Numerical schemes for nonlinear predictor feedback

Abstract

This paper focuses on a specific aspect of the implementation problem for predictor-based feedback laws: the problem of the approximation of the predictor mapping. It is shown that the numerical approximation of the predictor mapping by means of a numerical scheme in conjunction with a hybrid feedback law that uses sampled measurements can be used for the global stabilization of all forward complete nonlinear systems that are globally asymptotically stabilizable and locally exponentially stabilizable in the delay-free case. Explicit formulae are provided for the estimation of the parameters of the resulting hybrid control scheme.

Appendix

Proof of Lemma 2.3: Define the function:

$$\begin{aligned} g(\lambda )=W( {x_i +\lambda (x_{i+1} -x_i )}) \end{aligned}$$

(6.1)

for $\lambda \in [0,1]$. The following equalities hold for all $\lambda \in [0,1]$:

$$\begin{aligned} \begin{aligned} \frac{\hbox {d}g}{\hbox {d}\lambda }(\lambda )&=\nabla W( {x_i +\lambda (x_{i+1} -x_i )})(x_{i+1} -x_i ) \\ \frac{\hbox {d}^2g}{\hbox {d}\lambda ^2}(\lambda )&=(x_{i+1} -x_i {)}'\nabla ^2W( {x_i +\lambda (x_{i+1} -x_i )})(x_{i+1} -x_i ) \end{aligned} \end{aligned}$$

(6.2)

Moreover, notice that by virtue of (2.2) and (2.9), it holds that $|{x_{i+1} -x_i}|\le h( {|{x_i}|+\Vert u\Vert })L( {|{x_i} |+\Vert u\Vert })$. The previous inequality in conjunction with (2.5) and (6.2) gives:

$$\begin{aligned} \left| {\frac{\hbox {d}^2g}{\hbox {d}\lambda ^2}(\lambda )} \right| \le h^2P( {|{x_i}|+\Vert u\Vert }) \end{aligned}$$

(6.3)

where $P:\mathfrak {R}_+ \rightarrow \mathfrak {R}_+$ is the function involved in (2.5). Furthermore, inequality (2.3) in conjunction with (2.9) and (6.2) gives:

$$\begin{aligned} \frac{\hbox {d}g}{\hbox {d}\lambda }(0)=\nabla W( {x_i })\int \limits _{ih}^{(i+1)h} {f(x_i ,u(s))\,\hbox {d}s} \le chW( {x_i })+\int \limits _{ih}^{(i+1)h} {p(|{u(s)}|)\,\hbox {d}s} \end{aligned}$$

(6.4)

Combining (6.1), (6.3) and (6.4), we get:

$$\begin{aligned} W(x_{i+1} )\!=\!g(1)\le (1+ch)W(x_i )+\int \limits _{ih}^{(i+1)h} {p(|{u(s)}|)\,\hbox {d}s} +\frac{h^2}{2}P( {|{x_i}|+\Vert u\Vert })\qquad \end{aligned}$$

(6.5)

Inequality (6.5) in conjunction with the following inequality

$$\begin{aligned}&( {1+ch})W(x_i )+\int \limits _{ih}^{(i+1)h} {p(|{u(s)}|)\,\hbox {d}s} +\frac{h^2}{2}P( {|{x_i}|+\Vert u\Vert })\le \exp ( {2ch})W(x_i)\\&\quad +\int \limits _{ih}^{(i+1)h} {\exp ( {2c(ih+h-s)})p(|{u(s)}|)\,\hbox {d}s} \end{aligned}$$

which holds for all $h\le \frac{2cW(x_i )}{P({|{x_i}|+\Vert u\Vert })}$ imply that (2.12) holds. The proof is complete. $\square$

Proof of Lemma 2.4

We will first prove that if there exists $j\in \{0,\ldots ,N-1\}$ such that $\Vert u\Vert +\mathop {\min }\nolimits _{i=0,\ldots ,j} |{x_i}|>0$ and $h\le \frac{2c}{P( {Q(|{x_0}|+ \Vert u\Vert )+ \Vert u\Vert })}$ then (2.13) holds for all $i=0,\ldots ,j+1$. The proof is by induction.

First notice that (2.13) holds for $i=0$. Suppose that it holds for some $i\in \{0,\ldots ,j\}$. Clearly, inequality (2.13) implies $W(x_i )\le \exp ( {2c\,\tau })W(x_0 )+\frac{\exp ( {2c\,\tau })-1}{2c}p( {\Vert u\Vert })$. The previous inequality in conjunction with (2.6) implies $|{x_i}|\le Q(|{x_0}|+\Vert u\Vert )$. Consequently, the facts that $P:\mathfrak {R}_+ \rightarrow \mathfrak {R}_+$ is non-decreasing and $W(x_i )\ge 1$ imply $h\le \frac{2c}{P( {Q(|{x_0}|+\Vert u\Vert )+\Vert u\Vert })}\le \frac{2cW(x_i )}{P({|{x_i }|+\Vert u\Vert })}$. Since $|{x_i}|+\Vert u\Vert >0$ and $h\le \frac{2cW(x_i )}{P( {|{x_i}|+\Vert u\Vert })}$, Lemma 2.3 shows that:

$$\begin{aligned} W(x_{i+1} )\le \exp ( {2ch})W(x_i )+\int \limits _{ih}^{(i+1)h} {\exp ( {2c(ih+h-s)})p(|{u(s)}|)\,\hbox {d}s} \end{aligned}$$

The above inequality in conjunction with (2.13) shows that (2.13) holds for $i$ replaced by $i+1$.

