SCIENCE CHINA Information Sciences, Volume 64 , Issue 9 : 192201(2021) https://doi.org/10.1007/s11432-020-2979-0

## PID control of uncertain nonlinear stochastic systems with state observer

• ReceivedFeb 9, 2020
• AcceptedMay 29, 2020
• PublishedAug 23, 2021
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### Acknowledgment

This work was supported by the Youth Scholars Fund of Beijing Technology and Business University (Grant No. PXM2018_014213_000033) and National Natural Science Foundation of China (Grant No. 61973329). The authors would like to thank Professor Lei GUO from Academy of Mathematics and Systems Science, Chinese Academy of Sciences, for valuable discussion on PID control of nonlinear stochastic systems.

### Supplement

Appendix

Definition A1 (17) . Let $(\Omega,~\mathcal{F},P)$ be a probability space. A filtration is a family of $\sigma$-algebra $\{\mathcal{F}_t\}_{t\geq~0}$ satisfying $\mathcal{F}_{t}\subset~\mathcal{F}_{s}\subset~\mathcal{F}$ for all $0\leq~t<s<\infty$. The filtration is considered to be right continuous if $\mathcal{F}_{t}=\bigcap_{s>t}\mathcal{F}_{s}$ for all $t\geq~0$. When the probability space is complete, the filtration is considered to satisfy the usual conditions if it is right continuous and $\mathcal{F}_{0}$ contains all $P$-null sets.

We consider the following stochastic system: \begin{equation*} \text{d}x(t)=f(x(t),t)\text{d}t+\sigma(x(t),t)\text{d}B(t), \eqno \text{(A1)}\end{equation*} where $x~\in~\mathbb{R}^{n}$ is the state of the system, $f~\in~\mathbb{R}^{n}$, $\sigma~\in~\mathbb{R}^{n}$, and $B(t)$ is (standard one-dimensional) Brownian motion. The $f$ term is referred to as a drift or vector field, and the noise term $\sigma~\text{d}B({t})$ is an uncertainty model. The uncertainty of this model could be caused by external random influences or by fluctuating coefficients and parameters in a mathematical model. The $\sigma$ function is referred to as a diffusion coefficient.

Itô's formula 17 . We denote $C^{2,1}(\mathbb{R}^n\times\mathbb{R}^{+},\mathbb{R}^{+})$ as the space of all nonnegative functions $V(x,t)$ defined on $\mathbb{R}^n\times\mathbb{R}^{+}$ that are continuously twice differentiable in $x$ and once in $t$. We define the differential operator ${L}$ associated with (A1) as follows: $${L}=\frac{\partial}{\partial t}+\sum\limits_{i = 1}^n {f_{i}(x,t)\frac{\partial}{\partial x_{i}}}+\frac{1}{2} \sum\limits_{i,j = 1}^n [\sigma(x,t)\sigma^{\text{T}}(x,t)]_{ij} \frac{\partial^{2}}{\partial x_{i}\partial x_{j}}.$$ If ${L}$ acts on function $V~\in~C^{2,1}(\mathbb{R}^n\times\mathbb{R}^{+},\mathbb{R}^{+})$, then $${L}V(x,t)=\frac{\partial V}{\partial t}(x,t)+f^{\text{T}}\triangledown V(x,t)+\frac{1}{2}\text{Tr}[\sigma\sigma ^{\text{T}}{H}(V)](x,t).$$ By Itô's formula, we can obtain $$\text{d}V(x(t),t)={L}V(x,t)\text{d}t+(\triangledown V(x(t),t))^{\text{T}}\sigma(x(t),t)\text{d}B(t),$$ where $\triangledown~V$ is the gradient of $V$, ${H}(V)=V_{x_{i}x_{j}}$ is the $n~\times~n$ Hessian matrix of $V$, and $\text{Tr}(A)$ denotes the trace of a matrix $A$.

Lemma A1 (Barbalat) . Assume that function $f:\mathbb{R}^{+}\rightarrow~\mathbb{R}$ is uniformly continuous and $\lim_{t~\to~\infty}~\int_0^t~f(\tau)~\text{d}\tau$ exists and is finite. Then $$\lim_{t \to \infty}f(t)=0.$$

See Lemma A.6 in 20 for a detailed discussion.

