Observer-based event-triggered control for semilinear time-fractional diffusion systems with distributed feedback

Abstract

This paper is concerned with the observer-based distributed event-triggered feedback control for semilinear time-fractional diffusion systems under the Robin boundary conditions. To this end, an extended Luenberger-type observer is presented to solve the limitations caused by the impossible availability of full-state information that is needed for feedback control in practical applications due to the difficulties of measuring. With this, we propose the distributed output feedback event-triggered controllers via backstepping technique under which the considered systems admit Mittag–Leffler stability. It is shown that the given event-triggered control strategy could significantly reduce the amount of transmitted control inputs while guaranteeing the desired system performance with the Zeno phenomenon being excluded. A numerical illustration is finally presented to illustrate our theoretical results.

Appendices

Appendix

Proof of Proposition 1

Proof

Differentiating both sides of (11), we get that

$$\begin{aligned}&\omega _{xx} (x,t)\\&\quad = e_{xx}(x,t)-\frac{\mathrm{d}}{\mathrm{d}x}g(x,x) e(x,t)-g(x,x) e_x(x,t)\\&\qquad -g_x(x,x) e(x,t) -\int _0^x{g_{xx}(x,\varsigma ) e(\varsigma ,t)}\mathrm{d}\varsigma \end{aligned}$$

and

$$\begin{aligned}&{}^C_0D_t^\alpha \omega (x,t) \\&\quad ={}^C_0D_t^\alpha e(x,t)- \int _0^x{g(x,\varsigma ){}^C_0D_t^\alpha e(\varsigma ,t)}\mathrm{d}\varsigma \\&\quad ={}^C_0D_t^\alpha e(x,t)- \int _0^x{g_{\varsigma \varsigma }(x,\varsigma )e(\varsigma ,t)}\mathrm{d}\varsigma \\&\qquad -\,g(x,x)e_x(x,t)\\&\qquad +\,g(x,0)e_x(0,t) +g_\varsigma (x,x)e(x,t)\\&\qquad -\,g_\varsigma (x,0)e(0,t)\\&\qquad -\int _0^x{g(x,\varsigma )f(\varsigma ,t,e(\varsigma ,t)+{\hat{y}}(\varsigma ,t))}\mathrm{d}\varsigma \\&\qquad + \int _0^x{g(x,\varsigma )f(\varsigma ,t,{\hat{y}}(\varsigma ,t))}\mathrm{d}\varsigma \\&\qquad + \int _0^x{g(x,\varsigma )k_1(\varsigma )}\mathrm{d}\varsigma e(0,t). \end{aligned}$$

These, together with $e_x(0,t)=\frac{p_{2}- k_2}{p_{1}} e(0,t)$, $e(0,t)=\omega (0,t)$ and $\frac{\mathrm{d}}{\mathrm{d}x}G(x,x) =G_x(x,x) +G_\varsigma (x,x)$, yield that

$$\begin{aligned} 0= & {} {}^C_0D^{\alpha }_{t}\omega (x,t)- \omega _{xx}(x,t)+\mu \omega (x,t)-\varphi \nonumber \\= & {} f(x,t,e(x,t)+{\hat{y}}(x,t))-f({\hat{y}}(x,t),x,t)\nonumber \\&-\int _0^x{g(x,\varsigma )f(\varsigma ,t,e(\varsigma ,t)+{\hat{y}}(\varsigma ,t))}\mathrm{d}\varsigma \nonumber \\&+ \int _0^x{g(x,\varsigma )f(\varsigma ,t,{\hat{y}}(\varsigma ,t))}\mathrm{d}\varsigma -\varphi \nonumber \\&+\left( 2\frac{\mathrm{d}}{\mathrm{d}x}g(x,x) +\mu \right) e(x,t) \nonumber \\&+\int _0^x{\left\{ g_{xx}(x,\varsigma )-g_{\varsigma \varsigma }(x,\varsigma ) -\mu g(x,\varsigma )\right\} e(\varsigma ,t)}\mathrm{d}\varsigma \nonumber \\&+\left\{ \begin{array}{l}\int _0^x{g(x,\varsigma )k_1(\varsigma )}\mathrm{d}\varsigma -g_\varsigma (x,0)-k_1(x)\\ +g(x,0)\frac{p_{2}- k_2}{p_1}\end{array}\right\} e(0,t).\nonumber \\ \end{aligned}$$

(69)

Since this equation has to hold for all (x, t) $\in (0,l)\times (0,\infty )$, if g satisfies

$$\begin{aligned} \left\{ \begin{array}{l} g_{xx}(x,\varsigma )-g_{\varsigma \varsigma }(x,\varsigma )=\mu g(x,\varsigma ),~0<\varsigma< x< l,\\ 2\frac{\mathrm{d}}{\mathrm{d}x}g(x,x)=-\mu ,~0< x<l, \end{array}\right. \nonumber \\ \end{aligned}$$