The case that there exists $j\in \{0,\ldots ,N-1\}$ with $\Vert u\Vert +\mathop {\min }\nolimits _{i=0,\ldots ,j} |{x_i}|=0$ can be treated in the following way. Let $j\in \{0,\ldots ,N-1\}$ be the smallest integer with $\Vert u\Vert +\mathop {\min }\nolimits _{i=0,\ldots ,j} |{x_i}|=0$. This implies that $\Vert u\Vert =0$ and $|{x_j}|=0$. Since $f(0,0)=0$, (2.9) implies that $|{x_i}|=0$ for all $i=j+1,\ldots ,N$. Consequently, (2.13) holds for all $i=j+1,\ldots ,N$.

The proof is complete. $\square$

Proof of Lemma 2.5

Notice that, by virtue of (2.9), the following equation holds for all $i\in \{0,\ldots ,N-1\}$:

$$\begin{aligned} e_{i+1} =e_i +\int \limits _{ih}^{(i+1)h} {( {f(x_i ,u(s))-f(x(s),u(s))})\,\hbox {d}s} \end{aligned}$$

(6.6)

Inequality (2.1) implies the following inequality for all $i\in \{0,\ldots ,N-1\}$ and $s\in [ih,(i+1)h]$:

$$\begin{aligned} |{f(x_i ,u(s))-f(x(s),u(s))}|\le L( {|{x_i}|+|{x(s)}|+\Vert u\Vert })|{x_i -x(s)}| \end{aligned}$$

(6.7)

Using the definition $e_i :=x_i -x(ih)$ and inequalities (2.2), (2.4), we get for all $i\in \{0,\ldots ,N-1\}$ and $s\in [ih,(i+1)h]$:

$$\begin{aligned}&|{x_i -x(s)}|\le |{e_i}|+|{x(s)-x(ih)}| \nonumber \\&\quad \le |{e_i}|+(s-ih)\left( {\mathop {\max }\limits _{ih\le l\le s} ( {|{x(l)}|})+\Vert u\Vert }\right) L\left( {\mathop {\max }\limits _{ih\le l\le s} ( {|{x(l)}|})+\Vert u\Vert }\right) \nonumber \\&\quad \le |{e_i}|+(s-ih)\left( {a_\tau ( {|{x_0}|+\Vert u\Vert })+\Vert u\Vert }\right) L\left( {a_\tau ( {|{x_0}|+\Vert u\Vert })+\Vert u\Vert }\right) \end{aligned}$$

(6.8)

Notice that all hypotheses of Lemma 2.4 hold. Therefore, inequality (2.13) holds for all $i=0,\ldots ,N$. Clearly, inequality (2.13) implies $W(x_i )\le \exp ( {2c\,\tau })W(x_0 )+\frac{\exp ( {2c\,\tau })-1}{2c}p( {\Vert u\Vert })$. The previous inequality in conjunction with (2.6) implies $|{x_i } |\le Q(|{x_0}|+\Vert u\Vert )$ for all $i=0,\ldots ,N$. Exploiting the fact that $|{x_i}|\le Q(|{x_0}|+\Vert u\Vert )$ for all $i=0,\ldots ,N$ and (6.6), (6.7), (6.8), we obtain for all $i\in \{0,\ldots ,N-1\}$:

$$\begin{aligned}&|{e_{i+1} }|\le |{e_i}|+hL( {Q(|{x_0}|+\Vert u\Vert )+a_\tau ({|{x_0}|+\Vert u\Vert })+\Vert u\Vert })|{e_i}| \nonumber \\&\quad +\frac{h^2}{2}L( {Q(|{x_0}|+\Vert u\Vert )+a_\tau ({| {x_0 }|+\Vert u \Vert })+\Vert u \Vert })\,( {a_\tau ( {|{x_0}|+\Vert u\Vert })}\nonumber \\&\quad +{\Vert u\Vert })L( {a_\tau ( {|{x_0}|+\Vert u\Vert })+\Vert u\Vert }) \end{aligned}$$

(6.9)

Definitions (2.7), (2.8) in conjunction with inequality (6.9) shows that the following recursive relation holds for all $i\in \{0,\ldots ,N-1\}$

$$\begin{aligned} |{e_{i+1}}|\le \exp ( {hA(|{x_0}|+\Vert u\Vert )})|{e_i}|+\frac{h^2}{2}B(|{x_0}|+\Vert u\Vert ) \end{aligned}$$

(6.10)

Using the fact $e_0 =0$, in conjunction with relation (6.10), gives the desired inequality (2.14). The proof is complete. $\square$

Proof of claim 1

First, we show that for each partition $\{T_i \}_{i=0}^\infty$ of $\mathfrak {R}_+$ with $\mathop {\sup }\nolimits _{i\ge 0} ( {T_{i+1} -T_i })\le r$, for each $x_0 \in \mathfrak {R}^n$ and $u_0 \in L^\infty ( {[-\tau ,0);\mathfrak {R}^m})$, the solution of (1.2), (1.3), (1.4) and (1.5) with initial condition $x(0)=x_0$, $\breve{T} _\tau (0)u=u_0$ is unique and exists for all $t\ge 0$.