Theorem A1. Let $p\geq~2$ and $x_{0}\in~L^{p}(\Omega;\mathbb{R}^{d})$. Assume there exists a constant $\alpha>0$ such that, for all $(x,t)\in~\mathbb{R}^{d}\times~[t_{0},T]$: $$x^{\text{T}}f(x,t)+\frac{p-1}{2}|\sigma(x,t)|^{2}\leq \alpha(1+|x|^{2}).$$ Then, for the solution of (A1) on $t\in~[t_{0},~T]$, we obtain the following: $$\mathbb{E}|x(t)|^{p}\leq 2^{\frac{p-2}{2}}(1+\mathbb{E}|x_{0}|^{p}){\rm e}^{p\alpha (t-t_{0})}.$$

See Theorem 4.1 in 16 for a detailed discussion.

Lemma B1. Let the following linear growth condition hold for all $(x,t)\in\mathbb{R}^{d}\times[t_{0},\infty)$: \begin{equation*} \|f(x,t)\|\vee \|\sigma(x,t)\| \leq K\|x\|, \eqno \text{(B1)}\end{equation*} and let $x(t)$ be a solution to SDE (A1) on $[t_0,\infty)$. Let $h(t)=E\|x(t)\|^2$, and assume that $\sup_{t\geq~t_0}~h(t)<\infty$. Then, $h(t)$ is a uniformly continuous function of $t$ in $[t_0,\infty)$.

Proof. Let $C=\sup_{t\geq~t_0}~h(t)$, and assume that $t_0\le~t_1<t_2$. Then, we obtain the inequality $|h(t_2)-h(t_1)|\le~\mathbb{E}|\,\|x(t_2)\|^2-\|x(t_1)\|^2|$.

According to the Schwarz inequality, we obtain $$\mathbb{E}\big|\|x(t_2)\|^2-\|x(t_1)\|^2\big|\le \sqrt{\mathbb{E}(\|x(t_2)\|+\|x(t_1)\|)^2}\sqrt{\mathbb{E}(\|x(t_2)\|-\|x(t_1)\|)^2}.$$

Therefore, it is easy to see that \begin{equation*} \mathbb{E}(\|x(t_2)\|+\|x(t_1)\|)^2\le 2(\mathbb{E}\|x(t_2)\|^2+\mathbb{E}\|x(t_1)\|^2)\le 4\sup_{t\geq t_0} h(t)=4C, \eqno \text{(B2)}\end{equation*} where the RHS of (B2) is a constant that is independent of $t_1,t_2$.

In addition, we can obtain the following: \begin{align*}\mathbb{E}(\|x(t_2)\|-\|x(t_1)\|)^2&\le \mathbb{E}(\|x(t_2)-x(t_1)\|^2)=\mathbb{E}\left\|\int_{t_1}^{t_2} f(x(t),t)\text{d}t+\int_{t_1}^{t_2}\sigma(x(t),t)\text{d}B(t)\right\|^2 \\ &\le 2\left(\mathbb{E}\left\|\int_{t_1}^{t_2} f(x(t),t)\text{d}t\right\|^2+\mathbb{E}\left\|\int_{t_1}^{t_2}\sigma(x(t),t)\text{d}B(t)\right\|^2\right). \end{align*} From (B1), it is easy to obtain the following: \begin{align*}\mathbb{E}\left\|\int_{t_1}^{t_2} f(x(t),t)\text{d}t\right\|^2&\le \mathbb{E}\left[\int_{t_1}^{t_2} K\|x(t)\|\text{d}t\right]^2\le K^2(t_2-t_1)\mathbb{E}\left[\int_{t_1}^{t_2} \|x(t)\|^2\text{d}t\right] \\ &=K^2(t_2-t_1)\int_{t_1}^{t_2} h(t)\text{d}t\le K^2C(t_2-t_1)^2. \end{align*} From Itô's isometry, we obtain \begin{align*}\mathbb{E}\left\|\int_{t_1}^{t_2}\sigma(x(t),t)\text{d}B(t)\right\|^2 =\mathbb{E}\left[\int_{t_1}^{t_2}\|\sigma(x(t),t)\|^2\text{d}t\right]\le K^2\mathbb{E}\left[\int_{t_1}^{t_2}\|x(t)\|^2\text{d}t\right]=K^2\int_{t_1}^{t_2} h(t)\text{d}t\le K^2C(t_2-t_1). \end{align*} Therefore, we conclude that $$|h(t_2)-h(t_1)|\le \sqrt{4C}\sqrt{2K^2C[(t_2-t_1)^2+(t_2-t_1)]}<4KC\sqrt{(t_2-t_1)^2+(t_2-t_1)},$$ which implies $h(t)$ is uniformly continuous.

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