(70)

then the observer gain $k_2$ satisfying (13) can be determined according to the boundary conditions at $x=0$, and besides, $k_1(x)$ can be chosen as in Eq.(12). Here $r_1\ne 0$ is imposed in consistent with $p_1\ne 0$. In addition, the boundary conditions at $x=l$ yield that

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} g(L,\varsigma )=0, \text{ if } q_1,s_1=0;\\ \left\{ \begin{array}{l} s_1 g_x(l,\varsigma )+s_2 g(l,\varsigma )=0,\\ g(l,l)=\frac{q_2}{q_1}-\frac{s_2}{s_1}, \end{array}\right. \text{ if } q_1,s_1\ne 0. \end{array} \right. \end{array} \end{aligned}$$

(71)

With these preliminaries, by [30, 35], we see the following results.

Lemma 4

For the unique solution of the kernel function $g(x,\varsigma )$ governed by (14), the following conclusions hold true:

• 1. If $q_1,s_1=0$, we have

  $$\begin{aligned}&g(x,\varsigma )\nonumber \\&\quad =-\mu (l-x) \frac{I_1\left( \sqrt{\mu (2l-x-\varsigma )(x-\varsigma ) }\right) }{\sqrt{\mu (2l-x-\varsigma )(x-\varsigma ) }},\nonumber \\ \end{aligned}$$

  (72)

  where $I_1$ is the modified first-order Bessel function.

• 2. If $q_1,s_1\ne 0$, the methods of successive approximation lead to the solution

  $$\begin{aligned}&g(x,\varsigma )\nonumber \\&\quad =-\mu (l-\varsigma ) \frac{I_1\left( \sqrt{\mu (2l-x-\varsigma )(x-\varsigma ) }\right) }{\sqrt{\mu (2l-x-\varsigma )(x-\varsigma ) }}\nonumber \\&\qquad -\frac{c^*\mu }{\sqrt{\mu +{c^*}^2}}\int _0^{x-\varsigma }\mathrm{e}^{\frac{c^*s}{2}}\sinh \left( \frac{\sqrt{\mu +{c^*}^2}}{2}s\right) \nonumber \\&\qquad \times I_0\left( \sqrt{\mu (2l-x-\varsigma ) (x-\varsigma -s)}\right) \mathrm{d}s,\nonumber \\ \end{aligned}$$

  (73)

  where $c^*=\frac{s_2}{s_1}$ and $I_0$ is the modified zero-order Bessel function.

Moreover, we have

$$\begin{aligned} \left| g(x,\varsigma )\right| \leqslant N \mathrm{e}^{2 N x}\leqslant N \mathrm{e}^{2 N l}\triangleq C_g \end{aligned}$$

(74)

for all $0\leqslant \xi \leqslant x\leqslant l,$ where

$$\begin{aligned} N= \left\{ \begin{array}{l}\mu , \text{ if } s_{1}=0,\\ \mu (1+\mathrm{e}^{-s_{2}/s_{1}}), \text{ if } s_{1}\ne 0.\end{array}\right. \end{aligned}$$

(75)

Define the operator ${\mathcal {E}}: L^2(0,l)\rightarrow L^2(0,l)$ as

$$\begin{aligned} \omega (x,t)= & {} ({\mathcal {E}}e)(x,t) \triangleq e(x,t)\nonumber \\&-\int _0^x{g(x,\varsigma ) e(\varsigma ,t)}\mathrm{d}\varsigma , \end{aligned}$$

(76)

From Eq. (74), we get that ${\mathcal {E}}$ is bounded.

Furthermore, set $\varphi (x,t)=\int _0^x{g(x,\varsigma )e(\varsigma ,t)}\mathrm{d}\varsigma$. One has $\omega (x,t)= e(x,t)-\varphi (x,t)$ and

$$\begin{aligned} \varphi (x,t)= & {} \int _0^x{g(x,\varsigma ) \omega (\varsigma ,t)}\mathrm{d}\varsigma \nonumber \\&+\int _0^x{g(x,\varsigma )\varphi (\varsigma ,t)}\mathrm{d}\varsigma . \end{aligned}$$

(77)