The solution of (1.2), (1.3), (1.4) and (1.5) is determined by the following process:

Initial step: Given $x(0)=x_0 , \breve{T} _\tau (0)u=u_0$ we determine the solution $x(t)$ of (1.2) for $t\in [0,\tau ]$. Notice that the solution is unique. Inequality (2.4) implies the following estimate:

$$\begin{aligned} |{x(t)}|\le a_\tau ( {|{x_0}|+\Vert {u_0}\Vert _\tau }), \hbox { for all } t\in [0,\tau ] \end{aligned}$$

(6.11)

$i$-th step: Given $x(t)$ for $t\in [0,T_i +\tau ]$ and $u(t)$ for $t\in [-\tau ,T_i )$, we determine $x(t)$ for $t\in [0,T_{i+1} +\tau ]$ and $u(t)$ for $t\in [-\tau ,T_{i+1} )$. The solution $z(t)$ of (1.3) for $t\in [T_i ,T_{i+1} )$ with initial condition $z(T_i )=z_N$ is unique (by virtue of the fact that $f$ and $k$ are locally Lipschitz mappings). Inequality (3.1) implies:

$$\begin{aligned} V(z(t))\le V( {z(T_i )}), \hbox { for all } t\in [T_i ,T_{i+1} ) \end{aligned}$$

(6.12)

We determine $u(t)$ for $t\in [T_i ,T_{i+1} )$ by means of the equation $u(t)=k(z(t))$. Notice that inequalities (3.5), (3.6) in conjunction with (6.12) imply the following inequality for all $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} | {u(t)}|= |{k(z(t))}|\le a_4 ( {a_1^{-1} ( {a_2 ( {| {z(T_i )} |})})}) \end{aligned}$$

(6.13)

Finally, we determine the solution $x(t)$ of (1.2) for $t\in [T_i +\tau ,T_{i+1} +\tau ]$. Notice that the solution is unique. The fact that $T_{i+1} -T_i \le r$ in conjunction with inequality (2.4) with $\tau$ replaced by $r>0$ and inequality (6.13) implies the estimate:

$$\begin{aligned} |{x(t)}|\le a_r ( {|{x(T_i +\tau )}|+a_4 ( {a_1^{-1} ( {a_2 ( {|{z(T_i)}|})})})}), \hbox { for all } t\in [T_i +\tau ,T_{i+1} +\tau ]\nonumber \\ \end{aligned}$$

(6.14)

Next we evaluate the difference $z(t)-x(t+\tau )$ for $t\in [T_i ,T_{i+1} )$. Exploiting (2.1) we get:

$$\begin{aligned}&| {z(t)-x(t+\tau )}|=\left| {z(T_i )-x(T_i +\tau )+\int \limits _{T_i }^t {( {f(z(s),k(z(s)))-f(x(s+\tau ),k(z(s)))})\,\hbox {d}s}} \right| \\&\le \left| {z(T_i )-x(T_i +\tau )\,} \right| +\int \limits _{T_i }^t {L( {|{z(s)}|+|{x(s+\tau )}|+|{k(z(s))} |}) |{z(s)-x(s+\tau )}|\,\hbox {d}s} \\ \end{aligned}$$

Using the right inequality (3.5), inequalities (6.12), (6.13), (6.14), in conjunction with the above inequality, we obtain:

$$\begin{aligned}&| {z(t)-x(t+\tau )}|\le |{z(T_i )-x(T_i +\tau )}|+L ( {a_1^{-1} ( {a_2 ( {|{z(T_i )}|})})} \\&\quad {+a_4 ( {a_1^{-1} ( {a_2 ( {|{z(T_i)}|})})})+a_r ( {|{x(T_i +\tau )}|+a_4 ( {a_1^{-1} ( {a_2 ( {|{z(T_i)}|})})})})})\\&\quad \times \int \limits _{T_i }^t {|{z(s)-x(s+\tau )}|\,\hbox {d}s} \end{aligned}$$

Define $\varphi (s):=a_r (s)+s$. Using the Growall–Bellman lemma, the above inequality and the fact that $T_{i+1} -T_i \le r$, we get for all $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} |{z(t)-x(t+\tau )}|\le |{z(T_i )-x(T_i +\tau )}|\exp ( {rL( {\varphi ({|{x(T_i +\tau )}|+q({|{z(T_i)}|})})})})\nonumber \\ \end{aligned}$$

(6.15)

Next we evaluate the quantity $\nabla V(x(t+\tau ))f(x(t+\tau ),k(z(t)))$ for $t\in [T_i ,T_{i+1} )$. Using inequality (3.1) we get:

$$\begin{aligned}&\nabla V(x(t+\tau ))f(x(t+\tau ),k(z(t)))\le -\rho ( {V(x(t+\tau ))}) \\&\quad +\nabla V(x(t+\tau ))( {f(x(t+\tau ),k(z(t)))-f(x(t+\tau ),k(x(t+\tau )))}) \end{aligned}$$

The following estimate follows from (3.6), (3.8) and the above inequality:

$$\begin{aligned}&\nabla V(x(t+\tau ))f(x(t+\tau ),k(z(t)))\le -\rho ( {V(x(t+\tau ))}) \\&\quad +a_3 ( {|{x(t+\tau )}|})\,M({|{(t+\tau )}|+ |{z(t)}|})\, |{x(t+\tau )-z(t)}| \\ \end{aligned}$$

