{d}}{{\mathrm{d}t}}\int _{t_{kl_k - 2} }^{t_{kl_k - 1} } {\!\frac{{{\mathop {\mathrm{sgn}}} (\sigma (\tau ))}}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau }\\&\!\!\! = \frac{\mathrm{d}}{{\mathrm{d}t}}\int _{t_{kl_k - 2} }^{t_{kl_k - 1} } {\!\frac{0}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau } = 0, \nonumber \\&\!\!\! \frac{\mathrm{d}}{{\mathrm{d}t}}\int _{t_{kl_k - 3} }^{t_{kl_k - 2} } {\!\frac{{{\mathop {\mathrm{sgn}}} (\sigma (\tau ))}}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau } = (t - t_{kl_k - 3} )^{-{\bar{\beta }}}\\&\!\!\! - (t - t_{kl_k - 2} )^{-{\bar{\beta }}},\nonumber \end{aligned}$$

and so on, one can conclude

$$\begin{aligned} f_k (t)= & {} (t - t_{k0} )^{-{\bar{\beta }}} - (t - t_{k1} )^{-{\bar{\beta }}}+ (t - t_{k2} )^{-{\bar{\beta }}} \nonumber \\&- (t - t_{k3} )^{-{\bar{\beta }}} + \cdots +(t - t_{kl_k - 1} )^{-{\bar{\beta }}}. \end{aligned}$$

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Since $(t - t_{kj} )^{-{\bar{\beta }}}$ is an increasing function in $t_{kj}$, $f_k (t) \ge (t - t_{k} )^{-{\bar{\beta }}}$. Then, we discuss $f_i (t)$. When $\sigma (t') \ge 0,\forall t' \in (t_i,t_{i + 1} ],$ there exists a time partition $(t_i,t_{i + 1} ] = (t_{i0} ,t_{i1} ] \cup (t_{i1},t_{i2} ] \cup \cdots \cup (t_{il_i - 1} ,t_{il_i } ]$ in which $t_{i0} = t_i,t_{il_i } = t_{i + 1}$. Moreover, $\sigma (t')\ge 0, \forall t' \in (t_{i0},t_{i1} ]$ in which $\sigma (t') = 0$ just happen at some isolate points $t'$; $\sigma (t')\equiv 0, \forall t' \in (t_{i1},t_{i2} ]$; $\sigma (t')\ge 0, \forall t' \in (t_{i2},t_{i3} ]$ in which certain $\sigma (t') = 0$ just happen at some isolate points $t'$; and so on. We also claim that the initial time in $(t_{ij},t_{ij+1}],(j=0,1,\cdots ,l_{i}-1)$ is not zero. Considering $(t_{il_i - 1},t_{il_i } ]$, there are two possibilities (i.e., $\sigma (t') \ge 0,\forall t' \in (t_{il_i - 1},t_{il_i } ]$ or $\sigma (t') \equiv 0,\forall t' \in (t_{il_i - 1},t_{il_i } ]$). Hence, $f_i (t)$ can be calculated, similarly the calculation of $f_k (t)$, $f_i (t) \ge (t - t_i )^{-{\bar{\beta }}} - (t - t_{i + 1} )^{-{\bar{\beta }}}, \mathrm{or}\quad \!\!f_i (t) \ge (t - t_{i0} )^{-{\bar{\beta }}} - (t - t_{il_i -1} )^{-{\bar{\beta }}}$.

When $\sigma (t') \le 0,\forall t' \in (t_i,t_{i + 1} ]$, there exists $(t_i,t_{i + 1} ]$ such that $(t_i,t_{i + 1} ] = (t_{i0} ,t_{i1} ] \cup (t_{i1},t_{i2} ] \cup \cdots \cup (t_{il_i - 1} ,t_{il_i } ]$ in which $t_{i0} = t_i,t_{il_i } = t_{i + 1}$. Moreover, $\sigma (t')\le 0, \forall t' \in (t_{i0},t_{i1} ]$ in which $\sigma (t') = 0$ just happen at some isolate points $t'$; $\sigma (t')\equiv 0, \forall t' \in (t_{i1},t_{i2} ]$; $\sigma (t')\le 0, \forall t' \in (t_{i2},t_{i3} ]$ in which certain $\sigma (t') = 0$ just happen at some isolate points $t'$; and so on. We also claim that the initial time in $(t_{ij},t_{ij+1}],(j=0,1,\cdots ,l_{i}-1)$ is not zero. Considering $(t_{il_i - 1},t_{il_i } ]$, there are two possibilities (i.e., $\sigma (t') \le 0,\forall t' \in (t_{il_i - 1},t_{il_i } ]$ or $\sigma (t') \equiv 0,\forall t' \in (t_{il_i - 1},t_{il_i } ]$). Hence, one can conclude that

$$\begin{aligned} \frac{\hbox {d}}{{\mathrm{d}t}}\int _{t_{i0} }^{t_{i1} } {\frac{{{\mathop {\mathrm{sgn}}} (\sigma (\tau ))}}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau }= & {} (t - t_{i1} )^{-{\bar{\beta }}} - (t - t_{i0} )^{-{\bar{\beta }}},\nonumber \\ \end{aligned}$$

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$$\begin{aligned} \frac{\hbox {d}}{{\mathrm{d}t}}\int _{t_{i1} }^{t_{i2} } {\frac{{{\mathop {\mathrm{sgn}}} (\sigma (\tau ))}}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau }= & {} \frac{\hbox {d}}{{\mathrm{d}t}}\int _{t_{i1} }^{t_{i2} } {\frac{0}{{(t - \tau )^{\bar{\beta }} }}\mathrm{d}\tau } = 0,\nonumber \\ \end{aligned}$$

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and so on, one has

$$\begin{aligned} f_i (t)= & {} (t - t_{i1} )^{-{\bar{\beta }}} \!-\! (t - t_{i0} )^{-{\bar{\beta }}} \!+\! \cdots + (t - t_{il_i } )^{-{\bar{\beta }}}\\&-(t - t_{il_i - 1} )^{-{\bar{\beta }}} > 0, \end{aligned}$$

or

$$\begin{aligned} f_i (t)= & {} (t - t_{i1} )^{-{\bar{\beta }}} \!-\! (t - t_{i0} )^{-{\bar{\beta }}} \!+\! \cdots \!+\! (t \!-\! t_{il_i -1} )^{-{\bar{\beta }}}\nonumber \\&-(t - t_{il_i - 2} )^{-{\bar{\beta }}} > 0.\nonumber \end{aligned}$$

So, one can conclude $\sum \limits _{i = 0}^k {f_i (t)} > 0.$ Thus, we have

$$\begin{aligned} D^{\bar{\beta }} {\mathop {\mathrm{sgn}}} (\sigma (t)) = \frac{\left[ {f_0 (t) + f_1 (t) + \cdots + f_k (t)} \right] }{{\varGamma (1 - {\bar{\beta }})}}> 0. \end{aligned}$$

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Next, we consider the second case when $\sigma (t) < 0$. Similar to the first case, we have

$$\begin{aligned} D^{\bar{\beta }} {\mathop {\mathrm{sgn}}} (\sigma (t)) = \frac{\left[ {f_0 (t) + f_1 (t) + \cdots + f_k (t)} \right] }{{\varGamma (1 - {\bar{\beta }} )}} < 0. \end{aligned}$$

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