Based on this, let $\varphi _0(x,t)=\int _0^x{g(x,\varsigma ) \omega (\varsigma ,t)}\mathrm{d}\varsigma$ and $\varphi _n(x,t)=\int _0^x{g(x,\varsigma ) \varphi _{n-1}(\varsigma ,t)}\mathrm{d}\varsigma .$ It yields that

$$\begin{aligned} |\varphi _0(x,t)|\leqslant & {} C_g\sqrt{l}\Vert \omega (\cdot ,t)\Vert ,\\ |\varphi _1(x,t)|\leqslant & {} C_g^2\sqrt{l}\Vert \omega (\cdot ,t)\Vert x,\\ |\varphi _2(x,t)|\leqslant & {} \frac{C_g^3\sqrt{l}\Vert \omega (\cdot ,t)\Vert }{2!}x^2,\\ |\varphi _n(x,t)|\leqslant & {} \frac{C_g^{n+1}\sqrt{l} \Vert \omega (\cdot ,t)\Vert }{n!}x^n. \end{aligned}$$

Therefore, the series $\varphi (x,t)=\sum \nolimits _{n=0}^\infty \varphi _{n}(x,t)$ is absolutely and uniformly convergent. Moreover, since this series represents the solution of Eq. (77), the inverse of operator ${\mathcal {E}},$ denoted by ${\mathcal {E}}^{-1}$, is also a bounded operator. Then, we get that the error dynamic (10) and the target system (15) are equivalent and the proof is finished. $\square$

Proof of Proposition 2

Proof

Similar to the proof of Proposition 1, differentiating the transformation (19), we see

$$\begin{aligned} 0= & {} {}^C_0D^{\alpha }_{t}\rho (x,t)-\rho _{xx}(x,t)+\sigma \rho (x,t)\\&-H(x)e(0,t) -\psi \\= & {} \left[ \sigma +2\frac{\mathrm{d}}{\mathrm{d}x}h(x,x)\right] {\hat{y}}(x,t) \\&+\int _0^x{ {\hat{y}}(\varsigma ,t)\left\{ h_{xx}(x,\varsigma )-h_{\varsigma \varsigma }(x,\varsigma ) -\sigma h(x,\varsigma )\right\} }\mathrm{d}\varsigma \\&+\left( h(x,0)\frac{p_2}{p_1} -h_\varsigma (x,0)\right) {\hat{y}}(0,t)\\&+\left( k_1(x)-\int _0^x{h(x,\varsigma )k_1(\varsigma )}\mathrm{d}\varsigma +\frac{k_2}{p_1}-H(x) \right) e(0,t)\\&+f({\hat{y}}(x,t),x,t)-\int _0^x{h(x,\varsigma ) f({\hat{y}}(\varsigma ,t),\varsigma ,t)}\mathrm{d}\varsigma -\psi \\&+Bu(t_k)-\int _0^x{h(x,\varsigma ) Bu(t_k)}\mathrm{d}\varsigma . \end{aligned}$$

Then, system (17) can be converted into (23).

Lemma 5

The solution to the system (22) is

$$\begin{aligned}&h(x,\varsigma )\nonumber \\&\quad =\frac{{\hat{c}} \sigma }{\sqrt{\sigma +{{\hat{c}}}^2}} \int _0^{x-\varsigma }{\mathrm{e}^{-\frac{{\hat{c}} \tau }{2}} I_0\left( \sqrt{\sigma (x+\varsigma )(x-\varsigma -\tau ) }\right) } \nonumber \\&\qquad +\sinh \left( \frac{\sqrt{\sigma +{{\hat{c}}}^2}}{2} \tau \right) \mathrm{d}\tau \nonumber \\&\qquad -\,\sigma x \frac{I_1\left( \sqrt{\sigma \left( x^2-\varsigma ^2\right) }\right) }{\sqrt{\sigma \left( x^2-\varsigma ^2\right) }}, \end{aligned}$$

(78)

where ${\hat{c}} =p_1^{-1}p_2.$

Similar to the proof of Proposition 1, let operator ${\mathcal {E}}_2: L^2(0,l)\rightarrow L^2(0,l)$ be

$$\begin{aligned} \rho (x,t)= & {} ({\mathcal {E}}_2 {\hat{y}})(x,\cdot )\triangleq {\hat{y}} (x,t)\nonumber \\&\quad -\int _0^x{h(x,\varsigma ){\hat{y}}(\varsigma ,t)}\mathrm{d}\varsigma . \end{aligned}$$

(79)

Set

$$\begin{aligned} C_h=\max \limits _{0\leqslant \varsigma \leqslant x\leqslant l} \left| h(x,\varsigma )\right| , \end{aligned}$$

(80)

it is not difficult to deduce that both ${\mathcal {E}}_2$ and its inverse are bounded. Then, the transformation (19) and $h(0,0)=0$ yield that $\rho (0,t)={\hat{y}} (0,t)$ and $\rho _x(0,t)={\hat{y}}_x (0,t)$. These lead to the BCs of system (23). Therefore, the considered observer system (9) and the target system (23) are equivalent. The proof of Proposition 2 is finished. $\square$