Using the above inequality in conjunction with inequality (3.5), inequalities (6.12), (6.14) and definitions $q(s):=a_4 ( {a_1^{-1} ( {a_2 (s)})})+a_1^{-1} ( {a_2 (s)})$, $\varphi (s):=a_r (s)+s$, we get:

$$\begin{aligned}&\nabla V(x(t+\tau ))f(x(t+\tau ),k(z(t)))\le -\rho ( {V(x(t+\tau ))}) \nonumber \\&\quad +\,a_3 ({\varphi ( {|{x(T_i \!+\!\tau )}|\!+\!q( {|{z(T_i )} |})})})\,M( {\varphi ( {|{x(T_i \!+\!\tau )}|\!+\!q( {|{z(T_i)}|})})})\, |{x(t+\tau )-z(t)}|\nonumber \\ \end{aligned}$$

(6.16)

Combining inequalities (6.15), (6.16) and definition (3.9), we obtain for all $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned}&\nabla V(x(t+\tau ))f(x(t+\tau ),k(z(t)))\le -\rho ( {V(x(t+\tau ))})+D_r ( {|{x(T_i +\tau )}|}\nonumber \\&\quad +{q( {|{z(T_i)}|})})\,|{x(T_i +\tau )-z(T_i )}| \end{aligned}$$

(6.17)

Since $z(T_i )=z_N$ (recall (2.4)), it follows from (2.18), (2.19) and (1.2), (1.4) that the following inequalities hold for all $i=0,1,2,\ldots$:

$$\begin{aligned}&\left| {z(T_i )-x(T_i +\tau )} \right| \le R\left( {\,\left| {x(T_i )} \right| +\left\| {\breve{T}_\tau (T_i )u} \right\| _\tau \,}\right) \end{aligned}$$

(6.18)

$$\begin{aligned}&\left| {z(T_i )} \right| \le Q\left( {\,\left| {x(T_i )} \right| +\left\| {\breve{T} _\tau (T_i )u} \right\| _\tau \,}\right) \end{aligned}$$

(6.19)

Since $|{x(T_i +\tau )}|\le a_\tau ({|{x(T_i )}|+\Vert {\breve{T} _\tau (T_i )u}\Vert _\tau \,})$ (recall (2.4)), we obtain from (6.17), (6.18), (6.19) and definition (3.15) for all $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}V(x(t+\tau ))\le -\rho ( {V(x(t+\tau ))})+\gamma \end{aligned}$$

(6.20)

Using (6.20) and Lemma 2.14, page 82 in [7], we obtain for all $t\ge 0$:

$$\begin{aligned} V(x(t+\tau ))\le \max ( {\tilde{\sigma }( {V(x(\tau )),t})\,,\,\rho ^{-1}(2\gamma )}) \end{aligned}$$

(6.21)

for certain function $\tilde{\sigma }\in KL$. Combining (3.5), (6.11) and (6.21), we obtain inequality (3.16) with $\sigma (s,t):=\tilde{\sigma }( {a_2 ( {a_\tau ( s)}),t-\tau })\,$ for all $t>\tau$ and $\sigma (s,t):=\tilde{\sigma }( {a_2 ( {a_\tau ( s)}),0})\,$ for all $t\in [0,\tau ]$. The proof is complete. $\square$

Proof of Claim 2

Let arbitrary partition $\{T_i \}_{i=0}^\infty$ of $\mathfrak {R}_+$ with $\mathop {\sup }\nolimits _{i\ge 0} ( {T_{i+1} -T_i })\le r$, $x_0 \in \mathfrak {R}^n$, $u_0 \in L^\infty ( {[-\tau ,0);\mathfrak {R}^m})$ and consider the solution of (1.2), (1.3), (1.4) and (1.5) with (arbitrary) initial condition $x(0)=x_0$, $\breve{T} _\tau (0)u=u_0$. Inequalities (3.11) and (3.16) guarantee that there exists a unique smallest sampling time $T_j$ such that $V(x(T_j +\tau ))\le \delta$.

Moreover, inequalities (6.20), (3.11) and (3.5) allow us to conclude that

$$\begin{aligned} |{x(t)}|\le a_1^{-1} (\delta )\quad \hbox {and}\quad V(x(t))\le \delta , \hbox { for all } t\ge T_j +\tau \end{aligned}$$

(6.22)

Using (6.18), definition (3.15), (3.11) and (6.22), we obtain for all $i\ge j$:

$$\begin{aligned} |{z(T_i)}|\le |{z(T_i )-x(T_i +\tau )}|+|{x(T_i +\tau )}|\le \gamma +a_1^{-1} (\delta )\le 2a_1^{-1} (\delta )\qquad \end{aligned}$$

(6.23)

Using (6.12), (3.5) and (6.23), we get for all $t\ge T_j$:

$$\begin{aligned} |{z(t)}|\le a_1^{-1} ( {a_2 ( {2a_1^{-1} (\delta )})}) \end{aligned}$$

(6.24)

Next we evaluate the difference $z(t)-x(t+\tau )$ for $t\ge T_j$. Exploiting (2.1) and inequalities (3.6), (3.7), (3.10), (6.22), (6.24) and definition (3.13), we get for all $i\ge j$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} | {z(t)-x(t+\tau )} |&= \left| {z(T_i )-x(T_i +\tau )+\int \limits _{T_i }^t {( {f(z(s),k(z(s)))-f(x(s+\tau ),k(z(s)))})\,\hbox {d}s}} \right| \\&\le |{z(T_i )-x(T_i +\tau )}|+\tilde{L}\int \limits _{T_i }^t {|{z(s)-x(s+\tau )}|\,\hbox {d}s} \\ \end{aligned}$$

Using the Growall–Bellman lemma, the above inequality and the fact that $T_{i+1} -T_i \le r$ imply for all $i\ge j$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} |{z(t)-x(t+\tau )}|\le |{z(T_i )-x(T_i +\tau )}|\exp ( {\,r\,\tilde{L}}) \end{aligned}$$

(6.25)

Next we evaluate the quantity $\nabla V(x(t+\tau ))f(x(t+\tau ),k(z(t)))$ for $t\in [T_i ,T_{i+1} )$. Using inequalities (3.4), (3.5), (3.7), (6.22), (3.10), (3.8), (6.24) and (6.25) and definition (3.12), we get for all $i\ge j$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned}&\nabla V(x(t+\tau ))f(x(t+\tau ),k(z(t)))\le -\mu k_2^{-1} V(x(t+\tau ))\nonumber \\&\quad +\phi |{x(t+\tau )}| |{x(T_i +\tau )-z(T_i )}| \end{aligned}$$

(6.26)

Using (3.5), (3.7), (3.10), (6.22) and (6.26), we get for all $i\ge j$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} \dot{V}(t+\tau )\le -\frac{\mu }{2k_2 }V(t+\tau )+\frac{k_2 }{2\mu k_1 }\phi ^2\,\,\left| {x(T_i +\tau )-z(T_i )} \right| ^2 \end{aligned}$$

(6.27)

where $V(t)=V(x(t))$. Using (6.18), (3.15) and (6.27), we get for all $i\ge j$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} \dot{V}(t+\tau )\le -\frac{\mu }{2k_2 }V(t+\tau )\!+\!\frac{k_2 }{\mu k_1 }\phi ^2\tilde{R}^2\,|{x(T_i )}|^2\!+\!\frac{k_2 }{\mu k_1 }\phi ^2\tilde{R}^2\,\Vert {\breve{T} _\tau (T_i )u}\Vert _\tau ^2 \end{aligned}$$

(6.28)

Let $\omega <\frac{\mu }{4k_2 }$ be a positive constant sufficiently small so that

$$\begin{aligned}&k_4 \sqrt{\frac{k_2 }{k_1 }} \tilde{R}\,\exp (\omega \,(r+\tau ))<1 \hbox { and } \sqrt{2} \frac{k_2 }{k_1 }\phi \tilde{R}\frac{\exp ( {\omega (r+\tau )})}{\sqrt{\mu ^2-4\omega \mu k_2}}\nonumber \\&\quad \times \left( {\,1+\frac{k_4 \sqrt{k_2 } \exp ( {\omega (r+\tau )})( {\tilde{R}+\exp ( {-\omega \tau })})}{\sqrt{k_1 } -\tilde{R}k_4 \sqrt{k_2 } \exp ( {\omega (r+\tau )})}\,\,}\right) <1 \end{aligned}$$

(6.29)

The existence of $0<\omega <\frac{\mu }{4k_2 }$ satisfying (6.29) is guaranteed by (3.14). Using (6.28) and the fact that $\mathop {\sup }\nolimits _{i\ge 0} ( {T_{i+1} -T_i })\le r$, we obtain for all $i\ge j$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned}&\dot{V}(t+\tau )\le -\frac{\mu }{2k_2 }V(t+\tau )+\frac{k_2 }{\mu k_1 }\phi ^2\tilde{R}^2\,\exp ( {-2\omega t})\exp ( {2\omega r})\,\mathop {\sup }\limits _{T_i \le s\le t} ( {\exp ( {2\omega s})\left| {x(s)} \right| ^2})\nonumber \\&\quad +\frac{k_2 }{\mu k_1 }\phi ^2\tilde{R}^2\exp ( {-2\omega t})\exp ( {2\omega (r+\tau )})\,\,\mathop {\sup }\limits _{T_i -\tau \le s\le t} ( {\exp ( {2\omega s})\left| {u(s)} \right| ^2}) \end{aligned}$$

(6.30)

The differential inequality (6.30) allows us to conclude that the following differential inequality holds for $t\ge T_j$ a.e.:

$$\begin{aligned}&\dot{V}(t+\tau )\le -\frac{\mu }{2k_2 }V(t+\tau )+\frac{k_2 }{\mu k_1 }\phi ^2\tilde{R}^2\,\exp ( {-2\omega t})\exp ( {2\omega r})\,\mathop {\sup }\limits _{T_j \le s\le t} ( {\exp ( {2\omega s})|{x(s)}|^2})\nonumber \\&\quad +\frac{k_2 }{\mu k_1 }\phi ^2\tilde{R}^2\exp ( {-2\omega t})\exp ( {2\omega (r+\tau )})\,\,\mathop {\sup }\limits _{T_j -\tau \le s\le t} ( {\exp ( {2\omega s})|{u(s)}|^2}) \end{aligned}$$

(6.31)

Integrating (6.31) and since $\omega <\frac{\mu }{4k_2 }$, we obtain for all $t\ge T_j$:

$$\begin{aligned}&V(t\!+\!\tau )\le \exp ( {-2\omega ( {t-T_j })})V(T_j +\tau )\!+\!\frac{2k_2^2 }{\mu k_1 }\phi ^2\tilde{R}^2\,\frac{\exp ( {-2\omega t})}{\mu \!-\!4\omega k_2 }\exp ( {2\omega r})\,\mathop {\sup }\limits _{T_j \le s\le t}\nonumber \\&\quad \times ( {\exp ( {2\omega s})\left| {x(s)} \right| ^2})+\frac{2k_2^2 }{\mu k_1 }\phi ^2\tilde{R}^2\frac{\exp ( {-2\omega t})}{\mu -4\omega k_2 }\exp ( {2\omega (r+\tau )})\nonumber \\&\quad \mathop {\sup }\limits _{T_j -\tau \le s\le t} ( {\exp ( {2\omega s})\left| {u(s)} \right| ^2}) \end{aligned}$$

(6.32)

Using (3.5), (3.7), (6.21) and the fact that $\omega <\frac{\mu }{4k_2 }$, we obtain from (6.32) for all $t\ge T_j$:

$$\begin{aligned}&|{x(t+\tau )}|\exp ( {\omega (t+\tau )})\le \exp ( {\omega ( {T_j +\tau })})\sqrt{\frac{k_2 }{k_1 }} |{x(T_j +\tau )}| \nonumber \\&\quad +\sqrt{2} \frac{k_2 }{k_1 }\phi \tilde{R}\frac{\exp ( {\omega (r+\tau )})}{\sqrt{\mu ^2-4\omega \mu k_2 } }\,\mathop {\sup }\limits _{T_j \le s\le t} ( {\exp ( {\omega s}) |{x(s)}|})\nonumber \\&\quad +\sqrt{2} \frac{k_2 }{k_1 }\phi \tilde{R}\frac{\exp ( {\omega (r+2\tau )})}{\sqrt{\mu ^2-4\omega \mu k_2 } }\,\,\mathop {\sup }\limits _{T_j -\tau \le s\le t} ( {\exp ( {\omega s})\left| {u(s)} \right| }) \end{aligned}$$

(6.33)

Using (3.5), (3.6), (3.7), (3.10), (3.15), (6.12), (6.18) and (6.24) we obtain for all $i\ge j$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} |{u(t)}|&=|{k(z(t))}|\le k_4 | {z(t)} |\le k_4 \sqrt{\frac{k_2 }{k_1 }} | {z(T_i )}| \nonumber \\&\le k_4 \sqrt{\frac{k_2 }{k_1 }} |{z(T_i )-x(T_i +\tau )} |+k_4 \sqrt{\frac{k_2 }{k_1 }} |{x(T_i +\tau )}| \nonumber \\&\le \tilde{R}k_4 \sqrt{\frac{k_2 }{k_1 }} |{x(T_i )}|+\tilde{R}k_4 \sqrt{\frac{k_2 }{k_1 }} \Vert {\breve{T} _\tau (T_i )u} \Vert _\tau +k_4 \sqrt{\frac{k_2 }{k_1 }} |{x(T_i +\tau )} |\qquad \end{aligned}$$

(6.34)

Inequality (6.34) in conjunction with the fact that $\mathop {\sup }\nolimits _{i\ge 0} ( {T_{i+1} -T_i })\le r$ implies:

$$\begin{aligned}&| {u(t)} |\exp ( {\omega t})\le \tilde{R}k_4 \sqrt{\frac{k_2 }{k_1 }} \exp ( {\omega r}) |{x(T_i )}|\exp ( {\omega T_i }) \\&\quad +\tilde{R}k_4 \sqrt{\frac{k_2 }{k_1 }} \exp ( {\omega (r+\tau )})\mathop {\sup }\limits _{T_i -\tau \le s<T_i } ( {\exp ( {\omega s}) |{u(s)}|})\nonumber \\&\quad + k_4 \sqrt{\frac{k_2 }{k_1 }} \exp ( {\omega (r-\tau )})| {x(T_i +\tau )}|\exp ( {\omega (T_i +\tau )}) \end{aligned}$$

Therefore, we get from the above inequality for all $t\ge T_j$:

$$\begin{aligned}&|{u(t)} |\exp ( {\omega t})\le k_4 \exp ( {\omega r})\sqrt{\frac{k_2 }{k_1 }} ( {\tilde{R}\,+\exp ( {-\omega \tau })})\mathop {\sup }\limits _{T_j -\tau \le s\le t} ( {\exp ( {\omega (s+\tau )})| {x(s+\tau )}|})\nonumber \\&\quad +\tilde{R}k_4 \sqrt{\frac{k_2 }{k_1 }} \exp ( {\omega (r+\tau )})\mathop {\sup }\limits _{T_j -\tau \le s\le t} ( {\exp ( {\omega s})| {u(s)} |}) \end{aligned}$$

(6.35)

Distinguishing the cases $\mathop {\sup }\nolimits _{T_j -\tau \le s\le t} ( {\exp (\omega \,s) |{u(s)}|})\!=\!\mathop {\sup }\nolimits _{T_j \le s\le t} ( {\exp (\omega \,s) |{u(s)}|})$ and $\mathop {\sup }\nolimits _{T_j -\tau \le s\le t} ( {\exp (\omega \,s) |{u(s)}|})=\mathop {\sup }\nolimits _{T_j -\tau \le s<T_j } ( {\exp (\omega \,s) |{u(s)}|})$, we obtain from (6.35) for all $t\ge T_j$:

$$\begin{aligned}&|{u(t)}|\exp ( {\omega t})\le \frac{k_4 \exp ( {\omega r})\sqrt{k_2 } ( {\tilde{R}\,+\exp ( {-\omega \tau })})}{\sqrt{k_1 } -\tilde{R}k_4 \sqrt{k_2 } \exp ( {\omega (r+\tau )})}\mathop {\sup }\limits _{T_j -\tau \le s\le t} ( {\exp ( {\omega (s+\tau )})| {x(s+\tau )}|}) \nonumber \\&\quad + \tilde{R}k_4 \sqrt{\frac{k_2 }{k_1 }} \exp ( {\omega (r+\tau )})\mathop {\sup }\limits _{T_j -\tau \le s<T_j } ( {\exp ( {\omega s}) |{u(s)}|}) \end{aligned}$$

(6.36)

Combining (6.33) and (6.36), we get for all $t\ge T_j$:

$$\begin{aligned}&|{x(t+\tau )}|\exp ( {\omega (t+\tau )})\le \exp ( {\omega ( {T_j +\tau })})\sqrt{\frac{k_2 }{k_1 }} | {x(T_j +\tau )}| \\&\quad +\sqrt{2} \frac{k_2 }{k_1 }\phi \tilde{R}\frac{\exp ( {\omega (r+\tau )})}{\sqrt{\mu ^2-4\omega \mu k_2 } }\left( {\,1+\frac{k_4 \sqrt{k_2 } \exp ( {\omega (r+\tau )})( {\tilde{R}+\exp ( {-\omega \tau })})}{\sqrt{k_1 } -\tilde{R}k_4 \sqrt{k_2 } \exp ( {\omega (r+\tau )})}\,\,}\right) \\&\qquad \mathop {\sup }\limits _{T_j -\tau \le s\le t} ( {\exp ( {\omega (s+\tau )})| {x(s+\tau )} |}) \\&\quad +\sqrt{2} \frac{k_2 }{k_1 }\phi \tilde{R}\frac{\exp ( {\omega (r+2\tau )})}{\sqrt{\mu ^2-4\omega \mu k_2 } }\,\,\mathop {\sup }\limits _{T_j -\tau \le s<T_j } ( {\exp ( {\omega s}) |{u(s)}|}) \end{aligned}$$

Distinguishing the cases $\mathop {\sup }\nolimits _{T_j -\tau \le s\le t} \,( {\exp ( {\omega (s+\tau )})\,|{x(s+\tau )} |})=\mathop {\sup }\nolimits _{T_j -\tau \le s\le T_j } ( {\exp ( {\omega (s+\tau )})\,| {x(s+\tau )}|})$, $\mathop {\sup }\nolimits _{T_j -\tau \le s\le t} \,( {\exp ( {\omega (s+\tau )})\,|{x(s+\tau )} |})=\mathop {\sup }\nolimits _{T_j \le s\le t} ( {\exp ( {\omega (s+\tau )})\, |{x(s+\tau )} |})$ and using the above inequality, we obtain for all $t\ge T_j$:

$$\begin{aligned}&|{x(t+\tau )}|\exp ( {\omega (t+\tau )})\le \frac{\exp ( {\omega ( {T_j +\tau })})}{1-A}\sqrt{\frac{k_2 }{k_1 }} |{x(T_j +\tau )}|\nonumber \\&\quad +A\mathop {\sup }\nolimits _{T_j -\tau \le s \!\le \! T_j } ( {\exp ( {\omega (s+\tau )}) |{x(s+\tau )}|})\!+\!\sqrt{2} \frac{k_2 }{k_1 (1-A)}\phi \tilde{R}\frac{\exp ( {\omega (r+2\tau )})}{\sqrt{\mu ^2-4\omega \mu k_2 } }\nonumber \\&\qquad \times \mathop {\sup }\limits _{T_j -\tau \le s<T_j } ( {\exp ( {\omega s}) |{u(s)}|}) \end{aligned}$$

(6.37)

where $A=\sqrt{2} \frac{k_2 }{k_1 }\phi \tilde{R}\frac{\exp ( {\omega (r+\tau )})}{\sqrt{\mu ^2-4\omega \mu k_2 } }\left( {\,1+\frac{k_4 \sqrt{k_2 } \exp ( {\omega (r+\tau )})( {\tilde{R}+\exp ( {-\omega \tau })})}{\sqrt{k_1 } -\tilde{R}k_4 \sqrt{k_2 } \exp ( {\omega (r+\tau )})}\,\,}\right)$. Inequalities (6.36), (6.37) imply that there exist constants $Q_1 ,Q_2 >0$ such that (3.17), (3.18) hold.

The proof is complete. $\square$

Proof of Claim 3

Let arbitrary partition $\{T_i \}_{i=0}^\infty$ of $\mathfrak {R}_+$ with $\mathop {\sup }\nolimits _{i\ge 0} ( {T_{i+1} -T_i })\le r$, $x_0 \in \mathfrak {R}^n$, $u_0 \in L^\infty ( {[-\tau ,0);\mathfrak {R}^m})$ and consider the solution of (1.2), (1.3), (1.4) and (1.5) with (arbitrary) initial condition $x(0)=x_0$, $\breve{T} _\tau (0)u=u_0$.

Define:

$$\begin{aligned} b(s):=a_4 ( {a_1^{-1} ( {a_2 ( s)})}), \hbox { for all } s\ge 0 \end{aligned}$$

(6.38)

and notice that $b\in K_\infty$. Moreover, notice that definitions (6.38) and (3.15) imply that

$$\begin{aligned} R(s)\le \frac{1}{2}b^{-1}\left( {\frac{s}{2}}\right) \!, \hbox { for all } s\ge 0 \end{aligned}$$

(6.39)

Furthermore, definition (6.38) and inequality (6.13) imply the following inequality for all $i\in Z_+$ and $t\in [T_i ,T_{i+1} )$:

$$\begin{aligned} \left| {u(t)} \right| \le b( {\left| {z(T_i )} \right| }) \end{aligned}$$

(6.40)

Inequalities (3.5), (3.16) imply the existence of a non-decreasing function $g:\mathfrak {R}_+ \rightarrow \mathfrak {R}_+$ such that:

$$\begin{aligned} | {x(t)}|\le g( {|{x_0}|+ \Vert {u_0 }\Vert _\tau }), \hbox { for all } t\ge 0 \end{aligned}$$

(6.41)

By virtue of (6.18), (6.39) and (6.41) we get for all $i\in Z_+$:

$$\begin{aligned}&|{z(T_i )-x(T_i +\tau )}|\le R\left( { |{x(T_i)}|+\mathop {\sup }\limits _{T_i -\tau \le s<T_i } ( {|{u(s)}|})}\right) \\&\quad \le \frac{1}{2}b^{-1}\left( {\frac{1}{2} |{x(T_i )} |+\frac{1}{2}\mathop {\sup }\limits _{T_i -\tau \le s<T_i } ( {| {u(s)}|})}\right) \\&\quad \le \max \left( {\frac{1}{2}b^{-1}( {| {x(T_i )} |}),\ \frac{1}{2}b^{-1}\left( {\mathop {\sup }\limits _{T_i -\tau \le s<T_i } \left| {u(s)} \right| }\right) }\right) \\&\quad \le \max \left( {\frac{1}{2}b^{-1}( {g( {\left| {x_0 } \right| +\left\| {u_0 } \right\| _\tau })})\,,\,\,\frac{1}{2}b^{-1}\left( {\mathop {\sup }\limits _{T_i -\tau \le s<T_i } \left| {u(s)} \right| }\right) }\right) \end{aligned}$$

The above inequality in conjunction with (6.41) gives for all $i\in Z_+$:

$$\begin{aligned}&|{z(T_i )}|\le |{x(T_i +\tau )}|\!+\!\max \left( {\frac{1}{2}b^{-1} \left( {g( {|{x_0}|\!+\!\Vert {u_0}\Vert _\tau })}\right) ,\,\,\frac{1}{2}b^{-1} \left( {\mathop {\sup }\limits _{T_i -\tau \le s<T_i } |{u(s)}|}\right) }\right) \\&\quad \le g( {|{x_0}|+ \Vert {u_0}\Vert _\tau })+\max \left( {\frac{1}{2}b^{-1}( {g( {|{x_0}|+ \Vert {u_0}\Vert _\tau })}),\,\,\frac{1}{2}b^{-1}\left( {\mathop {\sup }\limits _{T_i -\tau \le s<T_i } | {u(s)}|}\right) }\right) \\&\quad \left. \le \max ( {2g( {|{x_0}|+ \Vert {u_0}\Vert _\tau }),\,b^{-1}( {g( {| {x_0 }|+\Vert {u_0}\Vert _\tau })}),\,\,b^{-1}\left( {\mathop {\sup }\limits _{T_i -\tau \le s<T_i } | {u(s)} |}\right) }\right) \\&\quad \left. \le \max ( {2g( {|{x_0}|+ \Vert {u_0}\Vert _\tau }),\,b^{-1}( {g( {|{x_0}|+ \Vert {u_0}\Vert _\tau })}),\,\,b^{-1}\left( {\mathop {\sup }\limits _{-\tau \le s<T_i } |{u(s)}|}\right) }\right) \end{aligned}$$

Furthermore, using (6.40) and the above inequality, we obtain for all $i\in Z_+$:

$$\begin{aligned} \mathop {\sup }\limits _{T_i \le s<T_{i+1} } |{u(s)}|\le \max ( {\tilde{g}( {|{x_0}|+ \Vert {u_0}\Vert _\tau }),\,\,\mathop {\sup }\limits _{-\tau \le s<T_i } |{u(s)}|}) \end{aligned}$$

(6.42)

where $\tilde{g}(s):=\max ( {g( s)\,,\,b( {2g( s)})})$ for all $s\ge 0$, is a non-decreasing function. Define the sequence:

$$\begin{aligned} F_i :=\mathop {\sup }\limits _{-\tau \le s<T_i } |{u(s)}| \end{aligned}$$

(6.43)

Notice that definition (6.43) and the fact that $\mathop {\sup }\nolimits _{-\tau \le s<T_{i+1} } |{u(s)}|=\max ( {\mathop {\sup }\nolimits _{T_i \le s<T_{i+1} } |{u(s)}|\,,\,\,\mathop {\sup }\nolimits _{-\tau \le s<T_i} |{u(s)}|})$ in conjunction with (6.42) imply the following inequality for all $i\in Z_+$:

$$\begin{aligned} F_{i+1} \le \max ( {\tilde{g}( {|{x_0}|+\Vert {u_0 } \Vert _\tau })\,,\,\,F_i }) \end{aligned}$$

(6.44)

Inequality (6.44) in conjunction with the fact that $F_0 :=\Vert {u_0}\Vert _\tau$ allows us to prove by induction that the following inequality holds for all $i\in Z_+$:

$$\begin{aligned} F_i \le \max ( {\tilde{g}( {|{x_0}|+\Vert {u_0}\Vert _\tau })\,,\,\,\Vert {u_0 }\Vert _\tau }) \end{aligned}$$

(6.45)

Inequality (6.41) in conjunction with inequality (6.45) and definition (6.43) implies that estimate (3.19) holds with $G(s):=g(s)+\max ( {\tilde{g}( s)\,,\,\,s})$ for all $s\ge 0$. The proof is complete. $\square$